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+test/tasks/Exam-21-T10/images/Exam-21-T10-header-1.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-21-T3/images/Exam-21-T3-3-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-21-T5/images/Exam-21-T5-2-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-21-T6/images/Exam-21-T6-2-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-21-T6/images/Exam-21-T6-3-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-22-T5/images/Exam-22-T5-10-1-prompt.png filter=lfs diff=lfs merge=lfs -text diff --git a/README.md b/README.md index e3a23a536aa15a97c4f95a0c60ef18924a3094c8..86932f177936659920ad7e63e2a63e5ac79c9eb4 100644 --- a/README.md +++ b/README.md @@ -36,15 +36,15 @@ credit, instead of answer endpoints. | | Share | Problems | Subquestions | Marks | Public images | Rubrics | | --- | ---: | ---: | ---: | ---: | ---: | --- | -| 🧪 **dev** · Public Dev Set | 25% | 48 | 210 | 636 | 268 | ✅ included | -| 🏁 **test** · Held-out Test Set | 55% | 105 | 445 | 1387 | 268 | ❌ stripped | -| 🔒 **private** · Private Set | 20% | 38 | 165 | 559 | — | 🔐 local only | -| **Total** | 100% | **191** | **820** | **2582** | **536** | | +| 🧪 **dev** · Public Dev Set | 25% | 48 | 209 | 676 | 350 | ✅ included | +| 🏁 **test** · Held-out Test Set | 55% | 105 | 391 | 1271 | 173 | ❌ stripped | +| 🔒 **private** · Private Set | 20% | 38 | 220 | 635 | — | 🔐 local only | +| **Total** | 100% | **191** | **820** | **2582** | **523** | | -> 🎲 **Seeded stratified split** — the distribution of *level* (exam series), -> *subfield* (knowledge category), and *modality* (with / without model-visible -> images) stays consistent across all three partitions and matches the full set -> (max deviation ≈ 3 percentage points). +> 🖼 **Balanced image split** — problems are assigned deterministically +> (largest-image-first) so that the total number of model-visible images is +> roughly equal across the three partitions (155 / 165 / 173 of a 164-per-partition +> average), while the 25/55/20 problem ratio is kept exact. ## 📁 Repository Layout @@ -135,43 +135,37 @@ structure/SMILES equivalence judged by chemical meaning. *Snapshot: September 2026 · Normalized Rubric Score (%). Runs are sorted by `full_problem` within each input mode.* -🥇 **`full_problem` leaders** — Multimodal: Claude Opus 5 53.8%† · -Text: Claude Opus 5 61.5%† · Text + reader: GLM-5.2 52.0% +🥇 **`full_problem` leaders** — Multimodal: Claude Opus 5 53.8% · +Text: Claude Opus 5 61.5% · Text + reader: GLM-5.2 52.0% | Model | Input | full_problem | sequential_carry | oracle_scaffolded | | --- | --- | ---: | ---: | ---: | -| Claude Opus 5 † | multimodal | 53.8 | 59.4 | 64.9 | -| Gemini 3.8 Flash § | multimodal | 48.5 | 58.5 | 60.3 | -| Grok 4.6 § | multimodal | 33.8 | 41.8 | 45.5 | +| Claude Opus 5 | multimodal | 53.8 | 59.4 | 64.9 | +| Gemini 3.8 Flash | multimodal | 48.5 | 58.5 | 60.3 | +| Grok 4.6 | multimodal | 33.8 | 41.8 | 45.5 | | Qwen3.8-Max | multimodal | 29.3 | 31.2 | 34.0 | | Kimi K3 | multimodal | 27.5 | 30.6 | 31.8 | | MiniMax M3 | multimodal | 27.3 | 27.5 | 27.9 | | GPT-5.6-Sol | multimodal | 23.2 | 28.8 | 30.4 | -| Claude Opus 5 † | text | 61.5 | 61.7 | 61.7 | -| Gemini 3.8 Flash § | text | 46.5 | 59.0 | 59.7 | +| Claude Opus 5 | text | 61.5 | 61.7 | 61.7 | +| Gemini 3.8 Flash | text | 46.5 | 59.0 | 59.7 | | Qwen3.8-Max | text | 36.0 | 39.0 | 41.9 | | Kimi K3 | text | 34.5 | 36.0 | 37.5 | -| Grok 4.6 § | text | 33.4 | 42.0 | 46.0 | +| Grok 4.6 | text | 33.4 | 42.0 | 46.0 | | GPT-5.6-Sol | text | 29.8 | 35.6 | 36.5 | | MiniMax M3 | text | 26.2 | 28.7 | 29.1 | | DeepSeek V4 Pro | text | 25.0 | 25.0 | 27.6 | | GLM-5.2 | text | 23.5 | 24.9 | 25.1 | | GLM-5.2 | text + reader | 52.0 | 52.2 | 52.4 | -| Claude Opus 5 † | text + reader | 46.2 | 53.9 | 54.7 | +| Claude Opus 5 | text + reader | 46.2 | 53.9 | 54.7 | | DeepSeek V4 Pro | text + reader | 36.7 | 36.8 | 37.0 | -| Gemini 3.8 Flash § | text + reader | 33.8 | 45.1 | 49.7 | +| Gemini 3.8 Flash | text + reader | 33.8 | 45.1 | 49.7 | | Kimi K3 | text + reader | 32.3 | 32.3 | 32.4 | | GPT-5.6-Sol | text + reader | 31.1 | 33.6 | 34.0 | -| Grok 4.6 § | text + reader | 28.6 | 29.0 | 31.1 | +| Grok 4.6 | text + reader | 28.6 | 29.0 | 31.1 | | Qwen3.8-Max | text + reader | 27.9 | 29.2 | 31.5 | | MiniMax M3 | text + reader | 27.2 | 27.3 | 27.3 | -> † Claude Opus 5 runs rest on a small problem prefix (39–103 points) and are -> provisional until the full set completes. -> § Gemini 3.8 Flash and Grok 4.6 runs sit on a 12-paper, 1136-point subset. -> Percentages are computed on each run's own scored problem set, so -> denominators differ across runs. - Live leaderboard: ## 📖 Citation @@ -185,6 +179,10 @@ Live leaderboard: Wenzhe Gao and Zhihao Yu and Qijun Wang and Yetian Zhu and Jiande Chen and Chuhang Pei and Yao Jiang and Erhao Chen and Pinhan Wang and Xiaorui Pan and Qianyi Zhu and Yang Liu and Mang Ye and Guancheng Wan}, - year = {2026} + year = {2026}, + eprint = {2610.20012}, + archivePrefix = {arXiv}, + primaryClass = {cs.LG}, + url = {https://arxiv.org/abs/2610.20012} } ``` diff --git a/dev/dataset.jsonl b/dev/dataset.jsonl index 4138414a0588a8d3c0693deb42819043ff54fdd4..39aac71a85d9b4aace09df36f3d404072c5ebbaa 100644 --- a/dev/dataset.jsonl +++ b/dev/dataset.jsonl @@ -1,48 +1,48 @@ {"problem_id": "Exam-01-T5", "header": "Simple Molecules, Complex Reactions", "header_images": [], "subquestions": [{"sub_id": "Exam-01-T5-1", "tags": ["1.4", "1.5"], "text_open": "Label D denotes $N_2O$.\nScientists have developed catalyst X, which converts the toxic gas CO to CO_2 under the action of D. Its formula mass is approximately 471.5, and its structure is shown below, where P represents a P(tBu)_2 group. Write the d-orbital electron configuration of the metal ion in X in an octahedral field, approximating the geometry as a regular octahedron.", "text_images": [{"file": "Exam-01-T5-1-1-prompt.png", "description": "Structural representation of catalyst X: '[H][Ru]12(C#O)(N3C(C4=CC=CC=[N]42)=CC=CC3=C[P]1(C(C)(C)C)C(C)(C)C)C#O'"}], "full_mark": 2, "rubric": ["t_{2g}^{6} e_{g}^{0} (2 points)"], "rubric_images": []}, {"sub_id": "Exam-01-T5-2", "tags": ["2.2", "2.3", "2.4"], "text_open": "Label D denotes $N_2O$.\nBased on experimental observations, the following principal catalytic pathway has been proposed: X reacts with D to form Y, whose formula mass is approximately 487.5. Y rapidly converts into its more stable isomer Z. Z then takes up one molecule of CO while releasing one molecule of CO_2, regenerating X. Write the SMILES structures of Y and Z.", "text_images": [], "full_mark": 4, "rubric": ["Y: 'O[Ru]12(C#O)(N3C(C4=CC=CC=[N]42)=CC=CC3=C[P]1(C(C)(C)C)C(C)(C)C)C#O' (2 points)", "Z: 'CC(C)([P]1(C=C2N(C(C3=CC=CC=[N]34)=CC=C2)[Ru]14(C(O)=O)C#O)C(C)(C)C)C' (2 points). Stereochemistry is not required."], "rubric_images": [{"file": "Exam-01-T5-2-1-rubric.png", "description": "Reference structure of Y: 'O[Ru]12(C#O)(N3C(C4=CC=CC=[N]42)=CC=CC3=C[P]1(C(C)(C)C)C(C)(C)C)C#O'"}, {"file": "Exam-01-T5-2-2-rubric.png", "description": "Reference structure of Z: 'CC(C)([P]1(C=C2N(C(C3=CC=CC=[N]34)=CC=C2)[Ru]14(C(O)=O)C#O)C(C)(C)C)C'"}]}, {"sub_id": "Exam-01-T5-3", "tags": ["2.3", "2.4"], "text_open": "The scientists also found that when THF is used as the solvent, another pathway may operate: X slowly isomerizes to isomer W, which contains two aromatic rings, and W then catalyzes exactly the same process as X. Write the SMILES structure of W.", "text_images": [], "full_mark": 3, "rubric": ["W: 'CC(C)([P]1(CC2=[N](C(C3=NC=CC=C43)=CC=C2)[Ru]14(C#O)(C#O)[H])C(C)(C)C)C' (3 points)."], "rubric_images": [{"file": "Exam-01-T5-3-1-rubric.png", "description": "Reference structure of W: 'CC(C)([P]1(CC2=[N](C(C3=NC=CC=C43)=CC=C2)[Ru]14(C#O)(C#O)[H])C(C)(C)C)C'"}]}]} -{"problem_id": "Exam-01-T7", "header": "Efficiency of Oxidative Phosphorylation", "header_images": [], "subquestions": [{"sub_id": "Exam-01-T7-1", "tags": ["3.2"], "text_open": "A molecular-level hypothesis has been proposed: for every $2e^-$ transferred through the electron-transfer chain, Complex III transports $2HSuc^-$ across the membrane. The opposing concentration gradients that must be overcome in the two transport events are approximately $1:7$ and $1:8$, respectively. Energy losses from other ion-transport processes at Complexes I and IV may be neglected. The inner-side condition is $pH=8$, the outer-side condition is $pH=5$, succinic acid has $pK_{a1,2}=4.2,5.6$, and the temperature is $310\\,\\mathrm{K}$.\nFor this redox reaction, $\\Delta_rG_m^{\\ominus *}\\approx-2.20\\times10^2\\ \\mathrm{kJ\\,mol^{-1}}$; take the magnitude of the available energy as $220\\ \\mathrm{kJ\\,mol^{-1}}$.\nATP synthase can transport only $HSuc^-$ ions on the outer side of the membrane; neither $H_2Suc$ nor $Suc^{2-}$ outside the membrane can pass through ATP synthase. Thus, of the energy stored in $H_2Suc$, $HSuc^-$, and $Suc^{2-}$ during the redox stage, only the fraction associated with external $HSuc^-$ can be utilized. On this basis, calculate $\\eta_{redox}$.", "text_images": [], "full_mark": 3, "rubric": ["$\\eta_{redox} \\approx 0.709 \\times [1 - (RT ln 7 + RT ln 8)/220] \\approx 0.676$ (3 points)"], "rubric_images": []}]} -{"problem_id": "Exam-02-T1", "header": "Problem 1: Carbon Coordination! Since Frankland reported Et_2Zn in 1849, molecules containing carbon-metal bonds have been known and have continually attracted intensive study. Carbon atoms have been found to exhibit a wide variety of electron-donation numbers and coordination numbers when coordinated to metals.", "header_images": [], "subquestions": [{"sub_id": "Exam-02-T1-1", "tags": ["1.4", "1.6"], "text_open": "1-1 For the ligands Ph_3PCPPh_3 (denoted L) and RCR (where R is an alkyl group), state the coordination mode by which each preferentially coordinates to a transition-metal atom.", "text_images": [], "full_mark": 2, "rubric": ["L preferentially coordinates to a π acceptor, whereas the carbene preferentially coordinates to a π donor. (2 points)"], "rubric_images": []}, {"sub_id": "Exam-02-T1-2", "tags": ["1.1", "1.4"], "text_open": "1-2 L reacts with ZnI_2 or CdI_2 in a 1:1 ratio, but the products have different structures. The ZnI_2 complex has 2 mutually perpendicular mirror planes and 1 twofold rotation axis but no inversion center. The CdI_2 complex has only 1 mirror plane and 1 twofold rotation axis but does possess an inversion center. Draw both structures.", "text_images": [], "full_mark": 4, "rubric": ["Structure of the ZnI_2 complex: 'I[Zn]([L])I' (2 points).", "Structure of the CdI_2 complex: '[L][Cd]1(I)I[Cd](I)([L])I1' (2 points)."], "rubric_images": [{"file": "Exam-02-T1-2-1-rubric.png", "description": "Schematic structure of the ZnI_2 complex: 'I[Zn]([L])I'"}, {"file": "Exam-02-T1-2-2-rubric.png", "description": "Schematic structure of the CdI_2 complex: '[L][Cd]1(I)I[Cd](I)([L])I1'"}]}, {"sub_id": "Exam-02-T1-3", "tags": ["2.1", "2.4", "1.4"], "text_open": "1-5 As stated in the introduction, carbon atoms can donate different numbers of electrons in complexes. Reaction of WCl_6 with (CH_3)_3CCH_2Li and Ph_2PCH_2CH_2PPh_2 gives a mononuclear W complex containing 3 carbon-tungsten bonds whose lengths are all distinctly different. Write the equation for its preparation.", "text_images": [], "full_mark": 3, "rubric": ["WCl_6 + 6(CH_3)_3CCH_2Li + Ph_2PCH_2CH_2PPh_2 → 3C(CH_3)_4 + 6LiCl + W(Ph_2PCH_2CH_2PPh_2)[(CH_3)_3CCH_2][(CH_3)_3CCH][(CH_3)_3CC] (3 points)."], "rubric_images": []}]} -{"problem_id": "Exam-02-T3", "header": "Problem 3: Bi-Ion Removal Battery. A bi-ion removal battery (BDI) can intercalate cations and anions from seawater into crystalline phases analogous to those in lithium-ion batteries and store them through conversion reactions in the cathode and anode, enabling efficient seawater desalination with high ion-removal capacity and high charge efficiency. Studies show that a $\\mathrm{NaBi_3O_4Cl_2}$ electrode can simultaneously remove $\\mathrm{Na^+}$ and $\\mathrm{Cl^-}$ from seawater.", "header_images": [{"file": "Exam-02-T3-header-1.png", "description": "On the left is a stacked structure of red blocks. Following the process indicated by the blue arrow “$\\mathrm{-Na^+, -Cl^-}$,” it becomes the central stacked structure of red and orange blocks. The process indicated by the red arrow “$\\mathrm{+Na^+, +Cl^-}$” then converts it to the structure on the right, composed of red and blue blocks. A further process indicated by the blue arrow “$\\mathrm{-Na^+, -Cl^-}$” returns it to the central structure. The legend identifies red blocks as $\\mathrm{NaBi_3O_4Cl_2}$, blue blocks as substance X, and orange blocks as $\\mathrm{Bi}$."}], "subquestions": [{"sub_id": "Exam-02-T3-1", "tags": ["2.1", "3.5"], "text_open": "Substance X is $BiOCl$.\n3-2-3 Write the electrode-reaction equations occurring at the $\\mathrm{NaBi_3O_4Cl_2}$ electrode during charging.", "text_images": [], "full_mark": 4, "rubric": ["$\\mathrm{Na^+ + 3Bi + 2Cl^- + 4H_2O - 9e^- = NaBi_3O_4Cl_2 + 8H^+}$ (2 points; $\\mathrm{Na^+ + 3Bi + 2Cl^- + 8OH^- - 9e^- = NaBi_3O_4Cl_2 + 4H_2O}$ receives 1 point).", "$\\mathrm{Bi + Cl^- + H_2O - 3e^- = BiOCl + 2H^+}$ (2 points)", "$\\mathrm{Bi + Cl^- + 2OH^- - 3e^- = BiOCl + H_2O}$ (1 point).", "Combining the two equations above gives:", "$\\mathrm{3xBi + yNa^+ + (3x-y)Cl^- + (3x+y)H_2O - 9e^- = yNaBi_3O_4Cl_2 + (3x - 3y)BiOCl + (6x + 2y)H^+}$ (4 points)"], "rubric_images": []}, {"sub_id": "Exam-02-T3-2", "tags": ["3.5"], "text_open": "3-3 To construct a complete battery, $\\mathrm{NaTi_2(PO_4)_3@C}$ is used as the electrode on the other side. 3-3-1 State the type of the $\\mathrm{NaTi_2(PO_4)_3@C}$ electrode during seawater desalination.", "text_images": [], "full_mark": 2, "rubric": ["Cathode (2 points)"], "rubric_images": []}, {"sub_id": "Exam-02-T3-3", "tags": ["2.1", "3.5"], "text_open": "3-3-2 Write the electrode-reaction equation occurring at the $\\mathrm{NaTi_2(PO_4)_3@C}$ electrode during seawater desalination.", "text_images": [], "full_mark": 2, "rubric": ["$\\mathrm{NaTi_2(PO_4)_3@C + 2Na^+ + 2e^- = Na_3Ti_2(PO_4)_3@C}$ (2 points)", "Alternatively, $\\mathrm{NaTi_2(PO_4)_3@C + xNa^+ + xe^- = Na(1+x)Ti_2(PO_4)_3@C}$ (2 points)"], "rubric_images": []}]} -{"problem_id": "Exam-02-T5", "header": "Problem 5: Simple Molecules, Complex Reactions. Precise modification of selected sites in a molecule is among the most elegant yet challenging transformations in chemistry, encompassing molecular editing at the single-atom level and biomimetic modification of small molecules.", "header_images": [], "subquestions": [{"sub_id": "Exam-02-T5-1", "tags": ["2.1", "2.2"], "text_open": "5-1 A stable reagent A for introducing a single carbon atom was recently reported. It is prepared as follows: excess PPh_3 reacts with CH_2Br_2 to give compound B (reaction 1); B reacts with OR^- (R is an alkyl group) to give key intermediate C (reaction 2); and C reacts with an equimolar amount of N_2O to ultimately give A (reaction 3). Molecular-orbital analysis confirms that the central carbon atom in A is a **zero-valent carbon atom** stabilized by common ligands. 5-1-1 Write the equations for reactions 1-3.", "text_images": [], "full_mark": 6, "rubric": ["2PPh_3 + CH_2Br_2 = [CH_2(PPh_3)_2]Br_2 (2 points).", "[CH_2(PPh_3)_2]^{2-} + 2OR^- = C(PPh_3) + 2ROH (2 points);", "C(PPh_3)_2 + N_2O = PPh_3CN_2 + OPPh_3 (2 points);"], "rubric_images": []}, {"sub_id": "Exam-02-T5-2", "tags": ["1.6", "1.5"], "text_open": "5-1-2 Write one SMILES representation of A, with the requirement that the carbon atom carry no formal charge in the corresponding Lewis structure. State whether the central P-C-N unit is linear or nonlinear.", "text_images": [], "full_mark": 3, "rubric": ["The SMILES representation of A is '[N-]=[N+]=C=P(C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3' (2 points).", "The central P-C-N unit is nonlinear (1 point)."], "rubric_images": [{"file": "Exam-02-T5-2-1-rubric.png", "description": "Schematic Lewis structure of A, with SMILES '[N-]=[N+]=C=P(C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3'."}]}, {"sub_id": "Exam-02-T5-3", "tags": ["1.5", "1.6"], "text_open": "5-1-3 Write the SMILES representations of all other reasonable resonance structures of A.", "text_images": [], "full_mark": 3, "rubric": ["Other reasonable resonance-structure SMILES for A include '[N-]=[N+]=[C-][P+](C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3', 'N#[N+][C-2][P+](C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3', and 'N#[N+][C-]=P(C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3'. Each reasonable resonance structure is worth 1 point. If extra structures are written, evaluate only the first three."], "rubric_images": [{"file": "Exam-02-T5-3-1-rubric.png", "description": "Other reasonable resonance-structure SMILES for A include '[N-]=[N+]=[C-][P+](C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3', 'N#[N+][C-2][P+](C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3', and 'N#[N+][C-]=P(C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3'. Each reasonable resonance structure is worth 1 point. If extra structures are written, evaluate only the first three."}]}, {"sub_id": "Exam-02-T5-4", "tags": ["2.1", "2.4", "2.2"], "text_open": "5-1-4 Reagent A reacts with benzaldehyde to first form intermediate X after carbon-atom transfer. X can rearrange to stable product X_1 while also giving another product X_2. The molecular mass of X_2 is twice that of X_1. Write the equation for the reaction of A with benzaldehyde to give X_1, and write the SMILES representations of X_1 and X_2.", "text_images": [], "full_mark": 4, "rubric": ["PPh_3CN_2 + PhCHO = PhC≡CH + OPPh_3 + N_2 (2 points).", "The correct SMILES representation of X_1 is 'C#CC1=CC=CC=C1' (1 point).", "The correct SMILES representation of X_2 is 'C1(C=C=C=CC2=CC=CC=C2)=CC=CC=C1' (1 point)."], "rubric_images": [{"file": "Exam-02-T5-4-1-rubric.png", "description": "Structure of X_1, with SMILES 'C#CC1=CC=CC=C1'."}, {"file": "Exam-02-T5-4-2-rubric.png", "description": "Structure of X_2, with SMILES 'C1(C=C=C=CC2=CC=CC=C2)=CC=CC=C1'."}]}, {"sub_id": "Exam-02-T5-5", "tags": ["1.4", "1.5"], "text_open": "5-2 To study activation and conversion of nitrosobenzene (PhNO) during biological metabolism and to achieve reversible redox interconversion among three oxidation states of PhNO in one system, a research group designed a dicobalt system that realizes a biomimetic stepwise four-electron reduction of PhNO. The catalytic cycle is shown below; Cp* denotes the C_5Me_5^- ionic ligand. 5-2-1 Give the oxidation state of Co in N and the electron count around each Co ion. The Co ion in N is low-spin. Draw the d-orbital splitting diagram for Co in a tetrahedral field, approximated as regular tetrahedral although the actual geometry deviates greatly, and give the d-electron configuration.", "text_images": [{"file": "Exam-02-T5-5-1-prompt.png", "description": "Catalytic-cycle diagram. The cycle begins with compound N at the top center. Its cation is '[Cp*][Co]1([N]#CC)[S]([Co]([Cp*])([N]#CC)[S]1C2=CC=CC=C2)C3=CC=CC=C3', with 2 PF_6^- anions. Along the arrow on the right, PhNO is added and 2 MeCN molecules depart to give complex D, represented below as {PhNO}^0. D gains 1 electron to form E, and E can reversibly lose an electron to regenerate D. E gains 1 electron to form F, and F can reversibly lose an electron to regenerate E. Along the left arrow from F, reaction with PhSH and NaBPh_4 eliminates NaSPh and gives complex G; loss of 1 H^+ from G gives F. G reacts with BH_3 to form by-product PhNH_2 and complex H. H reacts with H^+, MeCN, and NH_4PF_6, eliminates 1 H_2O molecule, and returns to the starting structure N, completing the cycle."}], "full_mark": 4, "rubric": ["Co is in the +3 oxidation state (1 point).", "18 electrons (1 point).", "In the schematic d-orbital splitting and electron-configuration diagram for the Co ion, 2 points are awarded only if the d-orbital splitting pattern, electron configuration, and orbital labels are all correct."], "rubric_images": [{"file": "Exam-02-T5-5-1-rubric.png", "description": "Schematic d-orbital splitting and electron configuration of the Co ion. The upper level contains three degenerate t_2 orbitals, two of which each contain one electron of the same spin. The lower level contains two degenerate e orbitals, each filled with two electrons of opposite spin."}]}, {"sub_id": "Exam-02-T5-6", "tags": ["1.4", "2.2"], "text_open": "5-2-2 In intermediates D-H, PhNO and its derived ligands are all bridging ligands and all satisfy the EAN rule. Only in complex F does the ligand adopt an unsymmetrical μ-η^1:η^2 coordination mode. Write the SMILES representations of complexes D, F, G, and H; the anions may be abbreviated.", "text_images": [], "full_mark": 8, "rubric": ["Cation SMILES of D: '[Cp*][Co]1([N](C2=CC=CC=C2)=O3)[S]([Co]3([Cp*])[S]1C4=CC=CC=C4)C5=CC=CC=C5'; the anions are 2PF_6^- (2 points).", "SMILES of F: '[Cp*][Co]([N](C1=CC=CC=C1)2O3)(SC4=CC=CC=C4)[S]([Co]23[Cp*])C5=CC=CC=C5' (2 points).", "Cation SMILES of G: '[Cp*][Co]1([N]([H])(C2=CC=CC=C2)O3)[S]([Co]3([Cp*])[S]1C4=CC=CC=C4)C5=CC=CC=C5'; the anion is BPh_4^- (2 points).", "Cation SMILES of H: '[Cp*][Co]12[S]([Co]([Cp*])([OH]2)[S]1C3=CC=CC=C3)C4=CC=CC=C4'; the anion is 1BPh_4^- (2 points).", "Deduct 1 point if the anion is omitted or incorrect.", "If F is written with both SPh^- ligands bridging, as '[Cp*][Co]([N](C1=CC=CC=C1)2O3)([S]4C5=CC=CC=C5)[S]([Co]234[Cp*])C6=CC=CC=C6', award 1 point."], "rubric_images": [{"file": "Exam-02-T5-6-1-rubric.png", "description": "Structure of D. The cation is '[Cp*][Co]1([N](C2=CC=CC=C2)=O3)[S]([Co]3([Cp*])[S]1C4=CC=CC=C4)C5=CC=CC=C5' and carries 2 units of positive charge; the anions are 2PF_6^-."}, {"file": "Exam-02-T5-6-2-rubric.png", "description": "The structure of F is '[Cp*][Co]([N](C1=CC=CC=C1)2O3)(SC4=CC=CC=C4)[S]([Co]23[Cp*])C5=CC=CC=C5'."}, {"file": "Exam-02-T5-6-3-rubric.png", "description": "Structure of G. The cation is '[Cp*][Co]1([N]([H])(C2=CC=CC=C2)O3)[S]([Co]3([Cp*])[S]1C4=CC=CC=C4)C5=CC=CC=C5' and carries 1 unit of positive charge; the anion is 1BPh_4^-."}, {"file": "Exam-02-T5-6-4-rubric.png", "description": "Structure of H. The cation is '[Cp*][Co]12[S]([Co]([Cp*])([OH]2)[S]1C3=CC=CC=C3)C4=CC=CC=C4' and carries 1 unit of positive charge; the anion is 1BPh_4^-."}, {"file": "Exam-02-T5-6-5-rubric.png", "description": "An alternative structure for F, worth only 1 point, is '[Cp*][Co]([N](C1=CC=CC=C1)2O3)([S]4C5=CC=CC=C5)[S]([Co]234[Cp*])C6=CC=CC=C6'."}]}, {"sub_id": "Exam-02-T5-7", "tags": ["3.5", "3.6", "3.2"], "text_open": "5-2-4 The research group measured reduction potentials, pK_a values, and other data for the relevant species at ambient temperature:\nD + e^- = E, E(1) = -0.075 V\nE + e^- = F, E(2) = -1.08 V\nG = F + H^+, pK_a = 13-16\nTaking the pK_a of G as 13, calculate the Gibbs energy changes for conversion of D to G (reaction a) and homolytic cleavage of the N-H bond in G (reaction b). The ionization energy of a hydrogen atom is 13.6 eV.", "text_images": [], "full_mark": 7, "rubric": ["From ΔG = -zFE: ΔG(1) = -zFE(1) = 7.23 kJ mol^-1 (1 point).", "ΔG(2) = -zFE(2) = 104.2 kJ mol^-1 (1 point).", "From ΔG = -RTlnK: ΔG(3) = -8.314 × 298.15 × ln(10^-13) = 74.2 kJ mol^-1 (1 point).", "Gibbs energy change for reaction a: ΔG(a) = ΔG(1) + ΔG(2) - ΔG(3) = 37.2 kJ mol^-1 (1 point).", "Gibbs energy change for ionization of H: ΔG(4) = I(H) = 13.6 eV × 96485 = 1312 kJ mol^-1 (1 point).", "Gibbs energy change for reaction b: ΔG(b) = ΔG(3) - ΔG(a) - ΔG(4) = -1275 kJ mol^-1 (2 points).", "No credit is awarded if the unit is omitted or incorrect."], "rubric_images": []}]} +{"problem_id": "Exam-01-T9", "header": "How Well Do You Really Know Carbon-Phosphorus Double Bonds?\nIn studying organic chemistry, we seem to restrict our understanding of the reactivity of carbon-phosphorus double bonds to their “metathesis” reactions with carbon-oxygen double bonds. With the assistance of transition metals, however, double bonds between carbon and trivalent phosphorus display distinctive modes of formation and reactivity. In this problem, whenever a subquestion asks for the SMILES of a complex, supplement it with a verbal description of the specific coordination mode.", "header_images": [], "subquestions": [{"sub_id": "Exam-01-T9-1", "tags": ["1.4", "2.2", "2.4"], "text_open": "9-1 Carbon-phosphorus double bonds can be constructed in many ways and can coordinate to transition metals in more than one mode. 9-1-1 In the reaction below, [Mn] denotes (η^{5}-C_{5}H_{4}CH_{3})Mn(CO)_{2}, and the reaction produces 4 molecules of THF. Given that the major product contains 2 phosphorus-oxygen single bonds, write the SMILES structure of the major product. 4 [Mn]·THF + 2 ^{t}BuCP + H_{2}O →", "text_images": [], "full_mark": 2, "rubric": ["The SMILES structure of the major product is '[H]/C(C(C)(C)C)=P\\O/P=C([H])/C(C)(C)C.[Mn].[Mn].[Mn].[Mn]' (2 points). The P atoms and P=C double bonds coordinate to the 4 [Mn] fragments, respectively."], "rubric_images": [{"file": "Exam-01-T9-1-1-rubric.png", "description": "The chemical structure drawn in red shows the coordination of one ligand to four '[Mn]' metal fragments. The ligand has the SMILES '[H]/C(C(C)(C)C)=P\\O/P=C([H])/C(C)(C)C'. A central oxygen atom (O) bridges two phosphorus atoms (P); each P forms a double bond (P=C) to a carbon atom bearing one hydrogen atom (H) and one tert-butyl group (tBu). Four arrows explicitly indicate the coordination sites: the two arrows on the right originate from the upper and lower P atoms and point to the corresponding two '[Mn]' fragments on the right, whereas the two arrows on the left originate from the centers of the upper and lower P=C double bonds and point to the corresponding two '[Mn]' fragments on the left."}]}, {"sub_id": "Exam-01-T9-2", "tags": ["1.4", "2.2", "2.3"], "text_open": "9-1-2 The following example provides an initial illustration of the role played by transition metals in forming carbon-phosphorus double bonds. Use SMILES structures and any necessary verbal descriptions to represent the three key electrically neutral intermediates in the reaction. Coordination of double bonds to Mo need not be shown in any of them.", "text_images": [{"file": "Exam-01-T9-2-1-prompt.png", "description": "The reaction in the figure shows the anionic dinuclear molybdenum complex 'O=C[Mo](C=O)[Mo](PC1=CC=CC=C1)(C=O)C=O' reacting with acryloyl chloride, 'ClC(C=C)=O', to form a product in which the phosphaalkene ligand 'C/C=P\\C2=CC=CC=C2' is bound to the dinuclear molybdenum framework 'O=C[Mo](C=O)[Mo](C=O)C=O'. In the reactant, a phenylphosphido group bridges two Mo atoms, each bearing two carbonyl ligands; on one side, a coordinate bond shown by an arrow points to the left Mo atom. After reaction, the newly formed phosphaalkene ligand coordinates in a bidentate bridging mode: the lone pair on phosphorus forms a coordinate bond, shown by an arrow, to the left Mo atom, while the $P=C$ double bond coordinates through a $\\pi$ bond, shown by an arrow, to the right Mo atom."}], "full_mark": 3, "rubric": ["From left to right, the figure shows three red dinuclear molybdenum complex structures. **First structure on the left:** its core is a dinuclear molybdenum carbonyl framework containing two mononuclear molybdenum '[Mo]' units and four carbonyl '[C-]#[O+]' ligands. A phenylphosphido ligand '[PH]C1=CC=CC=C1' bridges over the top, coordinating to the left Mo atom through a coordinate bond on the left and bonding to the right Mo atom through a single bond on the right. An acryloyl ligand 'C=CC=O' is also bonded below the right Mo atom. **Second structure in the middle:** this structure is similar to the first and likewise contains a dinuclear molybdenum carbonyl framework and a bridging phenylphosphido ligand '[PH]C1=CC=CC=C1', but a vinyl ligand 'C=C', formed after loss of a carbonyl group, is bonded below the right Mo atom. **Third structure on the right:** a phenyl(vinyl)phosphine ligand 'C=CP([H])C1=CC=CC=C1' coordinates above the dinuclear molybdenum carbonyl framework in a bidentate bridging mode. Specifically, the lone pair on phosphorus (P) forms a coordinate bond to the left Mo atom, while the $C=C$ double bond coordinates through a $\\pi$ bond to the right Mo atom. (1 point each) No credit is awarded for either of the first two intermediates if P-to-Mo coordination is omitted. No points are deducted if double-bond-to-Mo coordination is omitted from the third intermediate."], "rubric_images": [{"file": "Exam-01-T9-2-1-rubric.png", "description": "From left to right, the figure shows three red dinuclear molybdenum complex structures. **First structure on the left:** its core is a dinuclear molybdenum carbonyl framework containing two mononuclear molybdenum '[Mo]' units and four carbonyl '[C-]#[O+]' ligands. A phenylphosphido ligand '[PH]C1=CC=CC=C1' bridges over the top, coordinating to the left Mo atom through a coordinate bond on the left and bonding to the right Mo atom through a single bond on the right. An acryloyl ligand 'C=CC=O' is also bonded below the right Mo atom. **Second structure in the middle:** this structure is similar to the first and likewise contains a dinuclear molybdenum carbonyl framework and a bridging phenylphosphido ligand '[PH]C1=CC=CC=C1', but a vinyl ligand 'C=C', formed after loss of a carbonyl group, is bonded below the right Mo atom. **Third structure on the right:** a phenyl(vinyl)phosphine ligand 'C=CP([H])C1=CC=CC=C1' coordinates above the dinuclear molybdenum carbonyl framework in a bidentate bridging mode. Specifically, the lone pair on phosphorus (P) forms a coordinate bond to the left Mo atom, while the $C=C$ double bond coordinates through a $\\pi$ bond to the right Mo atom."}]}, {"sub_id": "Exam-01-T9-3", "tags": ["1.4", "2.2", "2.6"], "text_open": "9-1-3 Write the SMILES structures of the six key intermediates in the reaction shown below. Note: the step involving addition of tert-butyllithium corresponds to three key intermediates; each other step corresponds to one key intermediate.", "text_images": [{"file": "Exam-01-T9-3-1-prompt.png", "description": "Description of the structures/reaction in the figure: '[H]C#CC1=CC=CC=C1' undergoes steps involving NEt_3/PCl_3, ArLi, tBuLi, ML_n, and irradiation/O_2 to form a species whose ligand has the SMILES 'O=C(C(/C(C(C1=CC=CC=C1)=O)=P/C)=P\\C)C2=CC=CC=C2'. The ligand coordinates to [ML_n] through its two P atoms. Ar denotes an aryl substituent, and [ML_n] denotes a metal coordination fragment."}], "full_mark": 6, "rubric": ["6 points total for the six key intermediates in the grading figure. SMILES of the first intermediate: 'ClP(C#CC1=CC=CC=C1)Cl'; second intermediate: 'ClP(C#CC1=CC=CC=C1)[Ar]'; third intermediate: '[Ar]P(C#CC1=CC=CC=C1)P([Ar])C#CC2=CC=CC=C2'; fourth intermediate: '[Ar]P=C=C(C1=CC=CC=C1)C(C2=CC=CC=C2)=C=P[Ar]'; fifth intermediate: '[Ar]/P=C(C(C1=CC=CC=C1)=C2C3=CC=CC=C3)\\C2=P/[Ar]'; sixth intermediate: '[Ar]/P=C1/C(C(C2=CC=CC=C2)=C1C3=CC=CC=C3)=P/[Ar].[M]', with the ligand coordinating to [M] through both P atoms. Each intermediate is worth 1 point. Writing the lithium phosphinoalkyne intermediate 'Ph-C#C-P(Ar)Li' earns no credit but incurs no deduction."], "rubric_images": [{"file": "Exam-01-T9-3-1-rubric.png", "description": "Description of the structures/reaction in the figure. SMILES of the first intermediate: 'ClP(C#CC1=CC=CC=C1)Cl'; second intermediate: 'ClP(C#CC1=CC=CC=C1)[Ar]'; third intermediate: '[Ar]P(C#CC1=CC=CC=C1)P([Ar])C#CC2=CC=CC=C2'; fourth intermediate: '[Ar]P=C=C(C1=CC=CC=C1)C(C2=CC=CC=C2)=C=P[Ar]'; fifth intermediate: '[Ar]/P=C(C(C1=CC=CC=C1)=C2C3=CC=CC=C3)\\C2=P/[Ar]'; sixth intermediate: '[Ar]/P=C1/C(C(C2=CC=CC=C2)=C1C3=CC=CC=C3)=P/[Ar].[M]', with the ligand coordinating to [M] through both P atoms."}, {"file": "Exam-01-T9-3-2-rubric.png", "description": "Description of the structure/reaction in the figure: the additional structure is 'Ph-C#C-P(Ar)Li'. Writing this intermediate earns no credit but incurs no deduction."}]}, {"sub_id": "Exam-01-T9-4", "tags": ["2.2", "2.3", "1.4"], "text_open": "9-2 An important class of reactions of carbon-phosphorus double bonds involves metal carbenes. 9-2-1 In the following reaction, addition between the two substrates gives A_{1}, and A_{1} then converts into its isomer A_{2}. A_{2} undergoes a hydrogen migration to give A_{3}; A_{3} eliminates HNMe_{2} to give A_{4}; and a further elimination gives A_{5} together with byproduct A_{6}. A_{5} undergoes a simple reaction to afford the final product. W remains six-coordinate throughout the process. In A_{2}, W is coordinated by the carbon-phosphorus double bond, whereas in A_{3}, A_{4}, and A_{6}, W is coordinated by P. On this basis, write the SMILES structures of A_{1}~A_{6}.", "text_images": [{"file": "Exam-01-T9-4-1-prompt.png", "description": "Description of the structures/reaction in the figure: 'CN(C)C(=P)N(C)C' reacts with 'C1CC1C(O[Si](C)(C)C)=[W]([C]#O)([C]#O)([C]#O)([C]#O)[C]#O' to form 'O=C1C2(CC2)C(=O)C13CC3'."}], "full_mark": 6, "rubric": ["6 points total; A_{1}~A_{6} in the grading figure are worth 1 point each. **A_1**, at the far left of the first row, is a zwitterionic complex represented as 'CN(C)/C(N(C)C)=[PH+]/C(C1CC1)(O[Si](C)(C)C)[W-]([C-]#[O+])([C-]#[O+])([C-]#[O+])([C-]#[O+])[C-]#[O+]'; the negatively charged W center bearing five carbonyl ligands is bonded directly to the ligand carbon atom. **A_2**, second from the left in the first row, is an $\\eta^2$-coordinated complex represented as 'CN(C)C(N(C)C)=P=C(C2CC2)O[Si](C)(C)C.[W]([C-]#[O+])([C-]#[O+])([C-]#[O+])([C-]#[O+])[C-]#[O+]'; its $P=C$ double bond coordinates through a $\\pi$ bond to the pentacarbonyltungsten fragment. **A_3**, second from the right in the first row, is a $\\sigma$-coordinated complex represented as 'CN(C)C(N(C)C)/P([W]([C-]#[O+])([C-]#[O+])([C-]#[O+])([C-]#[O+])[C-]#[O+])=C(C3CC3)/O[Si](C)(C)C'; the P atom of the phosphaalkene ligand coordinates to the pentacarbonyltungsten center. **A_4**, at the far right of the first row, is likewise a $\\sigma$-coordinated complex bearing a cyclopropylidene substituent and is represented as 'CN(C)/C=P([W]([C-]#[O+])([C-]#[O+])([C-]#[O+])([C-]#[O+])[C-]#[O+])/C(O[Si](C)(C)C)=C4CC\\4'. **A_5**, on the left of the second row, is free cyclopropylketene represented as 'O=C=C5CC5'. **A_6**, on the right of the second row, is also a $\\sigma$-coordinated complex; its silyl-substituted coordinated structure is represented as 'CN(C)/C=P([W]([C-]#[O+])([C-]#[O+])([C-]#[O+])([C-]#[O+])[C-]#[O+])/[Si](C)(C)C', again with P coordinated to the pentacarbonyltungsten center. Credit is also awarded if the second intermediate is drawn with the other double bond coordinated to the metal."], "rubric_images": [{"file": "Exam-01-T9-4-1-rubric.png", "description": "The figure shows six structures in red, labeled A_1 through A_6. **A_1**, at the far left of the first row, is a zwitterionic complex represented as 'CN(C)/C(N(C)C)=[PH+]/C(C1CC1)(O[Si](C)(C)C)[W-]([C-]#[O+])([C-]#[O+])([C-]#[O+])([C-]#[O+])[C-]#[O+]'; the negatively charged W center bearing five carbonyl ligands is bonded directly to the ligand carbon atom. **A_2**, second from the left in the first row, is an $\\eta^2$-coordinated complex represented as 'CN(C)C(N(C)C)=P=C(C2CC2)O[Si](C)(C)C.[W]([C-]#[O+])([C-]#[O+])([C-]#[O+])([C-]#[O+])[C-]#[O+]'; its $P=C$ double bond coordinates through a $\\pi$ bond to the pentacarbonyltungsten fragment. **A_3**, second from the right in the first row, is a $\\sigma$-coordinated complex represented as 'CN(C)C(N(C)C)/P([W]([C-]#[O+])([C-]#[O+])([C-]#[O+])([C-]#[O+])[C-]#[O+])=C(C3CC3)/O[Si](C)(C)C'; the P atom of the phosphaalkene ligand coordinates to the pentacarbonyltungsten center. **A_4**, at the far right of the first row, is likewise a $\\sigma$-coordinated complex bearing a cyclopropylidene substituent and is represented as 'CN(C)/C=P([W]([C-]#[O+])([C-]#[O+])([C-]#[O+])([C-]#[O+])[C-]#[O+])/C(O[Si](C)(C)C)=C4CC\\4'. **A_5**, on the left of the second row, is free cyclopropylketene represented as 'O=C=C5CC5'. **A_6**, on the right of the second row, is also a $\\sigma$-coordinated complex; its silyl-substituted coordinated structure is represented as 'CN(C)/C=P([W]([C-]#[O+])([C-]#[O+])([C-]#[O+])([C-]#[O+])[C-]#[O+])/[Si](C)(C)C', again with P coordinated to the pentacarbonyltungsten center."}]}, {"sub_id": "Exam-01-T9-5", "tags": ["2.2", "2.3", "1.4"], "text_open": "9-2-2 When the phosphorus atom no longer bears a hydrogen atom, a hydrogen migration analogous to the process above cannot occur. In the following reaction, the two substrates undergo a process similar to that in the preceding subquestion, proceeding through B_{1} to B_{2}. Write the SMILES structure of one key electrically neutral intermediate in the conversion of B_{2} to the products.", "text_images": [{"file": "Exam-01-T9-5-1-prompt.png", "description": "Description of the structures/reaction in the figure: 'CN(C)C(=P[Fe])N(C)C' + 'CCOC(=[M])C' -> 'CN(C)/C=C/C(=[M])OCC' + 'CCOC(C)=P([Fe])[M]. Here M denotes a metal. The ligands on Fe and M are omitted in this question.'"}], "full_mark": 3, "rubric": ["SMILES structure of the key electrically neutral intermediate: '[Fe]P(=C(CC(OCC)=[M])N(C)C)=C(C1CC1)O[Si](C)(C)C.[M]' (3 points). The C=P double bond coordinates to metal M; award 2 points if coordination of the double bond to the metal is omitted."], "rubric_images": [{"file": "Exam-01-T9-5-1-rubric.png", "description": "The figure shows a transition-metal complex in red whose core coordination system may be described by the SMILES '[Fe]P(=C(CC(OCC)=[M])N(C)C)=C(C1CC1)O[Si](C)(C)C.[M]'. In the detailed structure, the central phosphorus (P) atom is bonded directly to an iron fragment '[Fe]'. The double-bonded carbon on the left of P bears a dimethylamino group ('-N(CH_3)_2') and a side chain containing an ethoxy group ('-OEt') that forms a double bond to a metal fragment '[M]', producing a Fischer carbene structure ($C=[M]$). The double-bonded carbon on the right of P bears cyclopropyl and siloxy ('-OTMS') groups, and the right-hand $P=C$ double bond acts as a $\\pi$ donor and coordinates in an $\\eta^2$ mode, indicated by a coordination arrow, to another independent metal fragment '[M]'."}]}]} {"problem_id": "Exam-02-T6", "header": "Problem 6: Crystal Structures of Metal Superhydrides. On 2023-2-9, researchers from California State University published a study of metal superhydrides.", "header_images": [{"file": "Exam-02-T6-header-1.png", "description": "The image consists of three side-by-side crystal-structure diagrams labeled A, B, and C in black capital letters at their upper left. Each contains a unit-cell frame drawn with thin black lines, a pink ball-and-stick framework, and translucent colored polyhedra or spheres. In panel A, the black lines form an approximately cubic unit cell containing a large translucent blue cage-like polyhedron. Large dark-blue spheres occupy the cube center and all eight vertices. At the center is a polyhedral cage framework made of small pale-pink spheres and pink cylindrical bonds, defining the translucent blue polyhedron. On each face of the cube, pink spheres appear at positions near the middle left, middle right, upper middle, and lower middle. In panel B, thin black lines form an approximately cubic unit cell whose center contains a large translucent reddish-brown cage-like polyhedron. Large green spheres occupy the cube center and the midpoint of each edge. A polyhedral cage framework made of small pale-pink spheres and pink cylindrical bonds defines the translucent reddish-brown polyhedron. The framework contains two characteristic types of structural units. In-plane square units: near the inner side of every unit-cell face is a planar structure parallel to that face; 4 small pale-pink spheres at the upper-left, lower-left, upper-right, and lower-right positions on the side nearer the cell center form a small square. Vertex-directed tetrahedral units: viewed along a unit-cell vertex direction, the 3 pale-pink spheres nearest that vertex among the small squares on three adjacent faces form a triangular base. On the side of this triangle toward the unit-cell vertex is 1 pale-pink sphere outside the center of the triangle and not coplanar with it, with another 1 pale-pink sphere extending outward toward the vertex and bonded to it. These 5 spheres together form a tetrahedron-like unit containing a central atom. In panel C, thin black lines form an approximately hexagonal unit cell containing two overlapping translucent green cage-like polyhedra arranged diagonally from lower left to upper right, one in each of those regions. Each green polyhedron contains one large bright-green sphere near its center. The outer part of panel C is a continuous three-dimensional framework of pale-pink spheres and pink cylindrical bonds. Pink spheres lie around the edges and junction of the two green cages, and the pink bonds form multiple pentagonal, hexagonal, and oblique polygonal rings, resembling two connected or partially overlapping cages. The pink framework of the upper-right cage extends toward the upper-right and right edges of the cell, while that of the lower-left cage extends toward the lower-left and foreground regions. Overall, panel A is dominated by a blue cage polyhedron and blue spheres; panel B by a reddish-brown cage polyhedron and peripheral green spheres; and panel C by two translucent green cage polyhedra and two bright-green interior spheres. All three show the spatial relationship between the pink ball-and-stick framework and the unit-cell frame."}], "subquestions": [{"sub_id": "Exam-02-T6-1", "tags": ["1.3"], "text_open": "6-2 Panel C above shows a Ce superhydride crystal. 6-2-1 Describe how the polyhedra shown are connected, for example, which vertices, edges, or faces they share.", "text_images": [], "full_mark": 3, "rubric": ["They are interconnected by sharing all hexagonal and pentagonal faces (3 points).", "No credit for stating that they share 1 hexagonal face or pentagonal face."], "rubric_images": []}, {"sub_id": "Exam-02-T6-2", "tags": ["1.3"], "text_open": "6-2-3 Give the polyhedral notation for the polyhedron shown in C.", "text_images": [], "full_mark": 2, "rubric": ["Polyhedral notation: 4^{6}5^{6}6^{6} (2 points)"], "rubric_images": []}]} -{"problem_id": "Exam-02-T9", "header": "Problem 9: Selectivity in Chemistry", "header_images": [], "subquestions": [{"sub_id": "Exam-02-T9-1", "tags": ["2.7", "1.2", "2.4"], "text_open": "9-1 Reduction of the following compound with different reducing agents gives different stereochemical outcomes. Write the SMILES representation of major product 9A1 obtained with LiAlH_{4} and that of major product 9A2 obtained with ^{i}Bu_{2}AlH.", "text_images": [{"file": "Exam-02-T9-1-1-prompt.png", "description": "The structure in the image has SMILES 'O=C(C1)C[C@H]2N([C@@H]1CC2)C'."}], "full_mark": 4, "rubric": ["The structures of 9A1 and 9A2 are worth 2 points each; stereochemistry is strictly required.", "The SMILES representation of 9A1 is 'CN1[C@H]2CC[C@@H]1C[C@@H](O)C2'.", "The SMILES representation of 9A2 is 'CN3[C@H]4CC[C@@H]3C[C@H](O)C4'."], "rubric_images": [{"file": "Exam-02-T9-1-1-rubric.png", "description": "The rubric image shows 9A1 and 9A2. The SMILES representation of 9A1 is 'CN1[C@H]2CC[C@@H]1C[C@@H](O)C2', and that of 9A2 is 'CN3[C@H]4CC[C@@H]3C[C@H](O)C4'."}]}, {"sub_id": "Exam-02-T9-2", "tags": ["2.8"], "text_open": "$Zn(OTf)_2$ is used in the selective esterification shown below. It acts as a Lewis acid to activate the ester group and coordinates separately to the two substrates to control the reaction site.\n9-2-2 What amount of Zn(OTf)_{2} should be used in this reaction?", "text_images": [{"file": "Exam-02-T9-2-1-prompt.png", "description": "The image shows a reaction proceeding from left to right. At the far left is 'O=C(O/N=C/c2ncccc2)CCc3ccccc3', labeled **9B** below, joined by a plus sign to 'O[C@@H]1[C@@H](O)[C@H](CO)O[C@@H](OCCCCCCCC)[C@H]1NC(CO)=O' in the center. Above the reaction arrow is “Zn(OTf)_2”; below it are “DMF/THF/DME,” “(1:5:9),” and “30°C, 18h,” in sequence."}], "full_mark": 2, "rubric": ["A catalytic amount (2 points; an equivalent answer is also acceptable)."], "rubric_images": []}, {"sub_id": "Exam-02-T9-3", "tags": ["2.4", "2.7", "1.2"], "text_open": "9-2-3 Write the SMILES representation of the major product, given that only one hydroxy group is esterified.", "text_images": [], "full_mark": 2, "rubric": ["The major product shown in the rubric image (2 points) has SMILES 'CCCCCCCCO[C@H]1[C@H]([C@@H]([C@H]([C@@H](O1)CO)O)O)NC(COC(CCc2ccccc2)=O)=O'."], "rubric_images": [{"file": "Exam-02-T9-3-1-rubric.png", "description": "The rubric image shows the monoesterified major product for 9-2-3, with SMILES 'CCCCCCCCO[C@H]1[C@H]([C@@H]([C@H]([C@@H](O1)CO)O)O)NC(COC(CCc2ccccc2)=O)=O'."}]}, {"sub_id": "Exam-02-T9-4", "tags": ["2.2", "2.3"], "text_open": "9-3-2 Compound 9D is the addition product of acetonitrile and hydrogen peroxide, and 9G contains a peroxide bond. Write the SMILES representations of small molecules 9D and 9F in the catalytic cycle below.", "text_images": [{"file": "Exam-02-T9-4-1-prompt.png", "description": "Structure of compound 9C, with SMILES 'CC1(OC[C@]2(OC[C@H]3OC(C)(O[C@H]3C2=O)C)O1)C'."}, {"file": "Exam-02-T9-4-2-prompt.png", "description": "The image shows a reaction segment in which the far-left compound 'C/C=C/c1ccccc1' reacts to the right under an arrow labeled “9C (15 mol%), H_2O_2 (4 equiv.)” above and “K_2CO_3, CH_3CN” and “0°C, 12h” below."}, {"file": "Exam-02-T9-4-3-prompt.png", "description": "The image is a catalytic cycle. At the top is 'CC1(OC[C@]2(OC[C@H]3OC(C)(O[C@H]3C2=O)C)O1)C', labeled “9C” below. On the right, `CH_3CN + H_2O_2` points left by a straight arrow to “9D.” A curved arrow from “9D” runs down and left into the main cycle, merging with the downward path from “9C,” and both lead to “9E.” Two curved arrows descend from “9E,” one down-right to “9F” and the other down-left to “9G.” A curved arrow from “9G” runs up and left, meeting an upward curved arrow from 'C/C=C/c1ccccc1' at the lower left; together they lead to `[transition state]^‡`. Finally, a curved arrow from `[transition state]^‡` runs up and right to close the cycle at “9C,” with a curved branch along this upward path pointing up-left to the word “product.”"}], "full_mark": 4, "rubric": ["The SMILES representation of 9D is 'CC(OO)=N', and that of 9F is 'CC(N)=O'; 2 points each."], "rubric_images": [{"file": "Exam-02-T9-4-1-rubric.png", "description": "The rubric image gives the SMILES representation of 9D as 'CC(OO)=N' and that of 9F as 'CC(N)=O'."}]}, {"sub_id": "Exam-02-T9-5", "tags": ["2.7", "1.2", "2.4"], "text_open": "9-3-3 Predict the stereochemistry and write the SMILES representation of the product.", "text_images": [], "full_mark": 2, "rubric": ["The product, with SMILES 'C[C@@H]1[C@H](O1)C2=CC=CC=C2', is worth 2 points. No credit if (±) is written or the product is identified as a racemate."], "rubric_images": [{"file": "Exam-02-T9-5-1-rubric.png", "description": "The rubric image shows the asymmetric epoxidation product, with SMILES 'C[C@@H]1[C@H](O1)C2=CC=CC=C2'."}]}]} -{"problem_id": "Exam-03-T2", "header": "Problem 2 Metallabenzenes and Aromaticity. Note: throughout this problem, n-coordinate means that the coordination number of the atom is n.", "header_images": [], "subquestions": [{"sub_id": "Exam-03-T2-1", "tags": ["1.4", "2.2", "2.4"], "text_open": "2-1 The first metallabenzene, A, was prepared from Os(CO)(CS)(PPh_3)_3 and acetylene. Draw the missing structures in the following transformation scheme and give their SMILES. In A-D^+, Os is 6-coordinate, and A contains a three-membered ring.", "text_images": [{"file": "Exam-03-T2-1-1-prompt.png", "description": "Transformation scheme for A-D^+. At the upper left, Os(CO)(CS)(PPh_3)_3 is converted into boldfaced A by a right-pointing arrow labeled \"2C_2H_2\" above and \"- PPh_3\" below. From A, the scheme branches horizontally and vertically. Along the horizontal branch, A is converted into boldfaced B by a right-pointing arrow labeled \"CO\". A downward arrow from B, labeled \"MeI, NaClO_4\" on the right, leads to D^+ClO_4^- at the lower right, with D boldfaced. Along the vertical branch, A and boldfaced C below it are connected by two opposing half-arrows: the downward half-arrow to C is labeled \"HCl\" on the right, and the upward half-arrow to A is labeled \"-HCl\" on the left."}], "full_mark": 8, "rubric": ["2 points for each structure, 8 points total. A: 'O=C[Os]12([P](C3=CC=CC=C3)(C4=CC=CC=C4)C5=CC=CC=C5)([P](C6=CC=CC=C6)(C7=CC=CC=C7)C8=CC=CC=C8)=CC=CC=C1S2'; B: 'S=C1C=CC=C[Os]1([P]2(C3=CC=CC=C34CC=CC=C24)C5=CC=CC=C5)(C=O)(C=O)[P](C6=CC=CC=C6)(C7=CC=CC=C7)C8=CC=CC=C8'; C: 'Cl[Os]1([P](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4)([P](C5=CC=CC=C5)(C6=CC=CC=C6)C7=CC=CC=C7)(C=O)=CC=CC=C1S'; D^+: 'CSC1=CC=CC=[Os]1([P](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4)([P+](C5=CC=CC=C5)(C6=CC=CC=C6)C7=CC=CC=C7)(C=O)C=O'.", "Deduct 0.5 point if the charge on D^+ is omitted."], "rubric_images": [{"file": "Exam-03-T2-1-1-rubric.png", "description": "2-1 Grading structures. Four red chemical structures are arranged horizontally. From left to right: A, O=C[Os]12([P](C3=CC=CC=C3)(C4=CC=CC=C4)C5=CC=CC=C5)([P](C6=CC=CC=C6)(C7=CC=CC=C7)C8=CC=CC=C8)=CC=CC=C1S2; B, S=C1C=CC=C[Os]1([P]2(C3=CC=CC=C34CC=CC=C24)C5=CC=CC=C5)(C=O)(C=O)[P](C6=CC=CC=C6)(C7=CC=CC=C7)C8=CC=CC=C8; C, Cl[Os]1([P](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4)([P](C5=CC=CC=C5)(C6=CC=CC=C6)C7=CC=CC=C7)(C=O)=CC=CC=C1S; and, at the far right, the bracketed structure D^+, CSC1=CC=CC=[Os]1([P](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4)([P+](C5=CC=CC=C5)(C6=CC=CC=C6)C7=CC=CC=C7)(C=O)C=O."}]}, {"sub_id": "Exam-03-T2-2", "tags": ["1.4", "2.2", "2.4"], "text_open": "2-2 Mixing OsCl_2(PPh_3)_3 with a large excess of HC≡CTMS in wet benzene gives a yellow precipitate E and a brown solution. E was identified as a carbyne complex. Three substances were isolated from the solution: F (22% yield), G (4.5% yield), and the osma-benzyne H (30% yield). The authors proposed the key intermediate I, an osmium vinylcarbene complex.\n2-2-1 Draw E and I and give their SMILES. I contains no silicon.", "text_images": [{"file": "Exam-03-T2-2-1-prompt.png", "description": "Reaction and structure scheme for E, F, G, H, and I. On the left, OsCl_2(PPh_3)_3 is converted into boldfaced I by a right-pointing arrow labeled \"H—≡—TMS\" above and \"H_2O\" below. I is then converted into boldfaced E on the right by another right-pointing arrow labeled \"HCl\" above."}], "full_mark": 4, "rubric": ["2 points each for E and I, 4 points total. I: Cl[Os](Cl)([P](C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3)([P](C4=CC=CC=C4)(C5=CC=CC=C5)C6=CC=CC=C6)=C=C; E: CC#[Os](Cl)(Cl)(Cl)([P](C7=CC=CC=C7)(C8=CC=CC=C8)C9=CC=CC=C9)[P](C%10=CC=CC=C%10)(C%11=CC=CC=C%11)C%12=CC=CC=C%12."], "rubric_images": [{"file": "Exam-03-T2-2-1-rubric.png", "description": "2-2-1 Grading structures. Two red chemical structures are arranged horizontally. From left to right: I, Cl[Os](Cl)([P](C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3)([P](C4=CC=CC=C4)(C5=CC=CC=C5)C6=CC=CC=C6)=C=C; and E, CC#[Os](Cl)(Cl)(Cl)([P](C7=CC=CC=C7)(C8=CC=CC=C8)C9=CC=CC=C9)[P](C%10=CC=CC=C%10)(C%11=CC=CC=C%11)C%12=CC=CC=C%12."}]}, {"sub_id": "Exam-03-T2-3", "tags": ["2.2", "2.3", "2.4"], "text_open": "2-2-2 H has molecular formula C_48H_52Si_2P_2Cl_2Os. Draw the reaction intermediates in the conversion of I into H and give their SMILES, together with the SMILES of H.", "text_images": [], "full_mark": 6, "rubric": ["Each red intermediate is worth 2 points. Their SMILES are 'Cl[Os](Cl)(C7=C)(C=C7[Si](C)(C)C)([P](C8=CC=CC=C8)(C9=CC=CC=C9)C%10=CC=CC=C%10)[P](C%11=CC=CC=C%11)(C%12=CC=CC=C%12)C%13=CC=CC=C%13' and 'Cl[Os]1(Cl)([P](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4)([P](C5=CC=CC=C5)(C6=CC=CC=C6)C7=CC=CC=C7)C=C([Si](C)(C)C)C(=C)C([Si](C)(C)C)=C1'. The structure of H is worth 2 points: 'Cl[Os]1(Cl)([P](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4)([P](C5=CC=CC=C5)(C6=CC=CC=C6)C7=CC=CC=C7)#CC([Si](C)(C)C)=C(C([Si](C)(C)C)=C1)C'.", "If the red intermediates are omitted but the blue structures are given, award 1 point. The two blue intermediates have SMILES 'Cl[Os](Cl)([P](C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3)([P](C4=CC=CC=C4)(C5=CC=CC=C5)C6=CC=CC=C6)=CC([Si](C)(C)C)=C=C' and 'Cl[Os]7(Cl)([P](C8=CC=CC=C8)(C9=CC=CC=C9)C%10=CC=CC=C%10)([P](C%11=CC=CC=C%11)(C%12=CC=CC=C%12)C%13=CC=CC=C%13)C=C([Si](C)(C)C)[C+](C)C([Si](C)(C)C)=C7'."], "rubric_images": [{"file": "Exam-03-T2-3-1-rubric.png", "description": "2-2-2 Grading scheme 1. This multistep reaction scheme shows the intermediates and H in red. At the far left is the black starting material I, 'Cl[Os](Cl)([P](C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3)([P](C4=CC=CC=C4)(C5=CC=CC=C5)C6=CC=CC=C6)=C=C'. A right-pointing arrow labeled with '[H]C#C[Si](C)(C)C' leads to the first red five-membered-ring intermediate, 'Cl[Os](Cl)(C7=C)(C=C7[Si](C)(C)C)([P](C8=CC=CC=C8)(C9=CC=CC=C9)C%10=CC=CC=C%10)[P](C%11=CC=CC=C%11)(C%12=CC=CC=C%12)C%13=CC=CC=C%13'. An unlabeled right-pointing arrow then leads to the second red six-membered-ring intermediate, 'Cl[Os]1(Cl)([P](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4)([P](C5=CC=CC=C5)(C6=CC=CC=C6)C7=CC=CC=C7)C=C([Si](C)(C)C)C(=C)C([Si](C)(C)C)=C1'. Finally, a right-pointing arrow labeled H^+ gives the red final product H, 'Cl[Os]1(Cl)([P](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4)([P](C5=CC=CC=C5)(C6=CC=CC=C6)C7=CC=CC=C7)#CC([Si](C)(C)C)=C(C([Si](C)(C)C)=C1)C'."}, {"file": "Exam-03-T2-3-2-rubric.png", "description": "2-2-2 Grading scheme 2. The two blue intermediates have SMILES 'Cl[Os](Cl)([P](C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3)([P](C4=CC=CC=C4)(C5=CC=CC=C5)C6=CC=CC=C6)=CC([Si](C)(C)C)=C=C' and 'Cl[Os]7(Cl)([P](C8=CC=CC=C8)(C9=CC=CC=C9)C%10=CC=CC=C%10)([P](C%11=CC=CC=C%11)(C%12=CC=CC=C%12)C%13=CC=CC=C%13)C=C([Si](C)(C)C)[C+](C)C([Si](C)(C)C)=C7'."}]}, {"sub_id": "Exam-03-T2-4", "tags": ["1.6", "3.9"], "text_open": "2-3 In 2020, researchers reported an example of a nonplanar aromatic transition-metal heterocycle. The HOMO of the butadiene ligand combines with the d orbitals of the two Fe atoms to form a molecular orbital, producing stabilization and aromatic character.\n2-3-1 State the total number of d electrons contributed by the two Fe atoms to the aromatic orbital.", "text_images": [{"file": "Exam-03-T2-4-1-prompt.png", "description": "Structure of a nonplanar aromatic transition-metal heterocycle (SMILES: 'CC1=C([Fe]23[N](C)(C)CC[N]3(C)C)([Si](C)(C)C)[Fe](Br)(Br)C2([Si](C)(C)C)=C1C'). The center is a complex network comprising a planar four-carbon butadiene skeleton and two Fe atoms positioned above and below the plane. Both solid and dashed connections are shown. In the conventional solid-line bonding, the upper Fe is bonded to both N atoms of a TMEDA ligand above it; short strokes on the N atoms denote methyl groups. The lower Fe is bonded to two Br atoms, directed downward and to the right. The dashed lines depict multicenter coordination: two dashed lines extend downward from the upper Fe to the two TMS-substituted carbon atoms, and two dashed lines extend upward from the lower Fe to the same two TMS-substituted carbon atoms."}], "full_mark": 1, "rubric": ["Each Fe contributes 1 electron, for a total of 2 electrons. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-03-T2-5", "tags": ["1.6", "1.7", "3.9"], "text_open": "2-3-3 Explain why the ground-state spin multiplicity of this tetrahedral Fe(II) complex is 0. (No more than 100 words.)", "text_images": [], "full_mark": 2, "rubric": ["The Fe(II) centers interact electronically through the butadiene dianion ligand by antiferromagnetic coupling, so their unpaired-electron spins are opposite. (2 points)", "An answer of \"broken symmetry\" also receives full credit. (2 points)"], "rubric_images": []}, {"sub_id": "Exam-03-T2-6", "tags": ["1.6", "3.9"], "text_open": "2-4 For rare-earth metals, however, synthesizing analogous nonplanar aromatic heterocycles is difficult because of the d^0 electron configuration and the energy mismatch of the d orbitals. In 2024, researchers further reduced the butadiene ligand and prepared the first nonplanar aromatic rare-earth heterocycle. (The left figure shows the 1,3-butadiene dianion; the right figure shows the 2-butene tetraanion obtained after reduction.) Explain how the reduced ligand overcomes each of the two difficulties above. (No more than 60 words for each point.)", "text_images": [{"file": "Exam-03-T2-6-1-prompt.png", "description": "Rare-earth heterocycle structure scheme. A right-pointing arrow labeled \"Two-electron reduction\" connects two molecular structures. The reactant on the left is enclosed in black brackets bearing a ^{2-} charge and is labeled \"1,3-butadiene dianion\" below. Inside is an s-cis 1,3-butadiene-core derivative whose core may be represented as '[*]C=C([*])C([*])=C[*]'. The carbon skeleton is highlighted by thick gray lines. The top and bottom carbon-carbon bonds are double bonds, each drawn as a thick gray line parallel to a thin gray line, whereas the central vertical carbon-carbon bond is a single thick gray line. A dashed line extends outward from each of the four carbon vertices to denote an unspecified substituent. The product on the right is similarly enclosed in brackets bearing a ^{4-} charge and labeled \"2-butene tetraanion\" below. Its spatial arrangement is the same, but the skeleton is highlighted in thick blue. The central vertical carbon-carbon bond is now double, and the top and bottom bonds are single, corresponding to the core SMILES '[*]CC([*])=C([*])C[*]'. Each of its four vertices likewise bears an outward dashed substituent marker."}, {"file": "Exam-03-T2-6-2-prompt.png", "description": "Nonplanar aromatic rare-earth heterocycle. A complex bimetallic structure is fully enclosed by a thick purple circle; directly below it, bold red text reads \"RE = Sc, Sm, Lu, Gd\". Inside the circle on the right are two metal centers, one above the other, each shown in bold red brackets as \"[RE]\". On the left, thick blue lines depict the ligand skeleton corresponding to the 2-butene tetraanion in the preceding figure (SMILES '[*]CC([*])=C([*])C[*]', where the asterisks denote unspecified substituents). The ligand contains a vertical carbon-carbon double bond on the left, shown by a thick blue line and a thin black line. Each carbon of this double bond is joined by a thick blue single bond extending rightward to a terminal carbon node. All four carbon nodes carry outward dashed substituent markers. Each of the upper and lower terminal carbon nodes is connected by two separate thick blue lines to both the upper and lower [RE] centers."}], "full_mark": 4, "rubric": ["The reduced ligand has two additional electrons and supplies 6 electrons to the molecular orbital, compensating for the inability of RE to contribute d electrons. (2 points)", "After reduction, the higher-energy Ψ_3 orbital interacts with the d orbitals and is better energy-matched to the higher-energy d orbitals of RE. (2 points)"], "rubric_images": []}]} -{"problem_id": "Exam-03-T10", "header": "Problem 10 Synthesis of Indigoid Compounds (25 points)", "header_images": [], "subquestions": [{"sub_id": "Exam-03-T10-1", "tags": ["2.4", "2.2"], "text_open": "10-1 In 1909, H. Friedländer noted that an excellent purple pigment mentioned in the Bible—the pigment that the Romans called Tyrian purple after the Phoenician port of Tyre in present-day Lebanon—was in fact 6,6'-dibromoindigo, F. This precious early dye was extracted from small purple sea snails, the murex; as many as 12000 snails were required to prepare 1.5 g of dye. After nearly a century of continuing research into its synthesis, Pinkney and Chammers proposed a new method, which Iming and coworkers improved into an efficient route. 10-1-1 Give the SMILES of intermediate E.", "text_images": [{"file": "Exam-03-T10-1-1-prompt.png", "description": "10-1 Synthetic route. This multistep scheme begins at the upper left with 'CC1=CC=C(N)C=C1', which reacts under HNO_3 and H_2SO_4 to give intermediate B. B reacts under 1) NaNO_2, HBr and 2) CuBr to give C. C then reacts under CrO_3 and (CH_3CO)_2O to give D. In the second row, D reacts under H_2SO_4 and H_2O to give E. Finally, E reacts under NaOH and H_2O, (CH_3)_2CO to give the product F at the lower right, with SMILES 'O=C1NC2=C(C=C(Br)C=C2)C1=C3NC4=C(C=C(Br)C=C4)C3=O'."}], "full_mark": 2, "rubric": ["The SMILES of intermediate E is 'BrC(C=C1[N+]([O-])=O)=CC=C1C=O'. (2 points)"], "rubric_images": [{"file": "Exam-03-T10-1-1-rubric.png", "description": "The grading figure gives intermediate E with SMILES 'BrC(C=C1[N+]([O-])=O)=CC=C1C=O'."}]}, {"sub_id": "Exam-03-T10-2", "tags": ["2.7", "1.6"], "text_open": "10-1-2 Explain the regioselectivity of the conversion of A into B. (No more than 50 words.)", "text_images": [], "full_mark": 2, "rubric": ["The amino group is protonated and becomes an electron-withdrawing meta-directing group. (1 point) The methyl group is an ortho/para-directing group. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-03-T10-3", "tags": ["2.1", "2.2"], "text_open": "10-2 In the route above, conversion of E into F is clearly the key step. In fact, as early as 1882, Baeyer used this reaction to synthesize indigo in one step from o-nitrobenzaldehyde. 10-2-1 Write the chemical equation for this reaction, representing o-nitrobenzaldehyde and indigo by their SMILES.", "text_images": [{"file": "Exam-03-T10-3-1-prompt.png", "description": "One-step synthesis. At the far left is 'O=CC1=CC=CC=C1[N+]([O-])=O', labeled G below. Under NaOH and H_2O, (CH_3)_2CO, it gives the product at the far right, 'O=C1NC(C=CC=C2)=C2/C1=C3NC4=C(C=CC=C4)C/3=O', labeled H below."}], "full_mark": 2, "rubric": ["2NaOH + 2(CH_{3})_{2}CO + 2'O=CC1=CC=CC=C1[N+]([O-])=O' → 2CH_{3}COONa + 4H_{2}O + 'O=C1NC(C=CC=C2)=C2/C1=C3NC4=C(C=CC=C4)C/3=O'. (2 points) An incorrectly balanced equation earns only 1 point. An ionic equation earns only 1 point. No credit is awarded if the acid-base conditions are incorrect."], "rubric_images": [{"file": "Exam-03-T10-3-1-rubric.png", "description": "The o-nitrobenzaldehyde in the equation has SMILES 'O=CC1=CC=CC=C1[N+]([O-])=O'."}, {"file": "Exam-03-T10-3-2-rubric.png", "description": "The indigo in the equation has SMILES 'O=C1NC(C=CC=C2)=C2/C1=C3NC4=C(C=CC=C4)C/3=O'."}]}, {"sub_id": "Exam-03-T10-4", "tags": ["2.2", "2.3"], "text_open": "10-2-2 The final step of the reaction is shown below. Considering the byproducts formed in the equation in 10-2-1, give the SMILES of the four key intermediates in the conversion of G into M. For intermediates related by proton gain, loss, or transfer, only one need be given.", "text_images": [{"file": "Exam-03-T10-4-1-prompt.png", "description": "10-2-2 Question figure. A reaction proceeds from left to right. At the far left is 'O=C1C=NC2=C1C=CC=C2', labeled M below, followed by a plus sign and the middle molecule 'OC1=C(NC2=C1C=CC=C2)O', labeled N below. A one-way right-pointing reaction arrow leads to the product at the far right, 'O=C5NC(C=CC=C6)=C6/C5=C7NC8=C(C=CC=C8)C/7=O', labeled H below, indicating that M and N react to form H."}], "full_mark": 6, "rubric": ["6 points total, 1.5 points for each of the four key intermediates. Their SMILES are 'CC(CC(O)C1=CC=CC=C1[N+]([O-])=O)=O', 'CC(C1C(O)C2=C(C=CC=C2)[N+]1(O)[O-])=O', 'CC(C1C(C2=C(C=CC=C2)N1O)=O)=O', and 'O=C1CN(C2=C1C=CC=C2)O'. The fourth intermediate may instead be drawn in its carbonyl form, with SMILES 'ON1C2=C(C=CC=C2)C(O)C1C(O)([O-])C'. For each intermediate, another form obtainable by proton transfer, gain, or loss also receives credit."], "rubric_images": [{"file": "Exam-03-T10-4-1-rubric.png", "description": "Five chemical structures are shown side by side from left to right. The first four are red and the fifth, at the far right, is blue. A blue \"or\" separates the fourth and fifth structures. The red structures have SMILES 'CC(CC(O)C1=CC=CC=C1[N+]([O-])=O)=O', 'CC(C1C(O)C2=C(C=CC=C2)[N+]1(O)[O-])=O', 'CC(C1C(C2=C(C=CC=C2)N1O)=O)=O', and 'O=C1CN(C2=C1C=CC=C2)O'. The blue structure has SMILES 'ON1C2=C(C=CC=C2)C(O)C1C(O)([O-])C'."}]}, {"sub_id": "Exam-03-T10-5", "tags": ["2.2", "2.5"], "text_open": "10-3 Nitrogen plays a key role in the reaction above, but analogous substrates and methods are unsuitable for obtaining the sulfur-substituted product. After the nineteen-eighties, Scottish chemists synthesized thioindigo, K, using the reaction below. The reaction still proceeds through an intermediate containing a fused six- and five-membered ring system. It is known that contraction of the six-membered ring into the five-membered ring does not occur by ring opening followed by ring closure. Answer the following questions. 10-3-1 In what form does the oxygen atom of the sulfoxide depart?", "text_images": [{"file": "Exam-03-T10-5-1-prompt.png", "description": "Structure/reaction description for synthesis of thioindigo K. Sulfoxide J gives thioindigo K under NaOAc/HOAc. The SMILES of J is 'BrC(C1=O)=C([S+](C2=C1C=CC=C2)[O-])Br'; thioindigo K may be represented as 'O=C1C2=C(S/C1=C3SC4=C(C\\3=O)C=CC=C4)C=CC=C2'."}], "full_mark": 1, "rubric": ["AcO^{-}. (1 point; AcOH is also acceptable.)"], "rubric_images": []}, {"sub_id": "Exam-03-T10-6", "tags": ["2.3", "2.7"], "text_open": "10-3-2 During ring contraction, which atom is lost? ( ) (a) the carbon atom nearer the carbonyl group (b) the carbon atom nearer the sulfoxide group", "text_images": [], "full_mark": 2, "rubric": ["(b) (2 points)"], "rubric_images": []}, {"sub_id": "Exam-03-T10-7", "tags": ["2.2", "2.3"], "text_open": "10-3-4 Give the SMILES of the six key intermediates in the reaction. For intermediates related by proton gain, loss, or transfer, only one need be given. A substitution proceeding by addition-elimination or by an S_{N1} process counts as only one step.", "text_images": [], "full_mark": 9, "rubric": ["9 points total; each of the six key intermediates in the grading figure is worth 1.5 points. Their SMILES are 'O=C1C(Br)=C(OC(C)=O)[S+]([O-])C2=C1C=CC=C2', 'O=C1C(Br)=C([O-])[S+](OC(C)=O)C2=C1C=CC=C2', 'O=C1C2=C(C=CC=C2)[S+]3C(C13Br)=O', 'O=C1C2=C(C=CC=C2)SC1(Br)C(OC(C)=O)=O', 'OC1=C(SC2=C1C=CC=C2)Br', and 'O=C1C(C2(Br)SC3=C(C=CC=C3)C2=O)SC4=C1C=CC=C4'. For each intermediate, another form obtainable by proton transfer, gain, or loss also receives credit. The fifth intermediate may also be drawn in its carbonyl form."], "rubric_images": [{"file": "Exam-03-T10-7-1-rubric.png", "description": "The six key intermediates in the synthesis of thioindigo K have SMILES 'O=C1C(Br)=C(OC(C)=O)[S+]([O-])C2=C1C=CC=C2', 'O=C1C(Br)=C([O-])[S+](OC(C)=O)C2=C1C=CC=C2', 'O=C1C2=C(C=CC=C2)[S+]3C(C13Br)=O', 'O=C1C2=C(C=CC=C2)SC1(Br)C(OC(C)=O)=O', 'OC1=C(SC2=C1C=CC=C2)Br', and 'O=C1C(C2(Br)SC3=C(C=CC=C3)C2=O)SC4=C1C=CC=C4'."}]}]} +{"problem_id": "Exam-02-T10", "header": "Problem 10: Forever Unnamed", "header_images": [], "subquestions": [{"sub_id": "Exam-02-T10-1", "tags": ["2.2"], "text_open": "10-1 A neutral reaction intermediate can be obtained by the process shown below. Write its SMILES representation.", "text_images": [{"file": "Exam-02-T10-1-1-prompt.png", "description": "A chemical reaction equation: the reactants on the left are [R1]CC(=O)Cl and R_2NC, joined by a plus sign `+`; a horizontal right-pointing arrow in the center leads to bold X on the right."}], "full_mark": 2, "rubric": ["Write the SMILES of X shown in the rubric image: '[R2]/N=C(C(C[R1])=O)\\Cl' (2 points)."], "rubric_images": [{"file": "Exam-02-T10-1-1-rubric.png", "description": "The rubric image shows X, with SMILES '[R2]/N=C(C(C[R1])=O)\\Cl'."}]}, {"sub_id": "Exam-02-T10-2", "tags": ["2.4", "2.2", "1.2"], "text_open": "10-2-1 Further reactions can proceed from intermediates of type X. Write the SMILES representations of products A and B in the reactions below, given that B contains a five-membered aromatic ring. 10-2-2 Label the absolute configurations of all chiral carbon atoms in the structure of A from the preceding question.", "text_images": [{"file": "Exam-02-T10-2-1-prompt.png", "description": "The image shows two parallel reactions. In the upper reaction, the starting material is 'CC1=C(O[Si](C)(C)C(C)(C)C)CC[C@@H]1C[N+]#[C-]'. Above the arrow is “(a)tBuCOCl,DCM,r.t.” and below it “(b)AgBF_4,DCM,-78∘C”; the arrow leads to product A. In the lower reaction, '[R]C(Cl)=O' + '[Ar]C[N+]#[C-]' reacts under “base” to give product B."}], "full_mark": 4, "rubric": ["Structure A is worth 1 point; its SMILES is 'O=C1[C@]2(C)[C@](CN=C2C(C(C)(C)C)=O)([H])CC1'.", "The chiral carbon bonded to H has the S configuration (1 point), and the chiral carbon bonded to the methyl group has the R configuration (1 point).", "Structure B is worth 1 point; its SMILES is '[R]C1=CN=C(O1)[Ar]'."], "rubric_images": [{"file": "Exam-02-T10-2-1-rubric.png", "description": "The rubric image shows the structure and configurations of A, with SMILES 'O=C1[C@]2(C)[C@](CN=C2C(C(C)(C)C)=O)([H])CC1'. The H-bearing chiral carbon is S, and the methyl-bearing chiral carbon is R."}, {"file": "Exam-02-T10-2-2-rubric.png", "description": "The rubric image shows B, with SMILES '[R]C1=CN=C(O1)[Ar]'."}]}, {"sub_id": "Exam-02-T10-3", "tags": ["2.2", "2.3"], "text_open": "10-2-3 Write the SMILES representation of another electrically neutral key intermediate Y, other than the type-X intermediate, involved in formation of B.", "text_images": [], "full_mark": 2, "rubric": ["Write the SMILES of Y shown in the rubric image: '[R]C(C#[N+][CH-][Ar])=O' (2 points). No credit if multiple structures are given."], "rubric_images": [{"file": "Exam-02-T10-3-1-rubric.png", "description": "The rubric image shows Y, with SMILES '[R]C(C#[N+][CH-][Ar])=O'."}]}, {"sub_id": "Exam-02-T10-4", "tags": ["2.4", "2.3", "2.7"], "text_open": "10-2-4 If an appropriate amount of CH_{2}=CHCO_{2}Me is added to the reaction system that forms B, Y is trapped to form two products, C_{1} and C_{2}. Write their SMILES representations.", "text_images": [], "full_mark": 4, "rubric": ["Write the SMILES representations of C_{1} and C_{2} shown in the rubric image; 2 points each.", "C1 can be represented as 'O=C(C1=CC(C(N1)[Ar])C(OC)=O)[R]'.", "C2 can be represented as 'O=C(C1=C(CC(N1)[Ar])C(OC)=O)[R]'.", "The position of the double bond within the ring is not specified; an aromatized pyrrole product also receives credit."], "rubric_images": [{"file": "Exam-02-T10-4-1-rubric.png", "description": "The rubric image shows C1, which can be represented as 'O=C(C1=CC(C(N1)[Ar])C(OC)=O)[R]', and C2, which can be represented as 'O=C(C1=C(CC(N1)[Ar])C(OC)=O)[R]'."}]}, {"sub_id": "Exam-02-T10-5", "tags": ["2.2"], "text_open": "10-3-1 Addition of other reagents changes the reaction pathway. Write the small-molecule by-product formed in the reaction below and the SMILES representation of another electrically neutral key intermediate, other than the type-X intermediate, involved in the reaction.", "text_images": [{"file": "Exam-02-T10-5-1-prompt.png", "description": "A continuous multistep reaction sequence. The substrate is `R_1COCl`; above the first arrow is R_2NC; above the second is `P(OR_3)_3`; and above the third is '[O-]/C([Ar])=C\\[N+]1=CC=CC=C1'. The final product is 'O=C([Ar])C1=C(C([R1])=C2C=CC=CN12)N[R2]'."}], "full_mark": 3, "rubric": ["The small-molecule by-product may be written as P(O)(OR_3)_3, or as P(O)(OH)(OR_3)_2 and R_3Cl (1 point).", "The key intermediate can be represented as '[R1]C(OP(O[R3])(O[R3])=O)=C=N[R2]' or '[R1]C(O[P+](O[R3])(O[R3])O[R3])=C=N[R2]' (2 points)."], "rubric_images": [{"file": "Exam-02-T10-5-1-rubric.png", "description": "The rubric image gives two representations of the key intermediate in 10-3-1. The general SMILES can be '[R1]C(OP(O[R3])(O[R3])=O)=C=N[R2]' or '[R1]C(O[P+](O[R3])(O[R3])O[R3])=C=N[R2]'."}]}, {"sub_id": "Exam-02-T10-6", "tags": ["2.4", "2.2"], "text_open": "10-3-2 In the presence of Et_{3}N, EtO_{2}CCOCl, CyNC, P(Oi-Pr)_{3}, NCCH_{2}CN, and CS_{2} undergo a multicomponent reaction in MeCN. Write the SMILES representation of product D.", "text_images": [], "full_mark": 3, "rubric": ["Write the SMILES representation of D: 'N#C/C(C#N)=C1SC(C(OCC)=O)=C(S/1)NC2CCCCC2' (3 points)."], "rubric_images": [{"file": "Exam-02-T10-6-1-rubric.png", "description": "The rubric image shows multicomponent-reaction product D, with SMILES 'N#C/C(C#N)=C1SC(C(OCC)=O)=C(S/1)NC2CCCCC2'."}]}, {"sub_id": "Exam-02-T10-7", "tags": ["2.2", "2.4"], "text_open": "10-4 Under certain conditions, an α-carbonyl diazo compound can replace an acyl chloride in a similar reaction with an isocyanide. Write the SMILES representations of all intermediates and products in the scheme below. W and all three products contain a five-membered ring, but only the five-membered ring in E is aromatic; F also contains a six-membered ring.", "text_images": [{"file": "Exam-02-T10-7-1-prompt.png", "description": "A synthetic reaction scheme: '[R1]C(C(C([R1])=O)=[N+]=[N-])=O' is the starting material and is converted by an arrow to Z. Intermediate Z then reacts with `R_2NC` to form W. Intermediate **W** branches into three transformations: the upper branch reacts with `MeNHNH_2` to form E; the middle branch reacts with '[R1]C(C([R1])=O)=C=O' to form F; and the lower branch forms H under H^+ conditions."}], "full_mark": 10, "rubric": ["Write the SMILES representations of the five structures Z, W, E, F, and H in the rubric image; 2 points each, 10 points total.", "SMILES of **Z**: '[R1]C(C([R1])=C=O)=O'", "SMILES of **W**: 'O=C1C([R1])=C(O/C1=N/[R2])[R1]'", "SMILES of **E**: 'CN2N=C(C([R1])=C2[R1])C(N[R2])=O'", "SMILES of **F**: 'O=C3C([R1])=C(OC34N(C(C([R1])=C(O4)[R1])=O)[R2])[R1]'", "SMILES of **H**: 'O=C5C(N(C([R1])=C5[R1])[R2])=O'"], "rubric_images": [{"file": "Exam-02-T10-7-1-rubric.png", "description": "The image shows five red intermediate and product structures from the transformations above, labeled Z, W, E, F, and H. **Z** is an acylketene intermediate with cumulated double bonds, with SMILES '[R1]C(C([R1])=C=O)=O'. **W** is a five-membered heterocyclic iminofuranone intermediate, with SMILES 'O=C1C([R1])=C(O/C1=N/[R2])[R1]'. **E** is the pyrazole carboxamide product formed with methylhydrazine, with SMILES 'CN2N=C(C([R1])=C2[R1])C(N[R2])=O'. **F** is a complex spirocyclic lactone derivative formed by condensation, with SMILES 'O=C3C([R1])=C(OC34N(C(C([R1])=C(O4)[R1])=O)[R2])[R1]'. **H** is a pyrroledione (maleimide-type) product formed by isomerization under acidic conditions, with SMILES 'O=C5C(N(C([R1])=C5[R1])[R2])=O'."}]}]} +{"problem_id": "Exam-03-T8", "header": "", "header_images": [], "subquestions": [{"sub_id": "Exam-03-T8-1", "tags": ["2.7"], "text_open": "8-3 The reaction shown below, 'OCCOCCOCCO' + 'ClCCOCCCl' → 'C1COCCOCCOCCOCCO1', is best conducted under basic conditions. Which base is preferred? ( ) (a) LiOH (b) NaOH (c) KOH", "text_images": [{"file": "Exam-03-T8-1-1-prompt.png", "description": "The figure shows the reaction 'OCCOCCOCCO' + 'ClCCOCCCl' → 'C1COCCOCCOCCOCCO1' in THF (tetrahydrofuran) and water."}], "full_mark": 2, "rubric": ["(b) (2 points)"], "rubric_images": []}, {"sub_id": "Exam-03-T8-2", "tags": ["1.2"], "text_open": "8-4 Describe the most stable conformation of the molecule shown below.", "text_images": [{"file": "Exam-03-T8-2-1-prompt.png", "description": "The figure shows an organic compound with SMILES 'CN1CN(CN(C1)C)C'."}], "full_mark": 2, "rubric": ["2 methyl groups occupy equatorial positions and 1 methyl group occupies an axial position. (2 points)"], "rubric_images": [{"file": "Exam-03-T8-2-1-rubric.png", "description": "The figure shows the most stable conformation of the organic compound with SMILES 'CN1CN(C)CN(C)C1'. The six-membered ring adopts a chair conformation, with two methyl groups equatorial and one methyl group axial."}]}, {"sub_id": "Exam-03-T8-3", "tags": ["2.2"], "text_open": "8-5 Ethylene glycol is commonly condensed with carbonyl compounds to protect carbonyl groups. As shown below, '[R]C([H])=O' + 'SCCCS' → '[R]C1([H])SCCCS1'; '[R]C1([H])SCCCS1' + 'CCCCLi' → '[R]C1([Li])SCCCS1'. Thioacetals invert the polarity of the aldehydic carbon, and propane-1,3-dithiol is often used for this purpose. Briefly explain why ethane-1,2-dithiol, the sulfur analogue of ethylene glycol, is not used instead. (No more than 50 words.)", "text_images": [{"file": "Exam-03-T8-3-1-prompt.png", "description": "The figure shows a two-step reaction. The first step is '[R]C([H])=O' + 'SCCCS' → '[R]C1([H])SCCCS1', in which the carbonyl carbon has δ^+ character. The second step is '[R]C1([H])SCCCS1' + 'CCCCLi' → '[R]C1([Li])SCCCS1', in which the carbon bonded to lithium has δ^- character; R denotes an alkyl group."}], "full_mark": 2, "rubric": ["The lithium salt obtained from ethane-1,2-dithiol readily undergoes fragmentation. (2 points; a correct reaction equation is also acceptable.)"], "rubric_images": []}, {"sub_id": "Exam-03-T8-4", "tags": ["2.4"], "text_open": "8-6 Organic compound A (C_4H_6O_2) is oxidized by acidified potassium permanganate to give a single organic product B without gas evolution. B reacts with two equivalents of CH_3Li and, after workup, gives the major product C. C reacts with sodium metal in benzene and, after workup, gives D. Under sulfuric acid, D rearranges to E (C_6H_12O). Write the SMILES structures of A-E.", "text_images": [], "full_mark": 5, "rubric": ["A is 'CC(=O)C(=O)C'; B is 'CC(=O)O'; C is 'CC(=O)C'; D is 'CC(C(C)(C)O)(O)C'; and E is 'CC(C)(C(C)=O)C'. (1 point each; 5 points total.)"], "rubric_images": [{"file": "Exam-03-T8-4-1-rubric.png", "description": "The figure shows five compounds: A is 'CC(=O)C(=O)C'; B is 'CC(=O)O'; C is 'CC(=O)C'; D is 'CC(C(C)(C)O)(O)C'; and E is 'CC(C)(C(C)=O)C'."}]}]} +{"problem_id": "Exam-03-T9", "header": "Problem 9 Hypervalent-Iodine-Induced Ring Contraction. In 2004, researchers at the University of Tokushima published a study of hypervalent-iodine-induced ring contraction.", "header_images": [{"file": "Exam-03-T9-header-1.png", "description": "'OC1CCCNC1' reacts under PhIO and H_2O conditions."}], "subquestions": [{"sub_id": "Exam-03-T9-1", "tags": ["2.2", "2.3", "4.3"], "text_open": "9-2 The reaction gives two amide products, A and B. Write the SMILES structures of the corresponding key intermediates; an intermediate common to both pathways need be written only once. Note: additional key intermediates may be included, but no incorrect structure may be given. The ^1H NMR data are as follows. A: ^1H NMR (CDCl_3): 6.68, 3.41, 2.31, 2.15; B: ^1H NMR (CDCl_3): 4.81, 3.58, 3.13, 2.41, 2.06.", "text_images": [], "full_mark": 9, "rubric": ["Common key intermediates: 'OC1CCCN(I(O)C2=CC=CC=C2)C1' (1 point), 'C=NCCCC=O' (1 point), and 'O=CCCCNCO' (1 point). Formation of A additionally proceeds through 'NCCCC=O' (2 points) and 'OC1NCCC1' (2 points). Formation of B additionally proceeds through 'OC1CCCN1CO' (2 points)."], "rubric_images": [{"file": "Exam-03-T9-1-1-rubric.png", "description": "The figure has 3 rows. The first row shows the common key intermediates: 'OC1CCCN(I(O)C2=CC=CC=C2)C1' (1 point), 'C=NCCCC=O' (1 point), and 'O=CCCCNCO' (1 point). The second row shows the additional key intermediates leading to A: 'NCCCC=O' (2 points) and 'OC1NCCC1' (2 points). The third row shows the additional key intermediate leading to B: 'OC1CCCN1CO' (2 points)."}]}, {"sub_id": "Exam-03-T9-2", "tags": ["2.7"], "text_open": "9-3 Identify the major and minor products among the two amides obtained in 9-2.", "text_images": [], "full_mark": 2, "rubric": ["Major product: B; minor product: A. (2 points only if both are correct.)"], "rubric_images": []}, {"sub_id": "Exam-03-T9-3", "tags": ["2.2", "2.3", "2.4"], "text_open": "9-4 Based on the mechanistic information above, write the SMILES structures of the key intermediates and product in the reaction shown below. The first step is known to form a zwitterion (internal salt); the acetonitrile was not predried; and the product has more than one plane of symmetry. Note: additional key intermediates may be included, but no incorrect structure may be given.", "text_images": [{"file": "Exam-03-T9-3-1-prompt.png", "description": "'O=C1C(O)=CC(C2=CC=CC=C21)=O' reacts first under PhI(OAc)_2 and DCM, and then under heat in CH_3CN."}], "full_mark": 5, "rubric": ["'O=C1C([O-])=C([I+]C2=CC=CC=C2)C(C3=CC=CC=C31)=O' (1 point), 'O=C(C1=CC=CC=C1C([C]2)=O)C2=O' (1 point), 'O=C(C1=CC=CC=C1C2=O)C2=C=O' (1 point), 'O=C1C(C(O)=O)C(C2=CC=CC=C21)=O' (1 point), and 'O=C1CC(C2=CC=CC=C21)=O' (1 point). Other additional correct intermediates receive neither credit nor a penalty."], "rubric_images": [{"file": "Exam-03-T9-3-1-rubric.png", "description": "The figure has 2 rows. The first row shows 3 key intermediates: 'O=C1C([O-])=C([I+]C2=CC=CC=C2)C(C3=CC=CC=C31)=O' (1 point), 'O=C1C2=CC=CC=C2C([C]C1=O)=O' (1 point), and 'O=C(C1=CC=CC=C1C2=O)C2=C=O' (1 point). The second row shows two key intermediates: 'O=C1C(C(O)=O)C(C2=CC=CC=C21)=O' (1 point) and 'O=C1CC(C2=CC=CC=C21)=O' (1 point)."}]}]} {"problem_id": "Exam-04-T6", "header": "Problem 6 Preparation of TPP. TPP is 5,10,15,20-tetraphenylporphyrin (M = 614.7 g·mol-1) and can be prepared rapidly in quantity by the following method: 'O=CC1=CC=CC=C1' + 'C1=CNC=C1' --('CCC(=O)O', reflux, 30 min)--> 'C1(/C=CC(/C(C2=CC=CC=C2)=C(N/3)/C=CC3=C(C(C=C/4)=NC4=C5\\C6=CC=CC=C6)/C7=CC=CC=C7)=N1)=C(C8=CC=CC=C8)\\C9=CC=C5N9'.", "header_images": [{"file": "Exam-04-T6-header-1.png", "description": "The image shows the synthesis of a porphyrin. PhCHO and pyrrole appear as reactants on the left, separated by a plus sign. Propionic acid is drawn above a right-pointing reaction arrow, and \"reflux, 30 min\" is written below it. The product on the right is 'C1(/C=CC(/C(C2=CC=CC=C2)=C(N/3)/C=CC3=C(C(C=C/4)=NC4=C5\\C6=CC=CC=C6)/C7=CC=CC=C7)=N1)=C(C8=CC=CC=C8)\\C9=CC=C5N9'."}], "subquestions": [{"sub_id": "Exam-04-T6-1", "tags": ["4.1"], "text_open": "6-1 Pretreatment of reagents: the benzaldehyde and pyrrole used in the reaction must be dried and purified. 6-1-1 Select the drying agents suitable for pyrrole. (a) P_4O_10 (b) anhydrous Na_2SO_4 (c) concentrated H_2SO_4 (d) CaH_2", "text_images": [], "full_mark": 2, "rubric": ["(b), (d). (1 point each, 2 points total; any incorrect selection receives no credit.)"], "rubric_images": []}, {"sub_id": "Exam-04-T6-2", "tags": ["4.1"], "text_open": "6-1-2 Select the solid drying agent that must be removed by filtration before distillation. (a) CaO (b) anhydrous Na_2SO_4 (c) Na metal (d) CaH_2", "text_images": [], "full_mark": 2, "rubric": ["(b) (2 points)"], "rubric_images": []}]} -{"problem_id": "Exam-05-T4", "header": "Problem 4 Phosphoric Acid and Phosphate Chemistry. Phosphates are important in industrial production, food additives, and other fields, so measuring their content and understanding their properties are significant.", "header_images": [], "subquestions": [{"sub_id": "Exam-05-T4-1", "tags": ["2.1"], "text_open": "4-1 Aqueous sodium phosphate dodecahydrate, Na_3PO_4·12H_2O, is strongly basic and is used in washing powders and fat-saponification agents. It is prepared by reacting calcium dihydrogen phosphate with sodium carbonate and then adding sodium hydroxide solution. Write the preparation equation.", "text_images": [], "full_mark": 2, "rubric": ["Ca(H_2PO_4)_2+Na_2CO_3 + 4NaOH + 20H_2O = 2[Na_3PO_4·12H_2O] + CaCO_3. (2 points)"], "rubric_images": []}, {"sub_id": "Exam-05-T4-2", "tags": ["4.5"], "text_open": "Titration notation: Add a volume $V_0$ of standard HCl of concentration $c_1$ to the sample, then titrate with standard NaOH of concentration $c_2$. The first jump near pH 4.0 consumes a NaOH volume $V_1$, and continuing to the second jump near pH 8.8 consumes an additional volume $V_2$.\n4-2-2 What relation among V_0, V_1, V_2, c_1, and c_2 indicates that the sample contains Na_2HPO_4? Express it symbolically.", "text_images": [], "full_mark": 2, "rubric": ["$V_0c_1-V_1c_2<2V_2c_2$, or equivalently $V_0c_1>CC2(C)OOOC(C)(C)OO2.CC(C)=O'. (1 point)", "Reaction 2: 'CC1(C)OOOC(C)(C)OO1>>C/C(C)=[O+]/[O-].CC(C)=O.O=O'. (1 point)", "No credit unless A and B are SMILES. 'CC1(OO1)C' and 'CC([O+])([O-])C' are not accepted for B."], "rubric_images": [{"file": "Exam-05-T7-2-1-rubric.png", "description": "Reaction 1: 'CC1(C)OOC(C)(C)OOC(C)(C)OO1>>CC2(C)OOOC(C)(C)OO2.CC(C)=O'."}, {"file": "Exam-05-T7-2-2-rubric.png", "description": "Reaction 2: 'CC1(C)OOOC(C)(C)OO1>>C/C(C)=[O+]/[O-].CC(C)=O.O=O'."}, {"file": "Exam-05-T7-2-3-rubric.png", "description": "Incorrect B: 'CC1(OO1)C'."}, {"file": "Exam-05-T7-2-4-rubric.png", "description": "Incorrect B: 'CC([O+])([O-])C'."}]}, {"sub_id": "Exam-05-T7-3", "tags": ["1.6", "1.7"], "text_open": "7-1-4 DADP is unexpectedly much more stable than M. Lone-pair donation into a particular antibonding orbital controls the difference. Give limiting resonance-form SMILES for each and explain. No more than 100 words.", "text_images": [{"file": "Exam-05-T7-3-1-prompt.png", "description": "'CC1(C)CCC(C)(C)OO1'."}], "full_mark": 4, "rubric": ["DADP: 'CC1(C)OOC(C)(C)OO1' ↔ '[O-]OC(C)(C)O/[O+]=C(C)/C'; M: 'CC1(C)CCC(C)(C)OO1' ↔ '[CH2-]CC(C)(C)O/[O+]=C(C)/C', or the listed methyl-anion alternative. (1 point each; carbocation forms accepted.) (2 points total.) Additional structure strings specified: 'CC1(C)CCC(C)(C)OO1', 'CC2=[O+]OC(C)(C)CC2.[CH3-]'.", "O is more electronegative than C, so the O^- limiting form of DADP is more stable and gives greater stabilization. Equivalently, σ*(C-O) is lower than σ*(C-C) and hyperconjugates better with the O lone pair, reducing lone-pair repulsion. (2 points)"], "rubric_images": [{"file": "Exam-05-T7-3-1-rubric.png", "description": "DADP and M resonance pairs as listed. SMILES/structure strings shown: 'CC1(C)OOC(C)(C)OO1', '[O-]OC(C)(C)O/[O+]=C(C)/C', 'CC1(C)CCC(C)(C)OO1', '[CH2-]CC(C)(C)O/[O+]=C(C)/C'."}, {"file": "Exam-05-T7-3-2-rubric.png", "description": "Alternative M resonance pair with 'CC2=[O+]OC(C)(C)CC2.[CH3-]'. SMILES/structure strings shown: 'CC1(C)CCC(C)(C)OO1'."}]}, {"sub_id": "Exam-05-T7-4", "tags": ["1.7"], "text_open": "7-2-2 If H_2O_2 is enclosed in the CL-20 cage, does sensitivity rise or fall? Explain briefly. (No more than 100 words.)", "text_images": [], "full_mark": 2, "rubric": ["7-2-2 It rises. (1 point) Heated H_2O_2 generates radicals that initiate rapid chain decomposition of CL-20. (1 point; radical generation or initiation of decomposition is sufficient.)"], "rubric_images": []}, {"sub_id": "Exam-05-T7-5", "tags": ["2.2"], "text_open": "7-3 Thermal decomposition of condensed-phase CL-20 may proceed through a series of complex transformations. Initially, cleavage of an N-N bond in CL-20 generates NO_2, which then reacts with CL-20 and its remaining fragments through several pathways: (1) NO_2 oxidizes CL-20 to G. G may be represented by the formula CN-CHO. Homolytic bond cleavage and hydrogen-radical transfer by G produce the common small molecules H and I, of which H contains nitrogen. (2) NO_2 reacts with CL-20 to form J, which decomposes to a hydroxyl radical and K; K contains no carbon. 7-3-1 Give the SMILES representations of G (without formal charges), I, and J.", "text_images": [], "full_mark": 5, "rubric": ["7-3-1 G: '[C]=NC([H])=O'; I: '[C-]#[O+]'; J: 'O=NO[H]'. 5 points total: G and J are worth 2 points each, and I is worth 1 point. Each structure earns credit only if it is entirely correct."], "rubric_images": [{"file": "Exam-05-T7-5-1-rubric.png", "description": "G SMILES. SMILES/structure strings shown: '[C]=NC([H])=O'."}, {"file": "Exam-05-T7-5-2-rubric.png", "description": "I SMILES. SMILES/structure strings shown: '[C-]#[O+]'."}, {"file": "Exam-05-T7-5-3-rubric.png", "description": "J SMILES. SMILES/structure strings shown: 'O=NO[H]'."}]}, {"sub_id": "Exam-05-T7-6", "tags": ["1.5"], "text_open": "7-3-2 Give two other reasonable resonance forms of G.", "text_images": [], "full_mark": 2, "rubric": ["'[C-]#[N+]C([H])=O' and '[C]=[N+]=C([H])[O-]'. (1 point each) (2 points total.)"], "rubric_images": [{"file": "Exam-05-T7-6-1-rubric.png", "description": "The two resonance forms. SMILES/structure strings shown: '[C-]#[N+]C([H])=O', '[C]=[N+]=C([H])[O-]'."}]}, {"sub_id": "Exam-05-T7-7", "tags": ["2.2", "2.3"], "text_open": "7-3-3 Intermediate X rearranges intramolecularly and then cleaves to two common small molecules. Give the other key intermediate and final products.", "text_images": [{"file": "Exam-05-T7-7-1-prompt.png", "description": "Lewis structure of X: 'C=N[N+]([O-])=O'."}], "full_mark": 3, "rubric": ["Intermediate '[O-]CN=[N+]=O'. (1 point)", "Products '[H]C([H])=O' and '[N-]=[N+]=O'. (1 point each)"], "rubric_images": [{"file": "Exam-05-T7-7-1-rubric.png", "description": "Red mechanism: X rearranges to '[O-]CN=[N+]=O', then cleaves to formaldehyde and nitrous oxide, with curved electron arrows shown. SMILES/structure strings shown: 'C=N[N+]([O-])=O', 'C=N[N+]([O-])=O', '[O-]CN=[N+]=O', '[H]C([H])=O', '[H]C([H])=O', '[N-]=[N+]=O', '[N-]=[N+]=O'."}]}, {"sub_id": "Exam-05-T7-8", "tags": ["3.7", "3.8", "4.7"], "text_open": "7-4 N-N bond cleavage is the rate-determining step in the early thermal decomposition of crystalline CL-20. To determine its activation energy $E_a$, researchers simulated CL-20 decomposition in a rigid container and obtained the initial $NO_2$ formation rates below. $T\\ (\\mathrm{K})$: 1488, 1969, 2547, 3124. $k\\ (\\mathrm{s^{-1}})$: $1.28\\times10^9$, $1.54\\times10^9$, $1.74\\times10^9$, $1.89\\times10^9$. 7-4-1 Treat N-N cleavage under the simulated conditions as a first-order reaction obeying the Arrhenius law. Calculate $E_a$ and the pre-exponential factor $A$, both to 2 significant figures.", "text_images": [], "full_mark": 6, "rubric": ["From the Arrhenius equation $k=Ae^{-E_a/(RT)}$, linearization gives $\\ln k=-\\dfrac{E_a}{RT}+\\ln A$. (1 point; either the original or linearized equation earns credit.)\n\n| \\(1/T \\, (\\text{K}^{-1})\\) | \\(6.720 \\times 10^{-4}\\) | \\(5.079 \\times 10^{-4}\\) | \\(3.926 \\times 10^{-4}\\) | \\(3.201 \\times 10^{-4}\\) |\n|---|---|---|---|---|\n| \\(\\ln k\\) | 20.97 | 21.16 | 21.28 | 21.36 |\n\n(1 point; at least one type of data must be processed, otherwise no credit is awarded for this step.) $\\ln k=-1.10\\times10^3/T+21.7$. (2 points) Therefore $E_a=9.2\\ \\mathrm{kJ\\cdot mol^{-1}}$ and $A=2.7\\times10^9\\ \\mathrm{s^{-1}}$. (2 points)"], "rubric_images": []}, {"sub_id": "Exam-05-T7-9", "tags": ["3.7", "4.7"], "text_open": "7-4-3 The simulated pre-exponential factor is four orders of magnitude below experiment for CL-20 under real explosion conditions. Give 1 principal reason based on the setup. No more than 100 words.", "text_images": [], "full_mark": 1, "rubric": ["The simulation uses a rigid container of constant volume, whereas the volume of a real explosion increases rapidly; equivalently, the fixed volume keeps the local concentration of gaseous products higher than in a real explosion. (1 point; the complete explanation is required.)"], "rubric_images": []}]} -{"problem_id": "Exam-05-T8", "header": "Problem 8 Basic Organic Chemistry", "header_images": [], "subquestions": [{"sub_id": "Exam-05-T8-1", "tags": ["1.2", "1.6"], "text_open": "8.2 Apply conformational analysis analogous to n-butane. 8.2.1 Rank the shown conformations of 1-butene from high to low energy.", "text_images": [{"file": "Exam-05-T8-1-1-prompt.png", "description": "3 conformations of C=CCC. A has a left double bond and a zigzag chain ending in an ordinary vertical-up bond. B has a similar skeleton but an upward solid-wedge terminal bond. C has a solid-wedge terminal bond directed lower right."}], "full_mark": 2, "rubric": ["B > A > C. (2 points; if n pairwise swaps give the correct order, award 2 - n.)"], "rubric_images": []}, {"sub_id": "Exam-05-T8-2", "tags": ["1.2", "1.6"], "text_open": "8.2.2 Rank the shown conformations of allyl alcohol from high to low energy.", "text_images": [{"file": "Exam-05-T8-2-1-prompt.png", "description": "4 conformations of C=CCO. A ends in an ordinary upward OH bond; B in an upward solid-wedge OH bond; C in a rightward solid-wedge OH bond of different orientation; D ends directly in OH by an ordinary line without wedge stereochemistry."}], "full_mark": 2, "rubric": ["D > B > C > A. (2 points; if n pairwise swaps give the correct order, award 2 - n.)"], "rubric_images": []}, {"sub_id": "Exam-05-T8-3", "tags": ["1.2", "1.6"], "text_open": "8.2.3 Rank the shown conformations of 2-methyl-1-butene from high to low energy.", "text_images": [{"file": "Exam-05-T8-3-1-prompt.png", "description": "4 conformations of 'C=C(C)CC'. A has the terminal methyl cis to the double bond by an ordinary upward bond; B has an upward solid wedge; C a lower-right solid wedge; D an ordinary trans arrangement."}], "full_mark": 2, "rubric": ["D > B > A > C. (2 points; if n pairwise swaps give the correct order, award 2 - n.)"], "rubric_images": []}, {"sub_id": "Exam-05-T8-4", "tags": ["1.2", "1.6"], "text_open": "8.2.4 Rank the shown conformations of methylallyl alcohol from high to low energy.", "text_images": [{"file": "Exam-05-T8-4-1-prompt.png", "description": "4 conformations of 'C=C(C)CO'. A has OH cis to the double bond by an ordinary line; B an upward solid-wedge OH; C a rightward solid-wedge OH; D an ordinary trans double-bond/OH arrangement."}], "full_mark": 2, "rubric": ["D > B > C > A. (2 points; if n pairwise swaps give the correct order, award 2 - n.)"], "rubric_images": []}, {"sub_id": "Exam-05-T8-5", "tags": ["1.5", "1.6"], "text_open": "8.3.3 Which has the largest X-C-O bond angle, X = CF_3, CH_3, or CBr_3?", "text_images": [], "full_mark": 1, "rubric": ["B. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-05-T8-6", "tags": ["1.6"], "text_open": "8.3.4 Which of X-C-O (X = CF_3, CH_3, CBr_3) has the smallest bond angle?", "text_images": [], "full_mark": 1, "rubric": ["A. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-05-T8-7", "tags": ["1.7"], "text_open": "8.4 Ring strain can strongly affect reaction rates and equilibria. 8.4.1 Rank the substances from greatest to least ring strain.", "text_images": [{"file": "Exam-05-T8-7-1-prompt.png", "description": "3 compounds: A, 'C1CC=CC1'; B, 'C1=CC=CC1'; C, 'C=C1CCCC1'."}], "full_mark": 2, "rubric": ["A > C > B (2 points). If the order is not completely correct and n swaps are required to obtain the correct order, award 2 - n points, with a minimum score of 0."], "rubric_images": []}, {"sub_id": "Exam-05-T8-8", "tags": ["1.7"], "text_open": "8.4.2 Select the substance that reacts faster with NaBH_4.", "text_images": [{"file": "Exam-05-T8-8-1-prompt.png", "description": "2 compounds: A, 'O=C1CCCC1'; B, 'O=C1CCCCC1'."}], "full_mark": 1, "rubric": ["B. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-05-T8-9", "tags": ["1.7"], "text_open": "8.4.3 Select the substance with the greater equilibrium enol fraction.", "text_images": [{"file": "Exam-05-T8-9-1-prompt.png", "description": "2 compounds: A, 'O=C1C(C(OCC)=O)CCC1'; B, 'O=C1C(C(OCC)=O)CCCC1'."}], "full_mark": 1, "rubric": ["B. (1 point)"], "rubric_images": []}]} -{"problem_id": "Exam-06-T4", "header": "Problem 4 Determination of Tin by Potassium Iodate Titration. Tin is commonly determined by potassium iodate titration. The procedure includes sample decomposition, masking and separation, reductive fusion with zinc powder, reduction with aluminum powder, and titration of tin(II) ions with potassium iodate.", "header_images": [], "subquestions": [{"sub_id": "Exam-06-T4-1", "tags": ["2.1"], "text_open": "4.2 Sample pretreatment. Place 0.2026 g of the sample in a 250 mL beaker, add 2 g of potassium chlorate and 20 mL of concentrated hydrochloric acid, and digest at low temperature for 20 min. Add 60 mL of hydrochloric acid (1+1), 1 g of hydrazinium sulfate, and 0.3 g of potassium bromide, then heat and evaporate for 50 min. Remove from the heat and allow to cool slightly. Cover with a watch glass and add hydrogen peroxide dropwise until brownish-red bromine appears. Add 10 mL of EDTA solution (100 g/L) and 5 mL of beryllium sulfate solution (40 g/L), and neutralize with aqueous ammonia to pH = 9-10. Add a small amount of paper pulp and filter. Wash the precipitate with an aqueous ammonia-ammonium nitrate wash solution. Transfer the precipitate together with the filter paper to a crucible, dry, ash, and cool. 4.2.1 Write the ionic equation for the reduction of arsenic acid by hydrazine in hydrochloric acid medium, by which arsenic is removed through volatilization.", "text_images": [], "full_mark": 2, "rubric": ["Ionic equation: 2H_3AsO_4 + N_2H_6^{2+} + 6Cl^{-} + 4H^{+} = 2AsCl_3↑ + N_2↑ + 8H_2O (2 points)", "An equation written using N_2H_5^{+} is also acceptable: 2H_3AsO_4 + N_2H_5^{+} + 6Cl^{-} + 5H^{+} = 2AsCl_3 + N_2 + 8H_2O."], "rubric_images": []}, {"sub_id": "Exam-06-T4-2", "tags": ["4.1", "4.7"], "text_open": "Reductive-fusion conditions: Add a zinc powder-borax-boric acid mixed flux to the crucible and cover it with additional mixed flux and NaCl. Place it in a furnace at 750 ℃, raise the temperature to 820 ℃, fuse for 30 min, then remove and cool.\n4.3.3 Explain how an excessively long fusion time or excessively high temperature affects the determination.", "text_images": [], "full_mark": 2, "rubric": ["① Zinc and tin may be lost by volatilization or oxidized by air.", "② Reduction may be incomplete, giving a result that is too low.", "③ The crucible may be corroded, introducing impurities.", "Full credit for 3 correct points, 1 point for 2 correct points, and no credit for only 1 correct point."], "rubric_images": []}, {"sub_id": "Exam-06-T4-3", "tags": ["2.1", "4.5"], "text_open": "4.4.2 Write the ionic equations for the titration process.", "text_images": [], "full_mark": 2, "rubric": ["IO_3^{-} + 5I^{-} + 6H^{+} = 3I_2 + 3H_2O (1 point)", "or 8I^{-} + IO_3^{-} + 6H^{+} = 3I_3^{-} + 3H_2O", "or HIO_3 + 5I^{-} + 5H^{+} = 3I_2 + 3H_2O", "I_2 + Sn^{2+} = Sn^{4+} + 2I^{-} (1 point)", "or Sn^{2+} +I_3^{-} = Sn^{4+} + 3I^{-}", "or SnCl_4^{2-} + I_2 + 2Cl^{-} = SnCl_6^{2-} + 2I^{-}", "or SnCl_4^{2-} + I_3^{-} + 2Cl^{-} = SnCl_6^{2-} + 3I^{-}", "A combined equation is also acceptable:", "3Sn^{2+} + IO_3^{-} + 6H^{+} = 3Sn^{4+} + I^{-} + 3H_2O (2 points)", "or 3SnCl_4^{2-} + IO_3^{-} + 6H^{+} +6Cl^{-} = 3SnCl_6^{2-} + I^{-} + 3H_2O", "or HIO_3 + 3Sn^{2+} + 5H^{+} = I^{-} + 3Sn^{4+} + 3H_2O"], "rubric_images": []}]} +{"problem_id": "Exam-06-T2", "header": "Problem 2: Molecular Compass. A compass is a simple yet powerful device whose most important component is a freely rotating magnetized needle that senses Earth's magnetic field. At the molecular level, this mechanical function can be mimicked through noncovalent interactions and dynamic behavior among groups. One molecular compass exploits the redox behavior of pillar[5]quinone (P5Q) and incorporates guest molecules bearing polar groups. The dipole moments of these polar groups act as “needles” that respond to changes in the electric field and reorient during redox processes, thereby regulating electronic communication between remote quinone units.", "header_images": [{"file": "Exam-06-T2-header-1.png", "description": "A macroscopic compass."}], "subquestions": [{"sub_id": "Exam-06-T2-1", "tags": ["2.4", "3.5"], "text_open": "2.1 Electrochemical behavior of pillar[5]quinone and synthesis of a molecular compass. The p-benzoquinone monomer Q undergoes two successive one-electron reductions in an aprotic solution. Linking 5 quinone units into a ring through methylene groups gives pillar[5]quinone (P5Q), which exhibits more complex electrochemical behavior. 2.1.1 Write the SMILES representations of the most stable resonance structures of Q^{·-} and Q^{2-}, the products of the two successive one-electron reductions of monomer Q.", "text_images": [{"file": "Exam-06-T2-1-1-prompt.png", "description": "Schematic two-step one-electron reduction of p-benzoquinone Q (SMILES: 'O=C(C=C1)C=CC1=O'). Q is first reversibly reduced by one electron to Q^{·-}, which is then reversibly reduced by another electron to Q^{2-}."}, {"file": "Exam-06-T2-1-2-prompt.png", "description": "SMILES of P5Q: 'C%10C1=CC(=O)C(=CC1=O)CC2=CC(=O)C(=CC2=O)CC3=CC(=O)C(=CC3=O)CC4=CC(=O)C(=CC4=O)CC5=CC(=O)C(=CC5=O)%10'"}], "full_mark": 2, "rubric": ["Q^{·-}: '[O-]C1=CC=C([O])C=C1'; Q^{2-}: '[O-]C2=CC=C([O-])C=C2'.", "2 points total, 1 point for each structure."], "rubric_images": [{"file": "Exam-06-T2-1-1-rubric.png", "description": "Rubric image: Q^{·-} is drawn as the semiquinone radical anion '[O-]C1=CC=C([O])C=C1', and Q^{2-} as the dianionic resonance structure '[O-]C2=CC=C([O-])C=C2' with both oxygen atoms negatively charged."}]}, {"sub_id": "Exam-06-T2-2", "tags": ["1.6"], "text_open": "2.1.2 The redox sequence of quinone units in pillar[5]quinone is regulated by the spatial charge distribution. The reduction pattern of P5Q is “2-1-2,” as shown below. White circles represent quinone monomers, black circles represent quinone monomers after one-electron reduction, and each quinone unit undergoes only one-electron reduction. Draw the SMILES structures of P5Q-2 and P5Q-3.", "text_images": [{"file": "Exam-06-T2-2-1-prompt.png", "description": "Schematic 2-1-2 reduction pattern of P5Q. A legend at the right identifies a white circle as Q, a quinone monomer, and a black circle as Q^{·-}. P5Q is simplified as a regular pentagon with one hollow white circle on each side. Two-electron reduction of P5Q gives P5Q-2; one-electron reduction of P5Q-2 gives P5Q-3; and two-electron reduction of P5Q-3 gives a product simplified as a regular pentagon with one solid black circle on each side."}], "full_mark": 4, "rubric": ["SMILES of P5Q-2: 'O=C(C(CC1=CC([O-])=C(C=C1[O])CC2=CC(C(CC3=CC4=O)=CC2=O)=O)=CC5=O)C=C5CC(C=C6[O])=C([O-])C=C6CC4=CC3=O' (2 points)", "SMILES of P5Q-3: 'O=C(C=C1CC(C=C2[O])=C(C=C2CC3=CC4=O)[O-])C(CC5=CC([O-])=C(CC6=C([O-])C=C(CC4=CC3=O)C([O])=C6)C=C5[O])=CC1=O' (2 points)"], "rubric_images": [{"file": "Exam-06-T2-2-1-rubric.png", "description": "The rubric image shows the reduction positions in P5Q-2 and P5Q-3. In the left diagram, P5Q-2 has two nonadjacent solid circles at pentagon vertices. In the right diagram, P5Q-3 has three solid circles, while the remaining two hollow circles are nonadjacent."}]}, {"sub_id": "Exam-06-T2-3", "tags": ["2.6", "2.4"], "text_open": "2.1.3 The synthetic route to a molecular compass Q-Rot-A containing amide “needles” is shown below. The molecular compass can be regarded as a supramolecule in which the amide “needle” is the guest molecule. Note the perspective relationships of the chemical bonds in the product. Write the SMILES structures of intermediates A and B.", "text_images": [{"file": "Exam-06-T2-3-1-prompt.png", "description": "The molecule 'C%10C1=CC(OC)=C(C=C1OC)CC2=CC(OC)=C(C=C2OC)CC3=CC(OC)=C(C=C3OC)CC4=CC(OC)=C(C=C4OC)CC5=CC(OC)=C(C=C5OC)%10' reacts with 'NCCCCCCN' using 'O=C(Cl)C1=CC(C(F)(F)F)=CC(C(F)(F)F)=C1' in CHCl_3 from 0°C to r.t. to form A. A reacts with BBr_3 in CHCl_3 at r.t. to form B. B then reacts with NaIO_4 in Me_2CO/H_2O/CH_2Cl_2 at r.t. to give a supramolecule in which 'C%10C1=CC(=O)C(=CC1=O)CC2=CC(=O)C(=CC2=O)CC3=CC(=O)C(=CC3=O)CC4=CC(=O)C(=CC4=O)CC5=CC(=O)C(=CC5=O)%10' encircles 'O=C(NCCCCCCNC(=O)C1=CC(C(F)(F)F)=CC(C(F)(F)F)=C1)C2=CC(C(F)(F)F)=CC(C(F)(F)F)=C2'."}], "full_mark": 2, "rubric": ["A: 'C%10C1=CC(OC)=C(C=C1OC)CC2=CC(OC)=C(C=C2OC)CC3=CC(OC)=C(C=C3OC)CC4=CC(OC)=C(C=C4OC)CC5=CC(OC)=C(C=C5OC)%10' encircles 'O=C(NCCCCCCNC(=O)C1=CC(C(F)(F)F)=CC(C(F)(F)F)=C1)C2=CC(C(F)(F)F)=CC(C(F)(F)F)=C2'; B: 'C%10C1=CC(O)=C(C=C1O)CC2=CC(O)=C(C=C2O)CC3=CC(O)=C(C=C3O)CC4=CC(O)=C(C=C4O)CC5=CC(O)=C(C=C5O)%10' encircles 'O=C(NCCCCCCNC(=O)C1=CC(C(F)(F)F)=CC(C(F)(F)F)=C1)C2=CC(C(F)(F)F)=CC(C(F)(F)F)=C2'.", "2 points total, 1 point for each structure.", "No credit if the methylene groups between the benzene rings are missing."], "rubric_images": [{"file": "Exam-06-T2-3-1-rubric.png", "description": "The image shows two molecules. On the left, A consists of 'C%10C1=CC(OC)=C(C=C1OC)CC2=CC(OC)=C(C=C2OC)CC3=CC(OC)=C(C=C3OC)CC4=CC(OC)=C(C=C4OC)CC5=CC(OC)=C(C=C5OC)%10' encircling 'O=C(NCCCCCCNC(=O)C1=CC(C(F)(F)F)=CC(C(F)(F)F)=C1)C2=CC(C(F)(F)F)=CC(C(F)(F)F)=C2'. On the right, B consists of 'C%10C1=CC(O)=C(C=C1O)CC2=CC(O)=C(C=C2O)CC3=CC(O)=C(C=C3O)CC4=CC(O)=C(C=C4O)CC5=CC(O)=C(C=C5O)%10' encircling 'O=C(NCCCCCCNC(=O)C1=CC(C(F)(F)F)=CC(C(F)(F)F)=C1)C2=CC(C(F)(F)F)=CC(C(F)(F)F)=C2'."}]}, {"sub_id": "Exam-06-T2-4", "tags": ["1.2", "1.1"], "text_open": "2.2 Structural characterization and chirality analysis. Owing to the asymmetry of the 2,5-substituents, P5Q-type structures theoretically have multiple atropisomers. The planar chirality of each unit is defined as shown. 2.2.1 State directly the theoretical number of atropisomers of a P5Q-type structure.", "text_images": [{"file": "Exam-06-T2-4-1-prompt.png", "description": "The image explains planar chirality. A vertical dashed mirror plane divides the image into left and right halves containing two enantiomeric substituted-benzene fragments. In the left structure, the bond on the right side of the benzene ring is drawn thick to project out of the page toward the viewer. Two ethyl and two “-OR” groups are arranged in alternating para positions, with the two -OR groups directed upper right and lower left. An equals sign connects this structure to a circled-dot symbol ($\\odot$), indicating projection out of the page, and it is labeled $R_{\\text{p}}$. In the mirror-image structure on the right, the bond on the left side of the benzene ring projects out of the page and the substituent positions are mirror-related. An equals sign connects it to a circled-cross symbol ($\\otimes$), indicating projection into the page away from the viewer, and it is labeled $S_{\\text{p}}$."}], "full_mark": 2, "rubric": ["8 (2 points; no credit for an incorrect answer)"], "rubric_images": []}, {"sub_id": "Exam-06-T2-5", "tags": ["1.2", "4.2"], "text_open": "2.2.2 ^{1}H-NMR measurements show that all 5 sets of methylene hydrogen atoms in the P5Q frameworks of A and Q-Rot-A appear as singlets, indicating that the planar rings of both are apparently homodirectional: either (R_{p}, R_{p}, R_{p}, R_{p}, R_{p}) or (S_{p}, S_{p}, S_{p}, S_{p}, S_{p}). However, the reasons for their apparently homodirectional configurations differ. Explain each case in terms of the different barriers to planar-ring rotation. Note: NMR measurements have a finite timescale.", "text_images": [], "full_mark": 4, "rubric": ["A: methyl groups hinder rotation of the planar rings, so the configuration is retained.", "2 points; “rotation is hindered,” “cannot rotate,” or any equivalent expression receives full credit.", "No credit for a thermodynamic explanation such as “methyl groups cause high steric hindrance, so the all-cis form has the least steric hindrance,” because it does not follow the hint.", "Q-Rot-A: the barrier to planar-ring rotation is low, so all quinone units are equivalent on the NMR timescale.", "2 points: 1 point for “low rotational barrier” or “rapid rotation,” and 1 point for “equivalent on the NMR timescale,” “the inversion time is shorter than the NMR timescale,” or an equivalent statement."], "rubric_images": []}, {"sub_id": "Exam-06-T2-6", "tags": ["4.2", "1.2"], "text_open": "2.2.3 For B, ^{1}H-NMR data show shifts and splitting of the methylene hydrogen signals. A partial spectrum and the corresponding hydrogen assignments are shown below; numbers beneath brackets are peak integrals, and the bottom scale gives chemical shift. Which of the following configurations are consistent with the integral data? ( ). (a) (R_{p}, S_{p}, R_{p}, R_{p}, S_{p}); (b) (R_{p}, R_{p}, R_{p}, S_{p}, S_{p}); (c) (S_{p}, R_{p}, R_{p}, R_{p}, R_{p}); (d) (R_{p}, S_{p}, S_{p}, R_{p}, S_{p})", "text_images": [{"file": "Exam-06-T2-6-1-prompt.png", "description": "This partial $^{1}$H-NMR spectrum of B shows three groups of peaks along a horizontal axis labeled 3.8, 3.4, and 3.0 from left to right. Braces above them are labeled “Group I,” “Group II,” and “Group III.” Under Group I are four peaks: the first and third are connected by dashed lines to a red label 4^1, and the second and fourth to a red label 4^2. A horizontal integral bracket beneath the four peaks is labeled 4.00. Group II contains one peak with a red label 4^3 above and an integral bracket labeled 2.13 below. Group III contains four peaks: the first and second are connected by dashed lines to a red label 4^4, and the third and fourth to a red label 4^5. An integral bracket beneath the four peaks is labeled 4.03."}, {"file": "Exam-06-T2-6-2-prompt.png", "description": "The image shows two molecular structures. On the left is 'CC1=C(O)C=C(C([H])([H])C2=C(O)C=C(C)C(O)=C2)C(O)=C1'. The left benzene ring is labeled “R_p” above and the right ring “S_p.” The two methylene hydrogens linking the rings are highlighted as blue “H” labels; a blue arrow labeled “Group I” points to the upper H and one labeled “Group III” to the lower H. On the right is 'CC3=C(O)C=C(C([H])([H])C4=CC(O)=C(C)C=C4O)C(O)=C3'. Blue arrows labeled “Group II” point to both the upper and lower blue H atoms of the central methylene group, and “(R_p, R_p) or (S_p, S_p)” appears below the structure."}], "full_mark": 3, "rubric": ["(a)(d) (3 points; full credit only if both are selected, and no credit if any incorrect choice is included)"], "rubric_images": []}]} +{"problem_id": "Exam-07-T4", "header": "Determination of Sulfur Dioxide in Wine\nThe sulfur dioxide content of a wine sample is determined spectrophotometrically as follows. I. Construct a calibration curve. Accurately transfer 0.00 mL, 0.20 mL, 0.50 mL, 1.00 mL, 2.00 mL, and 3.00 mL of sulfur dioxide working standard solutions (containing 0.0 μg, 2.0 μg, 5.0 μg, 10.0 μg, 20.0 μg, and 30.0 μg sulfur dioxide, respectively) into 25 mL stoppered test tubes, and add formaldehyde-buffered absorbing solution to 10.00 mL. Then add, in sequence, 0.5 mL of 3 g/L ammonium sulfamate solution, 0.5 mL of 1.5 mol/L sodium hydroxide solution, and 1.0 mL of 0.5 g/L pararosaniline hydrochloride solution (structure shown below). Mix thoroughly and allow to stand for 20 min. Measure the absorbance of each standard solution at 579 nm using a UV–visible spectrophotometer. The following data are obtained:\n\n| $\\text{SO}_2$ content $m / \\mu\\text{g}$ | 0.0 | 2.0 | 5.0 | 10.0 | 20.0 | 30.0 |\n|---|---|---|---|---|---|---|\n| Absorbance $A$ | 0.003 | 0.062 | 0.159 | 0.312 | 0.618 | 0.928 |\n\nII. Accurately weigh 0.5000 g of the wine sample and add formaldehyde-buffered absorbing solution to 100 mL. After shaking for 2 h, take 10 mL of the solution and perform the spectrophotometric determination as in Step I. Repeat the measurement three times; the absorbances A’ are 0.637, 0.638, and 0.635. In China, the required sulfur dioxide content of alcoholic beverages is below $0.2500 \\ \\mathrm{g\\,dm^{-3}}$. Take the density of the wine to be $0.9 \\ \\mathrm{g\\,cm^{-3}}$. The structure of pararosaniline hydrochloride is 'NC1=CC=C(/C(C2=CC=C(N)C=C2)=C3C=CC(C=C\\3)=[NH2+])C=C1.[Cl-]'.", "header_images": [{"file": "Exam-07-T4-header-1.png", "description": "Structural diagram of pararosaniline hydrochloride; the SMILES representation is 'NC1=CC=C(/C(C2=CC=C(N)C=C2)=C3C=CC(C=C\\3)=[NH2+])C=C1.[Cl-]'."}], "subquestions": [{"sub_id": "Exam-07-T4-1", "tags": ["2.4"], "text_open": "4-1 The principle used in this titration is that A, formed by the addition reaction of sulfur dioxide with formaldehyde, forms a blue-violet complex with pararosaniline hydrochloride. Its absorbance is proportional to the sulfur dioxide concentration. Write the SMILES representation of A.", "text_images": [], "full_mark": 2, "rubric": ["'[H]C(S(O)=O)=O' (2 points)"], "rubric_images": [{"file": "Exam-07-T4-1-1-rubric.png", "description": "The SMILES representation is '[H]C(S(O)=O)=O'."}]}, {"sub_id": "Exam-07-T4-2", "tags": ["4.1", "2.8"], "text_open": "4-2 Why must 0.5 mL of 1.5 mol/L sodium hydroxide solution be added to adjust the pH before measurement? (No more than 50 words.)", "text_images": [], "full_mark": 2, "rubric": ["① This ensures complete reaction between HCHO and $SO_2$, thereby ensuring the accuracy of the spectrophotometric determination (1 point). ② The absorption wavelength of the conjugated structure of pararosaniline hydrochloride varies with acidity; failure to adjust the pH to an appropriate range would affect the accuracy of the spectrophotometric determination (1 point)."], "rubric_images": []}]} {"problem_id": "Exam-07-T6", "header": "The Phases and Properties of Water and the Evolution of Temperature-Scale Definitions\nWater is the source of life, and its unique physicochemical properties play a crucial role in nature. The phase diagram below describes the phase equilibria of water at different temperatures and pressures. ", "header_images": [{"file": "Exam-07-T6-header-1.png", "description": "Schematic physicochemical phase diagram drawn with black lines on a white background. The origin of the rectangular coordinate system is labeled “O”; the vertical axis points upward and is labeled “p,” and the horizontal axis points rightward and is labeled “T/K.” The value “273.16” appears toward the left of the horizontal axis, with a vertical dashed line rising from it to the triple point, which is also labeled “O.” Four lines meet at triple point O: a solid solid-gas coexistence curve bending down and left to point B; a slightly curved solid-liquid coexistence line with negative slope extending up and left to point C; dashed line OD extending leftward; and a solid liquid-gas coexistence curve bending up and right to point A, the critical point. Point P lies on solid curve OA. A double-slash break symbol “//” appears midway along this curve, with a corresponding “//” break on the right side of the horizontal axis. A vertical dashed line fq passes through point P on curve OA, with f above and q below. Above and to the right of point A is a right-angled boundary extending rightward and upward; the top of its vertical segment is labeled “E,” and the right end of its horizontal segment is labeled “F.” The regions are labeled as follows: “ice” to the left of line OC, “water” between OC and OA, “water vapor” below OA and OB, and “supercritical water,” written vertically, inside the upper-right right-angled region bounded by E and F."}], "subquestions": [{"sub_id": "Exam-07-T6-1", "tags": ["3.1"], "text_open": "6.1.2 State the physical meaning of curve OD.", "text_images": [], "full_mark": 2, "rubric": ["Line OD is the metastable equilibrium line between supercooled liquid water and water vapor, or equivalently the vapor-pressure curve of supercooled water (2 points)."], "rubric_images": []}]} {"problem_id": "Exam-07-T10", "header": "The Remarkable Role of LiCl", "header_images": [], "subquestions": [{"sub_id": "Exam-07-T10-1", "tags": ["2.4"], "text_open": "10.1.1 Write the SMILES representation of the reaction product, given that the product contains 1 fewer carbon atom than the reactant.", "text_images": [{"file": "Exam-07-T10-1-1-prompt.png", "description": "'COC(C=C1)=CC=C1C2=CC=C(OC)N=C2' is reacted under LiCl/DMF conditions with microwave irradiation."}], "full_mark": 2, "rubric": ["'O=C1C=CC(C2=CC=C(OC)C=C2)=CN1' (2 points)"], "rubric_images": [{"file": "Exam-07-T10-1-1-rubric.png", "description": "'O=C1C=CC(C2=CC=C(OC)C=C2)=CN1'"}]}, {"sub_id": "Exam-07-T10-2", "tags": ["2.4", "2.2"], "text_open": "10.1.2 Based on 10.1.1, write the SMILES representation of the reaction product and the SMILES representation of 1 key intermediate that illustrates the role of LiCl, given that the product contains 1 fewer carbon atom than the reactant.", "text_images": [{"file": "Exam-07-T10-2-1-prompt.png", "description": "'COC1=C(OC)C=C(C=O)C=C1' is reacted under LiCl/DMF conditions with microwave irradiation."}], "full_mark": 4, "rubric": ["Product: 'OC1=C(OC)C=C(C=O)C=C1' (2 points)", "Intermediate: '[Cl-].[Li+].O=CC1=CC(OC)=C(OC)C=C1', in which the Li ion is coordinated simultaneously by the O atoms of the two methoxy groups and by Cl^-, giving a three-coordinate geometry. (2 points)"], "rubric_images": [{"file": "Exam-07-T10-2-1-rubric.png", "description": "'OC1=C(OC)C=C(C=O)C=C1'"}, {"file": "Exam-07-T10-2-2-rubric.png", "description": "Intermediate: the two ether O atoms coordinate to Li to form the complex '[Cl-].[Li+].O=CC1=CC(OC)=C(OC)C=C1'"}]}, {"sub_id": "Exam-07-T10-3", "tags": ["2.4", "1.2", "2.8"], "text_open": "10.1.3 The starting material is an optically active chiral ester. Write the SMILES representation of the reaction product. State the advantage of this method over methods such as hydrolysis with a strong base.", "text_images": [{"file": "Exam-07-T10-3-1-prompt.png", "description": "'O=C([C@@H](C1=CC2=CC=C(C=C2C=C1)OC)C)OC' is reacted under LiCl/DMF conditions with microwave irradiation."}], "full_mark": 4, "rubric": ["'OC([C@H](C)C1=CC2=CC=C(OC)C=C2C=C1)=O' (2 points)", "The configuration at the α stereocenter is retained without racemization. (2 points)"], "rubric_images": [{"file": "Exam-07-T10-3-1-rubric.png", "description": "'OC([C@H](C)C1=CC2=CC=C(OC)C=C2C=C1)=O'"}]}, {"sub_id": "Exam-07-T10-4", "tags": ["2.2", "2.3"], "text_open": "10.2 The following reaction proceeds through a radical process, in which the fluorinating reagent is a single-electron oxidant.\n 10.2.1 Write the SMILES representations of 1 key intermediate associated with the fluorinating reagent and 2 key radical intermediates that contain no halogen atoms and only one six-membered ring.", "text_images": [{"file": "Exam-07-T10-4-1-prompt.png", "description": "'NC1=CC=CC=N1' and 'ClC[N+]12CC[N+](CC2)(F)CC1' · $2\\text{BF}_4^{-}$) react under LiCl/DMF conditions to give the product with the SMILES representation 'NC1=CC=C(Cl)C=N1'."}], "full_mark": 6, "rubric": ["'ClC[N+]1(CC2)CC[N+]2CC1' (note: a nitrogen-centered radical cation) · $\\text{BF}_4^{-}\\text{F}^{-}$ (2 points).", "'NC1=C[CH+][C]C=N1' (2 points).", "'N=C1C=C[C]C=N1' (2 points)"], "rubric_images": [{"file": "Exam-07-T10-4-1-rubric.png", "description": "'ClC[N+]1(CC2)CC[N+]2CC1' (note: a nitrogen-centered radical cation) · $\\text{BF}_4^{-}\\text{F}^{-}$"}, {"file": "Exam-07-T10-4-2-rubric.png", "description": "'NC1=C[CH+][C]C=N1'."}, {"file": "Exam-07-T10-4-3-rubric.png", "description": "'N=C1C=C[C]C=N1'"}]}, {"sub_id": "Exam-07-T10-5", "tags": ["2.4"], "text_open": "10.2.2 Write all by-products of the reaction; give the SMILES representation for the organic product.", "text_images": [], "full_mark": 6, "rubric": ["'ClC[N+]1(CC2)CCN2CC1' · $\\text{BF}_4^{-}$ (2 points)", "HF (2 points)", "$LiBF_4$ (2 points)"], "rubric_images": [{"file": "Exam-07-T10-5-1-rubric.png", "description": "'ClC[N+]1(CC2)CCN2CC1' · $\\text{BF}_4^{-}$"}]}]} -{"problem_id": "Exam-08-T8", "header": "Conformational Stability and Electronic Effects\nThis problem tests a systematic understanding of basic conformational analysis and the origins of molecular stability. Compare conformations accessible through bond rotation and answer in terms of steric effects, hyperconjugation, dipole-dipole interactions, electronegativity differences, and related factors.", "header_images": [], "subquestions": [{"sub_id": "Exam-08-T8-1", "tags": ["1.2", "1.7"], "text_open": "For 8.1–8.3 and 8.5–8.6, describe in words the most stable conformation of each molecule shown, specifying in particular whether each substituent is equatorial or axial.\n8.1", "text_images": [{"file": "Exam-08-T8-1-1-prompt.png", "description": "'CC(C)(C)N1CN(C(C)(C)C)CN(C(C)(C)C)C1'"}], "full_mark": 2, "rubric": ["Most stable conformation: a trisubstituted nitrogen-containing six-membered ring whose skeleton lies approximately in the plane of the page. It contains three nitrogen atoms, 1 at the left, one at the upper right, and one at the lower right. Each nitrogen bears a tert-butyl (tBu) substituent, and all three tBu groups project outward from the ring. Two tBu groups are equatorial and one is axial. (2 points)"], "rubric_images": [{"file": "Exam-08-T8-1-1-rubric.png", "description": "A trisubstituted nitrogen-containing six-membered ring whose skeleton lies approximately in the plane of the page. It contains three nitrogen atoms, 1 at the left, one at the upper right, and one at the lower right. Each nitrogen bears a tert-butyl (tBu) substituent, and all three tBu groups project outward from the ring. Two tBu groups are equatorial and one is axial."}]}, {"sub_id": "Exam-08-T8-2", "tags": ["1.2", "1.7"], "text_open": "8.2", "text_images": [{"file": "Exam-08-T8-2-1-prompt.png", "description": "'CN1N(C)CN(C)N(C)C1'"}], "full_mark": 2, "rubric": ["Most stable conformation: a six-membered ring containing four nitrogen atoms in an approximately chair conformation. Two adjacent N atoms lie on the left and another adjacent pair on the right; carbon atoms at the top and bottom connect the pairs. Each N bears an outward-pointing substituent. The methyl substituents on N atoms at positions 1 and 4 are equatorial, whereas those on N atoms at positions 2 and 5 are axial. (2 points)"], "rubric_images": [{"file": "Exam-08-T8-2-1-rubric.png", "description": "A six-membered ring containing four nitrogen atoms in an approximately chair conformation. Two adjacent N atoms lie on the left and another adjacent pair on the right; carbon atoms at the top and bottom connect the pairs. Each N bears an outward-pointing substituent. The methyl substituents on N atoms at positions 1 and 4 are equatorial, whereas those on N atoms at positions 2 and 5 are axial."}]}, {"sub_id": "Exam-08-T8-3", "tags": ["1.2", "1.7"], "text_open": "8.3", "text_images": [{"file": "Exam-08-T8-3-1-prompt.png", "description": "'CN1N(CC2=CC=CC=C2)CN(C)N(CC3=CC=CC=C3)C1'"}], "full_mark": 2, "rubric": ["Most stable conformation: a six-membered chair containing four nitrogen atoms, arranged as adjacent pairs on the left and right sides. Both Bn substituents are axial, while the other two methyl substituents are equatorial. (2 points)"], "rubric_images": [{"file": "Exam-08-T8-3-1-rubric.png", "description": "A six-membered chair containing four nitrogen atoms, arranged as adjacent pairs on the left and right sides. Both Bn substituents are axial, while the other two methyl substituents are equatorial."}]}, {"sub_id": "Exam-08-T8-4", "tags": ["1.2", "1.6", "1.7"], "text_open": "8.4 Explain, in terms of steric hindrance, hyperconjugation, dipole-dipole interactions, electronegativity differences, and related effects, why the anti-staggered conformation predominates in 1,2-dihaloethanes when the halogen is Cl or Br, whereas the gauche-staggered conformation predominates when the halogen is F. (No more than 100 words.)", "text_images": [], "full_mark": 5, "rubric": ["Cl and Br have large atomic radii. Although C-Cl and C-Br bonds are long, the atoms themselves are bulky, and bringing the two halogens close together in the gauche conformation causes strong van der Waals repulsion. (1 point)", "Because F is strongly electronegative, the C-F antibonding orbital is low in energy and is a good electron acceptor. (1 point)", "In the gauche conformation, a C-H bond and a C-F bond are antiperiplanar, giving effective orbital overlap that substantially lowers the system's energy. (1 point)", "The two strongly polar bonds subtend an angle of 60°, producing substantial dipole-dipole repulsion. (1 point)", "The stabilization from hyperconjugation exceeds the destabilization from dipole-dipole repulsion. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-08-T8-5", "tags": ["1.2", "1.7"], "text_open": "8.5", "text_images": [{"file": "Exam-08-T8-5-1-prompt.png", "description": "'O=C(C)N1CCCC[C@H]1C'"}], "full_mark": 2, "rubric": ["Most stable conformation: an N-acyl-substituted nitrogen-containing cyclic skeleton. At the left is an acetyl fragment, $\\mathrm{CH_3}$-C(=O)-N-, and to the right is a nitrogen-containing ring attached through N. The ring adopts an overall chair conformation. The acetyl group is approximately equatorial, while the methyl substituent at the $\\alpha$ position to N is axial. (2 points)"], "rubric_images": [{"file": "Exam-08-T8-5-1-rubric.png", "description": "An N-acyl-substituted nitrogen-containing cyclic skeleton. At the left is an acetyl fragment, $\\mathrm{CH_3}$-C(=O)-N-, and to the right is a nitrogen-containing ring attached through N. The ring adopts an overall chair conformation. The acetyl group is approximately equatorial, while the methyl substituent at the $\\alpha$ position to N is axial."}]}, {"sub_id": "Exam-08-T8-6", "tags": ["1.2", "1.7"], "text_open": "8.6", "text_images": [{"file": "Exam-08-T8-6-1-prompt.png", "description": "'O=C(C)N1[C@@H](C)CCC[C@H]1C'"}], "full_mark": 2, "rubric": ["Most stable conformation: an N-acyl-substituted nitrogen-containing cyclic skeleton. At the left is an acetyl fragment, $\\mathrm{CH_3}$-C(=O)-N-, and to the right is a nitrogen-containing ring attached through N. The ring adopts an overall chair conformation. The acetyl group is approximately equatorial, and both methyl groups at positions $\\alpha$ to N are axial. (2 points)"], "rubric_images": [{"file": "Exam-08-T8-6-1-rubric.png", "description": "An N-acyl-substituted nitrogen-containing cyclic skeleton. At the left is an acetyl fragment, $\\mathrm{CH_3}$-C(=O)-N-, and to the right is a nitrogen-containing ring attached through N. The ring adopts an overall chair conformation. The acetyl group is approximately equatorial, and both methyl groups at positions $\\alpha$ to N are axial."}]}]} {"problem_id": "Exam-08-T10", "header": "Brook Rearrangement\nNote: Unless otherwise specified, cyclic structures involved in the Brook-rearrangement process need not be shown for intermediates in this problem. The Brook rearrangement is a nucleophilic rearrangement centered on silyl migration and is commonly used to construct carbon-oxygen and carbon-carbon bonds.", "header_images": [], "subquestions": [{"sub_id": "Exam-08-T10-1", "tags": ["1.2", "2.2", "2.4"], "text_open": "10.1 A typical Brook rearrangement proceeds as shown below.\n10.1.1 Give the SMILES structures of intermediates A and B.", "text_images": [{"file": "Exam-08-T10-1-1-prompt.png", "description": "Under basic conditions, '[R][C@@](O)([2H])[Si]([R])([R])[R]' reversibly forms [A]; [A] reversibly forms [B]; and protonation of [B] gives '[H][C@@]([R])([2H])O[Si]([R])([R])[R]'."}], "full_mark": 2, "rubric": ["A: '[O-][C@]([R])([2H])[Si]([R])([R])[R]'; B: '[R][C-]([2H])O[Si]([R])([R])[R]'. (1 point each.) No credit is awarded unless retention of configuration at the carbon atom is indicated."], "rubric_images": [{"file": "Exam-08-T10-1-1-rubric.png", "description": "A ('[O-][C@]([R])([2H])[Si]([R])([R])[R]') and B ('[R][C-]([2H])O[Si]([R])([R])[R]')."}]}, {"sub_id": "Exam-08-T10-2", "tags": ["2.4"], "text_open": "10.2 Complete the following reactions using the information above. Ignore stereochemistry.\n10.2.1 Give the SMILES structure of product C.", "text_images": [{"file": "Exam-08-T10-2-1-prompt.png", "description": "'O=C([Si](C)(C)C)C1=CC=CC=C1' reacts with 'FC(F)(F)[Si](C)(C)C' in the presence of $[\\mathrm{nBu}_4\\mathrm{N}][\\mathrm{Ph}_3\\mathrm{SnF}_2]$ to give C."}], "full_mark": 2, "rubric": ["C: 'F/C(F)=C(O[Si](C)(C)C)/C1=CC=CC=C1'. (2 points)"], "rubric_images": [{"file": "Exam-08-T10-2-1-rubric.png", "description": "C ('F/C(F)=C(O[Si](C)(C)C)/C1=CC=CC=C1')."}]}, {"sub_id": "Exam-08-T10-3", "tags": ["2.4"], "text_open": "10.2.2 Give the SMILES structure of product D. Hint: D contains a TMS group but no O-Si bond.", "text_images": [{"file": "Exam-08-T10-3-1-prompt.png", "description": "'BrC1CC1CC2=CC=CC=C2' is treated with tBuLi in $\\mathrm{Et_2O}$ at -78 °C, then with 'O=C([Si](C)(C)C)C1=CC=CC=C1' from -78 °C to room temperature for 30 min, followed by $\\mathrm{NH_4Cl}$ and $\\mathrm{H_2O}$, to give D."}], "full_mark": 2, "rubric": ["D: 'O=C(CC(CC1=CC=CC=C1)C[Si](C)(C)C)C2=CC=CC=C2'. (2 points)"], "rubric_images": [{"file": "Exam-08-T10-3-1-rubric.png", "description": "D ('O=C(CC(CC1=CC=CC=C1)C[Si](C)(C)C)C2=CC=CC=C2')."}]}, {"sub_id": "Exam-08-T10-4", "tags": ["2.4"], "text_open": "10.3 The Brook rearrangement has broad applications and can be used to construct an eight-membered ring stereoselectively. The Brook rearrangement occurs during formation of E, and G contains two fewer seven-membered rings than F.\n10.3.1 Ignoring stereochemistry, give the SMILES structures of intermediates E and F.", "text_images": [{"file": "Exam-08-T10-4-1-prompt.png", "description": "'O=C([Si](C)(C)C(C)(C)C)/C=C/C(C)(C)C' reacts with '[Li]OC1=CCCCC=C1' in THF at -80 °C to give E. E reacts with 'O=S(N1C(C2=CC=CC=C2)O1)(C3=CC=CC=C3)=O' under NaHMDS conditions to give F. F is treated with $\\mathrm{Pb(OAc)_4}$ in MeOH/benzene to give G ('O=C(C1CCCC(C=O)C(C(C)(C)C)/C=C1/O[Si](C)(C)C(C)(C)C)OC'; the stereochemistry of its three stereogenic carbon atoms is not shown)."}], "full_mark": 2, "rubric": ["E: 'O=C(C1)[C@H]2CCC[C@@H]1[C@@H](C(C)(C)C)C=C2O[Si](C)(C)C(C)(C)C'; F: 'O=C(C1O)C2CCCC1[C@H](C(C)(C)C)C=C2O[Si](C)(C)C(C)(C)C'. (1 point each; stereochemistry is not required.)"], "rubric_images": [{"file": "Exam-08-T10-4-1-rubric.png", "description": "E: 'O=C(C1)[C@H]2CCC[C@@H]1[C@@H](C(C)(C)C)C=C2O[Si](C)(C)C(C)(C)C'; F: 'O=C(C1O)C2CCCC1[C@H](C(C)(C)C)C=C2O[Si](C)(C)C(C)(C)C'."}]}, {"sub_id": "Exam-08-T10-5", "tags": ["1.2"], "text_open": "10.3.2 Give a SMILES structure that shows the relative stereochemistry of the stereogenic centers in G.", "text_images": [], "full_mark": 2, "rubric": ["G: 'O=C([C@H]1CCC[C@@H](C=O)[C@@H](C(C)(C)C)/C=C1/O[Si](C)(C)C(C)(C)C)OC'. (2 points; no credit if the stereochemistry is incorrect.)"], "rubric_images": [{"file": "Exam-08-T10-5-1-rubric.png", "description": "G ('O=C([C@H]1CCC[C@@H](C=O)[C@@H](C(C)(C)C)/C=C1/O[Si](C)(C)C(C)(C)C)OC')."}]}, {"sub_id": "Exam-08-T10-6", "tags": ["1.5", "2.3"], "text_open": "10.4 In recent years, research on Brook rearrangements has expanded beyond silicon systems, and phosphorus-mediated rearrangements have received increasing attention. The following is a new two-step, one-pot synthesis of tetrasubstituted furans using a Brook-like rearrangement.\n10.4.1 Briefly explain why P(V), rather than P(III), is used for the Brook-like rearrangement. (No more than 100 words.)", "text_images": [{"file": "Exam-08-T10-6-1-prompt.png", "description": "'[R1]C(O)(P(OC)(OC)=O)C#C[R2]' reacts with '[R3]C([H])=O' under tBuOK (10 mol%), DMF, -60 °C, 3 h to give H. H then reacts with 'O=C(N1I)CCC1=O' (NIS) to give '[R1]C1=C(I)C([R2])=C([R3])O1'."}], "full_mark": 1, "rubric": ["A phosphate ester formed by P(V) can accept electron attack to form a pentacoordinate intermediate. P(III), with its lone pair, is predominantly nucleophilic and is less able to accept electron attack. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-08-T10-7", "tags": ["2.2", "2.3", "2.4"], "text_open": "10.4.2 Give the SMILES structure of H and at least two charged intermediates formed on the pathway to H.", "text_images": [], "full_mark": 4, "rubric": ["H: '[R1]C(OP(OC)(OC)=O)=C=C(C(O)[R3])[R2]'. (2 points; no other structure is accepted.)", "Two charged intermediates in the formation of H are '[R1]C(C#C[R3])(P(OC)(OC)=O)[O-]' and '[R1][C-](OP(OC)(OC)=O)C#C[R3]'. (1 point each; reasonable resonance forms also earn credit.)"], "rubric_images": [{"file": "Exam-08-T10-7-1-rubric.png", "description": "H: '[R1]C(OP(OC)(OC)=O)=C=C(C(O)[R3])[R2]'."}, {"file": "Exam-08-T10-7-2-rubric.png", "description": "First: '[R1]C(C#C[R3])(P(OC)(OC)=O)[O-]'; second: '[R1][C-](OP(OC)(OC)=O)C#C[R3]'."}]}, {"sub_id": "Exam-08-T10-8", "tags": ["2.4"], "text_open": "10.4.3 I is a phenanthrene derivative. Using the information above, give the SMILES structure of product I.", "text_images": [{"file": "Exam-08-T10-8-1-prompt.png", "description": "'[Ar]C#CC1=CC=CC=C1C2=CC=CC=C2C(C(OCC)=O)=O' is treated with HPO(OiPr)_2 and P2-tBu (10 mol%) in refluxing EtOAc for 7 h to give I."}, {"file": "Exam-08-T10-8-2-prompt.png", "description": "P2-tBu: 'CN(P(N=P(N(C)C)(N(C)C)N(C)C)(N(C)C)=NC(C)(C)C)C'."}], "full_mark": 4, "rubric": ["I: '[Ar]C(OP(OC(C)C)(OC(C)C)=O)C1=C(C=CC=C2)C2=C(C=CC=C3)C3=C1C(OCC)=O'. (3 points)"], "rubric_images": [{"file": "Exam-08-T10-8-1-rubric.png", "description": "I: '[Ar]C(OP(OC(C)C)(OC(C)C)=O)C1=C(C=CC=C2)C2=C(C=CC=C3)C3=C1C(OCC)=O'."}]}]} +{"problem_id": "Exam-09-T4", "header": "Kjeldahl Nitrogen Determination\nDetermination of nitrogen content has long been an important topic in analytical chemistry. In 1883, the Danish chemist Johan G.C.T. Kjeldahl invented the Kjeldahl method. The method mainly comprises digestion, alkalization, and titration: sulfuric-acid digestion converts the nitrogen in a sample into an ammonium salt; alkalization releases ammonia by distillation; and the absorbed ammonia is quantified indirectly by acid-base titration. The organic nitrogen content of a solid sample is to be determined. Answer the following questions based on the procedures described.", "header_images": [], "subquestions": [{"sub_id": "Exam-09-T4-1", "tags": ["4.1", "4.5"], "text_open": "4-1-1 First operation: digestion. Weigh $5.000 \\ \\mathrm{g}$ of the sample into a dry $100 \\ \\mathrm{mL}$ Kjeldahl flask. Add $0.4 \\ \\mathrm{g}$ of copper sulfate, $6 \\ \\mathrm{g}$ of potassium sulfate, and $20 \\ \\mathrm{mL}$ of anhydrous sulfuric acid. Swirl gently, place a small funnel in the flask mouth, and set the flask at an angle of 45° on an asbestos gauze with a small opening. Heat slowly. When the contents turn black and foaming has completely ceased, increase the heat to keep the liquid gently boiling. After the liquid becomes blue-green, clear, and transparent, continue heating for another $1 \\ \\mathrm{h}$. Remove the Kjeldahl flask and cool it to room temperature. Carefully add $20 \\ \\mathrm{mL}$ of water, quantitatively transfer the contents to a $100 \\ \\mathrm{mL}$ volumetric flask, rinse the inner wall of the Kjeldahl flask with a small amount of water, and combine the rinsings in the volumetric flask. Dilute to volume with water, mix, and reserve for use. Conduct a blank experiment in parallel. From the apparatus shown below, select the Kjeldahl flask suitable for digestion. The choices are labeled (a), (b), (c), and (d).", "text_images": [{"file": "Exam-09-T4-1-1-prompt.png", "description": "Image corresponding to option (a): a pear-shaped flask with a short neck and a ground-glass stopper."}, {"file": "Exam-09-T4-1-2-prompt.png", "description": "Image corresponding to option (b): a flask with a long neck, a pear-shaped bottom, and a ground-glass stopper."}, {"file": "Exam-09-T4-1-3-prompt.png", "description": "Image corresponding to option (c): an Erlenmeyer flask with a conical body, a flat bottom, and a ground-glass stopper."}, {"file": "Exam-09-T4-1-4-prompt.png", "description": "Image corresponding to option (d): a volumetric flask with a narrow neck, pear-shaped bottom, an annular calibration mark on the neck, and a ground-glass stopper."}], "full_mark": 1, "rubric": ["(b) (1 point)"], "rubric_images": []}, {"sub_id": "Exam-09-T4-2", "tags": ["4.1", "4.5"], "text_open": "4-2 Alkalization. Assemble the Kjeldahl distillation apparatus as shown below. Fill vessel 1 with water to 2/3 of its volume. Add several glass beads, several drops of methyl red, and several milliliters of sulfuric acid until the solution is red. Heat the water in the steam generator to boiling and maintain boiling. Add $10.0 \\ \\mathrm{mL}$ of $20 \\ \\mathrm{g/L}$ boric acid solution and 3 drops of mixed indicator A to the receiving flask, and immerse the lower end of apparatus 8 beneath the liquid surface in vessel 9. Using a pipette, accurately transfer $5.0 \\ \\mathrm{mL}$ of sample digest through apparatus 4 into reaction chamber 5. Rinse the small cup with $10 \\ \\mathrm{mL}$ of water and allow the rinsing to flow into reaction chamber 5, then tightly insert the rod-shaped glass stopper in apparatus 4. Slowly add $400 \\ \\mathrm{g/L}$ sodium hydroxide solution through apparatus 4 into reaction chamber 5 and immediately seal the system. After distilling for $15 \\ \\mathrm{min}$, move the distillate receiver so that the liquid surface is below the lower end of the condenser. Continue distilling for $1 \\ \\mathrm{min}$. Verify with pH paper that the distillate is neutral, rinse the outside of the lower end of the condenser with a small amount of water, and remove the receiving flask. Which of the following statements about the experimental operations and their purposes are correct?\n(a). Methyl red is added to apparatus 1 to generate acidic vapor that reacts with ammonia in apparatus 5 to form an ammonium salt.\n(b). After sodium hydroxide is added, the glass stopper must be inserted tightly and a water seal formed immediately to prevent loss of ammonia.\n(c). The principal purpose of apparatus 6 is to separate liquid water from the steam generated in apparatus 1.\n(d). After alkalization and collection for one digest have been completed, the next reaction may be started by opening screw clamp 7, closing screw clamp 3, and replacing the solution in vessel 9 with distilled water, in that order, to discharge the waste liquid and rinse the reaction chamber.", "text_images": [{"file": "Exam-09-T4-2-1-prompt.png", "description": "This black-and-white line drawing shows an assembly of laboratory glassware. Three main sections on the left, center, and right are connected by tubing, and the numbers 1 through 9 identify specific components. Left section: number 1 points to the liquid inside a long-neck round-bottom flask, represented by a horizontal dashed line. The bottom of the flask rests on a square base. Number 2 points to this base, which has a circular knob at the center of its front face and four feet underneath. The mouth of the round-bottom flask is sealed with a stopper. A very long straight vertical glass tube passes through the center of the stopper, extending below the liquid surface inside the flask and far above the flask mouth. A short right-angle glass tube also passes through the stopper and is connected at its end to U-shaped flexible tubing. Center section: the U-shaped flexible tubing from the left connects to a narrow inlet above the central apparatus. The main central apparatus is inclined from lower left to upper right and contains an outer jacket and an inner structure. Number 4 points to an upward-facing vertical inlet at the upper-right end of the inclined apparatus; it has a stopper through which a short straight tube is inserted. Number 5 points to an elongated cavity or inner-tube structure inside the inclined apparatus. Number 6 points to the enlarged spherical lower-left end of the inclined apparatus. Number 7 points to a narrow tube extending from the spherical end below number 6, connected to a short length of flexible tubing fitted with a clamp (pinchcock). Number 3 points to another narrow tube extending from the upper-middle part of the outer jacket, likewise connected to a short length of flexible tubing fitted with a clamp (pinchcock). The upper-right end of the inclined apparatus connects to the right section through an inverted U-shaped bent glass tube. Right section: number 8 points to a vertically oriented straight condenser. The end of the inverted U-shaped bent glass tube enters the top of the condenser. Two horizontal side-arm connections extend to the right from the outer jacket of condenser 8. A right-pointing arrow beside the upper side arm indicates outward flow, and a left-pointing arrow beside the lower side arm indicates inward flow. Number 9 points to an Erlenmeyer flask directly beneath the condenser. The flask contains liquid, represented by a horizontal dashed line. The inner straight tube of condenser 8 extends downward through the mouth of the flask and below the liquid surface inside it."}], "full_mark": 2, "rubric": ["(b)(c) (2 points; 1 point if 1 correct option is selected; no points for an incorrect selection or an omitted correct option)"], "rubric_images": []}]} {"problem_id": "Exam-09-T5", "header": "Reviewing the Old to Learn the New: Revisiting a Hexagonal Laves Phase\nA magnesium-zinc alloy belongs to the hexagonal crystal system, with unit-cell parameters a = 0.5157 nm and c = 0.8373 nm. Projections of the unit cell along different directions are shown below. The large light-colored spheres represent Mg atoms, and the small dark-colored spheres represent Zn atoms. Interestingly, the projections along the a and b directions are identical. One Zn atom has coordinates (0.830, 0.170, 0.250). ", "header_images": [{"file": "Exam-09-T5-header-1.png", "description": "The image presents, side by side, three two-dimensional projections of a crystal structure on white backgrounds, each enclosed by thin solid gray lines and composed of small dark-gray spheres and large light-gray spheres. Each projection includes blue coordinate axes with arrows. The left projection is a vertically elongated rectangular unit-cell projection showing the bc plane. At its lower-left corner, the coordinate axes indicate the b axis horizontally to the right and the c axis vertically upward. A total of 10 small dark-gray spheres are visible in the rectangle: four at the vertices, two at the midpoints of the vertical edges, and 4 within the cell arranged from bottom to top in a staggered left-right-left-right zigzag along the vertical direction. Another 4 large light-gray spheres are distributed inside, arranged from bottom to top in the opposite right-left-right-left sequence. The middle projection immediately beside it is a rectangular projection of the ac plane. Its lower-left axes indicate the c axis vertically upward but the a axis horizontally to the left. This view is completely identical to the left view in cell shape and in the number, positions, and visual arrangement of all dark- and light-gray spheres: 4 dark spheres at the vertices, 2 at the midpoints of the vertical edges, 4 additional dark spheres staggered left and right inside, and 4 large light spheres staggered right and left. The rightmost projection shows the ab plane. Its lower-left coordinate axes indicate the a axis horizontally to the right and the b axis obliquely upward to the left, forming an obtuse angle and defining a parallelogram unit-cell outline with a horizontal base and sides inclined upward to the left. A small dark-gray sphere lies at each of the four vertices. Another 6 small dark-gray spheres are arranged in a ring inside the cell: one each near the left and right interiors of the top horizontal edge, one each near the left and right interiors of the bottom horizontal edge, and one near the middle interior of each inclined side. Among these 6 internal dark spheres, only 2 large light-gray spheres lie in diagonal relation. One occupies an open region in the upper-left part of the cell and is partially surrounded by 3 internal dark spheres; the other occupies the lower-right region and is likewise partially surrounded by 3 internal dark spheres."}], "subquestions": [{"sub_id": "Exam-09-T5-1", "tags": ["1.3"], "text_open": "5.2 Calculate the shortest Zn-Zn distance. ", "text_images": [], "full_mark": 6, "rubric": ["The possible shortest Zn-Zn distances may occur between Zn1-Zn2, Zn2-Zn3, or Zn2-Zn4. First write their coordinates: Zn1(0, 0, 0), Zn2(0.830, 0.170, 0.250), Zn3(0.830, 0.660, 0.250), Zn4(0.830, -0.340, 0.250) (1 point). Calculate the Zn1-Zn2 distance, retaining one extra significant figure during the calculation: $d_a = |0.830 - 1| \\times 0.5157 \\text{ nm} = 0.08767 \\text{ nm}$, $d_b = |0.170 - 0| \\times 0.5157 \\text{ nm} = 0.08767 \\text{ nm}$, $d_c = |0.250 - 0| \\times 0.8373 \\text{ nm} = 0.2093 \\text{ nm}$, and $d(\\text{Zn1-Zn2}) = \\sqrt{d_a^2 + d_b^2 - 2 d_a d_b \\cos \\theta + d_c^2}$ $= \\sqrt{0.08767^2 + 0.08767^2 - 2 \\times 0.08767^2 \\times (-1/2) + 0.2093^2} = 0.259 (\\text{nm})$ (2 points). Calculate the Zn2-Zn3 distance: $d(\\text{Zn2-Zn3}) = (0.660 - 0.170) \\times 0.5157 \\text{ nm} = 0.253 (\\text{nm})$ (1 point). Calculate the Zn2-Zn4 distance: $d(\\text{Zn2-Zn4}) = 2 \\times (0.170 - (-0.340)) \\times 0.5157 \\text{ nm} = 0.263 (\\text{nm})$ (1 point). Therefore, the shortest Zn-Zn distance is Zn2-Zn3 and equals 0.253nm (1 point)."], "rubric_images": []}]} -{"problem_id": "Exam-09-T8", "header": "Problem 8 Fundamentals of Organic Chemistry", "header_images": [], "subquestions": [{"sub_id": "Exam-09-T8-1", "tags": ["1.1", "1.5"], "text_open": "Let the bond angles shown below be α and β, respectively, and define γ = α - β. Rank the following substances in decreasing order of γ.", "text_images": [{"file": "Exam-09-T8-1-1-prompt.png", "description": "A: 'O=C1OCC1'; B: 'O=C1OCCC1'; C: 'O=C1OCCCC1'. The image labels the two bond angles α and β, where α is the carbonyl O-C-C bond angle and β is the ester O-C-O bond angle."}], "full_mark": 3, "rubric": [" ABC (3 points)", "Each blank is worth 1 point; award that 1 point if correct, and that 1 point is not awarded if incorrect."], "rubric_images": []}, {"sub_id": "Exam-09-T8-2", "tags": ["2.3", "3.1", "3.7"], "text_open": "Reaction background: The rate constant of the reaction shown below is significantly greater in acetone than in ethanol.\nCompared with the reaction in ethanol, how do the Gibbs free energies of the reactant ground state and transition state change relative to one another in acetone, thereby decreasing the activation energy? (a) Only the transition-state energy decreases. (b) Only the reactant energy increases. (c) The reactant energy decreases and the transition-state energy also decreases. (d) None of the above (write the correct answer as well).", "text_images": [{"file": "Exam-09-T8-2-1-prompt.png", "description": "1-Bromopropane reacts with sodium iodide in a solvent: 'BrCCC' gives 'ICCC' under NaI and Solv. conditions."}], "full_mark": 4, "rubric": [" (d) (2 points). The energies of both the reactant and transition state are higher than in ethanol, but the reactant energy increases by more than the transition-state energy (2 points)."], "rubric_images": []}]} -{"problem_id": "Exam-09-T10", "header": "Indium-Catalyzed Alkyne-Migration Reaction\n4-Alkynyl-2-quinolone derivatives occur widely in bioactive molecules with antibacterial, antitumor, and other activities. Indium-catalyzed alkyne migration provides a new route for efficiently synthesizing these compounds. Using ortho-alkynyl benzamides as substrates, the method efficiently constructs a 4-alkynyl-2-quinolone framework under $\\mathrm{In(OTf)_3}$ catalysis accompanied by selective alkyne migration. Analyze the reaction process and answer the questions.", "header_images": [], "subquestions": [{"sub_id": "Exam-09-T10-1", "tags": ["2.2", "2.3"], "text_open": "10.1 Alkyne migration involves transposition of an alkynyl group along a carbon chain, allowing interconversion between internal and terminal alkynes. A classic example is the Alkyne Zipper Reaction, which uses KAPA reagent to move an alkyne to the terminus. Write the SMILES structures of the 3 key uncharged intermediates in this reaction.", "text_images": [{"file": "Exam-09-T10-1-1-prompt.png", "description": "Substrate: 'CCC#CC1=CC=CC=C1'. The reagent is KAPA, represented by SMILES 'NCCCN[K]', in which the terminal -NH_2 chelates to K to form a six-membered ring. Product: 'C#CCCC1=CC=CC=C1'. The figure shows migration of the alkyne from an internal position to the terminus."}], "full_mark": 3, "rubric": ["3 points, 1 point each. Reference SMILES: 'CC=C=C(C1=CC=CC=C1)', 'CC#CCC1=CC=CC=C1', and 'C=C=CCC1=CC=CC=C1'."], "rubric_images": [{"file": "Exam-09-T10-1-1-rubric.png", "description": "'CC=C=C(C1=CC=CC=C1)', 'CC#CCC1=CC=CC=C1', and 'C=C=CCC1=CC=CC=C1'"}]}, {"sub_id": "Exam-09-T10-2", "tags": ["2.8"], "text_open": "10.4 The authors used the ortho-alkynyl benzamide substrate above to study the alkyne-migration process. Analyze the following conversion and answer the questions. 10.4.1 The authors proposed that In(III) also catalyzes isomerization of B to A. The process proceeds smoothly in 1,4-dioxane, giving A in 89 % NMR yield. In contrast, under otherwise identical conditions in chlorobenzene or toluene, only trace A is detected. Explain why.", "text_images": [{"file": "Exam-09-T10-2-1-prompt.png", "description": "Substrate: 'O=C(C(C(C)C)=O)N(C)C1=CC=CC=C1C#CCCCC'; conditions: In(OTf)_3 and 1,4-dioxane (1,4-dioxane). Product A: 'CN1C2=CC=CC=C2C(C#CCCC)=C(C(C)C)C1=O'; product B: 'CN(C([C@@]1(O)C(C)C)=O)C2=CC=CC=C2C1=[C@@]=C([H])CC'."}], "full_mark": 2, "rubric": ["1,4-Dioxane can stabilize the carbocation intermediate and promote the dehydration pathway from B to A (2 points)."], "rubric_images": []}, {"sub_id": "Exam-09-T10-3", "tags": ["1.6", "2.7"], "text_open": "10.4.2 If the catalyst is replaced by a stronger Lewis acid, such as $\\mathrm{Fe(OTf)_3}$ or $\\mathrm{Sm(OTf)_3}$, the yield of product A instead decreases substantially. Explain why.", "text_images": [], "full_mark": 2, "rubric": ["As oxophilic Lewis acids/hard acids, Fe(OTf)_3 and Sm(OTf)_3 coordinate strongly with 1,4-dioxane and thereby inhibit dehydrative isomerization (2 points)."], "rubric_images": []}, {"sub_id": "Exam-09-T10-4", "tags": ["2.4"], "text_open": "10.5 The authors also explored downstream transformations of 4-alkyl-2-quinolone products. Write the SMILES structures of the following 2 products. Pay attention to stereochemistry in reaction 1. Reaction 2 is a ring-expansion reaction of another quinolone derivative.", "text_images": [{"file": "Exam-09-T10-4-1-prompt.png", "description": "Reaction 1 reactant: 'O=C1N(C2=CC=CC=C2C(C#CCCC)=C1C)C'; conditions: C=C[R], Thioxanthen-9-one (0.3 eq.), 390 nm LEDs, CH_3CN, 6-9h; product unknown. Reaction 2 reactant: 'O=C1N(C)C2=CC=CC=C2C(C#CCCC)=C1C3=CC=CC=C3'; conditions: HOTf (2.0 eq.), PhMe, 100°C, 24h."}], "full_mark": 4, "rubric": ["4 points, 2 points each. Reference SMILES: product 1 is 'CN1C2=CC=CC=C2[C@@]3(C#CCCC)[C@@](C[C@@H]3[R])(C)C1=O' (the relative stereochemistry shown must be retained); product 2 is 'CN1C2=CC=CC=C2C3=C(C4=CC=CC=C4C(CCC)=C3)C1=O'."], "rubric_images": [{"file": "Exam-09-T10-4-1-rubric.png", "description": "Product 1 is 'CN1C2=CC=CC=C2[C@@]3(C#CCCC)[C@@](C[C@@H]3[R])(C)C1=O'; product 2 is 'CN1C2=CC=CC=C2C3=C(C4=CC=CC=C4C(CCC)=C3)C1=O'."}]}]} -{"problem_id": "Exam-12-T1", "header": "Problem 1 Metallic Vitamins. Rare-earth elements are core strategic resources in modern materials and chemical industries and are known as \"industrial vitamins.\" Their unique electron configurations give them distinctive physicochemical properties, making them essential materials for emerging industries such as modern semiconductors and new energy. A rare-earth element M has density $7.563 \\ \\mathrm{g \\cdot cm^{-3}}$ and forms an alloy with Co in a fixed ratio for manufacturing strong magnets. It has the following chemical properties. (1) When elemental M is left in air, its metallic luster rapidly disappears and a composite thin layer whose principal component is oxide A forms on its surface. M reacts with hydrochloric acid to form blood-red chloride B. After M has dissolved completely, the blood-red color rapidly fades, producing pale-yellow chloride C and colorless gas D. (2) If M is placed in a tetrahydrofuran solution of an organic compound E containing $90.05\\%$ iodine, reaction at room temperature gives unstable green iodide F and releases gas G. Addition of a small amount of water turns the solution yellow. Gas G is a common plant-ripening hormone.", "header_images": [], "subquestions": [{"sub_id": "Exam-12-T1-1", "tags": ["1.4", "2.4"], "text_open": "1-4-1 Complexes of M also have interesting structures. Addition of 0.5 equivalents of $\\mathrm{P_2Ph_4}$ to $\\mathrm{MCp*_2(THF)_2}$ gives a chain polymer X. After polymer X is separated, the pentane mother liquor is left for several weeks, yielding a yellow-brown complex Y. Y contains two M atoms and a 14-membered macrocycle. The mass fraction of M is $22.19\\%$ in both X and Y. Write the SMILES structure of the repeat unit of X, using [R] for the site connected to the next repeat unit (for example, polypropylene may be written as C([R])C(C)[R]), and the SMILES structure of Y ($\\mathrm{Cp*}$ = pentamethylcyclopentadienyl anion).", "text_images": [], "full_mark": 4, "rubric": ["(2 points each for the structures of X and Y)", "Structure or SMILES corresponding to the scoring figure: X: the polymer repeat-unit SMILES is '[R][Sm]([Cp*])([Cp*])OCCCC[P](C1=CC=CC=C1)(C2=CC=CC=C2)[R]', where R denotes the connection site between polymer monomers. The SMILES of Y is '[*Cp][Sm]1([*Cp])OCCCC[P](C2=CC=CC=C2)(C3=CC=CC=C3)[Sm]([Cp*])([Cp*])OCCCC[P]1(C4=CC=CC=C4)C5=CC=CC=C5'."], "rubric_images": [{"file": "Exam-12-T1-1-1-rubric.png", "description": "X: the polymer repeat-unit SMILES is '[R][Sm]([Cp*])([Cp*])OCCCC[P](C1=CC=CC=C1)(C2=CC=CC=C2)[R]', where R denotes the connection site between polymer monomers. The SMILES of Y is '[*Cp][Sm]1([*Cp])OCCCC[P](C2=CC=CC=C2)(C3=CC=CC=C3)[Sm]([Cp*])([Cp*])OCCCC[P]1(C4=CC=CC=C4)C5=CC=CC=C5'."}]}]} -{"problem_id": "Exam-12-T4", "header": "Problem 4 Titrimetric Analysis of Etching Acids\nWet chemical etching is a major process in the semiconductor-chip industry. It uses acidic substances to corrode a substrate surface and thereby achieve etching. A mixture of hydrofluoric acid, fluorosilicic acid, and nitric acid is a common etchant. A rapid and accurate laboratory method for determining the contents of these three acids in an etchant is described below.", "header_images": [], "subquestions": [{"sub_id": "Exam-12-T4-1", "tags": ["4.2"], "text_open": "Procedure background: Accurately weigh 3.028 g of the sample. Place 1.502 g of it in a plastic beaker to prepare test solution A, and use the remaining sample to prepare test solution B by the same procedure. During preparation, add 10 mL of saturated $KNO_3$ solution and anhydrous ethanol, mix, dilute with anhydrous ethanol, and let stand for approximately 20 min until the precipitate has completely settled into a separate layer. Label this as test solution A. Prepare test solution B from the remaining sample by the same procedure.\n4.1.2 State the purpose of adding saturated KNO_3 solution.", "text_images": [], "full_mark": 2, "rubric": ["Adding KNO_3 precipitates SiF_6^{2-}, preventing interference from hydrolysis of SiF_6^{2-} (2 points)."], "rubric_images": []}, {"sub_id": "Exam-12-T4-2", "tags": ["2.1", "4.2"], "text_open": "4.2 Titrimetric analysis. Add 30 mL of water and 3 drops of phenolphthalein indicator to test solution A. Titrate with 0.5018 $\\mathrm{mol \\cdot L^{-1}}$ standard NaOH solution until the solution becomes faint pink; 24.39 mL of the standard NaOH solution is consumed. Add 30 mL of water and 2 drops of methyl-orange indicator to test solution B. Adjust the pH with NaOH solution until the solution just turns bright yellow. Using a fluoride-ion combination electrode as the measuring electrode, titrate with 0.08273 $\\mathrm{mol \\cdot L^{-1}}$ standard La(NO_3)_3 solution; precipitate forms continuously during the titration. At the endpoint, 25.11 mL of La(NO_3)_3 solution is consumed. Weigh 0.5003 g of the original sample into a beaker containing 15 mL of 0.5 $\\mathrm{mol \\cdot L^{-1}}$ NaOH solution, add water to 60 mL, and verify with pH paper that the pH is greater than 13. Heat slowly to boiling, maintain a gentle boil for 5 min, and cool to room temperature. Add 2 drops of methyl-red indicator, adjust with nitric acid until red, and then adjust with NaOH solution until the solution just turns bright yellow. Using a fluoride-ion combination electrode as the indicator electrode, titrate with 0.08273 $\\mathrm{mol \\cdot L^{-1}}$ standard La(NO_3)_3 solution. At the endpoint, 15.28 mL is consumed. 4.2.1 Write all reaction equations occurring in Sections 4.1 and 4.2; write repeated reactions only once.", "text_images": [], "full_mark": 4, "rubric": ["$2KNO_3 + H_2SiF_6 = 2HNO_3 + K_2SiF_6$ (1 point) $NaOH + HNO_3 = NaNO_3 + H_2O$ (0.5 point) $NaOH + HF = NaF + H_2O$ (0.5 point) $La(NO_3)_3 + 3NaF = 3NaNO_3 + LaF_3$ (1 point) $H_2SiF_6 + 8NaOH = Na_2SiO_3 + 6NaF + 5H_2O$ (1 point)"], "rubric_images": []}, {"sub_id": "Exam-12-T4-3", "tags": ["4.7", "4.5"], "text_open": "4.2.2 Calculate the mass fractions of HF, H_2SiF_6, and HNO_3 in the sample. Note: the fluoride-ion electrode indicates the endpoint by the abrupt change in fluoride-ion concentration.", "text_images": [], "full_mark": 6, "rubric": ["Let the mass fractions of HF, $H_2SiF_6$, and $HNO_3$ be $w_1$, $w_2$, and $w_3$, respectively.", "$0.5018\\ \\mathrm{mol\\cdot L^{-1}}\\times24.39\\ \\mathrm{mL}=1.502\\ \\mathrm{g}\\times\\left(\\frac{w_1}{20.008}+\\frac{2w_2}{144.106}+\\frac{w_3}{63.018}\\right)$", "$3\\times0.08273\\ \\mathrm{mol\\cdot L^{-1}}\\times25.11\\ \\mathrm{mL}=(3.028-1.502)\\ \\mathrm{g}\\times\\frac{w_1}{20.008}$", "$3\\times0.08273\\ \\mathrm{mol\\cdot L^{-1}}\\times15.28\\ \\mathrm{mL}=0.5003\\ \\mathrm{g}\\times\\left(\\frac{w_1}{20.008}+\\frac{6w_2}{144.106}\\right)$", "The system of equations is worth 3 points, 1 point for each equation. Solving the 3 equations simultaneously gives $w_1=8.171\\%, w_2=8.397\\%, w_3=18.27\\%$. The answers are worth 3 points, 1 point each. Significant figures are required. Intermediate results must contain at least 4 significant figures; deduct 1 point for any significant-figure error, only once in total."], "rubric_images": []}]} -{"problem_id": "Exam-12-T10", "header": "Phenonium-Ion Chemistry\nPhenonium ions are highly reactive cationic intermediates whose applications were long limited by difficult preparation and poor reaction control. In recent years, however, physical-organic studies and new strategies have greatly expanded their synthetic potential.", "header_images": [], "subquestions": [{"sub_id": "Exam-12-T10-1", "tags": ["2.2"], "text_open": "10.1 Kinetic measurements provide an indirect method for studying phenonium-ion intermediates. Consider the following solvolysis in glacial acetic acid. 10.1.1 Two mechanisms are possible: one-step path A and two-step path B. Write the SMILES structure of the intermediate in path B.", "text_images": [{"file": "Exam-12-T10-1-1-prompt.png", "description": "The figure shows a reaction in which 'CC1=CC=C(S(OCCC2=CC=CC=C2)(=O)=O)C=C1' reacts in HOAc to give 'O=C(OCCC1=CC=CC=C1)C'."}], "full_mark": 2, "rubric": ["The intermediate in path B is shown in the rubric figure. (2 points; a resonance structure also earns credit)", "Structure or SMILES corresponding to the rubric figure: 'C1=CC=CC2(CC2)[CH+]1'"], "rubric_images": [{"file": "Exam-12-T10-1-1-rubric.png", "description": "'C1=CC=CC2(CC2)[CH+]1'"}]}, {"sub_id": "Exam-12-T10-2", "tags": ["2.3", "3.7"], "text_open": "10.1.3 Solvolysis kinetics were similarly studied for multiple precursors capable of forming phenonium ions, giving two different kinetic patterns, Figure 1 and Figure 2. Briefly explain mechanistically the behavior in Figure 2.", "text_images": [{"file": "Exam-12-T10-2-1-prompt.png", "description": "Figures 1 and 2 show two kinetic plots for phenonium-ion precursors. Each has only a solid left vertical axis and bottom horizontal axis, forming an L-shaped frame with no numerical ticks. Both vertical axes are labeled $\\lg(k_X/k_H)$ and both horizontal axes Hammett $\\sigma$ scale. The left plot contains two intersecting dashed straight lines. One has a steep negative slope, is labeled $k_\\Delta$-line at its upper start, and slope: $\\rho_\\Delta$ near its middle. The other has a very shallow negative slope, crosses the lower half and intersects the first line near the lower left; it is labeled $k_s$-line at its left end and slope: $\\rho_s$ on the right. The right plot contains only the steep negatively sloped dashed line, similarly labeled $k_\\Delta$-line and slope: $\\rho_\\Delta$."}], "full_mark": 2, "rubric": ["For both electron-withdrawing and electron-donating substituents, the mechanism proceeds through a phenonium-ion intermediate. (1 point)", "This indicates that the carbocation is relatively stable. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-12-T10-3", "tags": ["2.3", "3.7"], "text_open": "10.1.4 The following two substrates correspond to the two cases in 10.1.3. Which has the kinetic behavior in Figure 1? ( ). Which has the behavior in Figure 2? ( ).", "text_images": [{"file": "Exam-12-T10-3-1-prompt.png", "description": "Two phenonium-ion precursors, (a) and (b), are shown. With no benzene substituent, (a) is 'CC1=CC=C(S(OC(C)CC2=CC=CC=C2)(=O)=O)C=C1' and (b) is 'CC1=CC=C(S(OC(C)(C)CC2=CC=CC=C2)(=O)=O)C=C1'. Either may be substituted on the benzene ring, so benzene-X denotes a substituent of unspecified position on the ring."}], "full_mark": 2, "rubric": ["(b) (1 point)", "(a) (1 point)"], "rubric_images": []}, {"sub_id": "Exam-12-T10-4", "tags": ["2.2"], "text_open": "10.2 In recent years, chemists have developed various synthetic strategies to overcome the limitations on phenonium-ion formation. The use of hypervalent iodine reagents marked an important turning point for phenonium ions in synthesis; the core strategy is to introduce a “super” leaving group. The following highly regioselective difluorination has been reported. (Amine: an amine compound used to regulate the acid-base conditions of the system.) Hypervalent iodine reagent X is first generated in situ during the reaction.\n10.2.2 Write the SMILES structures of 2 key reaction intermediates, excluding X.", "text_images": [{"file": "Exam-12-T10-4-1-prompt.png", "description": "The figure shows a one-step reaction. Reactant 'C=C(C)C1=CC=C([N+]([O-])=O)C=C1' is treated with 'CC1=CC=C(I)C=C1' (10 mol%), 'ClC[N+]1(CC2)CC[N+]2(F)CC1' (1.5 eq.), Amine:HF ≈ 1:9, in CHCl_3 at r.t., giving 'CC(F)(F)CC1=CC=C([N+]([O-])=O)C=C1'."}], "full_mark": 2, "rubric": ["The two key intermediates are shown in the rubric figure. 2 points total, 1 point each.", "Structures or SMILES corresponding to the rubric figure: the 2 key intermediates excluding X are 'CC(CI(F)[Ar])(F)C1=CC=C([N+]([O-])=O)C=C1' on the left and 'CC1(F)CC21C=CC([N+]([O-])=O)=C[CH+]2' on the right."], "rubric_images": [{"file": "Exam-12-T10-4-1-rubric.png", "description": "The 2 key intermediates excluding X are 'CC(CI(F)[Ar])(F)C1=CC=C([N+]([O-])=O)C=C1' on the left and 'CC1(F)CC21C=CC([N+]([O-])=O)=C[CH+]2' on the right."}]}, {"sub_id": "Exam-12-T10-5", "tags": ["2.7"], "text_open": "10.2.3 Briefly explain the regioselectivity of the final step.", "text_images": [], "full_mark": 1, "rubric": ["$F^-$ attacks the electron-deficient carbon atom and therefore attacks the carbon bearing F. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-12-T10-6", "tags": ["2.4"], "text_open": "10.2.4 Complete the following reaction.", "text_images": [{"file": "Exam-12-T10-6-1-prompt.png", "description": "The figure shows a one-step reaction. Reactant 'O=[N+](C1=CC=C2C(CCC/C2=C\\C(OCC)=O)=C1)[O-]' is treated with 'CC1=CC=C(I)C=C1' (20 mol%), mCPBA (1.5 eq.), Amine:HF = 1:7, in CHCl_3 at r.t. The product is unknown and indicated by ?."}], "full_mark": 2, "rubric": ["The reaction product is shown in the rubric figure. (2 points)", "Structure or SMILES corresponding to the rubric figure: 'CC1=CC=C2C(CCCC(F)(F)C2C(OCC)=O)=C1'"], "rubric_images": [{"file": "Exam-12-T10-6-1-rubric.png", "description": "'CC1=CC=C2C(CCCC(F)(F)C2C(OCC)=O)=C1'"}]}, {"sub_id": "Exam-12-T10-7", "tags": ["2.2"], "text_open": "10.3 Diazo reagents have also been used to generate phenonium-ion intermediates. Examine the following reaction. 10.3.1 Write the SMILES structure of the phenonium-ion intermediate.", "text_images": [{"file": "Exam-12-T10-7-1-prompt.png", "description": "The figure shows a one-step reaction between 'BrCC1=CC=C(OC)C=C1' and 'CCC(C(OCC1=CC=CC=C1)=O)=[N+]=[N-]' under SnBr_4 (50 mol%) in CH_2Cl_2 from -78℃ to r.t. The product is unknown."}], "full_mark": 1, "rubric": ["The phenonium-ion intermediate is shown in the rubric figure. (1 point)", "Structure or SMILES corresponding to the rubric figure: 'COC1=C[CH+]C2(CC2(C(OCC3=CC=CC=C3)=O)CC)C=C1'"], "rubric_images": [{"file": "Exam-12-T10-7-1-rubric.png", "description": "'COC1=C[CH+]C2(CC2(C(OCC3=CC=CC=C3)=O)CC)C=C1'"}]}, {"sub_id": "Exam-12-T10-8", "tags": ["2.2", "2.4"], "text_open": "10.3.2 The favored transition state for opening the phenonium ion is known to be a six-membered cyclic transition state. Write the SMILES of the major product; stereochemistry is not required.", "text_images": [], "full_mark": 1, "rubric": ["The major product is shown in the rubric figure. (1 point)", "Structure or SMILES corresponding to the rubric figure: 'BrCC(CC)(C(OCC1=CC=CC=C1)=O)C2=CC=C(OC)C=C2'"], "rubric_images": [{"file": "Exam-12-T10-8-1-rubric.png", "description": "'BrCC(CC)(C(OCC1=CC=CC=C1)=O)C2=CC=C(OC)C=C2'"}]}, {"sub_id": "Exam-12-T10-9", "tags": ["1.1", "2.2"], "text_open": "10.4 Another strategy for obtaining phenonium ions is to synthesize long-lived carbocations. Examine the following reaction; configurations of stereogenic carbons are not marked in the product. 10.4.1 Write the SMILES structures of the key intermediates, including stereochemistry.", "text_images": [{"file": "Exam-12-T10-9-1-prompt.png", "description": "The figure shows a one-step reaction. Reactant 'O=C(/C(C(OC)=O)=C/C1=CC=CC=C1)/C(C)=C(C2=CC=CC=C2)\\CC' is treated with Cu(SbF_6)_2 (1 eq.) in CH_2Cl_2 from r.t. to 40°C, giving 'O=C1C(C(OC)=O)=CC(C2=CC=CC=C2)(CC)C1(C)C3=CC=CC=C3'."}], "full_mark": 6, "rubric": ["The key intermediates are shown in the rubric figure. 6 points total. Deduct 1 point for each incorrect organic skeleton, up to 4 points. Chelating coordination between the organic skeleton and copper must be shown, and it must be stated in words or otherwise indicated that copper bears additional ligands besides the organic skeleton; otherwise deduct 1 point once. For stereochemical errors, deduct 1 point once.", "Structures or SMILES corresponding to the rubric figure, from left to right: '[H][C@@]1(C2=CC=CC=C2)[C@](C3=CC=CC=C3)(CC)[C+](C)C4=C1C(OC)=O[#29]O4', '[H][C@@]1(C2=CC=CC=C2)[C@]3(CC)[C@](C34[CH+]C=CC=C4)(C)C5=C1C(OC)=O[#29]O5', '[H][C@]1([C+]([C@@](C2=CC=CC=C2)(C3=C1C(OC)=O[Cu]O3)C)CC)C4=CC=CC=C4', and '[H][C@@]12[C@@]([C@@]23[CH+]C=CC=C3)(CC)[C@@](C)(C4=CC=CC=C4)C5=C1C(OC)=O[#29]O5'. The Cu catalytic center is denoted [Cu], indicating that in addition to chelation by the organic skeleton, the Cu ion bears other ligands that are irrelevant to the reaction."], "rubric_images": [{"file": "Exam-12-T10-9-1-rubric.png", "description": "The four intermediates from left to right are '[H][C@@]1(C2=CC=CC=C2)[C@](C3=CC=CC=C3)(CC)[C+](C)C4=C1C(OC)=O[#29]O4', '[H][C@@]1(C2=CC=CC=C2)[C@]3(CC)[C@](C34[CH+]C=CC=C4)(C)C5=C1C(OC)=O[#29]O5', '[H][C@]1([C+]([C@@](C2=CC=CC=C2)(C3=C1C(OC)=O[Cu]O3)C)CC)C4=CC=CC=C4', and '[H][C@@]12[C@@]([C@@]23[CH+]C=CC=C3)(CC)[C@@](C)(C4=CC=CC=C4)C5=C1C(OC)=O[#29]O5'. The Cu catalytic center is abbreviated [Cu], indicating that Cu bears additional ligands irrelevant to the reaction."}]}, {"sub_id": "Exam-12-T10-10", "tags": ["1.1", "2.4", "2.7"], "text_open": "10.4.2 If the phenyl group on the right side of the substrate is replaced with 2,4,6-trimethoxyphenyl, abbreviated Ar, a product with different relative substituent positions is obtained. Write its SMILES, including stereochemistry.", "text_images": [{"file": "Exam-12-T10-10-1-prompt.png", "description": "Schematic of the 2,4,6-trimethoxyphenyl group, represented by SMILES 'COC1=CC(OC)=CC(OC)=C1[R]', where [R] is the connection to another group."}], "full_mark": 2, "rubric": ["The product structure using the Ar abbreviation is shown in the rubric figure. (2 points)", "Structure or SMILES corresponding to the rubric figure: '[Ar]C1=C(C(OC)=O)C([C@](C)(C2=CC=CC=C2)[C@H]1CC)=O'"], "rubric_images": [{"file": "Exam-12-T10-10-1-rubric.png", "description": "'[Ar]C1=C(C(OC)=O)C([C@](C)(C2=CC=CC=C2)[C@H]1CC)=O'"}]}]} +{"problem_id": "Exam-09-T7", "header": "Semiconductors and Electrochemistry\nSemiconductors are materials whose electrical conductivity at room temperature lies between that of conductors and insulators. Common semiconductor materials include silicon, germanium, and gallium arsenide. Based on band theory and Fermi statistics, doping with group V or group III elements, or with other materials that can donate electrons (donors) or accept electrons (acceptors), can regulate the Fermi level and carrier concentration of a semiconductor and thereby its various properties. Semiconductor carriers generally comprise electrons and holes. In this problem, involving an n-type semiconductor in which electrons are the majority carriers, only electrons need be considered. Nonuniform doping can create a carrier-concentration gradient in a semiconductor. The resulting diffusion causes charge separation and a built-in electric field; the built-in field opposes diffusion until equilibrium is reached. This is fundamentally similar to the electrochemical picture in which a chemical-potential difference drives charge transfer, which in turn produces a potential difference that opposes further charge transfer, ultimately establishing an electrode potential at equilibrium. By comparing a concentration-gradient-doped semiconductor with the electromotive force of an electrochemical concentration cell, this problem explores the universality of diffusion-electric-field equilibrium.\n7.1 Built-in electric field and diffusion equilibrium in a semiconductor\nConsider an n-type semiconductor at temperature T = 300 K with electron concentration varying with position x as $n(x)=N_{d0}\\exp(-x/L)$. Here $N_{d0}=1\\times10^{18}\\ \\mathrm{cm}^{-3}$, $L=0.1\\ \\mu\\mathrm{m}$, $D_n=225\\ \\mathrm{cm}^2\\mathrm{/s}$, $e=1.6\\times10^{-19}\\ \\mathrm{C}$, and $k=1.38\\times10^{-23}\\ \\mathrm{J/K}$. These quantities are, respectively, the donor concentration at x = 0 (which, under the approximation used here, may be regarded as the electron concentration at x = 0), characteristic length, electron diffusion coefficient, elementary charge, and Boltzmann constant.", "header_images": [], "subquestions": [{"sub_id": "Exam-09-T7-1", "tags": ["3.6"], "text_open": "7.2 Electrochemistry. 7.2.1 When an iron sheet is immersed in water, an electrical double layer forms near the iron electrode, as shown below. Briefly explain how the structure forms, and qualitatively sketch the absolute value of the potential as a function of distance from the metal surface, taking the potential infinitely far from the electrode to be zero.", "text_images": [{"file": "Exam-09-T7-1-1-prompt.png", "description": "This two-dimensional diagram, titled \"Figure 9.4 Schematic Structure of the Electrical Double Layer,\" shows the spatial distribution of an electrical double layer from left to right. At the far-left edge is a vertical row of minus signs (short horizontal lines). Immediately to their right is a vertical solid line shaded with diagonal hatching on its left side; the words \"Electrode surface\" are set vertically to the left of this line. A short distance to the right is a parallel vertical dashed line bearing a neat vertical series of small circles, each containing a \"+\" sign. The narrow interval between the solid line and this middle dashed line is marked at the bottom by a horizontal upward brace, beneath which the spacing symbol $d$ and the label \"Compact layer\" appear. Extending rightward from the middle dashed line to another parallel dashed boundary at the far right is a wider rectangular region. Several similar small circles containing \"+\" signs are scattered irregularly within it. This wider interval is likewise marked at the bottom by a horizontal brace, beneath which the spacing symbol $\\delta$ and the label \"Diffuse layer\" appear."}], "full_mark": 5, "rubric": ["Formation of the electrical double layer: highly polar water molecules attract iron ions in the lattice and hydrate them. Bonding of some iron ions to other iron atoms in the metal weakens, or the ions leave the metal and enter the water layer adjacent to the iron surface. The metal thus becomes negatively charged and the nearby solution positively charged, producing a local/built-in electric field that inhibits further dissolution of the metal. Dynamic equilibrium is reached when dissolution and deposition rates become equal, establishing a stable potential difference between the metal and solution. (3 points: 1 point for stating that iron ions leave the metal, making the metal negative and the nearby solution positive; 1 point for stating that the electric field inhibits further dissolution; and 1 point for stating that dynamic equilibrium is reached.)\nSketch: 1 point if the absolute potential decreases overall and approaches 0; 1 point for labeling the compact layer and diffuse layer and showing a faster decrease in the compact layer."], "rubric_images": [{"file": "Exam-09-T7-1-1-rubric.png", "description": "This two-dimensional coordinate plot, titled \"Figure 9.5 Schematic Potential Profile of the Electrical Double Layer,\" shows a solid curve that starts high on the vertical axis, decreases monotonically with a concave-upward shape, and gradually approaches the horizontal axis. The left vertical axis is a solid line shaded with diagonal hatching on its left side to represent the \"Metal surface\"; above it is the label \"(negative) potential\" with an upward arrow. Beneath the horizontal axis extending to the right is the label \"Distance from the electrode surface\" with a rightward arrow. A vertical dashed line within the plot divides the horizontal space into two regions. Horizontal brackets below the axis label the shorter region adjacent to the vertical axis as $d$ and the following longer region as $\\delta$. Two parallel horizontal dashed lines extend rightward from the starting point of the curve on the vertical axis and from the intersection of the curve with the vertical dashed line. On the right side of the plot, a vertical indicator with arrows at both ends labels three height parameters: the total height from the upper horizontal dashed line to the horizontal axis is $\\Delta\\varphi$; the upper drop between the two horizontal dashed lines is $\\varphi_1$; and the lower drop from the middle horizontal dashed line to the horizontal axis is $\\varphi_2$."}]}]} +{"problem_id": "Exam-09-T9", "header": "Gold-Catalyzed Construction of Indoles\nIndole is an important structural unit in numerous natural products and pharmaceuticals, and efficient, diverse strategies for indole synthesis remain an important challenge in synthetic chemistry. Most known indole syntheses use benzene derivatives as substrates to construct the pyrrole ring. In contrast, strategies that start from pyrrole derivatives and construct the benzene ring have been reported only rarely. Two examples of indole synthesis based on this reverse strategy are shown below. Analyze the reaction processes and answer the questions.", "header_images": [], "subquestions": [{"sub_id": "Exam-09-T9-1", "tags": ["2.4"], "text_open": "9.2 Cationic gold catalysts are highly active $\\pi$ Lewis acids that selectively activate carbon-carbon unsaturated bonds. 9.2.1 Write the SMILES structures of products A, B, and C in the following 3 reactions. Note: A, B, and C each contain one five-membered ring. Formation of C proceeds through a six-membered-ring intermediate, which finally undergoes ring contraction to give a five-membered-ring product.", "text_images": [{"file": "Exam-09-T9-1-1-prompt.png", "description": "The figure shows a general mechanism for gold-catalyzed cyclization. The substrate is an ortho-substituted aromatic ring (SMILES: 'C=C=CC(C=CC=C1N[R2])=C1C#C[R1]'). Under [Au]^+, it forms a gold-containing cationic intermediate, '[R1][C+]=C1C2=C(C=C(C1)[#79])C=CC=C2N[R2]' (Au represents the coordinated [Au] center; other ligands are not shown), and subsequent protodeauration gives the fused products '[R1]C1=C2C3=C(C=CC2)C=CC=C3N1[R2]' and '[R1]C(C1=CC=CC2=C1C3=CC=C2)N3[R2]'."}, {"file": "Exam-09-T9-1-2-prompt.png", "description": "Reaction 1: a naphthalene substrate (SMILES: '[R]C#CC1=C(C=C=C)C=CC2=C1C=CC=C2') gives product A under [Au]^+ catalysis."}, {"file": "Exam-09-T9-1-3-prompt.png", "description": "Reaction 2: a six-membered-ring substrate containing an allene and an alkynyl group (SMILES: '[R2]C(CCCC#C[R1])=C=C[R3]') gives product B under [Au]^+ and H_2O."}, {"file": "Exam-09-T9-1-4-prompt.png", "description": "Reaction 3: substrate '[R3]C#CCC(C([R1])=C=C)(O[Si]([R])([R])[R])[R2]' gives product C under [Au]^+ and i-PrOH."}], "full_mark": 6, "rubric": ["(6 points, 2 points for each structure.) Reference SMILES: A is '[R]C1C2=CC=CC3=C2C4=C(C=CC=C41)C=C3'; B is '[R2]C1(/C=C/[R3])C(C([R1])=O)CCC1'; C is 'C=C1C(C([R2])=O)(CC=C1[R3])[R1]'."], "rubric_images": [{"file": "Exam-09-T9-1-1-rubric.png", "description": "A is '[R]C1C2=CC=CC3=C2C4=C(C=CC=C41)C=C3'; B is '[R2]C1(/C=C/[R3])C(C([R1])=O)CCC1'; C is 'C=C1C(C([R2])=O)(CC=C1[R3])[R1]'."}]}, {"sub_id": "Exam-09-T9-2", "tags": ["2.2", "2.3"], "text_open": "9.2.2 Write the SMILES structures of the 2 intermediates D and E in the following reaction. Note: D and E each contain only two rings.", "text_images": [{"file": "Exam-09-T9-2-1-prompt.png", "description": "The figure shows the reaction pathway in 9.2.2. Substrate 'C=C=CC1=CC=CC(C)=C1C#C[R]' is converted under [Au]^+ successively into intermediates D and E and finally into the fused product '[R]C1C2=CC=CC3=CC=CC(C1)=C32'."}], "full_mark": 4, "rubric": ["The SMILES of D is 'CC1=C2C(C=C([#79])C/C2=[C+]/[R])=CC=C1', and the SMILES of E is '[CH2+]C1=C2C(C=C([#79])C/C2=C/[R])=CC=C1' (4 points, 2 points for each structure)."], "rubric_images": [{"file": "Exam-09-T9-2-1-rubric.png", "description": "The SMILES of D is 'CC1=C2C(C=C([#79])C/C2=[C+]/[R])=CC=C1', and the SMILES of E is '[CH2+]C1=C2C(C=C([#79])C/C2=C/[R])=CC=C1'."}]}, {"sub_id": "Exam-09-T9-3", "tags": ["2.6", "2.4"], "text_open": "9.3 Using a gold-catalyzed cyclization strategy similar to that described in 9.2, the authors developed a new method for synthesizing 4-acylindoles. A specific cyclization precursor must first be prepared. The synthetic route to this precursor is shown below. Write the SMILES structures of products F, G, and H from each step. Note: the isopropyl Grignard reagent and formaldehyde are added sequentially; the final step involves a retro-ene reaction that releases nitrogen gas.", "text_images": [{"file": "Exam-09-T9-3-1-prompt.png", "description": "The figure shows the structure of reagent IPNBSH (SMILES: 'O=S(C1=CC=CC=C1[N+]([O-])=O)(N/N=C(C)/C)=O')."}, {"file": "Exam-09-T9-3-2-prompt.png", "description": "Synthetic route: the starting substrate is 3-iodo-2-formyl-N-Ts pyrrole ('[H]C(C1=C(I)C=CN1S(C2=CC=C(C)C=C2)(=O)=O)=O'). In the first step, phenylacetylene, Pd(PPh_3)_2Cl_2, CuI, NEt_3, THF, and 50℃ give F in 92% yield. In the second step, treatment with 1) CBr_4/PPh_3 and 2) i-PrMgCl/(HCHO)_n gives G in 55% over 2 steps. In the third step, reaction with IPNBSH, DIAD, PPh_3, then TFE/H_2O gives H."}], "full_mark": 6, "rubric": ["(6 points, 2 points for each structure.) Reference SMILES: F is 'O=CC1=C(C#CC2=CC=CC=C2)C=CN1S(=O)(=O)C3=CC=C(C)C=C3'; G is 'OCC#CC1=C(C#CC2=CC=CC=C2)C=CN1S(=O)(=O)C3=CC=C(C)C=C3'; H is 'C=C=CC1=C(C=CN1S(=O)(C2=CC=C(C=C2)C)=O)C#CC3=CC=CC=C3'."], "rubric_images": [{"file": "Exam-09-T9-3-1-rubric.png", "description": "F is 'O=CC1=C(C#CC2=CC=CC=C2)C=CN1S(=O)(=O)C3=CC=C(C)C=C3'; G is 'OCC#CC1=C(C#CC2=CC=CC=C2)C=CN1S(=O)(=O)C3=CC=C(C)C=C3'; H is 'C=C=CC1=C(C=CN1S(=O)(C2=CC=C(C=C2)C)=O)C#CC3=CC=CC=C3'."}]}, {"sub_id": "Exam-09-T9-4", "tags": ["2.3", "2.8"], "text_open": "9.4 When the substrate synthesized in 9.3 is subjected to gold-catalyzed cyclization, cyclization occurs but the expected product is not observed; instead, a more highly oxidized product forms. 9.4.1 To investigate the mechanism, the authors performed $^{18}\\mathrm{O}$ isotope-labeling experiments in methyl ethyl ketone (MEK) and THF. In both solvents, water participates as a nucleophile, but the degree of $^{18}\\mathrm{O}$ incorporation differs markedly: product formed in one solvent, B, shows 57% $^{18}\\mathrm{O}$ incorporation, whereas product formed in the other solvent, A, shows only 10% $^{18}\\mathrm{O}$ incorporation. Determine whether MEK is solvent A or B.", "text_images": [{"file": "Exam-09-T9-4-1-prompt.png", "description": "The figure shows the isotope-labeling experiment. Under BrettPhosAuCl/AgNTf_2 catalysis, substrate H reacts with H_2^{18}O to form the 4-acylindole 'O=C(C1=CC=CC=C1)C2=CC=CC3=C2C=CN3S(C4=CC=C(C)C=C4)(=O)=O'. The solvent is 0.2 M MEK or THF. Solvent A gives 10% $^{18}\\mathrm{O}$ incorporation, and solvent B gives 57% $^{18}\\mathrm{O}$ incorporation."}], "full_mark": 2, "rubric": ["A (2 points)"], "rubric_images": []}, {"sub_id": "Exam-09-T9-5", "tags": ["2.2", "2.3"], "text_open": "9.4.2 The catalytic-cycle mechanism is shown below, where X denotes either solvent A or B. Write the SMILES structures of J, K, L, and M in the cycle, including necessary charge/coordination information.", "text_images": [{"file": "Exam-09-T9-5-1-prompt.png", "description": "The figure shows the complete catalytic cycle for formation of a 4-acylindole. Substrate H binds the gold catalyst [Au]^+ to form J. J reacts with H_2O and loses [Au]^+ to form product M. M reversibly forms 'O=C(C1=CC=CC=C1)C2CC=CC3=C2C=CN3S(C4=CC=C(C)C=C4)(=O)=O', which is converted by O_2 into product 'O=C(C1=CC=CC=C1)C2=CC=CC3=C2C=CN3S(C4=CC=C(C)C=C4)(=O)=O'. In another branch, J reacts with X to form K. K reacts with H_2O and loses [Au]^+ to form product L. L then loses X to give intermediate product 'O=C(C1=CC=CC=C1)C2CC=CC3=C2C=CN3S(C4=CC=C(C)C=C4)(=O)=O'."}], "full_mark": 8, "rubric": ["(8 points, 2 points for each structure.) J is '[#79]C(C/C1=[C+]/C2=CC=CC=C2)=CC3=C1C=CN3S(C4=CC=C(C)C=C4)(=O)=O'; K is '[#79]C(C/C1=C(\\[O+]=C(C)\\CC)C2=CC=CC=C2)=CC3=C1C=CN3S(C4=CC=C(C)C=C4)(=O)=O'; L is 'CC(O)(CC)O/C(C1=CC=CC=C1)=C2CC=CC3=C\\2C=CN3S(C4=CC=C(C=C4)C)(=O)=O'; M is 'O/C(C1=CC=CC=C1)=C2CC=CC3=C\\2C=CN3S(C4=CC=C(C=C4)C)(=O)=O'. Au represents the coordinated [Au] center; not all ligands are shown."], "rubric_images": [{"file": "Exam-09-T9-5-1-rubric.png", "description": "J is '[#79]C(C/C1=[C+]/C2=CC=CC=C2)=CC3=C1C=CN3S(C4=CC=C(C)C=C4)(=O)=O'; K is '[#79]C(C/C1=C(\\[O+]=C(C)\\CC)C2=CC=CC=C2)=CC3=C1C=CN3S(C4=CC=C(C)C=C4)(=O)=O'; L is 'CC(O)(CC)O/C(C1=CC=CC=C1)=C2CC=CC3=C\\2C=CN3S(C4=CC=C(C=C4)C)(=O)=O'; M is 'O/C(C1=CC=CC=C1)=C2CC=CC3=C\\2C=CN3S(C4=CC=C(C=C4)C)(=O)=O'."}]}]} +{"problem_id": "Exam-10-T5", "header": "Structural Chemistry of Two-Dimensional Transition-Metal Dichalcogenides and Their Derivatives (26 points, 12%). Two-dimensional transition-metal dichalcogenides, or TMDs, generally have layered structures held together by van der Waals forces. Their symmetry and physical properties change substantially as the number of layers decreases.", "header_images": [], "subquestions": [{"sub_id": "Exam-10-T5-1", "tags": ["1.3"], "text_open": "Structural background: $WSe_2$ has a molybdenite-type hexagonal layered structure. The figure below is a projection along $a$; large spheres represent W and small spheres Se. One unit cell contains two layers along $c$.\n5-1-2 Explain how W and Se occupy the hexagonal plane lattice in a projection along the c axis, and give the projected relationship within one unit cell.", "text_images": [{"file": "Exam-10-T5-1-1-prompt.png", "description": "A two-dimensional projection of a vertical rectangular unit cell outlined by thin black lines. Axes at the lower left show c vertically upward and b horizontally to the left. The figure contains two larger gray spheres and eight smaller green spheres joined by slender cylindrical bonds. An upper gray sphere lies inside the upper half of the cell near its right boundary and is connected to four green spheres at the upper-left, lower-left, upper-right, and lower-right positions. The two green spheres on the left lie within the cell, whereas the two on the right lie outside the right boundary. A lower gray sphere lies inside the lower half near the left boundary and is likewise connected to four green spheres at the upper-left, lower-left, upper-right, and lower-right positions. The two green spheres on the left lie outside the left boundary, whereas the two on the right lie within the cell. The two gray spheres are not directly connected; together they form two separate four-coordinate units offset vertically and oriented in opposite horizontal directions."}], "full_mark": 2, "rubric": ["In the projection along c, W and Se occupy the A and B sites, respectively, of the hexagonal plane lattice. Only one set of W sites and one set of Se sites need be shown in a unit cell (2 points). If the A and B sites are correct but the C sites are also occupied incorrectly, award 1 point. Occluded atoms need not be shown repeatedly."], "rubric_images": [{"file": "Exam-10-T5-1-1-rubric.png", "description": "A two-dimensional projection of a parallelogram unit cell outlined by thin black lines. The figure contains two larger gray spheres and five smaller green spheres joined by slender cylindrical bonds. A gray sphere lies toward the left inside the cell, with a green sphere directly in front of it so that only the left and upper parts of the gray sphere remain visible. This central green sphere connects in three directions: to a green sphere directly above by an approximately vertical bond, to a green sphere outside the left boundary at the lower left by a diagonal bond, and to another gray sphere at the lower right. The right gray sphere lies toward the lower-right interior of the cell and, besides the central green sphere, connects to two more green spheres: one outside the right boundary at the upper right and one outside the lower boundary directly below. Overall, the two gray spheres are staggered left and right within the cell, while the five green spheres occupy the left-center, upper, outer-left, outer-right, and lower-outer positions, forming a connected structure across the parallelogram boundaries."}]}]} +{"problem_id": "Exam-10-T7", "header": "Question 7 Synthesis and Study of Palladium Complexes", "header_images": [], "subquestions": [{"sub_id": "Exam-10-T7-1", "tags": ["1.4", "1.6", "2.4"], "text_open": "7-1 Adding excess 2,2'-bipyridine to a solution of $\\mathrm{Pd(NO_3)_2}$ gives a neutral mononuclear Pd complex A. In solution, A dimerizes to form a 1:2 ionic complex D containing a Pd-Pd single bond and $\\mathrm{NO_3^-}$ counterions. Write the SMILES strings of A and the cation of D, and determine whether each obeys the EAN rule.", "text_images": [], "full_mark": 4, "rubric": ["The SMILES of A is '[N][Pd]([N])(O[N+]([O-])=O)O[N+]([O-])=O', and that of the D cation is '[N][Pd](O[N]1=O)([N])(O2)[Pd](O1)([N])(O[N]2=O)[N]'. In both structures, the 2 N atoms bonded to the same Pd atom are connected by an arc. (1 point for A and 1 point for the D cation; deduct 1 point if an otherwise correct answer omits the charge.) For A, $EAN=8+4(\\mathrm{bipy})+2\\times2(\\mathrm{ONO_2^-})=16\\mathrm{e^-}$, so it does not obey the rule (1 point). For the D cation, $EAN=8+4(\\mathrm{bipy})+2\\times2(\\mathrm{ONO_2^-})+1(\\mathrm{Pd-Pd})=17\\mathrm{e^-}$, so it does not obey the rule (1 point)."], "rubric_images": [{"file": "Exam-10-T7-1-1-rubric.png", "description": "The SMILES of A is '[N][Pd]([N])(O[N+]([O-])=O)O[N+]([O-])=O', and that of the D cation is '[N][Pd](O[N]1=O)([N])(O2)[Pd](O1)([N])(O[N]2=O)[N]'. In both structures, the 2 N atoms bonded to the same Pd atom are connected by an arc."}]}, {"sub_id": "Exam-10-T7-2", "tags": ["1.4", "1.6", "2.4"], "text_open": "7-2 Researchers found that D efficiently converts $\\mathrm{C=S}$ double bonds in organic compounds into $\\mathrm{C=O}$ double bonds. During this process, D is converted into ionic compound E, which has a threefold axis and mirror planes. Mass spectrometry gives $m/z = 425.4$ for the E cation. The 2,2'-bipyridine ligand may be represented as N⌒N. 7-2-1 Write the SMILES representation of E.", "text_images": [], "full_mark": 3, "rubric": ["The E cation carries 2 positive charges and has the SMILES '[R][Pd]123([R])[S]4[Pd]15([R])([R])[Pd]2([R])([R])4[S]35'. Two R groups attached to the same Pd denote the chelating attachment sites of a 2,2'-bipyridine ligand. The anion of E is nitrate. (3 points: 1 point for the Pd_3S_2 cluster; deduct 1 point if an otherwise correct answer omits the charge.)"], "rubric_images": []}, {"sub_id": "Exam-10-T7-3", "tags": ["1.4", "2.1"], "text_open": "7-2-2 Write the equation for the formation of E. The substrate may be represented as $\\mathrm{CSR_2}$, and the symbols given in the question may be used for the complexes.", "text_images": [], "full_mark": 2, "rubric": ["$\\mathrm{4CSR_2 + 4H_2O + 3D = 4COR_2 + 8HNO_3 + 2E}$ (2 points)"], "rubric_images": []}, {"sub_id": "Exam-10-T7-4", "tags": ["1.4", "2.1"], "text_open": "7-2-3 Addition of nitric acid converts E back into A, enabling recycling. Write the equation for this step.", "text_images": [], "full_mark": 2, "rubric": ["$\\mathrm{E + 16HNO_3 = 2SO_2 + 12NO_2 + 8H_2O + 3A}$ (2 points)"], "rubric_images": []}, {"sub_id": "Exam-10-T7-5", "tags": ["1.4", "2.1"], "text_open": "7-2-4 Write the overall equation for conversion of $\\mathrm{CSR_2}$ into $\\mathrm{COR_2}$.", "text_images": [], "full_mark": 2, "rubric": ["$\\mathrm{CSR_2 + 6HNO_3 = COR_2 + 6NO_2 + SO_2 + 3H_2O}$ (2 points)"], "rubric_images": []}, {"sub_id": "Exam-10-T7-6", "tags": ["2.4"], "text_open": "7-3 Add 3 equivalents of $\\mathrm{Hacac}$ and potassium tert-butoxide to 1,3,5-tris(bromomethyl)benzene, and quench the reaction with water. Then add 3 equivalents of hydrazine; the reaction gives organic ligand $\\mathrm{H_3L}$. Adding $\\mathrm{H_3L}$ to D produces $\\mathrm{\\{[(bpy)Pd]_{12}L_x\\}(NO_3)_{12}}$. 7-3-1 A synthesis of a ligand analogous to $\\mathrm{H_3L}$ is shown below. Write the structure of J.", "text_images": [{"file": "Exam-10-T7-6-1-prompt.png", "description": "A rightward arrow follows 'BrCC1=CC(CBr)=CC(CBr)=C1'. The text above the arrow is 'CC(C)[N+](=O)[O-]' and that below is NaOEt; the arrow points to uppercase J. Another rightward arrow follows J, with 'COP1(OC)(OC)CC(C)=C(C)C1' above it and no text below, and points to 'CC(=O)C(=C(O)C)C1=CC(C(=C(O)C)C(C)=O)=CC(C(=C(O)C)C(C)=O)=C1'. A third rightward arrow follows 'CC(=O)C(=C(O)C)C1=CC(C(=C(O)C)C(C)=O)=CC(C(=C(O)C)C(C)=O)=C1', with $\\mathrm{N_2H_4\\dotH_2O}$ above it and no text below, and points to 'CC1=C(C2=CC(C3=C(NN=C3C)C)=CC(C4=C(NN=C4C)C)=C2)C(C)=NN1'."}], "full_mark": 2, "rubric": ["The SMILES of J is 'O=CC1=CC(C=O)=CC(C=O)=C1' (2 points)."], "rubric_images": [{"file": "Exam-10-T7-6-1-rubric.png", "description": "The SMILES of J is 'O=CC1=CC(C=O)=CC(C=O)=C1'."}]}]} +{"problem_id": "Exam-10-T9", "header": "", "header_images": [], "subquestions": [{"sub_id": "Exam-10-T9-1", "tags": ["2.3", "2.4", "2.7"], "text_open": "9-1-1 In radical reactions of allylic compounds using NBS or HBr, substitution and addition compete, and different reagents show different selectivities. Write the major products of the following two reactions.", "text_images": [{"file": "Exam-10-T9-1-1-prompt.png", "description": "Two organic reaction schemes. In the first, 'CC1(C)C=CCCC1' reacts with NBS under irradiation. In the second, 'CC(C1CCCCC1)=C' reacts with HBr and $\\mathrm{(BzO)_2}$."}], "full_mark": 4, "rubric": ["'CC1(C)C=C[C@@H](Br)CC1', as a racemate (±) (2 points: 1 point for the skeleton and 1 point for stereochemistry); 'C[C@](CBr)([H])C1CCCCC1', as a racemate (±) (2 points: 1 point for the skeleton and 1 point for stereochemistry)."], "rubric_images": [{"file": "Exam-10-T9-1-1-rubric.png", "description": "Organic structural formula with the SMILES 'CC1(C)C=C[C@@H](Br)CC1', shown as a racemate (±)."}, {"file": "Exam-10-T9-1-2-rubric.png", "description": "Organic structural formula with the SMILES 'C[C@](CBr)([H])C1CCCCC1', shown as a racemate (±)."}]}, {"sub_id": "Exam-10-T9-2", "tags": ["2.2", "2.3"], "text_open": "9-2-1 Radical reactions can also achieve efficient intramolecular N-C bond coupling. Write the key intermediates in the following reaction.", "text_images": [{"file": "Exam-10-T9-2-1-prompt.png", "description": "Organic reaction scheme in which 'C1(CN2CCCCC2)NCCCC1' reacts with NCS under irradiation to give 'N12CCCCC1CN3C2CCCC3'."}], "full_mark": 5, "rubric": ["The SMILES strings of the key intermediates are 'ClN1CCCCC1CN2CCCCC2', 'C5(CN6CCCCC6)[N]CCCC5', 'C7(CN8[C]CCCC8)NCCCC7', 'ClC(CCCC3)N3CC4NCCCC4', and 'C9(C[N+]%10=CCCCC%10)NCCCC9' (1 point each, 5 points total)."], "rubric_images": [{"file": "Exam-10-T9-2-1-rubric.png", "description": "Five organic structural formulas with the SMILES strings 'ClN1CCCCC1CN2CCCCC2', 'C5(CN6CCCCC6)[N]CCCC5', 'C7(CN8[C]CCCC8)NCCCC7', 'ClC(CCCC3)N3CC4NCCCC4', and 'C9(C[N+]%10=CCCCC%10)NCCCC9'."}]}, {"sub_id": "Exam-10-T9-3", "tags": ["2.4"], "text_open": "9-3-2 Write the structure of the major product of the following reaction. Disregard stereochemistry.", "text_images": [{"file": "Exam-10-T9-3-1-prompt.png", "description": "Organic reaction scheme in which 'IC1C(OC/C=C/C)OCCC1' reacts with triethylborane and oxygen in water and hypophosphorous acid."}], "full_mark": 2, "rubric": ["'CCC1COC2OCCCC21' (2 points)."], "rubric_images": [{"file": "Exam-10-T9-3-1-rubric.png", "description": "Organic structural formula with the SMILES 'CCC1COC2OCCCC21'."}]}, {"sub_id": "Exam-10-T9-4", "tags": ["2.2", "2.3", "2.4"], "text_open": "9-3-4 The following reaction undergoes efficient triethylborane-catalyzed cyclization to give two isomers, A1 and A2, of which A2 is the major product. Write the SMILES string of A2 and those of the key intermediates leading to A2.", "text_images": [{"file": "Exam-10-T9-4-1-prompt.png", "description": "Organic reaction scheme in which 'CC1(C)C(O1)/C=C/CCCC#C' reacts with triethylborane and triphenyltin hydride in benzene to give A1, 'OC(C)(C)/C=C/C(CCC/1)C1=C[Sn](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4', and A2."}], "full_mark": 10, "rubric": ["A2: 'OC(C)(C)/C=C/C1=C(C[Sn](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4)CCC1' (2 points). Intermediates: 'CC1(C)C(O1)/C=C/CCC/[C]=C[Sn](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4', 'CC5C(O5)[C]C(CCC/6)C6=C[Sn](C7=CC=CC=C7)(C8=CC=CC=C8)C9=CC=CC=C9', '[O]C(C)(C)/C=C/C(CCC/%10)C%10=C[Sn](C%11=CC=CC=C%11)(C%12=CC=CC=C%12)C%13=CC=CC=C%13', and 'OC(C)(C)/C=C/[C](CCC/%14)C%14=C[Sn](C%15=CC=CC=C%15)(C%16=CC=CC=C%16)C%17=CC=CC=C%17' (2 points each, 8 points total; a reasonable resonance form of the last intermediate is accepted)."], "rubric_images": [{"file": "Exam-10-T9-4-1-rubric.png", "description": "A2 and 4 key intermediates. A2 is 'OC(C)(C)/C=C/C1=C(C[Sn](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4)CCC1'. In sequence, the intermediates are 'CC1(C)C(O1)/C=C/CCC/[C]=C[Sn](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4', 'CC5C(O5)[C]C(CCC/6)C6=C[Sn](C7=CC=CC=C7)(C8=CC=CC=C8)C9=CC=CC=C9', '[O]C(C)(C)/C=C/C(CCC/%10)C%10=C[Sn](C%11=CC=CC=C%11)(C%12=CC=CC=C%12)C%13=CC=CC=C%13', and 'OC(C)(C)/C=C/[C](CCC/%14)C%14=C[Sn](C%15=CC=CC=C%15)(C%16=CC=CC=C%16)C%17=CC=CC=C%17'."}]}]} +{"problem_id": "Exam-10-T10", "header": "", "header_images": [], "subquestions": [{"sub_id": "Exam-10-T10-1", "tags": ["1.2", "2.4"], "text_open": "10-1-1 The [4+2] reaction is an important method for forming carbon-carbon bonds and is widely used in modern organic synthesis. It exhibits rich stereochemistry, including both stereoselectivity and regioselectivity, and generally gives the endo product. Complete the following reaction, showing the stereochemistry.", "text_images": [{"file": "Exam-10-T10-1-1-prompt.png", "description": "Organic reaction scheme between 'C=CC1=CCCCC1(C)C' and 'C/C=C/C=O'."}], "full_mark": 3, "rubric": ["'CC1(C)CCC[C@]2([H])[C@H](C=O)[C@@H](C)CC=C21' and 'CC1(C)CCC[C@@]2([H])[C@@H](C=O)[C@H](C)CC=C21' (3 points: 2 points for the skeleton and 1 point for fully correct stereochemistry)."], "rubric_images": [{"file": "Exam-10-T10-1-1-rubric.png", "description": "Two organic structural formulas with the SMILES strings 'CC1(C)CCC[C@]2([H])[C@H](C=O)[C@@H](C)CC=C21' and 'CC1(C)CCC[C@@]2([H])[C@@H](C=O)[C@H](C)CC=C21'."}]}, {"sub_id": "Exam-10-T10-2", "tags": ["2.2", "2.4"], "text_open": "10-3 The [4+2] reaction can also trap highly reactive intermediates, thereby demonstrating their existence or constructing complex carbon skeletons. 10-3-1 An unstable electrophilic intermediate is generated in the reaction below. Write the key intermediates in the reaction sequence.", "text_images": [{"file": "Exam-10-T10-2-1-prompt.png", "description": "Organic reaction scheme in which 'BrC1=CC2=C(OCO2)C=C1' and 'O=C1OC(C#N)C2=C1C=CC=C2' react with LDA in THF from -78 °C to 40 °C to give 'O=C1C2=CC=C(OCO3)C3=C2C(C4=C1C=CC=C4)=O'."}], "full_mark": 8, "rubric": ["The SMILES strings of the key intermediates are 'C%12(OCO%13)=C%13C#CC=C%12', '[O-]C1=C2C=CC=CC2=C(C#N)O1', '[O-]C34C5=C6C(OCO6)=CC=C5C(C7=C4C=CC=C7)(C#N)O3', and 'O=C(C(C=CC=C8)=C8C9(C#N)[O-])C%10=C9C=CC%11=C%10OCO%11' (2 points each, 8 points total)."], "rubric_images": [{"file": "Exam-10-T10-2-1-rubric.png", "description": "Four organic structural formulas: 'C%12(OCO%13)=C%13C#CC=C%12', '[O-]C1=C2C=CC=CC2=C(C#N)O1', '[O-]C34C5=C6C(OCO6)=CC=C5C(C7=C4C=CC=C7)(C#N)O3', and 'O=C(C(C=CC=C8)=C8C9(C#N)[O-])C%10=C9C=CC%11=C%10OCO%11'."}]}, {"sub_id": "Exam-10-T10-3", "tags": ["2.2", "2.4"], "text_open": "10-3-2 Non-neutral substrates can also be trapped through [4+2] reactions. Reactions in which an aromatic ring is oxidized before cycloaddition are often termed [5+2] reactions. The solvent-dependent, oxidative-dearomatization-induced divergent [5+2] cascade strategy below constructs a bicyclo[3.2.1]octane skeleton. Write the SMILES strings of the key intermediates leading to each of the two products.", "text_images": [{"file": "Exam-10-T10-3-1-prompt.png", "description": "Organic reaction scheme in which 'OC1=C(CCC/C=C\\CCO)C=CC=C1OC' reacts with PhI(OAc)_2 in HFIP at 0 °C to give [A] 'O=C1[C@]2([C@]3([H])CCC2)C=C[C@H]4[C@]1(OC)[C@]3([H])CCO4' and [B] '[H][C@]12[C@](CCC3)([H])[C@]34C([C@H](C=C4)[C@@]1(OC)OCC2)=O'."}], "full_mark": 6, "rubric": ["Intermediate leading to A: 'O=C1[C@]2([C@]3([H])CCC2)[CH+]C=C[C@]1(OC)[C@]3([H])CCO' (2 points). Intermediates leading to B: 'O=C1[C@]2([C@]3([H])CCC2)[CH+]C=C[C@]1(OC)[C@@]3([H])CCO' and 'O=C([C@@H]1C=C2)[C@]23[C@@](CCC3)([H])[C@]([H])(CCO)C1=[O+]C' (2 points each)."], "rubric_images": [{"file": "Exam-10-T10-3-1-rubric.png", "description": "Intermediate leading to A: 'O=C1[C@]2([C@]3([H])CCC2)[CH+]C=C[C@]1(OC)[C@]3([H])CCO' (2 points). Intermediates leading to B: 'O=C1[C@]2([C@]3([H])CCC2)[CH+]C=C[C@]1(OC)[C@@]3([H])CCO' and 'O=C([C@@H]1C=C2)[C@]23[C@@](CCC3)([H])[C@]([H])(CCO)C1=[O+]C'."}]}]} +{"problem_id": "Exam-11-T3", "header": "A photoresist is a polymeric material sensitive to light of specific wavelengths. Upon irradiation, it undergoes specific chemical reactions that substantially change its solubility in a developer, thereby forming the desired pattern on the surface of a semiconductor material. As a key patterning material in integrated-circuit fabrication, advances in photoresist chemistry directly determine the limits of process miniaturization and the industry's capacity for technological self-reliance.", "header_images": [], "subquestions": [{"sub_id": "Exam-11-T3-1", "tags": ["2.1", "2.5"], "text_open": "3.1 The novolak-diazonaphthoquinone (Novolak-DNQ) system is the most widely used photoresist for the g-line (436 nm) and i-line (365 nm). One component is a high-molecular-weight linear phenolic resin, and the other is diazonaphthoquinone, whose structure is shown below.\n3.1.1 The phenolic resin prepared under acidic conditions is a low-molecular-weight linear prepolymer. The curing agent hexamethylenetetramine may be added to an aqueous solution of the prepolymer to cure and cross-link it upon heating. Identify which substrate must be used in excess when preparing the acid-catalyzed phenolic resin, write the decomposition equation of the curing agent upon heating, and on this basis state the role of the curing agent.", "text_images": [{"file": "Exam-11-T3-1-1-prompt.png", "description": "'O=C1C2=CC=CC=C2C(S(=O)(O[Ar])=O)=CC1=[N+]=[N-]'"}], "full_mark": 5, "rubric": ["The substrate used in excess is phenol. (1 point)", "C_6H_12N_4 + 6H_2O = 6CH_2O + 4NH_3. (2 points)", "The formaldehyde produced further cross-links the prepolymer into a three-dimensional network polymer. (1 point)", "The NH_3 produced by decomposition acts as a base to catalyze the polymerization. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-11-T3-2", "tags": ["1.6"], "text_open": "3.1.2 State the principal intermolecular forces between the two components of the Novolak-DNQ system.", "text_images": [], "full_mark": 2, "rubric": ["Hydrogen bonding (1 point).", "Coulombic interaction (1 point; equivalent answers such as electrostatic interaction or ionic bonding also receive credit)."], "rubric_images": []}, {"sub_id": "Exam-11-T3-3", "tags": ["2.3", "2.4"], "text_open": "3.1.3 In the Novolak-DNQ system, the photosensitive component undergoes rearrangement upon irradiation to form unstable structure A, which then forms B in the alkaline developer N(CH_3)_4^+OH^-, changing the solubility of the photoresist. Write the SMILES representations of A and B.", "text_images": [], "full_mark": 3, "rubric": ["A: 'O=C=C1C=C(C2=CC=CC=C21)S(=O)(O[Ar])=O'; B: 'O=S(C1=CC(C2=CC=CC=C21)C(O)=O)(O[Ar])=O'. 3 points total: 2 points for structure A and 1 point for structure B."], "rubric_images": [{"file": "Exam-11-T3-3-1-rubric.png", "description": "A: 'O=C=C1C=C(C2=CC=CC=C21)S(=O)(O[Ar])=O'; B: 'O=S(C1=CC(C2=CC=CC=C21)C(O)=O)(O[Ar])=O'"}]}, {"sub_id": "Exam-11-T3-4", "tags": ["2.4", "2.5"], "text_open": "3.2 PVCi is a high-resolution photoresist. One method for synthesizing PVCi is shown below.\n3.2.1 The first step is a conventional free-radical polymerization of a double bond, and all subsequent steps are functional-group substitutions. Write the SMILES representations of X, Y, Z, and PVCi. The terminal-group structures need not be specified; use [*] to denote the connection sites at the two ends of the polymer backbone.", "text_images": [{"file": "Exam-11-T3-4-1-prompt.png", "description": "X (C_4H_6O_2) reacts under AIBN/CH_3OH conditions to give Y; Y reacts under CH_3OH/NaOH conditions to give Z; Z reacts with 'O=C(Cl)/C=C/C1=CC=CC=C1' to give PVCi."}], "full_mark": 6, "rubric": ["X: 'C=COC(=O)C'; repeating unit of Y: '[*]CC(OC(=O)C)[*]'; repeating unit of Z: '[*]CC(O)[*]'; repeating unit of PVCi: '[*]CC(OC(=O)/C=C/c1ccccc1)[*]'. 6 points total: 2 points each for X and PVCi and 1 point each for Y and Z. Explicit terminal groups are not required."], "rubric_images": [{"file": "Exam-11-T3-4-1-rubric.png", "description": "X: 'C=COC(=O)C'; repeating unit of Y: '[*]CC(OC(=O)C)[*]'; repeating unit of Z: '[*]CC(O)[*]'; repeating unit of PVCi: '[*]CC(OC(=O)/C=C/c1ccccc1)[*]'"}]}, {"sub_id": "Exam-11-T3-5", "tags": ["2.3", "2.5"], "text_open": "3.2.2 Write the SMILES representation of the structure formed by the interchain reaction of PVCi upon irradiation at 360 nm. The terminal-group structures and degrees of polymerization need not be specified; use [*] and [**] to denote connection sites in the polymer chains.", "text_images": [], "full_mark": 2, "rubric": ["The structure formed by the interchain reaction of PVCi upon irradiation at 360 nm is shown in the rubric image. 2 points; the degrees of polymerization need not be distinguished. SMILES: 'O=C(OC([*])C[*])C1C(c2ccccc2)C(C(=O)OC([**])C[**])C1c2ccccc2', where [*] denotes a connection site in a polymer chain."], "rubric_images": [{"file": "Exam-11-T3-5-1-rubric.png", "description": "This can be represented by the SMILES 'O=C(OC([*])C[*])C1C(c2ccccc2)C(C(=O)OC([*])C[*])C1c2ccccc2', where [*] denotes a connection site in a polymer chain; the first two [*] sites form one chain and the last two [*] sites form another chain."}]}]} +{"problem_id": "Exam-11-T5", "header": "Problem 5: Familiar Structures from an Unfamiliar Perspective (18 points, 7%)", "header_images": [], "subquestions": [{"sub_id": "Exam-11-T5-1", "tags": ["1.3"], "text_open": "5.2 Crystal Y has the chemical formula QF_y. Its projections along the a and c directions are shown below; black spheres represent Q ions and white spheres represent F^-. 5.2.1 Give the value of y.", "text_images": [{"file": "Exam-11-T5-1-1-prompt.png", "description": "The image contains two two-dimensional projections of a crystal structure. The left panel is a rectangular view of the b-c plane. Axes at the lower left show c vertically upward, b horizontally to the right, and a perpendicular to the page. A large black sphere is located at each of the four corners, with two-dimensional fractional coordinates (b, c) = (0.00, 0.00), (1.00, 0.00), (0.00, 1.00), and (1.00, 1.00). Two large black spheres lie inside the unit cell at approximately (0.33, 0.33) and (0.67, 0.67). The small white spheres in the unit cell include 4 single spheres at approximately (b, c) = (0.00, 0.50), (0.67, 0.17), (1.00, 0.50), and (0.33, 0.83), and 3 pairs of overlapping spheres at approximately (b, c) = (0.17, 0.17), (0.50, 0.50), and (0.83, 0.83). The right panel is a parallelogram view of the a-b plane. Axes at the lower left show a horizontally to the right, b diagonally toward the upper left, and c perpendicular to the page; the angle between a and b is approximately 120°. A large black sphere is located at each of the four corners, with two-dimensional fractional coordinates (a, b) = (0.00, 0.00), (1.00, 0.00), (0.00, 1.00), and (1.00, 1.00). Two large black spheres lie inside the unit cell at approximately (0.33, 0.67) and (0.67, 0.33). All the small white spheres in the unit cell are single spheres, 11 in total, at approximately (0.00, 0.50), (0.50, 0.00), (0.33, 0.17), (0.17, 0.33), (0.83, 0.17), (0.50, 0.50), (0.17, 0.83), (1.00, 0.50), (0.83, 0.67), (0.67, 0.83), and (0.50, 1.00)."}], "full_mark": 2, "rubric": ["3"], "rubric_images": []}]} +{"problem_id": "Exam-12-T5", "header": "Problem 5 Structure of a Barium Titanate\nA hexagonal-family form of $\\mathrm{BaTiO_3}$ has an interesting structure.\nIn this crystal, all Ba and O atoms together form a special stacking pattern. In a close-packed structure, designate a close-packed layer whose coordination environment above and below is the same as in hexagonal close packing as h, and one whose coordination environment is the same as in cubic close packing as c. For example, double hexagonal close packing (ABACABAC...) can be represented as hchchc.... The joint Ba/O stacking sequence is hcchcch...; considering only Ba gives the same hcchcch... sequence.\nEach Ba is coordinated by 12 O atoms to form a $\\mathrm{[BaO_{12}]}$ tetradecahedral unit. Ti occurs in two chemical environments. In one, two $\\mathrm{TiO_6}$ octahedra share a face to form a $\\mathrm{[Ti_2O_9]}$ unit; in the other, an isolated $\\mathrm{[TiO_6]}$ unit forms, meaning that it does not share a face with another $\\mathrm{TiO_6}$ octahedron, although corner sharing is possible. The ratio of the two types of Ti atoms is 2:1. In the conventional unit cell whose vertices are the Ti atoms of the isolated $\\mathrm{[TiO_6]}$ units, one Ba has coordinates (2/3, 1/3, 0.5967). The unit-cell parameters are a = b = 574.6 pm and c = 1399.9 pm.", "header_images": [], "subquestions": [{"sub_id": "Exam-12-T5-1", "tags": ["1.3"], "text_open": "5.1 Multiple selection: which of the following statements are correct?\n(a) The Ti arrangement resembles hexagonal diamond.\n(b) Each O layer forms a kagome-type \"six-pointed-star array\" like that shown in [Exam-12-T5-1-1-prompt.png].\n(c) If all Ba and O atoms are regarded as jointly forming the hcchcch... close-packed sequence and Ti atoms are treated as interstitial atoms, Ti occupies all octahedral voids formed solely by O atoms.\n(d) This $\\mathrm{BaTiO_3}$ contains 2 distinct Ti-O bonds and 2 distinct Ba-O bonds.", "text_images": [{"file": "Exam-12-T5-1-1-prompt.png", "description": "The image shows a translationally symmetric planar network made of black straight lines and circles and possessing additional symmetries. There are 3 sets of black lines in different directions; within each set, adjacent parallel lines are equally spaced. Together, the three sets partition the plane into closed regular hexagons and equilateral triangles of equal edge length, with a circle at every line intersection. Around each circle, the four polygons appear in the order triangle, hexagon, triangle, hexagon. In the overall connectivity, each of the three edges of every triangle is shared with a hexagonal region, and each of the 6 edges of every hexagon is shared with a hexagonal region."}], "full_mark": 4, "rubric": ["(b) and (c) (4 points; no credit for an incorrect selection or no selection)"], "rubric_images": []}, {"sub_id": "Exam-12-T5-2", "tags": ["1.3", "3.9"], "text_open": "5.4 Calculate the second-shortest Ba-Ba distance.", "text_images": [], "full_mark": 5, "rubric": ["Plot the Ba-atom positions in the projection along the a axis:\n[Exam-12-T5-2-1-rubric.png]\nFirst determine the coordinates of each Ba atom: Ba0(2/3, 1/3, 0.9033), Ba1(1/3, 2/3, 0.4033), Ba2(1/3, 2/3, 0.0967), Ba3(2/3, 1/3, 0.5967), Ba4(0, 0, 0.25), and Ba5(0, 0, 0.75). (1 point)\nCandidate distances for the second-shortest Ba-Ba separation are Ba0~Ba0, Ba0~Ba2, Ba0~Ba3, and Ba0~Ba5.\n$d_{\\text{Ba0-Ba0}}=a=574.6\\ \\mathrm{pm}$.\n$d_{\\text{Ba0-Ba2}}=\\sqrt{((1+0.0967-0.9033)c)^2+\\left(\\frac{\\sqrt{3}}{3}a\\right)^2}=\\sqrt{(0.1934\\times1399.9)^2+\\left(\\frac{\\sqrt{3}}{3}\\times574.6\\right)^2}=428.2\\ \\mathrm{pm}$. (1 point)\n$d_{\\text{Ba0-Ba3}}=(0.9033-0.5967)c=(0.9033-0.5967)\\times1399.9=429.2\\ \\mathrm{pm}$. (1 point)\n$d_{\\text{Ba0-Ba5}}=\\sqrt{((0.9033-0.75)c)^2+\\left(\\frac{\\sqrt{3}}{3}a\\right)^2}=\\sqrt{((0.9033-0.75)\\times1399.9)^2+\\left(\\frac{\\sqrt{3}}{3}\\times574.6\\right)^2}=395.1\\ \\mathrm{pm}$. (1 point)\nThe second-shortest Ba-Ba distance is $428.2\\ \\mathrm{pm}$. (1 point)\n(A correct result obtained by another method may receive credit, but the Ba0-Ba3 distance must be calculated to earn its 1 process point.)"], "rubric_images": [{"file": "Exam-12-T5-2-1-rubric.png", "description": "An unfilled rectangle with a black border and an aspect ratio of approximately 3:1 contains green circles labeled with black letters and numbers. Taking the lower-left, lower-right, and upper-right corners as (0,0), (1,0), and (1,1), respectively, the circles and labels are located at approximately: (1/3,0.90), Ba0; (2/3,0.40), Ba1; (2/3,0.10), Ba2; (1/3,0.60), Ba3; (0,0.25), Ba4; (1,0.25), Ba4; (0,0.75), Ba5; and (1,0.75), Ba5."}]}]} +{"problem_id": "Exam-13-T5", "header": "Question 5 Crystal Distortions. Under ideal conditions, several classic crystal structures have high symmetry, including cubic perovskite, fluorite, and spinel structures. Under certain conditions, however, some crystal structures distort, breaking the original high symmetry and lowering it to tetragonal, trigonal, orthorhombic, monoclinic, or another crystal system.", "header_images": [{"file": "Exam-13-T5-header-1.png", "description": "A perspective ball-and-stick model of a three-dimensional crystal structure. Axes at the lower left show a in red pointing toward the lower left and into the view, b in green pointing horizontally right, and c in blue pointing vertically upward. The outer framework is a cube with a large light-blue sphere at each of its 8 vertices and thin light-blue rods joining adjacent vertices. A smaller red sphere is embedded near the midpoint of each of the 12 cube edges. The largest green sphere is suspended at the center of the cube and connects radially to the 12 red spheres through 12 thin green rods. The 8 red spheres in the top layer, c = 1, and bottom layer, c = 0, are accompanied by broad white arrows with black borders and bold black coordinates indicating displacement. In the top layer, the left sphere has a rightward arrow and (0.5,0.0098,1); the right sphere has a leftward arrow and (0.5,0.9902,1); the front sphere has a lower-right arrow and (1.0098,0.5,1); and the rear sphere has an upper-right arrow and (-0.0098,0.5,1). In the bottom layer, the left sphere has an outward left arrow and (0.5,-0.0098,0); the right sphere has an outward right arrow and (0.5,1.0098,0); the front sphere has a lower-left arrow and (0.9902,0.5,0); and the rear sphere, seen in perspective below and to the right of the central green sphere, has an upper-left arrow and (0.0098,0.5,0)."}], "subquestions": [{"sub_id": "Exam-13-T5-1", "tags": ["1.3"], "text_open": "5.2 At low temperature, $NaNiO_2$ distorts from a high-symmetry structure into a base-centered monoclinic crystal, as shown below; the large, medium, and small spheres are Na, Ni, and O, respectively. One O atom has coordinates (0.222, 0.5, 0.205), both Ni and Na occupy vertex or face-center positions, and the unit-cell parameters are a = 5.33 Å, b = 2.86 Å, c = 5.59 Å, and β = 110.5°. 5.2.1 Before distortion, the structure follows a close-packing model: O forms the closest-packed array, while alternating Na and Ni layers occupy octahedral holes in the O array. Use ABC for the close-packed O layers, abc for the Na layers, and a’b’c’ for the Ni layers. Represent the close-packing sequence before distortion using letters, giving one repeat unit, and state the Bravais lattice of the undistorted crystal.", "text_images": [{"file": "Exam-13-T5-1-1-prompt.png", "description": "A perspective three-dimensional ball-and-stick model enclosed by an inclined parallelepiped outlined with thin black lines. Axes at the lower left show a in red pointing roughly left, b in green pointing into the page, and c in blue pointing steeply toward the upper right. Three sphere sizes and colors represent the elements: the largest golden spheres are Na, medium silver-gray spheres are Ni, and the smallest red spheres are O. Thin cylindrical bonds are split between the colors of the connected atoms, such as yellow-red Na-O and silver-red Ni-O bonds. The structure is a highly ordered alternating layered arrangement along c. Silver-gray Ni layers lie near the upper and lower faces of the cell. Each Ni connects diagonally to 6 O atoms through 6 silver-red bonds, forming a typical $NiO_6$ octahedron. These edge-sharing octahedra extend laterally in planes perpendicular to c to form robust two-dimensional $NiO_2$ slabs. A layer of large golden Na spheres occupies the central plane in the open space between the 2 $NiO_2$ slabs. Each Na radiates 6 yellow-red bonds in an octahedral geometry, connecting upward to 3 O atoms at the bottom of the upper $NiO_2$ slab and downward to 3 O atoms at the top of the lower $NiO_2$ slab. Thus, rather than forming simple vertical linear units, the interlayer Na atoms act as structural anchors through a three-dimensional network of 6 contacts, 3 above and 3 below, joining the separated silver-red slabs along c into a continuous crystal network."}], "full_mark": 5, "rubric": ["The repeat unit is Ac’BaCb’AcBa’Cb, or an equivalent representation (3 points); rhombohedrally centered hexagonal, or hR (2 points)."], "rubric_images": []}, {"sub_id": "Exam-13-T5-2", "tags": ["1.3"], "text_open": "5.2.2 Calculate all distinct Ni-O bond lengths after distortion.", "text_images": [], "full_mark": 4, "rubric": ["$d_1(\\mathrm{Ni-O})=\\left\\|0.222\\vec{a}+0.5\\vec{b}+0.205\\vec{c}\\right\\|=1.95\\ \\text{angstrom}$ (2 points); $d_2(\\mathrm{Ni-O})=\\left\\|-0.278\\vec{a}+0.205\\vec{c}\\right\\|=2.17\\ \\text{angstrom}$ (2 points)."], "rubric_images": []}]} {"problem_id": "Exam-13-T9", "header": "Question 9 Meldrum's Acid. Meldrum's acid is 2,2-dimethyl-1,3-dioxane-4,6-dione and has distinctive reactivity and broad synthetic applications.", "header_images": [], "subquestions": [{"sub_id": "Exam-13-T9-1", "tags": ["2.3", "3.9"], "text_open": "9.2 On heating, the following Meldrum's-acid derivative decomposes through fragmentation and a series of other reactions to give the products shown. Draw the mechanisms leading to the two products and identify the type of each pericyclic reaction.", "text_images": [{"file": "Exam-13-T9-1-1-prompt.png", "description": "Thermal decomposition of the Meldrum's-acid derivative O=C(OC(C)(C)OC/1=O)C1=C(O)/C gives products O=C2C=C(C)OC(C)(C)O2 and O=C3C=C(C)OC(C3C(C)=O)=O."}], "full_mark": 8, "rubric": ["Award 4 points for the mechanisms and 4 points for the reaction types.", "Answer SMILES: 'CC(O)=C1C(=O)OC(C)(C)OC1=O'; 'CC(=O)C1=C(O)OC(C)(C)OC1=O'; 'CC(=O)C'; 'CC(=O)C(=C=O)C(=O)O'; 'CC(=O)C=C=O'; 'CC1=CC(=O)OC(C)(C)O1'; 'CC(=O)C1C(=O)C=C(C)OC1=O'."], "rubric_images": [{"file": "Exam-13-T9-1-1-rubric.png", "description": "Three rows of reaction equations. Row 1: 'CC(O)=C1C(=O)OC(C)(C)OC1=O' undergoes a 1,5-hydrogen shift, also labeled or [1,5]-σ rearrangement, to give 'CC(=O)C1=C(O)OC(C)(C)OC1=O'. This intermediate undergoes a retro-Diels-Alder reaction, also labeled or retro-[4+2] cycloaddition, with loss of acetone, - Me_2CO, SMILES 'CC(=O)C', to give the open-chain ketene carboxylic acid 'CC(=O)C(=C=O)C(=O)O'. Loss of - CO_2 gives O=C=CH-C(=O)CH_3, SMILES 'CC(=O)C=C=O'. Row 2: one molecule of 'CC(=O)C=C=O' and one acetone molecule, 'CC(=O)C', undergo a Diels-Alder reaction, also labeled or [4+2] cycloaddition, to give 'CC1=CC(=O)OC(C)(C)O1'. Row 3: 2 molecules of 'CC(=O)C=C=O' undergo the same cycloaddition to give 'CC(=O)C1C(=O)C=C(C)OC1=O'."}]}, {"sub_id": "Exam-13-T9-2", "tags": ["1.2", "2.7"], "text_open": "9.3 As shown below, Cu/Ag bimetallic catalysis of a Meldrum-acid derivative and a terminal alkyne in the presence of chiral ligand L^{*} gives a stereospecific product that can be transformed into many derivatives. The reagent loadings are CuI 5 mol%, AgOAc 6.5 mol%, and L^{*} 6.5 mol%. 9.3.1 When R_{1} = R_{2} = Ph, give the absolute configuration of the stereogenic carbon in product P.", "text_images": [{"file": "Exam-13-T9-2-1-prompt.png", "description": "Reactants 'O=C(OC(C)(C)OC/1=O)C1=C\\[R1]' and '[R2]C#C', joined by a plus sign, react under CuI, AgOAc and H_2O above the arrow and DCM, L* below to give product P, SMILES '[R1][C@@H]1CC(OC([R2])=C1)=O'. A right brace branches P into 3 pathways. Path A uses excess LiAlH_4 at -78 °C with $R_1 = 3-BrC_6H_4, R_2 = CH_2OBn$. Path B uses mCPBA, p-TsOH with $R_1 = Ph, R_2 = CH_2OBn$. Path C uses $BnNH_2$, p-TsOH with $R_1 = Ph, R_2 = (CH_2)_4OTBS$."}], "full_mark": 1, "rubric": ["R (1 point)."], "rubric_images": []}, {"sub_id": "Exam-13-T9-3", "tags": ["1.4", "2.3", "2.4"], "text_open": "9.3.2 Write the SMILES strings of at least 3 key metal-containing intermediates in formation of P from the 2 substrates. Ligands on the metal may be omitted; each intermediate contains only 1 type of metal.", "text_images": [], "full_mark": 6, "rubric": ["Any 3 key metal-containing intermediate SMILES are sufficient, 2 points each.", "Answer SMILES: '[R2]C#C[Cu].O=C1OC(C)(C)OC(O)=C1C(C#C[R2])[R1]'; 'O=C(OC(C)(C)O2)C(C(C#C[R2])[R1])C2=O'; 'O=C3OC(C)(C)OC4=[O+]C(C)=C([Ag])C([R1])C43'."], "rubric_images": [{"file": "Exam-13-T9-3-1-rubric.png", "description": "Answer metal-containing intermediates, in order: '[R2]C#C[Cu]'; 'O=C1OC(C)(C)OC(O)=C1C(C#C[R2])[R1]'; 'O=C(OC(C)(C)O2)C(C(C#C[R2])[R1])C2=O', whose C-C triple bond coordinates [Ag]; and 'O=C3OC(C)(C)OC4=[O+]C(C)=C([Ag])C([R1])C43'."}]}, {"sub_id": "Exam-13-T9-4", "tags": ["2.4", "2.6"], "text_open": "9.3.3 A, C_{19}H_{21}BrO_{3}, contains a hemiketal; B, C_{19}H_{18}O_{4}, contains a five-membered ring; and C, C_{22}H_{25}NO_{2}, contains a spiro ring. Using the hints, write the SMILES strings of A, B, and C obtained from their corresponding P products.", "text_images": [], "full_mark": 6, "rubric": ["Award 2 points each for the SMILES strings of A, B, and C.", "Answer SMILES: A 'BrC1=CC=CC(C2CCOC(COCC3=CC=CC=C3)(O)C2)=C1'; C 'O=C(CC(CC45CCCCO5)C6=CC=CC=C6)N4CC7=CC=CC=C7'; B 'O=C(C8)OC(C(COCC9=CC=CC=C9)=O)C8C%10=CC=CC=C%10'."], "rubric_images": [{"file": "Exam-13-T9-4-1-rubric.png", "description": "Answer structures A, B, and C. A contains a hemiketal and bromoaryl group, B contains a five-membered lactone/carbonyl motif, and C contains a nitrogen spirocycle. SMILES: A 'BrC1=CC=CC(C2CCOC(COCC3=CC=CC=C3)(O)C2)=C1'; C 'O=C(CC(CC45CCCCO5)C6=CC=CC=C6)N4CC7=CC=CC=C7'; B 'O=C(C8)OC(C(COCC9=CC=CC=C9)=O)C8C%10=CC=CC=C%10'."}]}, {"sub_id": "Exam-13-T9-5", "tags": ["2.3", "2.4", "2.6"], "text_open": "9.4 The following multicomponent cascade reaction of Meldrum's acid gives the product with high diastereoselectivity. Because Meldrum's acid is acidic, its own acidic proton promotes the reaction spontaneously, including a protonation step. 9.4.1 Write the SMILES strings of 4 key intermediates; stereochemistry need not be shown.", "text_images": [{"file": "Exam-13-T9-5-1-prompt.png", "description": "Multicomponent cascade reaction of Meldrum acid, 'O=C(OC(C)(C)O5)CC5=O', an α,β-unsaturated aldehyde, 'O=C/C(C)=C/C=C/C1=CC=CC=C1', and MeOH at 75 °C in a sealed tube to give polycyclic product '[H]C2(C(C=C3)C4=CC=CC=C4)[C@@]3(OC([C@@H]2C(OC)=O)=O)C'. The product has 2 wavy bonds indicating unspecified stereochemistry."}], "full_mark": 8, "rubric": ["The rubric gives SMILES for 4 key intermediates, 2 points each; stereochemistry is not required.", "Answer SMILES: 'O=C1OC(C)(C)OC(/C1=C\\C(C)=C\\C=C\\C2=CC=CC=C2)=O'; '[OH+]=C3OC(C)(C)OC(/C3=C\\C(C)=C\\C=C\\C4=CC=CC=C4)=O'; 'CC(O5)(C)OC(O[H])=C(C6[C+](C)C=CC6C7=CC=CC=C7)C5=O'; 'CC(OC8=C9C%10C(C=CC%10C%11=CC=CC=C%11)(C)O8)(C)OC9=[OH+]'; 'CC(OC%12=C%13C%14C(C=CC%14C%15=CC=CC=C%15)(C)O%12)(C)OC%13=O'; 'O=C(OC%16(C)C%17C(C%18=CC=CC=C%18)C=C%16)C%17=C=O'; 'O=C(OC)C%19=C(OC%20(C)C%19C(C%21=CC=CC=C%21)C=C%20)OC(C)(O)C'."], "rubric_images": [{"file": "Exam-13-T9-5-1-rubric.png", "description": "At upper left, molecules 'O=C1OC(C)(C)OC(/C1=C\\C(C)=C\\C=C\\C2=CC=CC=C2)=O' and '[OH+]=C3OC(C)(C)OC(/C3=C\\C(C)=C\\C=C\\C4=CC=CC=C4)=O' are separated by the English-localized blue word or. At upper right is 'CC(O5)(C)OC(O[H])=C(C6[C+](C)C=CC6C7=CC=CC=C7)C5=O'. At lower left, 'CC(OC8=C9C%10C(C=CC%10C%11=CC=CC=C%11)(C)O8)(C)OC9=[OH+]' and 'CC(OC%12=C%13C%14C(C=CC%14C%15=CC=CC=C%15)(C)O%12)(C)OC%13=O' are separated by blue or. At lower right, 'O=C(OC%16(C)C%17C(C%18=CC=CC=C%18)C=C%16)C%17=C=O' and 'O=C(OC)C%19=C(OC%20(C)C%19C(C%21=CC=CC=C%21)C=C%20)OC(C)(O)C' are separated by blue or."}]}, {"sub_id": "Exam-13-T9-6", "tags": ["1.2", "2.7", "2.4"], "text_open": "9.4.2 Write the SMILES string of the product, including the stereochemistry indicated by the 2 wavy bonds.", "text_images": [], "full_mark": 4, "rubric": ["Write the product SMILES shown in the rubric and include the stereochemistry of the two wavy bonds (4 points).", "Answer SMILES: '[H][C@]1([C@@H](C=C2)C3=CC=CC=C3)[C@@]2(OC([C@@H]1C(OC)=O)=O)C'."], "rubric_images": [{"file": "Exam-13-T9-6-1-rubric.png", "description": "A bicyclic structure drawn mainly in blue and red. A five-membered lactone on the left is fused to a five-membered cyclopentene on the right. The lactone contains one O and one carbonyl; one carbon bears a blue solid-wedge methoxycarbonyl group, MeO-C=O. At the ring-junction bridgeheads, the upper carbon bears a blue solid-wedge methyl group, Me, and the lower carbon bears a red solid-wedge hydrogen, H. The adjacent cyclopentene carbon bears a red solid-wedge phenyl group, Ph. The SMILES with stereochemistry is '[H][C@]1([C@@H](C=C2)C3=CC=CC=C3)[C@@]2(OC([C@@H]1C(OC)=O)=O)C'."}]}]} {"problem_id": "Exam-14-T3", "header": "Frustrated Noncovalent Organic Glass\nAs amorphous materials crosslinked through noncovalent interactions, supramolecular glasses are constrained by a trade-off between glass-forming ability and mechanical properties. Inspired by frustrated Lewis pairs, researchers designed and prepared a “frustrated noncovalent organic glass” in which the noncovalent interaction sites were altered. This increased the mechanical toughness of the supramolecular glass by a factor of 4.3 while providing high interfacial adhesion, enabling its use as an impact-resistant coating.", "header_images": [], "subquestions": [{"sub_id": "Exam-14-T3-1", "tags": ["2.4", "2.6"], "text_open": "3.1 A preparation route for a common supramolecular glass A is shown below. Write the SMILES strings of intermediates X1 and X2 and the repeat unit of polymer A (terminal groups are not required; use [*] to mark the connection sites between repeat units). EDC is 1-ethyl-3-(3-dimethylaminopropyl)carbodiimide, an organic dehydrating agent.", "text_images": [{"file": "Exam-14-T3-1-1-prompt.png", "description": "Reaction sequence: 'O=C1C(OCC)=C(OCC)C1=O' reacts with 'NCCO' under [$\\mathrm{Et_3N}$, EtOH, 70°C] to form X1; X1 then reacts with 'C=CCCCCCCCCC(O)=O' under [DMAP, EDC, $\\mathrm{CH_2Cl_2}$] to form X2; finally, X2 reacts with 'SCOCCOCS' under [AIBN, DMF, 70°C] to form A."}], "full_mark": 6, "rubric": ["Structure of X1: 'O=C1C(NCCO)=C(C1=O)NCCO' (2 points)", "Structure of X2: 'O=C1C(NCCOC(CCCCCCCCC=C)=O)=C(C1=O)NCCOC(CCCCCCCCC=C)=O' (2 points)", "Structure of polymer A; the repeat unit is 'O=C1C(NCCOC(CCCCCCCCCCSCCOCCOCCS[*])=O)=C(NCCOC(CCCCCCCCCC[*])=O)C1=O', where * denotes a connection site between repeat units. (2 points)"], "rubric_images": [{"file": "Exam-14-T3-1-1-rubric.png", "description": "Structure of intermediate X1: 'O=C1C(NCCO)=C(C1=O)NCCO'"}, {"file": "Exam-14-T3-1-2-rubric.png", "description": "Structure of intermediate X2: 'O=C1C(NCCOC(CCCCCCCCC=C)=O)=C(C1=O)NCCOC(CCCCCCCCC=C)=O'"}, {"file": "Exam-14-T3-1-3-rubric.png", "description": "Structure of polymer A; the repeat unit is 'O=C1C(NCCOC(CCCCCCCCCCSCCOCCOCCS[*])=O)=C(NCCOC(CCCCCCCCCC[*])=O)C1=O', where * denotes a connection site between repeat units."}]}, {"sub_id": "Exam-14-T3-2", "tags": ["1.7", "1.6", "2.4"], "text_open": "3.2.1 Replacing 2-hydroxyethylamine (ethanolamine) in the above route with the structure shown below, while leaving all other steps unchanged, gives “frustrated noncovalent organic glass” B. The polymer chains of B and A differ only by the isopropyl substituent. The repeat units of both polymers A and B form rings through hydrogen bonding, but the numbers of hydrogen bonds and the groups involved differ: A forms a 10-membered ring and B a 7-membered ring. Write the SMILES strings corresponding to schematic hydrogen-bonded structures between repeat units of A and B, respectively; the hydrogen bonds may be represented as ordinary chemical bonds.", "text_images": [{"file": "Exam-14-T3-2-1-prompt.png", "description": "SMILES string of the structure that replaces 2-hydroxyethylamine (ethanolamine): 'N[C@H](C(C)C)CO'"}], "full_mark": 4, "rubric": ["Hydrogen-bonding diagram for A (2 points)", "Hydrogen-bonding diagram for B (2 points)", "SMILES strings shown: 'O=C1C(N2CCOC([*])=O)=C(N(CCOC([*])=O)[H]/O=C3C(NCCOC([*])=O)=C(C/3=O/[H]2)NCCOC([*])=O)C1=O', where * represents part of the polymer chain and may be omitted from the drawing; 'O=C1C(N2[H]/O([H]N3[C@@H](COC([*])=O)C(C)C)=C([*])\\OC[C@@H](C(C)C)N(C(C4=O)=C(N([C@@H](COC([*])=O)C(C)C)[H]/5)C4=O)[H]O5=C([*])/OC[C@H]2C(C)C)=C3C1=O', where * represents part of the polymer chain and may be omitted from the drawing."], "rubric_images": [{"file": "Exam-14-T3-2-1-rubric.png", "description": "Hydrogen-bonding diagram in which repeat units of polymer A form a 10-membered ring: 'O=C1C(N2CCOC([*])=O)=C(N(CCOC([*])=O)[H]/O=C3C(NCCOC([*])=O)=C(C/3=O/[H]2)NCCOC([*])=O)C1=O', where * represents part of the polymer chain and may be omitted from the drawing."}, {"file": "Exam-14-T3-2-2-rubric.png", "description": "Hydrogen-bonding diagram in which repeat units of polymer B form a 7-membered ring: 'O=C1C(N2[H]/O([H]N3[C@@H](COC([*])=O)C(C)C)=C([*])\\OC[C@@H](C(C)C)N(C(C4=O)=C(N([C@@H](COC([*])=O)C(C)C)[H]/5)C4=O)[H]O5=C([*])/OC[C@H]2C(C)C)=C3C1=O', where * represents part of the polymer chain and may be omitted from the drawing."}]}, {"sub_id": "Exam-14-T3-3", "tags": ["1.7"], "text_open": "3.2.2 The chain structures of polymers A and B can be represented by the figures below. Which structure corresponds to frustrated noncovalent organic glass B? ( )\n(a) Left figure;\t(b) Right figure", "text_images": [{"file": "Exam-14-T3-3-1-prompt.png", "description": "Multiple short gray rod-like elements and curved gray lines. A tightly spaced, approximately horizontal group of rods appears in the upper-left region, with parallel offsets between rods. Two inclined groups in the upper-right run overall from upper left to lower right, again with parallel offsets. An approximately horizontal group appears near the lower center. Each rod has conspicuously wider protrusions at both ends. Fine gray curves connect the rod groups, bending, crossing, and reversing direction; a longer loop-like curved line is visible on the right."}, {"file": "Exam-14-T3-3-2-prompt.png", "description": "Four groups of short gray rod-like elements connected by curved gray lines. The upper-left group contains approximately 5 vertically stacked parallel rods; the upper-right group contains approximately 4; the lower-left and lower-right groups each contain approximately 5 stacked parallel rods and are slightly inclined. Multiple gray curves connect the four groups, with crossing, bending, and branching line relationships visible near the center."}], "full_mark": 1, "rubric": ["(a) (1 point)"], "rubric_images": []}, {"sub_id": "Exam-14-T3-4", "tags": ["1.7"], "text_open": "3.2.3 If A and B have the same degree of polymerization, give the ratio of the numbers of hydrogen bonds they form (as the simplest integer ratio).", "text_images": [], "full_mark": 2, "rubric": ["1:2 (2 points)"], "rubric_images": []}]} {"problem_id": "Exam-14-T6", "header": "Kinetic Proofreading in Enzyme Reactions\nIncorrect enzyme-amino acid binding can occur during protein synthesis, so organisms require proofreading mechanisms to ensure accurate enzyme recognition. Studies show that aminoacyl-tRNA synthetases (aaRSs) use kinetic proofreading to reduce the fraction of incorrect products. In this problem, C denotes the correct amino acid and D the incorrect amino acid.", "header_images": [], "subquestions": [{"sub_id": "Exam-14-T6-1", "tags": ["2.9", "2.4"], "text_open": "6.2.1 In fact, aaRS uses a kinetic proofreading mechanism that can be represented by the elementary-reaction pathway below. ATP can activate amino-acid functional groups. For example, under enzyme catalysis ATP reacts with the carboxyl group of phenylalanine, releasing pyrophosphate and forming an intermediate in which AMP is linked to the carboxyl group. The structures of ATP and phenylalanine are shown below. Write the SMILES string of the ATP-activated phenylalanine product.", "text_images": [{"file": "Exam-14-T6-1-1-prompt.png", "description": "Reaction scheme on a white background with black text and arrows.\n\nE + aa + ATP on the left is connected to E·aa·ATP on the right by horizontal reversible arrows labeled k₁ above and k₋₁ below.\n\nA horizontal rightward arrow labeled k₂ connects E·aa·ATP to E·aa·AMP + PPi.\n\nAn upward vertical arrow above E·aa·AMP is labeled k₃ on the left and points to E + aa·AMP.\n\nA downward vertical arrow below E·aa·AMP is labeled k₄ on the left and tRNA on the right and points to E + aa-tRNA + AMP, with aa-tRNA in bold."}, {"file": "Exam-14-T6-1-2-prompt.png", "description": "Adenosine triphosphate, ATP: 'NC1=NC=NC2=C1N=CN2[C@H]3[C@H](O)[C@H](O)[C@@H](COP(OP(OP([O-])([O-])=O)([O-])=O)([O-])=O)O3'; phenylalanine, Phe: '[O-]C([C@@H]([NH3+])CC1=CC=CC=C1)=O'"}], "full_mark": 2, "rubric": ["'NC1=NC=NC2=C1N=CN2[C@H]3[C@H](O)[C@H](O)[C@@H](COP(OC([C@H](CC4=CC=CC=C4)[NH3+])=O)([O-])=O)O3' (2 points; 1 point if the charges are incorrect)"], "rubric_images": [{"file": "Exam-14-T6-1-1-rubric.png", "description": "SMILES string: 'NC1=NC=NC2=C1N=CN2[C@H]3[C@H](O)[C@H](O)[C@@H](COP(OC([C@H](CC4=CC=CC=C4)[NH3+])=O)([O-])=O)O3'"}]}]} -{"problem_id": "Exam-15-T1", "header": "Problem 1 The Immortal Mirror. In 1863, two chemists at the Freiberg Mining Academy in Germany unexpectedly observed a previously unknown spectral line while studying locally mined sphalerite. They assigned it to an undiscovered element M and successfully isolated elemental M from the mineral. At the subsequent Paris Exposition, elemental M was publicly displayed as \"the rarest metal,\" causing a sensation. At that time, M was used to make highly reflective mirrors; today it is more commonly an important semiconductor raw material used in conductive films.", "header_images": [], "subquestions": [{"sub_id": "Exam-15-T1-1", "tags": ["2.1"], "text_open": "1.1 Burning M in air gives a pale-yellow amorphous solid A. Heating A under hydrogen at 400~500°C reduces it to a black crystalline solid B (Reaction 1); dissolving B in dilute hydrochloric acid releases a colorless gas C. Above approximately 700°C, B decomposes to A and elemental M (Reaction 2). At high temperature, A reacts with ammonia to form a yellow semiconductor material D and releases a gas (Reaction 3).\n1.1.2 Write the chemical equations for Reactions 1~3.", "text_images": [], "full_mark": 3, "rubric": ["Reaction 1: $In_2O_3$ + 2$H_2$ = $In_2O$ + 2$H_2O$ (1 point)", "Reaction 2: 3$In_2O$ = $In_2O_3$ + 4In (1 point)", "Reaction 3: $In_2O_3$ + 2$NH_3$ = 2InN + 3$H_2O$ (1 point)"], "rubric_images": []}, {"sub_id": "Exam-15-T1-2", "tags": ["1.4"], "text_open": "1.2 Chloride E of M is a white flaky solid that is extremely hygroscopic and dissolves in water with substantial heat release. In ether solution, E reacts with LiH to form complex F. F decomposes above 0°C and can be used as an in situ reducing agent in organic synthesis. E forms complex E·3THF with the electron-donor ligand THF. Single-crystal X-ray diffraction shows that the M center in this complex is six-coordinate with octahedral geometry. E has a layered crystal structure consisting of close-packed chloride ions with layers of octahedrally coordinated M centers. By contrast, chloride G of the congener whose period differs by only 1 commonly crystallizes as dimer H. 1.2.1 Write the chemical formulas of E~G, write the SMILES of H, and describe its three-dimensional structure in words.", "text_images": [], "full_mark": 5, "rubric": ["E: $InCl_3$ (1 point)", "F: $LiInH_4$ (1 point)", "G: $GaCl_3$ (1 point)", "H: a chloride-bridged $Ga_2Cl_6$ dimer (2 points)"], "rubric_images": [{"file": "Exam-15-T1-2-1-rubric.png", "description": "Structure of H: a chloride-bridged $Ga_2Cl_6$ dimer. Two Ga centers are bridged by two Cl atoms, and each Ga also bears two terminal Cl atoms directed respectively out of and into the drawing plane. Each Ga is tetrahedral. Corresponding SMILES: 'Cl[Ga]1(Cl)Cl[Ga](Cl)(Cl)Cl1'."}]}, {"sub_id": "Exam-15-T1-3", "tags": ["1.5", "1.6"], "text_open": "1.3 In $E\\cdot3THF$, the $M$–Cl bond length is approximately 242 pm and the $M$–O bond length is approximately 225 pm. The ionic radius of $M$ is 80 pm, the ionic radius of $Cl^-$ is 181 pm, and the van der Waals radius of O in THF is approximately 152 pm.\n1.3.2 Use the bond-length data to compare the bonding character of the M-Cl and M-O bonds, including their relative ionic/covalent character, and explain.", "text_images": [], "full_mark": 3, "rubric": ["The In-Cl bond has relatively greater covalent character than the In-O bond. (1 point)", "The In-Cl bond length, 242 pm, is less than the sum of the ionic radii, 80 + 181 = 261 pm. The 19 pm difference is approximately 7.28%, indicating some covalent contribution to the In-Cl bond. The In-O bond length, 225 pm, is less than the sum of the relevant radii, 80 + 152 = 232 pm. The 7 pm difference is approximately 3.01%, also indicating some covalent contribution. Comparison of these differences shows that In-Cl is shortened to a greater relative extent and therefore has greater covalent character. (2 points)", "If the element is identified incorrectly but the explanation is reasonable, credit is still awarded."], "rubric_images": []}]} -{"problem_id": "Exam-15-T4", "header": "Problem 4 Determination of Chloride by the Volhard Method. ① Pipette 20.00 mL of 0.1 mol/L standard silver nitrate titrant into a 250 mL conical flask. Add 30 mL water, 5 mL nitric acid solution, and 2 mL saturated ferric ammonium sulfate solution. While shaking, titrate with standard potassium thiocyanate titrant until a pale brown-red color appears and persists for 1 min. Record the volume of standard potassium thiocyanate consumed: V_{1} = 19.88 mL. ② Weigh 0.10 g sodium chloride that has been ignited at 500 ℃ ~ 600 ℃ to constant mass, accurate to 0.1 mg. Dissolve it in approximately 40 mL water in a beaker and transfer to a 100 mL volumetric flask. Add 5 mL nitric acid solution (1+3), then add 25.00 mL of 0.1 mol/L standard silver nitrate solution while shaking vigorously. Dilute to the mark with water and mix. Let stand in the dark for 5 min, filter through rapid filter paper, and discard the first 10 mL of filtrate. Accurately pipette 50.00 mL filtrate into a 250 mL conical flask, add 2 mL of 1 mol/L ferric ammonium sulfate solution, and titrate with standard potassium thiocyanate while shaking until a pale brown-red color appears and persists for 1 min. Record V_{2} = 4.01 mL. ③ Measure 50 mL test solution into a 100 mL colorimetric tube and otherwise follow the same procedure as in ②. Record V_{3} = 2.98 mL.", "header_images": [], "subquestions": [{"sub_id": "Exam-15-T4-1", "tags": ["4.5", "2.1"], "text_open": "4.1 Calculate the concentrations of the standard silver nitrate and potassium thiocyanate titrant solutions.", "text_images": [], "full_mark": 5, "rubric": ["Let the exact concentration of $AgNO_3$ solution be c1 and that of KSCN solution be c2.", "In step ①, $Ag^+$ and $SCN^-$ react in a 1 : 1 ratio:", "\\[{{c}_{1}}\\times 20.00\\text{ mL}={{c}_{2}}\\times 19.88\\text{ mL}\\]", "Therefore:", "\\[{{c}_{2}}=\\frac{20.00}{19.88}{{c}_{1}}\\] (1 point)", "From step ②, the mass of the primary standard NaCl is m = 0.1000 g, and M(NaCl) = 58.44 g/mol.", "The amount of NaCl is \\[n(\\text{C}{{\\text{l}}^{\\text{-}}})=\\frac{0.1000\\text{ g}}{58.44\\text{ g/mol}}=1.71116\\text{ mmol}\\] (1 point).", "In the 100 mL volumetric flask, 25.00 mL $AgNO_3$ is added; the excess $Ag^+$ is determined by back-titration of the remaining solution.", "Titration of 50 mL filtrate consumes 4.01 mL KSCN solution, so the excess $Ag^+$ in the 100 mL solution consumes 8.02 mL KSCN solution.", "From the material balance:", "\\[25.00\\cdot {{c}_{1}}=1.71116+8.02\\cdot {{c}_{2}}\\] (1 point).", "Solving simultaneously gives:", "\\[c(\\text{AgN}{{\\text{O}}_{\\text{3}}})={{c}_{1}}\\approx 0.1011\\text{ mol/L}\\] (1 point).", "\\[c(\\text{KSCN})={{c}_{2}}=1.006036\\times 0.10106\\approx 0.1017\\text{ mol/L}\\] (1 point)."], "rubric_images": []}, {"sub_id": "Exam-15-T4-2", "tags": ["4.1", "4.7"], "text_open": "4.4.2 Explain why the first 10 mL of filtrate is discarded.", "text_images": [], "full_mark": 1, "rubric": ["To eliminate adsorption of solutes such as $Ag^+$ ions by the filter paper, ensuring that the subsequently pipetted 50.00 mL filtrate accurately represents the concentration of the diluted solution."], "rubric_images": []}, {"sub_id": "Exam-15-T4-3", "tags": ["4.2", "4.5"], "text_open": "4.5 If addition of standard silver nitrate produces a colloidal solution, what additional operation is required before filtration?", "text_images": [], "full_mark": 2, "rubric": ["Heat the solution in a boiling-water bath until a coagulated silver chloride precipitate appears. (1 point) Remove it and cool rapidly to room temperature in cold water. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-15-T4-4", "tags": ["4.5", "2.1", "4.7"], "text_open": "4.6 Calculate the $Cl^-$ concentration in the sample.", "text_images": [], "full_mark": 3, "rubric": ["In step ③, the total amount of $Ag^+$ added is \\[25.00\\text{ mL}\\times 0.10106\\text{ mol/L}=2.5265\\text{ mmol}\\] (1 point)", "Titration of 50.00 mL filtrate consumes 2.98 mL standard KSCN solution.", "Therefore, the total excess $Ag^+$ is \\[2\\times 2.98\\text{ mL}\\times 0.10167\\text{ mol/L}=0.6060\\text{ mmol}\\] (1 point)", "The total amount of $Cl^-$ in the test solution is \\[n(\\text{C}{{\\text{l}}^{\\text{-}}})=2.5265-0.6060=1.9205\\text{ mmol}\\]", "The original $Cl^-$ concentration in the test solution is:", "\\[c(\\text{C}{{\\text{l}}^{\\text{-}}})=\\frac{1.9205\\text{ mmol}}{50.00\\text{ mL}}\\approx 0.03841\\text{ mol/L}\\] (1 point)"], "rubric_images": []}]} +{"problem_id": "Exam-15-T3", "header": "Problem 3 Solid Electrolyte Interphase\nThe solid electrolyte interphase (SEI) is an important research area in modern electrochemistry. During the first charging of a lithium-ion battery, the electrode potential lies below the reduction potential of the electrolyte, so additives and lithium salts in the electrolyte undergo electrochemical reductive decomposition at the solid-liquid interface. Insoluble solid products deposit on the electrode surface and form a passivation film that is electronically insulating but ionically conducting, thereby slowing electrode corrosion. Vinylene carbonate (VC), a classic film-forming additive, reacts preferentially and promotes formation of a high-quality SEI.", "header_images": [{"file": "Exam-15-T3-header-1.png", "description": "The structures in the figure have the following SMILES: VC, 'O=C1OC=CO1'; EC, 'O=C2OCCO2'."}], "subquestions": [{"sub_id": "Exam-15-T3-1", "tags": ["2.2", "2.3", "2.4"], "text_open": "3.3 Studies show that VC preferentially undergoes cleavage of a C-O single bond and is reduced to radical anion A carrying 1 unit of negative charge. It then undergoes radical polymerization to form several polymeric products. 3.3.1 In one pathway, A acts as a chain initiator and reacts with VC to form acyclic polymer B. Write the SMILES of B.", "text_images": [], "full_mark": 2, "rubric": ["Rubric structure for B: the corresponding SMILES is '*C=COC(=O)O*', where * denotes a repeat-unit connection site. (2 points)"], "rubric_images": [{"file": "Exam-15-T3-1-1-rubric.png", "description": "Rubric structure for B: the corresponding SMILES is '*C=COC(=O)O*', where * denotes a repeat-unit connection site."}]}, {"sub_id": "Exam-15-T3-2", "tags": ["2.1", "2.2", "3.6"], "text_open": "3.3.4 In the electrolyte, VC can also be electrochemically reduced to lithium oxalate and an enolate. Write the electrochemical reaction equation; EC participates in the reaction.", "text_images": [], "full_mark": 2, "rubric": ["Complete reaction: 'O=C1OC=CO1' + 'O=C1OCCO1' + 4$Li^+$ + 4$e^-$ = $Li_2C_2O_4$ + 2$CH_2CHOLi$ (2 points; 1 point if all reactants are correct but the equation is not balanced)."], "rubric_images": [{"file": "Exam-15-T3-2-1-rubric.png", "description": "The EC structure in the rubric equation has SMILES 'O=C1OCCO1'."}, {"file": "Exam-15-T3-2-2-rubric.png", "description": "The VC structure in the rubric equation has SMILES 'O=C1OC=CO1'."}, {"file": "Exam-15-T3-2-3-rubric.png", "description": "The lithium enolate product in the rubric equation has SMILES 'C=CO[Li]'."}]}]} {"problem_id": "Exam-15-T8", "header": "Problem 8 Analysis of Substituent Effects. This problem compares the magnitudes of -ΔG° for a series of substituents or substrates in the conformational or isomerization reactions shown.", "header_images": [], "subquestions": [{"sub_id": "Exam-15-T8-1", "tags": ["1.2", "3.2"], "text_open": "Reaction (a) below is shared by 8.3–8.6. In 8.3–8.5, compare the shared Reaction (a) with Reaction (b) shown in the corresponding subproblem and select the reaction with the larger -ΔG°.\n8.3", "text_images": [{"file": "Exam-15-T8-1-1-prompt.png", "description": "Prompt figure for part a: the image shows a change from left to right. On the left is a six-membered ring in a chair conformation with an axial Me group on the right-hand carbon; its SMILES is 'CC1CCCCC1'. A one-way arrow points right. On the right is the same chair-shaped ring with an equatorial Me group on the right-hand carbon; its SMILES is likewise 'CC1CCCCC1'."}, {"file": "Exam-15-T8-1-2-prompt.png", "description": "Prompt figure for part b: the image shows a change from left to right. On the left is a six-membered ring in a chair conformation bearing a carbonyl (=O) at the lower-left vertex and an axial Me group on the right-hand carbon; its SMILES is 'CC1CC(=O)CCC1'. A one-way arrow points right. On the right is the same chair-shaped ring with the same lower-left carbonyl and an equatorial Me group on the right-hand carbon; its SMILES is likewise 'CC1CC(=O)CCC1'."}], "full_mark": 2, "rubric": ["a (2 points)"], "rubric_images": []}, {"sub_id": "Exam-15-T8-2", "tags": ["1.2", "3.2"], "text_open": "In 8.4, Reaction (b) is a base-catalyzed isomerization.\n8.4", "text_images": [{"file": "Exam-15-T8-2-1-prompt.png", "description": "The image shows a change from left to right. On the left is a six-membered chair bearing an equatorial tert-butyl group at the lower-left carbon, an axial Me group at the right-hand carbon, and a carbonyl (=O) at the lower-right vertex; its SMILES is 'CC1CCC(CC1=O)C(C)(C)C'. A one-way arrow points right. On the right is the same chair with the equatorial tert-butyl group and lower-right carbonyl unchanged, but with the Me group equatorial; its SMILES is likewise 'CC1CCC(CC1=O)C(C)(C)C'."}], "full_mark": 2, "rubric": ["a (2 points)"], "rubric_images": []}, {"sub_id": "Exam-15-T8-3", "tags": ["1.2", "3.2"], "text_open": "In 8.5, Reaction (b) is a base-catalyzed isomerization.\n8.5", "text_images": [{"file": "Exam-15-T8-3-1-prompt.png", "description": "Prompt figure for option (b): the image shows a change from left to right. On the left is a six-membered chair bearing an equatorial tert-butyl group at the lower-left carbon, an axial $CHMe_2$ group at the right-hand carbon, and a carbonyl (=O) at the lower-right vertex; its SMILES is 'CC(C)C1CCC(CC1=O)C(C)(C)C'. A one-way arrow points right. On the right is the same chair with the equatorial tert-butyl group and lower-right carbonyl unchanged, but with the $CHMe_2$ group equatorial; its SMILES is likewise 'CC(C)C1CCC(CC1=O)C(C)(C)C'."}], "full_mark": 2, "rubric": ["a (2 points)"], "rubric_images": []}, {"sub_id": "Exam-15-T8-4", "tags": ["1.2", "3.2"], "text_open": "8.6 Arrange -ΔG° for the following reactions in decreasing order.", "text_images": [{"file": "Exam-15-T8-4-1-prompt.png", "description": "Prompt figure for b: the image shows a change from left to right. On the left is a six-membered chair heterocycle containing a ring oxygen near the lower-right position and bearing an axial Me group on the right-hand carbon; its SMILES is 'C[C@H]1OCCCC1'. A one-way arrow points right. On the right is the same chair heterocycle with the ring oxygen in the same position and an equatorial Me group; its SMILES is 'C[C@@H]1OCCCC1'."}, {"file": "Exam-15-T8-4-2-prompt.png", "description": "Prompt figure for c: the image shows a change from left to right. On the left is a six-membered chair heterocycle containing a ring oxygen at the lower-left position and bearing an axial Me group on the right-hand carbon; its SMILES is 'C[C@H]1COCCC1'. A one-way arrow points right. On the right is the same chair heterocycle with the ring oxygen in the same position and an equatorial Me group; its SMILES is 'C[C@@H]1COCCC1'."}, {"file": "Exam-15-T8-4-3-prompt.png", "description": "Prompt figure for d: the image shows a change from left to right. On the left is a six-membered chair heterocycle containing a ring oxygen at the far-left position and bearing an axial Me group on the right-hand carbon; its SMILES is 'CC1CCOCC1'. A one-way arrow points right. On the right is the same chair heterocycle with the ring oxygen in the same position and an equatorial Me group; its SMILES is likewise 'CC1CCOCC1'."}, {"file": "Exam-15-T8-4-4-prompt.png", "description": "The image shows a change from left to right. On the left is a six-membered chair heterocycle containing an NH group near the lower-right position and bearing an axial Me group on the right-hand carbon; its SMILES is 'C[C@H]1NCCCC1'. A one-way arrow points right. On the right is the same chair heterocycle with the NH group in the same position and an equatorial Me group; its SMILES is 'C[C@@H]1NCCCC1'."}], "full_mark": 3, "rubric": ["bedac (3 points)", "3 points for a completely correct answer."], "rubric_images": []}]} {"problem_id": "Exam-15-T9", "header": "Problem 9 Ketene Chemistry. Ketenes are highly reactive, unstable intermediates with diverse synthetic methods and broad industrial applications.", "header_images": [], "subquestions": [{"sub_id": "Exam-15-T9-1", "tags": ["2.3", "2.4"], "text_open": "9.2 Thermal electrocyclic ring opening of cyclobutenones is an important method for generating conjugated polyenyl ketene intermediates. Under thermal conditions, the following substrate undergoes a cascade reaction to give fused-ring product A. 9.2.1 Write the SMILES structure of product A, including stereochemistry.", "text_images": [{"file": "Exam-15-T9-1-1-prompt.png", "description": "The substrate has structure SMILES 'OC(C1=O)(C(OC)=C1OC)C2=C(COC)CCC2'. It is a mixture of enantiomers whose stereocenter is the carbon bearing the hydroxy group, with a wavy bond between that carbon and the hydroxy group. Under 120°C microwave conditions it gives A."}], "full_mark": 2, "rubric": ["'[H][C@]1(CCC2)[C@]2(COC)C(C(OC)=C(C1=O)OC)=O' (2 points)."], "rubric_images": [{"file": "Exam-15-T9-1-1-rubric.png", "description": "Rubric structure for A: corresponding SMILES '[H][C@]1(CCC2)[C@]2(COC)C(C(OC)=C(C1=O)OC)=O'."}]}, {"sub_id": "Exam-15-T9-2", "tags": ["2.3", "2.4"], "text_open": "9.2.2 Write the SMILES structure of B.", "text_images": [{"file": "Exam-15-T9-2-1-prompt.png", "description": "The image shows a reaction sequence and reagent structure. At far left is compound 'OC(C1=O)(C(OC)=C1OC)C2=C(OC)C(C(C)(C)C)CCC2', a stereoisomeric mixture with 2 stereocenters: the carbon bearing hydroxy and the carbon bearing tert-butyl; wavy bonds connect the two stereogenic carbons respectively to hydroxy and tert-butyl. At `120 °C, microwave`, it forms **B**. **B** then reacts under `DMP` to give 'CC(C)(C)C(CCCC34C(C(OC)=C(C4=O)OC)=O)=C3OC', labeled **C**. At far right is the DMP reagent, with SMILES 'O=C(OI(OC(C)=O)(OC(C)=O)5OC(C)=O)C6=C5C=CC=C6' and label \"DMP\"."}], "full_mark": 2, "rubric": ["'CC(C)(C1=C(OC)C2(C(C(OC)=C(OC)C2O)=O)CCC1)C' (2 points)."], "rubric_images": [{"file": "Exam-15-T9-2-1-rubric.png", "description": "Rubric structure for B: corresponding SMILES 'CC(C)(C1=C(OC)C2(C(C(OC)=C(OC)C2O)=O)CCC1)C'."}]}, {"sub_id": "Exam-15-T9-3", "tags": ["2.3", "2.4"], "text_open": "A slight substrate modification leads through a different pathway to a product of another structure. Thermal treatment of the following substrate generates a ketene that reacts differently; product B is ultimately oxidized to spirocyclic C. 9.2.3 Write the SMILES structures of the 2 key intermediates in conversion of this substrate to B.", "text_images": [], "full_mark": 4, "rubric": ["'CC(C)(C)C1CCCC(/C(O)=C(C(OC)=C=O)/OC)=C1OC' and 'CC(C)(C)C(CCCC23C(O)=C(C(OC)=C3O)OC)=C2OC' (2 points each)."], "rubric_images": [{"file": "Exam-15-T9-3-1-rubric.png", "description": "The two key intermediates in conversion to B have SMILES 'CC(C)(C)C1CCCC(/C(O)=C(C(OC)=C=O)/OC)=C1OC' and 'CC(C)(C)C(CCCC23C(O)=C(C(OC)=C3O)OC)=C2OC', respectively."}]}, {"sub_id": "Exam-15-T9-4", "tags": ["2.3", "2.4"], "text_open": "9.3 The following photolysis gives mainly products D and E. Mechanistically, the substrate is photoexcited and then cleaves a C-C bond to form diradical F. Radical coupling then forms ketene intermediate G, which is ultimately converted to D and E. 9.3.1 Write the SMILES structures of F and G.", "text_images": [{"file": "Exam-15-T9-4-1-prompt.png", "description": "The image shows a photochemical reaction. Molecule 'O=C1[C@@](N(N([C@H]2C=C1)C(OCC)=O)C(OCC)=O)(C=C2)OC' reacts under hv in `MeOH` (SMILES 'CO') to give two products. The left product, 'O=C(C[C@@H]3[C@]([C@]3([H])N(N4C(OCC)=O)C(OCC)=O)([H])C=C4OC)OC', is labeled **D**. The right product, 'O=C5C[C@]6([H])[C@]([C@]6([H])N(N7C(OCC)=O)C(OCC)=O)([H])[C@@]5([H])C7(OC)OC', is labeled **E**. A plus sign separates the products."}], "full_mark": 4, "rubric": ["F: 'O=[C]/C=C\\[C@@H]1N(N([C](C=C1)OC)C(OCC)=O)C(OCC)=O' or '[H][C@@](N(N2C(OCC)=O)C(OCC)=O)([CH]C=C=O)[CH]C=C2OC' (2 points); G: '[H][C@@](N(N3C(OCC)=O)C(OCC)=O)(C4C=C=O)C4C=C3OC' (2 points)."], "rubric_images": [{"file": "Exam-15-T9-4-1-rubric.png", "description": "Rubric structures for F and G: F is a diradical intermediate with SMILES 'O=[C]/C=C\\[C@@H]1N(N([C](C=C1)OC)C(OCC)=O)C(OCC)=O' or '[H][C@@](N(N2C(OCC)=O)C(OCC)=O)([CH]C=C=O)[CH]C=C2OC'; G is the ketene intermediate with SMILES '[H][C@@](N(N3C(OCC)=O)C(OCC)=O)(C4C=C=O)C4C=C3OC'."}]}, {"sub_id": "Exam-15-T9-5", "tags": ["2.3", "2.4"], "text_open": "9.3.2 If MeOH is replaced with deuterated $CD_3OD$, write the SMILES structures of deuterated products D-d and E-d.", "text_images": [], "full_mark": 4, "rubric": ["D-d: 'O=C(C([2H])[C@]1([H])[C@]([C@]1([H])N(N2C(OCC)=O)C(OCC)=O)([H])C=C2OC)OC([2H])([2H])[2H]' (2 points); E-d: 'O=C3C([2H])[C@]4([H])[C@]([C@]4([H])N(N5C(OCC)=O)C(OCC)=O)([H])[C@@]3([H])C5(OC([2H])([2H])[2H])OC' (2 points)."], "rubric_images": [{"file": "Exam-15-T9-5-1-rubric.png", "description": "Rubric structures for D-d and E-d: D-d SMILES 'O=C(C([2H])[C@]1([H])[C@]([C@]1([H])N(N2C(OCC)=O)C(OCC)=O)([H])C=C2OC)OC([2H])([2H])[2H]'; E-d SMILES 'O=C3C([2H])[C@]4([H])[C@]([C@]4([H])N(N5C(OCC)=O)C(OCC)=O)([H])[C@@]3([H])C5(OC([2H])([2H])[2H])OC'."}]}, {"sub_id": "Exam-15-T9-6", "tags": ["2.3", "2.4"], "text_open": "9.4 Radical reactions can also construct fused-ring centers bearing a quaternary carbon, as shown below. The reaction is known to proceed through a ketene radical intermediate. Write the SMILES structures of any 2 key intermediates.", "text_images": [{"file": "Exam-15-T9-6-1-prompt.png", "description": "The image shows a reaction equation. 'C#CCCCCCC', drawn as $C_6H_13$ attached to a carbon-carbon triple bond, reacts with 'C45=NCCCN4CCCCC5'. Above the arrow are $^{n}Bu_3SnH$, AIBN; below are CO (80 atm) and benzene, 80°C. The product is 'O=C1N2C(N3CCC2)(CCCCC3)C=C1CCCCCC'."}], "full_mark": 4, "rubric": ["4 rubric intermediates for 9.4 (4 points; any 2 suffice): first, 'CCCCCC/[C]=C/[Sn](CCCC)(CCCC)CCCC'; second, 'O=[C]/C(CCCCCC)=C\\[Sn](CCCC)(CCCC)CCCC' (equivalent structure 'O=C=C([CH][Sn](CCCC)(CCCC)CCCC)CCCCCC'); third, '[O-]/C([N+]1=C(CCCCC2)N2CCC1)=C([CH][Sn](CCCC)(CCCC)CCCC)\\CCCCCC'; fourth, 'O=C(N3CCCN4C35CCCCC4)[C](C5[Sn](CCCC)(CCCC)CCCC)CCCCCC'."], "rubric_images": [{"file": "Exam-15-T9-6-1-rubric.png", "description": "4 rubric intermediates for 9.4: 'CCCCCC/[C]=C/[Sn](CCCC)(CCCC)CCCC'; 'O=[C]/C(CCCCCC)=C\\[Sn](CCCC)(CCCC)CCCC', with equivalent structure 'O=C=C([CH][Sn](CCCC)(CCCC)CCCC)CCCCCC'; '[O-]/C([N+]1=C(CCCCC2)N2CCC1)=C([CH][Sn](CCCC)(CCCC)CCCC)\\CCCCCC'; and 'O=C(N3CCCN4C35CCCCC4)[C](C5[Sn](CCCC)(CCCC)CCCC)CCCCCC'."}]}, {"sub_id": "Exam-15-T9-7", "tags": ["2.3", "2.4"], "text_open": "9.5 Ketenes may also react with other highly reactive intermediates. In the following reaction, active species I forms together with ketene H; H and I react with dimethyl acetylenedicarboxylate to give the product. 9.5.1 Write the SMILES structure of I.", "text_images": [{"file": "Exam-15-T9-7-1-prompt.png", "description": "The image shows a three-component reaction. Molecules 'O=C(C(C6=CC=CC=C6)=C7C8=CC=CC=C8)C7=O', 'CC(O1)(C)N=NC1(OC)OC', and vertically drawn 'O=C(C#CC(OC)=O)OC', shown as a carbon-carbon triple bond bearing $CO_2Me$ at both ends, are separated by plus signs. They react to form complex spirocyclic product 'O=C(C(C2=CC=CC=C2)=C(C3=CC=CC=C3)C45OC(C(C(OC)=O)=C5C(OC)=O)(OC)OC)C4(OC)OC'."}], "full_mark": 2, "rubric": ["'CO[C]OC' (2 points)."], "rubric_images": [{"file": "Exam-15-T9-7-1-rubric.png", "description": "Rubric structure for I: dimethoxycarbene, with SMILES 'CO[C]OC'."}]}, {"sub_id": "Exam-15-T9-8", "tags": ["2.3", "2.4"], "text_open": "9.5.2 Write the SMILES structures of the key intermediates in conversion of H and I to the product.", "text_images": [], "full_mark": 4, "rubric": ["Key intermediates for 9.5.2 (4 points; any 1 of the first 2 may be written): first, 'O=C=C(C1=CC=CC=C1)/C(C2=CC=CC=C2)=C([O-])\\C(OC)=[O+]C'; second, 'O=C(C(C3=CC=CC=C3)=C(C4=CC=CC=C4)C5=O)C5(OC)OC'; third, 'O=C(/C(C(OC)=[O+]C)=[C-]\\C(OC)=O)OC'."], "rubric_images": [{"file": "Exam-15-T9-8-1-rubric.png", "description": "Rubric key intermediates in conversion of H and I to product: first, 'O=C=C(C1=CC=CC=C1)/C(C2=CC=CC=C2)=C([O-])\\C(OC)=[O+]C'; second, 'O=C(C(C3=CC=CC=C3)=C(C4=CC=CC=C4)C5=O)C5(OC)OC'; third, 'O=C(/C(C(OC)=[O+]C)=[C-]\\C(OC)=O)OC'."}]}, {"sub_id": "Exam-15-T9-9", "tags": ["1.4", "2.2"], "text_open": "9.6 Ketenes may also be connected with transition-metal carbonyl complexes. 9.6.1 Write the by-product formed in the first conversion.", "text_images": [{"file": "Exam-15-T9-9-1-prompt.png", "description": "The image shows a multistep reaction and a group-equivalence definition. The initial reactant is formed by coordination of the carbon-carbon double bond in ligand 'O=C(/C=C/C1=NC=CC=C1)[Fc]' to iron carbonyl fragment '[Fe](C#O)(C#O)(C#O)(C#O)', i.e. $Fe(CO)_4$. Under `MeLi, CO`, it forms an intermediate in which two carbon-carbon double bonds of the ligand coordinate to iron carbonyl fragment '[Fe](C#O)(C#O)(C#O)', i.e. $Fe(CO)_3$; the full intermediate SMILES is '[Fc]C(/C=C/C1=NC=CC=C1)=C=O.[O+]#[C-][Fe]([C-]#[O+])[C-]#[O+]'. Heating under `Δ` then gives product **J**, with molecular formula $C_19H_15FeNO$. At far right, `Fc ≡` is defined as the ferrocenyl structure."}], "full_mark": 1, "rubric": ["$MeCO_2Li$ (1 point)"], "rubric_images": []}, {"sub_id": "Exam-15-T9-10", "tags": ["1.4", "2.4"], "text_open": "9.6.3 Using the molecular-formula clue, write the SMILES structure of final product J.", "text_images": [], "full_mark": 2, "rubric": ["'[Fc]C1=CC=C(C=CC=C2)N2C1=O' (2 points)."], "rubric_images": [{"file": "Exam-15-T9-10-1-rubric.png", "description": "Rubric structure for J: corresponding SMILES '[Fc]C1=CC=C(C=CC=C2)N2C1=O'."}]}]} -{"problem_id": "Exam-16-T3", "header": "Problem 3: Photopromoted Polymerization", "header_images": [], "subquestions": [{"sub_id": "Exam-16-T3-1", "tags": ["2.2", "2.3", "2.4"], "text_open": "3.1 On 2023-3-24, researchers from Istanbul Technical University published a study of visible-light-induced cationic step-growth polymerization. 3.1.1 In a direct initiation system for cationic polymerization, an onium salt can absorb light directly to initiate the reaction. An advantage of this initiation mode is that it generates an acidic environment in situ, further promoting the reaction. On this basis, give all products of the initiation reaction shown below.", "text_images": [{"file": "Exam-16-T3-1-1-prompt.png", "description": "The figure shows a reaction in which the species on the left is a salt composed of a circled negatively charged `PF_6^-` anion and a circled positively charged cation; its corresponding SMILES is '[Ar][I+][Ar].F[P-](F)(F)(F)(F)F'. The salt reacts to the right under `hv` and `300 nm` shown above the arrow and `R-H` shown below the arrow."}], "full_mark": 4, "rubric": ["ArI (1 point), HPF_{6} (1 point), R· (1 point), Ar· (1 point)", "If more than 4 products are listed, only the first 4 are considered; incorrect entries do not incur deductions."], "rubric_images": []}, {"sub_id": "Exam-16-T3-2", "tags": ["2.2", "2.4"], "text_open": "3.1.2 Give the SMILES structures of 1 key active intermediate derived from the monomer and of the product in this reaction.", "text_images": [{"file": "Exam-16-T3-2-1-prompt.png", "description": "The figure shows a reaction in which the leftmost molecule is 'C1=CSC=C1', followed by the label \"(2 equiv)\". It reacts with a salt shown at the upper right, composed of a circled negatively charged `PF_6^-` anion and a circled positively charged cation; the corresponding SMILES of the salt is 'C1([I+]C2=CC=CC=C2)=CC=CC=C1.F[P-](F)(F)(F)(F)F'. The reaction proceeds to the right under `hv` shown below the arrow."}], "full_mark": 4, "rubric": ["'C1=CC=C[S+]1' (2 points), 'C1(C2=CC=CS2)=CC=CS1' (2 points)"], "rubric_images": [{"file": "Exam-16-T3-2-1-rubric.png", "description": "Rubric structure 1: a thiophene radical cation/key active intermediate. The structure in the image has the SMILES 'C1=CC=C[S+]1'."}, {"file": "Exam-16-T3-2-2-rubric.png", "description": "Rubric structure 2: the bithiophene product. The structure in the image has the SMILES 'C1(C2=CC=CS2)=CC=CS1'."}]}]} -{"problem_id": "Exam-17-T2", "header": "Problem 2 Determination of Paracetamol by Cyclic Voltammetry and Study of Its Electrochemistry\nCyclic voltammetry (CV) is an important electrochemical technique for investigating electrochemical reaction mechanisms and performing quantitative analysis. It uses a three-electrode system consisting of a working electrode, a reference electrode, and an auxiliary electrode. A triangular-wave potential that varies linearly with time is applied to the working electrode while the response current is recorded. One triangular-wave scan completes both an oxidation scan and a reduction scan, and the resulting current–potential curve is called a cyclic voltammogram. A substance that can undergo oxidation or reduction at an electrode within the scanned potential range is called an electroactive substance. A typical cyclic voltammogram exhibits an oxidation peak (anodic peak) and a reduction peak (cathodic peak). Under suitable conditions, the peak currents ($I_{pa}$ and $I_{pc}$) are proportional to the concentration of the electroactive substance and form the basis of quantitative analysis. The peak potentials ($E_{pa}$ and $E_{pc}$) and the peak-potential separation ($\\Delta E_p = E_{pa} - E_{pc}$) can be used to assess reaction reversibility and the number of electrons transferred. Changing the scan rate ($v$) can distinguish diffusion control from adsorption control and reveal the kinetics of subsequent chemical reactions.\nParacetamol (APAP, also known as acetaminophen) is a commonly used antipyretic analgesic. Studies show that APAP undergoes electrochemical oxidation at an electrode surface, and its oxidation product NAPQI undergoes subsequent chemical reactions in acidic media. A research group used cyclic voltammetry to investigate the electrochemical behavior of APAP systematically.\nAt room temperature, the peak current ($I_p$, in A) for a reversible electrochemical reaction in a quiescent solution at a planar electrode is related to the scan rate ($v$, in V/s), the concentration of the electroactive substance ($C$, in $\\mathrm{mol/cm^3}$), and other parameters by the Randles–Sevcik equation:\n$$I_p = 2.69 \\times 10^5 n^{3/2} A C D^{1/2} v^{1/2}$$\nn: number of electrons transferred in the reaction.\nA: effective area of the working electrode (unit: $\\mathrm{cm^2}$).\nD: diffusion coefficient of the electroactive substance (unit: $\\mathrm{cm^2/s}$), reflecting how rapidly the substance moves through solution.\nThe figure below is a typical cyclic voltammogram. Curve 1 is the background signal, and curve 2 is the normal response signal.", "header_images": [{"file": "Exam-17-T2-header-1.png", "description": "Typical cyclic voltammogram. The horizontal axis is potential, with tick marks of approximately 0.2, 0.1, 0, -0.1, and -0.2 from left to right; the vertical axis is current. The peak-current and peak-potential symbols $I_{pc}$, $I_{pa}$, $E_{pa}$, and $E_{pc}$ are labeled. Two closed curves are labeled 1 and 2. As stated in the question, curve 1 is the background signal and curve 2 is the normal response signal."}], "subquestions": [{"sub_id": "Exam-17-T2-1", "tags": ["4.7"], "text_open": "2.2 At a scan rate of $40\\ \\mathrm{mV\\cdot s^{-1}}$ in a pH = 2.2 buffer, the research group obtained the following cyclic-voltammetry data for standard APAP solutions:\n\n| APAP concentration/(mmol/L) | 0.10 | 0.40 | 1.00 | 3.00 | 5.00 |\n| ----------------- | ----- | ----- | ----- | ----- | ----- |\n| Anodic peak current $I_{pa}$/μA | 2.15 | 5.82 | 13.46 | 38.74 | 63.52 |\n\nUsing concentration as the horizontal coordinate and anodic peak current as the vertical coordinate, obtain the linear regression equation of $I_{pa}$ against APAP concentration and calculate the expected peak current for an APAP solution with a concentration of $3.50\\ \\mathrm{mmol\\cdot L^{-1}}$.", "text_images": [], "full_mark": 5, "rubric": ["With concentration as $x$ (unit: $\\mathrm{mmol\\cdot L^{-1}}$) and peak current as $y$ (unit: μA), least-squares fitting gives the linear regression equation $y = 12.55x + 0.90$, $R^2 = 1.0000$ (3 points). When $x = 3.50\\ \\mathrm{mmol\\cdot L^{-1}}$: $I_{pa} = 12.55 \\times 3.50 + 0.90 = 44.8\\ \\mu\\mathrm{A}$ (2 points; the number of significant figures is not required)."], "rubric_images": []}, {"sub_id": "Exam-17-T2-2", "tags": ["2.3"], "text_open": "2.3 To investigate the oxidation mechanism of APAP, the group performed CV scans at $40\\ \\mathrm{mV\\cdot s^{-1}}$ in media of different acidity (pH = 6.0, pH = 2.2, and $1.8\\ \\mathrm{mol\\cdot L^{-1}}\\ H_2SO_4$) and observed the following:\nAt pH = 6.0, both oxidation and reduction peaks appear, but $I_{pa}$ ≠ $I_{pc}$; at pH = 2.2, only the oxidation peak appears and the reduction peak disappears; under strongly acidic conditions ($1.8\\ \\mathrm{mol\\cdot L^{-1}}\\ H_2SO_4$), the reduction peak reappears.\n2.3.1 Explain why the reduction peak disappears at pH = 2.2.", "text_images": [], "full_mark": 2, "rubric": ["At pH = 2.2, the APAP oxidation product NAPQI rapidly undergoes addition of water to form an electrochemically inactive NAPQI hydrate. This substance cannot be reduced within the scanned potential range, so the reduction peak disappears."], "rubric_images": []}, {"sub_id": "Exam-17-T2-3", "tags": ["2.3"], "text_open": "2.3.2 What chemical reaction is indicated by the reappearance of the reduction peak under strongly acidic conditions?", "text_images": [], "full_mark": 2, "rubric": ["Under strongly acidic conditions, the NAPQI hydrate is further hydrolyzed to benzoquinone, which can be reduced at the electrode; the reduction peak therefore reappears (2 points)."], "rubric_images": []}]} +{"problem_id": "Exam-16-T6", "header": "Problem 6: Polysulfide Anions\nA sodium-sulfur battery is a high-temperature molten secondary alkali-metal energy-storage battery that uses metallic sodium as the negative electrode, elemental sulfur as the positive electrode, and molten sodium beta-alumina as the solid electrolyte. It offers high energy density, inexpensive raw materials, and suitability for large-scale energy storage, and has become a research focus in electrochemical energy storage in recent years. Charging and discharging proceed through the formation of sodium polysulfides (Na_{2}S_{x}) with different chain lengths.", "header_images": [], "subquestions": [{"sub_id": "Exam-16-T6-1", "tags": ["2.1", "2.4"], "text_open": "6.2 Sodium polysulfides (Na_{2}S_{x}) are key electroactive intermediates in sodium-sulfur batteries, and their anions can be synthesized and observed in the laboratory using common reagents. Under an N_{2} atmosphere, excess sodium tert-butoxide is added to a DMF solution of S_{8}, reducing S_{8} to polysulfide anion A. A disproportionates into another polysulfide anion B containing fewer atoms. Electron paramagnetic resonance (EPR) measurements show that some B anions undergo homolysis in solution to form bent anion C. C is trapped in zeolite cages in lazurite crystals, [(Na,Ca)_{4-8}(AlSiO_{4})_{6}(SO_{4},S,Cl)_{2}], and appears blue in solution. The sodium salt corresponding to polysulfide anion B is a common component of sodium-sulfur batteries. If it is heated to 300°C and melted, anion C dimerizes again to give isomer D of polysulfide anion B; the sulfur atoms in D occupy only two distinct chemical environments. 6.2.1 Give the chemical formulas of A and B.", "text_images": [], "full_mark": 2, "rubric": ["A: S_{8}^{2-} (1 point)", "B: S_{6}^{2-} (1 point)"], "rubric_images": []}, {"sub_id": "Exam-16-T6-2", "tags": ["1.6", "2.4"], "text_open": "6.2.2 Give the SMILES structures of C and D, showing the Lewis connectivity, and indicate the formal charge on every sulfur atom.", "text_images": [], "full_mark": 3, "rubric": ["1 point for the Lewis structure of C and 2 points for that of D; a reasonable resonance structure is also acceptable."], "rubric_images": [{"file": "Exam-16-T6-2-1-rubric.png", "description": "Lewis structures of C and D in the rubric. C is S_3^- with SMILES 'S=[S][S-]'; D is an S_6^{2-} isomer with SMILES '[S-]S(S([S-])=S)=S'."}]}, {"sub_id": "Exam-16-T6-3", "tags": ["1.6", "2.4"], "text_open": "6.3 Anion C can also be observed when S_{8} is dissolved in liquid ammonia, together with another hydrogen-free anion E. Early studies proposed the following mechanism: first, liquid ammonia undergoes autoionization to form NH_{2}^{-} and NH_{4}^{+}; second, NH_{2}^{-} attacks S_{8}, giving S_{8}NH_{2}^{-} by addition; third, S_{8}NH_{2}^{-} decomposes into cyclic anion F and a stable dibasic acid; fourth, F rearranges to form anion E and S_{8}, while the dibasic acid reacts with S_{8} and NH_{4}^{+} to form (NH_{4})_{2}S_{x}, which is subsequently converted into C. EPR measurements, however, show that anion F is not formed during the reaction. Moreover, the tendency of liquid ammonia to autoionize is extremely small and cannot explain the high solubility of S_{8} in liquid ammonia. Researchers therefore proposed a more plausible mechanism: first, S_{8}NH_{2}^{-} and NH_{3} participate in a proton-transfer equilibrium, forming S_{8}NH_{3} while generating NH_{2}^{-}, which promotes dissolution; second, NH_{3} can also react by nucleophilic substitution to form the small molecule S_{2}NH_{3}; third, S_{8}NH_{2}^{-} decomposes to anion B and intermediate G; fourth, G transfers all of its protons to NH_{3} to form anion H; fifth, H reacts with S_{2}NH_{3} to give anion E and NH_{3}, with all species in this reaction having a stoichiometric coefficient of 1. 6.3.1 Give the SMILES structures of E, F, G, and H, showing the Lewis connectivity. The Lewis structure of E is known to contain two S-S single bonds, and its sulfur atoms occupy two distinct chemical environments.", "text_images": [], "full_mark": 5, "rubric": ["1 point each for the Lewis structures of E, G, and H, and 2 points for that of F; a reasonable resonance structure is also acceptable."], "rubric_images": [{"file": "Exam-16-T6-3-1-rubric.png", "description": "Lewis structures of E, F, G, and H in the rubric. The SMILES of E is '[S-]S/N=[S+]/[S-]'; the SMILES of F is 'S1SS[N-]SSSS1'; the SMILES of G is 'S=[S+]N([H])[H]'; and the SMILES of H is 'S=S=[N-]'."}]}, {"sub_id": "Exam-16-T6-4", "tags": ["2.1", "2.2"], "text_open": "6.3.2 In fact, the fifth reaction proceeds in three steps, and both the first and second steps generate SNH_{3}. Write the chemical equations for the three steps in order.", "text_images": [], "full_mark": 3, "rubric": ["Step 1: S_{2}N^{-} + S_{2}NH_{3} = S_{3}N^{-} + SNH_{3} (1 point)", "Step 2: S_{3}N^{-} + S_{2}NH_{3} = S_{4}N^{-} + SNH_{3} (1 point)", "Step 3: 2SNH_{3} = S_{2}NH_{3} + NH_{3} (1 point)"], "rubric_images": []}, {"sub_id": "Exam-16-T6-5", "tags": ["3.2", "3.4"], "text_open": "6.4 Anion C is also relatively common in nature and occurs mainly in ores formed under high-temperature, high-pressure conditions. Calculations show that at 350℃ and 0.5 GPa, anion C can exist as a stable form of sulfur. Reaction 1 (unbalanced) is: H_2S(aq.) + SO_4^{2-}(aq.) + H^+(aq.) → C(aq.) + O_2(g.) + H_2O(l.), \\lg K_1 = -9.6 (when the stoichiometric coefficients of C and H^+ are both 1). The standard enthalpies of formation and standard entropies of several species are listed below (ignore the temperature dependence of enthalpy and entropy and ignore microscopic effects): H_2S(aq.): \\Delta_fH_m^⊖ = -39.0 kJ·mol^{-1}, S_m^⊖ = 122.0 J·mol^{-1}·K^{-1}; HS^-(aq.): -17.7, 61.1; HSO_4^-(aq.): -885.8, 126.9; SO_4^{2-}(aq.): -907.5, 17.0; H^+(aq.): 0, 0; O_2(g.): 0, 205.1; H_2O(l.): -285.8, 70.0.\n6.4.2 Derive the equation of the bold sloping line l_1 in the figure below. Hint: f_O2 denotes the fugacity of oxygen, which physically represents the effective pressure of a real gas. The total dissolved sulfur concentration in the figure is 1 wt% (that is, all dissolved sulfur-containing species in 1 kg of solution contain a total of 10 g of S atoms). Assume that the density of water under these conditions is ρ = 1.0 × 10^3 kg/m^3.", "text_images": [{"file": "Exam-16-T6-5-1-prompt.png", "description": "A two-dimensional predominance-diagram-style plot with a white background, black lines, and a rectangular frame. The top and right sides of the frame also have short tick marks. The horizontal axis is labeled “pH,” with major tick labels 1, 2, 3, 4, 5, 6, 7, 8, and 9; the axis extends beyond 9 without a clearly labeled 10. The vertical axis is labeled “$\\log_{10} f_{O_2}$,” with major ticks −22, −24, −26, −28, −30, −32, and −34 from top to bottom. Several solid black lines divide the plot into regions labeled with formulas or letters. The upper-left region is labeled “HSO_4^-,” the upper-right “SO_4^2-,” the lower-left “H_2S,” and the lower-right “HS^-.” Near the center-left is a light-gray irregular polygon labeled with a capital “C.” It has approximately five sides: an upper-left sloping boundary from about pH 1 and $\\log_{10} f_{O_2}$ about −24 to about pH 5.3 and $\\log_{10} f_{O_2}$ about −26; an upper-right sloping boundary from about pH 5.3 and $\\log_{10} f_{O_2}$ about −26 to about pH 7.6 and $\\log_{10} f_{O_2}$ about −28.4; a thinner lower-right boundary from about pH 7.6 and $\\log_{10} f_{O_2}$ about −28.4, extending down-left to about pH 6.0 and $\\log_{10} f_{O_2}$ about −31; a noticeably bold lower boundary from about pH 6.0 and $\\log_{10} f_{O_2}$ about −31, extending up-left to about pH 1 and $\\log_{10} f_{O_2}$ about −26.8; and a left boundary adjoining the vertical axis. A vertical solid line at about pH 5.3 runs from the top border to near the upper vertex of the gray region and separates HSO_4^- from SO_4^2-. Another vertical solid line at about pH 6.0 runs downward from the lower vertex of the gray region to the bottom border and separates H_2S from HS^-. A thinner sloping line also extends down-right from the upper-right vertex of the gray region to near the right border, separating SO_4^2- from HS^-."}], "full_mark": 4, "rubric": ["First balance Reaction 1 with the coefficient of C (S_{3}^{-}) set to 1:", "Reaction 1: \\[\\text{2}{{\\text{H}}_{\\text{2}}}\\text{S + SO}_{\\text{4}}^{\\text{2-}}\\text{ + }{{\\text{H}}^{\\text{+}}}\\rightleftharpoons \\text{S}_{\\text{3}}^{\\text{-}}\\text{ + 0}\\text{.75}{{\\text{O}}_{\\text{2}}}\\text{ + 2}\\text{.5}{{\\text{H}}_{\\text{2}}}\\text{O}\\quad \\lg {{K}_{1}}=-9.6\\] (1 point)", "Line l_{1} is the boundary between H_{2}S and S_{3}^{-}, represented by:", "Reaction 2: \\[\\text{3}{{\\text{H}}_{\\text{2}}}\\text{S + 1}\\text{.25}{{\\text{O}}_{\\text{2}}}\\rightleftharpoons \\text{S}_{\\text{3}}^{\\text{-}}\\text{ + }{{\\text{H}}^{\\text{+}}}\\text{ + 2}\\text{.5}{{\\text{H}}_{\\text{2}}}\\text{O}\\]", "By Hess’s law, subtracting Reaction 2 from Reaction 1 gives:", "Reaction 3: \\[\\text{SO}_{\\text{4}}^{\\text{2-}}\\text{ + 2}{{\\text{H}}^{\\text{+}}}\\rightleftharpoons {{\\text{H}}_{\\text{2}}}\\text{S + 2}{{\\text{O}}_{\\text{2}}}\\]", "From the given data, \\[{{\\Delta }_{\\text{r}}}H{{_{\\text{m}}^{{}}}^{\\ominus }}=\\sum{{{\\mathbf{\\Delta }}_{\\mathbf{f}}}{{H}_{\\mathbf{m}}}^{\\ominus }-}\\text{ }\\sum{{{\\mathbf{\\Delta }}_{\\mathbf{f}}}{{H}_{\\mathbf{m}}}^{\\ominus }}=868.5\\text{ kJ/mol}\\];", "\\[{{\\Delta }_{\\text{r}}}{{S}_{\\text{m}}}^{\\ominus }=\\sum{{{S}_{\\mathbf{m}}}^{\\ominus }-}\\sum{{{S}_{\\mathbf{m}}}^{\\ominus }}=515.2\\text{ J/(mol}\\cdot \\text{K)}\\];", "\\[{{\\Delta }_{\\text{r}}}{{G}_{\\text{m}}}^{\\ominus }=868.5-623.15\\times \\text{(0}\\text{.5152)}=547.45\\text{ kJ/mol}\\]; \\[\\lg {{K}_{3}}\\text{(}{{\\text{H}}_{\\text{2}}}\\text{S}{{\\text{O}}_{\\text{4}}}\\text{)}=\\frac{-{{\\Delta }_{\\text{r}}}{{G}_{\\text{m}}}^{\\ominus }}{2.303RT}=-45.88\\]", "Thus, lg K_{2} = lg K_{1} - lg K_{3} = 36.28 (1 point)", "At a total sulfur concentration of 1 wt%, the molar concentration is 0.3125 mol/kg. At the boundary, elemental conservation gives:", "\\[[\\mathrm{H_2S}] = 3[\\mathrm{S_3^-}]\\]", "Therefore:", "\\[[\\mathrm{H_2S}] = 0.156\\ \\mathrm{mol\\,L^{-1}},\\quad [\\mathrm{S_3^-}] = 0.052\\ \\mathrm{mol\\,L^{-1}}\\] (1 point)", "Substitution into the equilibrium-constant expression gives:", "\\[36.28 = \\lg[\\mathrm{S_3^-}] - 3\\lg[\\mathrm{H_2S}] - \\mathrm{pH} - 1.25\\lg f_{\\mathrm{O_2}}\\]", "Therefore:", "\\[\\lg {{f}_{{{O}_{2}}}}=-28.11-0.8\\text{pH}\\]", "(1 point; \\[\\text{pH}=-1.25\\text{ }lg\\text{ }{{f}_{{{\\text{O}}_{\\text{2}}}}}-35.14\\] is also acceptable)"], "rubric_images": []}, {"sub_id": "Exam-16-T6-6", "tags": ["2.2", "4.2"], "text_open": "6.5 Anion C also has many applications in organic synthesis. Use the results of the following control experiments to answer Questions 6.5.1-6.5.4. 6.5.1 According to the control experiments, from which species formed by conversion of elemental sulfur might the sulfur in the product originate?", "text_images": [{"file": "Exam-16-T6-6-1-prompt.png", "description": "Control experiments for 6.5.1, showing two parallel reactions. In the upper reaction, 'C1=CC=C(C#CC#CC2=CC=CC=C2)C=C1' (0.2 mmol) reacts with `S` (0.6 mmol) under 1.8 mmol NaOtBu in DMF, N_2, at 25°C to give 'C1=CC=C(C2=CC=C(C3=CC=CC=C3)S2)C=C1' in 65% yield. In the lower reaction, 'C1=CC=C(C#CC#CC2=CC=CC=C2)C=C1' (0.2 mmol) reacts with `S` (0.6 mmol) under 1.8 mmol NaO^tBu and 4.0 equiv TEMPO in ^tBuOH+DMF, N_2, at 25°C to give 'C1=CC=C(C2=CC=C(C3=CC=CC=C3)S2)C=C1' in 5% yield."}], "full_mark": 1, "rubric": ["S_{3}^{-} (1 point)"], "rubric_images": []}, {"sub_id": "Exam-16-T6-7", "tags": ["2.3", "2.7"], "text_open": "6.5.2 In the control experiments above, no radical intermediate was trapped in the first reaction, indicating that it does not follow a radical mechanism, whereas the third reaction proceeds through a radical mechanism. Explain, from the perspective of molecular structure, why the two reactions follow different mechanisms.", "text_images": [{"file": "Exam-16-T6-7-1-prompt.png", "description": "Control experiments for 6.5.2, showing four parallel reactions. In the first, 'C1=CC=C(C#CC#CC2=CC=CC=C2)C=C1' (0.2 mmol) reacts with `Na_2S · 9H_2O` (0.6 mmol) in DMF, air, at 25°C to give 'C1=CC=C(C2=CC=C(C3=CC=CC=C3)S2)C=C1' in 99% yield. In the second, 'C1=CC=C(C#CC#CC2=CC=CC=C2)C=C1' (0.2 mmol) reacts with `Na_2S · 9H_2O` (0.6 mmol) with 4.0 equiv TEMPO in DMF, air, at 25°C to give 'C1=CC=C(C2=CC=C(C3=CC=CC=C3)S2)C=C1' in 92% yield. In the third, 'C1=CC=C(C#CC2=CC=CC=C2)C=C1' (0.6 mmol) reacts with `K_2S` (0.36 mmol) in DMF, Ar, at 130°C to give 'C1(C2=CC=CC=C2)=C(C3=CC=CC=C3)C(C4=CC=CC=C4)=C(C5=CC=CC=C5)S1' in 65% yield. In the fourth, 'C1=CC=C(C#CC2=CC=CC=C2)C=C1' (0.6 mmol) reacts with `K_2S` (0.36 mmol) with 4.0 equiv TEMPO in DMF, Ar, at 130°C to give only a trace of 'C1(C2=CC=CC=C2)=C(C3=CC=CC=C3)C(C4=CC=CC=C4)=C(C5=CC=CC=C5)S1'."}], "full_mark": 2, "rubric": ["The substrate in the first reaction, 1,4-diphenylbutadiyne, contains a conjugated diyne system. Its low-energy LUMO and strong electrophilicity make it highly susceptible to direct nucleophilic attack and polar addition by sulfide anions; it therefore follows an ionic mechanism. (1 point)", "The substrate in the third reaction, phenylacetylene, is an ordinary terminal alkyne with a higher-energy LUMO and weaker electrophilicity. It cannot be attacked directly by sulfide anions and must instead generate radicals through single-electron transfer or a related process at high temperature; it therefore follows a radical mechanism. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-16-T6-8", "tags": ["2.2", "2.3"], "text_open": "6.5.3 Give the SMILES structures of two key intermediates in the third reaction of the control experiments in 6.5.2.", "text_images": [], "full_mark": 4, "rubric": ["'C1=CC=C(C(SS[S-])=[C](C2=CC=CC=C2))C=C1' and '[S-]SSC(C1=CC=CC=C1)=C(C(C2=CC=CC=C2)=[C]C3=CC=CC=C3)C4=CC=CC=C4' (4 points; 2 points for each intermediate)"], "rubric_images": [{"file": "Exam-16-T6-8-1-rubric.png", "description": "Two key intermediates in the rubric. The structure on the left is a Ph-substituted vinyl radical bearing an -S-S-S^- polysulfide chain, with SMILES 'C1=CC=C(C(SS[S-])=[C](C2=CC=CC=C2))C=C1'. The structure on the right is a precyclization/cyclization-related intermediate with four Ph substituents, a radical center, and an -S-S-S^- group, with SMILES 'C1=CC=C(C(SS[S-])=C(C2=CC=CC=C2)C(C3=CC=CC=C3)=[C](C4=CC=CC=C4))C=C1'. In the figure, Ph denotes phenyl, whose SMILES is 'C1=CC=CC=C1'."}]}, {"sub_id": "Exam-16-T6-9", "tags": ["2.8", "2.3"], "text_open": "6.5.4 Explain separately why the yields of the following two reactions are substantially lower than that of the third reaction in 6.5.2.", "text_images": [{"file": "Exam-16-T6-9-1-prompt.png", "description": "Control experiments for 6.5.4, showing two parallel reactions. In the upper reaction, 'C1=CC=C(C#CC2=CC=CC=C2)C=C1' (0.6 mmol) reacts with `Na_2S · 9H_2O` (0.36 mmol) in DMF, Ar, at 130°C to give 'C1(C2=CC=CC=C2)=C(C3=CC=CC=C3)C(C4=CC=CC=C4)=C(C5=CC=CC=C5)S1' in 36% yield. In the lower reaction, 'C1=CC=C(C#CC2=CC=CC=C2)C=C1' (0.6 mmol) reacts with `S_8` (0.36 mmol) under NaO^tBu in DMF, Ar, at 130°C to give 'C1(C2=CC=CC=C2)=C(C3=CC=CC=C3)C(C4=CC=CC=C4)=C(C5=CC=CC=C5)S1' in less than 5% yield."}], "full_mark": 2, "rubric": ["First reaction: this reaction must proceed through a radical-anion pathway. Na_{2}S·9H_{2}O contains a large amount of water of crystallization, whose water molecules protonate and quench the highly reactive radical-anion intermediates, interrupting the radical chain that leads to ring closure. (1 point)", "Second reaction: elemental sulfur is converted into S_{3}^{-} under strongly basic conditions. At 130°C in a strongly basic medium, S_{3}^{-} readily undergoes extensive self-disproportionation or polymerization to form inert higher polysulfide complexes and therefore cannot participate effectively in the coupling. (1 point; full credit is awarded for identifying a side reaction of S_{3}^{-})"], "rubric_images": []}]} +{"problem_id": "Exam-16-T8", "header": "Problem 8: Asymmetric Catalytic Synthesis of Thiochromenes (8 points, 8%)\nOn 2016-9-26, researchers from Washington University published a study on the asymmetric catalytic synthesis of thiochromenes. Thiochromenes are sulfur-containing heterocycles. This reaction is highly stereospecific, meaning that the configuration of the chiral carbon bonded to the R group in the product is determined. Stereochemistry is not required in this problem; the SMILES or structural formulas in your answer need not specify the configuration of chiral carbons, alkenes, or other stereochemical features.", "header_images": [], "subquestions": [{"sub_id": "Exam-16-T8-1", "tags": ["2.2", "2.3", "2.4"], "text_open": "8.1 Give the SMILES structures of the key intermediates and the reaction product. Hint: the product contains a fused six-membered/six-membered ring system.", "text_images": [{"file": "Exam-16-T8-1-1-prompt.png", "description": "An organic reaction whose substrate has SMILES 'O=CC1=C(C=CC=C1)SC(C=C[R])=O', under the conditions 'C[C@@H]1[C@H](C2=CC=CC=C2)N=C3SC4=CC=CC=C4N3C1' (10 mol%), CHCl_3, room temperature."}], "full_mark": 6, "rubric": ["Key intermediates and reaction product:", "'O=C([Cat+])C=C[R]' (1 point); 'O=CC1=C([S-])C=CC=C1' or 'S=C1C=CC=CC1=C[O-]' (1 point); '[O-]C1C2=C(C=CC=C2)S[CH]([R])C1C([Cat+])=O' (1 point)", "'[O-]C1([Cat+])OC2C3=CC=CC=C3S[CH]([R])C12' (1 point); 'O=C1OC2C3=C(S[CH](C12)[R])C=CC=C3' (1 point); '[R][CH]1SC2=CC=CC=C2C=C1' (1 point)", "If the abbreviation Cat is used without specifying the structure of Cat, deduct 1 point from the total for this subquestion.", "The additional structure '[O-]C([Cat+])=C[CH]([R])SC1=CC=CC=C1C=O' earns no credit but incurs no deduction."], "rubric_images": [{"file": "Exam-16-T8-1-1-rubric.png", "description": "Rubric structure 1, with SMILES 'O=C([Cat+])/C=C/[R]'."}, {"file": "Exam-16-T8-1-2-rubric.png", "description": "Rubric structure 2, with SMILES 'O=CC1=C([S-])C=CC=C1'."}, {"file": "Exam-16-T8-1-3-rubric.png", "description": "Rubric structure 3, an equivalent answer to rubric structure 2, with SMILES 'S=C1C=CC=C/C1=C/[O-]'."}, {"file": "Exam-16-T8-1-4-rubric.png", "description": "Rubric structure 4, with SMILES '[O-]C1C2=C(C=CC=C2)S[C@H]([R])C1C([Cat+])=O'."}, {"file": "Exam-16-T8-1-5-rubric.png", "description": "Rubric structure 5, a peroxide/lactone-like cyclized intermediate, with SMILES '[O-]C1([Cat+])OC2C3=CC=CC=C3S[C@H]([R])C12'."}, {"file": "Exam-16-T8-1-6-rubric.png", "description": "Rubric structure 6, with SMILES 'O=C1OC2C3=C(S[C@@H](C12)[R])C=CC=C3'."}, {"file": "Exam-16-T8-1-7-rubric.png", "description": "Final product in the rubric, with SMILES '[R][C@H]1SC2=CC=CC=C2C=C1'."}, {"file": "Exam-16-T8-1-8-rubric.png", "description": "Additional rubric structure, with SMILES '[O-]/C([Cat+])=C/[C@@H]([R])SC1=CC=CC=C1C=O'."}]}]} +{"problem_id": "Exam-16-T9", "header": "Problem 9: Effects of Reaction Conditions and Mechanistic Investigation of Electrochemical Decarboxylative Alkylation\nElectrochemical anodic oxidation can use inexpensive and readily available quaternary-ammonium salts of alkyl carboxylic acids as alkyl sources, avoiding the need for additional oxidants and transition-metal catalysts during decarboxylation. The method therefore offers a simple system, mild conditions, and environmental advantages. Researchers used an unsaturated amide as a model substrate to investigate decarboxylative methylation/arylation. Tetrabutylammonium acetate served as both the methyl source and electrolyte, with carbon cloth as the anode, platinum as the cathode, and DMSO as the solvent. Under constant-current electrolysis at 3 mA with 6.0 equivalents of acetic acid as a proton source at 70°C for 9 h (5.0 F/mol), the target oxindole was obtained in 80% yield.", "header_images": [], "subquestions": [{"sub_id": "Exam-16-T9-1", "tags": ["3.5", "2.8", "2.3"], "text_open": "9.1 Further study showed that the acidity of the proton source strongly affects the reaction. When weaker proton sources, such as tert-butanol or hexafluoroisopropanol, or stronger proton sources, such as p-toluenesulfonic acid or trifluoroacetic acid, were used, the yield decreased substantially and the reaction was sometimes almost completely suppressed. Give a reasonable explanation for each observation.", "text_images": [{"file": "Exam-16-T9-1-1-prompt.png", "description": "Standard reaction in the prompt: 'FC(F)(C(C(N(C1=CC=CC=C1)C)=O)=C)F' reacts with `TBAOAc (1.5 equiv.)` using `C (+) / Pt (-)` under `HOAc (6.0 equiv.)`, `DMSO 4 mL, 70°C`, and `I = 3 mA, 5.0 F/mol`, producing 'CCC1(C(N(C2=CC=CC=C21)C)=O)C(F)(F)F'."}], "full_mark": 4, "rubric": ["Free alkyl carboxylate species must be present so that single-electron oxidation and decarboxylation can occur.", "Strong acid: H^{+} protonates the carboxylate and thereby suppresses its oxidation. (2 points)", "Weak acid: the system may become excessively nucleophilic or basic, which is unfavorable for the subsequent cyclization step. (2 points)"], "rubric_images": []}, {"sub_id": "Exam-16-T9-2", "tags": ["2.2", "2.3", "4.2"], "text_open": "9.3 The researchers performed the following experiments to investigate the reaction mechanism. Note: “standard condition” refers to the standard conditions described above, under which the yield is 80%. 9.3.1 To test for a free-radical intermediate, TEMPO was added as a radical trap. The target product did not form at all. Analysis of the reaction mixture together with high-resolution mass spectrometry revealed products formed by trapping with TEMPO. Give the SMILES structures of possible products.", "text_images": [{"file": "Exam-16-T9-2-1-prompt.png", "description": "Mechanistic experiment for 9.3.1: 'CN(C(C(C)=C)=O)C1=CC=CC=C1' reacts with the salt formed from anion 'CC([O-])=O' (shown with a circled minus sign) and cation 'CCCC[N+](CCCC)(CCCC)CCCC' (shown as `N^nBu_4` with a circled plus sign) under `standard condition` with *TEMPO (3.0 equiv.)*."}], "full_mark": 2, "rubric": ["Trapped product 1: 'CC1(C)CCCC(C)(C)N1OC' (1 point); trapped product 2 (TEMPO-OAc): 'CC1(C)CCCC(C)(C)N1OOC(C)=O' (1 point)"], "rubric_images": [{"file": "Exam-16-T9-2-1-rubric.png", "description": "Trapped product 1 in the rubric, with SMILES 'CC1(C)CCCC(C)(C)N1OC'."}, {"file": "Exam-16-T9-2-2-rubric.png", "description": "Trapped product 2 in the rubric, TEMPO-OAc, with SMILES 'CC1(C)CCCC(C)(C)N1OOC(C)=O'."}]}, {"sub_id": "Exam-16-T9-3", "tags": ["2.2", "2.3", "2.4"], "text_open": "9.3.3 On the basis of these mechanistic experiments, two pathways remain possible for the ring-closing step of the standard reaction. Give the SMILES structures of the key intermediates that may be involved in the standard reaction and identify the two possible pathways corresponding to the ring-closing step.", "text_images": [], "full_mark": 6, "rubric": ["Intermediate 1, the acetate radical: 'CC([O])=O' (1 point); intermediate 2, the methyl radical: '[CH3]' (1 point); intermediate 3: 'CN(C([C](CC)C(F)(F)F)=O)C1=CC=CC=C1' (1 point)", "The carbocationic-pathway intermediate: 'CN(C([C+](CC)C(F)(F)F)=O)C1=CC=CC=C1' (1 point)", "The radical-pathway intermediate: 'CN1C2=CC=C[C]C2C(CC)(C(F)(F)F)C1=O' (1 point)", "Subsequent intermediate: 'CN(C1=O)C2=CC=C[CH+]C2C1(CC)C(F)(F)F' (1 point)"], "rubric_images": [{"file": "Exam-16-T9-3-1-rubric.png", "description": "Rubric intermediate 1: the acetate radical, with SMILES 'CC([O])=O'."}, {"file": "Exam-16-T9-3-2-rubric.png", "description": "Rubric intermediate 2: the methyl radical ·CH_3, with SMILES '[CH3]'."}, {"file": "Exam-16-T9-3-3-rubric.png", "description": "Rubric intermediate 3: a CF_3-substituted oxindole-precursor radical formed after methyl-radical addition, containing a benzylic/amide-side-chain radical; its SMILES is 'CN(C([C](CC)C(F)(F)F)=O)C1=CC=CC=C1'."}, {"file": "Exam-16-T9-3-4-rubric.png", "description": "Carbocationic-pathway intermediate: a CF_3-substituted oxindole cation intermediate, with SMILES 'CN(C([C+](CC)C(F)(F)F)=O)C1=CC=CC=C1'."}, {"file": "Exam-16-T9-3-5-rubric.png", "description": "Radical-pathway intermediate: an aryl-radical/radical-cyclization intermediate whose benzene ring is depicted with a dashed circle and a radical, with SMILES 'CN1C2=CC=C[C]C2C(CC)(C(F)(F)F)C1=O'."}, {"file": "Exam-16-T9-3-6-rubric.png", "description": "Subsequent intermediate: a fused benzylic/aryl cation intermediate leading to the product, with SMILES 'CN(C1=O)C2=CC=C[CH+]C2C1(CC)C(F)(F)F'."}]}]} {"problem_id": "Exam-17-T3", "header": "Problem 3 Aqueous Self-Assembly and Competitive Recognition by Bowl-Shaped Molecules\nMolecular recognition in water is often governed simultaneously by electrostatic interactions, hydrogen bonding, solvation effects, and shape complementarity, so the actual outcome of supramolecular assembly can differ from a simple prediction based on charge complementarity. Tribenzotriquinacene (TBTQ) derivatives are rigid bowl-shaped molecules whose frameworks have distinct concave and convex faces. Introducing ionic groups around the rim of a TBTQ framework can confer water solubility and enable charge-controlled self-assembly.\nAs shown, compound A is a cationic TBTQ bowl bearing 6 ammonium groups, compound B is an anionic TBTQ bowl bearing 6 carboxylate groups, and compound C is mesitylene-α, α′, α″-trisulfonate, a trianionic guest bearing three $CH_2SO_3^-$ groups on an aromatic ring.", "header_images": [{"file": "Exam-17-T3-header-1.png", "description": "Compound A (TBTQ hexaammonium salt, labeled $A\\cdot Cl_6$): a bowl-shaped TBTQ framework in which each ring bears 3-aminopropoxy chains on 2 adjacent aromatic carbon atoms, for a total of 6 $NH_3^+$ groups and 6 $Cl^-$ ions. Its SMILES string is '[NH3+]CCCOC1=CC(C2C3(C4C5=C(C=C(C(OCCC[NH3+])=C5)OCCC[NH3+])C3C6=CC(OCCC[NH3+])=C(OCCC[NH3+])C=C62)C)=C4C=C1OCCC[NH3+].[Cl-].[Cl-].[Cl-].[Cl-].[Cl-].[Cl-]'."}, {"file": "Exam-17-T3-header-2.png", "description": "Compound B (TBTQ hexacarboxylate, labeled $B\\cdot (NH_4)_6$): a bowl-shaped TBTQ framework in which each ring bears carboxymethoxy chains on 2 adjacent aromatic carbon atoms, for a total of 6 $CO_2^-$ groups and 6 $NH_4^+$ ions. Its SMILES string is '[O-]C(COC(C=C1C2C3(C)C(C4=CC(OCC([O-])=O)=C(C=C4C3C5=CC(OCC([O-])=O)=C(C=C52)OCC([O-])=O)OCC([O-])=O)C1=C6)=C6OCC([O-])=O)=O.[NH4+].[NH4+].[NH4+].[NH4+].[NH4+].[NH4+]'."}, {"file": "Exam-17-T3-header-3.png", "description": "Compound C (trisodium benzene-1,3,5-tris(methylsulfonate), labeled $Na_3\\cdot C$): its SMILES string is '[O-]S(=O)(CC1=CC(CS(=O)([O-])=O)=CC(CS(=O)([O-])=O)=C1)=O.[Na+].[Na+].[Na+]'."}], "subquestions": [{"sub_id": "Exam-17-T3-1", "tags": ["1.7"], "text_open": "3.1 Based only on the charge complementarity and bowl-shaped structures of A and B, describe in words the AB capsule assembly that would be most readily expected. No specific structure needs to be drawn.", "text_images": [], "full_mark": 2, "rubric": ["The two bowl openings face each other, and the positive charges of the ammonium groups around the rim of A attract the negative charges of the carboxylate groups around the rim of B (2 points), as illustrated by:"], "rubric_images": [{"file": "Exam-17-T3-1-1-rubric.png", "description": "Schematic answer for the capsule assembly in 3.1. On a pure white background, two similarly sized hemispheres are arranged vertically, centered, and separated by a narrow horizontal white gap without touching. The upper hemisphere is sky blue, with its curved face upward and flat face downward, and bears a black plus sign (+) at the center of its front curved surface. The lower hemisphere is red, with its flat face upward and curved face downward like a bowl, and bears a black minus sign (-) at the center of its front curved surface. Their flat faces are parallel and face each other, together approximating a complete sphere. Both hemispheres have smooth three-dimensional shading, highlights, and gloss without outlines. The figure contains no words, letters, numbers, arrows, or explicit chemical structures."}]}]} -{"problem_id": "Exam-17-T4", "header": "Problem 4 Gravimetric Determination of Tungsten (17 points, 8%)\nTungsten is an important strategic metal that has an irreplaceable strengthening role in cemented carbides, high-speed steel, and specialty steels. Accurate determination of tungsten content is essential for quality control of alloy materials. In the classical gravimetric method, tungsten is converted to a tungstic-acid precipitate, ignited to tungsten trioxide, and weighed. Accurate tungsten content is obtained by correcting for interference from coexisting elements such as molybdenum, chromium, and vanadium. This problem concerns precipitation and separation of tungsten, correction for impurities, and special treatment of niobium-containing samples, and tests understanding of the principles of gravimetric analysis and strategies for eliminating interference.\nThe hydrochloric acid used in this problem is diluted 1:1, the nitric acid is concentrated, and every reported mass has been corrected for the tare mass of the crucible.", "header_images": [], "subquestions": [{"sub_id": "Exam-17-T4-1", "tags": ["2.1"], "text_open": "4.1 Procedure for samples without niobium\nA sample of mass $m$ is weighed into a beaker. Hydrochloric acid and nitric acid are added, and the mixture is heated gently to dissolve the sample and evaporated to a syrup to remove nitrogen oxides. Hydrochloric acid is added to dissolve the salts, the solution is diluted with water and brought to the boil, and cinchonine hydrochloride solution is added slowly to precipitate all tungsten as tungstic acid. The precipitate is aged at 70-80℃ for 4 h and then filtered and washed with a cinchonine hydrochloride wash solution.\nThe precipitate and filter paper are placed in a platinum crucible, carbonized, and ignited at 750-800℃. The residue is moistened with water, hydrofluoric acid and sulfuric acid are added, and the mixture is heated to volatilize silicon. It is then ignited to constant mass, giving a mass $m_1$ of impure $WO_3$.\nSodium carbonate is added to the platinum crucible for fusion. After cooling, the melt is extracted with hot water and filtered, and the residue is ignited and weighed to give $m_2$. The filtrate is analyzed spectrophotometrically for molybdenum, chromium, and vanadium, which are converted to the corresponding oxide masses $m_3$, $m_4$, and $m_5$.\n4.1.1 Write the equation for oxidation of elemental tungsten to tungstic acid, $H_2WO_4$, in hydrochloric acid/nitric acid.", "text_images": [], "full_mark": 2, "rubric": ["$W + 2HNO_3 = H_2WO_4 + 2NO\\uparrow$ (2 points)"], "rubric_images": []}, {"sub_id": "Exam-17-T4-2", "tags": ["4.7", "2.1"], "text_open": "4.1.6 During aqueous extraction after sodium carbonate fusion, $WO_3$ is converted to $Na_2WO_4$ and enters solution, whereas $TiO_2$ and $Fe_2O_3$ remain insoluble. Analysis of a tungsten-steel sample gives the following data: $m$ = 0.5123 g, $m_1$ = 0.0842 g, $m_2$ = 0.0018 g, $m_3$ = 0.0023 g, $m_4$ = 0.0015 g, and $m_5$ = 0.0012 g. Calculate the mass fraction of tungsten in the sample.", "text_images": [], "full_mark": 2, "rubric": ["$w_W = (m_1-m_2-m_3-m_4-m_5)/m \\times 0.7930 \\times 100\\% = 12.0\\%$ (2 points). Deduct 1 point if the answer is not reported to 3 significant figures."], "rubric_images": []}]} -{"problem_id": "Exam-17-T7", "header": "Problem 7 Low-Valent Cluster Compounds of Group 13 Metals\nLow-valent compounds of Group 13 elements such as aluminum and gallium can form metal clusters with diverse structures. Their bonding modes and electronic structures are important for understanding the nature of metal-metal bonds. This problem concerns low-valent gallium and aluminum halides and their assembly reactions with ligands, focusing on structural analysis, oxidation-state calculations, and cluster reactivity.", "header_images": [], "subquestions": [{"sub_id": "Exam-17-T7-1", "tags": ["2.1"], "text_open": "Synthesis and structures of $E_3$ units: Ga-Ga bonds are relatively stable and can support extended compounds.\n7.1 Chain-like $E_3$ structure: reaction of freshly prepared LiM with 1.67 eq GaI affords the trigallane C = $Ga_3M_3I_2$.\n7.1.1 Write the reaction equation for the formation of C from LiM and GaI.", "text_images": [{"file": "Exam-17-T7-1-1-prompt.png", "description": "Two-step reaction: '[dipp]N=C=N[dipp]'>t-BuLi (1eq), n-hexane, 0 °C>intermediate Li+M-, where M denotes the ligand and the SMILES for M- is '[dipp]/N=C([N-][dipp])/C(C)(C)C'. The M- anion has an N,N'-bis(dipp) amidinate framework with t-Bu bonded to the central carbon. The second step is '[dipp]/N=C([N-][dipp])/C(C)(C)C'>GaI (1.67eq), Toluene, -78 °C>C, where C = $Ga_3M_3I_2$ and its SMILES is 'I[Ga]1([Ga]2([Ga]3(I)N([dipp])C(C(C)(C)C)=[N]3[dipp])N([dipp])C(C(C)(C)C)=[N]2[dipp])N([dipp])C(C(C)(C)C)=[N]1[dipp]'. Three Ga atoms form a chain through Ga-Ga single bonds, each terminal Ga bears one I, and each Ga is chelated by one M ligand. No wedge or dashed bonds are shown."}], "full_mark": 2, "rubric": ["$3LiM + 5GaI = Ga_3M_3I_2 + 2Ga + 3LiI$ (2 points)"], "rubric_images": []}, {"sub_id": "Exam-17-T7-2", "tags": ["2.1"], "text_open": "7.1.2 Write the equation for the substitution reaction of trigallane C with $NaBEt_3H$ to form Et(M)Ga-Ga(M)-Ga(M)H.", "text_images": [], "full_mark": 2, "rubric": ["$Ga_3M_3I_2 + NaBEt_3H = Et(M)Ga-Ga(M)-Ga(M)H + NaI + BEt_2I$ (2 points)", "$Ga_3M_3I_2 + NaBEt_3H = Et(M)Ga-Ga(M)-Ga(M)H + NaBEt_2I_2$ is also acceptable."], "rubric_images": []}, {"sub_id": "Exam-17-T7-3", "tags": ["2.4", "1.6"], "text_open": "7.2 Cyclic $E_3$ structure: cyclotrigallanes have intriguing electronic properties. Reaction of $(Mes_2C_6H_3)Li$ with $GaCl_3$ gives $(Mes_2C_6H_3)GaCl_2$, which is reduced with potassium metal in diethyl ether to afford the cyclic gallium compound D = $K_2[(Mes_2C_6H_3)Ga]_3$. This compound has a nearly equilateral triangular $Ga_3$ core, with an average Ga-Ga bond length of 2.426 Å that is markedly shorter than a typical single bond, indicating a higher bond order. The two potassium ions lie on opposite sides of the $Ga_3$ plane. Write the SMILES string for D and determine whether its anion is aromatic.", "text_images": [{"file": "Exam-17-T7-3-1-prompt.png", "description": "Two-step reaction: $GaCl_3$ reacts with LiAr (1.0 eq) in $Et_2O$ at -78 °C; $K^0$ (4.0 eq) is then added and the mixture is reacted at rt. for 5 days to give $K_2(ArGa)_3·(Et_2O)_2$. The structure defining Ar = $(Mes_2C_6H_3)$ appears at the right: the top carbon of the central benzene ring, drawn in an alternating-double-bond Kekulé form, bears a wavy bond marking the attachment point to Ga, and each of the two ortho carbons bears a 2,4,6-trimethylphenyl group (Mes). The encodable structure of Ar is represented by the SMILES '[*]C1=C(C2=C(C)C=C(C)C=C2C)C=CC=C1C3=C(C)C=C(C)C=C3C', where [*] is the attachment point corresponding to the wavy bond."}], "full_mark": 3, "rubric": ["Structure of D: 'CC1=CC(C)=CC(C)=C1C(C=CC=C2C3=C(C)C=C(C)C=C3C)=C2[Ga-]4=[Ga-]([Ga]4C5=C(C6=C(C)C=C(C)C=C6C)C=CC=C5C7=C(C)C=C(C)C=C7C)C8=C(C9=C(C)C=C(C)C=C9C)C=CC=C8C%10=C(C)C=C(C)C=C%10C.[K+].[K+]'. The two K+ ions coordinate on opposite sides of the Ga3 triangular plane (2 points). The anion is aromatic (1 point)."], "rubric_images": [{"file": "Exam-17-T7-3-1-rubric.png", "description": "Reference structure of D: the anion [(Mes2C6H3)Ga]3 is represented by the SMILES 'CC1=CC(C)=CC(C)=C1C(C=CC=C2C3=C(C)C=C(C)C=C3C)=C2[Ga-]4=[Ga-]([Ga]4C5=C(C6=C(C)C=C(C)C=C6C)C=CC=C5C7=C(C)C=C(C)C=C7C)C8=C(C9=C(C)C=C(C)C=C9C)C=CC=C8C%10=C(C)C=C(C)C=C%10C.[K+].[K+]'. Three Ga atoms form a $Ga_3$ triangle, and each Ga is bonded to one Ar group; each central benzene ring bears Mes groups at the 2,6-positions, and all 6 Mes groups are 2,4,6-trimethylphenyl. The cluster has an overall 2- charge, shown at the center of the $Ga_3$ triangle and represented in the SMILES by two [Ga-] atoms. The two K+ ions coordinate on opposite sides of the Ga3 triangular plane, with their interactions with $Ga_3$ shown by dashed lines. In the SMILES, the Ga3 ring contains one Ga=Ga double bond as an aromatic Kekulé representation, whereas the original drawing shows all three Ga-Ga bonds as solid single lines. No wedge bonds are shown."}]}, {"sub_id": "Exam-17-T7-4", "tags": ["2.4", "1.4"], "text_open": "7.4 Eight-membered-ring structure: gaseous GaI is cocondensed with excess $PEt_3$ in toluene at -196 °C and then warmed slowly, giving F = $Ga_8I_8·6PEt_3$. This cluster has $C_{2h}$ symmetry and contains a nearly planar eight-membered $Ga_8$ ring. Two iodine atoms bridge across the ring to form a four-membered $Ga_2I_2$ ring. All eight Ga atoms are four-coordinate, and the average Ga-Ga bond length is 2.466 Å.\n7.4.1 Write the SMILES string for F.", "text_images": [], "full_mark": 4, "rubric": ["Structure of F = $Ga_8I_8·6PEt_3$: 'I[Ga@]1([P](CC)(CC)CC)[Ga@](I)([P](CC)(CC)CC)[Ga@@](I)([P](CC)(CC)CC)[Ga@@]2(I3)[Ga@](I)([P](CC)(CC)CC)[Ga@@](I)([P](CC)(CC)CC)[Ga@](I)([P](CC)(CC)CC)[Ga@]13I2' (4 points)"], "rubric_images": [{"file": "Exam-17-T7-4-1-rubric.png", "description": "Reference structure of F = $Ga_8I_8·6PEt_3$, represented by the SMILES 'I[Ga@]1([P](CC)(CC)CC)[Ga@](I)([P](CC)(CC)CC)[Ga@@](I)([P](CC)(CC)CC)[Ga@@]2(I3)[Ga@](I)([P](CC)(CC)CC)[Ga@@](I)([P](CC)(CC)CC)[Ga@](I)([P](CC)(CC)CC)[Ga@]13I2'. The 8 Ga atoms form an eight-membered ring through single bonds. Two I atoms bridge opposite Ga atoms across the ring; in the drawing, one bridging I is connected to the two Ga atoms by solid wedge bonds and the other by dashed bonds. Each of the remaining 6 Ga atoms bears one terminal I and one PEt3 (Et3P). The stereochemistry is retained by the @/@@ markers."}]}, {"sub_id": "Exam-17-T7-5", "tags": ["1.7", "2.4"], "text_open": "7.4.2 The Ga/I/$PEt_3$ system forms the eight-membered-ring cluster F ($Ga_8I_8$), whereas the Al/Br/$NEt_3$ system forms the four-membered-ring cluster E ($Al_4Br_4$) under similar conditions. Which explanations are the most reasonable? Select all that apply.\n(A) Ga has a smaller atomic radius than Al and a shorter Ga-Ga bond; formation of a four-membered ring would therefore cause greater ring strain and bond-angle distortion, so an eight-membered ring is favored to relieve strain.\n(B) I has a larger atomic radius than Br and is well suited for bridging; transannular bridges stabilize the eight-membered ring.\n(C) $PEt_3$ is a stronger electron donor than $NEt_3$, increasing the electron density at Ga and shortening the Ga-Ga bonds, which favors formation of a larger ring system.\n(D) The Ga-Ga bonds in the eight-membered ring have a higher bond order than the Al-Al bonds in the four-membered ring, giving greater thermodynamic stability.", "text_images": [], "full_mark": 2, "rubric": ["AB (2 points)"], "rubric_images": []}, {"sub_id": "Exam-17-T7-6", "tags": ["2.4", "1.2"], "text_open": "7.5 Fused-ring structure: attempts to synthesize amine-stabilized $Ga_8I_8·6NEt_3$ were unsuccessful, but partial oxidation unexpectedly gave G = $Ga_8I_8O_2·6NEt_3$. Its structure has $C_i$ symmetry and consists of two nearly planar $Ga_4O$ rings fused through a shared edge of a central four-membered $Ga_2O_2$ ring. All Ga atoms are four-coordinate. Write the SMILES string for G.", "text_images": [], "full_mark": 2, "rubric": ["Structure of G = $Ga_8I_8O_2·6NEt_3$: 'I[Ga@]1(O2[Ga@@](I)([N](CC)(CC)CC)[Ga@](I)([N](CC)(CC)CC)[Ga@]1([N](CC)(CC)CC)I)O3[Ga@@]2(I)[Ga@@](I)([N](CC)(CC)CC)[Ga@](I)([N](CC)(CC)CC)[Ga@@]3(I)[N](CC)(CC)CC' (2 points). Stereochemistry must be shown. The scoring guide includes a second acceptable structure: 'I[Ga@@]1(O2[Ga@@](I)([N](CC)(CC)CC)[Ga@](I)([N](CC)(CC)CC)[Ga@]1([N](CC)(CC)CC)I)O3[Ga@]2(I)[Ga@@](I)([N](CC)(CC)CC)[Ga@](I)([N](CC)(CC)CC)[Ga@@]3(I)[N](CC)(CC)CC'."], "rubric_images": [{"file": "Exam-17-T7-6-1-rubric.png", "description": "Reference structure of G = $Ga_8I_8O_2·6NEt_3$ in the first scoring-guide image, represented by the SMILES 'I[Ga@]1(O2[Ga@@](I)([N](CC)(CC)CC)[Ga@](I)([N](CC)(CC)CC)[Ga@]1([N](CC)(CC)CC)I)O3[Ga@@]2(I)[Ga@@](I)([N](CC)(CC)CC)[Ga@](I)([N](CC)(CC)CC)[Ga@@]3(I)[N](CC)(CC)CC'. Two $Ga_4O$ rings are fused through a shared edge of the central four-membered $Ga_2O_2$ ring. Each of the 8 Ga atoms bears one terminal I, and 6 of the Ga atoms additionally bear NEt3 (Et3N); the two Ga atoms in the central ring have no amine ligand. Wedge/dashed-bond stereochemistry is retained by the @/@@ markers."}, {"file": "Exam-17-T7-6-2-rubric.png", "description": "Reference structure of G = $Ga_8I_8O_2·6NEt_3$ in the second scoring-guide image, represented by the SMILES 'I[Ga@@]1(O2[Ga@@](I)([N](CC)(CC)CC)[Ga@](I)([N](CC)(CC)CC)[Ga@]1([N](CC)(CC)CC)I)O3[Ga@]2(I)[Ga@@](I)([N](CC)(CC)CC)[Ga@](I)([N](CC)(CC)CC)[Ga@@]3(I)[N](CC)(CC)CC'. The framework composition and connectivity are the same as in the first image, but the wedge/dashed orientations of I and NEt3 at some Ga atoms differ. One Ga vertex at the upper right is unlabeled in the image; this vertex is [Ga@@] in the SMILES."}]}, {"sub_id": "Exam-17-T7-7", "tags": ["1.1"], "text_open": "7.6 Tetrahedral type: high-temperature gaseous GaBr is cocondensed with a toluene solution of $NHEt_2$. The resulting dark solution is concentrated and allowed to stand at 25 °C, yielding crystals of cluster H = $Ga_5Br_7(NHEt_2)_5$. H contains a $Ga_5$ core in which the central Ga atom is tetrahedrally surrounded by four Ga atoms.\n7.6.2 In the sterically congested H molecule, the 12 heteroatoms—the halogens and the O/N atoms of the ligands—adopt an arrangement that maximizes their mutual separation. What polyhedron is formed by these 12 vertices?", "text_images": [{"file": "Exam-17-T7-7-1-prompt.png", "description": "Structure of H = $Ga_5Br_7(NHEt_2)_5$: the central Ga is connected by single bonds to four peripheral Ga atoms in a tetrahedral arrangement. The encodable structure of H is represented by the SMILES '[L][Ga](Br)(Br)[Ga]([Ga]([L])(Br)Br)([Ga]([L])([L])Br)[Ga]([L])(Br)Br', where [L] denotes the $NHEt_2$ ligand labeled in the image. A total of 5 L ligands and 7 Br atoms are distributed among the peripheral Ga atoms. The lower right of the image defines L = $NHEt_2$, and all bonds are solid single bonds."}], "full_mark": 1, "rubric": ["Distorted icosahedron or icosahedron (1 point). A regular icosahedron receives no credit."], "rubric_images": []}]} -{"problem_id": "Exam-17-T9", "header": "Problem 9 Stereochemistry\nCareful design of reactant structures and reaction conditions can provide highly selective stereochemical control.", "header_images": [], "subquestions": [{"sub_id": "Exam-17-T9-1", "tags": ["2.4", "1.2"], "text_open": "9.1 Garner's aldehyde is frequently used for stereochemical control in synthesis. The route below prepares it from the natural product S-serine.\n9.1.2 Garner's aldehyde can undergo further nucleophilic addition. Write the SMILES strings of the major products under the different reaction conditions below, representing them as anions and specifying stereochemistry.", "text_images": [{"file": "Exam-17-T9-1-1-prompt.png", "description": "Two-step reaction: 'OC[C@@H](C(O)=O)N'>>'OC[C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O'; 'OC[C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O'>DMSO, (COCl)2, DIPEA>'[H]C([C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O)=O'."}, {"file": "Exam-17-T9-1-2-prompt.png", "description": "Two parallel reactions: '[H]C([C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O)=O'>Nu^->[A]; '[H]C([C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O)=O'>Nu^-, [M]>[B]. The products are labeled only [A] and [B]; their structures are not shown."}], "full_mark": 4, "rubric": ["The SMILES for A is '[H][C@@]([O-])([Nu])[C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O', where Nu is the placeholder used in the original diagram.", "The SMILES for B is '[Nu][C@@]([O-])([H])[C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O', where Nu is the placeholder used in the original diagram.", "Each structure is worth 2 points."], "rubric_images": [{"file": "Exam-17-T9-1-1-rubric.png", "description": "The SMILES for A is '[H][C@@]([O-])([Nu])[C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O'; the SMILES for B is '[Nu][C@@]([O-])([H])[C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O'. Nu is the placeholder used in the original diagram."}]}, {"sub_id": "Exam-17-T9-2", "tags": ["2.7", "1.2"], "text_open": "9.1.3 In practice, the two reaction pathways compete. Determine which of the following conditions favor formation of product B, and briefly explain why.\n(1) Solvent: a. THF; b. toluene", "text_images": [{"file": "Exam-17-T9-2-1-prompt.png", "description": "'[H]C([C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O)=O'>nucleophile, solvent, additive>[A]+[B]. The products are labeled only [A] and [B]; their structures are not shown."}], "full_mark": 2, "rubric": ["b. (1 point). THF competitively coordinates the metal cation, reducing the amount of metal cation available for chelation (1 point)."], "rubric_images": []}, {"sub_id": "Exam-17-T9-3", "tags": ["2.7", "1.2"], "text_open": "(3) Additive: a. DMPU; b. $ZnCl_2$", "text_images": [], "full_mark": 2, "rubric": ["b. (1 point). As a Lewis base, DMPU binds the metal cation, reducing the amount of metal cation available for chelation (1 point)."], "rubric_images": []}, {"sub_id": "Exam-17-T9-4", "tags": ["1.2", "2.7"], "text_open": "9.1.4 Garner's aldehyde can also be synthesized from R-serine. In the reaction below, replacing Boc with Ts reverses the selectivity. Explain why.", "text_images": [{"file": "Exam-17-T9-4-1-prompt.png", "description": "'CC1(C)N([R])[C@@H](C=O)CO1'>Et-MgBr, THF>'CC1(C)N([R])[C@@H]([C@@H](O)CC)CO1' + 'CC1(C)N([R])[C@@H]([C@H](O)CC)CO1'. The image also states that the product ratio is 1:9 when R=Boc and 8:1 when R=Ts. R is the placeholder group used in the original diagram."}], "full_mark": 2, "rubric": ["The oxygen atoms of the $SO_2$ sulfonyl group in Ts have low electron density and extremely weak coordinating ability, so a stable chelate cannot form (1 point). The direction of attack therefore more closely follows a nonchelation model, giving the other diastereomer with high selectivity (1 point). Calculations show a larger LUMO energy difference between the two reactive Felkin-Anh conformations of the N-Ts analogue; the orientation of the C-NTs bond thus produces a stronger reactive-conformation preference and enhances anti selectivity."], "rubric_images": []}, {"sub_id": "Exam-17-T9-5", "tags": ["2.4", "1.2"], "text_open": "9.2 Garner's aldehyde can also participate in multicomponent reactions that construct new stereocenters. Write the SMILES string of the product of the reaction below.", "text_images": [{"file": "Exam-17-T9-5-1-prompt.png", "description": "Multicomponent reaction: 'C1(C=CC=C2)=C2NC=C1'.'CC(OC[C@H]1C=O)(C)N1C(OC(C)(C)C)=O'.'O=C1CC(OC(C)(C)O1)=O'>CH3CN, D,L-proline>. No product structure is shown to the right of the arrow."}], "full_mark": 2, "rubric": ["'O=C(OC(C)(C)O1)C(C1=O)[C@@H]([C@@H]2COC(C)(C)N2C(OC(C)(C)C)=O)C3=CNC4=C3C=CC=C4'"], "rubric_images": [{"file": "Exam-17-T9-5-1-rubric.png", "description": "The SMILES string is 'O=C(OC(C)(C)O1)C(C1=O)[C@@H]([C@@H]2COC(C)(C)N2C(OC(C)(C)C)=O)C3=CNC4=C3C=CC=C4'."}]}]} +{"problem_id": "Exam-17-T5", "header": "Problem 5 Dicarbon Anions", "header_images": [], "subquestions": [{"sub_id": "Exam-17-T5-1", "tags": ["1.3", "1.4"], "text_open": "5.2 Starting from crystalline $CaC_2$, inserting square Co layers into some of the interlayer spaces and replacing Ca with U gives a tetragonal $UCoC_2$ crystal. Its structure is shown below, where the large spheres represent U, the medium spheres represent Co, and the small spheres represent C. The lattice parameter c = 728 pm.\n5.2.1 Name the coordination polyhedron formed by the C atoms around Co.", "text_images": [{"file": "Exam-17-T5-1-1-prompt.png", "description": "An oblique three-dimensional ball-and-stick rendering of the tetragonal unit cell of $UCoC_2$ on a white background. Thin gray-black lines outline a rectangular-prismatic unit-cell frame elongated vertically along c. There are 10 large spheres, including spheres at face centers on the top and the upper and lower basal faces; 4 medium spheres lie on the side faces, one at an upper-middle position on each left/right face and one at a lower-middle position on each front/back face; and 8 small spheres form upper and lower groups, with two vertical pairs in each group along c joined by short brown sticks. The upper pair is on the front/back faces and the lower pair on the left/right faces. The small spheres are joined to adjacent large spheres by short sticks, forming a layered framework; the medium spheres have no sticks. At lower left, a blue arrow points upward and is labeled c, a green arrow points right and is labeled b, and a red arrow points lower-left and is labeled a. Bonding obscured by occlusion cannot be resolved. Apart from a, b, and c, the image contains no text or legend. According to the caption, the large, medium, and small spheres represent U, Co, and C, respectively. The coordinates of the two carbon atoms in the unit cell are (1/2, 0, 0.155) and (1/2, 0, 0.358)."}], "full_mark": 2, "rubric": ["Tetrahedron or distorted tetrahedron (2 points). A response of \"regular tetrahedron\" receives no credit."], "rubric_images": []}, {"sub_id": "Exam-17-T5-2", "tags": ["1.3"], "text_open": "5.3 A uranium carbide belongs to the cubic crystal system. Its unit-cell projections along the a, b, and c directions are all as shown below. The large spheres represent U and the small spheres represent C. All atoms of the same element are equivalent in the crystal, and the lattice parameter a = 808 pm.\n5.3.1 Write the chemical formula of this crystal.", "text_images": [{"file": "Exam-17-T5-2-1-prompt.png", "description": "Crystal-axis indicator on a white background: a solid red circle is at the lower left with a black letter a superimposed, indicating that the a axis points out of the page toward the viewer. A blue arrow extends upward from the circle and is labeled c above its tip; a green arrow extends rightward and is labeled b to the right of its tip. Apart from a, b, and c, the image contains no text or other graphics."}, {"file": "Exam-17-T5-2-2-prompt.png", "description": "Crystal-axis indicator on a white background: a solid green circle is at the lower left with a black letter b superimposed, indicating that the b axis points out of the page toward the viewer. A red arrow extends upward from the circle and is labeled a above its tip; a blue arrow extends rightward and is labeled c at its right end. Apart from a, b, and c, the image contains no text or other graphics."}, {"file": "Exam-17-T5-2-3-prompt.png", "description": "Crystal-axis indicator on a white background: a solid blue circle is at the lower left with a black letter c superimposed, indicating that the c axis points out of the page toward the viewer. A green arrow extends upward from the circle and is labeled b above its tip; a red arrow extends rightward and is labeled a at its right end. Apart from a, b, and c, the image contains no text or other graphics."}, {"file": "Exam-17-T5-2-4-prompt.png", "description": "Unit-cell projection of the uranium carbide in 5.3 (two-dimensional projection of a ball-and-stick model on a white background): thin black lines form a square unit-cell border, with some spheres lying on the border. The image contains only two sphere types: large light-blue highlighted spheres and small brown spheres. The large spheres occur at the four corners, at the middle of the top and bottom edges, and at several interior positions; some large spheres overlap in pairs. Some small spheres are joined by short brown sticks into paired or chain-like arrangements, while others are isolated. The image contains no text, letters, numbers, arrows, or legend. As specified in the prompt, the large spheres represent U and the small spheres represent C. The prompt states that the projections along the a, b, and c directions are all as shown; this image is that projection panel."}], "full_mark": 3, "rubric": ["$U_2C_3$ or $U_4(C_2)_3$ (3 points)"], "rubric_images": []}, {"sub_id": "Exam-17-T5-3", "tags": ["1.3"], "text_open": "5.3.4 One carbon atom in the unit cell has coordinates (1/2, 1/4, 0.536). Calculate the carbon-carbon bond length in this crystal.", "text_images": [], "full_mark": 2, "rubric": ["$d_{\\mathrm{C-C}} = [0.25 - (0.536 - 0.5) \\times 2] \\times 808\\ \\mathrm{pm} = 144\\ \\mathrm{pm}$ (2 points). Full credit requires both a calculation and the correct answer; an answer alone receives no credit."], "rubric_images": []}, {"sub_id": "Exam-17-T5-4", "tags": ["1.3", "1.7"], "text_open": "5.4 The carbon-carbon bond lengths in ethane, ethene, and ethyne are approximately 154 pm, 134 pm, and 120 pm, respectively. On this basis, predict the negative charge on the dicarbon anion in $CaC_2$, $UCoC_2$, and the uranium carbide in 5.3, respectively. The carbon-carbon bond orders in the three compounds are distinct integers.", "text_images": [], "full_mark": 2, "rubric": ["$CaC_2$: 2, -2, or $C_2^{2-}$", "$UCoC_2$: 6, -6, or $C_2^{6-}$", "Uranium carbide: 4, -4, or $C_2^{4-}$. Credit is awarded only if all responses are correct."], "rubric_images": []}, {"sub_id": "Exam-17-T5-5", "tags": ["1.3", "1.7"], "text_open": "5.5 Crystalline $DyCoC_2$ has a structure entirely different from that of $UCoC_2$. Its three-dimensional structure and projection along the c direction are shown below, where the large spheres represent Dy, the medium spheres represent Co, and the small spheres represent C. The only common stable positive oxidation state of Dy is +3. A small sample of $DyCoC_2$ is hydrolyzed in 2 mol/L dilute hydrochloric acid. Gas-chromatographic analysis gives the following mass fractions of the gaseous hydrolysis products: 74% ethane, 13% propane, 12% methane, and 1% ethene; no ethyne is formed.\n5.5.1 Predict the carbon-carbon bond order in $DyCoC_2$.", "text_images": [{"file": "Exam-17-T5-5-1-prompt.png", "description": "Three-dimensional ball-and-stick model of crystalline $DyCoC_2$ (oblique rendering on a white background): thin black lines outline a parallelepiped unit cell. There are 10 large blue-violet highlighted spheres of varying shades: 8 at the 8 corners and 2 at the centers of the top and bottom faces. There are 3 medium dark-blue spheres in a horizontal layer through the middle of the cell, positioned left of center, right of center, and at the center. Approximately 9 small brown spheres surround the medium spheres and are connected to them by dark-blue or brown short sticks; some small spheres bridge two medium spheres, forming an in-layer zigzag network. There are no direct sticks between large spheres or between medium spheres. Some spheres are obscured, so the exact number of brown small spheres cannot be fully determined. A three-axis indicator at the lower left has a blue upward arrow labeled c, a green rightward arrow labeled b, and a red arrow pointing down-left labeled a. Apart from a, b, and c, the image contains no text or legend. As specified in the prompt, the large, medium, and small spheres represent Dy, Co, and C, respectively."}, {"file": "Exam-17-T5-5-2-prompt.png", "description": "Projection of crystalline $DyCoC_2$ along the c direction (two-dimensional projection of a ball-and-stick model on a pale gray-white background): thin black lines form a rectangular unit-cell border. Four large blue spheres lie at the corners; 1 slightly smaller blue sphere lies at the center of the rectangle without any bonds; and 3 medium blue spheres lie at the midpoints of the left and right edges and at the center of the lower edge. There are 10 small brown spheres: 2 terminal spheres outside the rectangle at the upper left and upper right, while the remaining 8 form 4 horizontal dumbbell pairs joined by two parallel short brown sticks. These pairs lie at the upper center, middle left, middle right, and outside the rectangle at the lower center. Each medium blue sphere is connected to 4 brown spheres by two-colored blue-brown short sticks, and adjacent blue spheres and brown pairs form triangular units. A three-axis indicator at the lower left has a red rightward arrow labeled a, a green upward arrow labeled b, and a blue circle bearing the letter c, indicating that the c axis points out of the page toward the viewer. Apart from a, b, and c, the image contains no text or legend. As specified in the prompt, the large, medium, and small spheres represent Dy, Co, and C, respectively."}], "full_mark": 1, "rubric": ["1 or a single bond (1 point)"], "rubric_images": []}, {"sub_id": "Exam-17-T5-6", "tags": ["1.6"], "text_open": "5.5.3 Briefly explain why the experimentally measured carbon-carbon bond length in crystalline $DyCoC_2$ is 138 pm.", "text_images": [], "full_mark": 2, "rubric": ["Interaction between electrons in the π antibonding orbitals of the dicarbon anion and the d orbitals of $Co^{3+}$ increases the carbon-carbon bond order (2 points)."], "rubric_images": []}]} +{"problem_id": "Exam-17-T6", "header": "Problem 6 Synthesis of Triphenylmethanol Using a Grignard Reagent\nGrignard reagents are common nucleophiles in organic synthesis. They must be used under anhydrous conditions free of oxygen and $CO_2$, and reactions are usually performed under nitrogen. In a teaching laboratory, however, the synthesis can be carried out with a simpler procedure. In this problem, bromobenzene is used to prepare phenylmagnesium bromide, which is then reacted with methyl benzoate to produce triphenylmethanol.", "header_images": [{"file": "Exam-17-T6-header-1.png", "description": "Experimental apparatus: a three-neck flask is centered, with its left neck closed by a ground-glass stopper. The middle neck holds a funnel-shaped dropping device with a sidearm and stopcock, labeled \"Apparatus 1\". The angled right neck is connected to a spherical condenser sloping upward to the right. Liquid is present in the lower part of the three-neck flask. The labels in the image are Apparatus 1, spherical condenser, three-neck flask, and to drying tube at the top of the condenser."}], "subquestions": [{"sub_id": "Exam-17-T6-1", "tags": ["2.8"], "text_open": "6.2 Synthesis of triphenylmethanol: cool the flask in a cold-water bath while slowly adding a solution of methyl benzoate in anhydrous diethyl ether with stirring. After addition is complete, heat the mixture under reflux for 1 h, cool the flask again in a cold-water bath, and add saturated ammonium chloride solution dropwise to decompose the addition product. Separate the layers and perform steam distillation to obtain the crude product.\n6.2.1 State the purpose of the saturated ammonium chloride workup, and explain why water or hydrochloric acid is not used directly for the workup.", "text_images": [], "full_mark": 3, "rubric": ["Triphenylmethanol is sparingly soluble in water, and its solubility is further reduced in a saturated salt solution. This promotes transfer of the product to the organic phase and increases the yield (1 point; stating that high ionic strength reduces emulsification is also acceptable).\nUsing water directly produces magnesium salts such as $MgBrOH$, which form poorly soluble flocculent precipitates. The weak acidity of ammonium chloride solution helps dissolve or disperse these precipitates, facilitating their subsequent removal by filtration (1 point).\nTriphenylmethanol is unstable under strongly acidic conditions and readily forms the triphenylmethyl cation, leading to side reactions. Ammonium chloride is a weak acid, and its aqueous solution provides an acidic environment mild enough not to destroy the product (1 point)."], "rubric_images": []}]} {"problem_id": "Exam-17-T10", "header": "Problem 10 Cyclobutanation\nCyclobutane frameworks occur widely in bioactive natural products and pharmaceutical molecules. Because a four-membered ring combines appreciable ring strain with a rigid spatial geometry, the development of efficient and selective methods for constructing cyclobutanes has long been an important goal in organic synthesis.", "header_images": [], "subquestions": [{"sub_id": "Exam-17-T10-1", "tags": ["1.2"], "text_open": "10.1.2 A conformation of cyclobutane is shown below. For 1,3-dimethylcyclobutane and 1,2-dimethylcyclobutane, determine whether the cis or trans isomer is more stable, and write the SMILES string of its preferred conformation.", "text_images": [{"file": "Exam-17-T10-1-1-prompt.png", "description": "Cyclobutane is represented by the SMILES '[H][C@]([C@]([H])([H])[C@]([H])1[H])([H])[C@@]1([H])[H]'. The image is a perspective drawing of a puckered cyclobutane conformation, with two bold wedge bonds and 8 labeled H atoms."}], "full_mark": 4, "rubric": ["1,3-Dimethylcyclobutane: the cis isomer is more stable (1 point), with both substituents pseudo-equatorial in the preferred conformation (1 point). 1,2-Dimethylcyclobutane: the trans isomer is more stable (1 point), with both substituents pseudo-equatorial in the preferred conformation (1 point)."], "rubric_images": [{"file": "Exam-17-T10-1-1-rubric.png", "description": "Reference answer: the preferred conformation of cis-1,3-dimethylcyclobutane, with both substituents pseudo-equatorial, has the SMILES 'C[C@H]1C[C@@H](C)C1'. The image is an unlabeled skeletal conformation drawing."}, {"file": "Exam-17-T10-1-2-rubric.png", "description": "Reference answer: the preferred conformation of trans-1,2-dimethylcyclobutane, with both substituents pseudo-equatorial, has the SMILES 'C[C@H]1CC[C@@H]1C'. The image is an unlabeled skeletal conformation drawing."}]}, {"sub_id": "Exam-17-T10-2", "tags": ["2.2"], "text_open": "10.2 Werz and co-workers recently reported a method for constructing cyclobutanone frameworks by exploiting the intrinsic ring strain of cyclopropanol derivatives. Under mild conditions, the reaction couples release of small-ring strain with C-C bond formation, providing a new strategy for cyclobutane synthesis. To investigate the mechanism, the researchers designed deuterium-labeling and substituent-effect experiments. In the deuterium-labeling experiments, sulfonium salts bearing deuterium at different allylic positions were prepared and reacted under identical conditions; the resulting deuterium distributions are shown below. Using Experiment I as an example, write the SMILES strings of all key intermediates consistent with the observations.", "text_images": [{"file": "Exam-17-T10-2-1-prompt.png", "description": "Deuterium-labeling reaction in Experiment I: 'OC1(S(C2=CC=C(C)C=C2)(=O)=O)CC1'.'[2H]/C(C1=CC=CC=C1)=C\\C[S+]2C(C)(CCC2(C)C)C.F[B-](F)(F)F'>LiHMDS, CH2Cl2, -78°C to r.t.>'O=C1CCC1(C=C)C2=CC=CC=C2'. The product region shows the deuterium distribution: 89%, 0% D, 22% D, and 22% D."}, {"file": "Exam-17-T10-2-2-prompt.png", "description": "Deuterium-labeling reaction in Experiment II: 'OC1(S(C2=CC=C(C)C=C2)(=O)=O)CC1'.'CC1(C)CCC(C)(C)[S+]1C/C([2H])=C/C2=CC=CC=C2.F[B-](F)(F)F'>LiHMDS, CH2Cl2, -78°C to r.t.>'O=C1CCC1(C2=CC=CC=C2)C=C'. The product region shows the deuterium distribution: 94%, 94% D, 0% D, and 0% D."}], "full_mark": 7, "rubric": ["Key intermediates: '[2H]/C(C1=CC=CC=C1)=C\\[CH-][S+]2C(C)(C)CCC2(C)C' (1 point); 'O=C1CC1' (1 point); '[2H]C(/C=C/[S+]1C(C)(C)CCC1(C)C)(C2([O-])CC2)C3=CC=CC=C3' (1 point); 'CC1(C)CCC(C)(C)[S+]1[CH-]/C=C(C2(O[2H])CC2)/C3=CC=CC=C3' (2 points); 'CC1(C)CCC(C)(C)[S+]1C([2H])/C=C(C2([O-])CC2)/C3=CC=CC=C3' (2 points)"], "rubric_images": [{"file": "Exam-17-T10-2-1-rubric.png", "description": "Reference key intermediate 1, represented by the SMILES '[2H]/C(C1=CC=CC=C1)=C\\[CH-][S+]2C(C)(C)CCC2(C)C'."}, {"file": "Exam-17-T10-2-2-rubric.png", "description": "Reference key intermediate 2 (cyclopropanone), represented by the SMILES 'O=C1CC1'."}, {"file": "Exam-17-T10-2-3-rubric.png", "description": "Reference key intermediate 3, represented by the SMILES '[2H]C(/C=C/[S+]1C(C)(C)CCC1(C)C)(C2([O-])CC2)C3=CC=CC=C3'."}, {"file": "Exam-17-T10-2-4-rubric.png", "description": "Reference key intermediate 4, represented by the SMILES 'CC1(C)CCC(C)(C)[S+]1[CH-]/C=C(C2(O[2H])CC2)/C3=CC=CC=C3'."}, {"file": "Exam-17-T10-2-5-rubric.png", "description": "Reference key intermediate 5, represented by the SMILES 'CC1(C)CCC(C)(C)[S+]1C([2H])/C=C(C2([O-])CC2)/C3=CC=CC=C3'."}]}, {"sub_id": "Exam-17-T10-3", "tags": ["2.2", "2.4"], "text_open": "10.3 Inspired by this work, other research groups developed sulfur-ylide-mediated ring expansion of donor-acceptor spirocyclopropanes and applied it to the synthesis of spirocyclobutane frameworks.\n10.3.1 Based on the information above, write the SMILES strings of the key intermediate and final product in the reaction below. Stereochemistry is not required.", "text_images": [{"file": "Exam-17-T10-3-1-prompt.png", "description": "Single-step reaction: 'O=C1C2(C(C3=CC=CC=C3)C2)C(C4=CC5=CC=CC=C5C=C41)=O'.'O=C([CH-][S+](C)C)C1=CC=CC=C1'>EtOAc, reflux, 24 h (sulfur ylide 1.5 equiv)>; no product structure is drawn to the right of the arrow."}], "full_mark": 2, "rubric": ["Key intermediate: 'O=C1[C-](CC(C(C(C2=CC=CC=C2)=O)[S+](C)C)C3=CC=CC=C3)C(C4=CC5=CC=CC=C5C=C41)=O' (1 point). Final product: 'O=C1C2(CC(C3=CC=CC=C3)C2C(C4=CC=CC=C4)=O)C(C5=CC6=CC=CC=C6C=C51)=O' (1 point)."], "rubric_images": [{"file": "Exam-17-T10-3-1-rubric.png", "description": "Reference answer: the key intermediate in 10.3.1, represented by the SMILES 'O=C1[C-](CC(C(C(C2=CC=CC=C2)=O)[S+](C)C)C3=CC=CC=C3)C(C4=CC5=CC=CC=C5C=C41)=O'."}, {"file": "Exam-17-T10-3-2-rubric.png", "description": "Reference answer: the final product in 10.3.1, represented by the SMILES 'O=C1C2(CC(C3=CC=CC=C3)C2C(C4=CC=CC=C4)=O)C(C5=CC6=CC=CC=C6C=C51)=O'."}]}, {"sub_id": "Exam-17-T10-4", "tags": ["2.4"], "text_open": "10.3.2 If the spirocyclopropane substrate in 10.3.1 is replaced by the 2’, 3’-unsubstituted spirocyclopropane shown below, the reaction forms a fused-ring byproduct in addition to the desired ring-expanded product. Write the SMILES string of this byproduct.", "text_images": [{"file": "Exam-17-T10-4-1-prompt.png", "description": "The 2’, 3’-unsubstituted spirocyclopropane substrate has the SMILES 'O=C1C2(CC2)C(C3=CC4=CC=CC=C4C=C31)=O'."}], "full_mark": 2, "rubric": ["'O=C1C2=C(OC(C(C3=CC=CC=C3)=O)CC2)C4=CC5=CC=CC=C5C=C41' (2 points)"], "rubric_images": [{"file": "Exam-17-T10-4-1-rubric.png", "description": "Reference answer: the fused-ring byproduct in 10.3.2, represented by the SMILES 'O=C1C2=C(OC(C(C3=CC=CC=C3)=O)CC2)C4=CC5=CC=CC=C5C=C41'."}]}, {"sub_id": "Exam-17-T10-5", "tags": ["1.2", "2.7"], "text_open": "10.3.3 The reaction is in fact stereoselective. For the reaction in 10.3.1, assume that the substrate is racemic. Determine whether the major product is racemic, and determine whether the two substituents on the cyclobutane ring are cis or trans, explaining your answer.", "text_images": [], "full_mark": 3, "rubric": ["The product is racemic (1 point). The substituents are trans (1 point). In the cis pathway, Ph and COPh are crowded on the same face, whereas trans ring closure has less steric hindrance (1 point)."], "rubric_images": []}]} -{"problem_id": "Exam-18-T3", "header": "Problem 3 Latent Monomers and Controlled Metathesis Polymerization\nControlling the temporal and spatial sequence of polymerization is an important topic in conventional polymer synthesis. Chemists have generally developed “latent catalysts” that switch between inactive and active states in response to external stimuli such as light or heat. Recent research, however, has introduced a new “latent monomer” strategy: isomerization of a photoswitchable molecule generates an active monomer in situ and on demand, enabling efficient control of polymerization.\nThis problem concerns a classic class of highly reactive ring-opening metathesis polymerization (ROMP) monomers: norbornadiene (NBD) derivatives. Under ultraviolet irradiation of a specific wavelength, they undergo intramolecular cycloaddition to form highly strained quadricyclane (QC) structures.", "header_images": [], "subquestions": [{"sub_id": "Exam-18-T3-1", "tags": ["2.4"], "text_open": "3.1 Synthesis and Photothermal Isomerization of the Latent Monomer\nMonomer A, an NBD derivative, can be prepared by reacting cyclopentadiene with diethyl acetylenedicarboxylate at room temperature. Under 313 nm ultraviolet irradiation, A is quantitatively converted into latent monomer B, a QC derivative. Adding silica-coated gold nanobipyramids (AuBPs) to B and irradiating at 850 nm with infrared light regenerates A.\n3.1.1 Write the SMILES strings for A and B.", "text_images": [], "full_mark": 3, "rubric": ["'O=C(OCC)C1=C(C(OCC)=O)[C@@H]2C=C[C@H]1C2' and 'O=C(OCC)C12C3(C(OCC)=O)C4C3C1C2C4'. (1 point for the structure of A and 2 points for the structure of B)"], "rubric_images": [{"file": "Exam-18-T3-1-1-rubric.png", "description": "A has SMILES 'O=C(OCC)C1=C(C(OCC)=O)[C@@H]2C=C[C@H]1C2', and B has SMILES 'O=C(OCC)C12C3(C(OCC)=O)C4C3C1C2C4'."}]}, {"sub_id": "Exam-18-T3-2", "tags": ["2.4"], "text_open": "3.3 The complete inertness of latent monomers toward the catalyst at room temperature enables the synthesis of complex polymer topologies that are difficult to obtain by conventional metathesis polymerization. The researchers replaced the ethoxycarbonyl groups, $-COOC_2H_5$, in B with pent-4-enyloxycarbonyl groups, $-COO(CH_2)_3CH=CH_2$, to give monomer F. F was mixed with the Grubbs catalyst and reacted at 40℃, releasing gas X and giving a soluble linear polymer G with double bonds in its main chain. G was coated onto a microscope slide placed over a masked LED and then heated to 80℃, rapidly converting the film into an insoluble crosslinked network H. Write the SMILES strings for polymers G and H.", "text_images": [], "full_mark": 4, "rubric": ["G is a polymer whose repeat unit has the SMILES string 'O=C(OCCCC=C)C12C3(C(OCCCC=C)=O)C4C3C1C2C4'. Two repeat units are connected through their terminal double bonds, with one isolated double bond between each pair of repeat units. In the drawing, enclose the repeat unit in square brackets and place n as a subscript at the lower right. (2 points each for the structures of G and H; deduct 1 point if the degrees of polymerization of the upper and lower chains in H are both labeled m or both labeled n)"], "rubric_images": [{"file": "Exam-18-T3-2-1-rubric.png", "description": "G is a polymer with repeating-unit structure 'O=C(OCCCC=C)C12C3(C(OCCCC=C)=O)C4C3C1C2C4'. Two repeat units are connected through their terminal double bonds, with one isolated double bond between each pair of units. The repeating unit is enclosed in square brackets with a subscript n at the lower right."}, {"file": "Exam-18-T3-2-2-rubric.png", "description": "H is a polymer with repeating-unit structure 'C=CC(CC1C=C)C(C(OCCCC=C)=O)=C1C(OCCCC=C)=O'. Two repeat units are connected through double bonds attached to the five-membered ring, with one isolated double bond between each pair of units, and also through double bonds attached to the ester groups, again with one isolated double bond between each pair of units. The two connection modes have different degrees of polymerization: m for the former and n for the latter. Two pairs of square brackets enclose the two chains and carry subscripts n and m, respectively."}]}]} -{"problem_id": "Exam-18-T4", "header": "Problem 4 Complexometric Titration\nEthylenediaminetetraacetic acid (EDTA) is an effective chelating ligand widely used in chemical analysis. Approximately 20 g of a standard reagent is roughly weighed, 250mL of water is added, and the mixture is heated to dissolve the solid, cooled, and mixed thoroughly. The measured pH of the solution is 4.2. The stepwise ionization constants of EDTA are shown below.\n\n| | $K_{a1}$ | $K_{a2}$ | $K_{a3}$ | $K_{a4}$ | $K_{a5}$ | $K_{a6}$ |\n| --- | --- | --- | --- | --- | --- | --- |\n| | $10^{-0.9}$ | $10^{-1.6}$ | $10^{-2.0}$ | $10^{-2.67}$ | $10^{-6.16}$ | $10^{-10.26}$ |", "header_images": [], "subquestions": [{"sub_id": "Exam-18-T4-1", "tags": ["4.5", "3.4"], "text_open": "The standard solution is then standardized. Weigh 0.2500 g of working-reference ZnO that has been ignited to constant mass in a furnace at 800 ℃. Moisten it with a small amount of water, dissolve it in 2 mL of 20% hydrochloric acid solution, and add 100 mL of water. Adjust the solution to pH 8 with hexamethylenetetramine (HMTA) solution, add 10 mL of $(CH_2)_6N_4$-$(CH_2)_6N_4$·HCl buffer to adjust the pH to 10, and then add 5 drops of Eriochrome Black T indicator solution. Titrate with the prepared EDTA sodium-salt solution until the color changes from purple to pure blue. Conduct a blank determination in parallel. After blank correction, the volume of EDTA consumed is $V = 15.35\\ \\mathrm{mL}$.\n4.2.2 The buffering ability of a buffer solution is commonly measured by its buffer capacity, denoted β. It is defined as the amount db in mol of strong base required to raise the pH of 1L of buffer solution by dpH units, with the mathematical expression\n$\\beta = \\frac{db}{dpH} = -2.30[H^+] \\times \\frac{db}{d[H^+]}$\nGiven $pK_b((CH_2)_6N_4) = 4.9$, calculate the buffer capacity of a pH = 10 $(CH_2)_6N_4$-$(CH_2)_6N_4$·HCl buffer in which the analytical concentration of $(CH_2)_6N_4$ is 0.1000 mol/L. Hint: express $b$ as a function of $[H^+]$ and differentiate with respect to $[H^+]$ to obtain $\\frac{db}{d[H^+]}$. B may be used to represent $(CH_2)_6N_4$ in the calculation.", "text_images": [], "full_mark": 4, "rubric": ["In this problem, the HMTA-HMTA·HCl buffer may be treated as the sum of a $c$ mol/L HMTA·HCl solution and a $b$ mol/L monobasic strong-base solution. In the HMTA-HMTA·HCl buffer, addition of strong base causes $HB^+ + OH^- \\rightleftharpoons B + H_2O$. The proton-balance equation is\n$b = [OH^-] + [B] - [H^+]$.\nUsing the distribution fraction gives\n$b = -[H^+] + \\frac{K_w}{[H^+]} + \\frac{cK_a}{[H^+] + K_a}$. (1 point; either of the two equations above is sufficient.)\nDifferentiating with respect to $[H^+]$ gives\n$\\frac{db}{d[H^+]} = -1 - \\frac{K_w}{[H^+]^2} - \\frac{cK_a}{([H^+] + K_a)^2}$. (2 points)\nSubstitution into the original equation gives\n$\\beta = -2.30[H^+] \\times \\frac{db}{d[H^+]} = 2.30[H^+]\\{1 + \\frac{K_w}{[H^+]^2} + \\frac{cK_a}{([H^+] + K_a)^2}\\} = 0.02307$. (1 point)\nFull credit is also given for using $b \\approx [B] - [B]_0 = [HB^+]_0 - [HB^+]$ to obtain $\\frac{db}{d[H^+]} = \\frac{d([B] - [B]_0)}{d[H^+]} = -\\frac{cK_a}{([H^+] + K_a)^2}$.\nA correct result of $\\beta = 0.0228$ obtained using the Henderson-Hasselbalch equation is also awarded full credit."], "rubric_images": []}, {"sub_id": "Exam-18-T4-2", "tags": ["4.5", "3.4"], "text_open": "4.2.3 Neglecting volume changes during standardization, the change in solution pH over the entire process must not exceed 0.2 pH unit to ensure an accurate endpoint. Determine the minimum value of $[(CH_2)_6N_4]$.", "text_images": [], "full_mark": 4, "rubric": ["During standardization, $H_2Y^{2-} + Zn^{2+} = ZnY^{2-} + 2H^+$.\nThus, the amount of $H^+$ added to the solution during standardization is $n(H^+) = 2n(ZnO) = 0.006144\\ \\mathrm{mol}$. (1 point)\nFrom the definition of buffer capacity, the condition in the question becomes $\\beta \\geq -\\frac{da}{dpH} = \\frac{0.006144}{0.2} = 0.03072$. (1 point)\nBecause $[H^+]$ is very small under these conditions, $\\beta = 2.30[H^+]\\{1 + \\frac{K_w}{[H^+]^2} + \\frac{cK_a}{([H^+] + K_a)^2}\\} \\approx \\frac{2.30cK_a[H^+]}{([H^+] + K_a)^2}$. (1 point)\nSubstitution and solution of the equation give $c \\geq 0.09632\\ \\mathrm{mol/L}$. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-18-T4-3", "tags": ["4.1"], "text_open": "4.3 During standardization, why is the $(CH_2)_6N_4$-$(CH_2)_6N_4$·HCl buffer not added directly to adjust the solution to pH=10?", "text_images": [], "full_mark": 1, "rubric": ["The pH must be adjusted gradually to prevent formation of a ZnO precipitate that would interfere with standardization. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-18-T4-4", "tags": ["4.5", "1.4"], "text_open": "4.4 The EDTA standard solution standardized in this problem can be used to determine the concentrations of Cu2+ and Mg2+ in a solution by either of the following methods. Method 1: Masking Cu2+ ions. Adjust the solution to pH 5.5 with an HMTA buffer and add excess thiosulfate to mask Cu2+. In this reaction, thiosulfate acts as both a reducing agent and a ligand. Method 2: Selective determination of $Cu^{2+}$. Adjust the solution to pH 5 with an HMTA buffer. Using 5-Br-PADAT as the indicator, selectively determine $Cu^{2+}$ with an EDTP standard solution whose principal component is disodium ethylenediaminetetrapropionate. 4.4.2 Based on the data below, explain why EDTP shows high selectivity for $Cu^{2+}$.\n\n| | $Cu^{2+}$ | $Zn^{2+}$ | $Cd^{2+}$ | $Mg^{2+}$ |\n| --- | --- | --- | --- | --- |\n| Ionic radius/pm | 73 | 74 | 95 | 72 |\n| $\\lg K_{M-EDTP}$ | 15.4 | 7.8 | 6.0 | 1.8 |\n| $\\lg K_{M-EDTA}$ | 18.8 | 16.50 | 16.46 | 8.7 |", "text_images": [], "full_mark": 2, "rubric": ["$Cu^{2+}$ exhibits the Jahn-Teller effect in an octahedral field and adopts a distorted octahedral geometry. (1 point) EDTP is a chelating ligand that forms a relatively strained macrocycle on coordination to a metal ion, making the complexes less stable. (1 point) The distorted octahedral geometry relieves some of this strain, making Cu-EDTP more stable."], "rubric_images": []}]} +{"problem_id": "Exam-19-T6", "header": "Question 6 Boron Oxides\nBoron oxides possess advantageous properties and are used to prepare other boron compounds and to manufacture optical glass, electronic materials, ceramics, flame-retardant materials, and more. New boron-based 2D materials have been explored extensively in recent years. This problem examines the reactivity and structures of several boron oxides.", "header_images": [], "subquestions": [{"sub_id": "Exam-19-T6-1", "tags": ["1.3", "1.1"], "text_open": "6.1 $B_2O_3$ is the most common boron oxide and has some Lewis acidity. Reaction of $B_2O_3$ with $P_2O_5$ gives compound X, a catalyst for dehydration reactions. The figure below is a projection of a conventional unit cell of tetragonal X; the large spheres represent P, the medium spheres B, and the small spheres oxygen.\n6.1.1 Give the chemical formula of X and identify the characteristic symmetry element of the crystal.", "text_images": [{"file": "Exam-19-T6-1-1-prompt.png", "description": "The figure is a schematic of crystallographic axes. A small filled gray sphere at the lower left is labeled a. From this point, a thick horizontal arrow extends to the right and is labeled b at its tip, while a thick vertical arrow extends upward and is labeled c at its tip. No other text or numbers appear."}, {"file": "Exam-19-T6-1-2-prompt.png", "description": "A two-dimensional crystallographic projection of the unit cell. Fractional coordinates x and y are defined horizontally and vertically, with unit-cell boundaries at x=0,1 and y=0,1. Three visually distinct sphere types appear: large gray spheres, small gray spheres, and spheres labeled B or P. Approximate fractional coordinates of the large gray spheres are (0.0,1.0), (1.0,1.0), (0.0,0.75), (1.0,0.75), (0.0,0.0), (1.0,0.0), (0.0,0.25), and (1.0,0.25). Approximate fractional coordinates of the small gray spheres are (0.0,1.15), (0.3,1.15), (0.7,1.15), (1.0,1.15), (0.15,0.85), (0.15,0.65), (0.85,0.85), (0.85,0.65), (0.3,0.25), (0.7,0.25), (0.15,0.15), (0.15,-0.05), (0.85,0.15), and (0.85,-0.05), together with visible periodic images outside the unit-cell boundary. Approximate fractional coordinates of the labeled spheres are B at (0.0,0.75) and (0.5,0.25), and P at (0.5,0.5)."}], "full_mark": 2, "rubric": ["Chemical formula: $BPO_4$ (1 point).", "Characteristic symmetry element: an $I_4$ axis (1 point)."], "rubric_images": []}, {"sub_id": "Exam-19-T6-2", "tags": ["1.3"], "text_open": "6.1.2 How many distinct chemical environments does oxygen have in this crystal?", "text_images": [], "full_mark": 2, "rubric": ["1 (2 points)."], "rubric_images": []}, {"sub_id": "Exam-19-T6-3", "tags": ["1.3", "2.1"], "text_open": "6.1.3 The B–P distance is 272.9 pm. The B–P–B angle indicated in the figure is 111.7°, and the other B–P–B angle is 105.1°. Assuming that P and B are completely converted to their corresponding oxoacids, calculate the theoretical maximum mass of water that $1\\ \\mathrm{cm^3}$ of this crystal can absorb.", "text_images": [], "full_mark": 5, "rubric": ["$a=272.9\\times\\sin\\frac{105.1°}{2}\\times2=433.3\\ \\mathrm{pm}$ (1 point).", "$c=272.9\\times\\cos\\frac{105.1°}{2}\\times4=663.8\\ \\mathrm{pm}$ (1 point).", "Complete-hydrolysis equation: $BPO_4+3H_2O\\rightarrow H_3BO_3+H_3PO_4$ (1 point).", "Counting from the crystal structure gives $Z=2$; therefore, $m=\\frac{Z\\times M_r(H_2O)\\times3}{a^2cN_A}=1.439\\ \\mathrm{g}$ (2 points; 1.43–1.44 g is acceptable)."], "rubric_images": []}, {"sub_id": "Exam-19-T6-4", "tags": ["1.3"], "text_open": "6.2 The literature has reported formation of a crystalline compound with composition $B_2O$ by reacting B with $B_2O_3$ at high pressure and high temperature. One proposed structure belongs to the trigonal crystal system. It may be viewed as boron-only layers M and boron–oxygen layers N stacked in an …MNNMNN… sequence. Both M and N contain only juxtaposed six-membered rings. Adjacent N layers are connected by B–B covalent bonds, while B–O interactions parallel to the c axis occur between M and N layers, making both boron and oxygen nonplanar. The number of formula units per unit cell satisfies $Z\\leq3$. The shortest B–O distance within an N layer is 174.2 pm, and the shortest B–O distance between M and N layers is 199.6 pm. The various B–B bond lengths in the crystal are very similar and may all be taken as 169.3 pm in this problem.\n6.2.1 How many B–B covalent bonds are present in 1 mol of crystalline $B_2O$?", "text_images": [], "full_mark": 2, "rubric": ["$2N_A$ (2 points; 2 mol is also acceptable)."], "rubric_images": []}, {"sub_id": "Exam-19-T6-5", "tags": ["1.3"], "text_open": "6.2.3 The unit-cell parameter of this crystal is $c = 705.2\\ \\mathrm{pm}$. Calculate the unit-cell parameter $a$.", "text_images": [], "full_mark": 5, "rubric": ["Let $x$ be the corrugation height of an N layer and $y$ that of an M layer. Geometry gives $x^2+(\\frac{a}{\\sqrt{3}})^2=d_{B-O,horizontal}^2,\\ y^2+(\\frac{a}{\\sqrt{3}})^2=d_{B-B}^2$ (2 points).", "$c=2x+y+d_{B-B}+2d_{B-O,vertical}$ (1 point).", "Solving gives $a=287.9\\ \\mathrm{pm}$ (2 points; 287–289 pm is acceptable)."], "rubric_images": []}, {"sub_id": "Exam-19-T6-6", "tags": ["1.1"], "text_open": "6.3 BO was reported as early as 1940, but technological limitations long prevented determination of its structure. In 2023, researchers used multiple methods and analytical tools to propose several planar two-dimensional candidate models. BO can be formed by dehydration of compound A. A is synthesized as follows: reacting $BCl_3$ with Cu vapor on a liquid-nitrogen-cooled surface gives compound B (Reaction 1); hydrolysis of B gives A, which yields BO upon dehydration. When B reacts with an equimolar amount of hydrogen, the only products are two simple binary compounds (Reaction 2). B also reacts with cis-2-butene (Reaction 3).\n6.3.1 Compound B exists as a monomer in both the gas and solid phases, but the molecular structures have different symmetries. Write the SMILES strings for both structures; the gas/solid assignment need not be indicated.", "text_images": [], "full_mark": 2, "rubric": ["Planar 'ClB(Cl)B(Cl)Cl' and D2d 'ClB(Cl)B(Cl)Cl' (1 point each; either order)."], "rubric_images": [{"file": "Exam-19-T6-6-1-rubric.png", "description": "The figure shows two structures of compound B, both with the same SMILES 'ClB(Cl)B(Cl)Cl'. They differ in conformation/symmetry about the B–B bond. The left structure shows planar Cl2B–BCl2 using solid lines. In the right structure, the two Cl atoms on the right-hand B are shown with a solid wedge and a dashed bond, respectively, so the two BCl_2 planes are perpendicular. The gas/solid assignments are not labeled."}]}, {"sub_id": "Exam-19-T6-7", "tags": ["2.1", "2.4"], "text_open": "6.3.2 Write the equations for Reactions 1–3, paying attention to the stereochemistry of Reaction 3.", "text_images": [], "full_mark": 6, "rubric": ["Reaction 1: $2BCl_3+2Cu\\rightarrow B_2Cl_4+2CuCl$ (2 points).", "Reaction 2: $3B_2Cl_4+3H_2\\rightarrow B_2H_6+4BCl_3$ (2 points).", "Reaction 3: 'ClB(Cl)B(Cl)Cl'.'C/C=C\\C'>'C[C@H](B(Cl)Cl)[C@H](B(Cl)Cl)C' (2 points; only 1 point if stereochemistry is not shown)."], "rubric_images": [{"file": "Exam-19-T6-7-1-rubric.png", "description": "The figure shows the one-step reaction 'ClB(Cl)B(Cl)Cl'.'C/C=C\\C'>'C[C@H](B(Cl)Cl)[C@H](B(Cl)Cl)C'. The product is meso-2,3-bis(dichloroboryl)butane. The left BCl2 group is shown with a dashed wedge and the right BCl2 group with a solid wedge, placing the two BCl2 groups on opposite sides of the plane."}]}, {"sub_id": "Exam-19-T6-8", "tags": ["1.3", "1.1"], "text_open": "According to the hypothesis proposed in the 2023 study, both idealized ordered stacking models of BO can be regarded as different stackings of the same infinite planar BO sheet. The AAAA… stacking forms a primitive orthorhombic lattice, whereas ABCD stacking forms a body-centered tetragonal lattice. A single boron–oxygen sheet contains only 6-, 12-, and 16-membered rings; one actual two-dimensional unit cell contains 2, 1, and 1 of these rings, respectively. If the slight distortion of the 12-membered ring is ignored, the sheet has a fourfold axis; in the actual structure, the 12-membered ring has only a twofold axis.\n6.3.3 Write the SMILES string for the boron–oxygen sheet, and state the number of boron atoms in one smallest actual two-dimensional unit cell.", "text_images": [], "full_mark": 5, "rubric": ["'[*]B1B(OB2OB3OB4B5OB6OB7OB8B(OB9B(O8)OB(O%10)B(OB%11OB(B(OB%11OB%12OB(B(OB%12O9)[*])[*])[*])[*])OB(OB%13OB(B(OB%13OB2O1)[*])[*])B%10OB3O5)OB%14OB(B(OB%14OB%15OB(B(OB%15OB7OB6OB%16OB(B(OB%16OB%17OB(B(OB%17O4)[*])[*])[*])[*])[*])[*])[*])[*])[*]' (2 points for the correct $[B_4O_2]$ six-membered-ring fragment and 2 points for correctly depicting all three ring sizes; 4 points total).", "Number of boron atoms per unit cell: 8 (1 point)."], "rubric_images": [{"file": "Exam-19-T6-8-1-rubric.png", "description": "The figure shows a two-dimensional boron–oxygen network formed by $[B_4O_2]$ six-membered rings, with atom sequence B–B–O–B–B–O, connected through shared B–B bonds. Boron atoms at the network edges carry unlabeled short methyl termination lines. The full network SMILES is 'CB1B(C)OB(OB2OB3B(OB4OB(OB5B(OB6OB(C)B(C)OB6OB7OB(C)B(C)OB7O8)OB9B(O%10)O5)B8OB4OB%11OB(C)B(C)OB%11OB(OB(C)B(C)O%12)B%12O3)OB2OB%13B(OB(OB%14OB(C)B(C)OB%14OB(OB(C)B(C)O%15)B%15O9)B%10O%13)OB%16OB(C)B(C)OB%16O%17)B%17O1' and includes 16 methyl terminations."}]}, {"sub_id": "Exam-19-T6-9", "tags": ["4.3", "1.3"], "text_open": "6.3.5 To verify that the boron atoms in the crystal are bonded to one another and coplanar, meaning that both angle $\\theta$ and angle $\\psi$ in the right-hand figure are 0°, which experimental technique is most appropriate for a BO powder sample?\n(a) Electron paramagnetic resonance (b) Solid-state nuclear magnetic resonance (c) Electron diffraction (d) Thermogravimetric analysis", "text_images": [{"file": "Exam-19-T6-9-1-prompt.png", "description": "The figure shows two boron–oxygen structural fragments. In the upper fragment, a large central circle labeled B is connected by two short horizontal lines to two small circles labeled O on its right, by a diagonal line to one O at upper right, and by two diagonal lines to two O atoms on the left; the angle symbol θ is shown between the two left O–B lines. The lower fragment contains two B atoms joined by a single bond, each additionally bonded to two O atoms: the left B to upper-left and lower-left O, and the right B to upper-right and lower-right O. Arrows above the two B atoms point toward the upper left and upper right, and an arc between them is labeled ψ. The only text and symbols are B, O, θ, and ψ."}], "full_mark": 1, "rubric": ["(b) (1 point)."], "rubric_images": []}, {"sub_id": "Exam-19-T6-10", "tags": ["1.3"], "text_open": "In 2025, researchers prepared a $^{17}O$-labeled BO sample and used more precise instrumental analyses to establish that the BO crystal is not a stack of two-dimensional sheets but consists mainly of long one-dimensional chains, thereby disproving the planar two-dimensional model proposed in 2023.\n6.3.6 Write the SMILES string for the one-dimensional BO chain.", "text_images": [], "full_mark": 2, "rubric": ["'[*]OB1OB([*])B([*])OB1O[*]' (2 points; terminal groups need not be shown)."], "rubric_images": [{"file": "Exam-19-T6-10-1-rubric.png", "description": "The figure shows a one-dimensional long-chain BO polymer. Brackets enclose the repeating unit and a subscript n appears at the lower right. The repeating-unit SMILES is '[*]OB1OB([*])B([*])OB1O[*]', where [*] denotes a connection site to an adjacent repeating unit: each of the two left-hand B atoms is directly bonded to one [*], and each of the two right-hand B atoms connects to [*] through an O atom."}]}, {"sub_id": "Exam-19-T6-11", "tags": ["4.3", "1.3"], "text_open": "6.3.7 Powder X-ray diffraction once suggested that BO has two-dimensional character, apparently contradicting the one-dimensional long-chain structure. Briefly explain this observation.", "text_images": [], "full_mark": 2, "rubric": ["The one-dimensional B–O polymer chains exhibit random stacking or stacking faults, so their average structure produces diffraction features resembling an ordered two-dimensional layered structure (2 points: 1 point for random stacking and 1 point for averaging into a plane)."], "rubric_images": []}]} {"problem_id": "Exam-19-T7", "header": "Question 7 Fundamental Organic Chemistry", "header_images": [], "subquestions": [{"sub_id": "Exam-19-T7-1", "tags": ["1.2", "1.6"], "text_open": "7.2 Consider the two conformations of $FCH_2CH_2X^+$ shown below, which interconvert by rotation about the C–C single bond. The relative energy of the conformation on the right is defined as 0.0 kcal/mol. Define the relative energy of the left conformation with respect to the right as $\\Delta E=E_{left\\ conformation}-E_{right\\ conformation}$.\n7.2.2 Let the values of $\\Delta E$ for $X^+ = NH_3^+$, $OH_2^+$, and 1-pyridyl (1-pyridinium) be A, B, and C, respectively. Arrange A, B, and C in increasing order.", "text_images": [{"file": "Exam-19-T7-1-1-prompt.png", "description": "The figure shows two conformations of $FCH_2CH_2X^+$ connected by a reversible-equilibrium arrow. The left conformation has SMILES 'FCC[X+]', and the right conformation has SMILES 'FCC[X+]'. They are conformations of the same molecule and therefore share the same SMILES. In the left conformation, the C–X+ bond is shown as a thick wedge. The right conformation is an extended zigzag arrangement in which X+ and F are not anti, and 0.0 kcal/mol is written below it. The bottom label reads $X^+ = -NH_3^+, -OH_2^+, 1-pyridyl$."}], "full_mark": 2, "rubric": ["B < A < C (2 points).", "Credit is awarded only for the completely correct order."], "rubric_images": []}, {"sub_id": "Exam-19-T7-2", "tags": ["4.3"], "text_open": "7.3 Comparison of stretching wavenumbers\n7.3.1 Which of the following 2 compounds has the higher carbonyl stretching wavenumber in the infrared spectrum?", "text_images": [{"file": "Exam-19-T7-2-1-prompt.png", "description": "The SMILES of A is '[H]C(C1=CC=CC=C1)=O'; the SMILES of B is '[H]/C(C1=CC=CC=C1)=[O+]/[B-](F)(F)F'. B is the benzaldehyde–BF3 adduct. A ⊕ sign appears next to the carbonyl oxygen and a ⊖ sign next to F3B, corresponding to [O+] and [B-] in the SMILES. O–B is a single bond."}], "full_mark": 1, "rubric": ["A (1 point)."], "rubric_images": []}, {"sub_id": "Exam-19-T7-3", "tags": ["1.2", "1.7"], "text_open": "7.4 Drug molecules usually must adopt a specific conformation when binding to a target protein. If the dominant free-state conformation of a drug is already close to its bound conformation, the free-energy cost of conformational reorganization upon binding is reduced, which can increase binding affinity. Indinavir binds most effectively to HIV protease when its central carbon chain adopts an extended zigzag conformation. The figure below shows two epimers, A and B, of an indinavir analog. Determine which isomer is more likely to bind the target protein more strongly.", "text_images": [{"file": "Exam-19-T7-3-1-prompt.png", "description": "The SMILES of A is 'O=C(N[C@@H]1C(C=CC=C2)=C2C[C@H]1O)[C@H](CC3=CC=CC=C3)[C@@H](F)[C@H](O)CN(CC4)[C@H](C(NC(C)(C)C)=O)CN4CC5=CC=CN=C5'; the SMILES of B is 'O=C(N[C@@H]1C(C=CC=C2)=C2C[C@H]1O)[C@H](CC3=CC=CC=C3)[C@H](F)[C@H](O)CN(CC4)[C@H](C(NC(C)(C)C)=O)CN4CC5=CC=CN=C5'. A and B are epimers of an indinavir analog with the same skeleton, differing only in the orientation of F at the central-chain CH(F) carbon. F is shown with a wedge in A and a dashed bond in B; the SMILES preserve this stereochemical difference as [C@@H](F) and [C@H](F), respectively."}], "full_mark": 2, "rubric": ["A (2 points)."], "rubric_images": []}, {"sub_id": "Exam-19-T7-4", "tags": ["1.2"], "text_open": "7.5 Consider the three representative conformations A, B, and C shown below. Arrange their relative energies in increasing order.", "text_images": [{"file": "Exam-19-T7-4-1-prompt.png", "description": "The figure shows three representative conformations A, B, and C of dimethoxymethane ($CH_3OCH_2OCH_3$), with reversible-equilibrium arrows between A and B and between B and C. All three have the same SMILES: A is 'COCOC', B is 'COCOC', and C is 'COCOC'. Their conformational differences are expressed by bond orientations that SMILES cannot encode. A is a planar zigzag with both O–CH3 bonds drawn as ordinary single bonds. In B, the left O–CH3 bond is a solid wedge pointing downward and the right O–CH3 bond is an ordinary single bond extending toward the upper right. In C, the left O–CH3 bond is a solid wedge pointing downward and the right O–CH3 bond is a dashed bond pointing downward."}], "full_mark": 2, "rubric": ["C < B < A (2 points).", "Credit is awarded only for the completely correct order."], "rubric_images": []}]} {"problem_id": "Exam-20-T7", "header": "Nitric Oxide Synthase and Its Artificial Models\nNitric oxide, NO, is an important signaling molecule involved in regulating many physiological processes in mammalian cardiovascular, immune, and nervous systems. In the human body, NO is synthesized under the catalysis of the heme enzyme nitric oxide synthase (NOS). Researchers are also working to reproduce this biological synthetic pathway using artificial models outside the body.", "header_images": [], "subquestions": [{"sub_id": "Exam-20-T7-1", "tags": ["2.1", "1.6"], "text_open": "7.1 Arginine is a natural amino acid involved in multiple metabolic pathways in the human body, including synthesis of the signaling molecule nitric oxide and the urea cycle. The structure of arginine in the cellular environment is shown below.\n7.1.1 In the final step of the urea cycle, arginine is a reactant in a reaction that releases urea and forms amino acid A. No redox process occurs. Write the reaction equation, representing organic compounds by SMILES strings.", "text_images": [{"file": "Exam-20-T7-1-1-prompt.png", "description": "Arginine has the SMILES '[NH3+][C@@H](CCCNC(N)=[NH2+])C([O-])=O'. Stereochemistry at the α-carbon is shown by a solid wedge bond, and the overall structure is zwitterionic."}], "full_mark": 2, "rubric": ["'[NH3+][C@@H](CCCNC(N)=[NH2+])C([O-])=O' + H2O > '[NH3+][C@@H](CCC[NH3+])C([O-])=O' + 'NC(N)=O'. 2 points total; if the amino group of ornithine in the product is not protonated, award only 1 point."], "rubric_images": [{"file": "Exam-20-T7-1-1-rubric.png", "description": "Reference equation for 7.1.1: '[NH3+][C@@H](CCCNC(N)=[NH2+])C([O-])=O' + H2O > '[NH3+][C@@H](CCC[NH3+])C([O-])=O' + 'NC(N)=O'. The coefficient of every species is 1, and the α-carbon is shown with a solid wedge bond."}]}, {"sub_id": "Exam-20-T7-2", "tags": ["2.2", "2.1"], "text_open": "For the reaction equations below, electrically neutral citrulline in the cellular environment may be represented by the SMILES string '[NH3+][C@@H](CCCNC(N)=O)C([O-])=O'.\n7.1.3 In the biological pathway for nitric oxide synthesis, arginine is the starting amino acid and reacts with 1 equivalent of NADPH and 1 equivalent of $O_2$ to form intermediate B and water (reaction 1). B then reacts further with $O_2$ and NADPH to form citrulline and nitric oxide, NO (reaction 2). The oxidized form of NADPH in organisms is $NADP^+$. Based on this information, infer and write the SMILES string of B and write the equation for reaction 2. The detailed structures of NADPH/$NADP^+$ need not be represented by SMILES.", "text_images": [], "full_mark": 3, "rubric": ["Intermediate: '[NH3+][C@@H](CCCN/C(N)=N\\O)C([O-])=O' or '[NH3+][C@@H](CCCNC(NO)=[NH2+])C([O-])=O' (1 point).", "2'[NH3+][C@@H](CCCN/C(N)=N\\O)C([O-])=O' + 2O2 + NADPH + H+ > 2'[NH3+][C@@H](CCCNC(N)=O)C([O-])=O' + 2NO + 2H2O + NADP+ or 2'[NH3+][C@@H](CCCNC(NO)=[NH2+])C([O-])=O' + 2O2 + NADPH > 2'[NH3+][C@@H](CCCNC(N)=O)C([O-])=O' + 2NO + 2H2O + NADP+ + H+ (2 points total). Balancing with $OH^-$/$H_2O$ is also acceptable."], "rubric_images": [{"file": "Exam-20-T7-2-1-rubric.png", "description": "Intermediate B in the first representation has the SMILES '[NH3+][C@@H](CCCN/C(N)=N\\O)C([O-])=O'."}, {"file": "Exam-20-T7-2-2-rubric.png", "description": "Intermediate B in the second representation has the SMILES '[NH3+][C@@H](CCCNC(NO)=[NH2+])C([O-])=O'."}, {"file": "Exam-20-T7-2-3-rubric.png", "description": "Reference equation for reaction 2, first version: 2'[NH3+][C@@H](CCCN/C(N)=N\\O)C([O-])=O' + 2O2 + NADPH + H+ > 2'[NH3+][C@@H](CCCNC(N)=O)C([O-])=O' + 2NO + 2H2O + NADP+."}, {"file": "Exam-20-T7-2-4-rubric.png", "description": "Reference equation for reaction 2, second version: 2'[NH3+][C@@H](CCCNC(NO)=[NH2+])C([O-])=O' + 2O2 + NADPH > 2'[NH3+][C@@H](CCCNC(N)=O)C([O-])=O' + 2NO + 2H2O + NADP+ + H+."}]}, {"sub_id": "Exam-20-T7-3", "tags": ["2.4", "1.4"], "text_open": "7.2 The enzymatic mechanism above shows that guanidino groups and their derivatives play important roles in biological NO synthesis. Accordingly, oxidation of guanidine-related ligands by various oxidants to generate NO in vitro has been studied extensively as a strategy for artificial NOS models. Osmium-centered complexes can undergo pH-dependent aerobic oxidation at ambient temperature and pressure to release nitric oxide. The precursor complex is shown below.\n**7.2.1** Interestingly, the precursor complex reacts with a large excess of $CN^-$ (20 eq.) to form a new cyanamide complex **X**. During the process, the tridentate Schiff-base ligand isomerizes into a bidentate ligand containing a five-membered ring, accompanied by hydride transfer; finally, the Os center undergoes a one-electron oxidation. The resulting complex ion **X** has a meridional (mer) structure and an overall charge of two negative units. Draw the structure of complex ion **X** and state the oxidation state of the central Os.", "text_images": [{"file": "Exam-20-T7-3-1-prompt.png", "description": "The precursor complex has the SMILES 'Cl[Os]12([N](C3=CC=CC=C3O1)=CC4=C(O2)C=CC=C4)([OH2])#N'. The central Os is six-coordinate and bears chloride, water, and an N,O-containing ligand."}], "full_mark": 4, "rubric": ["Complex ion X has the SMILES '[H]N(C#N)[Os]1(C#N)(C#N)(C#N)OC2=CC=CC=C2C3=[N]1C4=CC=CC=C4O3' and is an overall 2- anion. The structure of X is worth 3 points: 1 point for the correct bidentate ligand, 1 point for the correct cyanamide ligand, and 1 point for the correct meridional arrangement of cyanide ligands. Deduct 1 point if the charge is not shown, and deduct 1 point if an unnecessary (±) label is added.", "Oxidation state of Os: +3 or III (1 point)"], "rubric_images": [{"file": "Exam-20-T7-3-1-rubric.png", "description": "Complex ion X has the SMILES '[H]N(C#N)[Os]1(C#N)(C#N)(C#N)OC2=CC=CC=C2C3=[N]1C4=CC=CC=C4O3' and is an overall 2- anion, with the total charge shown outside the brackets."}]}, {"sub_id": "Exam-20-T7-4", "tags": ["2.4", "2.2"], "text_open": "7.2.2 Complex ion X rapidly undergoes addition with $NH_2OH\\cdot HCl$ to form monoanionic complex C. Infrared spectroscopy indicates that C contains a C=N double bond outside an aromatic ring. C reacts with $O_2$ by different pathways at different pH values.\nAt pH = 6.8, C reacts with 0.5 equivalent of $O_2$ to form complex ion D, which undergoes linkage isomerization of the ligand to form E. Further oxidation of E by $O_2$ gives small molecule M and diamagnetic complex ion F containing a nitrosyl ligand. Single-crystal diffraction gives an N-O bond length of 1.393 Å in C and 1.158 Å in D.\nAt pH = 12, reaction of C with $O_2$ gives complex ion G. G is structurally similar to D and carries the same charge, but its relative molecular mass is 1 less than that of D. G can rapidly lose a proton to form X and release NO, or react with $OH^-$ to form dianionic complex ion H and likewise release NO.\nIn this problem, abbreviate ligand groups that do not participate in subsequent reactions. For example, the precursor complex in 7.2.1 may be written as [Os]≡N. Based on this information, infer and write the SMILES strings of C-H and M.", "text_images": [], "full_mark": 11, "rubric": ["C: '[#76]/[N]([H])=C(NO)\\N', an overall 1- anion (2 points).", "D: '[#76]/[N]([H])=C(N=O)\\N', an overall 1- anion (2 points).", "E: '[#76][N](/C(N)=N/[H])=O', an overall 1- anion (1 point).", "F: '[#76]N=O', an overall 1- anion (1 point).", "G: '[#76]/N=C(N=O)\\N', an overall 1- anion (2 points).", "H: '[#76]N([H])C(N)=O', an overall 2- anion (2 points).", "M: 'NC#N' (1 point).", "As required in the prompt, ligand groups that do not participate in subsequent reactions are abbreviated in these answers. Expanding [Os] to the full structure is also acceptable."], "rubric_images": [{"file": "Exam-20-T7-4-1-rubric.png", "description": "C has the SMILES '[#76]/[N]([H])=C(NO)\\N' and is an overall 1- anion, with the total charge shown outside the brackets."}, {"file": "Exam-20-T7-4-2-rubric.png", "description": "D has the SMILES '[#76]/[N]([H])=C(N=O)\\N' and is an overall 1- anion, with the total charge shown outside the brackets."}, {"file": "Exam-20-T7-4-3-rubric.png", "description": "E has the SMILES '[#76][N](/C(N)=N/[H])=O' and is an overall 1- anion, with the total charge shown outside the brackets."}, {"file": "Exam-20-T7-4-4-rubric.png", "description": "F has the SMILES '[#76]N=O' and is an overall 1- anion, with the total charge shown outside the brackets."}, {"file": "Exam-20-T7-4-5-rubric.png", "description": "G has the SMILES '[#76]/N=C(N=O)\\N' and is an overall 1- anion, with the total charge shown outside the brackets."}, {"file": "Exam-20-T7-4-6-rubric.png", "description": "H has the SMILES '[#76]N([H])C(N)=O' and is an overall 2- anion, with the total charge shown outside the brackets."}, {"file": "Exam-20-T7-4-7-rubric.png", "description": "M has the SMILES 'NC#N'."}, {"file": "Exam-20-T7-4-8-rubric.png", "description": "Reference structure with the [Os] abbreviation expanded, represented by the SMILES 'C[Os]1(C#N)(C#N)(C#N)OC2=CC=CC=C2C3=[N]1C4=CC=CC=C4O3'. The central Os is bonded to a methyl group, three cyanide ligands (C≡N), and an O,N-bidentate ligand, the deprotonated form of 2-hydroxyphenylbenzoxazole."}]}, {"sub_id": "Exam-20-T7-5", "tags": ["1.4"], "text_open": "7.2.3 Among complexes C-H, identify those in which the oxidation state of the central Os differs from that in X. Give their labels.", "text_images": [], "full_mark": 2, "rubric": ["F, G", "2 points total. Deduct 1 point for each extra or omitted choice, down to a minimum of zero."], "rubric_images": []}, {"sub_id": "Exam-20-T7-6", "tags": ["2.1"], "text_open": "7.2.4 Write the equation for formation of H from C with release of NO at pH = 12. Ligand groups not participating in the reaction may be abbreviated.", "text_images": [], "full_mark": 2, "rubric": ["4'[#76]/[N]([H])=C(NO)\\N' + 3O2 + 4OH- > 4'[#76]N([H])C(N)=O' + 4NO + 6H2O (2 points for the reaction equation)."], "rubric_images": [{"file": "Exam-20-T7-6-1-rubric.png", "description": "Reference equation for 7.2.4: 4'[#76]/[N]([H])=C(NO)\\N' + 3O2 + 4OH- > 4'[#76]N([H])C(N)=O' + 4NO + 6H2O. C is a 1- anion and H is a 2- anion."}]}]} -{"problem_id": "Exam-21-T3", "header": "Preparation and Properties of a Thermochromic Copper(II) Complex\nMaterials whose compounds exhibit thermochromism through structural phase transitions are widely used in anti-counterfeiting applications. To synthesize the thermochromic complex $[Cu(deen)_2](ClO_4)_2$ (deen denotes N,N-diethylethylenediamine), hydrated copper perchlorate must first be prepared, followed by introduction of the deen ligand.", "header_images": [], "subquestions": [{"sub_id": "Exam-21-T3-1", "tags": ["1.4", "4.3"], "text_open": "Hydrated copper perchlorate can be prepared by a precipitation–dissolution method. Dissolve $CuSO_4\\cdot 5H_2O$ in water, add $NaOH$ solution with stirring, and heat slowly on a hot plate until the precipitate turns black. Maintain gentle boiling for 5–10 min, allow the mixture to cool slightly, and collect the precipitate by suction filtration. Rinse it with a small amount of deionized water, then use a glass rod to transfer the precipitate into an evaporating dish containing 10 mL of water. Add perchloric acid, gently guide the precipitate into the solution with the glass rod, and stir until it dissolves completely to form a copper perchlorate solution. Concentrate the solution by gentle heating to about 2 mL. On cooling, crystals of hydrated copper perchlorate are obtained. The preparation of $[Cu(deen)_2](ClO_4)_2$ begins with the hydrated copper perchlorate crystals obtained above. Adding absolute ethanol to these crystals gives a clear blue solution. In a dry beaker, dissolve N,N-diethylethylenediamine in ethanol to obtain a clear pale-yellow solution. Slowly add the ethanolic deen solution to the ethanolic copper perchlorate solution with stirring; the mixture immediately turns deep blue-violet. Stir for 2–3min, then allow it to stand for about 10 min. A large amount of pink precipitate forms at the bottom of the vessel while the supernatant remains deep blue-violet. Collect the product by suction filtration and dry it to obtain $[Cu(deen)_2](ClO_4)_2$. 3.2.2 Dissolving the red $[Cu(deen)_2](ClO_4)_2$ in water again gives a deep-blue solution. Explain why. (No more than 50 words)", "text_images": [], "full_mark": 4, "rubric": ["$H_2O$ coordinates at the axial sites of $[Cu(deen)_2]^{2+}$, changing the coordination geometry of Cu(II) from square planar toward elongated octahedral. (2 points)", "The energy of the associated d–d transition decreases and the absorption undergoes a red shift, causing the complementary color to shift toward blue and giving a blue appearance. (2 points)"], "rubric_images": []}, {"sub_id": "Exam-21-T3-2", "tags": ["1.4"], "text_open": "3.3 In a thermochromism experiment on $[Cu(deen)_2](ClO_4)_2$, a small amount of the $[Cu(deen)_2](ClO_4)_2$ complex is held between pieces of adhesive tape and heated with a hair dryer. The sample changes from pink to blue-violet and returns to pink on cooling.\n3.3.1 Based on crystal field theory, describe in words as necessary the structural change of the $[Cu(deen)_2]^{2+}$ unit upon heating and cooling. The cooled unit may be approximated as centrosymmetric, whereas after heating it is no longer centrosymmetric.", "text_images": [], "full_mark": 3, "rubric": ["Before heating, the coordination geometry is square planar (1 point); in the high-temperature phase, the $CuN_4$ unit undergoes a slight tetrahedral distortion out of the plane. (2 points)", "The answer must show that the two ligands are not coplanar after heating. It need not be drawn as an ideal tetrahedron, although an ideal tetrahedral depiction is not penalized."], "rubric_images": [{"file": "Exam-21-T3-2-1-rubric.png", "description": "Rubric diagram for 3.3.1 showing the structures of the $[Cu(deen)_2]^{2+}$ unit after heating and cooling. Both the left and right structures contain a central Cu coordinated by four N atoms. Two N atoms are $-NH_2$ termini (each labeled $H_2$ next to N, at upper left and lower right), and two are $-NEt_2$ termini (each bearing two ethyl zigzag chains). The structure to the left of the reversible arrow is the preheating structure: all four Cu coordination bonds lie in one plane and Cu has square-planar geometry. In the heated structure on the right, the two bonds to the right of Cu are wedge bonds pointing behind and in front of the plane (a dashed wedge above and a solid bold wedge below), while the other two remain ordinary solid lines, indicating a slight tetrahedral distortion of four-coordinate Cu. A reversible arrow connects the structures, labeled \"Heating\" above and \"Cooling\" below. No charge is shown in the diagram."}]}, {"sub_id": "Exam-21-T3-3", "tags": ["4.6"], "text_open": "3.4 To investigate the thermochromism of $[Cu(deen)_2]X_2$ complexes formed with monovalent counteranion $X^-$, differential scanning calorimetry is performed on $[Cu(deen)_2](ClO_4)_2$ and $[Cu(deen)_2](NO_3)_2$. Multiple interactions occur between the cations and anions in both complexes. The results are shown below. For both complexes, the color change is associated with a phase-transition temperature, and the chemical composition remains unchanged across the color change. The temperature at the peak maximum is defined as the transition temperature.\n3.4.1 From the graph, directly give the thermochromic transition temperatures of $[Cu(deen)_2](ClO_4)_2$ and $[Cu(deen)_2](NO_3)_2$.", "text_images": [{"file": "Exam-21-T3-3-1-prompt.png", "description": "Untitled coordinate plot. The horizontal axis is labeled \"Temperature/°C\" with major ticks at 50, 100, 150, and 200 and minor ticks at 25, 75, 125, and 175. The vertical axis is labeled \"Thermogravimetry\" vertically and has no numerical ticks. Two solid black curves are shown. The legend at lower left uses short line segments for \"$ClO_4$(a)\" and \"$NO_3$(b),\" and the curves are labeled (a) and (b). Curve (a), corresponding to $[Cu(deen)_2](ClO_4)_2$, has a lower baseline and a sharp upward peak at about 45–50°C, remains flat thereafter, and drops steeply after about 200°C. Curve (b), corresponding to $[Cu(deen)_2](NO_3)_2$, has a higher baseline, a broad upward peak at about 110°C, a sharp upward peak at about 145°C, another smaller upward peak at about 180°C, and a steep drop after about 200°C. No arrows are shown."}], "full_mark": 4, "rubric": ["$[Cu(deen)_2](ClO_4)_2$ : 45 ℃ (any value from 40 ~ 50 ℃ is accepted) (2 points). No credit is awarded if any other peak is identified as the transition peak.", "$[Cu(deen)_2](NO_3)_2$: 145 ℃ (any value from 135–150 ℃ is accepted) (2 points)"], "rubric_images": []}, {"sub_id": "Exam-21-T3-4", "tags": ["1.4", "4.6"], "text_open": "3.4.2 Using the information above, explain the origin of the difference between the transition temperatures of $[Cu(deen)_2](ClO_4)_2$ and $[Cu(deen)_2](NO_3)_2$. (No more than 100 words)", "text_images": [], "full_mark": 4, "rubric": ["① Difference in coordination ability: $NO_3^-$ coordinates more strongly than $ClO_4^-$ and may partially coordinate to Cu(II), so structural distortion of $[Cu(deen)_2](NO_3)_2$ requires more energy and occurs at a higher transition temperature. (2 points)", "② The hydrogen-bond network formed by $NO_3^-$ and ligand N-H groups is stronger and more ordered than that in the $ClO_4^-$ system. This makes the low-temperature structure of $[Cu(deen)_2](NO_3)_2$ more stable, or raises the energy required for the phase transition, resulting in a higher transition temperature. (2 points)", "Anion size (steric effects) and lattice energy may also matter, but they were not addressed above. Answers based on these factors—for example, that bulky $ClO_4^-$ gives a loosely packed lattice and sufficient free space for configurational distortion of $[Cu(deen)_2]^{2+}$, whereas dense packing of $NO_3^-$ suppresses such distortion—may receive 1–2 points at the grader's discretion."], "rubric_images": []}]} -{"problem_id": "Exam-21-T4", "header": "Determination of Manganese by Ammonium Nitrate Oxidation Titrimetry\nManganese is an important element in steels and alloys, and its content directly affects their strength, hardness, and wear resistance. Ammonium nitrate oxidation titrimetry is a classical method for determining manganese: ammonium nitrate oxidizes Mn(II) to Mn(III) in a phosphoric acid medium, and the product is then titrated with a standard ammonium iron(II) sulfate solution. This problem addresses standardization of the titrant, control of the oxidation conditions, and calculation of the result.", "header_images": [], "subquestions": [{"sub_id": "Exam-21-T4-1", "tags": ["1.4"], "text_open": "4.2 Sample treatment and titration\nWeigh 0.2026 g of sample into a 300 mL conical flask. Add 15 mL of concentrated phosphoric acid and 5 mL of concentrated nitric acid, and heat until the sample dissolves completely. Continue heating and evaporating until the surface becomes still and faint fumes just begin to appear (about 220–240 °C), then remove the flask from the heat. Immediately add 3.0 g ± 0.2 g of ammonium nitrate. While swirling the flask, use a rubber-bulb pipette to blow tangentially across the mouth of the flask to remove nitrogen oxides until the yellow fumes disappear completely. Allow the mixture to cool slightly, add 60 mL of distilled water, mix thoroughly, and cool to room temperature. Titrate with the standard ammonium iron(II) sulfate solution until the solution is pale red. Add 2 drops of N-phenylanthranilic acid indicator solution (2 g/L) and continue titrating to a bright-yellow endpoint. The standard ammonium iron(II) sulfate solution consumed is 16.48 mL, and the blank consumes 0.84 mL.\n4.2.2 Give the predominant form of manganese after oxidation by ammonium nitrate. (Hint: the ligand is dihydrogen phosphate.)", "text_images": [], "full_mark": 2, "rubric": ["$Mn(H_2PO_4)_3$ (manganese(III) tris(dihydrogen phosphate), tris(dihydrogenphosphato)manganese(III), manganese dihydrogen phosphate, or another equivalent expression) (2 points)"], "rubric_images": []}, {"sub_id": "Exam-21-T4-2", "tags": ["2.1"], "text_open": "4.2.4 Write the net ionic equation for the titration. In the products, the metal ions are present as free ions rather than complexes.", "text_images": [], "full_mark": 2, "rubric": ["$Mn(H_2PO_4)_3 + Fe^{2+} + 3H^+ = Mn^{2+} + Fe^{3+} + 3H_3PO_4$ (2 points)"], "rubric_images": []}]} -{"problem_id": "Exam-21-T6", "header": "Pharmacokinetics and Molecular Design (II)\nPharmacokinetics quantitatively studies the absorption, distribution, metabolism, and excretion of drugs in living organisms and uses mathematical principles and methods to describe how plasma drug concentration changes with time. Chemical-kinetic modeling provides a systematic understanding of drug metabolism in vivo. This problem concerns the oxidative metabolism of diclofenac in the body.", "header_images": [], "subquestions": [{"sub_id": "Exam-21-T6-1", "tags": ["3.7"], "text_open": "The Michaelis equation is $r=r_{max}[S]/(K_m+[S])$.\nExperiments show that $[S]$ and $r$ for the 4′-hydroxylation of diclofenac catalyzed by human liver microsomes obey the following relationship:\n| Group | $[S]_1$ / $\\mu\\mathrm{M}$ | $r_1$ / $(\\mathrm{pmol \\cdot min^{-1} \\cdot mg^{-1}})$ | $[S]_2$ / $\\mu\\mathrm{M}$ | $r_2$ / $(\\mathrm{pmol \\cdot min^{-1} \\cdot mg^{-1}})$ |\n| --- | --- | --- | --- | --- |\n| Low concentration | 0.90 | 39.27 | 1.80 | 72.00 |\n| High concentration | 90.0 | 392.73 | 180.0 | 411.43 |\n\n6.2.4 Under the same conditions, the 5-hydroxylation has $K_m = 43\\ \\mu\\mathrm{M}$ and $r_{max} = 15.4\\ \\mathrm{pmol \\cdot min^{-1} \\cdot mg^{-1}}$. Intrinsic clearance $CL_{int}$ is defined by $CL_{int} = r_{max}/K_m$. Calculate $CL_{int}$ for the 4′- and 5-hydroxylation pathways and briefly explain the kinetic meaning of $CL_{int}$. Hint: the drug concentration in the liver is much lower than $K_m$.", "text_images": [], "full_mark": 5, "rubric": ["From the given Michaelis equation, $r = r_{max}[S]/(K_m + [S])$, which can be rearranged to $[S]/r = (1/r_{max})[S] + K_m/r_{max}$. Let $x = [S]$ and $y = [S]/r$; then $y = ax + b$, where $a = 1/r_{max}$ and $b = K_m/r_{max}$. (1 point)", "Using the experimental data provided:\n| x | 0.9 | 1.8 | 90 | 180 |\n| y | 0.02292 | 0.02500 | 0.2292 | 0.4375 |\nLinear regression gives $y = 0.00231480x + 0.0208339$, with $R^2 \\approx 1.000$. Therefore, $r_{max} = 1/a = 432\\ \\mathrm{pmol \\cdot min^{-1} \\cdot mg^{-1}}$ and $K_m = b/a = 9.00\\ \\mu\\mathrm{M}$. (1 point)", "$CL_{int}(4′) = 432/9.00 = 48.0\\ \\mu\\mathrm{L \\cdot min^{-1} \\cdot mg^{-1}}$ (1 point); $CL_{int}(5) = 15.4/43 = 0.358\\ \\mu\\mathrm{L \\cdot min^{-1} \\cdot mg^{-1}}$ (1 point).", "Kinetic meaning: when the drug concentration $[S]$ is extremely low, $[S]$ in the denominator of the Michaelis equation can be neglected, giving $r \\approx \\left(\\frac{r_{max}}{K_m}\\right) \\cdot [S]$. Under these conditions, $CL_{int}$ represents the first-order rate constant. (1 point; describing it as the clearance capacity per unit amount of enzyme at low concentration is also accepted.)"], "rubric_images": []}, {"sub_id": "Exam-21-T6-2", "tags": ["2.9", "3.7"], "text_open": "6.2.5 The kinetic parameters for the individual steps in the pathways above are shown below:\n| Pathway | Substrate → product | Enzyme | $CL_{int}$ / $(\\mu\\mathrm{L}\\cdot\\mathrm{min}^{-1}\\cdot\\mathrm{mg}^{-1})$ |\n| --- | --- | --- | --- |\n| 4′-hydroxylation pathway | DF → 4′-OH-DF | CYP2C9 | / |\n| 4′-hydroxylation pathway | 4′-OH-DF → A | CYP2C9 | 358 |\n| 5-hydroxylation pathway | DF → 5-OH-DF | CYP3A4 | / |\n| 5-hydroxylation pathway | 5-OH-DF → B | CYP2C9 | 103 |\n\nThe researchers also compared detoxification of the quinone products from the two hydroxylation pathways. The measured curves are shown below, where \"conjugate\" denotes a detoxification product.\nExplain why both steps of the 5-hydroxylation pathway have lower $CL_{int}$ values than those of the 4′-hydroxylation pathway, yet the 5-hydroxylation pathway is more toxic. (No more than 100 words)", "text_images": [{"file": "Exam-21-T6-2-1-prompt.png", "description": "Coordinate plot containing two side-by-side panels labeled A and B at upper left. Both horizontal axes are labeled 'GSH concentration (mM)' with ticks at 0, 2, 4, and 6; both vertical axes are labeled 'Conjugate formation (%)' with ticks at 0, 50, and 100. Each panel contains one solid black curve with filled circular data markers, some with vertical error bars. In panel A, the curve rises rapidly from near the origin as GSH concentration increases, approaches 100% at about 2 mM, and remains on a plateau. In panel B, the curve rises more gradually from near the origin and approaches 100% at about 5–6 mM, without a clear plateau. There is no legend or title."}], "full_mark": 2, "rubric": ["① In the 5-hydroxylation pathway, once 5-OH-DF is formed it is more readily converted to toxic B. In contrast, the first step of the 4′ pathway is relatively faster, so 4′-OH-DF is more likely to accumulate and can enter detoxification pathways such as conjugative metabolism. (1 point)", "② The graph shows that quinone A forms abundant adduct at a lower GSH concentration, indicating that it is captured more readily by GSH. Quinone B requires a higher GSH concentration to reach a similar level and is therefore more likely to escape GSH detoxification and bind covalently to cellular proteins. (1 point) Thus, although the 5′ pathway has lower metabolic flux, it is more toxic."], "rubric_images": []}, {"sub_id": "Exam-21-T6-3", "tags": ["1.7", "2.9"], "text_open": "6.3 Diclofenac has a half-life of about 1.0h in humans and is cleared mainly through CYP2C9-catalyzed 4′-hydroxylation of an aromatic ring. The structurally similar fenclorac is more resistant to aromatic hydroxylation and has a half-life exceeding 20h. Its structure is shown below.\nFrom a structural perspective, explain why fenclorac has a longer duration of action. (No more than 50 words)", "text_images": [{"file": "Exam-21-T6-3-1-prompt.png", "description": "The SMILES representation of fenclorac is 'OC(CC1=C(OC2=CC=C(Cl)C=C2Cl)C=CC=C1)=O'."}], "full_mark": 2, "rubric": ["In fenclorac, the principal 4′-hydroxylation site of diclofenac is occupied by a para Cl substituent, so there is no 4′-H available for hydroxylation. (2 points) Its oxidative metabolic clearance is therefore lower, giving a longer half-life and duration of action.", "An answer also noting that the bridging atom in fenclorac is O, so formation of a quinonoid structure by oxidation cannot eliminate the formal charge, whereas diclofenac has a bridging N atom that can form an imine upon oxidation, receives 1 point."], "rubric_images": []}]} -{"problem_id": "Exam-22-T1", "header": "Problem 1 Chemistry of the Rare-Earth Elements\nRare-earth elements are indispensable strategic resources for modern high technology, new energy, advanced materials, defense, and related fields. Often called the “vitamins of industry” and a “treasure house of new materials,” they comprise scandium, yttrium, and all the lanthanides. For convenience, the symbol Ln is commonly used in coordination and inorganic chemistry to denote the lanthanides or the entire rare-earth series. This problem examines rare-earth chemistry in detail.\nNote: Different letters or alphanumeric combinations in this problem may represent the same element or chemical formula.", "header_images": [], "subquestions": [{"sub_id": "Exam-22-T1-1", "tags": ["1.3"], "text_open": "1.2 Consider a rare-earth element M1 as an example of rare-earth chemistry. M1 has two common oxidation states in its compounds. Elemental M1 is a silvery-white metal that slowly tarnishes in air. M1 dissolves in dilute hydrochloric acid, releasing colorless gas A and giving a pale blue-green solution (Reaction 1). Adding dilute sulfuric acid dropwise to this solution immediately produces white precipitate B, which is insoluble in dilute nitric acid. After B is filtered, washed, and dried, it is ignited to constant mass at 800 °C under Ar; the white precipitate becomes dark-brown powder C (Reaction 2), an oxide of M1. If the mass of B is 2.480 g, 1.760 g of C is obtained after ignition. A separate portion of B is heated with dilute hydrochloric acid and bromine water until it dissolves (Reaction 3), during which the solution changes from pale blue-green to colorless (or very pale pink). The solution is divided into two portions. Excess aqueous ammonia and oxalic acid are added separately to give white gelatinous precipitate D (with aqueous ammonia) and white precipitate E (with oxalic acid), respectively. Filtering and igniting either D or E at 900 °C likewise gives C. C is mixed with Y2O3 in a certain ratio and calcined at high temperature; product F emits bright red fluorescence under ultraviolet irradiation. 1.2.3 Write ionic equations for Reactions 1 and 3 and a chemical equation for Reaction 2.", "text_images": [], "full_mark": 6, "rubric": ["Reaction 1: $Eu + 2H^+ \\rightarrow Eu^{2+} + H_2$ (2 points)\nReaction 2: $2EuSO_4 \\rightarrow Eu_2O_3 + SO_2 + SO_3$ (2 points)\nReaction 3: $2EuSO_4 + Br_2 \\rightarrow 2Eu^{3+} + 2Br^- + 2SO_4^{2-}$ (2 points)"], "rubric_images": []}, {"sub_id": "Exam-22-T1-2", "tags": ["1.4"], "text_open": "1.4 Coordination Compounds of Rare-Earth Elements\n1 equivalent of elemental rare-earth metal M5 and 2.8 equivalents of ethylene glycol monomethyl ether are added together to a toluene solution to form X. Addition of 2,6-dimethylphenol then precipitates orange crystals Y. Y is filtered and washed; upon slight heating it becomes a different crystal Z. Further characterization provides the following information:\n① The effective magnetic moment of X is 7.9 $\\mu_B$.\n② Crystal diffraction shows that Y belongs to the orthorhombic crystal system, with $a$ = 12.748 Å, $b$ = 12.900 Å, and $c$ = 29.220 Å. The unit cell contains 2 formula units, and the theoretical density of Y is 1.428 $\\mathrm{g/cm^3}$.\n③ Elemental analysis of Z gives the following experimental mass fractions: $\\omega(C)$ = 39.60%, $\\omega(O)$ = 18.86%, and $\\omega(H)$ = 5.74%. Crystal diffraction shows that Z belongs to the tetragonal crystal system, with $a$ = 14.837 Å and $c$ = 15.117 Å. The crystal contains 2 formula units, and the density of Z is 1.1854 times the theoretical density of Y.\n④ The anion of Z contains an $[M5_4O_4]$ cluster framework and exactly two types of ligands. One ligand adopts two different coordination modes. All M5 atoms are 7-coordinate, and a particular conformation has 1 twofold rotation axis.\n⑤ The oxidation state of M5 is the same in X, Y, and Z.\n1.4.2 Determine the chemical formulas of X, Y, and Z, showing your calculations. In your answer, the ethylene glycol monomethyl ether anion may be abbreviated as $mee^-$, and the 2,6-dimethylphenoxide anion as $dmp^-$.\n", "text_images": [], "full_mark": 9, "rubric": ["Y is orthorhombic, with unit-cell volume $V_Y = 12.748 \\times 12.900 \\times 29.220 = 4805.2\\ \\mathrm{Å}^3$. From the density equation $\\rho_Y = ZM / (N_AV)$, with $Z = 2$ for both Y and Z:\nFirst calculate $M_Y$: $\\frac{M_Z}{M_Y} = \\frac{\\rho_Z V_Z}{\\rho_Y V_Y} = \\frac{1.1854\\rho_Y V_Z}{\\rho_Y V_Y} = 1.1854 \\times \\frac{3327.80}{4805.2} = 0.8209$. (1 point for determining the molar mass of Y)\n$M_Y = \\frac{1.428 \\times 4805.2 \\times 10^{-24} \\times 6.022 \\times 10^{23}}{2} = 2066\\ \\mathrm{g/mol}$.\nThus, $M_Y - M_Z$ = 369 g/mol, approximately the mass of 4 toluene molecules, so Y is a toluene solvate of Z. (1 point)\nFor Z, $\\omega(C)$ = 39.63% and $\\omega(O)$ = 18.85%. Taking $M_Z$ = 1697 g/mol:\nNumber of C atoms: $n(C) = (1697 \\times 0.3960) / 12.01 \\approx 56$\nNumber of O atoms: $n(O) = (1697 \\times 0.1886) / 16.00 \\approx 20$\nNumber of H atoms: $n(H) = (1697 \\times 0.0574) / 1.008 = 96.6$\n(1 point for determining the numbers of C, O, and H atoms in Z; integers or decimals are accepted.)\nThe hydrogen mass fraction from elemental analysis often has a relatively large error. The system contains only two types of ligands: ① the ethylene glycol monomethyl ether anion, $mee^-$: $C_3H_7O_2^-$, containing 3 C and 2 O atoms; ② the 2,6-dimethylphenoxide anion, $dmp^-$: $C_8H_9O^-$, containing 8 C and 1 O atom.\nLet the numbers of $dmp^-$ and $mee^-$ ligands be x and y, respectively. Then $\\begin{cases} 8x + 3y = 56 \\\\ x + 2y = 20 \\end{cases}$, giving x = 4 and y = 8. Thus, Z contains 4 $dmp^-$ and 8 $mee^-$ ligands. (1 point for determining the numbers of ligands in Z; it is not necessary to distinguish dmp from dmpH or mee from meeH.)\nLet Z contain z metal atoms of atomic weight $M_{M5}$. From mass conservation:\n$z \\cdot M_{M5} = M_Z - 4M_{dmp} - 8M_{mee} - 4M_H = 1697 - 4 \\times 121.17 - 8 \\times 75.09 - 4 \\times 1.008 \\approx 608$ (g/mol).\nTry z = 4: $M_{M5}$ = 608/4 = 152. Atomic weight 152 corresponds to europium (Eu). In fact, the statement that the anion of Z contains an $[M5_4O_4]$ cluster framework establishes that z ≥ 4.\nMagnetic-moment verification: $\\mu_{eff}$ = 7.9 $\\mu_B$, in close agreement with the theoretical value of 7.94 $\\mu_B$ for $Eu^{2+}$ ($4f^7$); experimental values are generally 7.6~8.0 $\\mu_B$. Therefore, M5 = Eu in the +2 oxidation state. (2 points for identifying M5. Any reasonable procedure that identifies M5 earns at least 6 points for the entire subquestion.)\nZ = $H_4[Eu_4(mee)_8(dmp)_4]$ (1 point)\nY = $H_4[Eu_4(mee)_8(dmp)_4]\\cdot 4C_7H_8$ (1 point)\nX = $[Eu(mee)_2]_n$ (1 point; $Eu(mee)_2$ is also accepted)"], "rubric_images": [{"file": "Exam-22-T1-2-1-rubric.png", "description": "Equation: $\\frac{M_Z}{M_Y} = \\frac{\\rho_Z V_Z}{\\rho_Y V_Y} = 1.1854 \\times \\frac{3327.80}{4805.2} = 0.8209$."}, {"file": "Exam-22-T1-2-2-rubric.png", "description": "Equation: $M_Y = \\frac{1.428 \\times 4805.2 \\times 10^{-24} \\times 6.022 \\times 10^{23}}{2} = 2066\\ \\mathrm{g/mol}$."}, {"file": "Exam-22-T1-2-3-rubric.png", "description": "Equation: $\\begin{cases} 8x + 3y = 56 \\\\ x + 2y = 20 \\end{cases}$."}]}, {"sub_id": "Exam-22-T1-3", "tags": ["1.4"], "text_open": "1.4.3 Write the SMILES strings for the cation and anion of Z. You may treat all $\\sigma$ coordinate bonds as ordinary single bonds and represent the structures using SMILES.", "text_images": [], "full_mark": 4, "rubric": ["Structure: the cation consists of 4 hydrogen ions, $(H^+)$; the anion has SMILES 'CO1CCO[Eu]1234([dmp])O5(CCO6C)[Eu]789(OCCO8C)([dmp])O2(CCO9C)[Eu]%10%11%12(O(C)CCO%12)([dmp])O3(CCO4C)[Eu]5%136(O(C)CCO%13)([dmp])O%107CCO%11C', where dmp denotes a dmp ligand, and the anion carries a $4-$ charge. The anion has an approximately cubic [Eu_4O_4] core. All dmp ligands are terminal. The mee ligands fall into two classes: 4 each provide 1 O face-bridging group and 1 linear bridging group, with the terminal O bearing a methyl group, whereas the other 4 coordinate only through 2 terminal O atoms.", "This subquestion uses deductions. Deduct 1 point if, and only if, some mee ligands in the anion have the wrong “direction,” with the methyl-bearing O drawn as the face-bridging group and the non-methyl-bearing O as the linear bridge. Any other error in the anion structure loses all points assigned to the anion. Deduct 1 point if the cations are absent or incorrect, and 1 point if the charge of the anion is not shown. The minimum score is 0."], "rubric_images": [{"file": "Exam-22-T1-3-1-rubric.png", "description": "Scoring diagram. $(H^+)_4$ is shown on the left, and the anion is shown on the right. The anion has SMILES 'CO1CCO[Eu]1234([dmp])O5(CCO6C)[Eu]789(OCCO8C)([dmp])O2(CCO9C)[Eu]%10%11%12(O(C)CCO%12)([dmp])O3(CCO4C)[Eu]5%136(O(C)CCO%13)([dmp])O%107CCO%11C' and an approximately cubic [Eu_4O_4] core. A $4-$ charge is written at the upper right of the anion. The 4 dmp ligands are represented by the label 'dmp', with their structures omitted."}]}, {"sub_id": "Exam-22-T1-4", "tags": ["2.1"], "text_open": "1.5 Electrochemistry of the Rare-Earth Elements\nThe following standard electrode potentials are given for acidic medium at 25 °C:\n$O_2(g) + 4H^+ + 4e^- \\rightarrow 2H_2O$ $E° = +1.229\\ \\mathrm{V}$\n$M1^{n+} + e^- \\rightarrow M1^{(n-1)+}$ $E° = -0.35\\ \\mathrm{V}$\n1.5.1 At pH = 0 and 25 °C, calculate the common logarithm $\\lg K$ of the standard equilibrium constant for oxidation of $M1^{(n-1)+}$ to $M1^{n+}$ by $O_2$.", "text_images": [], "full_mark": 3, "rubric": ["Reaction: $4Eu^{2+} + O_2 + 4H^+ \\rightarrow 4Eu^{3+} + 2H_2O$\n$E° = E_+ - E_- = 1.229\\ \\mathrm{V} - (-0.35\\ \\mathrm{V}) = 1.579\\ \\mathrm{V}$ (1 point)\n$\\Delta_r G° = -nFE° = -4 \\times 96485\\ \\mathrm{C\\cdot mol^{-1}} \\times 1.579\\ \\mathrm{V} = -6.095 \\times 10^5\\ \\mathrm{J\\cdot mol^{-1}}$ (1 point)\n$\\ln K = -\\frac{\\Delta_r G°}{RT} = \\frac{6.095 \\times 10^5}{8.314 \\times 298} = 246$\n$\\lg K = \\lg e \\times \\ln K = 106.8$ (1 point). Writing M1 instead of Eu in the procedure and result incurs no deduction."], "rubric_images": [{"file": "Exam-22-T1-4-1-rubric.png", "description": "Equation: $E° = E_+ - E_- = 1.229\\ \\mathrm{V} - (-0.35\\ \\mathrm{V}) = 1.579\\ \\mathrm{V}$."}, {"file": "Exam-22-T1-4-2-rubric.png", "description": "Equation: $\\Delta_r G° = -nFE° = -4 \\times 96485\\ \\mathrm{C\\cdot mol^{-1}} \\times 1.579\\ \\mathrm{V} = -6.095 \\times 10^5\\ \\mathrm{J\\cdot mol^{-1}}$."}, {"file": "Exam-22-T1-4-3-rubric.png", "description": "Equation: $\\ln K = -\\frac{\\Delta_r G°}{RT} = \\frac{6.095 \\times 10^5}{8.314 \\times 298} = 246$."}, {"file": "Exam-22-T1-4-4-rubric.png", "description": "Equation: $\\lg K = \\lg e \\times \\ln K = 106.8$."}]}, {"sub_id": "Exam-22-T1-5", "tags": ["2.1"], "text_open": "1.5.2 The partial pressure of oxygen in air is $p(O_2)$ = 0.21 bar. Use a calculation to determine whether $M1^{(n-1)+}$ will be oxidized by oxygen in air when pH = 4.00 and $[M1^{(n-1)+}]$ = $[M1^{n+}]$ = 1.00 mol·L-1.", "text_images": [], "full_mark": 2, "rubric": ["$E(O_2) = E° + \\frac{RT}{4F}\\ln\\frac{[H^+]^4 p(O_2)}{p°} = 0.982\\ \\mathrm{V}$ (1 point)\nBecause the concentrations are equal, the $Eu^{3+}/Eu^{2+}$ electrode potential is $E = E° = -0.35\\ \\mathrm{V}$. The cell electromotive force is $\\Delta E = 0.982\\ \\mathrm{V} - (-0.35\\ \\mathrm{V}) = 1.33\\ \\mathrm{V} > 0$, and $\\Delta G < 0$. Therefore, under these conditions, $Eu^{2+}$ is still oxidized spontaneously by oxygen in air. (1 point for calculating the correct electromotive force and reaching the correct conclusion.)\nIf the calculated result is incorrect, no point is awarded for the conclusion even if it is correct. Writing M1 instead of Eu in the procedure and result incurs no deduction."], "rubric_images": [{"file": "Exam-22-T1-5-1-rubric.png", "description": "Equation: $E(O_2) = E° + \\frac{RT}{4F}\\ln\\frac{[H^+]^4 p(O_2)}{p°} = 0.982\\ \\mathrm{V}$."}, {"file": "Exam-22-T1-5-2-rubric.png", "description": "Equations: $\\Delta E = 0.982\\ \\mathrm{V} - (-0.35\\ \\mathrm{V}) = 1.33\\ \\mathrm{V} > 0$ and $\\Delta G < 0$."}]}]} -{"problem_id": "Exam-22-T5", "header": "Unlike most covalent compounds, whose structures can be described using 2c-2e bonds, polyhedral clusters are often electron-deficient systems and require multicenter bonding models.\nWade's rules, more generally known as the polyhedral skeletal electron pair theory (PSEPT), are an important tool for understanding cluster structures. Let $n$ denote the number of atoms or isolobal fragments that form the cluster framework. Subtracting the electrons used in terminal two-center two-electron bonds, nonskeletal lone pairs, and the internal bonding of ligands from the total valence-electron count gives the number of skeletal electrons. Skeletal electron-pair counts of $n+1$, $n+2$, $n+3$, and $n+4$ correspond, respectively, to closo, nido, arachno, and hypho structures. A closo cluster is a complete triangulated polyhedron with $n$ vertices, whereas nido, arachno, and hypho clusters may be regarded as the corresponding closo parent polyhedron with one, two, or three vertices removed, respectively. For example, in $B_5H_9$, 5 BH fragments contribute 10 electrons and 4 additional hydrogen atoms contribute another 4, giving 14 skeletal electrons or 7 skeletal electron pairs. Since the skeletal electron-pair count is $n+2$, the cluster is nido.\nThis problem uses low-valent clusters of Group 13 metals (Al, Ga, In, and Tl) to examine cluster synthesis, structure, bonding, and reactivity.", "header_images": [], "subquestions": [{"sub_id": "Exam-22-T5-1", "tags": ["1.6"], "text_open": "5.1 $B_4Cl_4$ and $B_4(t\\text{-}Bu)_4$ have tetrahedral cores, whereas the acetamido-substituted boron compound $B_4(NHAc)_4$ adopts a butterfly-shaped cyclic structure. Explain why.", "text_images": [], "full_mark": 2, "rubric": ["Because of B-N π bonding, the B atoms are relatively electron-rich. (2 points)"], "rubric_images": []}, {"sub_id": "Exam-22-T5-2", "tags": ["2.1"], "text_open": "5.2 Write equations for the following reactions.\n5.2.1 $AlCl\\cdot x\\,Et_2O$ ($x>1$) reacts with $MgCp^*_2$ at low temperature in toluene/diethyl ether to form $Al_4Cp^*_4$.", "text_images": [], "full_mark": 2, "rubric": ["$4\\,AlCl\\cdot x\\,Et_2O + 2\\,MgCp^*_2 = Al_4Cp^*_4 + 2\\,MgCl_2(OEt_2)_2 + (4x-4)\\,Et_2O$ (2 points)"], "rubric_images": []}, {"sub_id": "Exam-22-T5-3", "tags": ["2.1"], "text_open": "5.2.2 $GaI$ reacts with $Li(THF)_3Ge(SiMe_3)_3$ in toluene to form $[GaGe(SiMe_3)_3]_4$.", "text_images": [], "full_mark": 2, "rubric": ["$4\\,GaI + 4\\,Li(THF)_3Ge(SiMe_3)_3 = [GaGe(SiMe_3)_3]_4 + 4\\,LiI(THF)_3$ (2 points)"], "rubric_images": []}, {"sub_id": "Exam-22-T5-4", "tags": ["2.4"], "text_open": "5.4 Tetrahedral clusters not only have unusual structures but also show diverse oxidation chemistry, giving different products under different conditions. Draw the structures of products A, B, D, and E below. 5.4.1 Oxidation of [GaSi(t-Bu)_3]_4 with oxygen gives cubic cluster A = [OGaSi(t-Bu)_3]_4.", "text_images": [], "full_mark": 3, "rubric": ["A is the cubic cluster $[OGaSi(t\\text{-}Bu)_3]_4$. (3 points)"], "rubric_images": [{"file": "Exam-22-T5-4-1-rubric.png", "description": "The reference structure of A has SMILES '[Si]([Ga]12O3[Ga]4(O1[Ga]5([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)O4[Ga]3([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)O52)[Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)(C(C)(C)C)(C(C)(C)C)C(C)(C)C'. It has a $Ga_4O_4$ heterocubane framework in which 4 Ga and 4 O atoms alternately occupy the vertices of a cube, forming 12 Ga-O edges. Each Ga is bonded to one $Si(t\\text{-}Bu)_3$ group, giving $[OGaSi(t\\text{-}Bu)_3]_4$. All bonds are single bonds, and no charges are shown."}]}, {"sub_id": "Exam-22-T5-5", "tags": ["2.4"], "text_open": "5.4.2 Further hydrolysis of A gives product B, which contains only one ring and has only one chemical environment for its Ga atoms. Write the SMILES string for B. If necessary, describe its detailed structure in no more than 50 words. Multiple reasonable structures may be possible; writing any one reasonable SMILES string is sufficient.", "text_images": [], "full_mark": 2, "rubric": ["Draw either of the following structures. First: 'O[Ga]1([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)[OH][Ga](O)([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)[OH][Ga](O)([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)[OH][Ga](O)([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)[OH]1'. Second: '[OH2+][Ga-]1([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)O[Ga-]([OH2+])([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)O[Ga-]([OH2+])([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)O[Ga-]([OH2+])([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)O1'."], "rubric_images": [{"file": "Exam-22-T5-5-1-rubric.png", "description": "One reasonable structure of B, and the most accurate reference answer, has SMILES 'O[Ga]1([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)[OH][Ga](O)([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)[OH][Ga](O)([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)[OH][Ga](O)([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)[OH]1'. It is an eight-membered $Ga_4O_4$ ring in which 4 Ga atoms and 4 $\\mu_2$-OH groups alternate. Each Ga is bonded to one $Si(t\\text{-}Bu)_3$ group and one terminal OH group. All bonds are single bonds, and no charges are shown."}, {"file": "Exam-22-T5-5-2-rubric.png", "description": "Another acceptable structure of B has SMILES '[OH2+][Ga-]1([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)O[Ga-]([OH2+])([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)O[Ga-]([OH2+])([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)O[Ga-]([OH2+])([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)O1'. It is an eight-membered $Ga_4O_4$ ring, with the ring oxygen atoms labeled O. Each Ga is bonded to one $Si(t\\text{-}Bu)_3$ group and one water ligand. The SMILES uses charge-separated $[OH2+]$ and $[Ga-]$ forms; the 4 water ligands are labeled OH2/H2O in the figure."}]}, {"sub_id": "Exam-22-T5-6", "tags": ["2.4"], "text_open": "5.4.3 Oxidation of $C = [InC(SiMe_3)_3]_4$ with o-nitrosotoluene gives a cubic cluster and product D, which contains two benzene rings. Write the SMILES string for aromatic product D and, if necessary, describe its detailed structure in no more than 50 words.", "text_images": [], "full_mark": 2, "rubric": ["'CC1=C(/N=N/C2=C(C)C=CC=C2)C=CC=C1'"], "rubric_images": [{"file": "Exam-22-T5-6-1-rubric.png", "description": "The reference structure of D has SMILES 'CC1=C(/N=N/C2=C(C)C=CC=C2)C=CC=C1'. It has an azobenzene framework with an N=N double bond; /N=N/ in the SMILES denotes the trans configuration. Each benzene ring bears one methyl substituent drawn as a short line without a text label, and each methyl group is ortho to the position bonded to nitrogen. All remaining bonds are single bonds, and no charges are shown."}]}, {"sub_id": "Exam-22-T5-7", "tags": ["2.4"], "text_open": "5.4.4 Oxidation of C with 1,2-thiirane gives a trigonal-bipyramidal cluster E. Write the SMILES string for E and, if necessary, describe its detailed structure in no more than 50 words.", "text_images": [], "full_mark": 2, "rubric": ["'C[Si](C([In]12[In]34(C([Si](C)(C)C)([Si](C)(C)C)[Si](C)(C)C)[In]15(C([Si](C)(C)C)([Si](C)(C)C)[Si](C)(C)C)[In]23(C([Si](C)(C)C)([Si](C)(C)C)[Si](C)(C)C)[S]54)([Si](C)(C)C)[Si](C)(C)C)(C)C'"], "rubric_images": [{"file": "Exam-22-T5-7-1-rubric.png", "description": "The reference structure of E has SMILES 'C[Si](C([In]12[In]34(C([Si](C)(C)C)([Si](C)(C)C)[Si](C)(C)C)[In]15(C([Si](C)(C)C)([Si](C)(C)C)[Si](C)(C)C)[In]23(C([Si](C)(C)C)([Si](C)(C)C)[Si](C)(C)C)[S]54)([Si](C)(C)C)[Si](C)(C)C)(C)C'. It has an $In_4$ tetrahedral framework with 6 In-In edges and one S atom bridging a face in a $\\mu_3$ mode. Each In is bonded to one $C(SiMe_3)_3$ group, labeled C(SiMe3)3 in the figure, giving a trigonal-bipyramidal cluster. All bonds are single bonds, and no charges are shown."}]}, {"sub_id": "Exam-22-T5-8", "tags": ["2.2"], "text_open": "5.6 Reaction of $GaI$ with $Li(THF)_3R$ ($R = SiMe(SiMe_3)_2$) in toluene gives the octahedral cluster $G = [GaR]_6$. In contrast, reaction of $GaI$ with $Na(THF)_2R'$ ($R' = Si(t\\text{-}Bu)_3$) in toluene forms the octahedral cluster $H = [Ga_6R'_4Bn_2]^{2-}$, where Bn denotes benzyl.\n5.6.1 State the source of the benzyl groups in H.", "text_images": [], "full_mark": 1, "rubric": ["The solvent toluene is deprotonated to generate benzyl anions. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-22-T5-9", "tags": ["1.7"], "text_open": "5.6.3 Explain why G is a highly distorted octahedron whereas H is nearly a regular octahedron.", "text_images": [], "full_mark": 3, "rubric": ["$[GaR]_6$ has 6 skeletal electron pairs (sep), fewer than the 7 sep required for a closo octahedron (1 point). The electron deficiency removes the degeneracy and gives rise to a Jahn-Teller effect (2 points), ultimately causing severe distortion of the octahedron."], "rubric_images": []}, {"sub_id": "Exam-22-T5-10", "tags": ["1.7"], "text_open": "5.8 Electron-rich cluster compounds: When the number of ligands in a cluster is much smaller than the number of metal atoms, its structure approaches that of a bare metal-atom cluster; such species are called electron-rich cluster compounds. A representative example is the metalloid cluster $K = Ga_{84}[N(SiMe_3)_2]_{20}^{4-}$. This cluster contains 84 Ga atoms, 64 of which are “bare.” Its structure has three layers: a central $Ga_2$ dumbbell with a bond length of 235 pm, approaching a $Ga\\equiv Ga$ triple bond; a middle $Ga_{32}$ football-shaped tube; and an outer assembly of three $Ga_{10}$ rings surrounded by 20 $N(SiMe_3)_2$ ligands.\nCompare K with the fullerene $Sc_2@C_{84}$, and identify which structural units in K correspond to the encapsulated $Sc_2$ and to the $C_{84}$ carbon cage, respectively.", "text_images": [{"file": "Exam-22-T5-10-1-prompt.png", "description": "Unlabeled three-dimensional ball-and-stick molecular model. The overall structure is a vertically symmetric multilayer assembly. The upper and lower layers each contain a polyhedron made of dark-blue spheres, approximately 12 in total. A larger central polyhedron is formed by approximately 12 purple spheres, and the outermost framework contains approximately 24 orange-yellow spheres. Spheres of the same color are connected by solid lines of the same color, with radial connections between layers. The drawing represents the layered structure of the metalloid cluster $K = Ga_{84}[N(SiMe_3)_2]_{20}^{4-}$ described in the question."}], "full_mark": 2, "rubric": ["The central $Ga_2$ unit can be regarded as a “molecule” encapsulated within the $Ga_{32}$ cage, analogous to $Sc_2$ in $Sc_2@C_{84}$. (1 point)", "The $Ga_{32}$ unit is a hollow structure analogous to a fullerene carbon cage. (1 point)"], "rubric_images": []}]} +{"problem_id": "Exam-20-T9", "header": "Problem 9 Prins Cyclization (25 points, 12%)\nThe Prins cyclization is a powerful synthetic transformation that rapidly constructs tetrahydropyran rings. Tetrahydropyran is a common structural unit in many bioactive molecules, including spongistatins and bryostatins. Researchers have used the Prins cyclization in numerous elegant strategies for the total synthesis of natural-product targets.", "header_images": [], "subquestions": [{"sub_id": "Exam-20-T9-1", "tags": ["2.4", "2.2"], "text_open": "9.1 The following is a classic Prins cyclization that gives the final product with high stereoselectivity, where R = alkyl.\n9.1.1 Write the SMILES strings of A-D in the scheme, including stereochemistry. B is a highly strained intermediate.", "text_images": [{"file": "Exam-20-T9-1-1-prompt.png", "description": "Prins cyclization sequence: '[R]C1=CCCCC1'>H2C=O, H2SO4>[A⇌B]; [B]>H2O>[C]; [C]>H2C=O, H2SO4>[D]. In the image, the last arrow points leftward from C to D. A-D are shown only as letter labels and no structures are provided; R is the placeholder used in the original diagram."}], "full_mark": 5, "rubric": ["A: '[R][C+]1[C@H](CO)CCCC1', additionally labeled (±) in the image; B: '[R][C@@]12[C@@](C[OH+]2)([H])CCCC1', additionally labeled (±); C: 'O[C@@]1([R])[C@H](CO)CCCC1', additionally labeled (±) (1 point each). D: '[R][C@]1(OCOC2)[C@@]2([H])CCCC1', additionally labeled (±) (2 points).", "For each omitted (±) label, deduct 1 point from the score obtained, up to a maximum deduction of 3 points."], "rubric_images": [{"file": "Exam-20-T9-1-1-rubric.png", "description": "A has the SMILES '[R][C+]1[C@H](CO)CCCC1'. The image also shows (±), and R is the placeholder used in the original diagram."}, {"file": "Exam-20-T9-1-2-rubric.png", "description": "B has the SMILES '[R][C@@]12[C@@](C[OH+]2)([H])CCCC1'. The image also shows (±), and R is the placeholder used in the original diagram."}, {"file": "Exam-20-T9-1-3-rubric.png", "description": "C has the SMILES 'O[C@@]1([R])[C@H](CO)CCCC1'. The image also shows (±), and R is the placeholder used in the original diagram."}, {"file": "Exam-20-T9-1-4-rubric.png", "description": "D has the SMILES '[R][C@]1(OCOC2)[C@@]2([H])CCCC1'. The image also shows (±), and R is the placeholder used in the original diagram."}]}, {"sub_id": "Exam-20-T9-2", "tags": ["2.2", "2.4"], "text_open": "9.1.2 The following cascade comprises a Prins cyclization and a rearrangement and ultimately gives a product containing a five-membered ring. Write the SMILES string of the product and of at least 2 charged key intermediates, including stereochemistry.", "text_images": [{"file": "Exam-20-T9-2-1-prompt.png", "description": "Single-step reaction: 'C=C(C)[C@H]([C@H](C(C)=C)O)O'>Me2C=O, H2SO4>[product not shown]. Stereochemistry of the substrate's two OH groups is shown by wedge bonds and retained by the @ markers in the SMILES."}], "full_mark": 4, "rubric": ["Product structure: 'CC1(C)C[C@](C)(C=O)[C@H](C(C)=C)O1' (±) (2 points)", "Key intermediates: the first two SMILES strings are equivalent representations from different viewpoints. From left to right: 'C=C(C)[C@H]([C@H](C(C)=C)/[O+]=C(C)/C)O'; 'C=C(C)[C@]([C@H](O)C(C)=C)([H])/[O+]=C(C)/C'; 'C[C+]1[C@@H](O)[C@H](C(C)=C)OC(C)(C)C1'; 'CC1(C)C[C@](C)(C([H])=[OH+])[C@H](C(C)=C)O1'. Any 2 are worth 2 points total, 1 point each."], "rubric_images": [{"file": "Exam-20-T9-2-1-rubric.png", "description": "The product has the SMILES 'CC1(C)C[C@](C)(C=O)[C@H](C(C)=C)O1'. The image also shows (±), and stereochemistry is indicated by solid and dashed wedge bonds."}, {"file": "Exam-20-T9-2-2-rubric.png", "description": "The image shows 4 candidate key intermediates, of which any 2 are required for scoring. The first two SMILES strings are equivalent representations from different viewpoints. From left to right: 'C=C(C)[C@H]([C@H](C(C)=C)/[O+]=C(C)/C)O'; 'C=C(C)[C@]([C@H](O)C(C)=C)([H])/[O+]=C(C)/C'; 'C[C+]1[C@@H](O)[C@H](C(C)=C)OC(C)(C)C1'; 'CC1(C)C[C@](C)(C([H])=[OH+])[C@H](C(C)=C)O1'."}]}, {"sub_id": "Exam-20-T9-3", "tags": ["1.2", "2.7"], "text_open": "9.2 A key advantage of the Prins cyclization is that, starting from a single stereocenter and a defined alkene geometry, it can construct up to four stereocenters on a tetrahydropyran ring with high stereoselectivity. In some cases, however, enantiomeric excess (ee) is lost.\n9.2.1 In the Prins cyclization below, the ee of the major product is lower than that of the starting material, and small amounts of two byproducts are detected.\nPredict the configurations of the chiral carbons in the product by assigning the absolute configurations of chiral carbons a and b in the major product using S/R notation.", "text_images": [{"file": "Exam-20-T9-3-1-prompt.png", "description": "Single-step reaction: 'O[C@H](CC=C)C1=CC=CC=C1' reacts with PhCH2CH2CHO under BF3·OEt2 and AcOH conditions to give the major product, 'O=C(O[C@H]1C[C@@H](CCC2=CC=CC=C2)O[C@@H](C3=CC=CC=C3)C1)C', byproduct 1, 'O=C(OC1CC(C2=CC=CC=C2)OC(C3=CC=CC=C3)C1)C', and byproduct 2, 'O=C(OC1CC(CCC2=CC=CC=C2)OC(CCC3=CC=CC=C3)C1)C'. Starting material M is labeled 87% ee and the major product 67% ee. Chiral carbons a and b are labeled on the ring of the major product; OAc at a is shown on a solid wedge, while CH2CH2Ph at b is shown on a dashed bond."}], "full_mark": 2, "rubric": ["Chiral carbon a: S (1 point); chiral carbon b: R (1 point)"], "rubric_images": []}, {"sub_id": "Exam-20-T9-4", "tags": ["2.3", "2.2"], "text_open": "9.2.2 The pathway to the major product can undergo racemization through a rearrangement. Experimentally, the enantiomer M’ of substrate M is detected. If M and $PhCH_2CH_2CHO$ are replaced by structure N below, racemization through this rearrangement cannot occur under the same conditions. In light of the byproduct structures and the analysis above, write the SMILES strings of the reactants and products in each step of the mechanism by which M forms M’, and label the reaction conditions. Electron-flow arrows need not be shown.", "text_images": [{"file": "Exam-20-T9-4-1-prompt.png", "description": "Substrate N has the SMILES 'C=CC[C@@H](OC(OC(C)=O)CC1=CC=CC=C1)C2=CC=CC=C2'. N is the structure label in the image."}], "full_mark": 4, "rubric": ["The image shows the stepwise mechanism by which M forms M’ in two rows, labeled with conditions including PhCH2CH2CHO, -H2O, [3,3]-sigma, and H2O. Structures involved are M, 'O[C@H](CC=C)C1=CC=CC=C1'; the oxonium intermediates 'C=CC[C@@H](/[O+]=C/CCC1=CC=CC=C1)C2=CC=CC=C2' and 'C=CC[C@@H](CCC1=CC=CC=C1)/[O+]=C/C2=CC=CC=C2'; PhCHO, 'O=CC1=CC=CC=C1'; the alcohol product 'O[C@@H](CC=C)CCC1=CC=CC=C1'; and the enantiomer of M, 'O[C@@H](CC=C)C1=CC=CC=C1'. This subquestion is worth 4 points: 2 points for correctly including the [3,3]-sigma step, without needing to name the reaction type; 1 point for correctly forming the small molecule PhCHO; and 1 point for the reaction of PhCHO with the substrate to form the enantiomer. If the mechanism that forms PhCHO is complete, that step may be abbreviated as above."], "rubric_images": [{"file": "Exam-20-T9-4-1-rubric.png", "description": "The image shows the stepwise mechanism by which M forms M’ in two rows, labeled with conditions including PhCH2CH2CHO, -H2O, [3,3]-sigma, and H2O. Structures involved are M, 'O[C@H](CC=C)C1=CC=CC=C1'; the oxonium intermediates 'C=CC[C@@H](/[O+]=C/CCC1=CC=CC=C1)C2=CC=CC=C2' and 'C=CC[C@@H](CCC1=CC=CC=C1)/[O+]=C/C2=CC=CC=C2'; PhCHO, 'O=CC1=CC=CC=C1'; the alcohol product 'O[C@@H](CC=C)CCC1=CC=CC=C1'; and the enantiomer of M, 'O[C@@H](CC=C)C1=CC=CC=C1'."}]}, {"sub_id": "Exam-20-T9-5", "tags": ["2.3", "2.7"], "text_open": "9.2.3 Explain why a racemization process analogous to the one above cannot occur when substrate N is used.", "text_images": [], "full_mark": 2, "rubric": ["The oxocarbenium intermediate requires $H_2O$ in the system to hydrolyze into a small-molecule alcohol and aldehyde. After the substrate is changed, the system does not produce $H_2O$ (2 points; any equivalent statement that $H_2O$ is required for racemization is acceptable)."], "rubric_images": []}, {"sub_id": "Exam-20-T9-6", "tags": ["2.2", "2.4"], "text_open": "9.3 Researchers have developed many broadly useful Prins reactions using unconventional substrates such as alkynols, allenes, and small-ring compounds.\n9.3.1 The reaction below is based on a Prins-type cyclization of a 3,5-diynol with an aldehyde and proceeds through a domino sequence to synthesize a 2,3-dialkylidenetetrahydrofuran framework. The compound shown at the right was detected in the reaction mixture. This compound is not necessarily itself an intermediate and may arise from an intermediate. Based on this information, write the SMILES strings of the key intermediates. Stereochemistry is not required.", "text_images": [{"file": "Exam-20-T9-6-1-prompt.png", "description": "Single-step reaction: 'OCCC#CC#CC1=CC=CC=C1'.'O=CC1=CC=CC=C1'>HBF4·OEt2 (3 equiv), MeOH (3 equiv), DCM, rt, 8 h>'O=C(/C=C1OCCC\\1=C/C2=CC=CC=C2)C3=CC=CC=C3'."}, {"file": "Exam-20-T9-6-2-prompt.png", "description": "The image is labeled 'Detected in the reaction mixture:' and the detected compound has the SMILES 'OCC/C(C(C#CC1=CC=CC=C1)=O)=C\\C2=CC=CC=C2'."}], "full_mark": 4, "rubric": ["'CO/C(C#CC1=CC=CC=C1)=C2C(C3=CC=CC=C3)[OH+]CC/2'; 'OCC/C(C(C#CC1=CC=CC=C1)=[O+]C)=C\\C2=CC=CC=C2'; 'COC(OCC/1)(C#CC2=CC=CC=C2)C1=C\\C3=CC=CC=C3'; 'COC(C1=CC=CC=C1)=[C]=C(OCC/2)C2=C\\C3=CC=CC=C3'. 4 points total, 1 point for each intermediate highlighted in red. Except for the second red intermediate, reasonable protonated or deprotonated forms of the other intermediates are also acceptable."], "rubric_images": [{"file": "Exam-20-T9-6-1-rubric.png", "description": "This structure is not highlighted in red and is an alternative reasonable representation of one of the red intermediates. Its SMILES is 'C1(C#CC#CCC/[O+]=C/C2=CC=CC=C2)=CC=CC=C1'."}, {"file": "Exam-20-T9-6-2-rubric.png", "description": "The red intermediate has the SMILES 'CO/C(C#CC1=CC=CC=C1)=C2C(C3=CC=CC=C3)[OH+]CC/2'."}, {"file": "Exam-20-T9-6-3-rubric.png", "description": "The red intermediate has the SMILES 'OCC/C(C(C#CC1=CC=CC=C1)=[O+]C)=C\\C2=CC=CC=C2'."}, {"file": "Exam-20-T9-6-4-rubric.png", "description": "The red intermediate has the SMILES 'COC(OCC/1)(C#CC2=CC=CC=C2)C1=C\\C3=CC=CC=C3'."}, {"file": "Exam-20-T9-6-5-rubric.png", "description": "The red intermediate has the SMILES 'COC(C1=CC=CC=C1)=[C]=C(OCC/2)C2=C\\C3=CC=CC=C3'."}]}, {"sub_id": "Exam-20-T9-7", "tags": ["2.4", "2.7"], "text_open": "9.3.2 The following reaction proceeds through a three-membered-ring intermediate and gives the product with high stereoselectivity, governed primarily by stereoelectronic effects. Write the SMILES string of the product, including stereochemistry, and briefly explain the stereoselectivity.", "text_images": [{"file": "Exam-20-T9-7-1-prompt.png", "description": "Single-step reaction: 'C=C([Si](C)(C1=CC=CC=C1)C)C[C@@H](O)C'.'[R]C=O'>TMSOTf>[?]. The product structure is not shown and is represented by '?'. R is the placeholder used in the original diagram and was retained after manual confirmation."}], "full_mark": 4, "rubric": ["'[R][C@@H]1C[C@@]([Si](C)(O[Si](C)(C)C)C)(C2=CC=CC=C2)C[C@H](C)O1' (2 points total: 1 point for the correct framework and 1 point for correct stereochemistry)", "Explanation: during migration, the $\\sigma^*$ antibonding orbital of the newly forming C-C bond (1 point) overlaps effectively by hyperconjugation with the $\\sigma$ bonding orbital of the axial C-H bond on the adjacent carbon (1 point). This stabilizes the migration transition state and predominantly gives the product with phenyl in the axial position. 2 points total, with 1 point for correctly identifying each orbital participating in hyperconjugation. A reference schematic is shown below."], "rubric_images": [{"file": "Exam-20-T9-7-1-rubric.png", "description": "The product has the SMILES '[R][C@@H]1C[C@@]([Si](C)(O[Si](C)(C)C)C)(C2=CC=CC=C2)C[C@H](C)O1'. Stereochemistry of R, Ph, SiMe2OTMS, and other groups is indicated by solid and dashed wedge bonds. R is the placeholder used in the original diagram."}, {"file": "Exam-20-T9-7-2-rubric.png", "description": "Reference schematic for the stereoelectronic effect: a ring structure containing Si, Ph, R, O, and other atoms, with dumbbell-shaped orbital drawings and a curved arrow. The label 'Axial attack' appears below. This is a schematic and requires no SMILES."}]}]} +{"problem_id": "Exam-22-T4", "header": "Problem 4 Rotaxane Actuators and Mechanochemistry (28 points, 12%)\nIn the 38th Preliminary Round, we encountered the following reaction. Its principle is to apply mechanical force to a molecule so that four four-membered rings open sequentially: [Image: Exam-22-T4-header-1.png]\nMechanical force can do more than break covalent bonds: changing the force direction and mode of molecular deformation can select the reaction pathway. In 2024, Chen and co-workers constructed a rotaxane mechanical actuator using the sliding motion of a pillar[5]arene macrocycle relative to the axle. Mechanical action transmitted by ultrasonic cavitation allows the macrocycle to push mechanically responsive cargo sites along the axle and release the cargo molecules sequentially.", "header_images": [{"file": "Exam-22-T4-header-1.png", "description": "On the left is the structure of a polymer whose repeating-unit fragment has SMILES '[*]=CC1C(C2C3C4C=CC4C3C21)C=[*]'; the units are connected to one another through double bonds. On the right is a horizontal arrow pointing right, labeled “ultrasound, THF, 6~9 °C, Ar”."}], "subquestions": [{"sub_id": "Exam-22-T4-1", "tags": ["2.4", "2.6"], "text_open": "4.1 Synthesis of a rotaxane requires threading and end-capping. After threading, repeated sequences of dethreading → rethreading → dethreading may occur before end-capping is completed. In this study, the rotaxane structure was constructed by the reaction below. Threading can occur in two orientations, forming products in which the pillar[5]arene macrocycle and the 2-bromo-2-methylpropanoyloxy group on the axle lie on opposite sides (A$_{trans}$) or the same side (A$_{cis}$).\nBased on the reaction information above, write the SMILES string for B.", "text_images": [{"file": "Exam-22-T4-1-1-prompt.png", "description": "Rotaxane construction scheme. The reactants are 'COC(C=C(C(OC)=C1)CC2=C(C=C(C(OC)=C2)CC3=C(C=C(C(OC)=C3)CC4=C(C=C5C(OC)=C4)OC)OC)OC)=C1CC6=C(C=C(C(OC)=C6)C5)OCOC(C(Br)(C)C)=O' and 'CC(C)(Br)C(OCCCCCCCCCCCCO)=O'. Under BTBSCl, Et3N, CHCl3, -15 °C, 2 h, the products are [A$_{trans}$] (11%), [A$_{cis}$] (13%), and [B] (69%). The structure of the end-capping reagent BTBSCl, 'ClS(C1=CC(C(F)(F)F)=CC(C(F)(F)F)=C1)(=O)=O', is also shown. The structures of A$_{trans}$, A$_{cis}$, and B are not drawn; only their labels and yields are given."}], "full_mark": 3, "rubric": ["The SMILES of B is 'CC(C)(Br)C(OCCCCCCCCCCCCOS(C1=CC(C(F)(F)F)=CC(C(F)(F)F)=C1)(=O)=O)=O' (3 points)."], "rubric_images": [{"file": "Exam-22-T4-1-1-rubric.png", "description": "The SMILES of B is 'CC(C)(Br)C(OCCCCCCCCCCCCOS(C1=CC(C(F)(F)F)=CC(C(F)(F)F)=C1)(=O)=O)=O'."}]}, {"sub_id": "Exam-22-T4-2", "tags": ["1.2", "1.7"], "text_open": "4.2.3 In addition to the cis/trans stereoisomerism caused by the orientation of the long PMA polymer chains in the rotaxane, the [4+2] adduct fragment in the cargo molecule also exhibits endo/exo stereoisomerism. Using $^1$H NMR, the researchers measured the conversion of the Diels–Alder adduct, i.e., the cargo-release efficiency, for the four molecules and obtained the following data:\n\n| Species | cis-endo | cis-exo | trans-endo | trans-exo |\n| --- | --- | --- | --- | --- |\n| Cargo released/% | 62 | 51 | 71 | 71 |\n\nUnder the experimental conditions, the long PMA polymer chains experience mechanical forces from the extensional flow field generated by ultrasonic cavitation, causing the pillar[5]arene macrocycle to slide along the axle. Based on the data above, determine the force direction on the polymer chain attached to the pillar[5]arene macrocycle, either from the chain terminus toward the pillar[5]arene macrocycle or from the pillar[5]arene macrocycle toward the chain terminus, and briefly explain your answer.", "text_images": [], "full_mark": 4, "rubric": ["From the pillar[5]arene macrocycle toward the chain terminus. (2 points) The data show that for both endo and exo adducts, the trans rotaxane has a higher release efficiency than the corresponding cis rotaxane. (1 point) This indicates that pulling the two PMA chains apart in opposite directions, as in the trans configuration, more effectively activates the cargo site. (1 point) Therefore, the force direction is from the pillar[5]arene macrocycle toward the chain terminus."], "rubric_images": []}, {"sub_id": "Exam-22-T4-3", "tags": ["1.7", "2.5"], "text_open": "4.3 In addition, a cargo-bearing chain can carry multiple cargo molecules to form an oligomeric cargo chain. The pillar[5]arene macrocycle can then move continuously along the main axle and release cargo from multiple cargo sites in sequence by a “pushing” action from front to back. Researchers developed rotaxane compounds containing 3 cargo units ($n$ = 0 in the figure below) and 5 cargo units ($n$ = 1 in the figure below), as shown:\n4.3.2 The researchers found, however, that increasing the number of cargo molecules loaded into a rotaxane markedly decreased the release efficiency. Rotaxanes carrying 3 or 5 cargo molecules released fewer than 2 cargo molecules on average. For this calculation, the rotaxane composed of the pillar[5]arene macrocycle and axle is treated as one molecule.\n\n| Number of cargo molecules loaded | Average release per site/% | Internal-site release/% | Terminal-site release/% |\n| --- | --- | --- | --- |\n| 1 | 71 | / | / |\n| 3 | 44 | 45 | 42 |\n| 5 | 22 | 22 | 23 |\n\nBased on these data, propose one reasonable explanation for the decrease in release efficiency.", "text_images": [{"file": "Exam-22-T4-3-1-prompt.png", "description": "The figure shows a supramolecular assembly. One component is a pillar[5]arene with structure 'COC1=C(CC2=C(O[R])C=C(C3)C(OC)=C2)C=C(OC)C(CC4=C(OC)C=C(CC5=C(OC)C=C(CC6=C(OC)C=C3C(OC)=C6)C(OC)=C5)C(OC)=C4)=C1'. The guest is a polymeric structure with repeating units and has SMILES '[R]OCCCCCCCCCCCCOC(C1=CC=CC(OCCN2N=NC(CCCCOC[C@@]3(O4)C=C[C@@]4(COCCCCC5=CN(CCCCCCOC[C@@]6(O7)C=C[C@@]7(COCCCCCCN8N=NC(CCCCOC[C@@]9%10C=C[C@@](O%10)(OCCCCC%11=CN(CCCCCCOC[C@@]%12(O%13)C=C[C@@]%13(COCCCCCCN%14C=C(CCCCOC[C@@]%15%16C=C[C@H](O%16)C(C(N%17C(C%18=CC=CC=C%18)(C%19=CC=CC=C%19)C%20=CC=CC=C%20)=O)C%15C%17=O)N=N%14)C(C(N%21C(C%22=CC=CC=C%22)(C%23=CC=CC=C%23)C%24=CC=CC=C%24)=O)C%12C%21=O)N=N%11)C(C(N%25C(C%26=CC=CC=C%26)(C%27=CC=CC=C%27)C%28=CC=CC=C%28)=O)C9C%25=O)=C8)C(C(N%29C(C%30=CC=CC=C%30)(C%31=CC=CC=C%31)C%32=CC=CC=C%32)=O)C6C%29=O)N=N5)C(C(N%33C(C%34=CC=CC=C%34)(C%35=CC=CC=C%35)C%36=CC=CC=C%36)=O)C3C%33=O)=C2)=C1)=O'. The repeating-unit fragment has structure 'O=C1N(C(C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4)C(C(C51)[C@](O6)(OCCCCC7=CN(CCCCCCOC[C@@]8(O9)C=C[C@@]9(COCCCCCCN%10C=C(CCCC[*])N=N%10)C(C(N%11C(C%12=CC=CC=C%12)(C%13=CC=CC=C%13)C%14=CC=CC=C%14)=O)C8C%11=O)N=N7)C=C[C@]56CO[*])=O'. The R groups on the pillar[5]arene and guest are polymer chains whose SMILES may be represented as '[*]C(C(C)(C)CC(C(OC)=O)Br)=O', where * is the attachment point to another group. The repeating unit of R is '[*]CC(C(OC)=O)[*]', where * marks connections between repeat units."}], "full_mark": 4, "rubric": ["The long PMA chains may break under ultrasound and thus fail to drive the pillar[5]arene along the axle. (2 points; any answer involving chain breakage is accepted, such as breakage of the cargo chain preventing some rotaxanes from releasing cargo.) Consequently, some rotaxanes cannot release cargo, whereas others release cargo from nearly all sites. (2 points) No credit is given for stating that the pillar[5]arene macrocycle slips off the axle."], "rubric_images": []}, {"sub_id": "Exam-22-T4-4", "tags": ["2.4", "2.5"], "text_open": "4.4 In addition to cargo release driven by a “pushing” action, the researchers designed the molecule below to investigate the effect of a “pulling” action on cargo release.\nUnder ultrasound in aqueous solution, the molecule undergoes not only a retro-[4+2] reaction but also heterolytic C-N bond cleavage to give two additional products.\n4.4.1 Write the SMILES strings for the two products formed by heterolytic C-N bond cleavage, both of which contain long polymer chains.", "text_images": [{"file": "Exam-22-T4-4-1-prompt.png", "description": "The molecule used to study the “pulling” action has SMILES 'O=C1N(C(C2=CC=CC=C2)(C3=CC=C(OCC4=CN(CCCO[R])N=N4)C=C3)C5=CC=CC=C5)C([C@@]6([H])[C@]7(COC(C)=O)C=C[C@](O7)(CCCCCC8=CN(N=N8)CCCO[R])[C@]61[H])=O'. [R] is a placeholder for a PMA polymer chain, and both side chains are -CH2CH2CH2-O-[R]. The definition of R is also shown at the left: R = ~C(=O)C(CH3)2CH2CH(Br)(CH2)mCO2Me, where ~ is the connection point and m is a subscript."}], "full_mark": 4, "rubric": ["(2 points each; the order may be reversed, and drawing the R substituents explicitly is also accepted)"], "rubric_images": [{"file": "Exam-22-T4-4-1-rubric.png", "description": "The reference answer contains two products shown side by side; their order may be reversed. The left product has SMILES 'O=C1NC([C@@]2([H])[C@]3(COC(C)=O)C=C[C@](O3)(CCCCCC4=CN(N=N4)CCCO[R])[C@]21[H])=O'. The right product has SMILES 'OC(C1=CC=CC=C1)(C2=CC=C(OCC3=CN(CCCO[R])N=N3)C=C2)C4=CC=CC=C4'. Both products contain the [R] placeholder for a PMA chain."}]}]} +{"problem_id": "Exam-23-T3", "header": "Question 3 Reversible-Deactivation Anionic Polymerization (8 points, 4%)\nConventional anionic polymerization of methacrylates, such as ethyl methacrylate shown below, requires highly hazardous organolithium initiators. Because the chain end is extremely reactive, the reaction must be conducted at very low temperature, such as -78 ℃, under rigorously anhydrous and oxygen-free conditions; otherwise severe side reactions terminate the chains.", "header_images": [{"file": "Exam-23-T3-header-1.png", "description": "The SMILES representation of ethyl methacrylate (EMA) is 'O=C(OCC)C(C)=C'."}], "subquestions": [{"sub_id": "Exam-23-T3-1", "tags": ["2.2"], "text_open": "3.1 Chemists recently developed a $CO_2$-mediated reversible-deactivation anionic polymerization (CMAP). The method operates at 90 ℃, uses a stable and safe solid potassium salt as initiator, and produces polymers with a very narrow molecular-weight distribution, or very low dispersity $Đ$, under a $CO_2$ atmosphere. The key conditions are shown below.\n3.1.1 Anionic polymerization initiator A first decarboxylates to form an enolate that initiates polymerization. Write the SMILES string for A.", "text_images": [{"file": "Exam-23-T3-1-1-prompt.png", "description": "'O=C(C(C)=C)OCC'>1.0 equiv. A; 18-crown-6 (1.2 equiv.); TEMPO (0.1 equiv.); 1 atm $CO_2$; DMF, 90°C, 6 h>'O=C(C(CC(C(OCC)=O)(C)C(O[K])=O)(C)C)OCC'. The product is poly(ethyl methacrylate), poly-EMA. Its repeating unit is enclosed in brackets with subscript n at the lower right. The listed product SMILES schematically represents an EtO-C(=O)-C(CH3)2- initiating end connected through -CH2- to a dormant-end unit -C(CH3)(CO2Et)(CO2K)."}], "full_mark": 1, "rubric": ["Structure of A (1 point)."], "rubric_images": [{"file": "Exam-23-T3-1-1-rubric.png", "description": "The SMILES representation of initiator A is 'O=C(C(C)(C(O[K])=O)C)OCC'."}]}, {"sub_id": "Exam-23-T3-2", "tags": ["2.8"], "text_open": "3.1.2 State the principal chemical role of 18-crown-6 (shown as 18-crown-6 in the reaction scheme) in the system.", "text_images": [], "full_mark": 2, "rubric": ["By complexing potassium ions, 18-crown-6 increases the separation within the ion pair. This greatly lowers the activation energy for decarboxylation and promotes chain initiation (1 point), while also increasing the nucleophilicity of the resulting carbanion toward the monomer (1 point).", "1 point for promoting decarboxylation and 1 point for increasing carbanion nucleophilicity."], "rubric_images": []}, {"sub_id": "Exam-23-T3-3", "tags": ["2.4", "2.3"], "text_open": "3.2 Control of side reactions is also an important topic in anionic polymerization.\n3.2.1 In conventional high-temperature anionic polymerization without $CO_2$ protection, the active chain end of a polymethacrylate readily attacks an upstream carbonyl group in the polymer backbone. This intramolecular backbiting completely terminates the chain. Write the SMILES string for the final cyclic end group formed after backbiting termination of active poly(ethyl methacrylate), poly-EMA; only one chain end need be shown.", "text_images": [], "full_mark": 2, "rubric": ["2 points for the right-hand end-group structure. Errors in the polymer chain or left-hand termination are not penalized."], "rubric_images": [{"file": "Exam-23-T3-3-1-rubric.png", "description": "The figure shows a polymer whose repeating-unit SMILES is '[*]C(C)(C(OCC)=O)C[*]', where * denotes a connection between structural units. One polymer end group is methyl; the other is 'O=C(OCC)C1(C)CC(C(OCC)=O)(C)C(C(C[*])(C)C1)=O', where * denotes its connection to the repeating unit."}]}]} {"problem_id": "Exam-23-T8", "header": "Problem 8 Norbornane (24 points, 11%)\nBecause of their rigid frameworks and fixed conformations, norbornane and its derivatives have become classic models in physical organic chemistry for studying stereoelectronic effects, neighboring-group participation, and nonclassical carbocations. The structure and carbon numbering are shown below.", "header_images": [{"file": "Exam-23-T8-header-1.png", "description": "Skeletal structure of norbornane (bicyclo[2.2.1]heptane), with the carbon atoms labeled 1, 2, 3, 4, 5, 6, and 7. Its SMILES is '[C@@H]12CC[C@H](C2)CC1'."}], "subquestions": [{"sub_id": "Exam-23-T8-1", "tags": ["1.6", "2.4"], "text_open": "8.2 A substituent X can be introduced at C2 of norbornane, giving the exo-X and endo-X diastereomers.\n8.2.1 When the substituent is hydroxy, which is more stable, the exo or endo configuration? Explain.", "text_images": [{"file": "Exam-23-T8-1-1-prompt.png", "description": "Two norbornane frameworks with substituent X attached to a ring carbon. The left structure is labeled 'exo' below; X extends up-right by a single bond, and the halogen is equatorial. Its SMILES is '[F,Cl,Br,I][C@H]1[C@@H](C2)CC[C@@H]2C1'. The right structure is labeled 'endo'; X extends vertically downward, and the halogen is axial. Its SMILES is '[F,Cl,Br,I][C@@H]1[C@@H](C2)CC[C@@H]2C1'. Here [F,Cl,Br,I] is the ChemDraw export representation of substituent X as a list of halogen atoms."}], "full_mark": 3, "rubric": ["The exo configuration is more stable (1 point).", "In the endo configuration, the C–O bond and C6-endo-H have a 1,3-diaxial spatial relationship and repel each other (1 point).", "In the exo configuration, the C1–C6 bond can hyperconjugate with the C–O bond (1 point).", "Award 1 point for the steric difference and 1 point for the hyperconjugative effect. If the stability order is incorrect, award no explanation points."], "rubric_images": []}, {"sub_id": "Exam-23-T8-2", "tags": ["1.6", "2.4"], "text_open": "8.2.3 Compare the acidities of exo-2-norbornanecarboxylic acid and endo-2-norbornanecarboxylic acid in aqueous solution, and explain the difference.", "text_images": [], "full_mark": 3, "rubric": ["exo-2-Norbornanecarboxylic acid is more acidic (1 point).", "The endo carboxylate anion lies on the concave face of the molecule, where the surrounding framework hinders the approach of water. Poor solvation destabilizes the anion, so its conjugate acid is weaker (2 points).", "Award 1 point for identifying steric hindrance and 1 point for stating that the exo carboxylate is well solvated by water and its negative charge is effectively stabilized. If the acidity order is incorrect, award no explanation points."], "rubric_images": []}, {"sub_id": "Exam-23-T8-3", "tags": ["2.4", "2.3"], "text_open": "8.2.4 Write the SMILES string for the major product of the following reaction.", "text_images": [{"file": "Exam-23-T8-3-1-prompt.png", "description": "Two-step reaction. The substrate is a norbornane derivative with SMILES 'Br[C@@H]1[C@@H](C2)CC[C@@H]2[C@@H]1[2H]': bromine is attached to one ring carbon and deuterium [2H] to the adjacent ring carbon. Two successive rightward arrows follow the substrate; the first is labeled 'LDA' and the second 'm-CPBA'. No structure is drawn after either arrow."}], "full_mark": 2, "rubric": ["Major product: (2 points)"], "rubric_images": [{"file": "Exam-23-T8-3-1-rubric.png", "description": "Major product: an oxygen atom bridges the two adjacent ring carbons that originally bore Br and D in the norbornane framework, forming a three-membered epoxide ring. The oxygen is labeled 'O'. Its SMILES is 'C1C[C@H]2C[C@@H]1[C@@H]3O[C@H]23'."}]}, {"sub_id": "Exam-23-T8-4", "tags": ["2.3", "2.4"], "text_open": "8.3 The distinctive framework gives norbornane and norbornene derivatives exceptionally rich reactivity.\n8.3.1 The two compounds below give the same product upon solvolysis in acetic acid. Identify which compound solvolyzes faster and write the SMILES string for the product.", "text_images": [{"file": "Exam-23-T8-4-1-prompt.png", "description": "Two norbornane frameworks labeled A (left) and B (right). In A, OTs (p-toluenesulfonate) extends vertically downward by a single bond; its SMILES is 'CC1=CC=C(S(O[C@@H]2[C@@H](C3)CC[C@@H]3C2)(=O)=O)C=C1'. In B, OTs extends diagonally up-right; its SMILES is 'CC1=CC=C(S(O[C@H]2[C@@H](C3)CC[C@@H]3C2)(=O)=O)C=C1'. The OTs group occupies the same position in both structures but has a different stereochemical orientation."}], "full_mark": 3, "rubric": ["B (1 point)", "Product structure: (2 points; award only 1 point if the product is not identified as racemic; an endo product receives no credit)"], "rubric_images": [{"file": "Exam-23-T8-4-1-rubric.png", "description": "Product: a norbornane ring carbon bears OAc (acetoxy), with '(±)' at lower right indicating a racemate. Its SMILES is 'O=C(O[C@H]1[C@@H](C2)CC[C@@H]2C1)C'."}]}, {"sub_id": "Exam-23-T8-5", "tags": ["2.3", "2.4"], "text_open": "8.3.2 Which of the following compounds solvolyzes faster in acetic acid? Explain using an important intermediate.", "text_images": [{"file": "Exam-23-T8-5-1-prompt.png", "description": "Two norbornene frameworks labeled A (left) and B (right). Each ring contains one C=C bond, and p-toluenesulfonate is attached to bridge carbon C7. In A, TsO extends down-left by a single bond; its SMILES is 'CC1=CC=C(S(O[C@H]2[C@H]3CC[C@@H]2C=C3)(=O)=O)C=C1'. In B, OTs extends up-right; its SMILES is 'CC1=CC=C(S(O[C@@H]2[C@H]3CC[C@@H]2C=C3)(=O)=O)C=C1'. The structures otherwise differ only in the orientation of the sulfonate group."}], "full_mark": 2, "rubric": ["A (1 point)", "Key intermediate: (1 point; its enantiomer also receives credit)"], "rubric_images": [{"file": "Exam-23-T8-5-1-rubric.png", "description": "Key intermediate (first image, red): its SMILES can be written as '[C@H]12CC[C@@H]([CH+]2)C=C1'. The double bond conjugatively stabilizes the carbocation at the bridgehead carbon, as indicated by dashed lines."}, {"file": "Exam-23-T8-5-2-rubric.png", "description": "Key intermediate (second image, blue): its SMILES can be written as '[C@H]12CC[C@@H]([C@@H]23)C3[CH+]1'."}]}, {"sub_id": "Exam-23-T8-6", "tags": ["2.3", "2.4"], "text_open": "8.3.3 In the reaction below, an optically active substrate racemizes under the action of $KO^tBu$. Write the SMILES string for a key intermediate that explains the racemization.", "text_images": [{"file": "Exam-23-T8-6-1-prompt.png", "description": "One-step reaction. The substrate is a norbornanone derivative whose carbonyl oxygen is labeled 'O'; the ring carbon adjacent to the carbonyl carbon bears two methyl groups, shown as two short single bonds. Its SMILES is 'CC1(C)[C@H]2CC[C@H](C2)C1=O'. The arrow is labeled 'KOtBu, HOtBu' above (t is superscript) and '250°C' below. The product at right has the same framework as the substrate and is marked '(±)' at lower right to indicate a racemate. Thus, the substrate is chiral and the product is racemic."}], "full_mark": 2, "rubric": ["Key intermediate: (2 points)"], "rubric_images": [{"file": "Exam-23-T8-6-1-rubric.png", "description": "Key intermediate: in the norbornanone framework, the former carbonyl carbon is bonded to a negatively charged oxygen, labeled 'O' with a circled minus sign. That carbon also has an additional single bond to another ring carbon (C2), and C5 retains two methyl groups. Its SMILES is 'CC1(C)C2CC3C(C2)C13[O-]'."}]}, {"sub_id": "Exam-23-T8-7", "tags": ["2.4", "1.6"], "text_open": "8.3.4 Based on the structure in 8.3.3, predict which hydrogen in the substrate is the most acidic. Give a reasonable explanation.\n(a) $H_a$\n(b) $H_b$\n(c) $H_c$\n(d) $H_d$\n(e) $H_e$", "text_images": [{"file": "Exam-23-T8-7-1-prompt.png", "description": "The same substrate as in 8.3.3: the carbonyl oxygen is labeled 'O', and the ring carbon adjacent to the carbonyl carbon bears two methyl groups. Five hydrogen positions are additionally labeled: $H_a$ and $H_b$ are attached to the same ring carbon (C2), $H_c$ to bridgehead carbon C4, and $H_d$ and $H_e$ to bridge carbon C7. The hydrogen-labeled SMILES is 'CC1(C)[C@@]2([Hc])C[C@]([Ha])([Hb])[C@H]([C@@]2([He])[Hd])C1=O', where [Ha] through [He] are placeholders for the image labels $H_a$ through $H_e$."}], "full_mark": 4, "rubric": ["(a) (2 points)", "The bonding orbital of the C2–H bond overlaps effectively with the antibonding orbital of the C=O bond. Donation of the C2–H bonding electrons into the carbonyl π* orbital increases the acidity of this hydrogen (2 points)."], "rubric_images": []}]} +{"problem_id": "Exam-23-T10", "header": "Problem 10 Dearomatization of Benzene Rings (27 points, 13%)\nAromaticity makes the benzene ring structurally stable and provides substantial resonance stabilization. Directly disrupting its aromatic conjugation or cleaving a C–C bond in the ring framework normally requires considerable energy. Dearomatization or ring-opening cleavage of a benzene ring is therefore highly challenging.", "header_images": [], "subquestions": [{"sub_id": "Exam-23-T10-1", "tags": ["2.2"], "text_open": "10.1 In 2024, researchers developed a novel benzene-ring-opening strategy that directly converts a range of aryl azide derivatives into valuable multiply conjugated enynes.\n10.1.1 Write the SMILES string for intermediate A formed before $^tBuONa$ is added.", "text_images": [{"file": "Exam-23-T10-1-1-prompt.png", "description": "Single-step reaction: 'O=C(C1=CC=CC=C1N=[N+]=[N-])C2=CC=CC=C2'>TosNHNH2 (1.2 equiv), TFA (20 mol%), PhCl (0.2 M); then $^tBuONa$ (1.5 equiv), PhCl (0.1 M), 130°C>'N#C/C=C\\C=C/C#CC1=CC=CC=C1'."}], "full_mark": 1, "rubric": ["A: '[N-]=[N+]=NC1=CC=CC=C1/C(C2=CC=CC=C2)=N/NS(C3=CC=C(C)C=C3)(=O)=O' (1 point; E/Z geometry of the double bond is not required)"], "rubric_images": [{"file": "Exam-23-T10-1-1-rubric.png", "description": "The SMILES for A is '[N-]=[N+]=NC1=CC=CC=C1/C(C2=CC=CC=C2)=N/NS(C3=CC=C(C)C=C3)(=O)=O'."}]}, {"sub_id": "Exam-23-T10-2", "tags": ["2.2", "2.3"], "text_open": "10.1.2 To investigate the mechanism, the researchers performed the following experiments.\nCalculations indicate that after $^tBuONa$ is added, loss of one molecule of gas first produces highly reactive intermediate B. B then reacts with the azide group to form charge-separated tricyclic intermediate C. A concerted reaction of C expels another gas molecule to give the product. Write the SMILES strings for B and C.", "text_images": [{"file": "Exam-23-T10-2-1-prompt.png", "description": "Two parallel mechanistic experiments. Upper reaction: the starting material is labeled only A and its structure is not drawn; conditions are TEMPO (2.0 equiv), $^tBuONa$ (1.5 equiv), PhCl, heat, giving 'N#C/C=C\\C=C/C#CC1=CC=CC=C1' in 70% yield. Lower reaction: 'C1(C(C2=CC=CC=C2)=NN=N3)=C3C=CC=C1'>$^tBuONa$ (1.5 equiv), PhCl, heat>'N#C/C=C\\C=C/C#CC1=CC=CC=C1', 0% yield."}], "full_mark": 3, "rubric": ["B: '[N-]=[N+]=NC1=CC=CC=C1[C]C2=CC=CC=C2' (1 point); C: '[N-]=[N+]1C(C2=CC=CC=C2)=C(C=CC=C3)C3=N1' (2 points). Reasonable resonance forms of B and C also receive credit."], "rubric_images": [{"file": "Exam-23-T10-2-1-rubric.png", "description": "The SMILES for B is '[N-]=[N+]=NC1=CC=CC=C1[C]C2=CC=CC=C2', where [C] is a carbene carbon shown with two dots in the image. The SMILES for C is '[N-]=[N+]1C(C2=CC=CC=C2)=C(C=CC=C3)C3=N1'."}]}, {"sub_id": "Exam-23-T10-3", "tags": ["2.4"], "text_open": "10.1.3 Considering the reaction sequence above, write the SMILES strings for the products of the following reactions. (Hint: each reaction proceeds through a high-energy intermediate similar to B.)", "text_images": [{"file": "Exam-23-T10-3-1-prompt.png", "description": "Two parallel reactions, arranged vertically, with neither product drawn. Upper reaction: 'O=C(C1=CC=CC=C1N=[N+]=[N-])C2=CC=C(C)C=C2'>TosNHNH2 (1.2 equiv), TFA (20 mol%), PhCl; then $^tBuONa$ (1.5 equiv), PhCl, heat>[product not shown]. Lower reaction: 'O=C(C)/C(C(C)=O)=C/C#CC1=CC=CC=C1N=[N+]=[N-]'>PhCl (0.05 M), heat>[product not shown]."}], "full_mark": 3, "rubric": ["First product: 'CC(C=C1)=CC=C1C#C/C=C\\C=C/C#N' (1 point); second product: 'O=C(C)C1=C(C)OC(C#C/C=C\\C=C/C#N)=C1' (2 points)"], "rubric_images": [{"file": "Exam-23-T10-3-1-rubric.png", "description": "The SMILES for the left (first) product is 'CC(C=C1)=CC=C1C#C/C=C\\C=C/C#N'; the SMILES for the right (second) product is 'O=C(C)C1=C(C)OC(C#C/C=C\\C=C/C#N)=C1'."}]}, {"sub_id": "Exam-23-T10-4", "tags": ["2.2"], "text_open": "10.2 In 2025, the group of Professor Daniele Leonori reported oxidative cleavage of electron-rich aromatic rings using photoexcited nitroarenes. (λ = 390 nm denotes irradiation with 390 nm light.)\n10.2.1 Under irradiation, the two substrates react to form a five-membered-ring intermediate Int, whose fragmentation gives the product. Write the SMILES string for Int; the aromatic ring attached to the nitro group may be abbreviated as -Ar.", "text_images": [{"file": "Exam-23-T10-4-1-prompt.png", "description": "Photooxidative-cleavage reaction: 'COC1=CC=CC2=CC=CC=C21'.'O=[N+](C1=CC=CC=C1)[O-].[R]C'>CH2Cl2 (0.1 M), -65°C, 5h, λ = 390 nm>'O=C(OC)C1=CC=CC=C1/C=C\\C=O'. [R] in substrate 2 is the original R-label placeholder."}], "full_mark": 1, "rubric": ["Int: '[Ar]N(O1)OC2C=CC3=CC=CC=C3C21OC' (1 point; [Ar] is the original placeholder for the aromatic ring attached to the nitro group)"], "rubric_images": [{"file": "Exam-23-T10-4-1-rubric.png", "description": "The SMILES for Int is '[Ar]N(O1)OC2C=CC3=CC=CC=C3C21OC', where [Ar] is the original placeholder for the aromatic ring attached to the nitro group. The structure is a naphthalene-fused five-membered O–N–O ring bearing a methoxy substituent on a ring carbon."}]}, {"sub_id": "Exam-23-T10-5", "tags": ["3.9", "1.6"], "text_open": "10.2.2 The researchers reacted differently substituted nitroarenes with a substrate containing both an aromatic ring and an alkene and found that the substituents affect chemoselectivity. For example, nitroarenes N1 and N2 react with D to give selective oxidation of either the alkene or the aromatic ring.\nStudies of nitroarene excited states show that their lowest triplet state may have either a $^3(n,\\pi^*)$ or $^3(\\pi,\\pi^*)$ electronic configuration. These correspond, respectively, to an n to $\\pi^*$ transition, representable as a dioxygen-diradical species, and a $\\pi, \\pi^*$ transition, involving intramolecular electron transfer from the benzene unit to the nitro group with the unpaired electron delocalized over the entire $\\pi$ system. Based on the electronic properties of the substituents, which state is lower in energy for N1, and which is lower for N2?\n(a) n, $\\pi^*$ state; (b) $\\pi, \\pi^*$ state.", "text_images": [{"file": "Exam-23-T10-5-1-prompt.png", "description": "Three compounds, labeled D, N1, and N2 from left to right. The SMILES for N1 is 'O=[N+](C1=CC=C(OC)C(OC)=C1)[O-]'; for N2, 'O=[N+](C1=CC(C(F)(F)F)=CC(C(F)(F)F)=C1)[O-]'; and for D, 'C/C(C)=C\\CCOC1=CC=CC2=C1C=CC=C2'."}], "full_mark": 2, "rubric": ["(b), (a) (2 points only if both are correct)"], "rubric_images": []}, {"sub_id": "Exam-23-T10-6", "tags": ["2.7", "3.9"], "text_open": "10.2.3 A triplet excited state with an n, $\\pi^*$ configuration can undergo cycloaddition directly. A triplet excited state with a $\\pi, \\pi^*$ configuration cannot undergo direct concerted cycloaddition; calculations indicate the following pathway. It first forms a complex with the other substrate, in which dispersion is the principal intermolecular attraction. Partial electron transfer between the substrates then gives the nitro group a partial negative charge and the other substrate a partial positive charge, promoting formation of the first C–O bond. Finally, intersystem crossing and ring closure give the five-membered intermediate. Accordingly, a nitroarene excited state with an n, $\\pi^*$ configuration reacts more readily with ( ), whereas one with a $\\pi, \\pi^*$ configuration reacts more readily with ( ). Explain each choice.\n(a) An electron-rich aromatic ring; (b) an isolated double bond.", "text_images": [], "full_mark": 5, "rubric": ["(b), (a) (2 points only if both are correct. Award no explanation points for an incorrect choice; likewise, a correct choice without the corresponding correct explanation receives no explanation credit.)", "n, $\\pi^*$ configuration: cycloaddition to an aromatic ring would destroy aromaticity and therefore has a high activation energy, so the n, $\\pi^*$ nitroarene excited state reacts more readily with an isolated double bond (1 point for mentioning the difficulty of disrupting aromaticity).", "$\\pi, \\pi^*$ configuration: an aromatic ring has a larger polarizable $\\pi$ surface and broader contact area than an isolated alkene bond, providing stronger dispersion and favoring complex formation. An electron-rich aromatic ring also transfers an electron to the nitro group more readily than an isolated double bond because its HOMO is higher (2 points: 1 for stronger dispersion and 1 for the charge-transfer complex)."], "rubric_images": []}, {"sub_id": "Exam-23-T10-7", "tags": ["2.4"], "text_open": "10.2.4 Write the SMILES string for the product formed when D is oxidized by N2 under irradiation.", "text_images": [], "full_mark": 2, "rubric": ["Product: 'O=CCCOC1=CC=CC2=CC=CC=C21' (2 points)"], "rubric_images": [{"file": "Exam-23-T10-7-1-rubric.png", "description": "The SMILES for the product in 10.2.4 is 'O=CCCOC1=CC=CC2=CC=CC=C21'."}]}, {"sub_id": "Exam-23-T10-8", "tags": ["2.2", "2.3"], "text_open": "10.3 In 2024, Professor Youliang Wang's group achieved thermochemical dearomatization of a benzene ring, constructing a complex ring system in one step.\nExperiments show that when TFA is replaced by deuterated TFA and the reaction is run for 10 h, 99% of the circled hydrogen atom in the product is replaced by deuterium.\n10.3.1 The reaction includes a pericyclic step. Write the SMILES strings for 3–4 key intermediates and name the pericyclic reaction.", "text_images": [{"file": "Exam-23-T10-8-1-prompt.png", "description": "Single-step reaction: 'C=CCN(C1=CC=CC=C1)C(C#CC2=CC=CC=C2)=O'>TFA, 50°C, 12h>'O=C(N1C2)C([H])=C(C31C4C5C=CC3C2C4)C6=C5C=CC=C6'. The hydrogen at C([H]) in the product is circled, corresponding to the deuterium-labeling experiment in the prompt."}], "full_mark": 7, "rubric": ["Intermediate 1: 'O=C1N(CC=C)C2(C=CC=C[CH+]2)C(C3=CC=CC=C3)=C1[H]'; intermediate 2: 'O=C(N1C2)C([H])=C([C@@]31[C@H]4[CH+]C=C[C@@H]3C2C4)C5=CC=CC=C5'; intermediate 3: 'O=C1C([H])=C2C3=CC=C[CH+]C3C4C=CC5C62C4CC5CN16' (2 points each; award credit if connectivity is correct and the charge is represented reasonably. Consider only the first 4 intermediates drawn by the student).", "An intermediate formed by addition of $H^+$ to the triple bond receives neither credit nor a penalty.", "Pericyclic reaction: [5+2] cycloaddition (or [4π+2π] cycloaddition) (1 point)"], "rubric_images": [{"file": "Exam-23-T10-8-1-rubric.png", "description": "Three cationic intermediates, from left to right. Intermediate 1 has SMILES 'O=C1N(CC=C)C2(C=CC=C[CH+]2)C(C3=CC=CC=C3)=C1[H]'; intermediate 2, 'O=C(N1C2)C([H])=C([C@@]31[C@H]4[CH+]C=C[C@@H]3C2C4)C5=CC=CC=C5'; intermediate 3, 'O=C1C([H])=C2C3=CC=C[CH+]C3C4C=CC5C62C4CC5CN16'."}, {"file": "Exam-23-T10-8-2-rubric.png", "description": "The SMILES for the intermediate formed by addition of $H^+$ to the triple bond is 'C=CCN(C1=CC=CC=C1)C(/C([H])=[C+]/C2=CC=CC=C2)=O'. The positive charge is at [C+]. The rubric specifies that drawing this intermediate earns no credit but incurs no penalty."}]}, {"sub_id": "Exam-23-T10-9", "tags": ["2.4"], "text_open": "10.3.2 The reaction also forms two byproducts. E has a ring system similar to that of the major product, whereas F contains only three rings and no bridged ring. Write the SMILES strings for E and F.", "text_images": [], "full_mark": 3, "rubric": ["E: 'O=C(N1C2)C([H])=C([C@@]31[C@H]4C=C[C@@H]5[C@@H]3[C@@H]2C4)C6=C5C=CC=C6' (1 point); F: 'O=C1N(CC=C)C2=CC=CC=C2C(C3=CC=CC=C3)=C1' (2 points)"], "rubric_images": [{"file": "Exam-23-T10-9-1-rubric.png", "description": "The SMILES for E is 'O=C(N1C2)C([H])=C([C@@]31[C@H]4C=C[C@@H]5[C@@H]3[C@@H]2C4)C6=C5C=CC=C6'."}, {"file": "Exam-23-T10-9-2-rubric.png", "description": "The SMILES for F is 'O=C1N(CC=C)C2=CC=CC=C2C(C3=CC=CC=C3)=C1'."}]}]} diff --git a/dev/instance_ids.json b/dev/instance_ids.json index 06e4f2f9f4803d98d47098c0c942ddf12365c2b7..2a5ee3bd9e1cced8361a5e5525eb289792ef3bb6 100644 --- a/dev/instance_ids.json +++ b/dev/instance_ids.json @@ -1,50 +1,50 @@ [ "Exam-01-T5", - "Exam-01-T7", - "Exam-02-T1", - "Exam-02-T3", - "Exam-02-T5", + "Exam-01-T9", "Exam-02-T6", - "Exam-02-T9", - "Exam-03-T2", - "Exam-03-T10", + "Exam-02-T10", + "Exam-03-T8", + "Exam-03-T9", "Exam-04-T6", - "Exam-05-T4", + "Exam-04-T10", + "Exam-05-T5", "Exam-05-T7", - "Exam-05-T8", - "Exam-06-T4", + "Exam-06-T2", + "Exam-07-T4", "Exam-07-T6", "Exam-07-T10", - "Exam-08-T8", "Exam-08-T10", + "Exam-09-T4", "Exam-09-T5", - "Exam-09-T8", - "Exam-09-T10", - "Exam-12-T1", - "Exam-12-T4", - "Exam-12-T10", + "Exam-09-T7", + "Exam-09-T9", + "Exam-10-T5", + "Exam-10-T7", + "Exam-10-T9", + "Exam-10-T10", + "Exam-11-T3", + "Exam-11-T5", + "Exam-12-T5", + "Exam-13-T5", "Exam-13-T9", "Exam-14-T3", "Exam-14-T6", - "Exam-15-T1", - "Exam-15-T4", + "Exam-15-T3", "Exam-15-T8", "Exam-15-T9", - "Exam-16-T3", - "Exam-17-T2", + "Exam-16-T6", + "Exam-16-T8", + "Exam-16-T9", "Exam-17-T3", - "Exam-17-T4", - "Exam-17-T7", - "Exam-17-T9", + "Exam-17-T5", + "Exam-17-T6", "Exam-17-T10", - "Exam-18-T3", - "Exam-18-T4", + "Exam-19-T6", "Exam-19-T7", "Exam-20-T7", - "Exam-21-T3", - "Exam-21-T4", - "Exam-21-T6", - "Exam-22-T1", - "Exam-22-T5", - "Exam-23-T8" + "Exam-20-T9", + "Exam-22-T4", + "Exam-23-T3", + "Exam-23-T8", + "Exam-23-T10" 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restrict our understanding of the reactivity of carbon-phosphorus double bonds to their “metathesis” reactions with carbon-oxygen double bonds. With the assistance of transition metals, however, double bonds between carbon and trivalent phosphorus display distinctive modes of formation and reactivity. In this problem, whenever a subquestion asks for the SMILES of a complex, supplement it with a verbal description of the specific coordination mode.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-01-T9-1", + "tags": [ + "1.4", + "2.2", + "2.4" + ], + "text_open": "9-1 Carbon-phosphorus double bonds can be constructed in many ways and can coordinate to transition metals in more than one mode. 9-1-1 In the reaction below, [Mn] denotes (η^{5}-C_{5}H_{4}CH_{3})Mn(CO)_{2}, and the reaction produces 4 molecules of THF. Given that the major product contains 2 phosphorus-oxygen single bonds, write the SMILES structure of the major product. 4 [Mn]·THF + 2 ^{t}BuCP + H_{2}O →", + "text_images": [], + "full_mark": 2, + "rubric": [ + "The SMILES structure of the major product is '[H]/C(C(C)(C)C)=P\\O/P=C([H])/C(C)(C)C.[Mn].[Mn].[Mn].[Mn]' (2 points). The P atoms and P=C double bonds coordinate to the 4 [Mn] fragments, respectively." + ], + "rubric_images": [ + { + "file": "Exam-01-T9-1-1-rubric.png", + "description": "The chemical structure drawn in red shows the coordination of one ligand to four '[Mn]' metal fragments. The ligand has the SMILES '[H]/C(C(C)(C)C)=P\\O/P=C([H])/C(C)(C)C'. A central oxygen atom (O) bridges two phosphorus atoms (P); each P forms a double bond (P=C) to a carbon atom bearing one hydrogen atom (H) and one tert-butyl group (tBu). Four arrows explicitly indicate the coordination sites: the two arrows on the right originate from the upper and lower P atoms and point to the corresponding two '[Mn]' fragments on the right, whereas the two arrows on the left originate from the centers of the upper and lower P=C double bonds and point to the corresponding two '[Mn]' fragments on the left." + } + ] + }, + { + "sub_id": "Exam-01-T9-2", + "tags": [ + "1.4", + "2.2", + "2.3" + ], + "text_open": "9-1-2 The following example provides an initial illustration of the role played by transition metals in forming carbon-phosphorus double bonds. Use SMILES structures and any necessary verbal descriptions to represent the three key electrically neutral intermediates in the reaction. Coordination of double bonds to Mo need not be shown in any of them.", + "text_images": [ + { + "file": "Exam-01-T9-2-1-prompt.png", + "description": "The reaction in the figure shows the anionic dinuclear molybdenum complex 'O=C[Mo](C=O)[Mo](PC1=CC=CC=C1)(C=O)C=O' reacting with acryloyl chloride, 'ClC(C=C)=O', to form a product in which the phosphaalkene ligand 'C/C=P\\C2=CC=CC=C2' is bound to the dinuclear molybdenum framework 'O=C[Mo](C=O)[Mo](C=O)C=O'. In the reactant, a phenylphosphido group bridges two Mo atoms, each bearing two carbonyl ligands; on one side, a coordinate bond shown by an arrow points to the left Mo atom. After reaction, the newly formed phosphaalkene ligand coordinates in a bidentate bridging mode: the lone pair on phosphorus forms a coordinate bond, shown by an arrow, to the left Mo atom, while the $P=C$ double bond coordinates through a $\\pi$ bond, shown by an arrow, to the right Mo atom." + } + ], + "full_mark": 3, + "rubric": [ + "From left to right, the figure shows three red dinuclear molybdenum complex structures. **First structure on the left:** its core is a dinuclear molybdenum carbonyl framework containing two mononuclear molybdenum '[Mo]' units and four carbonyl '[C-]#[O+]' ligands. A phenylphosphido ligand '[PH]C1=CC=CC=C1' bridges over the top, coordinating to the left Mo atom through a coordinate bond on the left and bonding to the right Mo atom through a single bond on the right. An acryloyl ligand 'C=CC=O' is also bonded below the right Mo atom. **Second structure in the middle:** this structure is similar to the first and likewise contains a dinuclear molybdenum carbonyl framework and a bridging phenylphosphido ligand '[PH]C1=CC=CC=C1', but a vinyl ligand 'C=C', formed after loss of a carbonyl group, is bonded below the right Mo atom. **Third structure on the right:** a phenyl(vinyl)phosphine ligand 'C=CP([H])C1=CC=CC=C1' coordinates above the dinuclear molybdenum carbonyl framework in a bidentate bridging mode. Specifically, the lone pair on phosphorus (P) forms a coordinate bond to the left Mo atom, while the $C=C$ double bond coordinates through a $\\pi$ bond to the right Mo atom. (1 point each) No credit is awarded for either of the first two intermediates if P-to-Mo coordination is omitted. No points are deducted if double-bond-to-Mo coordination is omitted from the third intermediate." + ], + "rubric_images": [ + { + "file": "Exam-01-T9-2-1-rubric.png", + "description": "From left to right, the figure shows three red dinuclear molybdenum complex structures. **First structure on the left:** its core is a dinuclear molybdenum carbonyl framework containing two mononuclear molybdenum '[Mo]' units and four carbonyl '[C-]#[O+]' ligands. A phenylphosphido ligand '[PH]C1=CC=CC=C1' bridges over the top, coordinating to the left Mo atom through a coordinate bond on the left and bonding to the right Mo atom through a single bond on the right. An acryloyl ligand 'C=CC=O' is also bonded below the right Mo atom. **Second structure in the middle:** this structure is similar to the first and likewise contains a dinuclear molybdenum carbonyl framework and a bridging phenylphosphido ligand '[PH]C1=CC=CC=C1', but a vinyl ligand 'C=C', formed after loss of a carbonyl group, is bonded below the right Mo atom. **Third structure on the right:** a phenyl(vinyl)phosphine ligand 'C=CP([H])C1=CC=CC=C1' coordinates above the dinuclear molybdenum carbonyl framework in a bidentate bridging mode. Specifically, the lone pair on phosphorus (P) forms a coordinate bond to the left Mo atom, while the $C=C$ double bond coordinates through a $\\pi$ bond to the right Mo atom." + } + ] + }, + { + "sub_id": "Exam-01-T9-3", + "tags": [ + "1.4", + "2.2", + "2.6" + ], + "text_open": "9-1-3 Write the SMILES structures of the six key intermediates in the reaction shown below. Note: the step involving addition of tert-butyllithium corresponds to three key intermediates; each other step corresponds to one key intermediate.", + "text_images": [ + { + "file": "Exam-01-T9-3-1-prompt.png", + "description": "Description of the structures/reaction in the figure: '[H]C#CC1=CC=CC=C1' undergoes steps involving NEt_3/PCl_3, ArLi, tBuLi, ML_n, and irradiation/O_2 to form a species whose ligand has the SMILES 'O=C(C(/C(C(C1=CC=CC=C1)=O)=P/C)=P\\C)C2=CC=CC=C2'. The ligand coordinates to [ML_n] through its two P atoms. Ar denotes an aryl substituent, and [ML_n] denotes a metal coordination fragment." + } + ], + "full_mark": 6, + "rubric": [ + "6 points total for the six key intermediates in the grading figure. SMILES of the first intermediate: 'ClP(C#CC1=CC=CC=C1)Cl'; second intermediate: 'ClP(C#CC1=CC=CC=C1)[Ar]'; third intermediate: '[Ar]P(C#CC1=CC=CC=C1)P([Ar])C#CC2=CC=CC=C2'; fourth intermediate: '[Ar]P=C=C(C1=CC=CC=C1)C(C2=CC=CC=C2)=C=P[Ar]'; fifth intermediate: '[Ar]/P=C(C(C1=CC=CC=C1)=C2C3=CC=CC=C3)\\C2=P/[Ar]'; sixth intermediate: '[Ar]/P=C1/C(C(C2=CC=CC=C2)=C1C3=CC=CC=C3)=P/[Ar].[M]', with the ligand coordinating to [M] through both P atoms. Each intermediate is worth 1 point. Writing the lithium phosphinoalkyne intermediate 'Ph-C#C-P(Ar)Li' earns no credit but incurs no deduction." + ], + "rubric_images": [ + { + "file": "Exam-01-T9-3-1-rubric.png", + "description": "Description of the structures/reaction in the figure. SMILES of the first intermediate: 'ClP(C#CC1=CC=CC=C1)Cl'; second intermediate: 'ClP(C#CC1=CC=CC=C1)[Ar]'; third intermediate: '[Ar]P(C#CC1=CC=CC=C1)P([Ar])C#CC2=CC=CC=C2'; fourth intermediate: '[Ar]P=C=C(C1=CC=CC=C1)C(C2=CC=CC=C2)=C=P[Ar]'; fifth intermediate: '[Ar]/P=C(C(C1=CC=CC=C1)=C2C3=CC=CC=C3)\\C2=P/[Ar]'; sixth intermediate: '[Ar]/P=C1/C(C(C2=CC=CC=C2)=C1C3=CC=CC=C3)=P/[Ar].[M]', with the ligand coordinating to [M] through both P atoms." + }, + { + "file": "Exam-01-T9-3-2-rubric.png", + "description": "Description of the structure/reaction in the figure: the additional structure is 'Ph-C#C-P(Ar)Li'. Writing this intermediate earns no credit but incurs no deduction." + } + ] + }, + { + "sub_id": "Exam-01-T9-4", + "tags": [ + "2.2", + "2.3", + "1.4" + ], + "text_open": "9-2 An important class of reactions of carbon-phosphorus double bonds involves metal carbenes. 9-2-1 In the following reaction, addition between the two substrates gives A_{1}, and A_{1} then converts into its isomer A_{2}. A_{2} undergoes a hydrogen migration to give A_{3}; A_{3} eliminates HNMe_{2} to give A_{4}; and a further elimination gives A_{5} together with byproduct A_{6}. A_{5} undergoes a simple reaction to afford the final product. W remains six-coordinate throughout the process. In A_{2}, W is coordinated by the carbon-phosphorus double bond, whereas in A_{3}, A_{4}, and A_{6}, W is coordinated by P. On this basis, write the SMILES structures of A_{1}~A_{6}.", + "text_images": [ + { + "file": "Exam-01-T9-4-1-prompt.png", + "description": "Description of the structures/reaction in the figure: 'CN(C)C(=P)N(C)C' reacts with 'C1CC1C(O[Si](C)(C)C)=[W]([C]#O)([C]#O)([C]#O)([C]#O)[C]#O' to form 'O=C1C2(CC2)C(=O)C13CC3'." + } + ], + "full_mark": 6, + "rubric": [ + "6 points total; A_{1}~A_{6} in the grading figure are worth 1 point each. **A_1**, at the far left of the first row, is a zwitterionic complex represented as 'CN(C)/C(N(C)C)=[PH+]/C(C1CC1)(O[Si](C)(C)C)[W-]([C-]#[O+])([C-]#[O+])([C-]#[O+])([C-]#[O+])[C-]#[O+]'; the negatively charged W center bearing five carbonyl ligands is bonded directly to the ligand carbon atom. **A_2**, second from the left in the first row, is an $\\eta^2$-coordinated complex represented as 'CN(C)C(N(C)C)=P=C(C2CC2)O[Si](C)(C)C.[W]([C-]#[O+])([C-]#[O+])([C-]#[O+])([C-]#[O+])[C-]#[O+]'; its $P=C$ double bond coordinates through a $\\pi$ bond to the pentacarbonyltungsten fragment. **A_3**, second from the right in the first row, is a $\\sigma$-coordinated complex represented as 'CN(C)C(N(C)C)/P([W]([C-]#[O+])([C-]#[O+])([C-]#[O+])([C-]#[O+])[C-]#[O+])=C(C3CC3)/O[Si](C)(C)C'; the P atom of the phosphaalkene ligand coordinates to the pentacarbonyltungsten center. **A_4**, at the far right of the first row, is likewise a $\\sigma$-coordinated complex bearing a cyclopropylidene substituent and is represented as 'CN(C)/C=P([W]([C-]#[O+])([C-]#[O+])([C-]#[O+])([C-]#[O+])[C-]#[O+])/C(O[Si](C)(C)C)=C4CC\\4'. **A_5**, on the left of the second row, is free cyclopropylketene represented as 'O=C=C5CC5'. **A_6**, on the right of the second row, is also a $\\sigma$-coordinated complex; its silyl-substituted coordinated structure is represented as 'CN(C)/C=P([W]([C-]#[O+])([C-]#[O+])([C-]#[O+])([C-]#[O+])[C-]#[O+])/[Si](C)(C)C', again with P coordinated to the pentacarbonyltungsten center. Credit is also awarded if the second intermediate is drawn with the other double bond coordinated to the metal." + ], + "rubric_images": [ + { + "file": "Exam-01-T9-4-1-rubric.png", + "description": "The figure shows six structures in red, labeled A_1 through A_6. **A_1**, at the far left of the first row, is a zwitterionic complex represented as 'CN(C)/C(N(C)C)=[PH+]/C(C1CC1)(O[Si](C)(C)C)[W-]([C-]#[O+])([C-]#[O+])([C-]#[O+])([C-]#[O+])[C-]#[O+]'; the negatively charged W center bearing five carbonyl ligands is bonded directly to the ligand carbon atom. **A_2**, second from the left in the first row, is an $\\eta^2$-coordinated complex represented as 'CN(C)C(N(C)C)=P=C(C2CC2)O[Si](C)(C)C.[W]([C-]#[O+])([C-]#[O+])([C-]#[O+])([C-]#[O+])[C-]#[O+]'; its $P=C$ double bond coordinates through a $\\pi$ bond to the pentacarbonyltungsten fragment. **A_3**, second from the right in the first row, is a $\\sigma$-coordinated complex represented as 'CN(C)C(N(C)C)/P([W]([C-]#[O+])([C-]#[O+])([C-]#[O+])([C-]#[O+])[C-]#[O+])=C(C3CC3)/O[Si](C)(C)C'; the P atom of the phosphaalkene ligand coordinates to the pentacarbonyltungsten center. **A_4**, at the far right of the first row, is likewise a $\\sigma$-coordinated complex bearing a cyclopropylidene substituent and is represented as 'CN(C)/C=P([W]([C-]#[O+])([C-]#[O+])([C-]#[O+])([C-]#[O+])[C-]#[O+])/C(O[Si](C)(C)C)=C4CC\\4'. **A_5**, on the left of the second row, is free cyclopropylketene represented as 'O=C=C5CC5'. **A_6**, on the right of the second row, is also a $\\sigma$-coordinated complex; its silyl-substituted coordinated structure is represented as 'CN(C)/C=P([W]([C-]#[O+])([C-]#[O+])([C-]#[O+])([C-]#[O+])[C-]#[O+])/[Si](C)(C)C', again with P coordinated to the pentacarbonyltungsten center." + } + ] + }, + { + "sub_id": "Exam-01-T9-5", + "tags": [ + "2.2", + "2.3", + "1.4" + ], + "text_open": "9-2-2 When the phosphorus atom no longer bears a hydrogen atom, a hydrogen migration analogous to the process above cannot occur. In the following reaction, the two substrates undergo a process similar to that in the preceding subquestion, proceeding through B_{1} to B_{2}. Write the SMILES structure of one key electrically neutral intermediate in the conversion of B_{2} to the products.", + "text_images": [ + { + "file": "Exam-01-T9-5-1-prompt.png", + "description": "Description of the structures/reaction in the figure: 'CN(C)C(=P[Fe])N(C)C' + 'CCOC(=[M])C' -> 'CN(C)/C=C/C(=[M])OCC' + 'CCOC(C)=P([Fe])[M]. Here M denotes a metal. The ligands on Fe and M are omitted in this question.'" + } + ], + "full_mark": 3, + "rubric": [ + "SMILES structure of the key electrically neutral intermediate: '[Fe]P(=C(CC(OCC)=[M])N(C)C)=C(C1CC1)O[Si](C)(C)C.[M]' (3 points). The C=P double bond coordinates to metal M; award 2 points if coordination of the double bond to the metal is omitted." + ], + "rubric_images": [ + { + "file": "Exam-01-T9-5-1-rubric.png", + "description": "The figure shows a transition-metal complex in red whose core coordination system may be described by the SMILES '[Fe]P(=C(CC(OCC)=[M])N(C)C)=C(C1CC1)O[Si](C)(C)C.[M]'. In the detailed structure, the central phosphorus (P) atom is bonded directly to an iron fragment '[Fe]'. The double-bonded carbon on the left of P bears a dimethylamino group ('-N(CH_3)_2') and a side chain containing an ethoxy group ('-OEt') that forms a double bond to a metal fragment '[M]', producing a Fischer carbene structure ($C=[M]$). The double-bonded carbon on the right of P bears cyclopropyl and siloxy ('-OTMS') groups, and the right-hand $P=C$ double bond acts as a $\\pi$ donor and coordinates in an $\\eta^2$ mode, indicated by a coordination arrow, to another independent metal fragment '[M]'." + } + ] + } + ] +} diff --git a/dev/tasks/Exam-02-T10/images/Exam-02-T10-1-1-prompt.png b/dev/tasks/Exam-02-T10/images/Exam-02-T10-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..33c48a89da8759f1d3f7bf6ca5904b6cc2595cc9 --- /dev/null +++ b/dev/tasks/Exam-02-T10/images/Exam-02-T10-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:9c4cd3a51bfe2e4e94bc1b6335e7d9f936ccb4219a6ac5514deaadfe826b497a +size 2977 diff --git a/dev/tasks/Exam-02-T10/images/Exam-02-T10-1-1-rubric.png b/dev/tasks/Exam-02-T10/images/Exam-02-T10-1-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..f3eaf7cf8b71e69cad41c134eee1ea29b8e25bf9 --- /dev/null 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0000000000000000000000000000000000000000..222202db42db9b7eca5dfba2a160ebaa3168da36 --- /dev/null +++ b/dev/tasks/Exam-02-T10/problem.json @@ -0,0 +1,175 @@ +{ + "problem_id": "Exam-02-T10", + "header": "Problem 10: Forever Unnamed", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-02-T10-1", + "tags": [ + "2.2" + ], + "text_open": "10-1 A neutral reaction intermediate can be obtained by the process shown below. Write its SMILES representation.", + "text_images": [ + { + "file": "Exam-02-T10-1-1-prompt.png", + "description": "A chemical reaction equation: the reactants on the left are [R1]CC(=O)Cl and R_2NC, joined by a plus sign `+`; a horizontal right-pointing arrow in the center leads to bold X on the right." + } + ], + "full_mark": 2, + "rubric": [ + "Write the SMILES of X shown in the rubric image: '[R2]/N=C(C(C[R1])=O)\\Cl' (2 points)." + ], + "rubric_images": [ + { + "file": "Exam-02-T10-1-1-rubric.png", + "description": "The rubric image shows X, with SMILES '[R2]/N=C(C(C[R1])=O)\\Cl'." + } + ] + }, + { + "sub_id": "Exam-02-T10-2", + "tags": [ + "2.4", + "2.2", + "1.2" + ], + "text_open": "10-2-1 Further reactions can proceed from intermediates of type X. Write the SMILES representations of products A and B in the reactions below, given that B contains a five-membered aromatic ring. 10-2-2 Label the absolute configurations of all chiral carbon atoms in the structure of A from the preceding question.", + "text_images": [ + { + "file": "Exam-02-T10-2-1-prompt.png", + "description": "The image shows two parallel reactions. In the upper reaction, the starting material is 'CC1=C(O[Si](C)(C)C(C)(C)C)CC[C@@H]1C[N+]#[C-]'. Above the arrow is “(a)tBuCOCl,DCM,r.t.” and below it “(b)AgBF_4,DCM,-78∘C”; the arrow leads to product A. In the lower reaction, '[R]C(Cl)=O' + '[Ar]C[N+]#[C-]' reacts under “base” to give product B." + } + ], + "full_mark": 4, + "rubric": [ + "Structure A is worth 1 point; its SMILES is 'O=C1[C@]2(C)[C@](CN=C2C(C(C)(C)C)=O)([H])CC1'.", + "The chiral carbon bonded to H has the S configuration (1 point), and the chiral carbon bonded to the methyl group has the R configuration (1 point).", + "Structure B is worth 1 point; its SMILES is '[R]C1=CN=C(O1)[Ar]'." + ], + "rubric_images": [ + { + "file": "Exam-02-T10-2-1-rubric.png", + "description": "The rubric image shows the structure and configurations of A, with SMILES 'O=C1[C@]2(C)[C@](CN=C2C(C(C)(C)C)=O)([H])CC1'. The H-bearing chiral carbon is S, and the methyl-bearing chiral carbon is R." + }, + { + "file": "Exam-02-T10-2-2-rubric.png", + "description": "The rubric image shows B, with SMILES '[R]C1=CN=C(O1)[Ar]'." + } + ] + }, + { + "sub_id": "Exam-02-T10-3", + "tags": [ + "2.2", + "2.3" + ], + "text_open": "10-2-3 Write the SMILES representation of another electrically neutral key intermediate Y, other than the type-X intermediate, involved in formation of B.", + "text_images": [], + "full_mark": 2, + "rubric": [ + "Write the SMILES of Y shown in the rubric image: '[R]C(C#[N+][CH-][Ar])=O' (2 points). No credit if multiple structures are given." + ], + "rubric_images": [ + { + "file": "Exam-02-T10-3-1-rubric.png", + "description": "The rubric image shows Y, with SMILES '[R]C(C#[N+][CH-][Ar])=O'." + } + ] + }, + { + "sub_id": "Exam-02-T10-4", + "tags": [ + "2.4", + "2.3", + "2.7" + ], + "text_open": "10-2-4 If an appropriate amount of CH_{2}=CHCO_{2}Me is added to the reaction system that forms B, Y is trapped to form two products, C_{1} and C_{2}. Write their SMILES representations.", + "text_images": [], + "full_mark": 4, + "rubric": [ + "Write the SMILES representations of C_{1} and C_{2} shown in the rubric image; 2 points each.", + "C1 can be represented as 'O=C(C1=CC(C(N1)[Ar])C(OC)=O)[R]'.", + "C2 can be represented as 'O=C(C1=C(CC(N1)[Ar])C(OC)=O)[R]'.", + "The position of the double bond within the ring is not specified; an aromatized pyrrole product also receives credit." + ], + "rubric_images": [ + { + "file": "Exam-02-T10-4-1-rubric.png", + "description": "The rubric image shows C1, which can be represented as 'O=C(C1=CC(C(N1)[Ar])C(OC)=O)[R]', and C2, which can be represented as 'O=C(C1=C(CC(N1)[Ar])C(OC)=O)[R]'." + } + ] + }, + { + "sub_id": "Exam-02-T10-5", + "tags": [ + "2.2" + ], + "text_open": "10-3-1 Addition of other reagents changes the reaction pathway. Write the small-molecule by-product formed in the reaction below and the SMILES representation of another electrically neutral key intermediate, other than the type-X intermediate, involved in the reaction.", + "text_images": [ + { + "file": "Exam-02-T10-5-1-prompt.png", + "description": "A continuous multistep reaction sequence. The substrate is `R_1COCl`; above the first arrow is R_2NC; above the second is `P(OR_3)_3`; and above the third is '[O-]/C([Ar])=C\\[N+]1=CC=CC=C1'. The final product is 'O=C([Ar])C1=C(C([R1])=C2C=CC=CN12)N[R2]'." + } + ], + "full_mark": 3, + "rubric": [ + "The small-molecule by-product may be written as P(O)(OR_3)_3, or as P(O)(OH)(OR_3)_2 and R_3Cl (1 point).", + "The key intermediate can be represented as '[R1]C(OP(O[R3])(O[R3])=O)=C=N[R2]' or '[R1]C(O[P+](O[R3])(O[R3])O[R3])=C=N[R2]' (2 points)." + ], + "rubric_images": [ + { + "file": "Exam-02-T10-5-1-rubric.png", + "description": "The rubric image gives two representations of the key intermediate in 10-3-1. The general SMILES can be '[R1]C(OP(O[R3])(O[R3])=O)=C=N[R2]' or '[R1]C(O[P+](O[R3])(O[R3])O[R3])=C=N[R2]'." + } + ] + }, + { + "sub_id": "Exam-02-T10-6", + "tags": [ + "2.4", + "2.2" + ], + "text_open": "10-3-2 In the presence of Et_{3}N, EtO_{2}CCOCl, CyNC, P(Oi-Pr)_{3}, NCCH_{2}CN, and CS_{2} undergo a multicomponent reaction in MeCN. Write the SMILES representation of product D.", + "text_images": [], + "full_mark": 3, + "rubric": [ + "Write the SMILES representation of D: 'N#C/C(C#N)=C1SC(C(OCC)=O)=C(S/1)NC2CCCCC2' (3 points)." + ], + "rubric_images": [ + { + "file": "Exam-02-T10-6-1-rubric.png", + "description": "The rubric image shows multicomponent-reaction product D, with SMILES 'N#C/C(C#N)=C1SC(C(OCC)=O)=C(S/1)NC2CCCCC2'." + } + ] + }, + { + "sub_id": "Exam-02-T10-7", + "tags": [ + "2.2", + "2.4" + ], + "text_open": "10-4 Under certain conditions, an α-carbonyl diazo compound can replace an acyl chloride in a similar reaction with an isocyanide. Write the SMILES representations of all intermediates and products in the scheme below. W and all three products contain a five-membered ring, but only the five-membered ring in E is aromatic; F also contains a six-membered ring.", + "text_images": [ + { + "file": "Exam-02-T10-7-1-prompt.png", + "description": "A synthetic reaction scheme: '[R1]C(C(C([R1])=O)=[N+]=[N-])=O' is the starting material and is converted by an arrow to Z. Intermediate Z then reacts with `R_2NC` to form W. Intermediate **W** branches into three transformations: the upper branch reacts with `MeNHNH_2` to form E; the middle branch reacts with '[R1]C(C([R1])=O)=C=O' to form F; and the lower branch forms H under H^+ conditions." + } + ], + "full_mark": 10, + "rubric": [ + "Write the SMILES representations of the five structures Z, W, E, F, and H in the rubric image; 2 points each, 10 points total.", + "SMILES of **Z**: '[R1]C(C([R1])=C=O)=O'", + "SMILES of **W**: 'O=C1C([R1])=C(O/C1=N/[R2])[R1]'", + "SMILES of **E**: 'CN2N=C(C([R1])=C2[R1])C(N[R2])=O'", + "SMILES of **F**: 'O=C3C([R1])=C(OC34N(C(C([R1])=C(O4)[R1])=O)[R2])[R1]'", + "SMILES of **H**: 'O=C5C(N(C([R1])=C5[R1])[R2])=O'" + ], + "rubric_images": [ + { + "file": "Exam-02-T10-7-1-rubric.png", + "description": "The image shows five red intermediate and product structures from the transformations above, labeled Z, W, E, F, and H. **Z** is an acylketene intermediate with cumulated double bonds, with SMILES '[R1]C(C([R1])=C=O)=O'. **W** is a five-membered heterocyclic iminofuranone intermediate, with SMILES 'O=C1C([R1])=C(O/C1=N/[R2])[R1]'. **E** is the pyrazole carboxamide product formed with methylhydrazine, with SMILES 'CN2N=C(C([R1])=C2[R1])C(N[R2])=O'. **F** is a complex spirocyclic lactone derivative formed by condensation, with SMILES 'O=C3C([R1])=C(OC34N(C(C([R1])=C(O4)[R1])=O)[R2])[R1]'. **H** is a pyrroledione (maleimide-type) product formed by isomerization under acidic conditions, with SMILES 'O=C5C(N(C([R1])=C5[R1])[R2])=O'." + } + ] + } + ] +} diff --git a/dev/tasks/Exam-03-T8/images/Exam-03-T8-1-1-prompt.png b/dev/tasks/Exam-03-T8/images/Exam-03-T8-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..2793555185378075f36987c16e88c30deaff0f2e --- /dev/null +++ b/dev/tasks/Exam-03-T8/images/Exam-03-T8-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:84de1a10221ea3f45069cef05c710d01588d0dc231fb898274f387918f09e2d1 +size 31140 diff --git a/dev/tasks/Exam-03-T8/images/Exam-03-T8-2-1-prompt.png b/dev/tasks/Exam-03-T8/images/Exam-03-T8-2-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..1d65c6ce1d276a66cd2471fbffd75ba2b27bec4d --- /dev/null +++ b/dev/tasks/Exam-03-T8/images/Exam-03-T8-2-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:72d1fe93f80181f585f8d66cd99f8963cf986951cb82c67408d4e7977a19caa4 +size 12380 diff --git a/dev/tasks/Exam-03-T8/images/Exam-03-T8-2-1-rubric.png b/dev/tasks/Exam-03-T8/images/Exam-03-T8-2-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..c15d297a54f8f70102a0ef822c02a66dea90799c --- /dev/null +++ b/dev/tasks/Exam-03-T8/images/Exam-03-T8-2-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:370da9d3da5ae732419d1e6d3857e3e869f8eb7733e40b3a7b5bca92ad19542d +size 5781 diff --git a/dev/tasks/Exam-03-T8/images/Exam-03-T8-3-1-prompt.png b/dev/tasks/Exam-03-T8/images/Exam-03-T8-3-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..637e60414540dde977340f58eda303c57ad02271 --- /dev/null +++ b/dev/tasks/Exam-03-T8/images/Exam-03-T8-3-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:9ed2fb507d46d2294fe7f5c3b7ed80471df24a7b6d6a597cb08f87a11c16ca00 +size 63568 diff --git a/dev/tasks/Exam-03-T8/images/Exam-03-T8-4-1-rubric.png b/dev/tasks/Exam-03-T8/images/Exam-03-T8-4-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..9406e75d3b9ac16bd4ee727ec62a8d80290f4fda --- /dev/null +++ b/dev/tasks/Exam-03-T8/images/Exam-03-T8-4-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:34fc94b02d72deb84a71fc40cf44a2bcf419b1bc080a7ea882b4c49a6bf75d0e +size 14368 diff --git a/dev/tasks/Exam-03-T8/problem.json b/dev/tasks/Exam-03-T8/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..5d1335b54a76cabe00141049ae2f6521732cf745 --- /dev/null +++ b/dev/tasks/Exam-03-T8/problem.json @@ -0,0 +1,84 @@ +{ + "problem_id": "Exam-03-T8", + "header": "", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-03-T8-1", + "tags": [ + "2.7" + ], + "text_open": "8-3 The reaction shown below, 'OCCOCCOCCO' + 'ClCCOCCCl' → 'C1COCCOCCOCCOCCO1', is best conducted under basic conditions. Which base is preferred? ( ) (a) LiOH (b) NaOH (c) KOH", + "text_images": [ + { + "file": "Exam-03-T8-1-1-prompt.png", + "description": "The figure shows the reaction 'OCCOCCOCCO' + 'ClCCOCCCl' → 'C1COCCOCCOCCOCCO1' in THF (tetrahydrofuran) and water." + } + ], + "full_mark": 2, + "rubric": [ + "(b) (2 points)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-03-T8-2", + "tags": [ + "1.2" + ], + "text_open": "8-4 Describe the most stable conformation of the molecule shown below.", + "text_images": [ + { + "file": "Exam-03-T8-2-1-prompt.png", + "description": "The figure shows an organic compound with SMILES 'CN1CN(CN(C1)C)C'." + } + ], + "full_mark": 2, + "rubric": [ + "2 methyl groups occupy equatorial positions and 1 methyl group occupies an axial position. (2 points)" + ], + "rubric_images": [ + { + "file": "Exam-03-T8-2-1-rubric.png", + "description": "The figure shows the most stable conformation of the organic compound with SMILES 'CN1CN(C)CN(C)C1'. The six-membered ring adopts a chair conformation, with two methyl groups equatorial and one methyl group axial." + } + ] + }, + { + "sub_id": "Exam-03-T8-3", + "tags": [ + "2.2" + ], + "text_open": "8-5 Ethylene glycol is commonly condensed with carbonyl compounds to protect carbonyl groups. As shown below, '[R]C([H])=O' + 'SCCCS' → '[R]C1([H])SCCCS1'; '[R]C1([H])SCCCS1' + 'CCCCLi' → '[R]C1([Li])SCCCS1'. Thioacetals invert the polarity of the aldehydic carbon, and propane-1,3-dithiol is often used for this purpose. Briefly explain why ethane-1,2-dithiol, the sulfur analogue of ethylene glycol, is not used instead. (No more than 50 words.)", + "text_images": [ + { + "file": "Exam-03-T8-3-1-prompt.png", + "description": "The figure shows a two-step reaction. The first step is '[R]C([H])=O' + 'SCCCS' → '[R]C1([H])SCCCS1', in which the carbonyl carbon has δ^+ character. The second step is '[R]C1([H])SCCCS1' + 'CCCCLi' → '[R]C1([Li])SCCCS1', in which the carbon bonded to lithium has δ^- character; R denotes an alkyl group." + } + ], + "full_mark": 2, + "rubric": [ + "The lithium salt obtained from ethane-1,2-dithiol readily undergoes fragmentation. (2 points; a correct reaction equation is also acceptable.)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-03-T8-4", + "tags": [ + "2.4" + ], + "text_open": "8-6 Organic compound A (C_4H_6O_2) is oxidized by acidified potassium permanganate to give a single organic product B without gas evolution. B reacts with two equivalents of CH_3Li and, after workup, gives the major product C. C reacts with sodium metal in benzene and, after workup, gives D. Under sulfuric acid, D rearranges to E (C_6H_12O). Write the SMILES structures of A-E.", + "text_images": [], + "full_mark": 5, + "rubric": [ + "A is 'CC(=O)C(=O)C'; B is 'CC(=O)O'; C is 'CC(=O)C'; D is 'CC(C(C)(C)O)(O)C'; and E is 'CC(C)(C(C)=O)C'. (1 point each; 5 points total.)" + ], + "rubric_images": [ + { + "file": "Exam-03-T8-4-1-rubric.png", + "description": "The figure shows five compounds: A is 'CC(=O)C(=O)C'; B is 'CC(=O)O'; C is 'CC(=O)C'; D is 'CC(C(C)(C)O)(O)C'; and E is 'CC(C)(C(C)=O)C'." + } + ] + } + ] +} diff --git a/dev/tasks/Exam-03-T9/images/Exam-03-T9-1-1-rubric.png b/dev/tasks/Exam-03-T9/images/Exam-03-T9-1-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..f953030feb53acff6efa8b3a105cc28b12bc1656 --- /dev/null +++ b/dev/tasks/Exam-03-T9/images/Exam-03-T9-1-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:c5ab561c4a70ffcf131b6d4853cc03f426ff3823e4c6b47f133cbcbe5f4fd73e +size 781203 diff --git a/dev/tasks/Exam-03-T9/images/Exam-03-T9-3-1-prompt.png b/dev/tasks/Exam-03-T9/images/Exam-03-T9-3-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..447b397d84d6720f8c1599c45e36b551313fb031 --- /dev/null +++ b/dev/tasks/Exam-03-T9/images/Exam-03-T9-3-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:4cb1e9354c5ae39e321e9b61027e3404ed1c407c5390377835e2824a254553d8 +size 33021 diff --git a/dev/tasks/Exam-03-T9/images/Exam-03-T9-3-1-rubric.png b/dev/tasks/Exam-03-T9/images/Exam-03-T9-3-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..e993ea3118626665722fcec6a61d9d20e12b855d --- /dev/null +++ b/dev/tasks/Exam-03-T9/images/Exam-03-T9-3-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:00579c2f3f4aa62525928ee81814c8ec86ff015747ce1e7836c9158d1a50074f +size 783312 diff --git a/dev/tasks/Exam-03-T9/images/Exam-03-T9-header-1.png b/dev/tasks/Exam-03-T9/images/Exam-03-T9-header-1.png new file mode 100644 index 0000000000000000000000000000000000000000..55eab17ced5930edfdb567cd16c036646b4b068a --- /dev/null +++ b/dev/tasks/Exam-03-T9/images/Exam-03-T9-header-1.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:44c94c06ce99ff897b14bd67475f5d12e19b3efd2f3fe6e3bc27b84806dc68dc +size 7500 diff --git a/dev/tasks/Exam-03-T9/problem.json b/dev/tasks/Exam-03-T9/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..88dbb2f2a64856323b6347a94de4dff1a5dbebab --- /dev/null +++ b/dev/tasks/Exam-03-T9/problem.json @@ -0,0 +1,70 @@ +{ + "problem_id": "Exam-03-T9", + "header": "Problem 9 Hypervalent-Iodine-Induced Ring Contraction. In 2004, researchers at the University of Tokushima published a study of hypervalent-iodine-induced ring contraction.", + "header_images": [ + { + "file": "Exam-03-T9-header-1.png", + "description": "'OC1CCCNC1' reacts under PhIO and H_2O conditions." + } + ], + "subquestions": [ + { + "sub_id": "Exam-03-T9-1", + "tags": [ + "2.2", + "2.3", + "4.3" + ], + "text_open": "9-2 The reaction gives two amide products, A and B. Write the SMILES structures of the corresponding key intermediates; an intermediate common to both pathways need be written only once. Note: additional key intermediates may be included, but no incorrect structure may be given. The ^1H NMR data are as follows. A: ^1H NMR (CDCl_3): 6.68, 3.41, 2.31, 2.15; B: ^1H NMR (CDCl_3): 4.81, 3.58, 3.13, 2.41, 2.06.", + "text_images": [], + "full_mark": 9, + "rubric": [ + "Common key intermediates: 'OC1CCCN(I(O)C2=CC=CC=C2)C1' (1 point), 'C=NCCCC=O' (1 point), and 'O=CCCCNCO' (1 point). Formation of A additionally proceeds through 'NCCCC=O' (2 points) and 'OC1NCCC1' (2 points). Formation of B additionally proceeds through 'OC1CCCN1CO' (2 points)." + ], + "rubric_images": [ + { + "file": "Exam-03-T9-1-1-rubric.png", + "description": "The figure has 3 rows. The first row shows the common key intermediates: 'OC1CCCN(I(O)C2=CC=CC=C2)C1' (1 point), 'C=NCCCC=O' (1 point), and 'O=CCCCNCO' (1 point). The second row shows the additional key intermediates leading to A: 'NCCCC=O' (2 points) and 'OC1NCCC1' (2 points). The third row shows the additional key intermediate leading to B: 'OC1CCCN1CO' (2 points)." + } + ] + }, + { + "sub_id": "Exam-03-T9-2", + "tags": [ + "2.7" + ], + "text_open": "9-3 Identify the major and minor products among the two amides obtained in 9-2.", + "text_images": [], + "full_mark": 2, + "rubric": [ + "Major product: B; minor product: A. (2 points only if both are correct.)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-03-T9-3", + "tags": [ + "2.2", + "2.3", + "2.4" + ], + "text_open": "9-4 Based on the mechanistic information above, write the SMILES structures of the key intermediates and product in the reaction shown below. The first step is known to form a zwitterion (internal salt); the acetonitrile was not predried; and the product has more than one plane of symmetry. Note: additional key intermediates may be included, but no incorrect structure may be given.", + "text_images": [ + { + "file": "Exam-03-T9-3-1-prompt.png", + "description": "'O=C1C(O)=CC(C2=CC=CC=C21)=O' reacts first under PhI(OAc)_2 and DCM, and then under heat in CH_3CN." + } + ], + "full_mark": 5, + "rubric": [ + "'O=C1C([O-])=C([I+]C2=CC=CC=C2)C(C3=CC=CC=C31)=O' (1 point), 'O=C(C1=CC=CC=C1C([C]2)=O)C2=O' (1 point), 'O=C(C1=CC=CC=C1C2=O)C2=C=O' (1 point), 'O=C1C(C(O)=O)C(C2=CC=CC=C21)=O' (1 point), and 'O=C1CC(C2=CC=CC=C21)=O' (1 point). Other additional correct intermediates receive neither credit nor a penalty." + ], + "rubric_images": [ + { + "file": "Exam-03-T9-3-1-rubric.png", + "description": "The figure has 2 rows. The first row shows 3 key intermediates: 'O=C1C([O-])=C([I+]C2=CC=CC=C2)C(C3=CC=CC=C31)=O' (1 point), 'O=C1C2=CC=CC=C2C([C]C1=O)=O' (1 point), and 'O=C(C1=CC=CC=C1C2=O)C2=C=O' (1 point). The second row shows two key intermediates: 'O=C1C(C(O)=O)C(C2=CC=CC=C21)=O' (1 point) and 'O=C1CC(C2=CC=CC=C21)=O' (1 point)." + } + ] + } + ] +} diff --git a/dev/tasks/Exam-04-T10/images/Exam-04-T10-1-1-prompt.png b/dev/tasks/Exam-04-T10/images/Exam-04-T10-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..9708031c332c297660cef9f2955de841e14cc552 --- /dev/null +++ b/dev/tasks/Exam-04-T10/images/Exam-04-T10-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:c358c38c7dc791f75e30f3aa21a144c7c52bd31043179eaa34417e3a752a8331 +size 13049 diff --git a/dev/tasks/Exam-04-T10/images/Exam-04-T10-1-1-rubric.png b/dev/tasks/Exam-04-T10/images/Exam-04-T10-1-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..d3a9dd3af36eb0078bffc4e3e344e46f490f5bbf --- /dev/null +++ b/dev/tasks/Exam-04-T10/images/Exam-04-T10-1-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid 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The “second step” is conversion of the hydrazone intermediate to the final product under $H^+/H_2O$.\n10-1-2 Explain why the hydrazone is introduced, and give the SMILES of one key intermediate in the second step.", + "text_images": [ + { + "file": "Exam-04-T10-1-1-prompt.png", + "description": "Two-step reaction. The leftmost structure is '[*]C1C([*])CC[C@]2([H])[C@@H](Br)C(CC[C@@]21C)=O'. Each * is attached to a wavy line denoting an additional ring system that does not participate and need not be considered. Reaction with ethoxycarbonyl hydrazine 'NNC(OCC)=O' gives '[*]C1C(CC[C@@]2([C@H](/C(CC[C@@]21C)=N\\NC(OCC)=O)Br)[H])[*]', with the same meaning for * and the upper-right wavy line. H^+/H_2O then gives '[*]C1C(CCC2=CC(CC[C@@]21C)=O)[*]', again with the same meaning for *." + } + ], + "full_mark": 3, + "rubric": [ + "H and Br are cis and therefore cannot undergo direct elimination. (1 point)", + "Award 1 point for stating that the hydrazone promotes Br departure, stabilizes the carbocation at the carbonyl α position, or equivalent. Merely stating that the hydrazone causes polarity inversion at the α position receives no credit.", + "A key intermediate in the second step is '[*]C1C([*])CC[C@@]2(C=C(/N=N/C(OCC)=O)CC[C@@]21C)[H]'. Protonation of the N adjacent to the ester is also acceptable. (1 point)" + ], + "rubric_images": [ + { + "file": "Exam-04-T10-1-1-rubric.png", + "description": "Key intermediate in the second step, SMILES '[*]C1C([*])CC[C@@]2(C=C(/N=N/C(OCC)=O)CC[C@@]21C)[H]'. Each * is connected to a wavy line representing an additional ring system that does not participate and need not be considered." + } + ] + }, + { + "sub_id": "Exam-04-T10-2", + "tags": [ + "2.3" + ], + "text_open": "10-2 A hydrazone can be reduced under basic conditions to a C-C double bond. 10-2-1 A dianion is known to appear during the reaction. Which statements are correct? (a) The first proton removed is on N. (b) The first proton removed is on C. (c) The reaction rate is controlled by departure of Ts^-. (d) The reaction rate is controlled by the second deprotonation. (e) It cannot be determined.", + "text_images": [ + { + "file": "Exam-04-T10-2-1-prompt.png", + "description": "Reaction of hydrazone '[R1]/C(C[R2])=N/NS(C1=CC=C(C)C=C1)(=O)=O' under 1) Base and 2) TMSCl gives the C-C double-bond product '[R1]/C([Si](C)(C)C)=C/[R2]'." + } + ], + "full_mark": 1, + "rubric": [ + "(a), (d). (1 point) Other answers receive no credit." + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-04-T10-3", + "tags": [ + "2.7", + "2.8" + ], + "text_open": "Consider the reaction shown; Quench means quenching. With potassium carbonate as base, product A forms. With LDA, A and a small amount of B form. The substituent R also affects the outcome. 10-2-2 With LDA, arrange the following R groups in decreasing B yield/A yield: (a) R = H; (b) R = Me; (c) R = F.", + "text_images": [ + { + "file": "Exam-04-T10-3-1-prompt.png", + "description": "Multiple pathways from common substrate 'O=C(CC(C)(C)C1)C([R])C1=O'. Downward reaction with TsNHNH_2, 1 eq, gives bracketed intermediate M. M gives A, 'O=C1CC(C)(C)CC=C1[R]', under 1) K_2CO_3, heat; 2) Quench. Under 1) LDA; 2) Quench, M gives both A and B, where B is 'O=C1CC(C)(C)C=CC1[R]'. SMILES/structure strings shown: 'O=C1CC(C)(C)CC=C1[R]'." + } + ], + "full_mark": 1, + "rubric": [ + "(c)(b)(a) or (c)(a)(b). (1 point) Other answers receive no credit." + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-04-T10-4", + "tags": [ + "2.2", + "2.4" + ], + "text_open": "10-3 Similarly, the N-N bond of a hydrazone can be converted into a diazo group, which undergoes many interesting reactions. 10-3-1 Give the SMILES of intermediate C.", + "text_images": [ + { + "file": "Exam-04-T10-4-1-prompt.png", + "description": "Two-step reaction. The left substrate is 'BrC1=C(OC)C(OC)=C(C=NNS(C2=CC=C(C)C=C2)(=O)=O)C(COC)=C1'. It reacts with 'BrC1=C(OC)C=CC(B(O)O)=C1' under K_2CO_3, 1,4-dioxane, 110 ℃ to give unknown C. C then gives 'BrC1=C(OC)C(OC)=C(CC2=CC=C(OC)C(Br)=C2)C(CO)=C1' under BBr_3, CH_2Cl, then THF/H_2O." + } + ], + "full_mark": 2, + "rubric": [ + "C: 'BrC1=C(C(OC)=C(C(COC)=C1)CC2=CC=C(C(Br)=C2)OC)OC'. (2 points) Writing only 'BrC1=C(OC)C(OC)=C(C(B(O)O)C2=CC=C(OC)C(Br)=C2)C(COC)=C1' earns 1 point. A six-membered chelate between B and -OMe also earns 1 point." + ], + "rubric_images": [ + { + "file": "Exam-04-T10-4-1-rubric.png", + "description": "1-point answer: 'BrC1=C(OC)C(OC)=C(C(B(O)O)C2=CC=C(OC)C(Br)=C2)C(COC)=C1'." + }, + { + "file": "Exam-04-T10-4-2-rubric.png", + "description": "2-point answer: 'BrC1=C(C(OC)=C(C(COC)=C1)CC2=CC=C(C(Br)=C2)OC)OC'." + } + ] + }, + { + "sub_id": "Exam-04-T10-5", + "tags": [ + "1.2", + "2.4" + ], + "text_open": "10-3-2 Based on the reaction above, give the SMILES of products D, E, and F. Include stereochemistry.", + "text_images": [ + { + "file": "Exam-04-T10-5-1-prompt.png", + "description": "Reaction producing D. Substrate '[R2][C@@]([R3])([BPin])[C@](C/1)([R1])CC1=N\\NC2=C(C=C(C=C2C)C)C' gives unknown D under Cs_2CO_3, 3.0 equiv, in 1,4-dioxane at 100 ℃." + }, + { + "file": "Exam-04-T10-5-2-prompt.png", + "description": "Reaction producing E. 'CC1=CC=C(S(NN=C2[C@@H](CCN=[N+]=[N-])CCCC2)(=O)=O)C=C1' plus '[R1]B(O)O' gives unknown E under K_2CO_3, 3.0 equiv, in 1,4-dioxane at 150 ℃." + }, + { + "file": "Exam-04-T10-5-3-prompt.png", + "description": "Reaction producing F. 'CC1=CC=C(S(NN=C2CCN(CC#N)CC2)(=O)=O)C=C1' plus '[R2]/C=C/B(O)O' gives unknown F under K_2CO_3, 2.0 equiv, in 1,4-dioxane with MW at 120 ℃." + } + ], + "full_mark": 6, + "rubric": [ + "D: '[BPin][C@@]12[C@]([R3])([R2])[C@@](C2)([R1])C1'. (2 points)", + "E: '[R1][C@@]12CCCC[C@]1([H])CCN2'. (2 points)", + "F: '[R2]/C=C/[C@@]12CC[N@](CC2=O)CC1'. (2 points)" + ], + "rubric_images": [ + { + "file": "Exam-04-T10-5-1-rubric.png", + "description": "SMILES '[BPin][C@@]12[C@]([R3])([R2])[C@@](C2)([R1])C1'." + }, + { + "file": "Exam-04-T10-5-2-rubric.png", + "description": "SMILES '[R1][C@@]12CCCC[C@]1([H])CCN2'." + }, + { + "file": "Exam-04-T10-5-3-rubric.png", + "description": "SMILES '[R2]/C=C/[C@@]12CC[N@](CC2=O)CC1'." + } + ] + }, + { + "sub_id": "Exam-04-T10-6", + "tags": [ + "2.2" + ], + "text_open": "10-4 Under suitable conditions, a hydrazone can generate a highly reactive carbene. When first proposed, the reaction shown was called an \"unusual carbon-skeleton rearrangement\". After spectroscopic structure determination and isotope labeling, a mechanism was proposed that includes an intramolecular ring closure and a \"questionable\" addition to an sp-hybridized carbon. 10-4-1 Give the SMILES of 2 neutral key intermediates.", + "text_images": [ + { + "file": "Exam-04-T10-6-1-prompt.png", + "description": "One-step reaction. Reactant 'CC1=CC=C(C=C1)S(N/N=C(C2C(/C3=N/NS(C4=CC=C(C=C4)C)(=O)=O)C2)\\C5C3C5)(=O)=O' under excess NaH in diglyme at 190 ℃ gives 'C12=CC=NN1C=CC3C2C3'." + } + ], + "full_mark": 4, + "rubric": [ + "The two neutral key intermediates are 'C=C[C-]([N+]#N)C(C1)C1C#C' and 'C#CC1C(C1)C2=CCN=N2'. (2 points each, 4 total.) A reasonable resonance form of the diazo group also receives credit; other answers do not." + ], + "rubric_images": [ + { + "file": "Exam-04-T10-6-1-rubric.png", + "description": "Left intermediate: 'C=C[C-]([N+]#N)C(C1)C1C#C'. Right intermediate: 'C#CC1C(C1)C2=CCN=N2'." + } + ] + }, + { + "sub_id": "Exam-04-T10-7", + "tags": [ + "2.3", + "3.7" + ], + "text_open": "10-4-2 When the two three-membered rings of the reactant are on the same side, the yield is 60%; when on opposite sides, it is 40%. Explain this in terms of activation energy.", + "text_images": [], + "full_mark": 1, + "rubric": [ + "When on the same side, the two three-membered rings repel, raising the reactant energy and lowering the activation energy, thereby favoring ring opening. (1 point)" + ], + "rubric_images": [] + } + ] +} diff --git a/dev/tasks/Exam-05-T5/images/Exam-05-T5-1-1-prompt.png b/dev/tasks/Exam-05-T5/images/Exam-05-T5-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..7da96ee330da9323e80b557572bc4014de15bd00 --- /dev/null +++ b/dev/tasks/Exam-05-T5/images/Exam-05-T5-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:00284e70809c430ecd774717b13e8a7c683275f126d6fc6c54aea0c2d769c442 +size 8310 diff --git a/dev/tasks/Exam-05-T5/images/Exam-05-T5-1-2-prompt.png b/dev/tasks/Exam-05-T5/images/Exam-05-T5-1-2-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..828b0b83587362a801f1e68de33fc571a6d3c4de --- /dev/null +++ b/dev/tasks/Exam-05-T5/images/Exam-05-T5-1-2-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:2a622da3fa9618fe2abefcb6c6f7b7b1b785bb99ce945eaef67b21dcce513a75 +size 65753 diff --git a/dev/tasks/Exam-05-T5/images/Exam-05-T5-4-1-prompt.png b/dev/tasks/Exam-05-T5/images/Exam-05-T5-4-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..4b65051753d820615e959e30f916458b34d9cd63 --- /dev/null +++ b/dev/tasks/Exam-05-T5/images/Exam-05-T5-4-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:d66f6f1a43192a8a25addf43af0334d01beea14e8fb3212cc9a51a4e5668f76b +size 20073 diff --git a/dev/tasks/Exam-05-T5/images/Exam-05-T5-4-2-prompt.png b/dev/tasks/Exam-05-T5/images/Exam-05-T5-4-2-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..208609fe98b2b57785024b30154917e1d9c80516 --- /dev/null +++ b/dev/tasks/Exam-05-T5/images/Exam-05-T5-4-2-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:219a387fac883f65061b98411d28f60918424bb0a32b9274c378399bf72cd946 +size 19450 diff --git a/dev/tasks/Exam-05-T5/images/Exam-05-T5-5-1-prompt.png b/dev/tasks/Exam-05-T5/images/Exam-05-T5-5-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..e9e143835ac9413d3c3c0bf306e17dc1b9846d14 --- /dev/null +++ b/dev/tasks/Exam-05-T5/images/Exam-05-T5-5-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:b6aa954d99ba65a3d8add92aa6c8ab7b4e6d9f9e325c4348ec236fc81cd1a4dd +size 2444 diff --git a/dev/tasks/Exam-05-T5/problem.json b/dev/tasks/Exam-05-T5/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..9f4cc22c9bca2dc249268c936cbbb35a7fa4b464 --- /dev/null +++ b/dev/tasks/Exam-05-T5/problem.json @@ -0,0 +1,135 @@ +{ + "problem_id": "Exam-05-T5", + "header": "Problem 5 Structural Chemistry of Metal M. Formulas must use M and should be in approximately simplest ratios.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-05-T5-1", + "tags": [ + "1.3" + ], + "text_open": "5-1 The crystal structure of chloride A of M and its projection along c are shown; large spheres are M and small spheres Cl. 5-1-1 Name the coordination polyhedron formed by Cl around one M.", + "text_images": [ + { + "file": "Exam-05-T5-1-1-prompt.png", + "description": "Perspective view of crystal A inside a gray dashed hexagonal frame. Large dark-gray spheres are M and small light-gray spheres Cl. 2 M atoms at upper left and lower right share two bridging Cl atoms; each M also connects radially to 7 terminal Cl atoms. Around each M are three upper, three lower, and three equatorial Cl atoms. An axis triad has c upward, b lower right, and a left." + }, + { + "file": "Exam-05-T5-1-2-prompt.png", + "description": "Projection of A along c within a dashed parallelogram. Large spheres are M and small spheres Cl. Two M atoms, one at (1/3,2/3), share two bridging Cl atoms. Each M also has 4 radial terminal Cl atoms in projection, with three nearer and three farther Cl neighbors overall. The axis marker has a rightward, b upper-left, and c out of the page." + } + ], + "full_mark": 2, + "rubric": [ + "Tricapped trigonal prism. (2 points; equivalent wording is acceptable.)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-05-T5-2", + "tags": [ + "1.3" + ], + "text_open": "5-1-2 Give the number of distinct Cl environments and the coordination number of Cl.", + "text_images": [], + "full_mark": 2, + "rubric": [ + "One environment. (1 point) Coordination number 3. (1 point)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-05-T5-3", + "tags": [ + "1.3" + ], + "text_open": "5-1-3 Give the crystal system and formula of A.", + "text_images": [], + "full_mark": 2, + "rubric": [ + "Hexagonal; MCl_3. (1 point each)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-05-T5-4", + "tags": [ + "1.3" + ], + "text_open": "5-2 M forms complex compounds with Ni and O. Projections of a mixed oxide X along b and c are shown; large spheres are M, small spheres O, and medium spheres Ni. Give its formula.", + "text_images": [ + { + "file": "Exam-05-T5-4-1-prompt.png", + "description": "Crystallographic projection of the unit cell of composite oxide X viewed along b. The vertical a-axis and horizontal c-axis each range from 0 to 1. Large spheres are M, medium spheres Ni, and small spheres O. Approximate fractional projection coordinates (a,c) are: M, (1.00,0.25), (0.00,0.75), (0.50,0.25), (0.50,0.75); Ni, the four vertices (0.00,0.00), (1.00,0.00), (0.00,1.00), (1.00,1.00), the edge midpoints (1.00,0.50), (0.00,0.50), (0.50,0.00), (0.50,1.00), and the face center (0.50,0.50); O, approximately (0.76,0.04), (0.68,0.04), (0.42,0.25), (0.12,0.25), (0.76,0.50), (0.68,0.50), (0.30,0.50), (0.24,0.50), (0.88,0.75), (0.58,0.75), (0.30,0.96), and (0.24,0.96). The lower-left direction marker points a upward and c rightward." + }, + { + "file": "Exam-05-T5-4-2-prompt.png", + "description": "Crystallographic projection of the unit cell of composite oxide X viewed along c. The horizontal a-axis and vertical b-axis each range from 0 to 1. Large spheres are M, medium spheres Ni, and small spheres O. Approximate fractional projection coordinates (a,b) are: M, upper-left (0.02,0.90), lower-right (0.98,0.10), and the two vertically overlapping central sites (0.50,0.55) and (0.50,0.45); Ni, near the edge centers (0.00,0.47), (1.00,0.47), (0.53,1.00), and (0.53,0.00); O, next to boundary Ni sites at approximately (0.08,0.47), (0.92,0.47), (0.45,0.98), and (0.61,0.02), plus interior sites at (0.30,0.70), (0.78,0.78), (0.22,0.22), and (0.70,0.30). The lower-left direction marker points b upward and a rightward." + } + ], + "full_mark": 2, + "rubric": [ + "MNiO_3. (2 points; NiMO_3 is also acceptable.)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-05-T5-5", + "tags": [ + "1.3" + ], + "text_open": "5-3 Compound Y, prepared from an oxide of M, NiO, and B_2O_3, contains M, Ni, O, and B and has a primitive monoclinic lattice. It consists of M cations, Ni^2+, and a network borate built from spiro units sharing sp^3-hybridized B atoms. The network contains [B_3O_3] six-membered and [B_4O_4] 8-membered rings. In every [B_4O_4], all four B atoms are four-coordinate and shared with neighboring [B_3O_3] rings; two ring O atoms coordinate one metal each and two coordinate two metals each. In every [B_3O_3], two B atoms are four-coordinate and shared with adjacent rings, while one is three-coordinate. The three ring O atoms each coordinate one metal; half the terminal O atoms coordinate two metals and half three. A monoclinic unit cell has [B_4O_4] at the corners and body center, each connected to four of eight neighboring [B_4O_4] rings through two [B_3O_3] rings. 5-3-1 Determine the number of B atoms per unit cell.", + "text_images": [ + { + "file": "Exam-05-T5-5-1-prompt.png", + "description": "Schematic of two [B_4O_4] rings connected through two vertex-sharing [B_3O_3] rings, topologically like 'C1CCCC2(CCCC3(CC4(CCCCCCC4)CCC3)C2)CCC1'." + } + ], + "full_mark": 2, + "rubric": [ + "20. (2 points)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-05-T5-6", + "tags": [ + "1.3" + ], + "text_open": "5-3-2 Give the simplest formula of the borate anion.", + "text_images": [], + "full_mark": 2, + "rubric": [ + "BO_2^-. (2 points)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-05-T5-7", + "tags": [ + "1.4", + "2.1" + ], + "text_open": "5-3-3 M is 9-coordinate, Ni^2+ five-coordinate, and M:Ni = 1:1. Give the formula of Y.", + "text_images": [], + "full_mark": 3, + "rubric": [ + "MNiB_5O_10. (3 points; NiMB_5O_10, MNi(BO_2)_5, or NiM(BO_2)_5 is also acceptable.)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-05-T5-8", + "tags": [ + "1.4" + ], + "text_open": "5-3-4 State the oxidation state of M.", + "text_images": [], + "full_mark": 1, + "rubric": [ + "+3 (1 point)" + ], + "rubric_images": [] + } + ] +} diff --git a/dev/tasks/Exam-06-T2/images/Exam-06-T2-1-1-prompt.png b/dev/tasks/Exam-06-T2/images/Exam-06-T2-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..f7702fe8005214469d918fe31b7e5a60c143ed9c --- /dev/null +++ b/dev/tasks/Exam-06-T2/images/Exam-06-T2-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:b33d6f2fd20ace4634938a909beffc77a38c8f954be30d49ed7eaa41c3365fe4 +size 6717 diff --git a/dev/tasks/Exam-06-T2/images/Exam-06-T2-1-1-rubric.png b/dev/tasks/Exam-06-T2/images/Exam-06-T2-1-1-rubric.png new file mode 100644 index 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A compass is a simple yet powerful device whose most important component is a freely rotating magnetized needle that senses Earth's magnetic field. At the molecular level, this mechanical function can be mimicked through noncovalent interactions and dynamic behavior among groups. One molecular compass exploits the redox behavior of pillar[5]quinone (P5Q) and incorporates guest molecules bearing polar groups. The dipole moments of these polar groups act as “needles” that respond to changes in the electric field and reorient during redox processes, thereby regulating electronic communication between remote quinone units.", + "header_images": [ + { + "file": "Exam-06-T2-header-1.png", + "description": "A macroscopic compass." + } + ], + "subquestions": [ + { + "sub_id": "Exam-06-T2-1", + "tags": [ + "2.4", + "3.5" + ], + "text_open": "2.1 Electrochemical behavior of pillar[5]quinone and synthesis of a molecular compass. The p-benzoquinone monomer Q undergoes two successive one-electron reductions in an aprotic solution. Linking 5 quinone units into a ring through methylene groups gives pillar[5]quinone (P5Q), which exhibits more complex electrochemical behavior. 2.1.1 Write the SMILES representations of the most stable resonance structures of Q^{·-} and Q^{2-}, the products of the two successive one-electron reductions of monomer Q.", + "text_images": [ + { + "file": "Exam-06-T2-1-1-prompt.png", + "description": "Schematic two-step one-electron reduction of p-benzoquinone Q (SMILES: 'O=C(C=C1)C=CC1=O'). Q is first reversibly reduced by one electron to Q^{·-}, which is then reversibly reduced by another electron to Q^{2-}." + }, + { + "file": "Exam-06-T2-1-2-prompt.png", + "description": "SMILES of P5Q: 'C%10C1=CC(=O)C(=CC1=O)CC2=CC(=O)C(=CC2=O)CC3=CC(=O)C(=CC3=O)CC4=CC(=O)C(=CC4=O)CC5=CC(=O)C(=CC5=O)%10'" + } + ], + "full_mark": 2, + "rubric": [ + "Q^{·-}: '[O-]C1=CC=C([O])C=C1'; Q^{2-}: '[O-]C2=CC=C([O-])C=C2'.", + "2 points total, 1 point for each structure." + ], + "rubric_images": [ + { + "file": "Exam-06-T2-1-1-rubric.png", + "description": "Rubric image: Q^{·-} is drawn as the semiquinone radical anion '[O-]C1=CC=C([O])C=C1', and Q^{2-} as the dianionic resonance structure '[O-]C2=CC=C([O-])C=C2' with both oxygen atoms negatively charged." + } + ] + }, + { + "sub_id": "Exam-06-T2-2", + "tags": [ + "1.6" + ], + "text_open": "2.1.2 The redox sequence of quinone units in pillar[5]quinone is regulated by the spatial charge distribution. The reduction pattern of P5Q is “2-1-2,” as shown below. White circles represent quinone monomers, black circles represent quinone monomers after one-electron reduction, and each quinone unit undergoes only one-electron reduction. Draw the SMILES structures of P5Q-2 and P5Q-3.", + "text_images": [ + { + "file": "Exam-06-T2-2-1-prompt.png", + "description": "Schematic 2-1-2 reduction pattern of P5Q. A legend at the right identifies a white circle as Q, a quinone monomer, and a black circle as Q^{·-}. P5Q is simplified as a regular pentagon with one hollow white circle on each side. Two-electron reduction of P5Q gives P5Q-2; one-electron reduction of P5Q-2 gives P5Q-3; and two-electron reduction of P5Q-3 gives a product simplified as a regular pentagon with one solid black circle on each side." + } + ], + "full_mark": 4, + "rubric": [ + "SMILES of P5Q-2: 'O=C(C(CC1=CC([O-])=C(C=C1[O])CC2=CC(C(CC3=CC4=O)=CC2=O)=O)=CC5=O)C=C5CC(C=C6[O])=C([O-])C=C6CC4=CC3=O' (2 points)", + "SMILES of P5Q-3: 'O=C(C=C1CC(C=C2[O])=C(C=C2CC3=CC4=O)[O-])C(CC5=CC([O-])=C(CC6=C([O-])C=C(CC4=CC3=O)C([O])=C6)C=C5[O])=CC1=O' (2 points)" + ], + "rubric_images": [ + { + "file": "Exam-06-T2-2-1-rubric.png", + "description": "The rubric image shows the reduction positions in P5Q-2 and P5Q-3. In the left diagram, P5Q-2 has two nonadjacent solid circles at pentagon vertices. In the right diagram, P5Q-3 has three solid circles, while the remaining two hollow circles are nonadjacent." + } + ] + }, + { + "sub_id": "Exam-06-T2-3", + "tags": [ + "2.6", + "2.4" + ], + "text_open": "2.1.3 The synthetic route to a molecular compass Q-Rot-A containing amide “needles” is shown below. The molecular compass can be regarded as a supramolecule in which the amide “needle” is the guest molecule. Note the perspective relationships of the chemical bonds in the product. Write the SMILES structures of intermediates A and B.", + "text_images": [ + { + "file": "Exam-06-T2-3-1-prompt.png", + "description": "The molecule 'C%10C1=CC(OC)=C(C=C1OC)CC2=CC(OC)=C(C=C2OC)CC3=CC(OC)=C(C=C3OC)CC4=CC(OC)=C(C=C4OC)CC5=CC(OC)=C(C=C5OC)%10' reacts with 'NCCCCCCN' using 'O=C(Cl)C1=CC(C(F)(F)F)=CC(C(F)(F)F)=C1' in CHCl_3 from 0°C to r.t. to form A. A reacts with BBr_3 in CHCl_3 at r.t. to form B. B then reacts with NaIO_4 in Me_2CO/H_2O/CH_2Cl_2 at r.t. to give a supramolecule in which 'C%10C1=CC(=O)C(=CC1=O)CC2=CC(=O)C(=CC2=O)CC3=CC(=O)C(=CC3=O)CC4=CC(=O)C(=CC4=O)CC5=CC(=O)C(=CC5=O)%10' encircles 'O=C(NCCCCCCNC(=O)C1=CC(C(F)(F)F)=CC(C(F)(F)F)=C1)C2=CC(C(F)(F)F)=CC(C(F)(F)F)=C2'." + } + ], + "full_mark": 2, + "rubric": [ + "A: 'C%10C1=CC(OC)=C(C=C1OC)CC2=CC(OC)=C(C=C2OC)CC3=CC(OC)=C(C=C3OC)CC4=CC(OC)=C(C=C4OC)CC5=CC(OC)=C(C=C5OC)%10' encircles 'O=C(NCCCCCCNC(=O)C1=CC(C(F)(F)F)=CC(C(F)(F)F)=C1)C2=CC(C(F)(F)F)=CC(C(F)(F)F)=C2'; B: 'C%10C1=CC(O)=C(C=C1O)CC2=CC(O)=C(C=C2O)CC3=CC(O)=C(C=C3O)CC4=CC(O)=C(C=C4O)CC5=CC(O)=C(C=C5O)%10' encircles 'O=C(NCCCCCCNC(=O)C1=CC(C(F)(F)F)=CC(C(F)(F)F)=C1)C2=CC(C(F)(F)F)=CC(C(F)(F)F)=C2'.", + "2 points total, 1 point for each structure.", + "No credit if the methylene groups between the benzene rings are missing." + ], + "rubric_images": [ + { + "file": "Exam-06-T2-3-1-rubric.png", + "description": "The image shows two molecules. On the left, A consists of 'C%10C1=CC(OC)=C(C=C1OC)CC2=CC(OC)=C(C=C2OC)CC3=CC(OC)=C(C=C3OC)CC4=CC(OC)=C(C=C4OC)CC5=CC(OC)=C(C=C5OC)%10' encircling 'O=C(NCCCCCCNC(=O)C1=CC(C(F)(F)F)=CC(C(F)(F)F)=C1)C2=CC(C(F)(F)F)=CC(C(F)(F)F)=C2'. On the right, B consists of 'C%10C1=CC(O)=C(C=C1O)CC2=CC(O)=C(C=C2O)CC3=CC(O)=C(C=C3O)CC4=CC(O)=C(C=C4O)CC5=CC(O)=C(C=C5O)%10' encircling 'O=C(NCCCCCCNC(=O)C1=CC(C(F)(F)F)=CC(C(F)(F)F)=C1)C2=CC(C(F)(F)F)=CC(C(F)(F)F)=C2'." + } + ] + }, + { + "sub_id": "Exam-06-T2-4", + "tags": [ + "1.2", + "1.1" + ], + "text_open": "2.2 Structural characterization and chirality analysis. Owing to the asymmetry of the 2,5-substituents, P5Q-type structures theoretically have multiple atropisomers. The planar chirality of each unit is defined as shown. 2.2.1 State directly the theoretical number of atropisomers of a P5Q-type structure.", + "text_images": [ + { + "file": "Exam-06-T2-4-1-prompt.png", + "description": "The image explains planar chirality. A vertical dashed mirror plane divides the image into left and right halves containing two enantiomeric substituted-benzene fragments. In the left structure, the bond on the right side of the benzene ring is drawn thick to project out of the page toward the viewer. Two ethyl and two “-OR” groups are arranged in alternating para positions, with the two -OR groups directed upper right and lower left. An equals sign connects this structure to a circled-dot symbol ($\\odot$), indicating projection out of the page, and it is labeled $R_{\\text{p}}$. In the mirror-image structure on the right, the bond on the left side of the benzene ring projects out of the page and the substituent positions are mirror-related. An equals sign connects it to a circled-cross symbol ($\\otimes$), indicating projection into the page away from the viewer, and it is labeled $S_{\\text{p}}$." + } + ], + "full_mark": 2, + "rubric": [ + "8 (2 points; no credit for an incorrect answer)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-06-T2-5", + "tags": [ + "1.2", + "4.2" + ], + "text_open": "2.2.2 ^{1}H-NMR measurements show that all 5 sets of methylene hydrogen atoms in the P5Q frameworks of A and Q-Rot-A appear as singlets, indicating that the planar rings of both are apparently homodirectional: either (R_{p}, R_{p}, R_{p}, R_{p}, R_{p}) or (S_{p}, S_{p}, S_{p}, S_{p}, S_{p}). However, the reasons for their apparently homodirectional configurations differ. Explain each case in terms of the different barriers to planar-ring rotation. Note: NMR measurements have a finite timescale.", + "text_images": [], + "full_mark": 4, + "rubric": [ + "A: methyl groups hinder rotation of the planar rings, so the configuration is retained.", + "2 points; “rotation is hindered,” “cannot rotate,” or any equivalent expression receives full credit.", + "No credit for a thermodynamic explanation such as “methyl groups cause high steric hindrance, so the all-cis form has the least steric hindrance,” because it does not follow the hint.", + "Q-Rot-A: the barrier to planar-ring rotation is low, so all quinone units are equivalent on the NMR timescale.", + "2 points: 1 point for “low rotational barrier” or “rapid rotation,” and 1 point for “equivalent on the NMR timescale,” “the inversion time is shorter than the NMR timescale,” or an equivalent statement." + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-06-T2-6", + "tags": [ + "4.2", + "1.2" + ], + "text_open": "2.2.3 For B, ^{1}H-NMR data show shifts and splitting of the methylene hydrogen signals. A partial spectrum and the corresponding hydrogen assignments are shown below; numbers beneath brackets are peak integrals, and the bottom scale gives chemical shift. Which of the following configurations are consistent with the integral data? ( ). (a) (R_{p}, S_{p}, R_{p}, R_{p}, S_{p}); (b) (R_{p}, R_{p}, R_{p}, S_{p}, S_{p}); (c) (S_{p}, R_{p}, R_{p}, R_{p}, R_{p}); (d) (R_{p}, S_{p}, S_{p}, R_{p}, S_{p})", + "text_images": [ + { + "file": "Exam-06-T2-6-1-prompt.png", + "description": "This partial $^{1}$H-NMR spectrum of B shows three groups of peaks along a horizontal axis labeled 3.8, 3.4, and 3.0 from left to right. Braces above them are labeled “Group I,” “Group II,” and “Group III.” Under Group I are four peaks: the first and third are connected by dashed lines to a red label 4^1, and the second and fourth to a red label 4^2. A horizontal integral bracket beneath the four peaks is labeled 4.00. Group II contains one peak with a red label 4^3 above and an integral bracket labeled 2.13 below. Group III contains four peaks: the first and second are connected by dashed lines to a red label 4^4, and the third and fourth to a red label 4^5. An integral bracket beneath the four peaks is labeled 4.03." + }, + { + "file": "Exam-06-T2-6-2-prompt.png", + "description": "The image shows two molecular structures. On the left is 'CC1=C(O)C=C(C([H])([H])C2=C(O)C=C(C)C(O)=C2)C(O)=C1'. The left benzene ring is labeled “R_p” above and the right ring “S_p.” The two methylene hydrogens linking the rings are highlighted as blue “H” labels; a blue arrow labeled “Group I” points to the upper H and one labeled “Group III” to the lower H. On the right is 'CC3=C(O)C=C(C([H])([H])C4=CC(O)=C(C)C=C4O)C(O)=C3'. Blue arrows labeled “Group II” point to both the upper and lower blue H atoms of the central methylene group, and “(R_p, R_p) or (S_p, S_p)” appears below the structure." + } + ], + "full_mark": 3, + "rubric": [ + "(a)(d) (3 points; full credit only if both are selected, and no credit if any incorrect choice is included)" + ], + "rubric_images": [] + } + ] +} diff --git a/dev/tasks/Exam-07-T4/images/Exam-07-T4-1-1-rubric.png b/dev/tasks/Exam-07-T4/images/Exam-07-T4-1-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..8cf21334eeb8c5c7af3b4424390bffa99904c8ef --- /dev/null +++ b/dev/tasks/Exam-07-T4/images/Exam-07-T4-1-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:19647c49712aae1e631d63f20e572329bd234a8800cd9c60e67b700324f3859d +size 44095 diff --git a/dev/tasks/Exam-07-T4/images/Exam-07-T4-header-1.png b/dev/tasks/Exam-07-T4/images/Exam-07-T4-header-1.png new file mode 100644 index 0000000000000000000000000000000000000000..e81bb68ad190043e5e105c71ab27f504918d5028 --- /dev/null +++ b/dev/tasks/Exam-07-T4/images/Exam-07-T4-header-1.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:09665a863cbe84676312361df2404464844e8e1d25a0270b7c6007f9baebc079 +size 13193 diff --git a/dev/tasks/Exam-07-T4/problem.json b/dev/tasks/Exam-07-T4/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..40c5bf11bc9c29c24d2630b4664cb1e0be92fca1 --- /dev/null +++ b/dev/tasks/Exam-07-T4/problem.json @@ -0,0 +1,44 @@ +{ + "problem_id": "Exam-07-T4", + "header": "Determination of Sulfur Dioxide in Wine\nThe sulfur dioxide content of a wine sample is determined spectrophotometrically as follows. I. Construct a calibration curve. Accurately transfer 0.00 mL, 0.20 mL, 0.50 mL, 1.00 mL, 2.00 mL, and 3.00 mL of sulfur dioxide working standard solutions (containing 0.0 μg, 2.0 μg, 5.0 μg, 10.0 μg, 20.0 μg, and 30.0 μg sulfur dioxide, respectively) into 25 mL stoppered test tubes, and add formaldehyde-buffered absorbing solution to 10.00 mL. Then add, in sequence, 0.5 mL of 3 g/L ammonium sulfamate solution, 0.5 mL of 1.5 mol/L sodium hydroxide solution, and 1.0 mL of 0.5 g/L pararosaniline hydrochloride solution (structure shown below). Mix thoroughly and allow to stand for 20 min. Measure the absorbance of each standard solution at 579 nm using a UV–visible spectrophotometer. The following data are obtained:\n\n| $\\text{SO}_2$ content $m / \\mu\\text{g}$ | 0.0 | 2.0 | 5.0 | 10.0 | 20.0 | 30.0 |\n|---|---|---|---|---|---|---|\n| Absorbance $A$ | 0.003 | 0.062 | 0.159 | 0.312 | 0.618 | 0.928 |\n\nII. Accurately weigh 0.5000 g of the wine sample and add formaldehyde-buffered absorbing solution to 100 mL. After shaking for 2 h, take 10 mL of the solution and perform the spectrophotometric determination as in Step I. Repeat the measurement three times; the absorbances A’ are 0.637, 0.638, and 0.635. In China, the required sulfur dioxide content of alcoholic beverages is below $0.2500 \\ \\mathrm{g\\,dm^{-3}}$. Take the density of the wine to be $0.9 \\ \\mathrm{g\\,cm^{-3}}$. The structure of pararosaniline hydrochloride is 'NC1=CC=C(/C(C2=CC=C(N)C=C2)=C3C=CC(C=C\\3)=[NH2+])C=C1.[Cl-]'.", + "header_images": [ + { + "file": "Exam-07-T4-header-1.png", + "description": "Structural diagram of pararosaniline hydrochloride; the SMILES representation is 'NC1=CC=C(/C(C2=CC=C(N)C=C2)=C3C=CC(C=C\\3)=[NH2+])C=C1.[Cl-]'." + } + ], + "subquestions": [ + { + "sub_id": "Exam-07-T4-1", + "tags": [ + "2.4" + ], + "text_open": "4-1 The principle used in this titration is that A, formed by the addition reaction of sulfur dioxide with formaldehyde, forms a blue-violet complex with pararosaniline hydrochloride. Its absorbance is proportional to the sulfur dioxide concentration. Write the SMILES representation of A.", + "text_images": [], + "full_mark": 2, + "rubric": [ + "'[H]C(S(O)=O)=O' (2 points)" + ], + "rubric_images": [ + { + "file": "Exam-07-T4-1-1-rubric.png", + "description": "The SMILES representation is '[H]C(S(O)=O)=O'." + } + ] + }, + { + "sub_id": "Exam-07-T4-2", + "tags": [ + "4.1", + "2.8" + ], + "text_open": "4-2 Why must 0.5 mL of 1.5 mol/L sodium hydroxide solution be added to adjust the pH before measurement? (No more than 50 words.)", + "text_images": [], + "full_mark": 2, + "rubric": [ + "① This ensures complete reaction between HCHO and $SO_2$, thereby ensuring the accuracy of the spectrophotometric determination (1 point). ② The absorption wavelength of the conjugated structure of pararosaniline hydrochloride varies with acidity; failure to adjust the pH to an appropriate range would affect the accuracy of the spectrophotometric determination (1 point)." + ], + "rubric_images": [] + } + ] +} diff --git a/dev/tasks/Exam-09-T4/images/Exam-09-T4-1-1-prompt.png b/dev/tasks/Exam-09-T4/images/Exam-09-T4-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..b8b7e2983d02744e6c9349ce519cdc275a507bdc --- /dev/null +++ b/dev/tasks/Exam-09-T4/images/Exam-09-T4-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:a2e3ed4060119faeeb6cecba9cfd27be2cc69cc2eb050bc2c05690bee3d17551 +size 2927 diff --git a/dev/tasks/Exam-09-T4/images/Exam-09-T4-1-2-prompt.png b/dev/tasks/Exam-09-T4/images/Exam-09-T4-1-2-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..a6ef41ae4f3e5fffeac81b8535be98b2e4443723 --- /dev/null +++ b/dev/tasks/Exam-09-T4/images/Exam-09-T4-1-2-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:db7396d7d2f771563f499c6c43f3d074105091d64740238f0ea4096f529a9564 +size 3002 diff --git a/dev/tasks/Exam-09-T4/images/Exam-09-T4-1-3-prompt.png b/dev/tasks/Exam-09-T4/images/Exam-09-T4-1-3-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..406d74c5f4c9c9bee8d712303f4375a7f12b458d --- /dev/null +++ b/dev/tasks/Exam-09-T4/images/Exam-09-T4-1-3-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:0a5360f3b6d1dcdf3014e7bccccd7e4217cc920e53e11dd4b90e41d11843839c +size 3572 diff --git a/dev/tasks/Exam-09-T4/images/Exam-09-T4-1-4-prompt.png b/dev/tasks/Exam-09-T4/images/Exam-09-T4-1-4-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..160dada282e5c0811f7ca58904a00e50ed84aac7 --- /dev/null +++ b/dev/tasks/Exam-09-T4/images/Exam-09-T4-1-4-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:042cdfab09044f85d338475bb345de26fdb580726bacf892ccd58048ca62eeb3 +size 3845 diff --git a/dev/tasks/Exam-09-T4/images/Exam-09-T4-2-1-prompt.png b/dev/tasks/Exam-09-T4/images/Exam-09-T4-2-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..1134894d4177b5d064f2593234b2d0c26843e523 --- /dev/null +++ b/dev/tasks/Exam-09-T4/images/Exam-09-T4-2-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:d27ac108d57e9c762e662467b94b4873684ba70738bbdabbb14c9990f9a5b01c +size 74370 diff --git a/dev/tasks/Exam-09-T4/problem.json b/dev/tasks/Exam-09-T4/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..4bc32053701d3180dec096918b755d3a7ec79ad5 --- /dev/null +++ b/dev/tasks/Exam-09-T4/problem.json @@ -0,0 +1,57 @@ +{ + "problem_id": "Exam-09-T4", + "header": "Kjeldahl Nitrogen Determination\nDetermination of nitrogen content has long been an important topic in analytical chemistry. In 1883, the Danish chemist Johan G.C.T. Kjeldahl invented the Kjeldahl method. The method mainly comprises digestion, alkalization, and titration: sulfuric-acid digestion converts the nitrogen in a sample into an ammonium salt; alkalization releases ammonia by distillation; and the absorbed ammonia is quantified indirectly by acid-base titration. The organic nitrogen content of a solid sample is to be determined. Answer the following questions based on the procedures described.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-09-T4-1", + "tags": [ + "4.1", + "4.5" + ], + "text_open": "4-1-1 First operation: digestion. Weigh $5.000 \\ \\mathrm{g}$ of the sample into a dry $100 \\ \\mathrm{mL}$ Kjeldahl flask. Add $0.4 \\ \\mathrm{g}$ of copper sulfate, $6 \\ \\mathrm{g}$ of potassium sulfate, and $20 \\ \\mathrm{mL}$ of anhydrous sulfuric acid. Swirl gently, place a small funnel in the flask mouth, and set the flask at an angle of 45° on an asbestos gauze with a small opening. Heat slowly. When the contents turn black and foaming has completely ceased, increase the heat to keep the liquid gently boiling. After the liquid becomes blue-green, clear, and transparent, continue heating for another $1 \\ \\mathrm{h}$. Remove the Kjeldahl flask and cool it to room temperature. Carefully add $20 \\ \\mathrm{mL}$ of water, quantitatively transfer the contents to a $100 \\ \\mathrm{mL}$ volumetric flask, rinse the inner wall of the Kjeldahl flask with a small amount of water, and combine the rinsings in the volumetric flask. Dilute to volume with water, mix, and reserve for use. Conduct a blank experiment in parallel. From the apparatus shown below, select the Kjeldahl flask suitable for digestion. The choices are labeled (a), (b), (c), and (d).", + "text_images": [ + { + "file": "Exam-09-T4-1-1-prompt.png", + "description": "Image corresponding to option (a): a pear-shaped flask with a short neck and a ground-glass stopper." + }, + { + "file": "Exam-09-T4-1-2-prompt.png", + "description": "Image corresponding to option (b): a flask with a long neck, a pear-shaped bottom, and a ground-glass stopper." + }, + { + "file": "Exam-09-T4-1-3-prompt.png", + "description": "Image corresponding to option (c): an Erlenmeyer flask with a conical body, a flat bottom, and a ground-glass stopper." + }, + { + "file": "Exam-09-T4-1-4-prompt.png", + "description": "Image corresponding to option (d): a volumetric flask with a narrow neck, pear-shaped bottom, an annular calibration mark on the neck, and a ground-glass stopper." + } + ], + "full_mark": 1, + "rubric": [ + "(b) (1 point)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-09-T4-2", + "tags": [ + "4.1", + "4.5" + ], + "text_open": "4-2 Alkalization. Assemble the Kjeldahl distillation apparatus as shown below. Fill vessel 1 with water to 2/3 of its volume. Add several glass beads, several drops of methyl red, and several milliliters of sulfuric acid until the solution is red. Heat the water in the steam generator to boiling and maintain boiling. Add $10.0 \\ \\mathrm{mL}$ of $20 \\ \\mathrm{g/L}$ boric acid solution and 3 drops of mixed indicator A to the receiving flask, and immerse the lower end of apparatus 8 beneath the liquid surface in vessel 9. Using a pipette, accurately transfer $5.0 \\ \\mathrm{mL}$ of sample digest through apparatus 4 into reaction chamber 5. Rinse the small cup with $10 \\ \\mathrm{mL}$ of water and allow the rinsing to flow into reaction chamber 5, then tightly insert the rod-shaped glass stopper in apparatus 4. Slowly add $400 \\ \\mathrm{g/L}$ sodium hydroxide solution through apparatus 4 into reaction chamber 5 and immediately seal the system. After distilling for $15 \\ \\mathrm{min}$, move the distillate receiver so that the liquid surface is below the lower end of the condenser. Continue distilling for $1 \\ \\mathrm{min}$. Verify with pH paper that the distillate is neutral, rinse the outside of the lower end of the condenser with a small amount of water, and remove the receiving flask. Which of the following statements about the experimental operations and their purposes are correct?\n(a). Methyl red is added to apparatus 1 to generate acidic vapor that reacts with ammonia in apparatus 5 to form an ammonium salt.\n(b). After sodium hydroxide is added, the glass stopper must be inserted tightly and a water seal formed immediately to prevent loss of ammonia.\n(c). The principal purpose of apparatus 6 is to separate liquid water from the steam generated in apparatus 1.\n(d). After alkalization and collection for one digest have been completed, the next reaction may be started by opening screw clamp 7, closing screw clamp 3, and replacing the solution in vessel 9 with distilled water, in that order, to discharge the waste liquid and rinse the reaction chamber.", + "text_images": [ + { + "file": "Exam-09-T4-2-1-prompt.png", + "description": "This black-and-white line drawing shows an assembly of laboratory glassware. Three main sections on the left, center, and right are connected by tubing, and the numbers 1 through 9 identify specific components. Left section: number 1 points to the liquid inside a long-neck round-bottom flask, represented by a horizontal dashed line. The bottom of the flask rests on a square base. Number 2 points to this base, which has a circular knob at the center of its front face and four feet underneath. The mouth of the round-bottom flask is sealed with a stopper. A very long straight vertical glass tube passes through the center of the stopper, extending below the liquid surface inside the flask and far above the flask mouth. A short right-angle glass tube also passes through the stopper and is connected at its end to U-shaped flexible tubing. Center section: the U-shaped flexible tubing from the left connects to a narrow inlet above the central apparatus. The main central apparatus is inclined from lower left to upper right and contains an outer jacket and an inner structure. Number 4 points to an upward-facing vertical inlet at the upper-right end of the inclined apparatus; it has a stopper through which a short straight tube is inserted. Number 5 points to an elongated cavity or inner-tube structure inside the inclined apparatus. Number 6 points to the enlarged spherical lower-left end of the inclined apparatus. Number 7 points to a narrow tube extending from the spherical end below number 6, connected to a short length of flexible tubing fitted with a clamp (pinchcock). Number 3 points to another narrow tube extending from the upper-middle part of the outer jacket, likewise connected to a short length of flexible tubing fitted with a clamp (pinchcock). The upper-right end of the inclined apparatus connects to the right section through an inverted U-shaped bent glass tube. Right section: number 8 points to a vertically oriented straight condenser. The end of the inverted U-shaped bent glass tube enters the top of the condenser. Two horizontal side-arm connections extend to the right from the outer jacket of condenser 8. A right-pointing arrow beside the upper side arm indicates outward flow, and a left-pointing arrow beside the lower side arm indicates inward flow. Number 9 points to an Erlenmeyer flask directly beneath the condenser. The flask contains liquid, represented by a horizontal dashed line. The inner straight tube of condenser 8 extends downward through the mouth of the flask and below the liquid surface inside it." + } + ], + "full_mark": 2, + "rubric": [ + "(b)(c) (2 points; 1 point if 1 correct option is selected; no points for an incorrect selection or an omitted correct option)" + ], + "rubric_images": [] + } + ] +} diff --git a/dev/tasks/Exam-09-T7/images/Exam-09-T7-1-1-prompt.png b/dev/tasks/Exam-09-T7/images/Exam-09-T7-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..810ad77102b8a1b0fd2d51b36f1d6a39d7b1df1e --- /dev/null +++ b/dev/tasks/Exam-09-T7/images/Exam-09-T7-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:0a475ef004946a67b9c8e31d58504513382f56618e82d340ffc20e345861826a +size 995507 diff --git a/dev/tasks/Exam-09-T7/images/Exam-09-T7-1-1-rubric.png b/dev/tasks/Exam-09-T7/images/Exam-09-T7-1-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..0b1ebfde094062e3831913aaa2d9dadfb651e2d5 --- /dev/null +++ b/dev/tasks/Exam-09-T7/images/Exam-09-T7-1-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:eb2aeec29b73c01f8b760813342d3b46bf6d17bf574ec84e1d78c31ac7b1360c +size 1130015 diff --git a/dev/tasks/Exam-09-T7/problem.json b/dev/tasks/Exam-09-T7/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..c262574887219b59acb7882cf9654cdf03ba8ee6 --- /dev/null +++ b/dev/tasks/Exam-09-T7/problem.json @@ -0,0 +1,30 @@ +{ + "problem_id": "Exam-09-T7", + "header": "Semiconductors and Electrochemistry\nSemiconductors are materials whose electrical conductivity at room temperature lies between that of conductors and insulators. Common semiconductor materials include silicon, germanium, and gallium arsenide. Based on band theory and Fermi statistics, doping with group V or group III elements, or with other materials that can donate electrons (donors) or accept electrons (acceptors), can regulate the Fermi level and carrier concentration of a semiconductor and thereby its various properties. Semiconductor carriers generally comprise electrons and holes. In this problem, involving an n-type semiconductor in which electrons are the majority carriers, only electrons need be considered. Nonuniform doping can create a carrier-concentration gradient in a semiconductor. The resulting diffusion causes charge separation and a built-in electric field; the built-in field opposes diffusion until equilibrium is reached. This is fundamentally similar to the electrochemical picture in which a chemical-potential difference drives charge transfer, which in turn produces a potential difference that opposes further charge transfer, ultimately establishing an electrode potential at equilibrium. By comparing a concentration-gradient-doped semiconductor with the electromotive force of an electrochemical concentration cell, this problem explores the universality of diffusion-electric-field equilibrium.\n7.1 Built-in electric field and diffusion equilibrium in a semiconductor\nConsider an n-type semiconductor at temperature T = 300 K with electron concentration varying with position x as $n(x)=N_{d0}\\exp(-x/L)$. Here $N_{d0}=1\\times10^{18}\\ \\mathrm{cm}^{-3}$, $L=0.1\\ \\mu\\mathrm{m}$, $D_n=225\\ \\mathrm{cm}^2\\mathrm{/s}$, $e=1.6\\times10^{-19}\\ \\mathrm{C}$, and $k=1.38\\times10^{-23}\\ \\mathrm{J/K}$. These quantities are, respectively, the donor concentration at x = 0 (which, under the approximation used here, may be regarded as the electron concentration at x = 0), characteristic length, electron diffusion coefficient, elementary charge, and Boltzmann constant.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-09-T7-1", + "tags": [ + "3.6" + ], + "text_open": "7.2 Electrochemistry. 7.2.1 When an iron sheet is immersed in water, an electrical double layer forms near the iron electrode, as shown below. Briefly explain how the structure forms, and qualitatively sketch the absolute value of the potential as a function of distance from the metal surface, taking the potential infinitely far from the electrode to be zero.", + "text_images": [ + { + "file": "Exam-09-T7-1-1-prompt.png", + "description": "This two-dimensional diagram, titled \"Figure 9.4 Schematic Structure of the Electrical Double Layer,\" shows the spatial distribution of an electrical double layer from left to right. At the far-left edge is a vertical row of minus signs (short horizontal lines). Immediately to their right is a vertical solid line shaded with diagonal hatching on its left side; the words \"Electrode surface\" are set vertically to the left of this line. A short distance to the right is a parallel vertical dashed line bearing a neat vertical series of small circles, each containing a \"+\" sign. The narrow interval between the solid line and this middle dashed line is marked at the bottom by a horizontal upward brace, beneath which the spacing symbol $d$ and the label \"Compact layer\" appear. Extending rightward from the middle dashed line to another parallel dashed boundary at the far right is a wider rectangular region. Several similar small circles containing \"+\" signs are scattered irregularly within it. This wider interval is likewise marked at the bottom by a horizontal brace, beneath which the spacing symbol $\\delta$ and the label \"Diffuse layer\" appear." + } + ], + "full_mark": 5, + "rubric": [ + "Formation of the electrical double layer: highly polar water molecules attract iron ions in the lattice and hydrate them. Bonding of some iron ions to other iron atoms in the metal weakens, or the ions leave the metal and enter the water layer adjacent to the iron surface. The metal thus becomes negatively charged and the nearby solution positively charged, producing a local/built-in electric field that inhibits further dissolution of the metal. Dynamic equilibrium is reached when dissolution and deposition rates become equal, establishing a stable potential difference between the metal and solution. (3 points: 1 point for stating that iron ions leave the metal, making the metal negative and the nearby solution positive; 1 point for stating that the electric field inhibits further dissolution; and 1 point for stating that dynamic equilibrium is reached.)\nSketch: 1 point if the absolute potential decreases overall and approaches 0; 1 point for labeling the compact layer and diffuse layer and showing a faster decrease in the compact layer." + ], + "rubric_images": [ + { + "file": "Exam-09-T7-1-1-rubric.png", + "description": "This two-dimensional coordinate plot, titled \"Figure 9.5 Schematic Potential Profile of the Electrical Double Layer,\" shows a solid curve that starts high on the vertical axis, decreases monotonically with a concave-upward shape, and gradually approaches the horizontal axis. The left vertical axis is a solid line shaded with diagonal hatching on its left side to represent the \"Metal surface\"; above it is the label \"(negative) potential\" with an upward arrow. Beneath the horizontal axis extending to the right is the label \"Distance from the electrode surface\" with a rightward arrow. A vertical dashed line within the plot divides the horizontal space into two regions. Horizontal brackets below the axis label the shorter region adjacent to the vertical axis as $d$ and the following longer region as $\\delta$. Two parallel horizontal dashed lines extend rightward from the starting point of the curve on the vertical axis and from the intersection of the curve with the vertical dashed line. On the right side of the plot, a vertical indicator with arrows at both ends labels three height parameters: the total height from the upper horizontal dashed line to the horizontal axis is $\\Delta\\varphi$; the upper drop between the two horizontal dashed lines is $\\varphi_1$; and the lower drop from the middle horizontal dashed line to the horizontal axis is $\\varphi_2$." + } + ] + } + ] +} diff --git a/dev/tasks/Exam-09-T9/images/Exam-09-T9-1-1-prompt.png b/dev/tasks/Exam-09-T9/images/Exam-09-T9-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..9a4484d0335d0f19d5a6a5655e7f05a04ff1b57d --- /dev/null +++ b/dev/tasks/Exam-09-T9/images/Exam-09-T9-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:6aba5b45b270a669ead30996fac239369919b98c1a58813ae4777b1adb17bfe4 +size 9190 diff --git a/dev/tasks/Exam-09-T9/images/Exam-09-T9-1-1-rubric.png b/dev/tasks/Exam-09-T9/images/Exam-09-T9-1-1-rubric.png new file mode 100644 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Most known indole syntheses use benzene derivatives as substrates to construct the pyrrole ring. In contrast, strategies that start from pyrrole derivatives and construct the benzene ring have been reported only rarely. Two examples of indole synthesis based on this reverse strategy are shown below. Analyze the reaction processes and answer the questions.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-09-T9-1", + "tags": [ + "2.4" + ], + "text_open": "9.2 Cationic gold catalysts are highly active $\\pi$ Lewis acids that selectively activate carbon-carbon unsaturated bonds. 9.2.1 Write the SMILES structures of products A, B, and C in the following 3 reactions. Note: A, B, and C each contain one five-membered ring. Formation of C proceeds through a six-membered-ring intermediate, which finally undergoes ring contraction to give a five-membered-ring product.", + "text_images": [ + { + "file": "Exam-09-T9-1-1-prompt.png", + "description": "The figure shows a general mechanism for gold-catalyzed cyclization. The substrate is an ortho-substituted aromatic ring (SMILES: 'C=C=CC(C=CC=C1N[R2])=C1C#C[R1]'). Under [Au]^+, it forms a gold-containing cationic intermediate, '[R1][C+]=C1C2=C(C=C(C1)[#79])C=CC=C2N[R2]' (Au represents the coordinated [Au] center; other ligands are not shown), and subsequent protodeauration gives the fused products '[R1]C1=C2C3=C(C=CC2)C=CC=C3N1[R2]' and '[R1]C(C1=CC=CC2=C1C3=CC=C2)N3[R2]'." + }, + { + "file": "Exam-09-T9-1-2-prompt.png", + "description": "Reaction 1: a naphthalene substrate (SMILES: '[R]C#CC1=C(C=C=C)C=CC2=C1C=CC=C2') gives product A under [Au]^+ catalysis." + }, + { + "file": "Exam-09-T9-1-3-prompt.png", + "description": "Reaction 2: a six-membered-ring substrate containing an allene and an alkynyl group (SMILES: '[R2]C(CCCC#C[R1])=C=C[R3]') gives product B under [Au]^+ and H_2O." + }, + { + "file": "Exam-09-T9-1-4-prompt.png", + "description": "Reaction 3: substrate '[R3]C#CCC(C([R1])=C=C)(O[Si]([R])([R])[R])[R2]' gives product C under [Au]^+ and i-PrOH." + } + ], + "full_mark": 6, + "rubric": [ + "(6 points, 2 points for each structure.) Reference SMILES: A is '[R]C1C2=CC=CC3=C2C4=C(C=CC=C41)C=C3'; B is '[R2]C1(/C=C/[R3])C(C([R1])=O)CCC1'; C is 'C=C1C(C([R2])=O)(CC=C1[R3])[R1]'." + ], + "rubric_images": [ + { + "file": "Exam-09-T9-1-1-rubric.png", + "description": "A is '[R]C1C2=CC=CC3=C2C4=C(C=CC=C41)C=C3'; B is '[R2]C1(/C=C/[R3])C(C([R1])=O)CCC1'; C is 'C=C1C(C([R2])=O)(CC=C1[R3])[R1]'." + } + ] + }, + { + "sub_id": "Exam-09-T9-2", + "tags": [ + "2.2", + "2.3" + ], + "text_open": "9.2.2 Write the SMILES structures of the 2 intermediates D and E in the following reaction. Note: D and E each contain only two rings.", + "text_images": [ + { + "file": "Exam-09-T9-2-1-prompt.png", + "description": "The figure shows the reaction pathway in 9.2.2. Substrate 'C=C=CC1=CC=CC(C)=C1C#C[R]' is converted under [Au]^+ successively into intermediates D and E and finally into the fused product '[R]C1C2=CC=CC3=CC=CC(C1)=C32'." + } + ], + "full_mark": 4, + "rubric": [ + "The SMILES of D is 'CC1=C2C(C=C([#79])C/C2=[C+]/[R])=CC=C1', and the SMILES of E is '[CH2+]C1=C2C(C=C([#79])C/C2=C/[R])=CC=C1' (4 points, 2 points for each structure)." + ], + "rubric_images": [ + { + "file": "Exam-09-T9-2-1-rubric.png", + "description": "The SMILES of D is 'CC1=C2C(C=C([#79])C/C2=[C+]/[R])=CC=C1', and the SMILES of E is '[CH2+]C1=C2C(C=C([#79])C/C2=C/[R])=CC=C1'." + } + ] + }, + { + "sub_id": "Exam-09-T9-3", + "tags": [ + "2.6", + "2.4" + ], + "text_open": "9.3 Using a gold-catalyzed cyclization strategy similar to that described in 9.2, the authors developed a new method for synthesizing 4-acylindoles. A specific cyclization precursor must first be prepared. The synthetic route to this precursor is shown below. Write the SMILES structures of products F, G, and H from each step. Note: the isopropyl Grignard reagent and formaldehyde are added sequentially; the final step involves a retro-ene reaction that releases nitrogen gas.", + "text_images": [ + { + "file": "Exam-09-T9-3-1-prompt.png", + "description": "The figure shows the structure of reagent IPNBSH (SMILES: 'O=S(C1=CC=CC=C1[N+]([O-])=O)(N/N=C(C)/C)=O')." + }, + { + "file": "Exam-09-T9-3-2-prompt.png", + "description": "Synthetic route: the starting substrate is 3-iodo-2-formyl-N-Ts pyrrole ('[H]C(C1=C(I)C=CN1S(C2=CC=C(C)C=C2)(=O)=O)=O'). In the first step, phenylacetylene, Pd(PPh_3)_2Cl_2, CuI, NEt_3, THF, and 50℃ give F in 92% yield. In the second step, treatment with 1) CBr_4/PPh_3 and 2) i-PrMgCl/(HCHO)_n gives G in 55% over 2 steps. In the third step, reaction with IPNBSH, DIAD, PPh_3, then TFE/H_2O gives H." + } + ], + "full_mark": 6, + "rubric": [ + "(6 points, 2 points for each structure.) Reference SMILES: F is 'O=CC1=C(C#CC2=CC=CC=C2)C=CN1S(=O)(=O)C3=CC=C(C)C=C3'; G is 'OCC#CC1=C(C#CC2=CC=CC=C2)C=CN1S(=O)(=O)C3=CC=C(C)C=C3'; H is 'C=C=CC1=C(C=CN1S(=O)(C2=CC=C(C=C2)C)=O)C#CC3=CC=CC=C3'." + ], + "rubric_images": [ + { + "file": "Exam-09-T9-3-1-rubric.png", + "description": "F is 'O=CC1=C(C#CC2=CC=CC=C2)C=CN1S(=O)(=O)C3=CC=C(C)C=C3'; G is 'OCC#CC1=C(C#CC2=CC=CC=C2)C=CN1S(=O)(=O)C3=CC=C(C)C=C3'; H is 'C=C=CC1=C(C=CN1S(=O)(C2=CC=C(C=C2)C)=O)C#CC3=CC=CC=C3'." + } + ] + }, + { + "sub_id": "Exam-09-T9-4", + "tags": [ + "2.3", + "2.8" + ], + "text_open": "9.4 When the substrate synthesized in 9.3 is subjected to gold-catalyzed cyclization, cyclization occurs but the expected product is not observed; instead, a more highly oxidized product forms. 9.4.1 To investigate the mechanism, the authors performed $^{18}\\mathrm{O}$ isotope-labeling experiments in methyl ethyl ketone (MEK) and THF. In both solvents, water participates as a nucleophile, but the degree of $^{18}\\mathrm{O}$ incorporation differs markedly: product formed in one solvent, B, shows 57% $^{18}\\mathrm{O}$ incorporation, whereas product formed in the other solvent, A, shows only 10% $^{18}\\mathrm{O}$ incorporation. Determine whether MEK is solvent A or B.", + "text_images": [ + { + "file": "Exam-09-T9-4-1-prompt.png", + "description": "The figure shows the isotope-labeling experiment. Under BrettPhosAuCl/AgNTf_2 catalysis, substrate H reacts with H_2^{18}O to form the 4-acylindole 'O=C(C1=CC=CC=C1)C2=CC=CC3=C2C=CN3S(C4=CC=C(C)C=C4)(=O)=O'. The solvent is 0.2 M MEK or THF. Solvent A gives 10% $^{18}\\mathrm{O}$ incorporation, and solvent B gives 57% $^{18}\\mathrm{O}$ incorporation." + } + ], + "full_mark": 2, + "rubric": [ + "A (2 points)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-09-T9-5", + "tags": [ + "2.2", + "2.3" + ], + "text_open": "9.4.2 The catalytic-cycle mechanism is shown below, where X denotes either solvent A or B. Write the SMILES structures of J, K, L, and M in the cycle, including necessary charge/coordination information.", + "text_images": [ + { + "file": "Exam-09-T9-5-1-prompt.png", + "description": "The figure shows the complete catalytic cycle for formation of a 4-acylindole. Substrate H binds the gold catalyst [Au]^+ to form J. J reacts with H_2O and loses [Au]^+ to form product M. M reversibly forms 'O=C(C1=CC=CC=C1)C2CC=CC3=C2C=CN3S(C4=CC=C(C)C=C4)(=O)=O', which is converted by O_2 into product 'O=C(C1=CC=CC=C1)C2=CC=CC3=C2C=CN3S(C4=CC=C(C)C=C4)(=O)=O'. In another branch, J reacts with X to form K. K reacts with H_2O and loses [Au]^+ to form product L. L then loses X to give intermediate product 'O=C(C1=CC=CC=C1)C2CC=CC3=C2C=CN3S(C4=CC=C(C)C=C4)(=O)=O'." + } + ], + "full_mark": 8, + "rubric": [ + "(8 points, 2 points for each structure.) J is '[#79]C(C/C1=[C+]/C2=CC=CC=C2)=CC3=C1C=CN3S(C4=CC=C(C)C=C4)(=O)=O'; K is '[#79]C(C/C1=C(\\[O+]=C(C)\\CC)C2=CC=CC=C2)=CC3=C1C=CN3S(C4=CC=C(C)C=C4)(=O)=O'; L is 'CC(O)(CC)O/C(C1=CC=CC=C1)=C2CC=CC3=C\\2C=CN3S(C4=CC=C(C=C4)C)(=O)=O'; M is 'O/C(C1=CC=CC=C1)=C2CC=CC3=C\\2C=CN3S(C4=CC=C(C=C4)C)(=O)=O'. Au represents the coordinated [Au] center; not all ligands are shown." + ], + "rubric_images": [ + { + "file": "Exam-09-T9-5-1-rubric.png", + "description": "J is '[#79]C(C/C1=[C+]/C2=CC=CC=C2)=CC3=C1C=CN3S(C4=CC=C(C)C=C4)(=O)=O'; K is '[#79]C(C/C1=C(\\[O+]=C(C)\\CC)C2=CC=CC=C2)=CC3=C1C=CN3S(C4=CC=C(C)C=C4)(=O)=O'; L is 'CC(O)(CC)O/C(C1=CC=CC=C1)=C2CC=CC3=C\\2C=CN3S(C4=CC=C(C=C4)C)(=O)=O'; M is 'O/C(C1=CC=CC=C1)=C2CC=CC3=C\\2C=CN3S(C4=CC=C(C=C4)C)(=O)=O'." + } + ] + } + ] +} diff --git a/dev/tasks/Exam-10-T10/images/Exam-10-T10-1-1-prompt.png b/dev/tasks/Exam-10-T10/images/Exam-10-T10-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..5339ed611ef0fdb5b4cfd69a47319a97c4245e25 --- /dev/null +++ b/dev/tasks/Exam-10-T10/images/Exam-10-T10-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:cc3b81a9f7cfa8d26f61fa8ddb41292d9e4ef4299ea6e96f5421faa7d044a5e0 +size 9448 diff --git a/dev/tasks/Exam-10-T10/images/Exam-10-T10-1-1-rubric.png b/dev/tasks/Exam-10-T10/images/Exam-10-T10-1-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..593edde962b18df47f2411e825d666fb865613db --- /dev/null +++ b/dev/tasks/Exam-10-T10/images/Exam-10-T10-1-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:42f968b6aedcc7a1dcc4e7eced78820cc33e53b7bd5e5f1dd409007798e2c056 +size 14309 diff --git a/dev/tasks/Exam-10-T10/images/Exam-10-T10-2-1-prompt.png b/dev/tasks/Exam-10-T10/images/Exam-10-T10-2-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..1de412973791afe4dc1ba85f98a9ea9b0cc5f79e --- /dev/null +++ b/dev/tasks/Exam-10-T10/images/Exam-10-T10-2-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:b4dad9a53ce394fddd5200c69d0f91a8e757a539aa42fe0563553480fdf849bb +size 24101 diff --git a/dev/tasks/Exam-10-T10/images/Exam-10-T10-2-1-rubric.png b/dev/tasks/Exam-10-T10/images/Exam-10-T10-2-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..942e2f9740fdc41a88f03b0df07c788a3870ba23 --- /dev/null +++ b/dev/tasks/Exam-10-T10/images/Exam-10-T10-2-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:effc204c4ec88c419feec19bba4524a8a86f30c6d833f7088becbe48e55aaf8e +size 24301 diff --git a/dev/tasks/Exam-10-T10/images/Exam-10-T10-3-1-prompt.png b/dev/tasks/Exam-10-T10/images/Exam-10-T10-3-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..ac797f3c7369c4c2671fe5dafbdb6601fe2c7124 --- /dev/null +++ b/dev/tasks/Exam-10-T10/images/Exam-10-T10-3-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:40690900d4d410d2b3d0ce4752428a9d4b86870f7311be5164083e93d81599c2 +size 88686 diff --git a/dev/tasks/Exam-10-T10/images/Exam-10-T10-3-1-rubric.png b/dev/tasks/Exam-10-T10/images/Exam-10-T10-3-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..2761e9efd35c7064b58100169717d7b213cefaec --- /dev/null +++ b/dev/tasks/Exam-10-T10/images/Exam-10-T10-3-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:a91ab53d330aaa361a257f10de4fd64dcaa286ac3830f673d28f326286c69058 +size 271321 diff --git a/dev/tasks/Exam-10-T10/problem.json b/dev/tasks/Exam-10-T10/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..9ab050c2b135d3fc2349cac365f2527941b36fce --- /dev/null +++ b/dev/tasks/Exam-10-T10/problem.json @@ -0,0 +1,79 @@ +{ + "problem_id": "Exam-10-T10", + "header": "", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-10-T10-1", + "tags": [ + "1.2", + "2.4" + ], + "text_open": "10-1-1 The [4+2] reaction is an important method for forming carbon-carbon bonds and is widely used in modern organic synthesis. It exhibits rich stereochemistry, including both stereoselectivity and regioselectivity, and generally gives the endo product. Complete the following reaction, showing the stereochemistry.", + "text_images": [ + { + "file": "Exam-10-T10-1-1-prompt.png", + "description": "Organic reaction scheme between 'C=CC1=CCCCC1(C)C' and 'C/C=C/C=O'." + } + ], + "full_mark": 3, + "rubric": [ + "'CC1(C)CCC[C@]2([H])[C@H](C=O)[C@@H](C)CC=C21' and 'CC1(C)CCC[C@@]2([H])[C@@H](C=O)[C@H](C)CC=C21' (3 points: 2 points for the skeleton and 1 point for fully correct stereochemistry)." + ], + "rubric_images": [ + { + "file": "Exam-10-T10-1-1-rubric.png", + "description": "Two organic structural formulas with the SMILES strings 'CC1(C)CCC[C@]2([H])[C@H](C=O)[C@@H](C)CC=C21' and 'CC1(C)CCC[C@@]2([H])[C@@H](C=O)[C@H](C)CC=C21'." + } + ] + }, + { + "sub_id": "Exam-10-T10-2", + "tags": [ + "2.2", + "2.4" + ], + "text_open": "10-3 The [4+2] reaction can also trap highly reactive intermediates, thereby demonstrating their existence or constructing complex carbon skeletons. 10-3-1 An unstable electrophilic intermediate is generated in the reaction below. Write the key intermediates in the reaction sequence.", + "text_images": [ + { + "file": "Exam-10-T10-2-1-prompt.png", + "description": "Organic reaction scheme in which 'BrC1=CC2=C(OCO2)C=C1' and 'O=C1OC(C#N)C2=C1C=CC=C2' react with LDA in THF from -78 °C to 40 °C to give 'O=C1C2=CC=C(OCO3)C3=C2C(C4=C1C=CC=C4)=O'." + } + ], + "full_mark": 8, + "rubric": [ + "The SMILES strings of the key intermediates are 'C%12(OCO%13)=C%13C#CC=C%12', '[O-]C1=C2C=CC=CC2=C(C#N)O1', '[O-]C34C5=C6C(OCO6)=CC=C5C(C7=C4C=CC=C7)(C#N)O3', and 'O=C(C(C=CC=C8)=C8C9(C#N)[O-])C%10=C9C=CC%11=C%10OCO%11' (2 points each, 8 points total)." + ], + "rubric_images": [ + { + "file": "Exam-10-T10-2-1-rubric.png", + "description": "Four organic structural formulas: 'C%12(OCO%13)=C%13C#CC=C%12', '[O-]C1=C2C=CC=CC2=C(C#N)O1', '[O-]C34C5=C6C(OCO6)=CC=C5C(C7=C4C=CC=C7)(C#N)O3', and 'O=C(C(C=CC=C8)=C8C9(C#N)[O-])C%10=C9C=CC%11=C%10OCO%11'." + } + ] + }, + { + "sub_id": "Exam-10-T10-3", + "tags": [ + "2.2", + "2.4" + ], + "text_open": "10-3-2 Non-neutral substrates can also be trapped through [4+2] reactions. Reactions in which an aromatic ring is oxidized before cycloaddition are often termed [5+2] reactions. The solvent-dependent, oxidative-dearomatization-induced divergent [5+2] cascade strategy below constructs a bicyclo[3.2.1]octane skeleton. Write the SMILES strings of the key intermediates leading to each of the two products.", + "text_images": [ + { + "file": "Exam-10-T10-3-1-prompt.png", + "description": "Organic reaction scheme in which 'OC1=C(CCC/C=C\\CCO)C=CC=C1OC' reacts with PhI(OAc)_2 in HFIP at 0 °C to give [A] 'O=C1[C@]2([C@]3([H])CCC2)C=C[C@H]4[C@]1(OC)[C@]3([H])CCO4' and [B] '[H][C@]12[C@](CCC3)([H])[C@]34C([C@H](C=C4)[C@@]1(OC)OCC2)=O'." + } + ], + "full_mark": 6, + "rubric": [ + "Intermediate leading to A: 'O=C1[C@]2([C@]3([H])CCC2)[CH+]C=C[C@]1(OC)[C@]3([H])CCO' (2 points). Intermediates leading to B: 'O=C1[C@]2([C@]3([H])CCC2)[CH+]C=C[C@]1(OC)[C@@]3([H])CCO' and 'O=C([C@@H]1C=C2)[C@]23[C@@](CCC3)([H])[C@]([H])(CCO)C1=[O+]C' (2 points each)." + ], + "rubric_images": [ + { + "file": "Exam-10-T10-3-1-rubric.png", + "description": "Intermediate leading to A: 'O=C1[C@]2([C@]3([H])CCC2)[CH+]C=C[C@]1(OC)[C@]3([H])CCO' (2 points). Intermediates leading to B: 'O=C1[C@]2([C@]3([H])CCC2)[CH+]C=C[C@]1(OC)[C@@]3([H])CCO' and 'O=C([C@@H]1C=C2)[C@]23[C@@](CCC3)([H])[C@]([H])(CCO)C1=[O+]C'." + } + ] + } + ] +} diff --git a/dev/tasks/Exam-10-T5/images/Exam-10-T5-1-1-prompt.png b/dev/tasks/Exam-10-T5/images/Exam-10-T5-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..134e55cd54ece2e2f97c6ddbce30d6e2f8772cf2 --- /dev/null +++ b/dev/tasks/Exam-10-T5/images/Exam-10-T5-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:5f160fab215e9e57d4cb4835b0da843485ac01a26f111288ef4710bb2709c6a0 +size 39957 diff --git a/dev/tasks/Exam-10-T5/images/Exam-10-T5-1-1-rubric.png b/dev/tasks/Exam-10-T5/images/Exam-10-T5-1-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..3130b813d14261147d416c6b848e584488d9ccd0 --- /dev/null +++ b/dev/tasks/Exam-10-T5/images/Exam-10-T5-1-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:173061ff3652a651a696213dc60645d3d98728bf4f4f910393bf4fb10b7812bb +size 22920 diff --git a/dev/tasks/Exam-10-T5/problem.json b/dev/tasks/Exam-10-T5/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..22cf3bac84360cae8649ac91f3f54cbf24547977 --- /dev/null +++ b/dev/tasks/Exam-10-T5/problem.json @@ -0,0 +1,30 @@ +{ + "problem_id": "Exam-10-T5", + "header": "Structural Chemistry of Two-Dimensional Transition-Metal Dichalcogenides and Their Derivatives (26 points, 12%). Two-dimensional transition-metal dichalcogenides, or TMDs, generally have layered structures held together by van der Waals forces. Their symmetry and physical properties change substantially as the number of layers decreases.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-10-T5-1", + "tags": [ + "1.3" + ], + "text_open": "Structural background: $WSe_2$ has a molybdenite-type hexagonal layered structure. The figure below is a projection along $a$; large spheres represent W and small spheres Se. One unit cell contains two layers along $c$.\n5-1-2 Explain how W and Se occupy the hexagonal plane lattice in a projection along the c axis, and give the projected relationship within one unit cell.", + "text_images": [ + { + "file": "Exam-10-T5-1-1-prompt.png", + "description": "A two-dimensional projection of a vertical rectangular unit cell outlined by thin black lines. Axes at the lower left show c vertically upward and b horizontally to the left. The figure contains two larger gray spheres and eight smaller green spheres joined by slender cylindrical bonds. An upper gray sphere lies inside the upper half of the cell near its right boundary and is connected to four green spheres at the upper-left, lower-left, upper-right, and lower-right positions. The two green spheres on the left lie within the cell, whereas the two on the right lie outside the right boundary. A lower gray sphere lies inside the lower half near the left boundary and is likewise connected to four green spheres at the upper-left, lower-left, upper-right, and lower-right positions. The two green spheres on the left lie outside the left boundary, whereas the two on the right lie within the cell. The two gray spheres are not directly connected; together they form two separate four-coordinate units offset vertically and oriented in opposite horizontal directions." + } + ], + "full_mark": 2, + "rubric": [ + "In the projection along c, W and Se occupy the A and B sites, respectively, of the hexagonal plane lattice. Only one set of W sites and one set of Se sites need be shown in a unit cell (2 points). If the A and B sites are correct but the C sites are also occupied incorrectly, award 1 point. Occluded atoms need not be shown repeatedly." + ], + "rubric_images": [ + { + "file": "Exam-10-T5-1-1-rubric.png", + "description": "A two-dimensional projection of a parallelogram unit cell outlined by thin black lines. The figure contains two larger gray spheres and five smaller green spheres joined by slender cylindrical bonds. A gray sphere lies toward the left inside the cell, with a green sphere directly in front of it so that only the left and upper parts of the gray sphere remain visible. This central green sphere connects in three directions: to a green sphere directly above by an approximately vertical bond, to a green sphere outside the left boundary at the lower left by a diagonal bond, and to another gray sphere at the lower right. The right gray sphere lies toward the lower-right interior of the cell and, besides the central green sphere, connects to two more green spheres: one outside the right boundary at the upper right and one outside the lower boundary directly below. Overall, the two gray spheres are staggered left and right within the cell, while the five green spheres occupy the left-center, upper, outer-left, outer-right, and lower-outer positions, forming a connected structure across the parallelogram boundaries." + } + ] + } + ] +} diff --git a/dev/tasks/Exam-10-T7/images/Exam-10-T7-1-1-rubric.png b/dev/tasks/Exam-10-T7/images/Exam-10-T7-1-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..d481a63f3097012f6543118f8e28789bf980d2cd --- /dev/null +++ b/dev/tasks/Exam-10-T7/images/Exam-10-T7-1-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:e3cb222143977a63b6aa938f2dbf434fadf8071ff83e7fb5d3e8ba2f9b5628fd +size 41447 diff --git a/dev/tasks/Exam-10-T7/images/Exam-10-T7-6-1-prompt.png b/dev/tasks/Exam-10-T7/images/Exam-10-T7-6-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..7ecd05576fde3aab2d540d7661209606dfa6bf22 --- /dev/null +++ b/dev/tasks/Exam-10-T7/images/Exam-10-T7-6-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:e012cf7318d1215da8513538169e408277440221cf3dbf6255e1e74dab40c641 +size 21667 diff --git a/dev/tasks/Exam-10-T7/images/Exam-10-T7-6-1-rubric.png b/dev/tasks/Exam-10-T7/images/Exam-10-T7-6-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..2a3a188bcc9f49f643a281e786efa43e35ce4c1d --- /dev/null +++ b/dev/tasks/Exam-10-T7/images/Exam-10-T7-6-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:67dcf328a959d2795d76f9c12f58eb670017a483911d8f06ee098939eb2b4eaa +size 82763 diff --git a/dev/tasks/Exam-10-T7/problem.json b/dev/tasks/Exam-10-T7/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..26765791ae8b350418ca51cb489346f0e36e127a --- /dev/null +++ b/dev/tasks/Exam-10-T7/problem.json @@ -0,0 +1,107 @@ +{ + "problem_id": "Exam-10-T7", + "header": "Question 7 Synthesis and Study of Palladium Complexes", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-10-T7-1", + "tags": [ + "1.4", + "1.6", + "2.4" + ], + "text_open": "7-1 Adding excess 2,2'-bipyridine to a solution of $\\mathrm{Pd(NO_3)_2}$ gives a neutral mononuclear Pd complex A. In solution, A dimerizes to form a 1:2 ionic complex D containing a Pd-Pd single bond and $\\mathrm{NO_3^-}$ counterions. Write the SMILES strings of A and the cation of D, and determine whether each obeys the EAN rule.", + "text_images": [], + "full_mark": 4, + "rubric": [ + "The SMILES of A is '[N][Pd]([N])(O[N+]([O-])=O)O[N+]([O-])=O', and that of the D cation is '[N][Pd](O[N]1=O)([N])(O2)[Pd](O1)([N])(O[N]2=O)[N]'. In both structures, the 2 N atoms bonded to the same Pd atom are connected by an arc. (1 point for A and 1 point for the D cation; deduct 1 point if an otherwise correct answer omits the charge.) For A, $EAN=8+4(\\mathrm{bipy})+2\\times2(\\mathrm{ONO_2^-})=16\\mathrm{e^-}$, so it does not obey the rule (1 point). For the D cation, $EAN=8+4(\\mathrm{bipy})+2\\times2(\\mathrm{ONO_2^-})+1(\\mathrm{Pd-Pd})=17\\mathrm{e^-}$, so it does not obey the rule (1 point)." + ], + "rubric_images": [ + { + "file": "Exam-10-T7-1-1-rubric.png", + "description": "The SMILES of A is '[N][Pd]([N])(O[N+]([O-])=O)O[N+]([O-])=O', and that of the D cation is '[N][Pd](O[N]1=O)([N])(O2)[Pd](O1)([N])(O[N]2=O)[N]'. In both structures, the 2 N atoms bonded to the same Pd atom are connected by an arc." + } + ] + }, + { + "sub_id": "Exam-10-T7-2", + "tags": [ + "1.4", + "1.6", + "2.4" + ], + "text_open": "7-2 Researchers found that D efficiently converts $\\mathrm{C=S}$ double bonds in organic compounds into $\\mathrm{C=O}$ double bonds. During this process, D is converted into ionic compound E, which has a threefold axis and mirror planes. Mass spectrometry gives $m/z = 425.4$ for the E cation. The 2,2'-bipyridine ligand may be represented as N⌒N. 7-2-1 Write the SMILES representation of E.", + "text_images": [], + "full_mark": 3, + "rubric": [ + "The E cation carries 2 positive charges and has the SMILES '[R][Pd]123([R])[S]4[Pd]15([R])([R])[Pd]2([R])([R])4[S]35'. Two R groups attached to the same Pd denote the chelating attachment sites of a 2,2'-bipyridine ligand. The anion of E is nitrate. (3 points: 1 point for the Pd_3S_2 cluster; deduct 1 point if an otherwise correct answer omits the charge.)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-10-T7-3", + "tags": [ + "1.4", + "2.1" + ], + "text_open": "7-2-2 Write the equation for the formation of E. The substrate may be represented as $\\mathrm{CSR_2}$, and the symbols given in the question may be used for the complexes.", + "text_images": [], + "full_mark": 2, + "rubric": [ + "$\\mathrm{4CSR_2 + 4H_2O + 3D = 4COR_2 + 8HNO_3 + 2E}$ (2 points)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-10-T7-4", + "tags": [ + "1.4", + "2.1" + ], + "text_open": "7-2-3 Addition of nitric acid converts E back into A, enabling recycling. Write the equation for this step.", + "text_images": [], + "full_mark": 2, + "rubric": [ + "$\\mathrm{E + 16HNO_3 = 2SO_2 + 12NO_2 + 8H_2O + 3A}$ (2 points)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-10-T7-5", + "tags": [ + "1.4", + "2.1" + ], + "text_open": "7-2-4 Write the overall equation for conversion of $\\mathrm{CSR_2}$ into $\\mathrm{COR_2}$.", + "text_images": [], + "full_mark": 2, + "rubric": [ + "$\\mathrm{CSR_2 + 6HNO_3 = COR_2 + 6NO_2 + SO_2 + 3H_2O}$ (2 points)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-10-T7-6", + "tags": [ + "2.4" + ], + "text_open": "7-3 Add 3 equivalents of $\\mathrm{Hacac}$ and potassium tert-butoxide to 1,3,5-tris(bromomethyl)benzene, and quench the reaction with water. Then add 3 equivalents of hydrazine; the reaction gives organic ligand $\\mathrm{H_3L}$. Adding $\\mathrm{H_3L}$ to D produces $\\mathrm{\\{[(bpy)Pd]_{12}L_x\\}(NO_3)_{12}}$. 7-3-1 A synthesis of a ligand analogous to $\\mathrm{H_3L}$ is shown below. Write the structure of J.", + "text_images": [ + { + "file": "Exam-10-T7-6-1-prompt.png", + "description": "A rightward arrow follows 'BrCC1=CC(CBr)=CC(CBr)=C1'. The text above the arrow is 'CC(C)[N+](=O)[O-]' and that below is NaOEt; the arrow points to uppercase J. Another rightward arrow follows J, with 'COP1(OC)(OC)CC(C)=C(C)C1' above it and no text below, and points to 'CC(=O)C(=C(O)C)C1=CC(C(=C(O)C)C(C)=O)=CC(C(=C(O)C)C(C)=O)=C1'. A third rightward arrow follows 'CC(=O)C(=C(O)C)C1=CC(C(=C(O)C)C(C)=O)=CC(C(=C(O)C)C(C)=O)=C1', with $\\mathrm{N_2H_4\\dotH_2O}$ above it and no text below, and points to 'CC1=C(C2=CC(C3=C(NN=C3C)C)=CC(C4=C(NN=C4C)C)=C2)C(C)=NN1'." + } + ], + "full_mark": 2, + "rubric": [ + "The SMILES of J is 'O=CC1=CC(C=O)=CC(C=O)=C1' (2 points)." + ], + "rubric_images": [ + { + "file": "Exam-10-T7-6-1-rubric.png", + "description": "The SMILES of J is 'O=CC1=CC(C=O)=CC(C=O)=C1'." + } + ] + } + ] +} diff --git a/dev/tasks/Exam-10-T9/images/Exam-10-T9-1-1-prompt.png b/dev/tasks/Exam-10-T9/images/Exam-10-T9-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..38a60b9147f9e22920719846f2208ba192378368 --- /dev/null +++ b/dev/tasks/Exam-10-T9/images/Exam-10-T9-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:1140c8b5518abdd3016e0fe07ba6af59bec6a9996420dc57ea98e42e31cb8c95 +size 19584 diff --git 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"header_images": [], + "subquestions": [ + { + "sub_id": "Exam-10-T9-1", + "tags": [ + "2.3", + "2.4", + "2.7" + ], + "text_open": "9-1-1 In radical reactions of allylic compounds using NBS or HBr, substitution and addition compete, and different reagents show different selectivities. Write the major products of the following two reactions.", + "text_images": [ + { + "file": "Exam-10-T9-1-1-prompt.png", + "description": "Two organic reaction schemes. In the first, 'CC1(C)C=CCCC1' reacts with NBS under irradiation. In the second, 'CC(C1CCCCC1)=C' reacts with HBr and $\\mathrm{(BzO)_2}$." + } + ], + "full_mark": 4, + "rubric": [ + "'CC1(C)C=C[C@@H](Br)CC1', as a racemate (±) (2 points: 1 point for the skeleton and 1 point for stereochemistry); 'C[C@](CBr)([H])C1CCCCC1', as a racemate (±) (2 points: 1 point for the skeleton and 1 point for stereochemistry)." + ], + "rubric_images": [ + { + "file": "Exam-10-T9-1-1-rubric.png", + "description": "Organic structural formula with the SMILES 'CC1(C)C=C[C@@H](Br)CC1', shown as a racemate (±)." + }, + { + "file": "Exam-10-T9-1-2-rubric.png", + "description": "Organic structural formula with the SMILES 'C[C@](CBr)([H])C1CCCCC1', shown as a racemate (±)." + } + ] + }, + { + "sub_id": "Exam-10-T9-2", + "tags": [ + "2.2", + "2.3" + ], + "text_open": "9-2-1 Radical reactions can also achieve efficient intramolecular N-C bond coupling. Write the key intermediates in the following reaction.", + "text_images": [ + { + "file": "Exam-10-T9-2-1-prompt.png", + "description": "Organic reaction scheme in which 'C1(CN2CCCCC2)NCCCC1' reacts with NCS under irradiation to give 'N12CCCCC1CN3C2CCCC3'." + } + ], + "full_mark": 5, + "rubric": [ + "The SMILES strings of the key intermediates are 'ClN1CCCCC1CN2CCCCC2', 'C5(CN6CCCCC6)[N]CCCC5', 'C7(CN8[C]CCCC8)NCCCC7', 'ClC(CCCC3)N3CC4NCCCC4', and 'C9(C[N+]%10=CCCCC%10)NCCCC9' (1 point each, 5 points total)." + ], + "rubric_images": [ + { + "file": "Exam-10-T9-2-1-rubric.png", + "description": "Five organic structural formulas with the SMILES strings 'ClN1CCCCC1CN2CCCCC2', 'C5(CN6CCCCC6)[N]CCCC5', 'C7(CN8[C]CCCC8)NCCCC7', 'ClC(CCCC3)N3CC4NCCCC4', and 'C9(C[N+]%10=CCCCC%10)NCCCC9'." + } + ] + }, + { + "sub_id": "Exam-10-T9-3", + "tags": [ + "2.4" + ], + "text_open": "9-3-2 Write the structure of the major product of the following reaction. Disregard stereochemistry.", + "text_images": [ + { + "file": "Exam-10-T9-3-1-prompt.png", + "description": "Organic reaction scheme in which 'IC1C(OC/C=C/C)OCCC1' reacts with triethylborane and oxygen in water and hypophosphorous acid." + } + ], + "full_mark": 2, + "rubric": [ + "'CCC1COC2OCCCC21' (2 points)." + ], + "rubric_images": [ + { + "file": "Exam-10-T9-3-1-rubric.png", + "description": "Organic structural formula with the SMILES 'CCC1COC2OCCCC21'." + } + ] + }, + { + "sub_id": "Exam-10-T9-4", + "tags": [ + "2.2", + "2.3", + "2.4" + ], + "text_open": "9-3-4 The following reaction undergoes efficient triethylborane-catalyzed cyclization to give two isomers, A1 and A2, of which A2 is the major product. Write the SMILES string of A2 and those of the key intermediates leading to A2.", + "text_images": [ + { + "file": "Exam-10-T9-4-1-prompt.png", + "description": "Organic reaction scheme in which 'CC1(C)C(O1)/C=C/CCCC#C' reacts with triethylborane and triphenyltin hydride in benzene to give A1, 'OC(C)(C)/C=C/C(CCC/1)C1=C[Sn](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4', and A2." + } + ], + "full_mark": 10, + "rubric": [ + "A2: 'OC(C)(C)/C=C/C1=C(C[Sn](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4)CCC1' (2 points). Intermediates: 'CC1(C)C(O1)/C=C/CCC/[C]=C[Sn](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4', 'CC5C(O5)[C]C(CCC/6)C6=C[Sn](C7=CC=CC=C7)(C8=CC=CC=C8)C9=CC=CC=C9', '[O]C(C)(C)/C=C/C(CCC/%10)C%10=C[Sn](C%11=CC=CC=C%11)(C%12=CC=CC=C%12)C%13=CC=CC=C%13', and 'OC(C)(C)/C=C/[C](CCC/%14)C%14=C[Sn](C%15=CC=CC=C%15)(C%16=CC=CC=C%16)C%17=CC=CC=C%17' (2 points each, 8 points total; a reasonable resonance form of the last intermediate is accepted)." + ], + "rubric_images": [ + { + "file": "Exam-10-T9-4-1-rubric.png", + "description": "A2 and 4 key intermediates. A2 is 'OC(C)(C)/C=C/C1=C(C[Sn](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4)CCC1'. In sequence, the intermediates are 'CC1(C)C(O1)/C=C/CCC/[C]=C[Sn](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4', 'CC5C(O5)[C]C(CCC/6)C6=C[Sn](C7=CC=CC=C7)(C8=CC=CC=C8)C9=CC=CC=C9', '[O]C(C)(C)/C=C/C(CCC/%10)C%10=C[Sn](C%11=CC=CC=C%11)(C%12=CC=CC=C%12)C%13=CC=CC=C%13', and 'OC(C)(C)/C=C/[C](CCC/%14)C%14=C[Sn](C%15=CC=CC=C%15)(C%16=CC=CC=C%16)C%17=CC=CC=C%17'." + } + ] + } + ] +} diff --git a/dev/tasks/Exam-11-T3/images/Exam-11-T3-1-1-prompt.png b/dev/tasks/Exam-11-T3/images/Exam-11-T3-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..5efe84df8aa39b0e057a8b30476f1cad0ec32712 --- /dev/null +++ b/dev/tasks/Exam-11-T3/images/Exam-11-T3-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:736a833f6686b22910859568f37fd186b2922d0d38f13b786ce6d2180ff4b783 +size 3918 diff --git a/dev/tasks/Exam-11-T3/images/Exam-11-T3-3-1-rubric.png b/dev/tasks/Exam-11-T3/images/Exam-11-T3-3-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..6799491f4b1a47fe2a12a07654046c34103b95ca --- /dev/null +++ b/dev/tasks/Exam-11-T3/images/Exam-11-T3-3-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:dcb4f83eae09d08f5a052b1e0aeef2f9c70ed295e80b742dcf5f6304ca0a7c8d +size 8225 diff --git a/dev/tasks/Exam-11-T3/images/Exam-11-T3-4-1-prompt.png b/dev/tasks/Exam-11-T3/images/Exam-11-T3-4-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..935819c16d062044db35ed85c8bca0cca80c70a5 --- /dev/null +++ b/dev/tasks/Exam-11-T3/images/Exam-11-T3-4-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:1bef8f56c7d88063683ca348f4e5fcea5fec5bf4497b43bf073e0c0243e6bab2 +size 9281 diff --git a/dev/tasks/Exam-11-T3/images/Exam-11-T3-4-1-rubric.png b/dev/tasks/Exam-11-T3/images/Exam-11-T3-4-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..aece03045967bfb24d72c7c987b4f8a968f5b9b1 --- /dev/null +++ b/dev/tasks/Exam-11-T3/images/Exam-11-T3-4-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:adda2974ab904dc02f53a8603a982694f38e883507193c9955c540a323743bc8 +size 6889 diff --git a/dev/tasks/Exam-11-T3/images/Exam-11-T3-5-1-rubric.png b/dev/tasks/Exam-11-T3/images/Exam-11-T3-5-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..e79881b2039167768945c4dbb3197279b7458fa2 --- /dev/null +++ b/dev/tasks/Exam-11-T3/images/Exam-11-T3-5-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:2b05ce79c4bf9d090fa5adcc4ac014052f1fc1aa8e399c62baee4011f12ad05c +size 3918 diff --git a/dev/tasks/Exam-11-T3/problem.json b/dev/tasks/Exam-11-T3/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..4747c4f01ea2f26db66da538c666f0cc11df0224 --- /dev/null +++ b/dev/tasks/Exam-11-T3/problem.json @@ -0,0 +1,105 @@ +{ + "problem_id": "Exam-11-T3", + "header": "A photoresist is a polymeric material sensitive to light of specific wavelengths. Upon irradiation, it undergoes specific chemical reactions that substantially change its solubility in a developer, thereby forming the desired pattern on the surface of a semiconductor material. As a key patterning material in integrated-circuit fabrication, advances in photoresist chemistry directly determine the limits of process miniaturization and the industry's capacity for technological self-reliance.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-11-T3-1", + "tags": [ + "2.1", + "2.5" + ], + "text_open": "3.1 The novolak-diazonaphthoquinone (Novolak-DNQ) system is the most widely used photoresist for the g-line (436 nm) and i-line (365 nm). One component is a high-molecular-weight linear phenolic resin, and the other is diazonaphthoquinone, whose structure is shown below.\n3.1.1 The phenolic resin prepared under acidic conditions is a low-molecular-weight linear prepolymer. The curing agent hexamethylenetetramine may be added to an aqueous solution of the prepolymer to cure and cross-link it upon heating. Identify which substrate must be used in excess when preparing the acid-catalyzed phenolic resin, write the decomposition equation of the curing agent upon heating, and on this basis state the role of the curing agent.", + "text_images": [ + { + "file": "Exam-11-T3-1-1-prompt.png", + "description": "'O=C1C2=CC=CC=C2C(S(=O)(O[Ar])=O)=CC1=[N+]=[N-]'" + } + ], + "full_mark": 5, + "rubric": [ + "The substrate used in excess is phenol. (1 point)", + "C_6H_12N_4 + 6H_2O = 6CH_2O + 4NH_3. (2 points)", + "The formaldehyde produced further cross-links the prepolymer into a three-dimensional network polymer. (1 point)", + "The NH_3 produced by decomposition acts as a base to catalyze the polymerization. (1 point)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-11-T3-2", + "tags": [ + "1.6" + ], + "text_open": "3.1.2 State the principal intermolecular forces between the two components of the Novolak-DNQ system.", + "text_images": [], + "full_mark": 2, + "rubric": [ + "Hydrogen bonding (1 point).", + "Coulombic interaction (1 point; equivalent answers such as electrostatic interaction or ionic bonding also receive credit)." + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-11-T3-3", + "tags": [ + "2.3", + "2.4" + ], + "text_open": "3.1.3 In the Novolak-DNQ system, the photosensitive component undergoes rearrangement upon irradiation to form unstable structure A, which then forms B in the alkaline developer N(CH_3)_4^+OH^-, changing the solubility of the photoresist. Write the SMILES representations of A and B.", + "text_images": [], + "full_mark": 3, + "rubric": [ + "A: 'O=C=C1C=C(C2=CC=CC=C21)S(=O)(O[Ar])=O'; B: 'O=S(C1=CC(C2=CC=CC=C21)C(O)=O)(O[Ar])=O'. 3 points total: 2 points for structure A and 1 point for structure B." + ], + "rubric_images": [ + { + "file": "Exam-11-T3-3-1-rubric.png", + "description": "A: 'O=C=C1C=C(C2=CC=CC=C21)S(=O)(O[Ar])=O'; B: 'O=S(C1=CC(C2=CC=CC=C21)C(O)=O)(O[Ar])=O'" + } + ] + }, + { + "sub_id": "Exam-11-T3-4", + "tags": [ + "2.4", + "2.5" + ], + "text_open": "3.2 PVCi is a high-resolution photoresist. One method for synthesizing PVCi is shown below.\n3.2.1 The first step is a conventional free-radical polymerization of a double bond, and all subsequent steps are functional-group substitutions. Write the SMILES representations of X, Y, Z, and PVCi. The terminal-group structures need not be specified; use [*] to denote the connection sites at the two ends of the polymer backbone.", + "text_images": [ + { + "file": "Exam-11-T3-4-1-prompt.png", + "description": "X (C_4H_6O_2) reacts under AIBN/CH_3OH conditions to give Y; Y reacts under CH_3OH/NaOH conditions to give Z; Z reacts with 'O=C(Cl)/C=C/C1=CC=CC=C1' to give PVCi." + } + ], + "full_mark": 6, + "rubric": [ + "X: 'C=COC(=O)C'; repeating unit of Y: '[*]CC(OC(=O)C)[*]'; repeating unit of Z: '[*]CC(O)[*]'; repeating unit of PVCi: '[*]CC(OC(=O)/C=C/c1ccccc1)[*]'. 6 points total: 2 points each for X and PVCi and 1 point each for Y and Z. Explicit terminal groups are not required." + ], + "rubric_images": [ + { + "file": "Exam-11-T3-4-1-rubric.png", + "description": "X: 'C=COC(=O)C'; repeating unit of Y: '[*]CC(OC(=O)C)[*]'; repeating unit of Z: '[*]CC(O)[*]'; repeating unit of PVCi: '[*]CC(OC(=O)/C=C/c1ccccc1)[*]'" + } + ] + }, + { + "sub_id": "Exam-11-T3-5", + "tags": [ + "2.3", + "2.5" + ], + "text_open": "3.2.2 Write the SMILES representation of the structure formed by the interchain reaction of PVCi upon irradiation at 360 nm. The terminal-group structures and degrees of polymerization need not be specified; use [*] and [**] to denote connection sites in the polymer chains.", + "text_images": [], + "full_mark": 2, + "rubric": [ + "The structure formed by the interchain reaction of PVCi upon irradiation at 360 nm is shown in the rubric image. 2 points; the degrees of polymerization need not be distinguished. SMILES: 'O=C(OC([*])C[*])C1C(c2ccccc2)C(C(=O)OC([**])C[**])C1c2ccccc2', where [*] denotes a connection site in a polymer chain." + ], + "rubric_images": [ + { + "file": "Exam-11-T3-5-1-rubric.png", + "description": "This can be represented by the SMILES 'O=C(OC([*])C[*])C1C(c2ccccc2)C(C(=O)OC([*])C[*])C1c2ccccc2', where [*] denotes a connection site in a polymer chain; the first two [*] sites form one chain and the last two [*] sites form another chain." + } + ] + } + ] +} diff --git a/dev/tasks/Exam-11-T5/images/Exam-11-T5-1-1-prompt.png b/dev/tasks/Exam-11-T5/images/Exam-11-T5-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..8fedf55e7976f5667c8cfac4b8e480c08ec9f727 --- /dev/null +++ b/dev/tasks/Exam-11-T5/images/Exam-11-T5-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:59db528abcc9b9382676ccf88ca846754d01c191fb1f6936ecd06e37746d759d +size 79534 diff --git a/dev/tasks/Exam-11-T5/problem.json b/dev/tasks/Exam-11-T5/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..69c0b425d6fc8107cca0cf602d73932c4a8847bc --- /dev/null +++ b/dev/tasks/Exam-11-T5/problem.json @@ -0,0 +1,25 @@ +{ + "problem_id": "Exam-11-T5", + "header": "Problem 5: Familiar Structures from an Unfamiliar Perspective (18 points, 7%)", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-11-T5-1", + "tags": [ + "1.3" + ], + "text_open": "5.2 Crystal Y has the chemical formula QF_y. Its projections along the a and c directions are shown below; black spheres represent Q ions and white spheres represent F^-. 5.2.1 Give the value of y.", + "text_images": [ + { + "file": "Exam-11-T5-1-1-prompt.png", + "description": "The image contains two two-dimensional projections of a crystal structure. The left panel is a rectangular view of the b-c plane. Axes at the lower left show c vertically upward, b horizontally to the right, and a perpendicular to the page. A large black sphere is located at each of the four corners, with two-dimensional fractional coordinates (b, c) = (0.00, 0.00), (1.00, 0.00), (0.00, 1.00), and (1.00, 1.00). Two large black spheres lie inside the unit cell at approximately (0.33, 0.33) and (0.67, 0.67). The small white spheres in the unit cell include 4 single spheres at approximately (b, c) = (0.00, 0.50), (0.67, 0.17), (1.00, 0.50), and (0.33, 0.83), and 3 pairs of overlapping spheres at approximately (b, c) = (0.17, 0.17), (0.50, 0.50), and (0.83, 0.83). The right panel is a parallelogram view of the a-b plane. Axes at the lower left show a horizontally to the right, b diagonally toward the upper left, and c perpendicular to the page; the angle between a and b is approximately 120°. A large black sphere is located at each of the four corners, with two-dimensional fractional coordinates (a, b) = (0.00, 0.00), (1.00, 0.00), (0.00, 1.00), and (1.00, 1.00). Two large black spheres lie inside the unit cell at approximately (0.33, 0.67) and (0.67, 0.33). All the small white spheres in the unit cell are single spheres, 11 in total, at approximately (0.00, 0.50), (0.50, 0.00), (0.33, 0.17), (0.17, 0.33), (0.83, 0.17), (0.50, 0.50), (0.17, 0.83), (1.00, 0.50), (0.83, 0.67), (0.67, 0.83), and (0.50, 1.00)." + } + ], + "full_mark": 2, + "rubric": [ + "3" + ], + "rubric_images": [] + } + ] +} diff --git a/dev/tasks/Exam-12-T5/images/Exam-12-T5-1-1-prompt.png b/dev/tasks/Exam-12-T5/images/Exam-12-T5-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..9ddf84ba59f1c4b2ae48d01dd8575667ba8e01ea --- /dev/null +++ b/dev/tasks/Exam-12-T5/images/Exam-12-T5-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:7ad91126319f6715fb9499600c7def968887cb6d1cc6441bd3031f86851d2ef6 +size 27403 diff --git a/dev/tasks/Exam-12-T5/images/Exam-12-T5-2-1-rubric.png b/dev/tasks/Exam-12-T5/images/Exam-12-T5-2-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..005d0233d8d539b677f61b7b9dd85418ec0e72da --- /dev/null +++ b/dev/tasks/Exam-12-T5/images/Exam-12-T5-2-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:0ed3010ad1770b6a954b9d675ed1939fcac1f2ebbbb3edda6a3d209baa8b4569 +size 58685 diff --git a/dev/tasks/Exam-12-T5/problem.json b/dev/tasks/Exam-12-T5/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..11da2888a2f141ab43552dc9688985f43487fd71 --- /dev/null +++ b/dev/tasks/Exam-12-T5/problem.json @@ -0,0 +1,44 @@ +{ + "problem_id": "Exam-12-T5", + "header": "Problem 5 Structure of a Barium Titanate\nA hexagonal-family form of $\\mathrm{BaTiO_3}$ has an interesting structure.\nIn this crystal, all Ba and O atoms together form a special stacking pattern. In a close-packed structure, designate a close-packed layer whose coordination environment above and below is the same as in hexagonal close packing as h, and one whose coordination environment is the same as in cubic close packing as c. For example, double hexagonal close packing (ABACABAC...) can be represented as hchchc.... The joint Ba/O stacking sequence is hcchcch...; considering only Ba gives the same hcchcch... sequence.\nEach Ba is coordinated by 12 O atoms to form a $\\mathrm{[BaO_{12}]}$ tetradecahedral unit. Ti occurs in two chemical environments. In one, two $\\mathrm{TiO_6}$ octahedra share a face to form a $\\mathrm{[Ti_2O_9]}$ unit; in the other, an isolated $\\mathrm{[TiO_6]}$ unit forms, meaning that it does not share a face with another $\\mathrm{TiO_6}$ octahedron, although corner sharing is possible. The ratio of the two types of Ti atoms is 2:1. In the conventional unit cell whose vertices are the Ti atoms of the isolated $\\mathrm{[TiO_6]}$ units, one Ba has coordinates (2/3, 1/3, 0.5967). The unit-cell parameters are a = b = 574.6 pm and c = 1399.9 pm.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-12-T5-1", + "tags": [ + "1.3" + ], + "text_open": "5.1 Multiple selection: which of the following statements are correct?\n(a) The Ti arrangement resembles hexagonal diamond.\n(b) Each O layer forms a kagome-type \"six-pointed-star array\" like that shown in [Exam-12-T5-1-1-prompt.png].\n(c) If all Ba and O atoms are regarded as jointly forming the hcchcch... close-packed sequence and Ti atoms are treated as interstitial atoms, Ti occupies all octahedral voids formed solely by O atoms.\n(d) This $\\mathrm{BaTiO_3}$ contains 2 distinct Ti-O bonds and 2 distinct Ba-O bonds.", + "text_images": [ + { + "file": "Exam-12-T5-1-1-prompt.png", + "description": "The image shows a translationally symmetric planar network made of black straight lines and circles and possessing additional symmetries. There are 3 sets of black lines in different directions; within each set, adjacent parallel lines are equally spaced. Together, the three sets partition the plane into closed regular hexagons and equilateral triangles of equal edge length, with a circle at every line intersection. Around each circle, the four polygons appear in the order triangle, hexagon, triangle, hexagon. In the overall connectivity, each of the three edges of every triangle is shared with a hexagonal region, and each of the 6 edges of every hexagon is shared with a hexagonal region." + } + ], + "full_mark": 4, + "rubric": [ + "(b) and (c) (4 points; no credit for an incorrect selection or no selection)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-12-T5-2", + "tags": [ + "1.3", + "3.9" + ], + "text_open": "5.4 Calculate the second-shortest Ba-Ba distance.", + "text_images": [], + "full_mark": 5, + "rubric": [ + "Plot the Ba-atom positions in the projection along the a axis:\n[Exam-12-T5-2-1-rubric.png]\nFirst determine the coordinates of each Ba atom: Ba0(2/3, 1/3, 0.9033), Ba1(1/3, 2/3, 0.4033), Ba2(1/3, 2/3, 0.0967), Ba3(2/3, 1/3, 0.5967), Ba4(0, 0, 0.25), and Ba5(0, 0, 0.75). (1 point)\nCandidate distances for the second-shortest Ba-Ba separation are Ba0~Ba0, Ba0~Ba2, Ba0~Ba3, and Ba0~Ba5.\n$d_{\\text{Ba0-Ba0}}=a=574.6\\ \\mathrm{pm}$.\n$d_{\\text{Ba0-Ba2}}=\\sqrt{((1+0.0967-0.9033)c)^2+\\left(\\frac{\\sqrt{3}}{3}a\\right)^2}=\\sqrt{(0.1934\\times1399.9)^2+\\left(\\frac{\\sqrt{3}}{3}\\times574.6\\right)^2}=428.2\\ \\mathrm{pm}$. (1 point)\n$d_{\\text{Ba0-Ba3}}=(0.9033-0.5967)c=(0.9033-0.5967)\\times1399.9=429.2\\ \\mathrm{pm}$. (1 point)\n$d_{\\text{Ba0-Ba5}}=\\sqrt{((0.9033-0.75)c)^2+\\left(\\frac{\\sqrt{3}}{3}a\\right)^2}=\\sqrt{((0.9033-0.75)\\times1399.9)^2+\\left(\\frac{\\sqrt{3}}{3}\\times574.6\\right)^2}=395.1\\ \\mathrm{pm}$. (1 point)\nThe second-shortest Ba-Ba distance is $428.2\\ \\mathrm{pm}$. (1 point)\n(A correct result obtained by another method may receive credit, but the Ba0-Ba3 distance must be calculated to earn its 1 process point.)" + ], + "rubric_images": [ + { + "file": "Exam-12-T5-2-1-rubric.png", + "description": "An unfilled rectangle with a black border and an aspect ratio of approximately 3:1 contains green circles labeled with black letters and numbers. Taking the lower-left, lower-right, and upper-right corners as (0,0), (1,0), and (1,1), respectively, the circles and labels are located at approximately: (1/3,0.90), Ba0; (2/3,0.40), Ba1; (2/3,0.10), Ba2; (1/3,0.60), Ba3; (0,0.25), Ba4; (1,0.25), Ba4; (0,0.75), Ba5; and (1,0.75), Ba5." + } + ] + } + ] +} diff --git a/dev/tasks/Exam-13-T5/images/Exam-13-T5-1-1-prompt.png b/dev/tasks/Exam-13-T5/images/Exam-13-T5-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..68c56b066f1ac3ce4bfdc00e47fe73b73f8cf190 --- /dev/null +++ b/dev/tasks/Exam-13-T5/images/Exam-13-T5-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:e6d3d25fc2fdd4f1b6fa9f307452b7ef3972b92793183d0218bd4ae6c477f225 +size 101112 diff --git a/dev/tasks/Exam-13-T5/images/Exam-13-T5-header-1.png b/dev/tasks/Exam-13-T5/images/Exam-13-T5-header-1.png new file mode 100644 index 0000000000000000000000000000000000000000..081791167c66e0d94676045c76c5b883102f3e19 --- /dev/null +++ b/dev/tasks/Exam-13-T5/images/Exam-13-T5-header-1.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:043e972ccdc9b8132cd15e57565dc6e9d936f0b0fdf51d33cb23af594d1599b8 +size 167587 diff --git a/dev/tasks/Exam-13-T5/problem.json b/dev/tasks/Exam-13-T5/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..ba48dd60612544257615d81068a56f040484dd0c --- /dev/null +++ b/dev/tasks/Exam-13-T5/problem.json @@ -0,0 +1,43 @@ +{ + "problem_id": "Exam-13-T5", + "header": "Question 5 Crystal Distortions. Under ideal conditions, several classic crystal structures have high symmetry, including cubic perovskite, fluorite, and spinel structures. Under certain conditions, however, some crystal structures distort, breaking the original high symmetry and lowering it to tetragonal, trigonal, orthorhombic, monoclinic, or another crystal system.", + "header_images": [ + { + "file": "Exam-13-T5-header-1.png", + "description": "A perspective ball-and-stick model of a three-dimensional crystal structure. Axes at the lower left show a in red pointing toward the lower left and into the view, b in green pointing horizontally right, and c in blue pointing vertically upward. The outer framework is a cube with a large light-blue sphere at each of its 8 vertices and thin light-blue rods joining adjacent vertices. A smaller red sphere is embedded near the midpoint of each of the 12 cube edges. The largest green sphere is suspended at the center of the cube and connects radially to the 12 red spheres through 12 thin green rods. The 8 red spheres in the top layer, c = 1, and bottom layer, c = 0, are accompanied by broad white arrows with black borders and bold black coordinates indicating displacement. In the top layer, the left sphere has a rightward arrow and (0.5,0.0098,1); the right sphere has a leftward arrow and (0.5,0.9902,1); the front sphere has a lower-right arrow and (1.0098,0.5,1); and the rear sphere has an upper-right arrow and (-0.0098,0.5,1). In the bottom layer, the left sphere has an outward left arrow and (0.5,-0.0098,0); the right sphere has an outward right arrow and (0.5,1.0098,0); the front sphere has a lower-left arrow and (0.9902,0.5,0); and the rear sphere, seen in perspective below and to the right of the central green sphere, has an upper-left arrow and (0.0098,0.5,0)." + } + ], + "subquestions": [ + { + "sub_id": "Exam-13-T5-1", + "tags": [ + "1.3" + ], + "text_open": "5.2 At low temperature, $NaNiO_2$ distorts from a high-symmetry structure into a base-centered monoclinic crystal, as shown below; the large, medium, and small spheres are Na, Ni, and O, respectively. One O atom has coordinates (0.222, 0.5, 0.205), both Ni and Na occupy vertex or face-center positions, and the unit-cell parameters are a = 5.33 Å, b = 2.86 Å, c = 5.59 Å, and β = 110.5°. 5.2.1 Before distortion, the structure follows a close-packing model: O forms the closest-packed array, while alternating Na and Ni layers occupy octahedral holes in the O array. Use ABC for the close-packed O layers, abc for the Na layers, and a’b’c’ for the Ni layers. Represent the close-packing sequence before distortion using letters, giving one repeat unit, and state the Bravais lattice of the undistorted crystal.", + "text_images": [ + { + "file": "Exam-13-T5-1-1-prompt.png", + "description": "A perspective three-dimensional ball-and-stick model enclosed by an inclined parallelepiped outlined with thin black lines. Axes at the lower left show a in red pointing roughly left, b in green pointing into the page, and c in blue pointing steeply toward the upper right. Three sphere sizes and colors represent the elements: the largest golden spheres are Na, medium silver-gray spheres are Ni, and the smallest red spheres are O. Thin cylindrical bonds are split between the colors of the connected atoms, such as yellow-red Na-O and silver-red Ni-O bonds. The structure is a highly ordered alternating layered arrangement along c. Silver-gray Ni layers lie near the upper and lower faces of the cell. Each Ni connects diagonally to 6 O atoms through 6 silver-red bonds, forming a typical $NiO_6$ octahedron. These edge-sharing octahedra extend laterally in planes perpendicular to c to form robust two-dimensional $NiO_2$ slabs. A layer of large golden Na spheres occupies the central plane in the open space between the 2 $NiO_2$ slabs. Each Na radiates 6 yellow-red bonds in an octahedral geometry, connecting upward to 3 O atoms at the bottom of the upper $NiO_2$ slab and downward to 3 O atoms at the top of the lower $NiO_2$ slab. Thus, rather than forming simple vertical linear units, the interlayer Na atoms act as structural anchors through a three-dimensional network of 6 contacts, 3 above and 3 below, joining the separated silver-red slabs along c into a continuous crystal network." + } + ], + "full_mark": 5, + "rubric": [ + "The repeat unit is Ac’BaCb’AcBa’Cb, or an equivalent representation (3 points); rhombohedrally centered hexagonal, or hR (2 points)." + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-13-T5-2", + "tags": [ + "1.3" + ], + "text_open": "5.2.2 Calculate all distinct Ni-O bond lengths after distortion.", + "text_images": [], + "full_mark": 4, + "rubric": [ + "$d_1(\\mathrm{Ni-O})=\\left\\|0.222\\vec{a}+0.5\\vec{b}+0.205\\vec{c}\\right\\|=1.95\\ \\text{angstrom}$ (2 points); $d_2(\\mathrm{Ni-O})=\\left\\|-0.278\\vec{a}+0.205\\vec{c}\\right\\|=2.17\\ \\text{angstrom}$ (2 points)." + ], + "rubric_images": [] + } + ] +} diff --git a/dev/tasks/Exam-15-T3/images/Exam-15-T3-1-1-rubric.png b/dev/tasks/Exam-15-T3/images/Exam-15-T3-1-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..60a2d409a5b886d476a58d4a54d666e861513068 --- /dev/null +++ b/dev/tasks/Exam-15-T3/images/Exam-15-T3-1-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:00052b4b7f29b08b625f2084b7526fab3b1d51c4ac61fd7a59dc670ebd414502 +size 9911 diff --git a/dev/tasks/Exam-15-T3/images/Exam-15-T3-2-1-rubric.png b/dev/tasks/Exam-15-T3/images/Exam-15-T3-2-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..fac7041254c89d80940c9eb9f44a59b5c5dc4d99 --- /dev/null +++ b/dev/tasks/Exam-15-T3/images/Exam-15-T3-2-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:c98fed81ed79c9028f83d33217e192516861b5472a6821712e5ea27744b0f3c7 +size 6995 diff --git a/dev/tasks/Exam-15-T3/images/Exam-15-T3-2-2-rubric.png b/dev/tasks/Exam-15-T3/images/Exam-15-T3-2-2-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..4a120e26aac454361da5181928cf5c8432feec15 --- /dev/null +++ b/dev/tasks/Exam-15-T3/images/Exam-15-T3-2-2-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:ea491828073ab719aa97dcbd174103a4207f0013951a3e8c4262a7ed7d5161a4 +size 7128 diff --git a/dev/tasks/Exam-15-T3/images/Exam-15-T3-2-3-rubric.png b/dev/tasks/Exam-15-T3/images/Exam-15-T3-2-3-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..025b3b9a50db194feb8fe033dccf7747e026ddde --- /dev/null +++ b/dev/tasks/Exam-15-T3/images/Exam-15-T3-2-3-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:2bd44c93c3501db674bfe41a07e84c87961381860128a9535cb003bbe071532c +size 3521 diff --git a/dev/tasks/Exam-15-T3/images/Exam-15-T3-header-1.png b/dev/tasks/Exam-15-T3/images/Exam-15-T3-header-1.png new file mode 100644 index 0000000000000000000000000000000000000000..5775034a5e60a16656c63798adef471443f8fd91 --- /dev/null +++ b/dev/tasks/Exam-15-T3/images/Exam-15-T3-header-1.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:dec1f226a1037107602ee722b624f1782c0d177ed644ae7fd130bff03ed8f074 +size 17957 diff --git a/dev/tasks/Exam-15-T3/problem.json b/dev/tasks/Exam-15-T3/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..78b9a8756d7ad797fa990c0ab36d7c1beb2c86d6 --- /dev/null +++ b/dev/tasks/Exam-15-T3/problem.json @@ -0,0 +1,60 @@ +{ + "problem_id": "Exam-15-T3", + "header": "Problem 3 Solid Electrolyte Interphase\nThe solid electrolyte interphase (SEI) is an important research area in modern electrochemistry. During the first charging of a lithium-ion battery, the electrode potential lies below the reduction potential of the electrolyte, so additives and lithium salts in the electrolyte undergo electrochemical reductive decomposition at the solid-liquid interface. Insoluble solid products deposit on the electrode surface and form a passivation film that is electronically insulating but ionically conducting, thereby slowing electrode corrosion. Vinylene carbonate (VC), a classic film-forming additive, reacts preferentially and promotes formation of a high-quality SEI.", + "header_images": [ + { + "file": "Exam-15-T3-header-1.png", + "description": "The structures in the figure have the following SMILES: VC, 'O=C1OC=CO1'; EC, 'O=C2OCCO2'." + } + ], + "subquestions": [ + { + "sub_id": "Exam-15-T3-1", + "tags": [ + "2.2", + "2.3", + "2.4" + ], + "text_open": "3.3 Studies show that VC preferentially undergoes cleavage of a C-O single bond and is reduced to radical anion A carrying 1 unit of negative charge. It then undergoes radical polymerization to form several polymeric products. 3.3.1 In one pathway, A acts as a chain initiator and reacts with VC to form acyclic polymer B. Write the SMILES of B.", + "text_images": [], + "full_mark": 2, + "rubric": [ + "Rubric structure for B: the corresponding SMILES is '*C=COC(=O)O*', where * denotes a repeat-unit connection site. (2 points)" + ], + "rubric_images": [ + { + "file": "Exam-15-T3-1-1-rubric.png", + "description": "Rubric structure for B: the corresponding SMILES is '*C=COC(=O)O*', where * denotes a repeat-unit connection site." + } + ] + }, + { + "sub_id": "Exam-15-T3-2", + "tags": [ + "2.1", + "2.2", + "3.6" + ], + "text_open": "3.3.4 In the electrolyte, VC can also be electrochemically reduced to lithium oxalate and an enolate. Write the electrochemical reaction equation; EC participates in the reaction.", + "text_images": [], + "full_mark": 2, + "rubric": [ + "Complete reaction: 'O=C1OC=CO1' + 'O=C1OCCO1' + 4$Li^+$ + 4$e^-$ = $Li_2C_2O_4$ + 2$CH_2CHOLi$ (2 points; 1 point if all reactants are correct but the equation is not balanced)." + ], + "rubric_images": [ + { + "file": "Exam-15-T3-2-1-rubric.png", + "description": "The EC structure in the rubric equation has SMILES 'O=C1OCCO1'." + }, + { + "file": "Exam-15-T3-2-2-rubric.png", + "description": "The VC structure in the rubric equation has SMILES 'O=C1OC=CO1'." + }, + { + "file": "Exam-15-T3-2-3-rubric.png", + "description": "The lithium enolate product in the rubric equation has SMILES 'C=CO[Li]'." + } + ] + } + ] +} diff --git a/dev/tasks/Exam-16-T6/images/Exam-16-T6-2-1-rubric.png b/dev/tasks/Exam-16-T6/images/Exam-16-T6-2-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..bd6f1e3aa7b53ee36e5837938c6322f7b467bc3a --- /dev/null +++ b/dev/tasks/Exam-16-T6/images/Exam-16-T6-2-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:eda9a71c0cddec35e244b50856efdaf8f64c974e2e65a50cf1dd58c909e9355d +size 31099 diff --git a/dev/tasks/Exam-16-T6/images/Exam-16-T6-3-1-rubric.png b/dev/tasks/Exam-16-T6/images/Exam-16-T6-3-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..6d3de523797f17ad709d3892b9ff9499478bd8ba --- /dev/null +++ b/dev/tasks/Exam-16-T6/images/Exam-16-T6-3-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:39d90fc353a01feabed8354f87e3cb889429cc80356c356208c84b3f6b2acabc +size 55777 diff --git a/dev/tasks/Exam-16-T6/images/Exam-16-T6-5-1-prompt.png b/dev/tasks/Exam-16-T6/images/Exam-16-T6-5-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..8159fe5a5d0bdb3fd88afbab90c0c54cea4a20da --- /dev/null +++ b/dev/tasks/Exam-16-T6/images/Exam-16-T6-5-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:6dc1c19c5782801be166b7fc404b729212e830619c10fb94711a38219d47f828 +size 71458 diff --git a/dev/tasks/Exam-16-T6/images/Exam-16-T6-6-1-prompt.png b/dev/tasks/Exam-16-T6/images/Exam-16-T6-6-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..f9db9b7932a6796a8810e10a6e462e70ec6d1b86 --- /dev/null +++ b/dev/tasks/Exam-16-T6/images/Exam-16-T6-6-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:6f5df3ac44420d958f7c1645ca4c79d1b403cb0f079c894bdab0b58e0b75d24a +size 112743 diff --git a/dev/tasks/Exam-16-T6/images/Exam-16-T6-7-1-prompt.png b/dev/tasks/Exam-16-T6/images/Exam-16-T6-7-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..a68bab8c877888850a5b4356f7bc278ff7358390 --- /dev/null +++ b/dev/tasks/Exam-16-T6/images/Exam-16-T6-7-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:49e4ea6b4b026c6d6861f1fdaf60527071c57b05921318df3a7fe7592a69039d +size 213437 diff --git a/dev/tasks/Exam-16-T6/images/Exam-16-T6-8-1-rubric.png b/dev/tasks/Exam-16-T6/images/Exam-16-T6-8-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..b0c20ac02c0468b0a648cdf60ebcfe1101a72f1d --- /dev/null +++ b/dev/tasks/Exam-16-T6/images/Exam-16-T6-8-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:65cf1b731a5ed73a84ef0c144dac308b9e9435263c02615dba95d3fa2b2a92d2 +size 30761 diff --git a/dev/tasks/Exam-16-T6/images/Exam-16-T6-9-1-prompt.png b/dev/tasks/Exam-16-T6/images/Exam-16-T6-9-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..7a83f4e7e59e6bdabeb1f18d10473b664310a2da --- /dev/null +++ b/dev/tasks/Exam-16-T6/images/Exam-16-T6-9-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:89edb0885eb639c56f1d03c70c7a8dc1818ebac5c2a59e56cde6da7f73660c85 +size 107285 diff --git a/dev/tasks/Exam-16-T6/problem.json b/dev/tasks/Exam-16-T6/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..ea896dd65dfb345a8b1512e482eb3e7ed5752c8d --- /dev/null +++ b/dev/tasks/Exam-16-T6/problem.json @@ -0,0 +1,191 @@ +{ + "problem_id": "Exam-16-T6", + "header": "Problem 6: Polysulfide Anions\nA sodium-sulfur battery is a high-temperature molten secondary alkali-metal energy-storage battery that uses metallic sodium as the negative electrode, elemental sulfur as the positive electrode, and molten sodium beta-alumina as the solid electrolyte. It offers high energy density, inexpensive raw materials, and suitability for large-scale energy storage, and has become a research focus in electrochemical energy storage in recent years. Charging and discharging proceed through the formation of sodium polysulfides (Na_{2}S_{x}) with different chain lengths.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-16-T6-1", + "tags": [ + "2.1", + "2.4" + ], + "text_open": "6.2 Sodium polysulfides (Na_{2}S_{x}) are key electroactive intermediates in sodium-sulfur batteries, and their anions can be synthesized and observed in the laboratory using common reagents. Under an N_{2} atmosphere, excess sodium tert-butoxide is added to a DMF solution of S_{8}, reducing S_{8} to polysulfide anion A. A disproportionates into another polysulfide anion B containing fewer atoms. Electron paramagnetic resonance (EPR) measurements show that some B anions undergo homolysis in solution to form bent anion C. C is trapped in zeolite cages in lazurite crystals, [(Na,Ca)_{4-8}(AlSiO_{4})_{6}(SO_{4},S,Cl)_{2}], and appears blue in solution. The sodium salt corresponding to polysulfide anion B is a common component of sodium-sulfur batteries. If it is heated to 300°C and melted, anion C dimerizes again to give isomer D of polysulfide anion B; the sulfur atoms in D occupy only two distinct chemical environments. 6.2.1 Give the chemical formulas of A and B.", + "text_images": [], + "full_mark": 2, + "rubric": [ + "A: S_{8}^{2-} (1 point)", + "B: S_{6}^{2-} (1 point)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-16-T6-2", + "tags": [ + "1.6", + "2.4" + ], + "text_open": "6.2.2 Give the SMILES structures of C and D, showing the Lewis connectivity, and indicate the formal charge on every sulfur atom.", + "text_images": [], + "full_mark": 3, + "rubric": [ + "1 point for the Lewis structure of C and 2 points for that of D; a reasonable resonance structure is also acceptable." + ], + "rubric_images": [ + { + "file": "Exam-16-T6-2-1-rubric.png", + "description": "Lewis structures of C and D in the rubric. C is S_3^- with SMILES 'S=[S][S-]'; D is an S_6^{2-} isomer with SMILES '[S-]S(S([S-])=S)=S'." + } + ] + }, + { + "sub_id": "Exam-16-T6-3", + "tags": [ + "1.6", + "2.4" + ], + "text_open": "6.3 Anion C can also be observed when S_{8} is dissolved in liquid ammonia, together with another hydrogen-free anion E. Early studies proposed the following mechanism: first, liquid ammonia undergoes autoionization to form NH_{2}^{-} and NH_{4}^{+}; second, NH_{2}^{-} attacks S_{8}, giving S_{8}NH_{2}^{-} by addition; third, S_{8}NH_{2}^{-} decomposes into cyclic anion F and a stable dibasic acid; fourth, F rearranges to form anion E and S_{8}, while the dibasic acid reacts with S_{8} and NH_{4}^{+} to form (NH_{4})_{2}S_{x}, which is subsequently converted into C. EPR measurements, however, show that anion F is not formed during the reaction. Moreover, the tendency of liquid ammonia to autoionize is extremely small and cannot explain the high solubility of S_{8} in liquid ammonia. Researchers therefore proposed a more plausible mechanism: first, S_{8}NH_{2}^{-} and NH_{3} participate in a proton-transfer equilibrium, forming S_{8}NH_{3} while generating NH_{2}^{-}, which promotes dissolution; second, NH_{3} can also react by nucleophilic substitution to form the small molecule S_{2}NH_{3}; third, S_{8}NH_{2}^{-} decomposes to anion B and intermediate G; fourth, G transfers all of its protons to NH_{3} to form anion H; fifth, H reacts with S_{2}NH_{3} to give anion E and NH_{3}, with all species in this reaction having a stoichiometric coefficient of 1. 6.3.1 Give the SMILES structures of E, F, G, and H, showing the Lewis connectivity. The Lewis structure of E is known to contain two S-S single bonds, and its sulfur atoms occupy two distinct chemical environments.", + "text_images": [], + "full_mark": 5, + "rubric": [ + "1 point each for the Lewis structures of E, G, and H, and 2 points for that of F; a reasonable resonance structure is also acceptable." + ], + "rubric_images": [ + { + "file": "Exam-16-T6-3-1-rubric.png", + "description": "Lewis structures of E, F, G, and H in the rubric. The SMILES of E is '[S-]S/N=[S+]/[S-]'; the SMILES of F is 'S1SS[N-]SSSS1'; the SMILES of G is 'S=[S+]N([H])[H]'; and the SMILES of H is 'S=S=[N-]'." + } + ] + }, + { + "sub_id": "Exam-16-T6-4", + "tags": [ + "2.1", + "2.2" + ], + "text_open": "6.3.2 In fact, the fifth reaction proceeds in three steps, and both the first and second steps generate SNH_{3}. Write the chemical equations for the three steps in order.", + "text_images": [], + "full_mark": 3, + "rubric": [ + "Step 1: S_{2}N^{-} + S_{2}NH_{3} = S_{3}N^{-} + SNH_{3} (1 point)", + "Step 2: S_{3}N^{-} + S_{2}NH_{3} = S_{4}N^{-} + SNH_{3} (1 point)", + "Step 3: 2SNH_{3} = S_{2}NH_{3} + NH_{3} (1 point)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-16-T6-5", + "tags": [ + "3.2", + "3.4" + ], + "text_open": "6.4 Anion C is also relatively common in nature and occurs mainly in ores formed under high-temperature, high-pressure conditions. Calculations show that at 350℃ and 0.5 GPa, anion C can exist as a stable form of sulfur. Reaction 1 (unbalanced) is: H_2S(aq.) + SO_4^{2-}(aq.) + H^+(aq.) → C(aq.) + O_2(g.) + H_2O(l.), \\lg K_1 = -9.6 (when the stoichiometric coefficients of C and H^+ are both 1). The standard enthalpies of formation and standard entropies of several species are listed below (ignore the temperature dependence of enthalpy and entropy and ignore microscopic effects): H_2S(aq.): \\Delta_fH_m^⊖ = -39.0 kJ·mol^{-1}, S_m^⊖ = 122.0 J·mol^{-1}·K^{-1}; HS^-(aq.): -17.7, 61.1; HSO_4^-(aq.): -885.8, 126.9; SO_4^{2-}(aq.): -907.5, 17.0; H^+(aq.): 0, 0; O_2(g.): 0, 205.1; H_2O(l.): -285.8, 70.0.\n6.4.2 Derive the equation of the bold sloping line l_1 in the figure below. Hint: f_O2 denotes the fugacity of oxygen, which physically represents the effective pressure of a real gas. The total dissolved sulfur concentration in the figure is 1 wt% (that is, all dissolved sulfur-containing species in 1 kg of solution contain a total of 10 g of S atoms). Assume that the density of water under these conditions is ρ = 1.0 × 10^3 kg/m^3.", + "text_images": [ + { + "file": "Exam-16-T6-5-1-prompt.png", + "description": "A two-dimensional predominance-diagram-style plot with a white background, black lines, and a rectangular frame. The top and right sides of the frame also have short tick marks. The horizontal axis is labeled “pH,” with major tick labels 1, 2, 3, 4, 5, 6, 7, 8, and 9; the axis extends beyond 9 without a clearly labeled 10. The vertical axis is labeled “$\\log_{10} f_{O_2}$,” with major ticks −22, −24, −26, −28, −30, −32, and −34 from top to bottom. Several solid black lines divide the plot into regions labeled with formulas or letters. The upper-left region is labeled “HSO_4^-,” the upper-right “SO_4^2-,” the lower-left “H_2S,” and the lower-right “HS^-.” Near the center-left is a light-gray irregular polygon labeled with a capital “C.” It has approximately five sides: an upper-left sloping boundary from about pH 1 and $\\log_{10} f_{O_2}$ about −24 to about pH 5.3 and $\\log_{10} f_{O_2}$ about −26; an upper-right sloping boundary from about pH 5.3 and $\\log_{10} f_{O_2}$ about −26 to about pH 7.6 and $\\log_{10} f_{O_2}$ about −28.4; a thinner lower-right boundary from about pH 7.6 and $\\log_{10} f_{O_2}$ about −28.4, extending down-left to about pH 6.0 and $\\log_{10} f_{O_2}$ about −31; a noticeably bold lower boundary from about pH 6.0 and $\\log_{10} f_{O_2}$ about −31, extending up-left to about pH 1 and $\\log_{10} f_{O_2}$ about −26.8; and a left boundary adjoining the vertical axis. A vertical solid line at about pH 5.3 runs from the top border to near the upper vertex of the gray region and separates HSO_4^- from SO_4^2-. Another vertical solid line at about pH 6.0 runs downward from the lower vertex of the gray region to the bottom border and separates H_2S from HS^-. A thinner sloping line also extends down-right from the upper-right vertex of the gray region to near the right border, separating SO_4^2- from HS^-." + } + ], + "full_mark": 4, + "rubric": [ + "First balance Reaction 1 with the coefficient of C (S_{3}^{-}) set to 1:", + "Reaction 1: \\[\\text{2}{{\\text{H}}_{\\text{2}}}\\text{S + SO}_{\\text{4}}^{\\text{2-}}\\text{ + }{{\\text{H}}^{\\text{+}}}\\rightleftharpoons \\text{S}_{\\text{3}}^{\\text{-}}\\text{ + 0}\\text{.75}{{\\text{O}}_{\\text{2}}}\\text{ + 2}\\text{.5}{{\\text{H}}_{\\text{2}}}\\text{O}\\quad \\lg {{K}_{1}}=-9.6\\] (1 point)", + "Line l_{1} is the boundary between H_{2}S and S_{3}^{-}, represented by:", + "Reaction 2: \\[\\text{3}{{\\text{H}}_{\\text{2}}}\\text{S + 1}\\text{.25}{{\\text{O}}_{\\text{2}}}\\rightleftharpoons \\text{S}_{\\text{3}}^{\\text{-}}\\text{ + }{{\\text{H}}^{\\text{+}}}\\text{ + 2}\\text{.5}{{\\text{H}}_{\\text{2}}}\\text{O}\\]", + "By Hess’s law, subtracting Reaction 2 from Reaction 1 gives:", + "Reaction 3: \\[\\text{SO}_{\\text{4}}^{\\text{2-}}\\text{ + 2}{{\\text{H}}^{\\text{+}}}\\rightleftharpoons {{\\text{H}}_{\\text{2}}}\\text{S + 2}{{\\text{O}}_{\\text{2}}}\\]", + "From the given data, \\[{{\\Delta }_{\\text{r}}}H{{_{\\text{m}}^{{}}}^{\\ominus }}=\\sum{{{\\mathbf{\\Delta }}_{\\mathbf{f}}}{{H}_{\\mathbf{m}}}^{\\ominus }-}\\text{ }\\sum{{{\\mathbf{\\Delta }}_{\\mathbf{f}}}{{H}_{\\mathbf{m}}}^{\\ominus }}=868.5\\text{ kJ/mol}\\];", + "\\[{{\\Delta }_{\\text{r}}}{{S}_{\\text{m}}}^{\\ominus }=\\sum{{{S}_{\\mathbf{m}}}^{\\ominus }-}\\sum{{{S}_{\\mathbf{m}}}^{\\ominus }}=515.2\\text{ J/(mol}\\cdot \\text{K)}\\];", + "\\[{{\\Delta }_{\\text{r}}}{{G}_{\\text{m}}}^{\\ominus }=868.5-623.15\\times \\text{(0}\\text{.5152)}=547.45\\text{ kJ/mol}\\]; \\[\\lg {{K}_{3}}\\text{(}{{\\text{H}}_{\\text{2}}}\\text{S}{{\\text{O}}_{\\text{4}}}\\text{)}=\\frac{-{{\\Delta }_{\\text{r}}}{{G}_{\\text{m}}}^{\\ominus }}{2.303RT}=-45.88\\]", + "Thus, lg K_{2} = lg K_{1} - lg K_{3} = 36.28 (1 point)", + "At a total sulfur concentration of 1 wt%, the molar concentration is 0.3125 mol/kg. At the boundary, elemental conservation gives:", + "\\[[\\mathrm{H_2S}] = 3[\\mathrm{S_3^-}]\\]", + "Therefore:", + "\\[[\\mathrm{H_2S}] = 0.156\\ \\mathrm{mol\\,L^{-1}},\\quad [\\mathrm{S_3^-}] = 0.052\\ \\mathrm{mol\\,L^{-1}}\\] (1 point)", + "Substitution into the equilibrium-constant expression gives:", + "\\[36.28 = \\lg[\\mathrm{S_3^-}] - 3\\lg[\\mathrm{H_2S}] - \\mathrm{pH} - 1.25\\lg f_{\\mathrm{O_2}}\\]", + "Therefore:", + "\\[\\lg {{f}_{{{O}_{2}}}}=-28.11-0.8\\text{pH}\\]", + "(1 point; \\[\\text{pH}=-1.25\\text{ }lg\\text{ }{{f}_{{{\\text{O}}_{\\text{2}}}}}-35.14\\] is also acceptable)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-16-T6-6", + "tags": [ + "2.2", + "4.2" + ], + "text_open": "6.5 Anion C also has many applications in organic synthesis. Use the results of the following control experiments to answer Questions 6.5.1-6.5.4. 6.5.1 According to the control experiments, from which species formed by conversion of elemental sulfur might the sulfur in the product originate?", + "text_images": [ + { + "file": "Exam-16-T6-6-1-prompt.png", + "description": "Control experiments for 6.5.1, showing two parallel reactions. In the upper reaction, 'C1=CC=C(C#CC#CC2=CC=CC=C2)C=C1' (0.2 mmol) reacts with `S` (0.6 mmol) under 1.8 mmol NaOtBu in DMF, N_2, at 25°C to give 'C1=CC=C(C2=CC=C(C3=CC=CC=C3)S2)C=C1' in 65% yield. In the lower reaction, 'C1=CC=C(C#CC#CC2=CC=CC=C2)C=C1' (0.2 mmol) reacts with `S` (0.6 mmol) under 1.8 mmol NaO^tBu and 4.0 equiv TEMPO in ^tBuOH+DMF, N_2, at 25°C to give 'C1=CC=C(C2=CC=C(C3=CC=CC=C3)S2)C=C1' in 5% yield." + } + ], + "full_mark": 1, + "rubric": [ + "S_{3}^{-} (1 point)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-16-T6-7", + "tags": [ + "2.3", + "2.7" + ], + "text_open": "6.5.2 In the control experiments above, no radical intermediate was trapped in the first reaction, indicating that it does not follow a radical mechanism, whereas the third reaction proceeds through a radical mechanism. Explain, from the perspective of molecular structure, why the two reactions follow different mechanisms.", + "text_images": [ + { + "file": "Exam-16-T6-7-1-prompt.png", + "description": "Control experiments for 6.5.2, showing four parallel reactions. In the first, 'C1=CC=C(C#CC#CC2=CC=CC=C2)C=C1' (0.2 mmol) reacts with `Na_2S · 9H_2O` (0.6 mmol) in DMF, air, at 25°C to give 'C1=CC=C(C2=CC=C(C3=CC=CC=C3)S2)C=C1' in 99% yield. In the second, 'C1=CC=C(C#CC#CC2=CC=CC=C2)C=C1' (0.2 mmol) reacts with `Na_2S · 9H_2O` (0.6 mmol) with 4.0 equiv TEMPO in DMF, air, at 25°C to give 'C1=CC=C(C2=CC=C(C3=CC=CC=C3)S2)C=C1' in 92% yield. In the third, 'C1=CC=C(C#CC2=CC=CC=C2)C=C1' (0.6 mmol) reacts with `K_2S` (0.36 mmol) in DMF, Ar, at 130°C to give 'C1(C2=CC=CC=C2)=C(C3=CC=CC=C3)C(C4=CC=CC=C4)=C(C5=CC=CC=C5)S1' in 65% yield. In the fourth, 'C1=CC=C(C#CC2=CC=CC=C2)C=C1' (0.6 mmol) reacts with `K_2S` (0.36 mmol) with 4.0 equiv TEMPO in DMF, Ar, at 130°C to give only a trace of 'C1(C2=CC=CC=C2)=C(C3=CC=CC=C3)C(C4=CC=CC=C4)=C(C5=CC=CC=C5)S1'." + } + ], + "full_mark": 2, + "rubric": [ + "The substrate in the first reaction, 1,4-diphenylbutadiyne, contains a conjugated diyne system. Its low-energy LUMO and strong electrophilicity make it highly susceptible to direct nucleophilic attack and polar addition by sulfide anions; it therefore follows an ionic mechanism. (1 point)", + "The substrate in the third reaction, phenylacetylene, is an ordinary terminal alkyne with a higher-energy LUMO and weaker electrophilicity. It cannot be attacked directly by sulfide anions and must instead generate radicals through single-electron transfer or a related process at high temperature; it therefore follows a radical mechanism. (1 point)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-16-T6-8", + "tags": [ + "2.2", + "2.3" + ], + "text_open": "6.5.3 Give the SMILES structures of two key intermediates in the third reaction of the control experiments in 6.5.2.", + "text_images": [], + "full_mark": 4, + "rubric": [ + "'C1=CC=C(C(SS[S-])=[C](C2=CC=CC=C2))C=C1' and '[S-]SSC(C1=CC=CC=C1)=C(C(C2=CC=CC=C2)=[C]C3=CC=CC=C3)C4=CC=CC=C4' (4 points; 2 points for each intermediate)" + ], + "rubric_images": [ + { + "file": "Exam-16-T6-8-1-rubric.png", + "description": "Two key intermediates in the rubric. The structure on the left is a Ph-substituted vinyl radical bearing an -S-S-S^- polysulfide chain, with SMILES 'C1=CC=C(C(SS[S-])=[C](C2=CC=CC=C2))C=C1'. The structure on the right is a precyclization/cyclization-related intermediate with four Ph substituents, a radical center, and an -S-S-S^- group, with SMILES 'C1=CC=C(C(SS[S-])=C(C2=CC=CC=C2)C(C3=CC=CC=C3)=[C](C4=CC=CC=C4))C=C1'. In the figure, Ph denotes phenyl, whose SMILES is 'C1=CC=CC=C1'." + } + ] + }, + { + "sub_id": "Exam-16-T6-9", + "tags": [ + "2.8", + "2.3" + ], + "text_open": "6.5.4 Explain separately why the yields of the following two reactions are substantially lower than that of the third reaction in 6.5.2.", + "text_images": [ + { + "file": "Exam-16-T6-9-1-prompt.png", + "description": "Control experiments for 6.5.4, showing two parallel reactions. In the upper reaction, 'C1=CC=C(C#CC2=CC=CC=C2)C=C1' (0.6 mmol) reacts with `Na_2S · 9H_2O` (0.36 mmol) in DMF, Ar, at 130°C to give 'C1(C2=CC=CC=C2)=C(C3=CC=CC=C3)C(C4=CC=CC=C4)=C(C5=CC=CC=C5)S1' in 36% yield. In the lower reaction, 'C1=CC=C(C#CC2=CC=CC=C2)C=C1' (0.6 mmol) reacts with `S_8` (0.36 mmol) under NaO^tBu in DMF, Ar, at 130°C to give 'C1(C2=CC=CC=C2)=C(C3=CC=CC=C3)C(C4=CC=CC=C4)=C(C5=CC=CC=C5)S1' in less than 5% yield." + } + ], + "full_mark": 2, + "rubric": [ + "First reaction: this reaction must proceed through a radical-anion pathway. Na_{2}S·9H_{2}O contains a large amount of water of crystallization, whose water molecules protonate and quench the highly reactive radical-anion intermediates, interrupting the radical chain that leads to ring closure. (1 point)", + "Second reaction: elemental sulfur is converted into S_{3}^{-} under strongly basic conditions. At 130°C in a strongly basic medium, S_{3}^{-} readily undergoes extensive self-disproportionation or polymerization to form inert higher polysulfide complexes and therefore cannot participate effectively in the coupling. (1 point; full credit is awarded for identifying a side reaction of S_{3}^{-})" + ], + "rubric_images": [] + } + ] +} diff --git a/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-1-prompt.png b/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..52e8dc57e094a80943ea0c65de1875a06ef26e96 --- /dev/null +++ b/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:fb6db7d08ec8e9e2f98d83b75b557ce8a91c6256bc41b583821b585668de037f +size 26973 diff --git a/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-1-rubric.png b/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..f4242a938c75c6e2c5f1e5aae53fbd08721defcf --- /dev/null +++ b/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:10ffc2a4430ae7534a1cd98ae1065fca2ad44f97701a0a72fbd7afabeeaa4bf6 +size 5621 diff --git a/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-2-rubric.png b/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-2-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..31df6f1d82a2522fce6ad9ef14abb523d7579bb7 --- /dev/null +++ b/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-2-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:5444acfd280bf2ead5243580e3cf87b0c3293d8137b92742e32f0fa045fad69a +size 5711 diff --git a/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-3-rubric.png b/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-3-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..9349f963172a071a2d71d75abf83cc7cc8e904cd --- /dev/null +++ b/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-3-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:eabfb1e45523cab24b9a44c39ff6ec8badbbdbef4e7bd8b55da2b2b06720a735 +size 6292 diff --git a/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-4-rubric.png b/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-4-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..8c5077b4aa71ea7cf6550fd2f4060f3bf9fa3cc7 --- /dev/null +++ b/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-4-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:d543da80c9532219fdcb7f3665b37fb8dd010759969fd9d6ab173a2d87d3b3d2 +size 10580 diff --git a/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-5-rubric.png b/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-5-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..802491b89d40fab55e39019c81a582a7aab69b3f --- /dev/null +++ b/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-5-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:967ef30f4a443939c862880905f281bd25ad7054d9c7d7de2bedad299f20f445 +size 11374 diff --git a/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-6-rubric.png b/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-6-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..29a632017deaf096e48258603d0ac4986acc4682 --- /dev/null +++ b/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-6-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:1f3d9b8988d2697be6335d4fc9ebb7dfecbd314f5e35cf25e10a09e5c3c28f28 +size 8330 diff --git a/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-7-rubric.png b/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-7-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..f50f904b0d1b9b5c10f02028e4ac1fe156edfa6b --- /dev/null +++ b/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-7-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:393a6db69eb0aed9ffe5b8500858b92c5d738b214484d0ef03a4bc0084c0ecc4 +size 5702 diff --git a/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-8-rubric.png b/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-8-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..728c5735dfa7e211699d35ad426d217a8528d05d --- /dev/null +++ b/dev/tasks/Exam-16-T8/images/Exam-16-T8-1-8-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:c26d3b34671fc9ad663ddd375cc4e81a1849153bc252648dabf7446621cedc09 +size 10563 diff --git a/dev/tasks/Exam-16-T8/problem.json b/dev/tasks/Exam-16-T8/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..3b79c4f19b12bbb1ec979ac04f11317dbc2622a2 --- /dev/null +++ b/dev/tasks/Exam-16-T8/problem.json @@ -0,0 +1,64 @@ +{ + "problem_id": "Exam-16-T8", + "header": "Problem 8: Asymmetric Catalytic Synthesis of Thiochromenes (8 points, 8%)\nOn 2016-9-26, researchers from Washington University published a study on the asymmetric catalytic synthesis of thiochromenes. Thiochromenes are sulfur-containing heterocycles. This reaction is highly stereospecific, meaning that the configuration of the chiral carbon bonded to the R group in the product is determined. Stereochemistry is not required in this problem; the SMILES or structural formulas in your answer need not specify the configuration of chiral carbons, alkenes, or other stereochemical features.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-16-T8-1", + "tags": [ + "2.2", + "2.3", + "2.4" + ], + "text_open": "8.1 Give the SMILES structures of the key intermediates and the reaction product. Hint: the product contains a fused six-membered/six-membered ring system.", + "text_images": [ + { + "file": "Exam-16-T8-1-1-prompt.png", + "description": "An organic reaction whose substrate has SMILES 'O=CC1=C(C=CC=C1)SC(C=C[R])=O', under the conditions 'C[C@@H]1[C@H](C2=CC=CC=C2)N=C3SC4=CC=CC=C4N3C1' (10 mol%), CHCl_3, room temperature." + } + ], + "full_mark": 6, + "rubric": [ + "Key intermediates and reaction product:", + "'O=C([Cat+])C=C[R]' (1 point); 'O=CC1=C([S-])C=CC=C1' or 'S=C1C=CC=CC1=C[O-]' (1 point); '[O-]C1C2=C(C=CC=C2)S[CH]([R])C1C([Cat+])=O' (1 point)", + "'[O-]C1([Cat+])OC2C3=CC=CC=C3S[CH]([R])C12' (1 point); 'O=C1OC2C3=C(S[CH](C12)[R])C=CC=C3' (1 point); '[R][CH]1SC2=CC=CC=C2C=C1' (1 point)", + "If the abbreviation Cat is used without specifying the structure of Cat, deduct 1 point from the total for this subquestion.", + "The additional structure '[O-]C([Cat+])=C[CH]([R])SC1=CC=CC=C1C=O' earns no credit but incurs no deduction." + ], + "rubric_images": [ + { + "file": "Exam-16-T8-1-1-rubric.png", + "description": "Rubric structure 1, with SMILES 'O=C([Cat+])/C=C/[R]'." + }, + { + "file": "Exam-16-T8-1-2-rubric.png", + "description": "Rubric structure 2, with SMILES 'O=CC1=C([S-])C=CC=C1'." + }, + { + "file": "Exam-16-T8-1-3-rubric.png", + "description": "Rubric structure 3, an equivalent answer to rubric structure 2, with SMILES 'S=C1C=CC=C/C1=C/[O-]'." + }, + { + "file": "Exam-16-T8-1-4-rubric.png", + "description": "Rubric structure 4, with SMILES '[O-]C1C2=C(C=CC=C2)S[C@H]([R])C1C([Cat+])=O'." + }, + { + "file": "Exam-16-T8-1-5-rubric.png", + "description": "Rubric structure 5, a peroxide/lactone-like cyclized intermediate, with SMILES '[O-]C1([Cat+])OC2C3=CC=CC=C3S[C@H]([R])C12'." + }, + { + "file": "Exam-16-T8-1-6-rubric.png", + "description": "Rubric structure 6, with SMILES 'O=C1OC2C3=C(S[C@@H](C12)[R])C=CC=C3'." + }, + { + "file": "Exam-16-T8-1-7-rubric.png", + "description": "Final product in the rubric, with SMILES '[R][C@H]1SC2=CC=CC=C2C=C1'." + }, + { + "file": "Exam-16-T8-1-8-rubric.png", + "description": "Additional rubric structure, with SMILES '[O-]/C([Cat+])=C/[C@@H]([R])SC1=CC=CC=C1C=O'." + } + ] + } + ] +} diff --git a/dev/tasks/Exam-16-T9/images/Exam-16-T9-1-1-prompt.png b/dev/tasks/Exam-16-T9/images/Exam-16-T9-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..cf86df383f3ae76a46ab10aace2e34ab91f61730 --- /dev/null +++ b/dev/tasks/Exam-16-T9/images/Exam-16-T9-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:b9419b5d55e7aff509f0f294ffb52d896d0e591c7d55db00d0d720805c439afa +size 110416 diff --git a/dev/tasks/Exam-16-T9/images/Exam-16-T9-2-1-prompt.png b/dev/tasks/Exam-16-T9/images/Exam-16-T9-2-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..8cedeffe3fb138b2b17660a29142ac3e8628323d --- /dev/null +++ b/dev/tasks/Exam-16-T9/images/Exam-16-T9-2-1-prompt.png @@ -0,0 +1,3 @@ +version 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0000000000000000000000000000000000000000..1c5f265ea2787d2511d767fee3fe05c1201da8a4 --- /dev/null +++ b/dev/tasks/Exam-16-T9/problem.json @@ -0,0 +1,101 @@ +{ + "problem_id": "Exam-16-T9", + "header": "Problem 9: Effects of Reaction Conditions and Mechanistic Investigation of Electrochemical Decarboxylative Alkylation\nElectrochemical anodic oxidation can use inexpensive and readily available quaternary-ammonium salts of alkyl carboxylic acids as alkyl sources, avoiding the need for additional oxidants and transition-metal catalysts during decarboxylation. The method therefore offers a simple system, mild conditions, and environmental advantages. Researchers used an unsaturated amide as a model substrate to investigate decarboxylative methylation/arylation. Tetrabutylammonium acetate served as both the methyl source and electrolyte, with carbon cloth as the anode, platinum as the cathode, and DMSO as the solvent. Under constant-current electrolysis at 3 mA with 6.0 equivalents of acetic acid as a proton source at 70°C for 9 h (5.0 F/mol), the target oxindole was obtained in 80% yield.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-16-T9-1", + "tags": [ + "3.5", + "2.8", + "2.3" + ], + "text_open": "9.1 Further study showed that the acidity of the proton source strongly affects the reaction. When weaker proton sources, such as tert-butanol or hexafluoroisopropanol, or stronger proton sources, such as p-toluenesulfonic acid or trifluoroacetic acid, were used, the yield decreased substantially and the reaction was sometimes almost completely suppressed. Give a reasonable explanation for each observation.", + "text_images": [ + { + "file": "Exam-16-T9-1-1-prompt.png", + "description": "Standard reaction in the prompt: 'FC(F)(C(C(N(C1=CC=CC=C1)C)=O)=C)F' reacts with `TBAOAc (1.5 equiv.)` using `C (+) / Pt (-)` under `HOAc (6.0 equiv.)`, `DMSO 4 mL, 70°C`, and `I = 3 mA, 5.0 F/mol`, producing 'CCC1(C(N(C2=CC=CC=C21)C)=O)C(F)(F)F'." + } + ], + "full_mark": 4, + "rubric": [ + "Free alkyl carboxylate species must be present so that single-electron oxidation and decarboxylation can occur.", + "Strong acid: H^{+} protonates the carboxylate and thereby suppresses its oxidation. (2 points)", + "Weak acid: the system may become excessively nucleophilic or basic, which is unfavorable for the subsequent cyclization step. (2 points)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-16-T9-2", + "tags": [ + "2.2", + "2.3", + "4.2" + ], + "text_open": "9.3 The researchers performed the following experiments to investigate the reaction mechanism. Note: “standard condition” refers to the standard conditions described above, under which the yield is 80%. 9.3.1 To test for a free-radical intermediate, TEMPO was added as a radical trap. The target product did not form at all. Analysis of the reaction mixture together with high-resolution mass spectrometry revealed products formed by trapping with TEMPO. Give the SMILES structures of possible products.", + "text_images": [ + { + "file": "Exam-16-T9-2-1-prompt.png", + "description": "Mechanistic experiment for 9.3.1: 'CN(C(C(C)=C)=O)C1=CC=CC=C1' reacts with the salt formed from anion 'CC([O-])=O' (shown with a circled minus sign) and cation 'CCCC[N+](CCCC)(CCCC)CCCC' (shown as `N^nBu_4` with a circled plus sign) under `standard condition` with *TEMPO (3.0 equiv.)*." + } + ], + "full_mark": 2, + "rubric": [ + "Trapped product 1: 'CC1(C)CCCC(C)(C)N1OC' (1 point); trapped product 2 (TEMPO-OAc): 'CC1(C)CCCC(C)(C)N1OOC(C)=O' (1 point)" + ], + "rubric_images": [ + { + "file": "Exam-16-T9-2-1-rubric.png", + "description": "Trapped product 1 in the rubric, with SMILES 'CC1(C)CCCC(C)(C)N1OC'." + }, + { + "file": "Exam-16-T9-2-2-rubric.png", + "description": "Trapped product 2 in the rubric, TEMPO-OAc, with SMILES 'CC1(C)CCCC(C)(C)N1OOC(C)=O'." + } + ] + }, + { + "sub_id": "Exam-16-T9-3", + "tags": [ + "2.2", + "2.3", + "2.4" + ], + "text_open": "9.3.3 On the basis of these mechanistic experiments, two pathways remain possible for the ring-closing step of the standard reaction. Give the SMILES structures of the key intermediates that may be involved in the standard reaction and identify the two possible pathways corresponding to the ring-closing step.", + "text_images": [], + "full_mark": 6, + "rubric": [ + "Intermediate 1, the acetate radical: 'CC([O])=O' (1 point); intermediate 2, the methyl radical: '[CH3]' (1 point); intermediate 3: 'CN(C([C](CC)C(F)(F)F)=O)C1=CC=CC=C1' (1 point)", + "The carbocationic-pathway intermediate: 'CN(C([C+](CC)C(F)(F)F)=O)C1=CC=CC=C1' (1 point)", + "The radical-pathway intermediate: 'CN1C2=CC=C[C]C2C(CC)(C(F)(F)F)C1=O' (1 point)", + "Subsequent intermediate: 'CN(C1=O)C2=CC=C[CH+]C2C1(CC)C(F)(F)F' (1 point)" + ], + "rubric_images": [ + { + "file": "Exam-16-T9-3-1-rubric.png", + "description": "Rubric intermediate 1: the acetate radical, with SMILES 'CC([O])=O'." + }, + { + "file": "Exam-16-T9-3-2-rubric.png", + "description": "Rubric intermediate 2: the methyl radical ·CH_3, with SMILES '[CH3]'." + }, + { + "file": "Exam-16-T9-3-3-rubric.png", + "description": "Rubric intermediate 3: a CF_3-substituted oxindole-precursor radical formed after methyl-radical addition, containing a benzylic/amide-side-chain radical; its SMILES is 'CN(C([C](CC)C(F)(F)F)=O)C1=CC=CC=C1'." + }, + { + "file": "Exam-16-T9-3-4-rubric.png", + "description": "Carbocationic-pathway intermediate: a CF_3-substituted oxindole cation intermediate, with SMILES 'CN(C([C+](CC)C(F)(F)F)=O)C1=CC=CC=C1'." + }, + { + "file": "Exam-16-T9-3-5-rubric.png", + "description": "Radical-pathway intermediate: an aryl-radical/radical-cyclization intermediate whose benzene ring is depicted with a dashed circle and a radical, with SMILES 'CN1C2=CC=C[C]C2C(CC)(C(F)(F)F)C1=O'." + }, + { + "file": "Exam-16-T9-3-6-rubric.png", + "description": "Subsequent intermediate: a fused benzylic/aryl cation intermediate leading to the product, with SMILES 'CN(C1=O)C2=CC=C[CH+]C2C1(CC)C(F)(F)F'." + } + ] + } + ] +} diff --git a/dev/tasks/Exam-17-T5/images/Exam-17-T5-1-1-prompt.png b/dev/tasks/Exam-17-T5/images/Exam-17-T5-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..e6d90162c2a93796843a0fb1c1e2e616d4d1a36c --- /dev/null +++ b/dev/tasks/Exam-17-T5/images/Exam-17-T5-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:8310669ec639b10c8fb05e4657785a410106cc6d39b493d3ad086d2271109d1e +size 72723 diff --git a/dev/tasks/Exam-17-T5/images/Exam-17-T5-2-1-prompt.png b/dev/tasks/Exam-17-T5/images/Exam-17-T5-2-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..10877b175519a59c551fba8e22650322abbf50a3 --- /dev/null +++ b/dev/tasks/Exam-17-T5/images/Exam-17-T5-2-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:41c0c5a13b1e013552e77761b30f8af95ba0e2baad94a1a72a2e6217aaf25729 +size 2225 diff --git a/dev/tasks/Exam-17-T5/images/Exam-17-T5-2-2-prompt.png b/dev/tasks/Exam-17-T5/images/Exam-17-T5-2-2-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..5434fddb081cb88ad36abf236657684af84b44f1 --- /dev/null +++ b/dev/tasks/Exam-17-T5/images/Exam-17-T5-2-2-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:369e515a93f72df7e2238eb307d843641a2c7fbc4440f2de3469ae237d20e2ea +size 2418 diff --git a/dev/tasks/Exam-17-T5/images/Exam-17-T5-2-3-prompt.png b/dev/tasks/Exam-17-T5/images/Exam-17-T5-2-3-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..9faa351c69a6ff903c81cc2429154b489fcb5f9d --- /dev/null +++ b/dev/tasks/Exam-17-T5/images/Exam-17-T5-2-3-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:13484b3323d44a7e55f1eb236c6c9db5b1bcb8782bb67014d76c3eddd16583f5 +size 2209 diff --git a/dev/tasks/Exam-17-T5/images/Exam-17-T5-2-4-prompt.png b/dev/tasks/Exam-17-T5/images/Exam-17-T5-2-4-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..73bd1e9259e7125898f453f96cece0ecde3fd81c --- /dev/null +++ b/dev/tasks/Exam-17-T5/images/Exam-17-T5-2-4-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:87d39e19f74d3c9a7e227553e32e774a0a6853ba6e79bd88ac07c19fa653f647 +size 85869 diff --git a/dev/tasks/Exam-17-T5/images/Exam-17-T5-5-1-prompt.png b/dev/tasks/Exam-17-T5/images/Exam-17-T5-5-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..05d6bf8868117d3cfe61c0e0d5d9dff5a8d523bd --- /dev/null +++ b/dev/tasks/Exam-17-T5/images/Exam-17-T5-5-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:64c73abef178aeeda51878008a3b0e7291e51dd5590ea3ad4f2e35995ebf1e49 +size 76794 diff --git a/dev/tasks/Exam-17-T5/images/Exam-17-T5-5-2-prompt.png b/dev/tasks/Exam-17-T5/images/Exam-17-T5-5-2-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..4dd2089d76520c4587bda28693b294e69960701f --- /dev/null +++ b/dev/tasks/Exam-17-T5/images/Exam-17-T5-5-2-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:519088c62d25c4ec4790b61b549106d5ef86ffacfe3179374392fd64cf3c876e +size 69800 diff --git a/dev/tasks/Exam-17-T5/problem.json b/dev/tasks/Exam-17-T5/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..fc59399f59cd928a00f9b9f17ef7ae4ba64a899d --- /dev/null +++ b/dev/tasks/Exam-17-T5/problem.json @@ -0,0 +1,121 @@ +{ + "problem_id": "Exam-17-T5", + "header": "Problem 5 Dicarbon Anions", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-17-T5-1", + "tags": [ + "1.3", + "1.4" + ], + "text_open": "5.2 Starting from crystalline $CaC_2$, inserting square Co layers into some of the interlayer spaces and replacing Ca with U gives a tetragonal $UCoC_2$ crystal. Its structure is shown below, where the large spheres represent U, the medium spheres represent Co, and the small spheres represent C. The lattice parameter c = 728 pm.\n5.2.1 Name the coordination polyhedron formed by the C atoms around Co.", + "text_images": [ + { + "file": "Exam-17-T5-1-1-prompt.png", + "description": "An oblique three-dimensional ball-and-stick rendering of the tetragonal unit cell of $UCoC_2$ on a white background. Thin gray-black lines outline a rectangular-prismatic unit-cell frame elongated vertically along c. There are 10 large spheres, including spheres at face centers on the top and the upper and lower basal faces; 4 medium spheres lie on the side faces, one at an upper-middle position on each left/right face and one at a lower-middle position on each front/back face; and 8 small spheres form upper and lower groups, with two vertical pairs in each group along c joined by short brown sticks. The upper pair is on the front/back faces and the lower pair on the left/right faces. The small spheres are joined to adjacent large spheres by short sticks, forming a layered framework; the medium spheres have no sticks. At lower left, a blue arrow points upward and is labeled c, a green arrow points right and is labeled b, and a red arrow points lower-left and is labeled a. Bonding obscured by occlusion cannot be resolved. Apart from a, b, and c, the image contains no text or legend. According to the caption, the large, medium, and small spheres represent U, Co, and C, respectively. The coordinates of the two carbon atoms in the unit cell are (1/2, 0, 0.155) and (1/2, 0, 0.358)." + } + ], + "full_mark": 2, + "rubric": [ + "Tetrahedron or distorted tetrahedron (2 points). A response of \"regular tetrahedron\" receives no credit." + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-17-T5-2", + "tags": [ + "1.3" + ], + "text_open": "5.3 A uranium carbide belongs to the cubic crystal system. Its unit-cell projections along the a, b, and c directions are all as shown below. The large spheres represent U and the small spheres represent C. All atoms of the same element are equivalent in the crystal, and the lattice parameter a = 808 pm.\n5.3.1 Write the chemical formula of this crystal.", + "text_images": [ + { + "file": "Exam-17-T5-2-1-prompt.png", + "description": "Crystal-axis indicator on a white background: a solid red circle is at the lower left with a black letter a superimposed, indicating that the a axis points out of the page toward the viewer. A blue arrow extends upward from the circle and is labeled c above its tip; a green arrow extends rightward and is labeled b to the right of its tip. Apart from a, b, and c, the image contains no text or other graphics." + }, + { + "file": "Exam-17-T5-2-2-prompt.png", + "description": "Crystal-axis indicator on a white background: a solid green circle is at the lower left with a black letter b superimposed, indicating that the b axis points out of the page toward the viewer. A red arrow extends upward from the circle and is labeled a above its tip; a blue arrow extends rightward and is labeled c at its right end. Apart from a, b, and c, the image contains no text or other graphics." + }, + { + "file": "Exam-17-T5-2-3-prompt.png", + "description": "Crystal-axis indicator on a white background: a solid blue circle is at the lower left with a black letter c superimposed, indicating that the c axis points out of the page toward the viewer. A green arrow extends upward from the circle and is labeled b above its tip; a red arrow extends rightward and is labeled a at its right end. Apart from a, b, and c, the image contains no text or other graphics." + }, + { + "file": "Exam-17-T5-2-4-prompt.png", + "description": "Unit-cell projection of the uranium carbide in 5.3 (two-dimensional projection of a ball-and-stick model on a white background): thin black lines form a square unit-cell border, with some spheres lying on the border. The image contains only two sphere types: large light-blue highlighted spheres and small brown spheres. The large spheres occur at the four corners, at the middle of the top and bottom edges, and at several interior positions; some large spheres overlap in pairs. Some small spheres are joined by short brown sticks into paired or chain-like arrangements, while others are isolated. The image contains no text, letters, numbers, arrows, or legend. As specified in the prompt, the large spheres represent U and the small spheres represent C. The prompt states that the projections along the a, b, and c directions are all as shown; this image is that projection panel." + } + ], + "full_mark": 3, + "rubric": [ + "$U_2C_3$ or $U_4(C_2)_3$ (3 points)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-17-T5-3", + "tags": [ + "1.3" + ], + "text_open": "5.3.4 One carbon atom in the unit cell has coordinates (1/2, 1/4, 0.536). Calculate the carbon-carbon bond length in this crystal.", + "text_images": [], + "full_mark": 2, + "rubric": [ + "$d_{\\mathrm{C-C}} = [0.25 - (0.536 - 0.5) \\times 2] \\times 808\\ \\mathrm{pm} = 144\\ \\mathrm{pm}$ (2 points). Full credit requires both a calculation and the correct answer; an answer alone receives no credit." + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-17-T5-4", + "tags": [ + "1.3", + "1.7" + ], + "text_open": "5.4 The carbon-carbon bond lengths in ethane, ethene, and ethyne are approximately 154 pm, 134 pm, and 120 pm, respectively. On this basis, predict the negative charge on the dicarbon anion in $CaC_2$, $UCoC_2$, and the uranium carbide in 5.3, respectively. The carbon-carbon bond orders in the three compounds are distinct integers.", + "text_images": [], + "full_mark": 2, + "rubric": [ + "$CaC_2$: 2, -2, or $C_2^{2-}$", + "$UCoC_2$: 6, -6, or $C_2^{6-}$", + "Uranium carbide: 4, -4, or $C_2^{4-}$. Credit is awarded only if all responses are correct." + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-17-T5-5", + "tags": [ + "1.3", + "1.7" + ], + "text_open": "5.5 Crystalline $DyCoC_2$ has a structure entirely different from that of $UCoC_2$. Its three-dimensional structure and projection along the c direction are shown below, where the large spheres represent Dy, the medium spheres represent Co, and the small spheres represent C. The only common stable positive oxidation state of Dy is +3. A small sample of $DyCoC_2$ is hydrolyzed in 2 mol/L dilute hydrochloric acid. Gas-chromatographic analysis gives the following mass fractions of the gaseous hydrolysis products: 74% ethane, 13% propane, 12% methane, and 1% ethene; no ethyne is formed.\n5.5.1 Predict the carbon-carbon bond order in $DyCoC_2$.", + "text_images": [ + { + "file": "Exam-17-T5-5-1-prompt.png", + "description": "Three-dimensional ball-and-stick model of crystalline $DyCoC_2$ (oblique rendering on a white background): thin black lines outline a parallelepiped unit cell. There are 10 large blue-violet highlighted spheres of varying shades: 8 at the 8 corners and 2 at the centers of the top and bottom faces. There are 3 medium dark-blue spheres in a horizontal layer through the middle of the cell, positioned left of center, right of center, and at the center. Approximately 9 small brown spheres surround the medium spheres and are connected to them by dark-blue or brown short sticks; some small spheres bridge two medium spheres, forming an in-layer zigzag network. There are no direct sticks between large spheres or between medium spheres. Some spheres are obscured, so the exact number of brown small spheres cannot be fully determined. A three-axis indicator at the lower left has a blue upward arrow labeled c, a green rightward arrow labeled b, and a red arrow pointing down-left labeled a. Apart from a, b, and c, the image contains no text or legend. As specified in the prompt, the large, medium, and small spheres represent Dy, Co, and C, respectively." + }, + { + "file": "Exam-17-T5-5-2-prompt.png", + "description": "Projection of crystalline $DyCoC_2$ along the c direction (two-dimensional projection of a ball-and-stick model on a pale gray-white background): thin black lines form a rectangular unit-cell border. Four large blue spheres lie at the corners; 1 slightly smaller blue sphere lies at the center of the rectangle without any bonds; and 3 medium blue spheres lie at the midpoints of the left and right edges and at the center of the lower edge. There are 10 small brown spheres: 2 terminal spheres outside the rectangle at the upper left and upper right, while the remaining 8 form 4 horizontal dumbbell pairs joined by two parallel short brown sticks. These pairs lie at the upper center, middle left, middle right, and outside the rectangle at the lower center. Each medium blue sphere is connected to 4 brown spheres by two-colored blue-brown short sticks, and adjacent blue spheres and brown pairs form triangular units. A three-axis indicator at the lower left has a red rightward arrow labeled a, a green upward arrow labeled b, and a blue circle bearing the letter c, indicating that the c axis points out of the page toward the viewer. Apart from a, b, and c, the image contains no text or legend. As specified in the prompt, the large, medium, and small spheres represent Dy, Co, and C, respectively." + } + ], + "full_mark": 1, + "rubric": [ + "1 or a single bond (1 point)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-17-T5-6", + "tags": [ + "1.6" + ], + "text_open": "5.5.3 Briefly explain why the experimentally measured carbon-carbon bond length in crystalline $DyCoC_2$ is 138 pm.", + "text_images": [], + "full_mark": 2, + "rubric": [ + "Interaction between electrons in the π antibonding orbitals of the dicarbon anion and the d orbitals of $Co^{3+}$ increases the carbon-carbon bond order (2 points)." + ], + "rubric_images": [] + } + ] +} diff --git a/dev/tasks/Exam-17-T6/images/Exam-17-T6-header-1.png b/dev/tasks/Exam-17-T6/images/Exam-17-T6-header-1.png new file mode 100644 index 0000000000000000000000000000000000000000..c86b0cd859cb5f1932ae59eaa6ef512b3f5012e9 --- /dev/null +++ b/dev/tasks/Exam-17-T6/images/Exam-17-T6-header-1.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:18f3dbabdb18b3a848664eff6e5a3ff11e3af57124d93e2ab5f7ea1122a4cda5 +size 1421739 diff --git a/dev/tasks/Exam-17-T6/problem.json b/dev/tasks/Exam-17-T6/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..609aff144f9e8f7fe24475dd883cfba2cc9b9b9f --- /dev/null +++ b/dev/tasks/Exam-17-T6/problem.json @@ -0,0 +1,25 @@ +{ + "problem_id": "Exam-17-T6", + "header": "Problem 6 Synthesis of Triphenylmethanol Using a Grignard Reagent\nGrignard reagents are common nucleophiles in organic synthesis. They must be used under anhydrous conditions free of oxygen and $CO_2$, and reactions are usually performed under nitrogen. In a teaching laboratory, however, the synthesis can be carried out with a simpler procedure. In this problem, bromobenzene is used to prepare phenylmagnesium bromide, which is then reacted with methyl benzoate to produce triphenylmethanol.", + "header_images": [ + { + "file": "Exam-17-T6-header-1.png", + "description": "Experimental apparatus: a three-neck flask is centered, with its left neck closed by a ground-glass stopper. The middle neck holds a funnel-shaped dropping device with a sidearm and stopcock, labeled \"Apparatus 1\". The angled right neck is connected to a spherical condenser sloping upward to the right. Liquid is present in the lower part of the three-neck flask. The labels in the image are Apparatus 1, spherical condenser, three-neck flask, and to drying tube at the top of the condenser." + } + ], + "subquestions": [ + { + "sub_id": "Exam-17-T6-1", + "tags": [ + "2.8" + ], + "text_open": "6.2 Synthesis of triphenylmethanol: cool the flask in a cold-water bath while slowly adding a solution of methyl benzoate in anhydrous diethyl ether with stirring. After addition is complete, heat the mixture under reflux for 1 h, cool the flask again in a cold-water bath, and add saturated ammonium chloride solution dropwise to decompose the addition product. Separate the layers and perform steam distillation to obtain the crude product.\n6.2.1 State the purpose of the saturated ammonium chloride workup, and explain why water or hydrochloric acid is not used directly for the workup.", + "text_images": [], + "full_mark": 3, + "rubric": [ + "Triphenylmethanol is sparingly soluble in water, and its solubility is further reduced in a saturated salt solution. This promotes transfer of the product to the organic phase and increases the yield (1 point; stating that high ionic strength reduces emulsification is also acceptable).\nUsing water directly produces magnesium salts such as $MgBrOH$, which form poorly soluble flocculent precipitates. The weak acidity of ammonium chloride solution helps dissolve or disperse these precipitates, facilitating their subsequent removal by filtration (1 point).\nTriphenylmethanol is unstable under strongly acidic conditions and readily forms the triphenylmethyl cation, leading to side reactions. Ammonium chloride is a weak acid, and its aqueous solution provides an acidic environment mild enough not to destroy the product (1 point)." + ], + "rubric_images": [] + } + ] +} diff --git a/dev/tasks/Exam-19-T6/images/Exam-19-T6-1-1-prompt.png b/dev/tasks/Exam-19-T6/images/Exam-19-T6-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..d5314bdcffde3d3b25f234ab252076a825b7ed12 --- /dev/null +++ b/dev/tasks/Exam-19-T6/images/Exam-19-T6-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:afccc28e4713951a7a012f4500a1fa62fc08e834170252d0310e48b670a411e6 +size 30916 diff --git a/dev/tasks/Exam-19-T6/images/Exam-19-T6-1-2-prompt.png b/dev/tasks/Exam-19-T6/images/Exam-19-T6-1-2-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..08176925507ac442ebdbddd052963ed3f5471983 --- /dev/null +++ b/dev/tasks/Exam-19-T6/images/Exam-19-T6-1-2-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:a629541f901d6048f175cc4ddc0360963dfbfecaf3e17f677ad0914d2841a745 +size 303406 diff --git a/dev/tasks/Exam-19-T6/images/Exam-19-T6-10-1-rubric.png b/dev/tasks/Exam-19-T6/images/Exam-19-T6-10-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..4d5aca825674cef126458a5b8694658dded1b0d0 --- /dev/null +++ b/dev/tasks/Exam-19-T6/images/Exam-19-T6-10-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:06ee38f1fca522a64af2d2e5e56689d0bd5b32cc71bb6902471a6d84a173931e +size 26246 diff --git a/dev/tasks/Exam-19-T6/images/Exam-19-T6-6-1-rubric.png b/dev/tasks/Exam-19-T6/images/Exam-19-T6-6-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..ebfde9629a3b4f751e5631ee837d37e2b17b2f31 --- /dev/null +++ b/dev/tasks/Exam-19-T6/images/Exam-19-T6-6-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:f8f084b7fb2d794866b94c28a02f964f6250d0e450befc343e52d404db14d802 +size 30091 diff --git a/dev/tasks/Exam-19-T6/images/Exam-19-T6-7-1-rubric.png b/dev/tasks/Exam-19-T6/images/Exam-19-T6-7-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..e2b3f8945669183bc61dd9b251144f21287b5b46 --- /dev/null +++ b/dev/tasks/Exam-19-T6/images/Exam-19-T6-7-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:55a118cfd8bc5ce74c9f14dd4952764e4414057b6b14af2d0e6404fd5f59789b +size 38213 diff --git a/dev/tasks/Exam-19-T6/images/Exam-19-T6-8-1-rubric.png b/dev/tasks/Exam-19-T6/images/Exam-19-T6-8-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..1b67c2350b29d0c310c0c94696a0083049a9286d --- /dev/null +++ b/dev/tasks/Exam-19-T6/images/Exam-19-T6-8-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:681f06d61e198368568833222331eb986f91e98c576ff4435b8090653bab2318 +size 235101 diff --git a/dev/tasks/Exam-19-T6/images/Exam-19-T6-9-1-prompt.png b/dev/tasks/Exam-19-T6/images/Exam-19-T6-9-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..89ad35cd77073b022228d79e1ea79c5244c5e903 --- /dev/null +++ b/dev/tasks/Exam-19-T6/images/Exam-19-T6-9-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:6dc5ee0c25a08eee99722978aeea58573b3157b4d2a37b6af901b9f27cf50535 +size 144673 diff --git a/dev/tasks/Exam-19-T6/problem.json b/dev/tasks/Exam-19-T6/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..dbcd1d0598533bf20b33a23082afe460ca1cee80 --- /dev/null +++ b/dev/tasks/Exam-19-T6/problem.json @@ -0,0 +1,199 @@ +{ + "problem_id": "Exam-19-T6", + "header": "Question 6 Boron Oxides\nBoron oxides possess advantageous properties and are used to prepare other boron compounds and to manufacture optical glass, electronic materials, ceramics, flame-retardant materials, and more. New boron-based 2D materials have been explored extensively in recent years. This problem examines the reactivity and structures of several boron oxides.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-19-T6-1", + "tags": [ + "1.3", + "1.1" + ], + "text_open": "6.1 $B_2O_3$ is the most common boron oxide and has some Lewis acidity. Reaction of $B_2O_3$ with $P_2O_5$ gives compound X, a catalyst for dehydration reactions. The figure below is a projection of a conventional unit cell of tetragonal X; the large spheres represent P, the medium spheres B, and the small spheres oxygen.\n6.1.1 Give the chemical formula of X and identify the characteristic symmetry element of the crystal.", + "text_images": [ + { + "file": "Exam-19-T6-1-1-prompt.png", + "description": "The figure is a schematic of crystallographic axes. A small filled gray sphere at the lower left is labeled a. From this point, a thick horizontal arrow extends to the right and is labeled b at its tip, while a thick vertical arrow extends upward and is labeled c at its tip. No other text or numbers appear." + }, + { + "file": "Exam-19-T6-1-2-prompt.png", + "description": "A two-dimensional crystallographic projection of the unit cell. Fractional coordinates x and y are defined horizontally and vertically, with unit-cell boundaries at x=0,1 and y=0,1. Three visually distinct sphere types appear: large gray spheres, small gray spheres, and spheres labeled B or P. Approximate fractional coordinates of the large gray spheres are (0.0,1.0), (1.0,1.0), (0.0,0.75), (1.0,0.75), (0.0,0.0), (1.0,0.0), (0.0,0.25), and (1.0,0.25). Approximate fractional coordinates of the small gray spheres are (0.0,1.15), (0.3,1.15), (0.7,1.15), (1.0,1.15), (0.15,0.85), (0.15,0.65), (0.85,0.85), (0.85,0.65), (0.3,0.25), (0.7,0.25), (0.15,0.15), (0.15,-0.05), (0.85,0.15), and (0.85,-0.05), together with visible periodic images outside the unit-cell boundary. Approximate fractional coordinates of the labeled spheres are B at (0.0,0.75) and (0.5,0.25), and P at (0.5,0.5)." + } + ], + "full_mark": 2, + "rubric": [ + "Chemical formula: $BPO_4$ (1 point).", + "Characteristic symmetry element: an $I_4$ axis (1 point)." + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-19-T6-2", + "tags": [ + "1.3" + ], + "text_open": "6.1.2 How many distinct chemical environments does oxygen have in this crystal?", + "text_images": [], + "full_mark": 2, + "rubric": [ + "1 (2 points)." + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-19-T6-3", + "tags": [ + "1.3", + "2.1" + ], + "text_open": "6.1.3 The B–P distance is 272.9 pm. The B–P–B angle indicated in the figure is 111.7°, and the other B–P–B angle is 105.1°. Assuming that P and B are completely converted to their corresponding oxoacids, calculate the theoretical maximum mass of water that $1\\ \\mathrm{cm^3}$ of this crystal can absorb.", + "text_images": [], + "full_mark": 5, + "rubric": [ + "$a=272.9\\times\\sin\\frac{105.1°}{2}\\times2=433.3\\ \\mathrm{pm}$ (1 point).", + "$c=272.9\\times\\cos\\frac{105.1°}{2}\\times4=663.8\\ \\mathrm{pm}$ (1 point).", + "Complete-hydrolysis equation: $BPO_4+3H_2O\\rightarrow H_3BO_3+H_3PO_4$ (1 point).", + "Counting from the crystal structure gives $Z=2$; therefore, $m=\\frac{Z\\times M_r(H_2O)\\times3}{a^2cN_A}=1.439\\ \\mathrm{g}$ (2 points; 1.43–1.44 g is acceptable)." + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-19-T6-4", + "tags": [ + "1.3" + ], + "text_open": "6.2 The literature has reported formation of a crystalline compound with composition $B_2O$ by reacting B with $B_2O_3$ at high pressure and high temperature. One proposed structure belongs to the trigonal crystal system. It may be viewed as boron-only layers M and boron–oxygen layers N stacked in an …MNNMNN… sequence. Both M and N contain only juxtaposed six-membered rings. Adjacent N layers are connected by B–B covalent bonds, while B–O interactions parallel to the c axis occur between M and N layers, making both boron and oxygen nonplanar. The number of formula units per unit cell satisfies $Z\\leq3$. The shortest B–O distance within an N layer is 174.2 pm, and the shortest B–O distance between M and N layers is 199.6 pm. The various B–B bond lengths in the crystal are very similar and may all be taken as 169.3 pm in this problem.\n6.2.1 How many B–B covalent bonds are present in 1 mol of crystalline $B_2O$?", + "text_images": [], + "full_mark": 2, + "rubric": [ + "$2N_A$ (2 points; 2 mol is also acceptable)." + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-19-T6-5", + "tags": [ + "1.3" + ], + "text_open": "6.2.3 The unit-cell parameter of this crystal is $c = 705.2\\ \\mathrm{pm}$. Calculate the unit-cell parameter $a$.", + "text_images": [], + "full_mark": 5, + "rubric": [ + "Let $x$ be the corrugation height of an N layer and $y$ that of an M layer. Geometry gives $x^2+(\\frac{a}{\\sqrt{3}})^2=d_{B-O,horizontal}^2,\\ y^2+(\\frac{a}{\\sqrt{3}})^2=d_{B-B}^2$ (2 points).", + "$c=2x+y+d_{B-B}+2d_{B-O,vertical}$ (1 point).", + "Solving gives $a=287.9\\ \\mathrm{pm}$ (2 points; 287–289 pm is acceptable)." + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-19-T6-6", + "tags": [ + "1.1" + ], + "text_open": "6.3 BO was reported as early as 1940, but technological limitations long prevented determination of its structure. In 2023, researchers used multiple methods and analytical tools to propose several planar two-dimensional candidate models. BO can be formed by dehydration of compound A. A is synthesized as follows: reacting $BCl_3$ with Cu vapor on a liquid-nitrogen-cooled surface gives compound B (Reaction 1); hydrolysis of B gives A, which yields BO upon dehydration. When B reacts with an equimolar amount of hydrogen, the only products are two simple binary compounds (Reaction 2). B also reacts with cis-2-butene (Reaction 3).\n6.3.1 Compound B exists as a monomer in both the gas and solid phases, but the molecular structures have different symmetries. Write the SMILES strings for both structures; the gas/solid assignment need not be indicated.", + "text_images": [], + "full_mark": 2, + "rubric": [ + "Planar 'ClB(Cl)B(Cl)Cl' and D2d 'ClB(Cl)B(Cl)Cl' (1 point each; either order)." + ], + "rubric_images": [ + { + "file": "Exam-19-T6-6-1-rubric.png", + "description": "The figure shows two structures of compound B, both with the same SMILES 'ClB(Cl)B(Cl)Cl'. They differ in conformation/symmetry about the B–B bond. The left structure shows planar Cl2B–BCl2 using solid lines. In the right structure, the two Cl atoms on the right-hand B are shown with a solid wedge and a dashed bond, respectively, so the two BCl_2 planes are perpendicular. The gas/solid assignments are not labeled." + } + ] + }, + { + "sub_id": "Exam-19-T6-7", + "tags": [ + "2.1", + "2.4" + ], + "text_open": "6.3.2 Write the equations for Reactions 1–3, paying attention to the stereochemistry of Reaction 3.", + "text_images": [], + "full_mark": 6, + "rubric": [ + "Reaction 1: $2BCl_3+2Cu\\rightarrow B_2Cl_4+2CuCl$ (2 points).", + "Reaction 2: $3B_2Cl_4+3H_2\\rightarrow B_2H_6+4BCl_3$ (2 points).", + "Reaction 3: 'ClB(Cl)B(Cl)Cl'.'C/C=C\\C'>'C[C@H](B(Cl)Cl)[C@H](B(Cl)Cl)C' (2 points; only 1 point if stereochemistry is not shown)." + ], + "rubric_images": [ + { + "file": "Exam-19-T6-7-1-rubric.png", + "description": "The figure shows the one-step reaction 'ClB(Cl)B(Cl)Cl'.'C/C=C\\C'>'C[C@H](B(Cl)Cl)[C@H](B(Cl)Cl)C'. The product is meso-2,3-bis(dichloroboryl)butane. The left BCl2 group is shown with a dashed wedge and the right BCl2 group with a solid wedge, placing the two BCl2 groups on opposite sides of the plane." + } + ] + }, + { + "sub_id": "Exam-19-T6-8", + "tags": [ + "1.3", + "1.1" + ], + "text_open": "According to the hypothesis proposed in the 2023 study, both idealized ordered stacking models of BO can be regarded as different stackings of the same infinite planar BO sheet. The AAAA… stacking forms a primitive orthorhombic lattice, whereas ABCD stacking forms a body-centered tetragonal lattice. A single boron–oxygen sheet contains only 6-, 12-, and 16-membered rings; one actual two-dimensional unit cell contains 2, 1, and 1 of these rings, respectively. If the slight distortion of the 12-membered ring is ignored, the sheet has a fourfold axis; in the actual structure, the 12-membered ring has only a twofold axis.\n6.3.3 Write the SMILES string for the boron–oxygen sheet, and state the number of boron atoms in one smallest actual two-dimensional unit cell.", + "text_images": [], + "full_mark": 5, + "rubric": [ + "'[*]B1B(OB2OB3OB4B5OB6OB7OB8B(OB9B(O8)OB(O%10)B(OB%11OB(B(OB%11OB%12OB(B(OB%12O9)[*])[*])[*])[*])OB(OB%13OB(B(OB%13OB2O1)[*])[*])B%10OB3O5)OB%14OB(B(OB%14OB%15OB(B(OB%15OB7OB6OB%16OB(B(OB%16OB%17OB(B(OB%17O4)[*])[*])[*])[*])[*])[*])[*])[*])[*]' (2 points for the correct $[B_4O_2]$ six-membered-ring fragment and 2 points for correctly depicting all three ring sizes; 4 points total).", + "Number of boron atoms per unit cell: 8 (1 point)." + ], + "rubric_images": [ + { + "file": "Exam-19-T6-8-1-rubric.png", + "description": "The figure shows a two-dimensional boron–oxygen network formed by $[B_4O_2]$ six-membered rings, with atom sequence B–B–O–B–B–O, connected through shared B–B bonds. Boron atoms at the network edges carry unlabeled short methyl termination lines. The full network SMILES is 'CB1B(C)OB(OB2OB3B(OB4OB(OB5B(OB6OB(C)B(C)OB6OB7OB(C)B(C)OB7O8)OB9B(O%10)O5)B8OB4OB%11OB(C)B(C)OB%11OB(OB(C)B(C)O%12)B%12O3)OB2OB%13B(OB(OB%14OB(C)B(C)OB%14OB(OB(C)B(C)O%15)B%15O9)B%10O%13)OB%16OB(C)B(C)OB%16O%17)B%17O1' and includes 16 methyl terminations." + } + ] + }, + { + "sub_id": "Exam-19-T6-9", + "tags": [ + "4.3", + "1.3" + ], + "text_open": "6.3.5 To verify that the boron atoms in the crystal are bonded to one another and coplanar, meaning that both angle $\\theta$ and angle $\\psi$ in the right-hand figure are 0°, which experimental technique is most appropriate for a BO powder sample?\n(a) Electron paramagnetic resonance (b) Solid-state nuclear magnetic resonance (c) Electron diffraction (d) Thermogravimetric analysis", + "text_images": [ + { + "file": "Exam-19-T6-9-1-prompt.png", + "description": "The figure shows two boron–oxygen structural fragments. In the upper fragment, a large central circle labeled B is connected by two short horizontal lines to two small circles labeled O on its right, by a diagonal line to one O at upper right, and by two diagonal lines to two O atoms on the left; the angle symbol θ is shown between the two left O–B lines. The lower fragment contains two B atoms joined by a single bond, each additionally bonded to two O atoms: the left B to upper-left and lower-left O, and the right B to upper-right and lower-right O. Arrows above the two B atoms point toward the upper left and upper right, and an arc between them is labeled ψ. The only text and symbols are B, O, θ, and ψ." + } + ], + "full_mark": 1, + "rubric": [ + "(b) (1 point)." + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-19-T6-10", + "tags": [ + "1.3" + ], + "text_open": "In 2025, researchers prepared a $^{17}O$-labeled BO sample and used more precise instrumental analyses to establish that the BO crystal is not a stack of two-dimensional sheets but consists mainly of long one-dimensional chains, thereby disproving the planar two-dimensional model proposed in 2023.\n6.3.6 Write the SMILES string for the one-dimensional BO chain.", + "text_images": [], + "full_mark": 2, + "rubric": [ + "'[*]OB1OB([*])B([*])OB1O[*]' (2 points; terminal groups need not be shown)." + ], + "rubric_images": [ + { + "file": "Exam-19-T6-10-1-rubric.png", + "description": "The figure shows a one-dimensional long-chain BO polymer. Brackets enclose the repeating unit and a subscript n appears at the lower right. The repeating-unit SMILES is '[*]OB1OB([*])B([*])OB1O[*]', where [*] denotes a connection site to an adjacent repeating unit: each of the two left-hand B atoms is directly bonded to one [*], and each of the two right-hand B atoms connects to [*] through an O atom." + } + ] + }, + { + "sub_id": "Exam-19-T6-11", + "tags": [ + "4.3", + "1.3" + ], + "text_open": "6.3.7 Powder X-ray diffraction once suggested that BO has two-dimensional character, apparently contradicting the one-dimensional long-chain structure. Briefly explain this observation.", + "text_images": [], + "full_mark": 2, + "rubric": [ + "The one-dimensional B–O polymer chains exhibit random stacking or stacking faults, so their average structure produces diffraction features resembling an ordered two-dimensional layered structure (2 points: 1 point for random stacking and 1 point for averaging into a plane)." + ], + "rubric_images": [] + } + ] +} diff --git a/dev/tasks/Exam-20-T9/images/Exam-20-T9-1-1-prompt.png b/dev/tasks/Exam-20-T9/images/Exam-20-T9-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..d84c34a8d36a15d825dad1781f75cd29d1091766 --- /dev/null +++ b/dev/tasks/Exam-20-T9/images/Exam-20-T9-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:65b1bff4afb220ac134bfcf97db79ecbc9decfd4c3ed137d1b9ce0ca370088b4 +size 72532 diff --git a/dev/tasks/Exam-20-T9/images/Exam-20-T9-1-1-rubric.png b/dev/tasks/Exam-20-T9/images/Exam-20-T9-1-1-rubric.png new file mode 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+version https://git-lfs.github.com/spec/v1 +oid sha256:be5abe9f729db63d86ff53dcd3d8f09faa404c22a5c43da776e0ca2c76d681e6 +size 16493 diff --git a/dev/tasks/Exam-20-T9/problem.json b/dev/tasks/Exam-20-T9/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..99ca57891c7e96e35711ce8844c576dbf9073121 --- /dev/null +++ b/dev/tasks/Exam-20-T9/problem.json @@ -0,0 +1,203 @@ +{ + "problem_id": "Exam-20-T9", + "header": "Problem 9 Prins Cyclization (25 points, 12%)\nThe Prins cyclization is a powerful synthetic transformation that rapidly constructs tetrahydropyran rings. Tetrahydropyran is a common structural unit in many bioactive molecules, including spongistatins and bryostatins. Researchers have used the Prins cyclization in numerous elegant strategies for the total synthesis of natural-product targets.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-20-T9-1", + "tags": [ + "2.4", + "2.2" + ], + "text_open": "9.1 The following is a classic Prins cyclization that gives the final product with high stereoselectivity, where R = alkyl.\n9.1.1 Write the SMILES strings of A-D in the scheme, including stereochemistry. B is a highly strained intermediate.", + "text_images": [ + { + "file": "Exam-20-T9-1-1-prompt.png", + "description": "Prins cyclization sequence: '[R]C1=CCCCC1'>H2C=O, H2SO4>[A⇌B]; [B]>H2O>[C]; [C]>H2C=O, H2SO4>[D]. In the image, the last arrow points leftward from C to D. A-D are shown only as letter labels and no structures are provided; R is the placeholder used in the original diagram." + } + ], + "full_mark": 5, + "rubric": [ + "A: '[R][C+]1[C@H](CO)CCCC1', additionally labeled (±) in the image; B: '[R][C@@]12[C@@](C[OH+]2)([H])CCCC1', additionally labeled (±); C: 'O[C@@]1([R])[C@H](CO)CCCC1', additionally labeled (±) (1 point each). D: '[R][C@]1(OCOC2)[C@@]2([H])CCCC1', additionally labeled (±) (2 points).", + "For each omitted (±) label, deduct 1 point from the score obtained, up to a maximum deduction of 3 points." + ], + "rubric_images": [ + { + "file": "Exam-20-T9-1-1-rubric.png", + "description": "A has the SMILES '[R][C+]1[C@H](CO)CCCC1'. The image also shows (±), and R is the placeholder used in the original diagram." + }, + { + "file": "Exam-20-T9-1-2-rubric.png", + "description": "B has the SMILES '[R][C@@]12[C@@](C[OH+]2)([H])CCCC1'. The image also shows (±), and R is the placeholder used in the original diagram." + }, + { + "file": "Exam-20-T9-1-3-rubric.png", + "description": "C has the SMILES 'O[C@@]1([R])[C@H](CO)CCCC1'. The image also shows (±), and R is the placeholder used in the original diagram." + }, + { + "file": "Exam-20-T9-1-4-rubric.png", + "description": "D has the SMILES '[R][C@]1(OCOC2)[C@@]2([H])CCCC1'. The image also shows (±), and R is the placeholder used in the original diagram." + } + ] + }, + { + "sub_id": "Exam-20-T9-2", + "tags": [ + "2.2", + "2.4" + ], + "text_open": "9.1.2 The following cascade comprises a Prins cyclization and a rearrangement and ultimately gives a product containing a five-membered ring. Write the SMILES string of the product and of at least 2 charged key intermediates, including stereochemistry.", + "text_images": [ + { + "file": "Exam-20-T9-2-1-prompt.png", + "description": "Single-step reaction: 'C=C(C)[C@H]([C@H](C(C)=C)O)O'>Me2C=O, H2SO4>[product not shown]. Stereochemistry of the substrate's two OH groups is shown by wedge bonds and retained by the @ markers in the SMILES." + } + ], + "full_mark": 4, + "rubric": [ + "Product structure: 'CC1(C)C[C@](C)(C=O)[C@H](C(C)=C)O1' (±) (2 points)", + "Key intermediates: the first two SMILES strings are equivalent representations from different viewpoints. From left to right: 'C=C(C)[C@H]([C@H](C(C)=C)/[O+]=C(C)/C)O'; 'C=C(C)[C@]([C@H](O)C(C)=C)([H])/[O+]=C(C)/C'; 'C[C+]1[C@@H](O)[C@H](C(C)=C)OC(C)(C)C1'; 'CC1(C)C[C@](C)(C([H])=[OH+])[C@H](C(C)=C)O1'. Any 2 are worth 2 points total, 1 point each." + ], + "rubric_images": [ + { + "file": "Exam-20-T9-2-1-rubric.png", + "description": "The product has the SMILES 'CC1(C)C[C@](C)(C=O)[C@H](C(C)=C)O1'. The image also shows (±), and stereochemistry is indicated by solid and dashed wedge bonds." + }, + { + "file": "Exam-20-T9-2-2-rubric.png", + "description": "The image shows 4 candidate key intermediates, of which any 2 are required for scoring. The first two SMILES strings are equivalent representations from different viewpoints. From left to right: 'C=C(C)[C@H]([C@H](C(C)=C)/[O+]=C(C)/C)O'; 'C=C(C)[C@]([C@H](O)C(C)=C)([H])/[O+]=C(C)/C'; 'C[C+]1[C@@H](O)[C@H](C(C)=C)OC(C)(C)C1'; 'CC1(C)C[C@](C)(C([H])=[OH+])[C@H](C(C)=C)O1'." + } + ] + }, + { + "sub_id": "Exam-20-T9-3", + "tags": [ + "1.2", + "2.7" + ], + "text_open": "9.2 A key advantage of the Prins cyclization is that, starting from a single stereocenter and a defined alkene geometry, it can construct up to four stereocenters on a tetrahydropyran ring with high stereoselectivity. In some cases, however, enantiomeric excess (ee) is lost.\n9.2.1 In the Prins cyclization below, the ee of the major product is lower than that of the starting material, and small amounts of two byproducts are detected.\nPredict the configurations of the chiral carbons in the product by assigning the absolute configurations of chiral carbons a and b in the major product using S/R notation.", + "text_images": [ + { + "file": "Exam-20-T9-3-1-prompt.png", + "description": "Single-step reaction: 'O[C@H](CC=C)C1=CC=CC=C1' reacts with PhCH2CH2CHO under BF3·OEt2 and AcOH conditions to give the major product, 'O=C(O[C@H]1C[C@@H](CCC2=CC=CC=C2)O[C@@H](C3=CC=CC=C3)C1)C', byproduct 1, 'O=C(OC1CC(C2=CC=CC=C2)OC(C3=CC=CC=C3)C1)C', and byproduct 2, 'O=C(OC1CC(CCC2=CC=CC=C2)OC(CCC3=CC=CC=C3)C1)C'. Starting material M is labeled 87% ee and the major product 67% ee. Chiral carbons a and b are labeled on the ring of the major product; OAc at a is shown on a solid wedge, while CH2CH2Ph at b is shown on a dashed bond." + } + ], + "full_mark": 2, + "rubric": [ + "Chiral carbon a: S (1 point); chiral carbon b: R (1 point)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-20-T9-4", + "tags": [ + "2.3", + "2.2" + ], + "text_open": "9.2.2 The pathway to the major product can undergo racemization through a rearrangement. Experimentally, the enantiomer M’ of substrate M is detected. If M and $PhCH_2CH_2CHO$ are replaced by structure N below, racemization through this rearrangement cannot occur under the same conditions. In light of the byproduct structures and the analysis above, write the SMILES strings of the reactants and products in each step of the mechanism by which M forms M’, and label the reaction conditions. Electron-flow arrows need not be shown.", + "text_images": [ + { + "file": "Exam-20-T9-4-1-prompt.png", + "description": "Substrate N has the SMILES 'C=CC[C@@H](OC(OC(C)=O)CC1=CC=CC=C1)C2=CC=CC=C2'. N is the structure label in the image." + } + ], + "full_mark": 4, + "rubric": [ + "The image shows the stepwise mechanism by which M forms M’ in two rows, labeled with conditions including PhCH2CH2CHO, -H2O, [3,3]-sigma, and H2O. Structures involved are M, 'O[C@H](CC=C)C1=CC=CC=C1'; the oxonium intermediates 'C=CC[C@@H](/[O+]=C/CCC1=CC=CC=C1)C2=CC=CC=C2' and 'C=CC[C@@H](CCC1=CC=CC=C1)/[O+]=C/C2=CC=CC=C2'; PhCHO, 'O=CC1=CC=CC=C1'; the alcohol product 'O[C@@H](CC=C)CCC1=CC=CC=C1'; and the enantiomer of M, 'O[C@@H](CC=C)C1=CC=CC=C1'. This subquestion is worth 4 points: 2 points for correctly including the [3,3]-sigma step, without needing to name the reaction type; 1 point for correctly forming the small molecule PhCHO; and 1 point for the reaction of PhCHO with the substrate to form the enantiomer. If the mechanism that forms PhCHO is complete, that step may be abbreviated as above." + ], + "rubric_images": [ + { + "file": "Exam-20-T9-4-1-rubric.png", + "description": "The image shows the stepwise mechanism by which M forms M’ in two rows, labeled with conditions including PhCH2CH2CHO, -H2O, [3,3]-sigma, and H2O. Structures involved are M, 'O[C@H](CC=C)C1=CC=CC=C1'; the oxonium intermediates 'C=CC[C@@H](/[O+]=C/CCC1=CC=CC=C1)C2=CC=CC=C2' and 'C=CC[C@@H](CCC1=CC=CC=C1)/[O+]=C/C2=CC=CC=C2'; PhCHO, 'O=CC1=CC=CC=C1'; the alcohol product 'O[C@@H](CC=C)CCC1=CC=CC=C1'; and the enantiomer of M, 'O[C@@H](CC=C)C1=CC=CC=C1'." + } + ] + }, + { + "sub_id": "Exam-20-T9-5", + "tags": [ + "2.3", + "2.7" + ], + "text_open": "9.2.3 Explain why a racemization process analogous to the one above cannot occur when substrate N is used.", + "text_images": [], + "full_mark": 2, + "rubric": [ + "The oxocarbenium intermediate requires $H_2O$ in the system to hydrolyze into a small-molecule alcohol and aldehyde. After the substrate is changed, the system does not produce $H_2O$ (2 points; any equivalent statement that $H_2O$ is required for racemization is acceptable)." + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-20-T9-6", + "tags": [ + "2.2", + "2.4" + ], + "text_open": "9.3 Researchers have developed many broadly useful Prins reactions using unconventional substrates such as alkynols, allenes, and small-ring compounds.\n9.3.1 The reaction below is based on a Prins-type cyclization of a 3,5-diynol with an aldehyde and proceeds through a domino sequence to synthesize a 2,3-dialkylidenetetrahydrofuran framework. The compound shown at the right was detected in the reaction mixture. This compound is not necessarily itself an intermediate and may arise from an intermediate. Based on this information, write the SMILES strings of the key intermediates. Stereochemistry is not required.", + "text_images": [ + { + "file": "Exam-20-T9-6-1-prompt.png", + "description": "Single-step reaction: 'OCCC#CC#CC1=CC=CC=C1'.'O=CC1=CC=CC=C1'>HBF4·OEt2 (3 equiv), MeOH (3 equiv), DCM, rt, 8 h>'O=C(/C=C1OCCC\\1=C/C2=CC=CC=C2)C3=CC=CC=C3'." + }, + { + "file": "Exam-20-T9-6-2-prompt.png", + "description": "The image is labeled 'Detected in the reaction mixture:' and the detected compound has the SMILES 'OCC/C(C(C#CC1=CC=CC=C1)=O)=C\\C2=CC=CC=C2'." + } + ], + "full_mark": 4, + "rubric": [ + "'CO/C(C#CC1=CC=CC=C1)=C2C(C3=CC=CC=C3)[OH+]CC/2'; 'OCC/C(C(C#CC1=CC=CC=C1)=[O+]C)=C\\C2=CC=CC=C2'; 'COC(OCC/1)(C#CC2=CC=CC=C2)C1=C\\C3=CC=CC=C3'; 'COC(C1=CC=CC=C1)=[C]=C(OCC/2)C2=C\\C3=CC=CC=C3'. 4 points total, 1 point for each intermediate highlighted in red. Except for the second red intermediate, reasonable protonated or deprotonated forms of the other intermediates are also acceptable." + ], + "rubric_images": [ + { + "file": "Exam-20-T9-6-1-rubric.png", + "description": "This structure is not highlighted in red and is an alternative reasonable representation of one of the red intermediates. Its SMILES is 'C1(C#CC#CCC/[O+]=C/C2=CC=CC=C2)=CC=CC=C1'." + }, + { + "file": "Exam-20-T9-6-2-rubric.png", + "description": "The red intermediate has the SMILES 'CO/C(C#CC1=CC=CC=C1)=C2C(C3=CC=CC=C3)[OH+]CC/2'." + }, + { + "file": "Exam-20-T9-6-3-rubric.png", + "description": "The red intermediate has the SMILES 'OCC/C(C(C#CC1=CC=CC=C1)=[O+]C)=C\\C2=CC=CC=C2'." + }, + { + "file": "Exam-20-T9-6-4-rubric.png", + "description": "The red intermediate has the SMILES 'COC(OCC/1)(C#CC2=CC=CC=C2)C1=C\\C3=CC=CC=C3'." + }, + { + "file": "Exam-20-T9-6-5-rubric.png", + "description": "The red intermediate has the SMILES 'COC(C1=CC=CC=C1)=[C]=C(OCC/2)C2=C\\C3=CC=CC=C3'." + } + ] + }, + { + "sub_id": "Exam-20-T9-7", + "tags": [ + "2.4", + "2.7" + ], + "text_open": "9.3.2 The following reaction proceeds through a three-membered-ring intermediate and gives the product with high stereoselectivity, governed primarily by stereoelectronic effects. Write the SMILES string of the product, including stereochemistry, and briefly explain the stereoselectivity.", + "text_images": [ + { + "file": "Exam-20-T9-7-1-prompt.png", + "description": "Single-step reaction: 'C=C([Si](C)(C1=CC=CC=C1)C)C[C@@H](O)C'.'[R]C=O'>TMSOTf>[?]. The product structure is not shown and is represented by '?'. R is the placeholder used in the original diagram and was retained after manual confirmation." + } + ], + "full_mark": 4, + "rubric": [ + "'[R][C@@H]1C[C@@]([Si](C)(O[Si](C)(C)C)C)(C2=CC=CC=C2)C[C@H](C)O1' (2 points total: 1 point for the correct framework and 1 point for correct stereochemistry)", + "Explanation: during migration, the $\\sigma^*$ antibonding orbital of the newly forming C-C bond (1 point) overlaps effectively by hyperconjugation with the $\\sigma$ bonding orbital of the axial C-H bond on the adjacent carbon (1 point). This stabilizes the migration transition state and predominantly gives the product with phenyl in the axial position. 2 points total, with 1 point for correctly identifying each orbital participating in hyperconjugation. A reference schematic is shown below." + ], + "rubric_images": [ + { + "file": "Exam-20-T9-7-1-rubric.png", + "description": "The product has the SMILES '[R][C@@H]1C[C@@]([Si](C)(O[Si](C)(C)C)C)(C2=CC=CC=C2)C[C@H](C)O1'. Stereochemistry of R, Ph, SiMe2OTMS, and other groups is indicated by solid and dashed wedge bonds. R is the placeholder used in the original diagram." + }, + { + "file": "Exam-20-T9-7-2-rubric.png", + "description": "Reference schematic for the stereoelectronic effect: a ring structure containing Si, Ph, R, O, and other atoms, with dumbbell-shaped orbital drawings and a curved arrow. The label 'Axial attack' appears below. This is a schematic and requires no SMILES." + } + ] + } + ] +} diff --git a/dev/tasks/Exam-22-T4/images/Exam-22-T4-1-1-prompt.png b/dev/tasks/Exam-22-T4/images/Exam-22-T4-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..a1ef54dc134a7b9c42f6030d10170f9e03df3f4f --- /dev/null +++ b/dev/tasks/Exam-22-T4/images/Exam-22-T4-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:2ee32a8fae353d1b6dcc1001780c83443d89e1c4c4a1c923abc13ef7fc742728 +size 153660 diff --git a/dev/tasks/Exam-22-T4/images/Exam-22-T4-1-1-rubric.png b/dev/tasks/Exam-22-T4/images/Exam-22-T4-1-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..fdb6dacc743c93c6c9db10d491e4114a83928524 --- /dev/null +++ b/dev/tasks/Exam-22-T4/images/Exam-22-T4-1-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:f775c778c3e6bc36a179cde10c484df2bdd074ae99c7ad36a3f073b31aba0395 +size 29812 diff --git a/dev/tasks/Exam-22-T4/images/Exam-22-T4-3-1-prompt.png b/dev/tasks/Exam-22-T4/images/Exam-22-T4-3-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..9b1e2809b2007202912b5ac1c0995b1a7d1f97ab --- /dev/null +++ b/dev/tasks/Exam-22-T4/images/Exam-22-T4-3-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:3f9b22b1fdb5910a55a08e6565158afd99da461cf1316db9aac4f8b296ae279a +size 131588 diff --git a/dev/tasks/Exam-22-T4/images/Exam-22-T4-4-1-prompt.png b/dev/tasks/Exam-22-T4/images/Exam-22-T4-4-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..9bff508719b8a3ef3b44e3e8ba4f9c1ff9a17416 --- /dev/null +++ b/dev/tasks/Exam-22-T4/images/Exam-22-T4-4-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:2037f94ace36b8521eab425d929d234e0573b46d82dcb6cb14bfcc06f4d77adc +size 94044 diff --git a/dev/tasks/Exam-22-T4/images/Exam-22-T4-4-1-rubric.png b/dev/tasks/Exam-22-T4/images/Exam-22-T4-4-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..888c6441dbdecc50d68e888cdb66aa76e80f0d91 --- /dev/null +++ b/dev/tasks/Exam-22-T4/images/Exam-22-T4-4-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:4e9e4535a1df652f59ef9d8f79ca9ac94dfedbfe9deeadbb8f828ed7a017bd19 +size 78493 diff --git a/dev/tasks/Exam-22-T4/images/Exam-22-T4-header-1.png b/dev/tasks/Exam-22-T4/images/Exam-22-T4-header-1.png new file mode 100644 index 0000000000000000000000000000000000000000..16a596ea70f4999af2f18931f9fbdb924d724a82 --- /dev/null +++ b/dev/tasks/Exam-22-T4/images/Exam-22-T4-header-1.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:e4ac23d8ac5f5070be0e94f4caecf34e9330cc17297d441f09b5589acd2bd3dc +size 743974 diff --git a/dev/tasks/Exam-22-T4/problem.json b/dev/tasks/Exam-22-T4/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..26abeaaff55aa3c4599ed2f797976d1054919da3 --- /dev/null +++ b/dev/tasks/Exam-22-T4/problem.json @@ -0,0 +1,93 @@ +{ + "problem_id": "Exam-22-T4", + "header": "Problem 4 Rotaxane Actuators and Mechanochemistry (28 points, 12%)\nIn the 38th Preliminary Round, we encountered the following reaction. Its principle is to apply mechanical force to a molecule so that four four-membered rings open sequentially: [Image: Exam-22-T4-header-1.png]\nMechanical force can do more than break covalent bonds: changing the force direction and mode of molecular deformation can select the reaction pathway. In 2024, Chen and co-workers constructed a rotaxane mechanical actuator using the sliding motion of a pillar[5]arene macrocycle relative to the axle. Mechanical action transmitted by ultrasonic cavitation allows the macrocycle to push mechanically responsive cargo sites along the axle and release the cargo molecules sequentially.", + "header_images": [ + { + "file": "Exam-22-T4-header-1.png", + "description": "On the left is the structure of a polymer whose repeating-unit fragment has SMILES '[*]=CC1C(C2C3C4C=CC4C3C21)C=[*]'; the units are connected to one another through double bonds. On the right is a horizontal arrow pointing right, labeled “ultrasound, THF, 6~9 °C, Ar”." + } + ], + "subquestions": [ + { + "sub_id": "Exam-22-T4-1", + "tags": [ + "2.4", + "2.6" + ], + "text_open": "4.1 Synthesis of a rotaxane requires threading and end-capping. After threading, repeated sequences of dethreading → rethreading → dethreading may occur before end-capping is completed. In this study, the rotaxane structure was constructed by the reaction below. Threading can occur in two orientations, forming products in which the pillar[5]arene macrocycle and the 2-bromo-2-methylpropanoyloxy group on the axle lie on opposite sides (A$_{trans}$) or the same side (A$_{cis}$).\nBased on the reaction information above, write the SMILES string for B.", + "text_images": [ + { + "file": "Exam-22-T4-1-1-prompt.png", + "description": "Rotaxane construction scheme. The reactants are 'COC(C=C(C(OC)=C1)CC2=C(C=C(C(OC)=C2)CC3=C(C=C(C(OC)=C3)CC4=C(C=C5C(OC)=C4)OC)OC)OC)=C1CC6=C(C=C(C(OC)=C6)C5)OCOC(C(Br)(C)C)=O' and 'CC(C)(Br)C(OCCCCCCCCCCCCO)=O'. Under BTBSCl, Et3N, CHCl3, -15 °C, 2 h, the products are [A$_{trans}$] (11%), [A$_{cis}$] (13%), and [B] (69%). The structure of the end-capping reagent BTBSCl, 'ClS(C1=CC(C(F)(F)F)=CC(C(F)(F)F)=C1)(=O)=O', is also shown. The structures of A$_{trans}$, A$_{cis}$, and B are not drawn; only their labels and yields are given." + } + ], + "full_mark": 3, + "rubric": [ + "The SMILES of B is 'CC(C)(Br)C(OCCCCCCCCCCCCOS(C1=CC(C(F)(F)F)=CC(C(F)(F)F)=C1)(=O)=O)=O' (3 points)." + ], + "rubric_images": [ + { + "file": "Exam-22-T4-1-1-rubric.png", + "description": "The SMILES of B is 'CC(C)(Br)C(OCCCCCCCCCCCCOS(C1=CC(C(F)(F)F)=CC(C(F)(F)F)=C1)(=O)=O)=O'." + } + ] + }, + { + "sub_id": "Exam-22-T4-2", + "tags": [ + "1.2", + "1.7" + ], + "text_open": "4.2.3 In addition to the cis/trans stereoisomerism caused by the orientation of the long PMA polymer chains in the rotaxane, the [4+2] adduct fragment in the cargo molecule also exhibits endo/exo stereoisomerism. Using $^1$H NMR, the researchers measured the conversion of the Diels–Alder adduct, i.e., the cargo-release efficiency, for the four molecules and obtained the following data:\n\n| Species | cis-endo | cis-exo | trans-endo | trans-exo |\n| --- | --- | --- | --- | --- |\n| Cargo released/% | 62 | 51 | 71 | 71 |\n\nUnder the experimental conditions, the long PMA polymer chains experience mechanical forces from the extensional flow field generated by ultrasonic cavitation, causing the pillar[5]arene macrocycle to slide along the axle. Based on the data above, determine the force direction on the polymer chain attached to the pillar[5]arene macrocycle, either from the chain terminus toward the pillar[5]arene macrocycle or from the pillar[5]arene macrocycle toward the chain terminus, and briefly explain your answer.", + "text_images": [], + "full_mark": 4, + "rubric": [ + "From the pillar[5]arene macrocycle toward the chain terminus. (2 points) The data show that for both endo and exo adducts, the trans rotaxane has a higher release efficiency than the corresponding cis rotaxane. (1 point) This indicates that pulling the two PMA chains apart in opposite directions, as in the trans configuration, more effectively activates the cargo site. (1 point) Therefore, the force direction is from the pillar[5]arene macrocycle toward the chain terminus." + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-22-T4-3", + "tags": [ + "1.7", + "2.5" + ], + "text_open": "4.3 In addition, a cargo-bearing chain can carry multiple cargo molecules to form an oligomeric cargo chain. The pillar[5]arene macrocycle can then move continuously along the main axle and release cargo from multiple cargo sites in sequence by a “pushing” action from front to back. Researchers developed rotaxane compounds containing 3 cargo units ($n$ = 0 in the figure below) and 5 cargo units ($n$ = 1 in the figure below), as shown:\n4.3.2 The researchers found, however, that increasing the number of cargo molecules loaded into a rotaxane markedly decreased the release efficiency. Rotaxanes carrying 3 or 5 cargo molecules released fewer than 2 cargo molecules on average. For this calculation, the rotaxane composed of the pillar[5]arene macrocycle and axle is treated as one molecule.\n\n| Number of cargo molecules loaded | Average release per site/% | Internal-site release/% | Terminal-site release/% |\n| --- | --- | --- | --- |\n| 1 | 71 | / | / |\n| 3 | 44 | 45 | 42 |\n| 5 | 22 | 22 | 23 |\n\nBased on these data, propose one reasonable explanation for the decrease in release efficiency.", + "text_images": [ + { + "file": "Exam-22-T4-3-1-prompt.png", + "description": "The figure shows a supramolecular assembly. One component is a pillar[5]arene with structure 'COC1=C(CC2=C(O[R])C=C(C3)C(OC)=C2)C=C(OC)C(CC4=C(OC)C=C(CC5=C(OC)C=C(CC6=C(OC)C=C3C(OC)=C6)C(OC)=C5)C(OC)=C4)=C1'. The guest is a polymeric structure with repeating units and has SMILES '[R]OCCCCCCCCCCCCOC(C1=CC=CC(OCCN2N=NC(CCCCOC[C@@]3(O4)C=C[C@@]4(COCCCCC5=CN(CCCCCCOC[C@@]6(O7)C=C[C@@]7(COCCCCCCN8N=NC(CCCCOC[C@@]9%10C=C[C@@](O%10)(OCCCCC%11=CN(CCCCCCOC[C@@]%12(O%13)C=C[C@@]%13(COCCCCCCN%14C=C(CCCCOC[C@@]%15%16C=C[C@H](O%16)C(C(N%17C(C%18=CC=CC=C%18)(C%19=CC=CC=C%19)C%20=CC=CC=C%20)=O)C%15C%17=O)N=N%14)C(C(N%21C(C%22=CC=CC=C%22)(C%23=CC=CC=C%23)C%24=CC=CC=C%24)=O)C%12C%21=O)N=N%11)C(C(N%25C(C%26=CC=CC=C%26)(C%27=CC=CC=C%27)C%28=CC=CC=C%28)=O)C9C%25=O)=C8)C(C(N%29C(C%30=CC=CC=C%30)(C%31=CC=CC=C%31)C%32=CC=CC=C%32)=O)C6C%29=O)N=N5)C(C(N%33C(C%34=CC=CC=C%34)(C%35=CC=CC=C%35)C%36=CC=CC=C%36)=O)C3C%33=O)=C2)=C1)=O'. The repeating-unit fragment has structure 'O=C1N(C(C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4)C(C(C51)[C@](O6)(OCCCCC7=CN(CCCCCCOC[C@@]8(O9)C=C[C@@]9(COCCCCCCN%10C=C(CCCC[*])N=N%10)C(C(N%11C(C%12=CC=CC=C%12)(C%13=CC=CC=C%13)C%14=CC=CC=C%14)=O)C8C%11=O)N=N7)C=C[C@]56CO[*])=O'. The R groups on the pillar[5]arene and guest are polymer chains whose SMILES may be represented as '[*]C(C(C)(C)CC(C(OC)=O)Br)=O', where * is the attachment point to another group. The repeating unit of R is '[*]CC(C(OC)=O)[*]', where * marks connections between repeat units." + } + ], + "full_mark": 4, + "rubric": [ + "The long PMA chains may break under ultrasound and thus fail to drive the pillar[5]arene along the axle. (2 points; any answer involving chain breakage is accepted, such as breakage of the cargo chain preventing some rotaxanes from releasing cargo.) Consequently, some rotaxanes cannot release cargo, whereas others release cargo from nearly all sites. (2 points) No credit is given for stating that the pillar[5]arene macrocycle slips off the axle." + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-22-T4-4", + "tags": [ + "2.4", + "2.5" + ], + "text_open": "4.4 In addition to cargo release driven by a “pushing” action, the researchers designed the molecule below to investigate the effect of a “pulling” action on cargo release.\nUnder ultrasound in aqueous solution, the molecule undergoes not only a retro-[4+2] reaction but also heterolytic C-N bond cleavage to give two additional products.\n4.4.1 Write the SMILES strings for the two products formed by heterolytic C-N bond cleavage, both of which contain long polymer chains.", + "text_images": [ + { + "file": "Exam-22-T4-4-1-prompt.png", + "description": "The molecule used to study the “pulling” action has SMILES 'O=C1N(C(C2=CC=CC=C2)(C3=CC=C(OCC4=CN(CCCO[R])N=N4)C=C3)C5=CC=CC=C5)C([C@@]6([H])[C@]7(COC(C)=O)C=C[C@](O7)(CCCCCC8=CN(N=N8)CCCO[R])[C@]61[H])=O'. [R] is a placeholder for a PMA polymer chain, and both side chains are -CH2CH2CH2-O-[R]. The definition of R is also shown at the left: R = ~C(=O)C(CH3)2CH2CH(Br)(CH2)mCO2Me, where ~ is the connection point and m is a subscript." + } + ], + "full_mark": 4, + "rubric": [ + "(2 points each; the order may be reversed, and drawing the R substituents explicitly is also accepted)" + ], + "rubric_images": [ + { + "file": "Exam-22-T4-4-1-rubric.png", + "description": "The reference answer contains two products shown side by side; their order may be reversed. The left product has SMILES 'O=C1NC([C@@]2([H])[C@]3(COC(C)=O)C=C[C@](O3)(CCCCCC4=CN(N=N4)CCCO[R])[C@]21[H])=O'. The right product has SMILES 'OC(C1=CC=CC=C1)(C2=CC=C(OCC3=CN(CCCO[R])N=N3)C=C2)C4=CC=CC=C4'. Both products contain the [R] placeholder for a PMA chain." + } + ] + } + ] +} diff --git a/dev/tasks/Exam-23-T10/images/Exam-23-T10-1-1-prompt.png b/dev/tasks/Exam-23-T10/images/Exam-23-T10-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..ca285942c16b3213e2ace8648bb93064bee7819f --- /dev/null +++ b/dev/tasks/Exam-23-T10/images/Exam-23-T10-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:40b58d9ca754b1ade822609d4eb1536bbc81216163abc4f94c1bd97205548b63 +size 245964 diff --git a/dev/tasks/Exam-23-T10/images/Exam-23-T10-1-1-rubric.png b/dev/tasks/Exam-23-T10/images/Exam-23-T10-1-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..f9945467473d4e5e75b9afb26f836ab2ef4000d1 --- /dev/null +++ b/dev/tasks/Exam-23-T10/images/Exam-23-T10-1-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:a71466c9085a5b8a162971b1d0914d94ff88f685164a0df7d0e11eb2e50b77e9 +size 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Directly disrupting its aromatic conjugation or cleaving a C–C bond in the ring framework normally requires considerable energy. Dearomatization or ring-opening cleavage of a benzene ring is therefore highly challenging.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-23-T10-1", + "tags": [ + "2.2" + ], + "text_open": "10.1 In 2024, researchers developed a novel benzene-ring-opening strategy that directly converts a range of aryl azide derivatives into valuable multiply conjugated enynes.\n10.1.1 Write the SMILES string for intermediate A formed before $^tBuONa$ is added.", + "text_images": [ + { + "file": "Exam-23-T10-1-1-prompt.png", + "description": "Single-step reaction: 'O=C(C1=CC=CC=C1N=[N+]=[N-])C2=CC=CC=C2'>TosNHNH2 (1.2 equiv), TFA (20 mol%), PhCl (0.2 M); then $^tBuONa$ (1.5 equiv), PhCl (0.1 M), 130°C>'N#C/C=C\\C=C/C#CC1=CC=CC=C1'." + } + ], + "full_mark": 1, + "rubric": [ + "A: '[N-]=[N+]=NC1=CC=CC=C1/C(C2=CC=CC=C2)=N/NS(C3=CC=C(C)C=C3)(=O)=O' (1 point; E/Z geometry of the double bond is not required)" + ], + "rubric_images": [ + { + "file": "Exam-23-T10-1-1-rubric.png", + "description": "The SMILES for A is '[N-]=[N+]=NC1=CC=CC=C1/C(C2=CC=CC=C2)=N/NS(C3=CC=C(C)C=C3)(=O)=O'." + } + ] + }, + { + "sub_id": "Exam-23-T10-2", + "tags": [ + "2.2", + "2.3" + ], + "text_open": "10.1.2 To investigate the mechanism, the researchers performed the following experiments.\nCalculations indicate that after $^tBuONa$ is added, loss of one molecule of gas first produces highly reactive intermediate B. B then reacts with the azide group to form charge-separated tricyclic intermediate C. A concerted reaction of C expels another gas molecule to give the product. Write the SMILES strings for B and C.", + "text_images": [ + { + "file": "Exam-23-T10-2-1-prompt.png", + "description": "Two parallel mechanistic experiments. Upper reaction: the starting material is labeled only A and its structure is not drawn; conditions are TEMPO (2.0 equiv), $^tBuONa$ (1.5 equiv), PhCl, heat, giving 'N#C/C=C\\C=C/C#CC1=CC=CC=C1' in 70% yield. Lower reaction: 'C1(C(C2=CC=CC=C2)=NN=N3)=C3C=CC=C1'>$^tBuONa$ (1.5 equiv), PhCl, heat>'N#C/C=C\\C=C/C#CC1=CC=CC=C1', 0% yield." + } + ], + "full_mark": 3, + "rubric": [ + "B: '[N-]=[N+]=NC1=CC=CC=C1[C]C2=CC=CC=C2' (1 point); C: '[N-]=[N+]1C(C2=CC=CC=C2)=C(C=CC=C3)C3=N1' (2 points). Reasonable resonance forms of B and C also receive credit." + ], + "rubric_images": [ + { + "file": "Exam-23-T10-2-1-rubric.png", + "description": "The SMILES for B is '[N-]=[N+]=NC1=CC=CC=C1[C]C2=CC=CC=C2', where [C] is a carbene carbon shown with two dots in the image. The SMILES for C is '[N-]=[N+]1C(C2=CC=CC=C2)=C(C=CC=C3)C3=N1'." + } + ] + }, + { + "sub_id": "Exam-23-T10-3", + "tags": [ + "2.4" + ], + "text_open": "10.1.3 Considering the reaction sequence above, write the SMILES strings for the products of the following reactions. (Hint: each reaction proceeds through a high-energy intermediate similar to B.)", + "text_images": [ + { + "file": "Exam-23-T10-3-1-prompt.png", + "description": "Two parallel reactions, arranged vertically, with neither product drawn. Upper reaction: 'O=C(C1=CC=CC=C1N=[N+]=[N-])C2=CC=C(C)C=C2'>TosNHNH2 (1.2 equiv), TFA (20 mol%), PhCl; then $^tBuONa$ (1.5 equiv), PhCl, heat>[product not shown]. Lower reaction: 'O=C(C)/C(C(C)=O)=C/C#CC1=CC=CC=C1N=[N+]=[N-]'>PhCl (0.05 M), heat>[product not shown]." + } + ], + "full_mark": 3, + "rubric": [ + "First product: 'CC(C=C1)=CC=C1C#C/C=C\\C=C/C#N' (1 point); second product: 'O=C(C)C1=C(C)OC(C#C/C=C\\C=C/C#N)=C1' (2 points)" + ], + "rubric_images": [ + { + "file": "Exam-23-T10-3-1-rubric.png", + "description": "The SMILES for the left (first) product is 'CC(C=C1)=CC=C1C#C/C=C\\C=C/C#N'; the SMILES for the right (second) product is 'O=C(C)C1=C(C)OC(C#C/C=C\\C=C/C#N)=C1'." + } + ] + }, + { + "sub_id": "Exam-23-T10-4", + "tags": [ + "2.2" + ], + "text_open": "10.2 In 2025, the group of Professor Daniele Leonori reported oxidative cleavage of electron-rich aromatic rings using photoexcited nitroarenes. (λ = 390 nm denotes irradiation with 390 nm light.)\n10.2.1 Under irradiation, the two substrates react to form a five-membered-ring intermediate Int, whose fragmentation gives the product. Write the SMILES string for Int; the aromatic ring attached to the nitro group may be abbreviated as -Ar.", + "text_images": [ + { + "file": "Exam-23-T10-4-1-prompt.png", + "description": "Photooxidative-cleavage reaction: 'COC1=CC=CC2=CC=CC=C21'.'O=[N+](C1=CC=CC=C1)[O-].[R]C'>CH2Cl2 (0.1 M), -65°C, 5h, λ = 390 nm>'O=C(OC)C1=CC=CC=C1/C=C\\C=O'. [R] in substrate 2 is the original R-label placeholder." + } + ], + "full_mark": 1, + "rubric": [ + "Int: '[Ar]N(O1)OC2C=CC3=CC=CC=C3C21OC' (1 point; [Ar] is the original placeholder for the aromatic ring attached to the nitro group)" + ], + "rubric_images": [ + { + "file": "Exam-23-T10-4-1-rubric.png", + "description": "The SMILES for Int is '[Ar]N(O1)OC2C=CC3=CC=CC=C3C21OC', where [Ar] is the original placeholder for the aromatic ring attached to the nitro group. The structure is a naphthalene-fused five-membered O–N–O ring bearing a methoxy substituent on a ring carbon." + } + ] + }, + { + "sub_id": "Exam-23-T10-5", + "tags": [ + "3.9", + "1.6" + ], + "text_open": "10.2.2 The researchers reacted differently substituted nitroarenes with a substrate containing both an aromatic ring and an alkene and found that the substituents affect chemoselectivity. For example, nitroarenes N1 and N2 react with D to give selective oxidation of either the alkene or the aromatic ring.\nStudies of nitroarene excited states show that their lowest triplet state may have either a $^3(n,\\pi^*)$ or $^3(\\pi,\\pi^*)$ electronic configuration. These correspond, respectively, to an n to $\\pi^*$ transition, representable as a dioxygen-diradical species, and a $\\pi, \\pi^*$ transition, involving intramolecular electron transfer from the benzene unit to the nitro group with the unpaired electron delocalized over the entire $\\pi$ system. Based on the electronic properties of the substituents, which state is lower in energy for N1, and which is lower for N2?\n(a) n, $\\pi^*$ state; (b) $\\pi, \\pi^*$ state.", + "text_images": [ + { + "file": "Exam-23-T10-5-1-prompt.png", + "description": "Three compounds, labeled D, N1, and N2 from left to right. The SMILES for N1 is 'O=[N+](C1=CC=C(OC)C(OC)=C1)[O-]'; for N2, 'O=[N+](C1=CC(C(F)(F)F)=CC(C(F)(F)F)=C1)[O-]'; and for D, 'C/C(C)=C\\CCOC1=CC=CC2=C1C=CC=C2'." + } + ], + "full_mark": 2, + "rubric": [ + "(b), (a) (2 points only if both are correct)" + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-23-T10-6", + "tags": [ + "2.7", + "3.9" + ], + "text_open": "10.2.3 A triplet excited state with an n, $\\pi^*$ configuration can undergo cycloaddition directly. A triplet excited state with a $\\pi, \\pi^*$ configuration cannot undergo direct concerted cycloaddition; calculations indicate the following pathway. It first forms a complex with the other substrate, in which dispersion is the principal intermolecular attraction. Partial electron transfer between the substrates then gives the nitro group a partial negative charge and the other substrate a partial positive charge, promoting formation of the first C–O bond. Finally, intersystem crossing and ring closure give the five-membered intermediate. Accordingly, a nitroarene excited state with an n, $\\pi^*$ configuration reacts more readily with ( ), whereas one with a $\\pi, \\pi^*$ configuration reacts more readily with ( ). Explain each choice.\n(a) An electron-rich aromatic ring; (b) an isolated double bond.", + "text_images": [], + "full_mark": 5, + "rubric": [ + "(b), (a) (2 points only if both are correct. Award no explanation points for an incorrect choice; likewise, a correct choice without the corresponding correct explanation receives no explanation credit.)", + "n, $\\pi^*$ configuration: cycloaddition to an aromatic ring would destroy aromaticity and therefore has a high activation energy, so the n, $\\pi^*$ nitroarene excited state reacts more readily with an isolated double bond (1 point for mentioning the difficulty of disrupting aromaticity).", + "$\\pi, \\pi^*$ configuration: an aromatic ring has a larger polarizable $\\pi$ surface and broader contact area than an isolated alkene bond, providing stronger dispersion and favoring complex formation. An electron-rich aromatic ring also transfers an electron to the nitro group more readily than an isolated double bond because its HOMO is higher (2 points: 1 for stronger dispersion and 1 for the charge-transfer complex)." + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-23-T10-7", + "tags": [ + "2.4" + ], + "text_open": "10.2.4 Write the SMILES string for the product formed when D is oxidized by N2 under irradiation.", + "text_images": [], + "full_mark": 2, + "rubric": [ + "Product: 'O=CCCOC1=CC=CC2=CC=CC=C21' (2 points)" + ], + "rubric_images": [ + { + "file": "Exam-23-T10-7-1-rubric.png", + "description": "The SMILES for the product in 10.2.4 is 'O=CCCOC1=CC=CC2=CC=CC=C21'." + } + ] + }, + { + "sub_id": "Exam-23-T10-8", + "tags": [ + "2.2", + "2.3" + ], + "text_open": "10.3 In 2024, Professor Youliang Wang's group achieved thermochemical dearomatization of a benzene ring, constructing a complex ring system in one step.\nExperiments show that when TFA is replaced by deuterated TFA and the reaction is run for 10 h, 99% of the circled hydrogen atom in the product is replaced by deuterium.\n10.3.1 The reaction includes a pericyclic step. Write the SMILES strings for 3–4 key intermediates and name the pericyclic reaction.", + "text_images": [ + { + "file": "Exam-23-T10-8-1-prompt.png", + "description": "Single-step reaction: 'C=CCN(C1=CC=CC=C1)C(C#CC2=CC=CC=C2)=O'>TFA, 50°C, 12h>'O=C(N1C2)C([H])=C(C31C4C5C=CC3C2C4)C6=C5C=CC=C6'. The hydrogen at C([H]) in the product is circled, corresponding to the deuterium-labeling experiment in the prompt." + } + ], + "full_mark": 7, + "rubric": [ + "Intermediate 1: 'O=C1N(CC=C)C2(C=CC=C[CH+]2)C(C3=CC=CC=C3)=C1[H]'; intermediate 2: 'O=C(N1C2)C([H])=C([C@@]31[C@H]4[CH+]C=C[C@@H]3C2C4)C5=CC=CC=C5'; intermediate 3: 'O=C1C([H])=C2C3=CC=C[CH+]C3C4C=CC5C62C4CC5CN16' (2 points each; award credit if connectivity is correct and the charge is represented reasonably. Consider only the first 4 intermediates drawn by the student).", + "An intermediate formed by addition of $H^+$ to the triple bond receives neither credit nor a penalty.", + "Pericyclic reaction: [5+2] cycloaddition (or [4π+2π] cycloaddition) (1 point)" + ], + "rubric_images": [ + { + "file": "Exam-23-T10-8-1-rubric.png", + "description": "Three cationic intermediates, from left to right. Intermediate 1 has SMILES 'O=C1N(CC=C)C2(C=CC=C[CH+]2)C(C3=CC=CC=C3)=C1[H]'; intermediate 2, 'O=C(N1C2)C([H])=C([C@@]31[C@H]4[CH+]C=C[C@@H]3C2C4)C5=CC=CC=C5'; intermediate 3, 'O=C1C([H])=C2C3=CC=C[CH+]C3C4C=CC5C62C4CC5CN16'." + }, + { + "file": "Exam-23-T10-8-2-rubric.png", + "description": "The SMILES for the intermediate formed by addition of $H^+$ to the triple bond is 'C=CCN(C1=CC=CC=C1)C(/C([H])=[C+]/C2=CC=CC=C2)=O'. The positive charge is at [C+]. The rubric specifies that drawing this intermediate earns no credit but incurs no penalty." + } + ] + }, + { + "sub_id": "Exam-23-T10-9", + "tags": [ + "2.4" + ], + "text_open": "10.3.2 The reaction also forms two byproducts. E has a ring system similar to that of the major product, whereas F contains only three rings and no bridged ring. Write the SMILES strings for E and F.", + "text_images": [], + "full_mark": 3, + "rubric": [ + "E: 'O=C(N1C2)C([H])=C([C@@]31[C@H]4C=C[C@@H]5[C@@H]3[C@@H]2C4)C6=C5C=CC=C6' (1 point); F: 'O=C1N(CC=C)C2=CC=CC=C2C(C3=CC=CC=C3)=C1' (2 points)" + ], + "rubric_images": [ + { + "file": "Exam-23-T10-9-1-rubric.png", + "description": "The SMILES for E is 'O=C(N1C2)C([H])=C([C@@]31[C@H]4C=C[C@@H]5[C@@H]3[C@@H]2C4)C6=C5C=CC=C6'." + }, + { + "file": "Exam-23-T10-9-2-rubric.png", + "description": "The SMILES for F is 'O=C1N(CC=C)C2=CC=CC=C2C(C3=CC=CC=C3)=C1'." + } + ] + } + ] +} diff --git a/dev/tasks/Exam-23-T3/images/Exam-23-T3-1-1-prompt.png b/dev/tasks/Exam-23-T3/images/Exam-23-T3-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..3b1c85e73d5edcba1d3bad47fb319ed1a0020769 --- /dev/null +++ b/dev/tasks/Exam-23-T3/images/Exam-23-T3-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:3e9ceec3a7234c9db42d5a9475549bfff7ba2c6a126b42d11c687e02569e31bf +size 228895 diff --git a/dev/tasks/Exam-23-T3/images/Exam-23-T3-1-1-rubric.png b/dev/tasks/Exam-23-T3/images/Exam-23-T3-1-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..ffa3c384e5ca7ce514706980c561766dbc8c1229 --- /dev/null +++ b/dev/tasks/Exam-23-T3/images/Exam-23-T3-1-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:1d8d4e2e34d953d8d1db317492ef80b86814bdb5f8487decaf06d9a48c3ceef7 +size 28333 diff --git a/dev/tasks/Exam-23-T3/images/Exam-23-T3-3-1-rubric.png b/dev/tasks/Exam-23-T3/images/Exam-23-T3-3-1-rubric.png new file mode 100644 index 0000000000000000000000000000000000000000..75461e38f00957cb375f7af387d9af51e2fc940a --- /dev/null +++ b/dev/tasks/Exam-23-T3/images/Exam-23-T3-3-1-rubric.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:dfdf116f6b6ef6130b08a65151ec6d0130db68039940bcdad84ac69a40f2340d +size 44433 diff --git a/dev/tasks/Exam-23-T3/images/Exam-23-T3-header-1.png b/dev/tasks/Exam-23-T3/images/Exam-23-T3-header-1.png new file mode 100644 index 0000000000000000000000000000000000000000..e3951c9bc978b885c55e51074de331b1fd644748 --- /dev/null +++ b/dev/tasks/Exam-23-T3/images/Exam-23-T3-header-1.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:e07ca94c5dae96695c814160034f147660902763264662b7f794a2b6cdc106f0 +size 15909 diff --git a/dev/tasks/Exam-23-T3/problem.json b/dev/tasks/Exam-23-T3/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..cb2e8cacb846025814f3daadcd1b284d0ff59eb4 --- /dev/null +++ b/dev/tasks/Exam-23-T3/problem.json @@ -0,0 +1,68 @@ +{ + "problem_id": "Exam-23-T3", + "header": "Question 3 Reversible-Deactivation Anionic Polymerization (8 points, 4%)\nConventional anionic polymerization of methacrylates, such as ethyl methacrylate shown below, requires highly hazardous organolithium initiators. Because the chain end is extremely reactive, the reaction must be conducted at very low temperature, such as -78 ℃, under rigorously anhydrous and oxygen-free conditions; otherwise severe side reactions terminate the chains.", + "header_images": [ + { + "file": "Exam-23-T3-header-1.png", + "description": "The SMILES representation of ethyl methacrylate (EMA) is 'O=C(OCC)C(C)=C'." + } + ], + "subquestions": [ + { + "sub_id": "Exam-23-T3-1", + "tags": [ + "2.2" + ], + "text_open": "3.1 Chemists recently developed a $CO_2$-mediated reversible-deactivation anionic polymerization (CMAP). The method operates at 90 ℃, uses a stable and safe solid potassium salt as initiator, and produces polymers with a very narrow molecular-weight distribution, or very low dispersity $Đ$, under a $CO_2$ atmosphere. The key conditions are shown below.\n3.1.1 Anionic polymerization initiator A first decarboxylates to form an enolate that initiates polymerization. Write the SMILES string for A.", + "text_images": [ + { + "file": "Exam-23-T3-1-1-prompt.png", + "description": "'O=C(C(C)=C)OCC'>1.0 equiv. A; 18-crown-6 (1.2 equiv.); TEMPO (0.1 equiv.); 1 atm $CO_2$; DMF, 90°C, 6 h>'O=C(C(CC(C(OCC)=O)(C)C(O[K])=O)(C)C)OCC'. The product is poly(ethyl methacrylate), poly-EMA. Its repeating unit is enclosed in brackets with subscript n at the lower right. The listed product SMILES schematically represents an EtO-C(=O)-C(CH3)2- initiating end connected through -CH2- to a dormant-end unit -C(CH3)(CO2Et)(CO2K)." + } + ], + "full_mark": 1, + "rubric": [ + "Structure of A (1 point)." + ], + "rubric_images": [ + { + "file": "Exam-23-T3-1-1-rubric.png", + "description": "The SMILES representation of initiator A is 'O=C(C(C)(C(O[K])=O)C)OCC'." + } + ] + }, + { + "sub_id": "Exam-23-T3-2", + "tags": [ + "2.8" + ], + "text_open": "3.1.2 State the principal chemical role of 18-crown-6 (shown as 18-crown-6 in the reaction scheme) in the system.", + "text_images": [], + "full_mark": 2, + "rubric": [ + "By complexing potassium ions, 18-crown-6 increases the separation within the ion pair. This greatly lowers the activation energy for decarboxylation and promotes chain initiation (1 point), while also increasing the nucleophilicity of the resulting carbanion toward the monomer (1 point).", + "1 point for promoting decarboxylation and 1 point for increasing carbanion nucleophilicity." + ], + "rubric_images": [] + }, + { + "sub_id": "Exam-23-T3-3", + "tags": [ + "2.4", + "2.3" + ], + "text_open": "3.2 Control of side reactions is also an important topic in anionic polymerization.\n3.2.1 In conventional high-temperature anionic polymerization without $CO_2$ protection, the active chain end of a polymethacrylate readily attacks an upstream carbonyl group in the polymer backbone. This intramolecular backbiting completely terminates the chain. Write the SMILES string for the final cyclic end group formed after backbiting termination of active poly(ethyl methacrylate), poly-EMA; only one chain end need be shown.", + "text_images": [], + "full_mark": 2, + "rubric": [ + "2 points for the right-hand end-group structure. Errors in the polymer chain or left-hand termination are not penalized." + ], + "rubric_images": [ + { + "file": "Exam-23-T3-3-1-rubric.png", + "description": "The figure shows a polymer whose repeating-unit SMILES is '[*]C(C)(C(OCC)=O)C[*]', where * denotes a connection between structural units. One polymer end group is methyl; the other is 'O=C(OCC)C1(C)CC(C(OCC)=O)(C)C(C(C[*])(C)C1)=O', where * denotes its connection to the repeating unit." + } + ] + } + ] +} diff --git a/test/dataset.jsonl b/test/dataset.jsonl index 40770111a8ac597548d2874135e8e40879145371..89a5a4bee9f7aba28b47de0357b5dbcd7bfafc3a 100644 --- a/test/dataset.jsonl +++ b/test/dataset.jsonl @@ -1,105 +1,105 @@ +{"problem_id": "Exam-01-T1", "header": "Divine Gold!", "header_images": [], "subquestions": [{"sub_id": "Exam-01-T1-1", "tags": ["2.1", "1.4", "2.4"], "text_open": "1-2-2 Anhydrous AuCl_3 reacts with a small amount of CO to produce “AuCl_2,” which exists as a tetramer. The molecule possesses only a C_2 axis and a mirror plane perpendicular to the C_2 axis. Write the chemical equation for the reaction and the SMILES structure of tetrameric AuCl_2.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-01-T1-2", "tags": ["1.4", "2.4"], "text_open": "1-3 Au can form cluster compounds with unusual frameworks. Based on the descriptions, write reasonable SMILES structures for the following complexes. 1-3-1 Complex Au_4(dppm)_3I_2: the numbers of Au atoms in two chemical environments are in a ratio of 1:3; every Au is coordinated by one I atom; and all dppm ligands have the same chemical environment. (dppm=Ph_2PCH_2PPh_2)", "text_images": [], "full_mark": 2}]} {"problem_id": "Exam-01-T2", "header": "Determination of Ionic Constituents\nDuring the preparation of K_3[Al(C_2O_4)_3]·3H_2O crystals, some Cr^3+ ions were introduced to replace Al^3+. The following titration experiments were designed to determine the Al and Cr contents of the crystal sample. Titration experiment I: dissolve a 0.5062 g crystal sample in water and transfer it to a 100 mL conical flask. Add a buffer to adjust the pH, then add a solution containing 2.024×10^-3 mol of EDTA. Heat at a gentle boil for 10 min, cool, and add indicator A. Titrate to the endpoint with 0.09142 mol/L standard Zn^2+ solution; 10.40 mL of the standard Zn^2+ solution is consumed. Titration experiment II: dissolve a separate 2.5015 g crystal sample in water and transfer it to a 100 mL conical flask. Acidify with sulfuric acid, add excess (NH_4)_2S_2O_8 and a catalytic amount of Ag^+, then heat at boiling for 10 min. Cool, add indicator B, and titrate with 0.1515 mol/L standard Fe^2+ solution; 17.94 mL of the standard Fe^2+ solution is consumed. Note: assume that the two samples have the same composition and that impurities do not participate in the titration reactions. Report calculated results to 3 significant figures.", "header_images": [], "subquestions": [{"sub_id": "Exam-01-T2-1", "tags": ["2.1", "4.5", "4.2"], "text_open": "Write the ionic equations for the principal reactions involved after acidification with sulfuric acid in titration experiment II.", "text_images": [], "full_mark": 8}, {"sub_id": "Exam-01-T2-2", "tags": ["4.5", "4.7", "2.1"], "text_open": "Calculate the mass fractions of Al and Cr in the sample.", "text_images": [], "full_mark": 6}]} -{"problem_id": "Exam-01-T3", "header": "Lithium Redox-Flow Battery\nA lithium redox-flow battery based on redox-targeting reactions is shown below. Redox targeting links an active material to an electrode by introducing into the electrolyte a highly soluble redox mediator paired with that active material. This approach both circumvents the solubility limitation of the active material and allows low-solubility active materials to be stored in tanks.", "header_images": [{"file": "Exam-01-T3-header-1.png", "description": "Schematic of an electrochemical flow apparatus. At the center is a battery stack composed of multiple plate-like layers, with blue outer plates and black and green layers between them. The left blue plate is labeled “+” and the right blue plate “-”. A wire at the top connects to a glowing light bulb; \\(e\\) is labeled above the wire, and an arrow shows electron transfer from the blue plate labeled “-” to the blue plate labeled “+”. To the left of the stack is a cylindrical vessel containing a pale-blue liquid and a cylindrical material labeled \\(\\mathrm{LiFePO_4}\\). \\(\\mathrm{Fe(CN)_6^{4-}}\\) is labeled above it and \\(\\mathrm{Fe(CN)_6^{3-}}\\) below it. Blue tubing connects the left vessel to the central stack and forms a circulation loop through a bottom pump labeled \\(\\mathrm{Pump}\\). The circulation direction is: blue plate labeled “+” → lower blue tubing → \\(\\mathrm{Pump}\\) → bottom of the left vessel (\\(\\mathrm{Fe(CN)_6^{3-}}\\)) → \\(\\mathrm{LiFePO_4}\\) → upper part of the left vessel (\\(\\mathrm{Fe(CN)_6^{4-}}\\)) → upper blue tubing → blue plate on the side labeled “+”. To the right of the stack is a cylindrical vessel containing a pale-yellow liquid and a cylindrical material labeled \\(\\mathrm{LiTi_2(PO_4)_3}\\). \\(\\mathrm{S_2^{2-}}\\) is labeled above it and \\(\\mathrm{S^{2-}}\\) below it. Yellow tubing connects the right vessel to the central stack and forms a circulation loop through a bottom pump labeled \\(\\mathrm{Pump}\\). The circulation direction is: blue plate labeled “-” → lower yellow tubing → \\(\\mathrm{Pump}\\) → bottom of the right vessel (\\(\\mathrm{S^{2-}}\\)) → \\(\\mathrm{LiTi_2(PO_4)_3}\\) → upper part of the right vessel (\\(\\mathrm{S_2^{2-}}\\)) → upper yellow tubing → blue plate labeled “-”."}], "subquestions": [{"sub_id": "Exam-01-T3-1", "tags": ["2.1", "3.5"], "text_open": "3-2 Write the electrode reactions at the cathode and anode during charging.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-01-T3-2", "tags": ["2.1", "3.5"], "text_open": "3-3 Write the reactions occurring in the catholyte and anolyte tanks during charging.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-01-T3-3", "tags": ["3.5", "3.6"], "text_open": "3-4 This lithium redox-flow battery operates at a constant current of 2.0 A for 24 minutes, during which the mass of solid material in the catholyte tank changes by 1.52 g. Calculate the efficiency with which the battery converts chemical energy into electrical energy.", "text_images": [], "full_mark": 4}]} {"problem_id": "Exam-01-T4", "header": "Preparation of a Dinuclear Ti Complex\nIn benzene solution, 10.5 mmol of TiBr_4 is mixed with 5 mmol of B_2(NMe_2)_4. A dark-red solution forms immediately, followed by a brown precipitate. The reaction mixture is stirred vigorously at room temperature for 5 h, after which the volatile components are removed under vacuum and passed successively through cold traps at -22 ℃ and -196 ℃. The -196 ℃ trap contains the benzene solvent, whereas the substance collected in the -22 ℃ trap is identified as B(NMe_2)_3. The remaining liquid is removed by distillation, and the insoluble solid is washed with DCM to give a sample whose principal component is X. A 219.7 mg portion of the sample is treated with 25 % sulfuric acid solution, immediately producing a dark-purple solution with no gas evolution at room temperature. Addition of sufficient NaOH solution, from which dissolved air has been removed, to this dark-purple solution produces a blue-black precipitate that gradually evolves gas and is ultimately converted into a white precipitate. On the basis of IR, 1H-NMR, magnetic-moment, and other studies, a structure is proposed for X. In this structure, only one type of coordinating atom and two types of ligands surround Ti. The B-containing ligand, denoted L, is bidentate, and the three atoms bonded to each B atom are all different. The relative molecular mass of the molecule is 1114.3 g/mol.", "header_images": [], "subquestions": [{"sub_id": "Exam-01-T4-1", "tags": ["1.4", "2.1"], "text_open": "4-1 Determine the molecular formula of X by calculation and analysis.", "text_images": [], "full_mark": 6}, {"sub_id": "Exam-01-T4-2", "tags": ["1.4", "2.4"], "text_open": "4-2 Using the information above, write the SMILES structure of X.", "text_images": [], "full_mark": 2}]} -{"problem_id": "Exam-01-T8", "header": "Fundamental Concepts and Reactions of Organic Compounds\nFor Problems 8-1 through 8-5, make the requested judgment and provide a reasonable explanation. For Problems 8-6 and 8-7, answer according to the stated conditions and requirements.", "header_images": [], "subquestions": [{"sub_id": "Exam-01-T8-1", "tags": ["2.3"], "text_open": "Does Dewar benzene, whose structure is shown below, convert to benzene under irradiation or heating? Limit your answer to no more than 50 words.", "text_images": [{"file": "Exam-01-T8-1-1-prompt.png", "description": "Structure of Dewar benzene: 'C12C=CC1C=C2'"}], "full_mark": 3}, {"sub_id": "Exam-01-T8-2", "tags": ["1.2", "2.7"], "text_open": "Which compound is oxidized more rapidly by periodic acid, cis-1,2-cyclohexanediol or trans-1,2-cyclohexanediol? Limit your answer to no more than 50 words.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-01-T8-3", "tags": ["1.5", "1.6"], "text_open": "For the carbene shown below, is the singlet or triplet state more stable? Limit your answer to no more than 100 words.", "text_images": [{"file": "Exam-01-T8-3-1-prompt.png", "description": "Structure of the carbene in the question: '[C@@H]12CCC([C]2)C=C1'"}], "full_mark": 3}, {"sub_id": "Exam-01-T8-4", "tags": ["2.3", "2.4"], "text_open": "The compound shown below has an intense floral fragrance and is a useful perfume ingredient, but under basic conditions it is converted into an odorless isomer. Write the SMILES structure of this isomer.", "text_images": [{"file": "Exam-01-T8-4-1-prompt.png", "description": "Structure of the fragrance compound in the question: 'CC(C)(O)[C@H]1CC[C@@H](C=O)CC1'"}], "full_mark": 2}]} -{"problem_id": "Exam-01-T9", "header": "How Well Do You Really Know Carbon-Phosphorus Double Bonds?\nIn studying organic chemistry, we seem to restrict our understanding of the reactivity of carbon-phosphorus double bonds to their “metathesis” reactions with carbon-oxygen double bonds. With the assistance of transition metals, however, double bonds between carbon and trivalent phosphorus display distinctive modes of formation and reactivity. In this problem, whenever a subquestion asks for the SMILES of a complex, supplement it with a verbal description of the specific coordination mode.", "header_images": [], "subquestions": [{"sub_id": "Exam-01-T9-1", "tags": ["1.4", "2.2", "2.4"], "text_open": "9-1 Carbon-phosphorus double bonds can be constructed in many ways and can coordinate to transition metals in more than one mode. 9-1-1 In the reaction below, [Mn] denotes (η^{5}-C_{5}H_{4}CH_{3})Mn(CO)_{2}, and the reaction produces 4 molecules of THF. Given that the major product contains 2 phosphorus-oxygen single bonds, write the SMILES structure of the major product. 4 [Mn]·THF + 2 ^{t}BuCP + H_{2}O →", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-01-T9-2", "tags": ["1.4", "2.2", "2.3"], "text_open": "9-1-2 The following example provides an initial illustration of the role played by transition metals in forming carbon-phosphorus double bonds. Use SMILES structures and any necessary verbal descriptions to represent the three key electrically neutral intermediates in the reaction. Coordination of double bonds to Mo need not be shown in any of them.", "text_images": [{"file": "Exam-01-T9-2-1-prompt.png", "description": "The reaction in the figure shows the anionic dinuclear molybdenum complex 'O=C[Mo](C=O)[Mo](PC1=CC=CC=C1)(C=O)C=O' reacting with acryloyl chloride, 'ClC(C=C)=O', to form a product in which the phosphaalkene ligand 'C/C=P\\C2=CC=CC=C2' is bound to the dinuclear molybdenum framework 'O=C[Mo](C=O)[Mo](C=O)C=O'. In the reactant, a phenylphosphido group bridges two Mo atoms, each bearing two carbonyl ligands; on one side, a coordinate bond shown by an arrow points to the left Mo atom. After reaction, the newly formed phosphaalkene ligand coordinates in a bidentate bridging mode: the lone pair on phosphorus forms a coordinate bond, shown by an arrow, to the left Mo atom, while the $P=C$ double bond coordinates through a $\\pi$ bond, shown by an arrow, to the right Mo atom."}], "full_mark": 3}, {"sub_id": "Exam-01-T9-3", "tags": ["1.4", "2.2", "2.6"], "text_open": "9-1-3 Write the SMILES structures of the six key intermediates in the reaction shown below. Note: the step involving addition of tert-butyllithium corresponds to three key intermediates; each other step corresponds to one key intermediate.", "text_images": [{"file": "Exam-01-T9-3-1-prompt.png", "description": "Description of the structures/reaction in the figure: '[H]C#CC1=CC=CC=C1' undergoes steps involving NEt_3/PCl_3, ArLi, tBuLi, ML_n, and irradiation/O_2 to form a species whose ligand has the SMILES 'O=C(C(/C(C(C1=CC=CC=C1)=O)=P/C)=P\\C)C2=CC=CC=C2'. The ligand coordinates to [ML_n] through its two P atoms. Ar denotes an aryl substituent, and [ML_n] denotes a metal coordination fragment."}], "full_mark": 6}, {"sub_id": "Exam-01-T9-4", "tags": ["2.2", "2.3", "1.4"], "text_open": "9-2 An important class of reactions of carbon-phosphorus double bonds involves metal carbenes. 9-2-1 In the following reaction, addition between the two substrates gives A_{1}, and A_{1} then converts into its isomer A_{2}. A_{2} undergoes a hydrogen migration to give A_{3}; A_{3} eliminates HNMe_{2} to give A_{4}; and a further elimination gives A_{5} together with byproduct A_{6}. A_{5} undergoes a simple reaction to afford the final product. W remains six-coordinate throughout the process. In A_{2}, W is coordinated by the carbon-phosphorus double bond, whereas in A_{3}, A_{4}, and A_{6}, W is coordinated by P. On this basis, write the SMILES structures of A_{1}~A_{6}.", "text_images": [{"file": "Exam-01-T9-4-1-prompt.png", "description": "Description of the structures/reaction in the figure: 'CN(C)C(=P)N(C)C' reacts with 'C1CC1C(O[Si](C)(C)C)=[W]([C]#O)([C]#O)([C]#O)([C]#O)[C]#O' to form 'O=C1C2(CC2)C(=O)C13CC3'."}], "full_mark": 6}, {"sub_id": "Exam-01-T9-5", "tags": ["2.2", "2.3", "1.4"], "text_open": "9-2-2 When the phosphorus atom no longer bears a hydrogen atom, a hydrogen migration analogous to the process above cannot occur. In the following reaction, the two substrates undergo a process similar to that in the preceding subquestion, proceeding through B_{1} to B_{2}. Write the SMILES structure of one key electrically neutral intermediate in the conversion of B_{2} to the products.", "text_images": [{"file": "Exam-01-T9-5-1-prompt.png", "description": "Description of the structures/reaction in the figure: 'CN(C)C(=P[Fe])N(C)C' + 'CCOC(=[M])C' -> 'CN(C)/C=C/C(=[M])OCC' + 'CCOC(C)=P([Fe])[M]. Here M denotes a metal. The ligands on Fe and M are omitted in this question.'"}], "full_mark": 3}]} +{"problem_id": "Exam-01-T6", "header": "Crystal Structures of Layered Alloys", "header_images": [], "subquestions": [{"sub_id": "Exam-01-T6-1", "tags": ["1.3", "1.4"], "text_open": "Structural background: Crystal X is a layered Al-Ni-U alloy in which Ni and U form layer M and Al forms layer N, with an infinite $\\cdots MN\\cdots$ sequence. In M, Ni has a graphene-like arrangement and U occupies the center of every Ni hexagon. Along $c$, all M-layer projections coincide and all N-layer projections coincide. In N, Al forms regular hexagons and equilateral triangles; Ni lies at every Al-triangle center and U at every Al-hexagon center.\nConsidering the coordination of Ni by Al and U, name the coordination polyhedron of Ni and state how many coordinating atoms are U and how many are Al.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-01-T6-2", "tags": ["1.3"], "text_open": "State the crystal system of crystal X and write its chemical formula.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-01-T6-3", "tags": ["1.1", "1.3"], "text_open": "Crystal Y is an Al-Rh-U alloy in which all U atoms and some Rh atoms form layer P, while all Al atoms and the remaining Rh atoms form layer Q. The crystal extends infinitely in a …PQ… sequence to form a three-dimensional structure. The arrangement of U in layer P resembles that of Al in layer N of crystal X, while Rh occupies the center of every hexagon. Of the six U atoms in each hexagon, three are closer to the central Rh atom, causing distortion of layer P; correspondingly, the other three U atoms are farther from Rh, such that all triangles in layer P remain equilateral. Each U atom is shared by two hexagons, is closer to the central Rh atom of one and farther from that of the other, and lies on the line connecting those two Rh atoms. In layer Q, all Al atoms are arranged as equilateral Al_3 triangles. In projection along the c axis, the projections of all P layers coincide, as do those of all Q layers. The centers of the Al_3 triangles coincide with the Rh atoms in layer P, while the centers of the U_3 triangles coincide with the Rh atoms in layer Q. The projection of each Al atom lies on the line connecting a Rh atom in layer P and the more distant U atom in its hexagon. Define a unit cell with the Rh atoms in layer P as vertices. In this unit cell, one U atom has coordinates (0.5806, 0.0000, 0.5000). Give the coordinates of all other U atoms in the unit cell, to four decimal places, and the two types of U-U-U bond angles in a U_6 hexagon, to one decimal place.", "text_images": [], "full_mark": 6}, {"sub_id": "Exam-01-T6-4", "tags": ["1.3"], "text_open": "State the crystal system of crystal Y and write its chemical formula.", "text_images": [], "full_mark": 3}]} +{"problem_id": "Exam-01-T7", "header": "Efficiency of Oxidative Phosphorylation", "header_images": [], "subquestions": [{"sub_id": "Exam-01-T7-1", "tags": ["3.2"], "text_open": "A molecular-level hypothesis has been proposed: for every $2e^-$ transferred through the electron-transfer chain, Complex III transports $2HSuc^-$ across the membrane. The opposing concentration gradients that must be overcome in the two transport events are approximately $1:7$ and $1:8$, respectively. Energy losses from other ion-transport processes at Complexes I and IV may be neglected. The inner-side condition is $pH=8$, the outer-side condition is $pH=5$, succinic acid has $pK_{a1,2}=4.2,5.6$, and the temperature is $310\\,\\mathrm{K}$.\nFor this redox reaction, $\\Delta_rG_m^{\\ominus *}\\approx-2.20\\times10^2\\ \\mathrm{kJ\\,mol^{-1}}$; take the magnitude of the available energy as $220\\ \\mathrm{kJ\\,mol^{-1}}$.\nATP synthase can transport only $HSuc^-$ ions on the outer side of the membrane; neither $H_2Suc$ nor $Suc^{2-}$ outside the membrane can pass through ATP synthase. Thus, of the energy stored in $H_2Suc$, $HSuc^-$, and $Suc^{2-}$ during the redox stage, only the fraction associated with external $HSuc^-$ can be utilized. On this basis, calculate $\\eta_{redox}$.", "text_images": [], "full_mark": 3}]} {"problem_id": "Exam-01-T10", "header": "Light", "header_images": [], "subquestions": [{"sub_id": "Exam-01-T10-1", "tags": ["2.2", "2.4"], "text_open": "10-1 Light can excite certain chemical bonds. Write the SMILES structures of [C] and D. Stereochemistry is not required.", "text_images": [{"file": "Exam-01-T10-1-1-prompt.png", "description": "In the figure, 'O=C(C1=CC=CC=C1C)C2=CC=CC=C2' forms [A] under hv; [A] is converted to [B]; [B] is converted to [C]; and [C] reacts with 'O=C(/C=C/C(OCC)=O)OCC' to form D. A, B, and C are enclosed in square brackets to indicate highly reactive intermediates."}], "full_mark": 8}, {"sub_id": "Exam-01-T10-2", "tags": ["2.2", "2.3", "2.4"], "text_open": "10-3 The figure above is a simplified diagram of a photoredox catalytic mechanism. Such reactions use a photocatalyst capable of both single-electron oxidation and reduction under visible-light irradiation to initiate radical reactions and, after a series of complex steps, quench the radical process. 10-3-1 Based on the information above, write the SMILES structures of the 3 key intermediates in the reaction shown and of the other product in addition to the product depicted. PMP denotes p-methoxyphenyl.", "text_images": [{"file": "Exam-01-T10-2-1-prompt.png", "description": "Schematic of a photocatalyst quenching cycle. “P-Cat” is at the bottom center and “P-Cat*” at the top, with an upward vertical arrow between them labeled “hν.” Blue circular arrows connect “P-Cat*,” “P-Cat^+•,” “P-Cat^-•,” and “P-Cat”: the left blue arrow points downward from P-Cat* to P-Cat^+•, and the lower-left blue arrow points down and right from P-Cat^+• to P-Cat; the right blue arrow points downward from P-Cat* to P-Cat^-•, and the lower-right blue arrow points down and left from P-Cat^-• to P-Cat. “Oxidative Quenching” is labeled at the upper left, near “A” and “A^-•,” with a curved red arrow from A to A^-•. At the lower left are “D” and “D^+•,” with a curved red arrow from D to D^+•. “Reductive Quenching” is labeled at the upper right, near “D” and “D^+•,” with a curved red arrow from D to D^+•. At the lower right are “A” and “A^-•,” with a curved red arrow from A to A^-•."}, {"file": "Exam-01-T10-2-2-prompt.png", "description": "'CC1=CC=C(S(N=CC2=CC=CC=C2)(=O)=O)C=C1' and 'C[Si](COC1=CC=C(OC)C=C1)(C)C' react under 10 W blue LEDs, P-Cat (0.5 mol %), MeOH (Solv.), rt to form 'COC(C=C1)=CC=C1OCC(NS(C2=CC=C(C)C=C2)(=O)=O)C3=CC=CC=C3'."}], "full_mark": 6}, {"sub_id": "Exam-01-T10-3", "tags": ["2.2", "2.4", "2.5"], "text_open": "10-3-2 Under these conditions, the product can react further to remove the PMP protecting group and reveal an alcoholic hydroxy group. Write the SMILES structures of the 2 key intermediates in this reaction.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-01-T10-4", "tags": ["2.2", "2.3"], "text_open": "10-3-3 The photosensitizer P-Cat can capture O_2 from air. Write all species that may be formed.", "text_images": [], "full_mark": 6}]} +{"problem_id": "Exam-02-T1", "header": "Problem 1: Carbon Coordination! Since Frankland reported Et_2Zn in 1849, molecules containing carbon-metal bonds have been known and have continually attracted intensive study. Carbon atoms have been found to exhibit a wide variety of electron-donation numbers and coordination numbers when coordinated to metals.", "header_images": [], "subquestions": [{"sub_id": "Exam-02-T1-1", "tags": ["1.4", "1.6"], "text_open": "1-1 For the ligands Ph_3PCPPh_3 (denoted L) and RCR (where R is an alkyl group), state the coordination mode by which each preferentially coordinates to a transition-metal atom.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-02-T1-2", "tags": ["1.1", "1.4"], "text_open": "1-2 L reacts with ZnI_2 or CdI_2 in a 1:1 ratio, but the products have different structures. The ZnI_2 complex has 2 mutually perpendicular mirror planes and 1 twofold rotation axis but no inversion center. The CdI_2 complex has only 1 mirror plane and 1 twofold rotation axis but does possess an inversion center. Draw both structures.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-02-T1-3", "tags": ["2.1", "2.4", "1.4"], "text_open": "1-5 As stated in the introduction, carbon atoms can donate different numbers of electrons in complexes. Reaction of WCl_6 with (CH_3)_3CCH_2Li and Ph_2PCH_2CH_2PPh_2 gives a mononuclear W complex containing 3 carbon-tungsten bonds whose lengths are all distinctly different. Write the equation for its preparation.", "text_images": [], "full_mark": 3}]} +{"problem_id": "Exam-02-T3", "header": "Problem 3: Bi-Ion Removal Battery. A bi-ion removal battery (BDI) can intercalate cations and anions from seawater into crystalline phases analogous to those in lithium-ion batteries and store them through conversion reactions in the cathode and anode, enabling efficient seawater desalination with high ion-removal capacity and high charge efficiency. Studies show that a $\\mathrm{NaBi_3O_4Cl_2}$ electrode can simultaneously remove $\\mathrm{Na^+}$ and $\\mathrm{Cl^-}$ from seawater.", "header_images": [{"file": "Exam-02-T3-header-1.png", "description": "On the left is a stacked structure of red blocks. Following the process indicated by the blue arrow “$\\mathrm{-Na^+, -Cl^-}$,” it becomes the central stacked structure of red and orange blocks. The process indicated by the red arrow “$\\mathrm{+Na^+, +Cl^-}$” then converts it to the structure on the right, composed of red and blue blocks. A further process indicated by the blue arrow “$\\mathrm{-Na^+, -Cl^-}$” returns it to the central structure. The legend identifies red blocks as $\\mathrm{NaBi_3O_4Cl_2}$, blue blocks as substance X, and orange blocks as $\\mathrm{Bi}$."}], "subquestions": [{"sub_id": "Exam-02-T3-1", "tags": ["2.1", "3.5"], "text_open": "Substance X is $BiOCl$.\n3-2-3 Write the electrode-reaction equations occurring at the $\\mathrm{NaBi_3O_4Cl_2}$ electrode during charging.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-02-T3-2", "tags": ["3.5"], "text_open": "3-3 To construct a complete battery, $\\mathrm{NaTi_2(PO_4)_3@C}$ is used as the electrode on the other side. 3-3-1 State the type of the $\\mathrm{NaTi_2(PO_4)_3@C}$ electrode during seawater desalination.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-02-T3-3", "tags": ["2.1", "3.5"], "text_open": "3-3-2 Write the electrode-reaction equation occurring at the $\\mathrm{NaTi_2(PO_4)_3@C}$ electrode during seawater desalination.", "text_images": [], "full_mark": 2}]} {"problem_id": "Exam-02-T8", "header": "Problem 8: Familiar Named Reactions (14 points)", "header_images": [], "subquestions": [{"sub_id": "Exam-02-T8-1", "tags": ["2.7", "1.2"], "text_open": "8-1 The Baeyer-Villiger oxidation should be familiar, and migrating groups are known to follow an order of migratory aptitude. Explain why the following reaction gives a single product in very high yield.", "text_images": [{"file": "Exam-02-T8-1-1-prompt.png", "description": "A reaction proceeds from left to right. The molecule 'O=C1[C@@H]2C3C4CC2[C@H]5C4CC3[C@H]51' on the left reacts under a one-way arrow labeled “m-CPBA” above and “DCM,93%” below to give 'O=C1[C@@H]2C3C4CC2[C@H]5C4CC3[C@H]5O1' on the far right."}], "full_mark": 2}, {"sub_id": "Exam-02-T8-2", "tags": ["2.3", "2.7", "2.2"], "text_open": "8-3 The Schmidt rearrangement should be familiar. The following two substrates give different products under identical conditions. Explain the outcome for the second substrate.", "text_images": [{"file": "Exam-02-T8-2-1-prompt.png", "description": "The image shows two parallel reaction equations. In the upper reaction, the reactant on the left has SMILES 'O=C1CCCC1CCC(c2ccccc2)N=[N+]=[N-]' and reacts under an arrow labeled “CF_3SO_2OH” above and “0°C” below to give 'O=C3CCCC4N3C(c5ccccc5)CC4' on the right. In the lower reaction, the reactant on the left has SMILES 'O=C6CCCC6CCCC(c7ccccc7)N=[N+]=[N-]' and reacts under an arrow labeled “CF_3SO_2OH” above and “0°C” below to give 'O=C8CCCC89C(CCC9)Nc%10ccccc%10' on the right."}], "full_mark": 2}, {"sub_id": "Exam-02-T8-3", "tags": ["2.2", "2.3", "1.2"], "text_open": "8-4 At first glance, the following reaction appears simple and seemingly requires only elimination, hydride migration, and addition. Closer examination shows that this interpretation is inconsistent with the stereochemical information. Write the SMILES representations of all key intermediates actually involved, including their stereochemistry.", "text_images": [{"file": "Exam-02-T8-3-1-prompt.png", "description": "'C[C@@H]1CO[C@@]2([H])N1[C@](CC(OC)=C)(C#N)CCC2' and 'Cl[Mg]CCCC1OCCO1' react under tBuMe2SiOTf, DCM/Et2O/THF, then NH4Cl/H2O conditions to give 'C[C@@H]1CO[C@]2(CCCC3OCCO3)N1[C@]([H])(CC(OC)=C)CCC2'."}], "full_mark": 8}]} -{"problem_id": "Exam-02-T10", "header": "Problem 10: Forever Unnamed", "header_images": [], "subquestions": [{"sub_id": "Exam-02-T10-1", "tags": ["2.2"], "text_open": "10-1 A neutral reaction intermediate can be obtained by the process shown below. Write its SMILES representation.", "text_images": [{"file": "Exam-02-T10-1-1-prompt.png", "description": "A chemical reaction equation: the reactants on the left are [R1]CC(=O)Cl and R_2NC, joined by a plus sign `+`; a horizontal right-pointing arrow in the center leads to bold X on the right."}], "full_mark": 2}, {"sub_id": "Exam-02-T10-2", "tags": ["2.4", "2.2", "1.2"], "text_open": "10-2-1 Further reactions can proceed from intermediates of type X. Write the SMILES representations of products A and B in the reactions below, given that B contains a five-membered aromatic ring. 10-2-2 Label the absolute configurations of all chiral carbon atoms in the structure of A from the preceding question.", "text_images": [{"file": "Exam-02-T10-2-1-prompt.png", "description": "The image shows two parallel reactions. In the upper reaction, the starting material is 'CC1=C(O[Si](C)(C)C(C)(C)C)CC[C@@H]1C[N+]#[C-]'. Above the arrow is “(a)tBuCOCl,DCM,r.t.” and below it “(b)AgBF_4,DCM,-78∘C”; the arrow leads to product A. In the lower reaction, '[R]C(Cl)=O' + '[Ar]C[N+]#[C-]' reacts under “base” to give product B."}], "full_mark": 4}, {"sub_id": "Exam-02-T10-3", "tags": ["2.2", "2.3"], "text_open": "10-2-3 Write the SMILES representation of another electrically neutral key intermediate Y, other than the type-X intermediate, involved in formation of B.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-02-T10-4", "tags": ["2.4", "2.3", "2.7"], "text_open": "10-2-4 If an appropriate amount of CH_{2}=CHCO_{2}Me is added to the reaction system that forms B, Y is trapped to form two products, C_{1} and C_{2}. Write their SMILES representations.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-02-T10-5", "tags": ["2.2"], "text_open": "10-3-1 Addition of other reagents changes the reaction pathway. Write the small-molecule by-product formed in the reaction below and the SMILES representation of another electrically neutral key intermediate, other than the type-X intermediate, involved in the reaction.", "text_images": [{"file": "Exam-02-T10-5-1-prompt.png", "description": "A continuous multistep reaction sequence. The substrate is `R_1COCl`; above the first arrow is R_2NC; above the second is `P(OR_3)_3`; and above the third is '[O-]/C([Ar])=C\\[N+]1=CC=CC=C1'. The final product is 'O=C([Ar])C1=C(C([R1])=C2C=CC=CN12)N[R2]'."}], "full_mark": 3}, {"sub_id": "Exam-02-T10-6", "tags": ["2.4", "2.2"], "text_open": "10-3-2 In the presence of Et_{3}N, EtO_{2}CCOCl, CyNC, P(Oi-Pr)_{3}, NCCH_{2}CN, and CS_{2} undergo a multicomponent reaction in MeCN. Write the SMILES representation of product D.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-02-T10-7", "tags": ["2.2", "2.4"], "text_open": "10-4 Under certain conditions, an α-carbonyl diazo compound can replace an acyl chloride in a similar reaction with an isocyanide. Write the SMILES representations of all intermediates and products in the scheme below. W and all three products contain a five-membered ring, but only the five-membered ring in E is aromatic; F also contains a six-membered ring.", "text_images": [{"file": "Exam-02-T10-7-1-prompt.png", "description": "A synthetic reaction scheme: '[R1]C(C(C([R1])=O)=[N+]=[N-])=O' is the starting material and is converted by an arrow to Z. Intermediate Z then reacts with `R_2NC` to form W. Intermediate **W** branches into three transformations: the upper branch reacts with `MeNHNH_2` to form E; the middle branch reacts with '[R1]C(C([R1])=O)=C=O' to form F; and the lower branch forms H under H^+ conditions."}], "full_mark": 10}]} +{"problem_id": "Exam-02-T9", "header": "Problem 9: Selectivity in Chemistry", "header_images": [], "subquestions": [{"sub_id": "Exam-02-T9-1", "tags": ["2.7", "1.2", "2.4"], "text_open": "9-1 Reduction of the following compound with different reducing agents gives different stereochemical outcomes. Write the SMILES representation of major product 9A1 obtained with LiAlH_{4} and that of major product 9A2 obtained with ^{i}Bu_{2}AlH.", "text_images": [{"file": "Exam-02-T9-1-1-prompt.png", "description": "The structure in the image has SMILES 'O=C(C1)C[C@H]2N([C@@H]1CC2)C'."}], "full_mark": 4}, {"sub_id": "Exam-02-T9-2", "tags": ["2.8"], "text_open": "$Zn(OTf)_2$ is used in the selective esterification shown below. It acts as a Lewis acid to activate the ester group and coordinates separately to the two substrates to control the reaction site.\n9-2-2 What amount of Zn(OTf)_{2} should be used in this reaction?", "text_images": [{"file": "Exam-02-T9-2-1-prompt.png", "description": "The image shows a reaction proceeding from left to right. At the far left is 'O=C(O/N=C/c2ncccc2)CCc3ccccc3', labeled **9B** below, joined by a plus sign to 'O[C@@H]1[C@@H](O)[C@H](CO)O[C@@H](OCCCCCCCC)[C@H]1NC(CO)=O' in the center. Above the reaction arrow is “Zn(OTf)_2”; below it are “DMF/THF/DME,” “(1:5:9),” and “30°C, 18h,” in sequence."}], "full_mark": 2}, {"sub_id": "Exam-02-T9-3", "tags": ["2.4", "2.7", "1.2"], "text_open": "9-2-3 Write the SMILES representation of the major product, given that only one hydroxy group is esterified.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-02-T9-4", "tags": ["2.2", "2.3"], "text_open": "9-3-2 Compound 9D is the addition product of acetonitrile and hydrogen peroxide, and 9G contains a peroxide bond. Write the SMILES representations of small molecules 9D and 9F in the catalytic cycle below.", "text_images": [{"file": "Exam-02-T9-4-1-prompt.png", "description": "Structure of compound 9C, with SMILES 'CC1(OC[C@]2(OC[C@H]3OC(C)(O[C@H]3C2=O)C)O1)C'."}, {"file": "Exam-02-T9-4-2-prompt.png", "description": "The image shows a reaction segment in which the far-left compound 'C/C=C/c1ccccc1' reacts to the right under an arrow labeled “9C (15 mol%), H_2O_2 (4 equiv.)” above and “K_2CO_3, CH_3CN” and “0°C, 12h” below."}, {"file": "Exam-02-T9-4-3-prompt.png", "description": "The image is a catalytic cycle. At the top is 'CC1(OC[C@]2(OC[C@H]3OC(C)(O[C@H]3C2=O)C)O1)C', labeled “9C” below. On the right, `CH_3CN + H_2O_2` points left by a straight arrow to “9D.” A curved arrow from “9D” runs down and left into the main cycle, merging with the downward path from “9C,” and both lead to “9E.” Two curved arrows descend from “9E,” one down-right to “9F” and the other down-left to “9G.” A curved arrow from “9G” runs up and left, meeting an upward curved arrow from 'C/C=C/c1ccccc1' at the lower left; together they lead to `[transition state]^‡`. Finally, a curved arrow from `[transition state]^‡` runs up and right to close the cycle at “9C,” with a curved branch along this upward path pointing up-left to the word “product.”"}], "full_mark": 4}, {"sub_id": "Exam-02-T9-5", "tags": ["2.7", "1.2", "2.4"], "text_open": "9-3-3 Predict the stereochemistry and write the SMILES representation of the product.", "text_images": [], "full_mark": 2}]} {"problem_id": "Exam-03-T1", "header": "", "header_images": [], "subquestions": [{"sub_id": "Exam-03-T1-1", "tags": ["1.3"], "text_open": "1-3 The following structure is an As-S polyanion. Write its chemical formula, including the charge.", "text_images": [{"file": "Exam-03-T1-1-1-prompt.png", "description": "This schematic of an As-S polyanion shows a two-dimensional network of solid red circles (As atoms) and hollow white circles (S atoms) connected by black line segments. Its repeating motifs are cyclic clusters, each typically an eight-membered ring formed by alternating 4 red and four white nodes. Visually, each eight-membered ring contains a special terminal white node that is connected by only one line to a red node on the ring and points toward the center of the ring. The rings are not isolated; adjacent cyclic clusters are linked by individual bridging white nodes. Red nodes on adjacent clusters connect to a shared bridging white node between them. Every red node not connected to an inward-pointing white node participates in bridging, whereas a red node connected to an inward-pointing white node does not. Repetition in all directions forms a flexible, wavy-looking infinite two-dimensional network. At the image boundaries, some nodes have short black lines extending outward without terminating at another circle. Near the upper part of the image, the labels \"As\" and \"S\" point respectively to a specific red node and an adjacent white node."}], "full_mark": 2}, {"sub_id": "Exam-03-T1-2", "tags": ["2.4", "1.4"], "text_open": "1-4 Under illumination, Fe(CO)_5 reacts with (C_6F_5As)_4 to form complex X with ω(Fe) = 8.57%. 1-4-1 Write the chemical formula of complex X.", "text_images": [], "full_mark": 2}]} -{"problem_id": "Exam-03-T3", "header": "Problem 3 Polymers. Global plastics production reached 3.907 hundred million tonnes in 2021, yet only 8.3% of plastics were recycled, resulting in severe accumulation of plastic waste and environmental problems. Developing next-generation chemically recyclable polymers that can be efficiently and selectively depolymerized back to their original monomers is therefore a viable route toward a closed-loop, sustainable plastics economy and a solution to end-of-life problems. Owing to their distinctive biorenewability and biodegradability, polyesters are among the most promising recyclable polymers.", "header_images": [], "subquestions": [{"sub_id": "Exam-03-T3-1", "tags": ["2.2", "2.3"], "text_open": "3-2 In 2024, researchers developed a polymer containing cyclobutane rings in its main chain. The polymer is thermally stable, resistant to hydrolysis, has tunable mechanical properties, and exhibits excellent chemical recyclability. It is prepared from monomer 4R-BL (R = Bu or Ph) using catalyst Y_1. Denote the resulting polymer as P(4^R-BL). 3-2-1 Draw and give the SMILES of the key intermediates in the initiation and chain-propagation steps of Y_1-catalyzed polymerization of 4R-BL. Ligands on the central Y atom that do not participate in the transformation may be omitted.", "text_images": [{"file": "Exam-03-T3-1-1-prompt.png", "description": "Structure of 4^R-BL: '[R]C(CC1O2)(C1)C2=O'."}, {"file": "Exam-03-T3-1-2-prompt.png", "description": "Structure of Y1: 'C[Si](C)(C[Y]123(O4CCCC4)OC5=C(C[N]1(CCO3C)CC6=C(O2)C(C(C)(C)C)=CC(C(C)(C)C)=C6)C=C(C(C)(C)C)C=C5C(C)(C)C)C'."}], "full_mark": 4}, {"sub_id": "Exam-03-T3-2", "tags": ["2.3", "2.8"], "text_open": "3-2-3 Adding one equivalent of isopropanol relative to Y_1 increases the initiation rate of polymerization. Explain why. (No more than 100 words.)", "text_images": [], "full_mark": 4}]} -{"problem_id": "Exam-03-T5", "header": "Problem 5 Simple Substances, Complex Crystals", "header_images": [], "subquestions": [{"sub_id": "Exam-03-T5-1", "tags": ["1.1", "1.3"], "text_open": "5.1 Projections of a crystal along the b and c axes are shown below. Large black spheres represent A, white spheres represent B, and small black spheres represent O. 5.1.1 Name the coordination polyhedron of A and that of B.", "text_images": [{"file": "Exam-03-T5-1-1-prompt.png", "description": "Crystallographic projection of the unit cell viewed approximately along the b axis. The horizontal direction is the c axis and the vertical direction is the a axis. A black rectangle represents one unit cell, with a varying from 0 to 1 vertically and c from 0 to 1 horizontally. Blue, yellow, and red atoms are shown. Their approximate projected fractional coordinates (a,c), relative to the unit-cell rectangle, are as follows. Blue atoms: (0.85,0.50), (0.65,0.00), (0.30,0.00), and (0.15,0.50), together with periodic copies at the corresponding unit-cell boundaries. Yellow atoms: (0.85,0.00), (0.65,0.50), (0.30,0.50), and (0.15,0.00), together with periodic copies at c=1. Red atoms: approximately (1.20,0.50), (1.02,-0.17), (1.02,0.17), (1.02,0.83), (1.02,1.17), (0.70,0.00), (0.70,1.00), (0.48,-0.33), (0.48,1.33), (0.48,0.33), (0.48,0.67), (0.30,0.00), (0.30,1.00), (0.15,-0.17), (0.15,0.17), (0.15,0.83), and (0.15,1.17). Some atoms are partially obscured or overlap larger blue atoms. Coordinates outside the [0,1]×[0,1] rectangle represent visible periodic images. Red bonds connect neighboring red, blue, and yellow atoms."}, {"file": "Exam-03-T5-1-2-prompt.png", "description": "Crystallographic projection of the unit cell viewed approximately along the c axis. The horizontal direction is the a axis and the vertical direction is the b axis. A black rectangle represents one unit cell, with a varying from 0 to 1 horizontally and b from 0 to 1 vertically. Blue, yellow, and red atoms are shown. Their approximate projected fractional coordinates (a,b), relative to the unit-cell rectangle, are as follows. Blue atoms: (0.35,0.75), (0.85,0.75), (0.15,0.25), and (0.65,0.25). Yellow atoms: (0.85,0.75) and (0.15,0.25). Red atoms: approximately (0.30,1.08), (0.80,1.08), (0.05,0.92), (0.55,0.92), (1.05,0.92), (0.30,0.75), (0.80,0.75), (0.05,0.50), (0.55,0.50), (1.05,0.50), (0.30,0.25), (0.80,0.25), (0.05,0.08), (0.55,0.08), and (1.05,0.08), together with visible periodic images outside the unit-cell boundaries. Several red atoms visually overlap blue or yellow atoms. Red bonds connect neighboring red, blue, and yellow atoms."}], "full_mark": 4}, {"sub_id": "Exam-03-T5-2", "tags": ["1.3"], "text_open": "5.1.2 Using the coordination numbers obtained above, write the chemical formula of the crystal.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-03-T5-3", "tags": ["1.3"], "text_open": "5.1.3 State the lattice type of the crystal.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-03-T5-4", "tags": ["1.3"], "text_open": "5.2 Consider an ideal hexagonal crystal composed of X and O. The coordinates of three X atoms in the unit cell are (1/2, 1/2, 0), (0, 1/2, 1/3), and (1/2, 0, 2/3). The unit cell may be viewed as containing helical X-O-X-O chains extending along c. For example, when viewed along c, rotate the X atom at (1/2, 1/2, 0) clockwise by 120° about the line a = 1/3, b = 2/3 and then translate it upward by 1/3 of the c-axis length; it reaches (0, 1/2, 1/3). Applying the same clockwise rotation and upward translation to the X atom at (0, 1/2, 1/3) gives (1/2, 1, 2/3). Applying them once more gives (1/2, 1/2, 1), equivalent to (1/2, 1/2, 0). This is illustrated below; the numbers in the figure are the c-axis coordinates of the X atoms. 5.2.1 Call the helical sense described above left-handed. Correspondingly, a helix that rotates counterclockwise while advancing along c is called right-handed. The unit cell contains another helical chain whose axis lies at a = 2/3, b = 1/3. State whether this chain is left- or right-handed.", "text_images": [{"file": "Exam-03-T5-4-1-prompt.png", "description": "Schematic of the helical X-O-X-O chain, projected along c. The main outline is a solid parallelogram. Slightly left of center inside it is a solid black point surrounded by a circular path bearing three clockwise arrows. Three light-gray circles with black outlines are evenly spaced on this path. Their labels increase clockwise: \"0\" at the lower right, \"1/3\" at the lower left, and \"2/3\" at the top. At the lower left of the image is an axis legend: a horizontal right-pointing arrow labeled a, an arrow slanting up and left labeled b, and the letter c at their common origin."}], "full_mark": 2}, {"sub_id": "Exam-03-T5-5", "tags": ["1.1", "1.3"], "text_open": "Structural background: A crystal composed of C, D, and O has a body-centered lattice and formula $CDO_x$, with $a=b=433\\ \\mathrm{pm}$ and $c=664\\ \\mathrm{pm}$. C occupies the cell corners and body center, and one D atom has fractional coordinates $(0.50,0,0.75)$.\n5.3.2 Considering only C and D atoms, identify the polyhedron formed by D atoms around C and the polyhedron formed by C atoms around D.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-03-T5-6", "tags": ["1.3"], "text_open": "5.3.3 One O atom in the unit cell has fractional coordinates (0.138, 0.260, 0.131). Calculate its distance to the nearest C atom, d_{O-C}, and to the nearest D atom, d_{O-D}.", "text_images": [], "full_mark": 4}]} -{"problem_id": "Exam-03-T6", "header": "Problem 6 Exploring Our Little Broken Planet. Rocks are generally mixtures of one or more minerals and can be classified as igneous, metamorphic, or sedimentary. Geological processes can transform each type into the others. Igneous rocks form when magma cools. Magma is a complex mixture without a fixed melting point; melting occurs over a temperature range, beginning at the solidus and becoming complete at the liquidus. Most rocks undergo partial melting. During magma cooling and crystallization, changes in physical and chemical conditions cause minerals of different compositions to crystallize in sequence. Consequently, the solid and residual melt often have different chemical compositions. This phenomenon is called chemical differentiation. Temperature, chemical composition, and gravity are important factors affecting chemical differentiation. Metamorphic rocks form when pre-existing rocks in the crust—igneous, sedimentary, or metamorphic—undergo changes in mineral composition, texture, structure, and chemical composition under the influence of temperature, pressure, and chemically active fluids. Metamorphism generally occurs below the solidus temperature of the rock (600-1000 ℃), so melting does not occur in the great majority of cases; pressure, however, has no strict upper limit.", "header_images": [{"file": "Exam-03-T6-header-1.png", "description": "Pressure-temperature (P-T) phase diagram labeled \"A\". The horizontal axis is temperature $T$ in $^\\circ C$, ranging from $1000\\sim2000^\\circ C$. The vertical axis is pressure $P$ in $GPa$, ranging from $0\\sim10GPa$. The diagram contains a curve labeled solidus, which is the thickest curve, a curve labeled liquidus to its right, and several nearly parallel thin curves between them. All curves trend downward as temperature and pressure increase and converge in the high-pressure, high-temperature region near $9GPa$ and $2000^\\circ C$."}], "subquestions": [{"sub_id": "Exam-03-T6-1", "tags": ["4.2"], "text_open": "6-1 As magma passes through the crust, some cools and solidifies within the crust to form intrusive rocks, while some reaches the surface to form extrusive rocks. Crystallinity mainly reflects the extent to which a substance has transformed from a disordered amorphous state into an ordered crystalline state. A rock has high crystallinity if its mineral grains are large and well crystallized, with regular and complete particle arrangements and few defects or disorders; otherwise, it has low crystallinity. State whether intrusive or extrusive rocks have higher crystallinity and briefly explain why. (No more than 50 words.)", "text_images": [], "full_mark": 2}]} +{"problem_id": "Exam-03-T7", "header": "Problem 7 Li-CO_2 Batteries Based on Dimerization with a Solid Redox Mediator. 7-1 As energy demand and global greenhouse-gas emissions increase, efficiently storing energy while reducing CO_2 emissions has become a major scientific and technological challenge. Li-CO_2 batteries are a focus of next-generation energy-storage research because of their high theoretical energy density, approximately 1876 Wh/kg, and their ability to convert the greenhouse gas CO_2 into electrical energy. Conventional Li-CO_2 batteries, however, face three bottlenecks: (1) low discharge voltage because direct electrochemical reduction of CO_2 is kinetically slow; (2) high charging overpotential because the main discharge product is difficult to decompose and the charging voltage is typically above 4.3 V; and (3) short cycle life because conventional catalysts cannot stabilize the discharge product, so cycle life is usually fewer than 100 cycles.", "header_images": [], "subquestions": [{"sub_id": "Exam-03-T7-1", "tags": ["2.1", "3.5"], "text_open": "7-1-1 Write the typical cathodic discharge reaction of a Li-CO_2 battery. The products are Li_2CO_3 and an elemental substance in a 2:1 ratio.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-03-T7-2", "tags": ["1.7", "3.5"], "text_open": "Earlier soluble redox mediators had three problems: they diffused between the cathode and anode and underwent side reactions; their transport through the electrolyte was slow at high current density; and they decomposed during cycling to form irreversible byproducts.\n7-3 To address these problems, researchers developed a Li-CO_2 battery based on a solid redox mediator. The mediator, RM(II)-BTC, is a microporous MOF formed by coordination of Cu(II) with trimesic acid (BTC). It is combined with carbon nanotubes to form a three-dimensional conductive network and has a high specific surface area and a pore diameter of 9 Å. The electrochemical performance of the battery is greatly improved: (1) the discharge voltage increases to 2.8 V and the charging voltage decreases to 3.7 V; (2) cycle life exceeds 400 cycles; and (3) excellent kinetics are obtained at high current density. 7-3-1 Based on the information in 7-3, state at least two advantages of the proposed Li-CO_2 battery over earlier Li-CO_2 batteries. Use no more than 20 words for each point.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-03-T7-3", "tags": ["3.8"], "text_open": "The DFT pathway is defined as follows. A is the initial state. A reacts with $Li^+$, $e^-$, and $CO_2$ to form B, with $\\Delta E_1=-1.50\\ \\mathrm{eV}$. B dimerizes to C, with $\\Delta E_2=+0.80\\ \\mathrm{eV}$. C binds Cu(I) with an oxidation-state change to give D, the final state containing stably adsorbed $Li_2C_2O_4$, with $\\Delta E_3=-2.72\\ \\mathrm{eV}$. $E(A)=-300.00\\ \\mathrm{eV}$; $E(TS_2)-E(C)=+0.45\\ \\mathrm{eV}$ and $E(TS_3)-E(D)=+3.62\\ \\mathrm{eV}$.\n7-4-2 Calculate and compare the maximum energy barriers of Steps 2 and 3, defined as the difference between the transition state and the corresponding precursor intermediate, and identify the rate-controlling step of the pathway.", "text_images": [], "full_mark": 3}]} +{"problem_id": "Exam-03-T10", "header": "Problem 10 Synthesis of Indigoid Compounds (25 points)", "header_images": [], "subquestions": [{"sub_id": "Exam-03-T10-1", "tags": ["2.4", "2.2"], "text_open": "10-1 In 1909, H. Friedländer noted that an excellent purple pigment mentioned in the Bible—the pigment that the Romans called Tyrian purple after the Phoenician port of Tyre in present-day Lebanon—was in fact 6,6'-dibromoindigo, F. This precious early dye was extracted from small purple sea snails, the murex; as many as 12000 snails were required to prepare 1.5 g of dye. After nearly a century of continuing research into its synthesis, Pinkney and Chammers proposed a new method, which Iming and coworkers improved into an efficient route. 10-1-1 Give the SMILES of intermediate E.", "text_images": [{"file": "Exam-03-T10-1-1-prompt.png", "description": "10-1 Synthetic route. This multistep scheme begins at the upper left with 'CC1=CC=C(N)C=C1', which reacts under HNO_3 and H_2SO_4 to give intermediate B. B reacts under 1) NaNO_2, HBr and 2) CuBr to give C. C then reacts under CrO_3 and (CH_3CO)_2O to give D. In the second row, D reacts under H_2SO_4 and H_2O to give E. Finally, E reacts under NaOH and H_2O, (CH_3)_2CO to give the product F at the lower right, with SMILES 'O=C1NC2=C(C=C(Br)C=C2)C1=C3NC4=C(C=C(Br)C=C4)C3=O'."}], "full_mark": 2}, {"sub_id": "Exam-03-T10-2", "tags": ["2.7", "1.6"], "text_open": "10-1-2 Explain the regioselectivity of the conversion of A into B. (No more than 50 words.)", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-03-T10-3", "tags": ["2.1", "2.2"], "text_open": "10-2 In the route above, conversion of E into F is clearly the key step. In fact, as early as 1882, Baeyer used this reaction to synthesize indigo in one step from o-nitrobenzaldehyde. 10-2-1 Write the chemical equation for this reaction, representing o-nitrobenzaldehyde and indigo by their SMILES.", "text_images": [{"file": "Exam-03-T10-3-1-prompt.png", "description": "One-step synthesis. At the far left is 'O=CC1=CC=CC=C1[N+]([O-])=O', labeled G below. Under NaOH and H_2O, (CH_3)_2CO, it gives the product at the far right, 'O=C1NC(C=CC=C2)=C2/C1=C3NC4=C(C=CC=C4)C/3=O', labeled H below."}], "full_mark": 2}, {"sub_id": "Exam-03-T10-4", "tags": ["2.2", "2.3"], "text_open": "10-2-2 The final step of the reaction is shown below. Considering the byproducts formed in the equation in 10-2-1, give the SMILES of the four key intermediates in the conversion of G into M. For intermediates related by proton gain, loss, or transfer, only one need be given.", "text_images": [{"file": "Exam-03-T10-4-1-prompt.png", "description": "10-2-2 Question figure. A reaction proceeds from left to right. At the far left is 'O=C1C=NC2=C1C=CC=C2', labeled M below, followed by a plus sign and the middle molecule 'OC1=C(NC2=C1C=CC=C2)O', labeled N below. A one-way right-pointing reaction arrow leads to the product at the far right, 'O=C5NC(C=CC=C6)=C6/C5=C7NC8=C(C=CC=C8)C/7=O', labeled H below, indicating that M and N react to form H."}], "full_mark": 6}, {"sub_id": "Exam-03-T10-5", "tags": ["2.2", "2.5"], "text_open": "10-3 Nitrogen plays a key role in the reaction above, but analogous substrates and methods are unsuitable for obtaining the sulfur-substituted product. After the nineteen-eighties, Scottish chemists synthesized thioindigo, K, using the reaction below. The reaction still proceeds through an intermediate containing a fused six- and five-membered ring system. It is known that contraction of the six-membered ring into the five-membered ring does not occur by ring opening followed by ring closure. Answer the following questions. 10-3-1 In what form does the oxygen atom of the sulfoxide depart?", "text_images": [{"file": "Exam-03-T10-5-1-prompt.png", "description": "Structure/reaction description for synthesis of thioindigo K. Sulfoxide J gives thioindigo K under NaOAc/HOAc. The SMILES of J is 'BrC(C1=O)=C([S+](C2=C1C=CC=C2)[O-])Br'; thioindigo K may be represented as 'O=C1C2=C(S/C1=C3SC4=C(C\\3=O)C=CC=C4)C=CC=C2'."}], "full_mark": 1}, {"sub_id": "Exam-03-T10-6", "tags": ["2.3", "2.7"], "text_open": "10-3-2 During ring contraction, which atom is lost? ( ) (a) the carbon atom nearer the carbonyl group (b) the carbon atom nearer the sulfoxide group", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-03-T10-7", "tags": ["2.2", "2.3"], "text_open": "10-3-4 Give the SMILES of the six key intermediates in the reaction. For intermediates related by proton gain, loss, or transfer, only one need be given. A substitution proceeding by addition-elimination or by an S_{N1} process counts as only one step.", "text_images": [], "full_mark": 9}]} {"problem_id": "Exam-04-T1", "header": "Problem 1 Minerals and Chemistry. In the 18 century, Austrian chemist F. J. Müller discovered a substance X resembling antimony in a gold-bearing ore. It was not until 16 years later that German mineralogist M. H. Klaproth formally discovered and named it. Roasting an X-containing mineral A in air gives B and elemental silver. Theoretically, roasting each 1.000 g of A produces 0.628 g of silver.", "header_images": [], "subquestions": [{"sub_id": "Exam-04-T1-1", "tags": ["2.1", "2.2"], "text_open": "It is known that element $X=Te$, mineral $A=Ag_2Te$, and product $B=TeO_2$.\n1-2 B can be further converted into other compounds of X. (i) B dissolves in a concentrated solution of C to give a solution containing D, while base G is also formed. D reacts with E in a 1:2 ratio, precipitating elemental X and producing C and F. (ii) B reacts with base G to form H. A solution in glacial acetic acid of the acid corresponding to the anion of excess H reacts with F in a 1:4 ratio to give the X-containing compound I and the X-free compound J. Reaction of I with G precipitates elemental X and also forms F and H. (iii) Under acidic conditions, B reacts with hydrogen peroxide to give K. K is reduced by a common monoprotic strong acid L, producing B, X, and elemental M. Interestingly, M reacts with F to form J, a reaction commonly used to titrate M in analytical chemistry. C is a binary compound, and C, D, E, F, H, I, and J are all 2:1 salts. Determine the formulas of compounds D, E, I, and J.", "text_images": [], "full_mark": 10}, {"sub_id": "Exam-04-T1-2", "tags": ["3.4", "3.6"], "text_open": "1-3 The standard electrode potential for conversion of K to B is $+1.02\\mathrm{V}$, that of $\\mathrm{Au^{3+}/Au}$ is $+1.50\\mathrm{V}$, and the logarithmic cumulative stability constant for $\\mathrm{Au^{3+}+4Cl^- = AuCl_4^-}$ is $\\lg \\beta_4 = 25.6$. By a brief calculation, determine whether a mixture of K and concentrated hydrochloric acid can dissolve gold at room temperature.", "text_images": [], "full_mark": 5}]} +{"problem_id": "Exam-04-T2", "header": "Problem 2 Fundamentals of Iodometric Titration", "header_images": [], "subquestions": [{"sub_id": "Exam-04-T2-1", "tags": ["3.6", "3.4"], "text_open": "2-1-2 Calculate the conditional reduction potential of $\\mathrm{Fe(CN)_6^{3-}}$ to $\\mathrm{Fe(CN)_6^{4-}}$ in $\\mathrm{mol}\\cdot\\mathrm{L}^{-1}$ HCl. Given $\\varphi^\\Theta_{\\mathrm{Fe(CN)_6^{3-}/Fe(CN)_6^{4-}}}=0.36\\ \\mathrm{V}$; $\\mathrm{H_3Fe(CN)_6}$ is a strong acid; $\\mathrm{H_4Fe(CN)_6}$ has $\\mathrm{p}K_{\\mathrm{a3}}=2.2$ and $\\mathrm{p}K_{\\mathrm{a4}}=4.2$; neglect ionic effects.", "text_images": [], "full_mark": 5}, {"sub_id": "Exam-04-T2-2", "tags": ["2.1", "4.1"], "text_open": "2-2 Dissolve a 1.000 g sample containing sodium iodate and sodium periodate in a 250 mL volumetric flask and dilute to the mark. Transfer 50.00 mL to an Erlenmeyer flask, adjust to weakly basic pH, add excess potassium iodide, and titrate with 0.04000 mol·L^{-1} sodium thiosulfate; 10.00 mL is consumed. Transfer another 20.00 mL, acidify with hydrochloric acid, add excess potassium iodide, and titrate with sodium thiosulfate of the same concentration; 30.00 mL is consumed. 2-2-1 What reaction occurs in the first titration? Write its ionic equation and explain why the solution must be adjusted to weakly basic conditions.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-04-T2-3", "tags": ["4.5"], "text_open": "2-2-2 Calculate the mass percentages of sodium iodate and sodium periodate.", "text_images": [], "full_mark": 9}, {"sub_id": "Exam-04-T2-4", "tags": ["4.5"], "text_open": "2-2-3 In theory, this method can measure the mass percentages of sodium iodate and sodium periodate fairly accurately. In a student laboratory, however, it is difficult to obtain an accurate result, defined as an error within 0.1%. Explain the principal reason in terms of experimental readings. (No more than 20 words.)", "text_images": [], "full_mark": 3}]} {"problem_id": "Exam-04-T5", "header": "Rutile is a typical tetragonal crystal structure. One unit cell is shown, with large light-blue spheres representing Ti and small red spheres representing O. This structure occurs widely among transition-metal compounds and gives rise to many functional materials.", "header_images": [{"file": "Exam-04-T5-header-1.png", "description": "Three-dimensional ball-and-stick model of a TiO_2 unit cell. Red a, green b, and blue c axes are labeled at the lower left. Large light-blue spheres and small red spheres are connected by light-blue bonds inside a tetragonal unit-cell frame outlined in black. The large light-blue spheres occupy the corners and center at (0, 0, 0) and (0.5, 0.5, 0.5). Nearby red spheres surround them in coordination environments at (0.305, 0.305, 0), (0.695, 0.695, 0), (0.805, 0.195, 0.5), and (0.195, 0.805, 0.5)."}], "subquestions": [{"sub_id": "Exam-04-T5-1", "tags": ["1.3"], "text_open": "5.2 Potassium tetrachlorocuprate dihydrate, $\\mathrm{K_2CuCl_4·2H_2O}$, is a bright-blue deliquescent crystal used mainly as a copper-ion source and catalyst in laboratories and industry. Its crystal symmetry resembles that of rutile and may be compared with the structure above. $\\mathrm{[Cu(H_2O)Cl_4]^(2-)}$ units occupy the corners and body center, while $\\mathrm{K^+}$ ions lie on the c = 0.250 and c = 0.750 planes. Cu-O bonds are parallel to c, and each Cu-Cl bond is parallel to a corresponding basal-plane diagonal. All identical atoms except Cl have equivalent chemical environments. The cell parameters are a = b = 745 pm, c = 788 pm, and α = β = γ = 90°. 5-2-1 Two Cl atoms have coordinates (0.720, 0.280, 0) and (0.720, 0.280, 0.500). Give the coordinates of the remaining 6 Cl atoms.", "text_images": [{"file": "Exam-04-T5-1-1-prompt.png", "description": "Unit-cell model of $\\mathrm{K_2CuCl_4·2H_2O}$ containing blue, purple, green, red, and white spheres. Blue spheres occupy the corners and center. Purple spheres lie on the four side faces, two vertically separated on each face. Green spheres lie on the top, bottom, and middle planes, four evenly arranged at the upper-left, lower-left, upper-right, and lower-right positions of each plane. Red spheres lie above and below each blue sphere, approximately at the same height as the purple spheres. White spheres are bonded beside the red spheres, two per red sphere, forming V shapes."}], "full_mark": 6}, {"sub_id": "Exam-04-T5-2", "tags": ["1.3"], "text_open": "5.3 See the color figure for this part. A tetragonal crystal composed of Na, H, and Ru also has a rutile-like structure. Its projection along c is shown. Yellow spheres are Na, blue spheres Ru, and pink and white spheres H. Light-yellow Na lies in the c = 0 plane and dark-yellow Na in the c = 1/2 plane; light-blue Ru lies in c = 0 and dark-blue Ru in c = 1/2. White H lies in c = 0, while pink H is distributed within the unit cell. All Ru atoms have equivalent environments. H occurs with coordination numbers 3, 4, and 5, with more four-coordinate H than three-coordinate H; Na has only one coordination number. All atoms shown lie partly or wholly inside the unit cell. Hint: mirror planes occur at c = 0 and c = 1/2. 5-3-1 Give the chemical formula of the crystal.", "text_images": [{"file": "Exam-04-T5-2-1-prompt.png", "description": "Color ball-and-stick projection with a thin dark-gray square marking the unit-cell or projection boundary and no axes or text labels. 40 distinguishable spheres are visible: 16 yellow or gold, 14 light pink, 6 white, and 4 dark blue. Some lie partly outside the square. The yellow/gold spheres are distributed around the top, sides, center, and bottom; the light-pink spheres surround and connect the blue and yellow spheres; the six white spheres occur mainly near the upper right, center right, lower left, and bottom; and the four dark-blue spheres lie at left center, upper right, lower left, and lower right-center. Short rods connect the spheres, with colors generally transitioning between adjacent atoms, such as yellow-pink, yellow-white, pink-blue, and white-blue. Together they form a continuous 3-dimensional network passing through the square boundary, with some spheres and bonds extending outside it. The numbered sphere positions are explicitly enumerated as follows: yellow/gold 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, and 16; light pink 1 through 14; white 1 through 6; and dark blue 1 through 4."}], "full_mark": 2}, {"sub_id": "Exam-04-T5-3", "tags": ["1.3"], "text_open": "5-3-2 Name the coordination polyhedron around Ru in the crystal.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-04-T5-4", "tags": ["1.3"], "text_open": "5-3-3 Give the number ratio of three-, 4-, and five-coordinate H atoms in the crystal.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-04-T5-5", "tags": ["1.3"], "text_open": "5-3-4 Give the coordination number of Na in the crystal.", "text_images": [], "full_mark": 2}]} {"problem_id": "Exam-04-T7", "header": "Problem 7 Microbial Fuel Cells. Data that may be used in this problem, all at 25 ℃ unless otherwise stated: $\\varphi^\\Theta_{\\mathrm{HCO_3^-}/\\mathrm{CH_3COO^-}} = 0.187\\ \\mathrm{V}$; $\\varphi^\\Theta_{\\mathrm{K_3Fe(CN)_6/K_4Fe(CN)_6}} = 0.361\\ \\mathrm{V}$; $\\mathrm{p}K_{\\mathrm{a1},\\mathrm{a2},\\mathrm{a3}}(\\mathrm{H_3PO_4}) = 2.16,\\ 7.21,\\ 12.3$; $\\mathrm{p}K_\\mathrm{a}(\\mathrm{CH_3COOH}) = 4.76$; $\\mathrm{p}K_{\\mathrm{a1},\\mathrm{a2}}(\\mathrm{H_2CO_3}) = 6.37,\\ 10.33$; $\\Delta_\\mathrm{r}H_\\mathrm{c}(\\mathrm{CH_3COOH}) = -875\\ \\mathrm{kJ\\cdot mol^{-1}}$; $\\Delta_\\mathrm{r}H_\\mathrm{c}(n-\\mathrm{C_3H_7COOH}) = -2184\\ \\mathrm{kJ\\cdot mol^{-1}}$. A microbial fuel cell (MFC) uses microorganisms as catalysts to convert the chemical energy of organic and inorganic substances directly into electrical energy. MFCs oxidatively decompose organic matter in wastewater, such as glucose and acetic acid, while generating electricity. This treatment both reduces the chemical oxygen demand of the wastewater and converts organic matter into useful energy.", "header_images": [], "subquestions": [{"sub_id": "Exam-04-T7-1", "tags": ["3.4", "3.6"], "text_open": "The electrode-solution compositions, as analytical concentrations, are: electrode A, 0.0223 M $Na_2HPO_4$, 0.0359 M $NaH_2PO_4$, 0.0504 M $K_3[Fe(CN)_6]$, and 0.0167 M $K_4[Fe(CN)_6]$; electrode B, 0.0223 M $Na_2HPO_4$, 0.0359 M $NaH_2PO_4$, 0.0132 M $CH_3COONa$, and 0.0075 M $NaHCO_3$. The pH at A is 7.00 and B may be treated as having the same pH.\n7-1-2 Assuming the pH values at electrodes A and B are approximately equal, calculate the electrode potentials of A and B under these conditions. Give 3 significant figures.", "text_images": [], "full_mark": 4}]} -{"problem_id": "Exam-04-T10", "header": "Problem 10 Hydrazones and Their Salts: Our Advantage", "header_images": [], "subquestions": [{"sub_id": "Exam-04-T10-1", "tags": ["2.2", "2.3"], "text_open": "Reaction background: The reaction shown below first converts the ketone into a hydrazone intermediate and then effects elimination at the carbonyl $\\alpha$ carbon. The “second step” is conversion of the hydrazone intermediate to the final product under $H^+/H_2O$.\n10-1-2 Explain why the hydrazone is introduced, and give the SMILES of one key intermediate in the second step.", "text_images": [{"file": "Exam-04-T10-1-1-prompt.png", "description": "Two-step reaction. The leftmost structure is '[*]C1C([*])CC[C@]2([H])[C@@H](Br)C(CC[C@@]21C)=O'. Each * is attached to a wavy line denoting an additional ring system that does not participate and need not be considered. Reaction with ethoxycarbonyl hydrazine 'NNC(OCC)=O' gives '[*]C1C(CC[C@@]2([C@H](/C(CC[C@@]21C)=N\\NC(OCC)=O)Br)[H])[*]', with the same meaning for * and the upper-right wavy line. H^+/H_2O then gives '[*]C1C(CCC2=CC(CC[C@@]21C)=O)[*]', again with the same meaning for *."}], "full_mark": 3}, {"sub_id": "Exam-04-T10-2", "tags": ["2.3"], "text_open": "10-2 A hydrazone can be reduced under basic conditions to a C-C double bond. 10-2-1 A dianion is known to appear during the reaction. Which statements are correct? (a) The first proton removed is on N. (b) The first proton removed is on C. (c) The reaction rate is controlled by departure of Ts^-. (d) The reaction rate is controlled by the second deprotonation. (e) It cannot be determined.", "text_images": [{"file": "Exam-04-T10-2-1-prompt.png", "description": "Reaction of hydrazone '[R1]/C(C[R2])=N/NS(C1=CC=C(C)C=C1)(=O)=O' under 1) Base and 2) TMSCl gives the C-C double-bond product '[R1]/C([Si](C)(C)C)=C/[R2]'."}], "full_mark": 1}, {"sub_id": "Exam-04-T10-3", "tags": ["2.7", "2.8"], "text_open": "Consider the reaction shown; Quench means quenching. With potassium carbonate as base, product A forms. With LDA, A and a small amount of B form. The substituent R also affects the outcome. 10-2-2 With LDA, arrange the following R groups in decreasing B yield/A yield: (a) R = H; (b) R = Me; (c) R = F.", "text_images": [{"file": "Exam-04-T10-3-1-prompt.png", "description": "Multiple pathways from common substrate 'O=C(CC(C)(C)C1)C([R])C1=O'. Downward reaction with TsNHNH_2, 1 eq, gives bracketed intermediate M. M gives A, 'O=C1CC(C)(C)CC=C1[R]', under 1) K_2CO_3, heat; 2) Quench. Under 1) LDA; 2) Quench, M gives both A and B, where B is 'O=C1CC(C)(C)C=CC1[R]'. SMILES/structure strings shown: 'O=C1CC(C)(C)CC=C1[R]'."}], "full_mark": 1}, {"sub_id": "Exam-04-T10-4", "tags": ["2.2", "2.4"], "text_open": "10-3 Similarly, the N-N bond of a hydrazone can be converted into a diazo group, which undergoes many interesting reactions. 10-3-1 Give the SMILES of intermediate C.", "text_images": [{"file": "Exam-04-T10-4-1-prompt.png", "description": "Two-step reaction. The left substrate is 'BrC1=C(OC)C(OC)=C(C=NNS(C2=CC=C(C)C=C2)(=O)=O)C(COC)=C1'. It reacts with 'BrC1=C(OC)C=CC(B(O)O)=C1' under K_2CO_3, 1,4-dioxane, 110 ℃ to give unknown C. C then gives 'BrC1=C(OC)C(OC)=C(CC2=CC=C(OC)C(Br)=C2)C(CO)=C1' under BBr_3, CH_2Cl, then THF/H_2O."}], "full_mark": 2}, {"sub_id": "Exam-04-T10-5", "tags": ["1.2", "2.4"], "text_open": "10-3-2 Based on the reaction above, give the SMILES of products D, E, and F. Include stereochemistry.", "text_images": [{"file": "Exam-04-T10-5-1-prompt.png", "description": "Reaction producing D. Substrate '[R2][C@@]([R3])([BPin])[C@](C/1)([R1])CC1=N\\NC2=C(C=C(C=C2C)C)C' gives unknown D under Cs_2CO_3, 3.0 equiv, in 1,4-dioxane at 100 ℃."}, {"file": "Exam-04-T10-5-2-prompt.png", "description": "Reaction producing E. 'CC1=CC=C(S(NN=C2[C@@H](CCN=[N+]=[N-])CCCC2)(=O)=O)C=C1' plus '[R1]B(O)O' gives unknown E under K_2CO_3, 3.0 equiv, in 1,4-dioxane at 150 ℃."}, {"file": "Exam-04-T10-5-3-prompt.png", "description": "Reaction producing F. 'CC1=CC=C(S(NN=C2CCN(CC#N)CC2)(=O)=O)C=C1' plus '[R2]/C=C/B(O)O' gives unknown F under K_2CO_3, 2.0 equiv, in 1,4-dioxane with MW at 120 ℃."}], "full_mark": 6}, {"sub_id": "Exam-04-T10-6", "tags": ["2.2"], "text_open": "10-4 Under suitable conditions, a hydrazone can generate a highly reactive carbene. When first proposed, the reaction shown was called an \"unusual carbon-skeleton rearrangement\". After spectroscopic structure determination and isotope labeling, a mechanism was proposed that includes an intramolecular ring closure and a \"questionable\" addition to an sp-hybridized carbon. 10-4-1 Give the SMILES of 2 neutral key intermediates.", "text_images": [{"file": "Exam-04-T10-6-1-prompt.png", "description": "One-step reaction. Reactant 'CC1=CC=C(C=C1)S(N/N=C(C2C(/C3=N/NS(C4=CC=C(C=C4)C)(=O)=O)C2)\\C5C3C5)(=O)=O' under excess NaH in diglyme at 190 ℃ gives 'C12=CC=NN1C=CC3C2C3'."}], "full_mark": 4}, {"sub_id": "Exam-04-T10-7", "tags": ["2.3", "3.7"], "text_open": "10-4-2 When the two three-membered rings of the reactant are on the same side, the yield is 60%; when on opposite sides, it is 40%. Explain this in terms of activation energy.", "text_images": [], "full_mark": 1}]} +{"problem_id": "Exam-04-T8", "header": "Problem 8 Basic Organic Chemistry", "header_images": [], "subquestions": [{"sub_id": "Exam-04-T8-1", "tags": ["1.7"], "text_open": "8-2 Arrange the following radicals in decreasing order of stability.", "text_images": [{"file": "Exam-04-T8-1-1-prompt.png", "description": "Option a, SMILES '[CH]1(C2)CC[C]2CC1'."}, {"file": "Exam-04-T8-1-2-prompt.png", "description": "Option b, SMILES '[C]1(CC2)CCC2CC1'."}, {"file": "Exam-04-T8-1-3-prompt.png", "description": "Option c, SMILES '[CH]1(C[CH](C2)C3)C[C]2C[CH]3C1'."}, {"file": "Exam-04-T8-1-4-prompt.png", "description": "Option d, SMILES 'C12CCCC[C]1CCCC2'."}], "full_mark": 2}, {"sub_id": "Exam-04-T8-2", "tags": ["4.1"], "text_open": "8-3 Identify the three compounds shown by chemical tests. Available reagents are: (a) acetone (b) NaI (c) I_2 (d) concentrated hydrochloric acid (e) CuSO4 solution (f) NaOH solution.", "text_images": [{"file": "Exam-04-T8-2-1-prompt.png", "description": "Compound to be identified, SMILES 'CC(C)CCCl'."}, {"file": "Exam-04-T8-2-2-prompt.png", "description": "Compound to be identified, SMILES 'CC(C)CCO'."}, {"file": "Exam-04-T8-2-3-prompt.png", "description": "Compound to be identified, SMILES 'CC(C)CC=O'."}], "full_mark": 4}, {"sub_id": "Exam-04-T8-3", "tags": ["1.7"], "text_open": "8-4 At room temperature, X-ray diffraction shows that D-glucose crystallized from water contains only the α anomer, whereas its specific rotation after dissolution in water shows that the solution contains a mixture of the α and β anomers. 8-4-2 State which anomer is more abundant in aqueous solution and explain why.", "text_images": [{"file": "Exam-04-T8-3-1-prompt.png", "description": "Structure of the α anomer, SMILES 'O[C@H]1[C@H](O)[C@@H](CO)O[C@H](O)[C@@H]1O'."}], "full_mark": 3}]} {"problem_id": "Exam-05-T1", "header": "Problem 1 Chemistry of Element M. Historically, the mining and use of minerals largely determined the progress of elemental research. Element M and another element are named after the same country. M is about as abundant as nitrogen in Earth's crust but is extremely dispersed. Minerals containing M alone are uncommon, but binary compound A of M and element N often occurs with sphalerite. A contains 48.20% M by mass and is widely used in optoelectronics.", "header_images": [], "subquestions": [{"sub_id": "Exam-05-T1-1", "tags": ["2.1"], "text_open": "Given $M=Ga$, $B=GaCl$, $C=GaCl_3$, and $N=As$.\n1-1-2 Write the equation for the reaction of B with N.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-05-T1-2", "tags": ["1.4", "1.5"], "text_open": "1-1-3 D can also be prepared in benzene without heating. Explain why D forms readily in benzene by giving the SMILES of the relevant structure. No more than 100 words.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-05-T1-3", "tags": ["1.4", "2.4"], "text_open": "1-2 At 250 K, chloride C reacts with $\\mathrm{Me_3SiH}$ to form E. E commonly exists as a dimer in which all chlorine atoms have equivalent environments. 1-2-1 Give the SMILES of the actual structure of E.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-05-T1-4", "tags": ["1.4", "2.4", "2.7"], "text_open": "1-2-2 At low temperature, the different reactions of E with ammonia and trimethylamine attracted attention. IR and other data show that the products have similar compositions but different coordination numbers of M; the coordination number in the ammonia product equals that in the E dimer. Neither product is dimeric or polymeric. Give the product SMILES for reaction with trimethylamine and ammonia, and explain the difference in no more than 20 words.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-05-T1-5", "tags": ["3.2"], "text_open": "1-3 The physical properties of element M are equally distinctive. At standard atmospheric pressure, the thermodynamic data for M and related compounds are given below. Assume that all $\\Delta H$ and $\\Delta S$ values are independent of temperature. Throughout the temperature range considered, the most stable oxide of M remains in the solid state.\n| Substance | $\\Delta_\\mathrm{f} H_m^\\circ \\, (\\mathrm{kJ\\cdot mol^{-1}})$ | $S_m^\\circ \\, (\\mathrm{J\\cdot mol^{-1}\\cdot K^{-1}})$ |\n|---|---:|---:|\n| $M(s)$ | 0 | 40.9 |\n| $M(l)$ | 5.6 | 56.0 |\n| $M(g)$ | 277.0 | 169.0 |\n| $O_2(g)$ | 0 | 205.2 |\n| Most stable oxide of M $(s)$ | -1089.0 | 88.0 |\n1-3-2 Determine the temperature range in which M reacts spontaneously with oxygen at standard atmospheric pressure.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-05-T1-6", "tags": ["3.2", "3.4"], "text_open": "1-3-3 Calculate the ambient pressure when the boiling point of M is 2200 K.", "text_images": [], "full_mark": 3}]} -{"problem_id": "Exam-05-T5", "header": "Problem 5 Structural Chemistry of Metal M. Formulas must use M and should be in approximately simplest ratios.", "header_images": [], "subquestions": [{"sub_id": "Exam-05-T5-1", "tags": ["1.3"], "text_open": "5-1 The crystal structure of chloride A of M and its projection along c are shown; large spheres are M and small spheres Cl. 5-1-1 Name the coordination polyhedron formed by Cl around one M.", "text_images": [{"file": "Exam-05-T5-1-1-prompt.png", "description": "Perspective view of crystal A inside a gray dashed hexagonal frame. Large dark-gray spheres are M and small light-gray spheres Cl. 2 M atoms at upper left and lower right share two bridging Cl atoms; each M also connects radially to 7 terminal Cl atoms. Around each M are three upper, three lower, and three equatorial Cl atoms. An axis triad has c upward, b lower right, and a left."}, {"file": "Exam-05-T5-1-2-prompt.png", "description": "Projection of A along c within a dashed parallelogram. Large spheres are M and small spheres Cl. Two M atoms, one at (1/3,2/3), share two bridging Cl atoms. Each M also has 4 radial terminal Cl atoms in projection, with three nearer and three farther Cl neighbors overall. The axis marker has a rightward, b upper-left, and c out of the page."}], "full_mark": 2}, {"sub_id": "Exam-05-T5-2", "tags": ["1.3"], "text_open": "5-1-2 Give the number of distinct Cl environments and the coordination number of Cl.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-05-T5-3", "tags": ["1.3"], "text_open": "5-1-3 Give the crystal system and formula of A.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-05-T5-4", "tags": ["1.3"], "text_open": "5-2 M forms complex compounds with Ni and O. Projections of a mixed oxide X along b and c are shown; large spheres are M, small spheres O, and medium spheres Ni. Give its formula.", "text_images": [{"file": "Exam-05-T5-4-1-prompt.png", "description": "Crystallographic projection of the unit cell of composite oxide X viewed along b. The vertical a-axis and horizontal c-axis each range from 0 to 1. Large spheres are M, medium spheres Ni, and small spheres O. Approximate fractional projection coordinates (a,c) are: M, (1.00,0.25), (0.00,0.75), (0.50,0.25), (0.50,0.75); Ni, the four vertices (0.00,0.00), (1.00,0.00), (0.00,1.00), (1.00,1.00), the edge midpoints (1.00,0.50), (0.00,0.50), (0.50,0.00), (0.50,1.00), and the face center (0.50,0.50); O, approximately (0.76,0.04), (0.68,0.04), (0.42,0.25), (0.12,0.25), (0.76,0.50), (0.68,0.50), (0.30,0.50), (0.24,0.50), (0.88,0.75), (0.58,0.75), (0.30,0.96), and (0.24,0.96). The lower-left direction marker points a upward and c rightward."}, {"file": "Exam-05-T5-4-2-prompt.png", "description": "Crystallographic projection of the unit cell of composite oxide X viewed along c. The horizontal a-axis and vertical b-axis each range from 0 to 1. Large spheres are M, medium spheres Ni, and small spheres O. Approximate fractional projection coordinates (a,b) are: M, upper-left (0.02,0.90), lower-right (0.98,0.10), and the two vertically overlapping central sites (0.50,0.55) and (0.50,0.45); Ni, near the edge centers (0.00,0.47), (1.00,0.47), (0.53,1.00), and (0.53,0.00); O, next to boundary Ni sites at approximately (0.08,0.47), (0.92,0.47), (0.45,0.98), and (0.61,0.02), plus interior sites at (0.30,0.70), (0.78,0.78), (0.22,0.22), and (0.70,0.30). The lower-left direction marker points b upward and a rightward."}], "full_mark": 2}, {"sub_id": "Exam-05-T5-5", "tags": ["1.3"], "text_open": "5-3 Compound Y, prepared from an oxide of M, NiO, and B_2O_3, contains M, Ni, O, and B and has a primitive monoclinic lattice. It consists of M cations, Ni^2+, and a network borate built from spiro units sharing sp^3-hybridized B atoms. The network contains [B_3O_3] six-membered and [B_4O_4] 8-membered rings. In every [B_4O_4], all four B atoms are four-coordinate and shared with neighboring [B_3O_3] rings; two ring O atoms coordinate one metal each and two coordinate two metals each. In every [B_3O_3], two B atoms are four-coordinate and shared with adjacent rings, while one is three-coordinate. The three ring O atoms each coordinate one metal; half the terminal O atoms coordinate two metals and half three. A monoclinic unit cell has [B_4O_4] at the corners and body center, each connected to four of eight neighboring [B_4O_4] rings through two [B_3O_3] rings. 5-3-1 Determine the number of B atoms per unit cell.", "text_images": [{"file": "Exam-05-T5-5-1-prompt.png", "description": "Schematic of two [B_4O_4] rings connected through two vertex-sharing [B_3O_3] rings, topologically like 'C1CCCC2(CCCC3(CC4(CCCCCCC4)CCC3)C2)CCC1'."}], "full_mark": 2}, {"sub_id": "Exam-05-T5-6", "tags": ["1.3"], "text_open": "5-3-2 Give the simplest formula of the borate anion.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-05-T5-7", "tags": ["1.4", "2.1"], "text_open": "5-3-3 M is 9-coordinate, Ni^2+ five-coordinate, and M:Ni = 1:1. Give the formula of Y.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-05-T5-8", "tags": ["1.4"], "text_open": "5-3-4 State the oxidation state of M.", "text_images": [], "full_mark": 1}]} +{"problem_id": "Exam-05-T4", "header": "Problem 4 Phosphoric Acid and Phosphate Chemistry. Phosphates are important in industrial production, food additives, and other fields, so measuring their content and understanding their properties are significant.", "header_images": [], "subquestions": [{"sub_id": "Exam-05-T4-1", "tags": ["2.1"], "text_open": "4-1 Aqueous sodium phosphate dodecahydrate, Na_3PO_4·12H_2O, is strongly basic and is used in washing powders and fat-saponification agents. It is prepared by reacting calcium dihydrogen phosphate with sodium carbonate and then adding sodium hydroxide solution. Write the preparation equation.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-05-T4-2", "tags": ["4.5"], "text_open": "Titration notation: Add a volume $V_0$ of standard HCl of concentration $c_1$ to the sample, then titrate with standard NaOH of concentration $c_2$. The first jump near pH 4.0 consumes a NaOH volume $V_1$, and continuing to the second jump near pH 8.8 consumes an additional volume $V_2$.\n4-2-2 What relation among V_0, V_1, V_2, c_1, and c_2 indicates that the sample contains Na_2HPO_4? Express it symbolically.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-05-T4-3", "tags": ["4.5", "4.7"], "text_open": "4-2-3 Calculate the mass fraction ω(%) of Na_3PO_4 in the sample. Significant figures are required.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-05-T4-4", "tags": ["1.7", "2.1"], "text_open": "4-3-2 Phosphorus atoms in different oxidation states can also be linked in catenated phosphoric acids. H_4P_2O_5 contains linked P(Ⅱ) and P(Ⅳ). It has two hydrolysis pathways under basic conditions. Only the first generates gas, and each gives only one phosphorus-containing product. Write the ionic equations for both pathways.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-05-T4-5", "tags": ["2.4"], "text_open": "4-3-3 Hydrolysis of cyclic H_6P_6O_12 is complex. First, one P-P bond cleaves to form a linear catenated hexaphosphoric acid by disproportionation; then terminal P-P bonds cleave stepwise. How many ways can the linear hexaphosphoric acid hydrolyze to catenated pentaphosphoric acid? Give the products of each. Use P for a phosphorus-containing group and a superscript to denote oxidation state; for example, catenated diphosphoric acid is P^4-P^4.", "text_images": [], "full_mark": 4}]} {"problem_id": "Exam-05-T9", "header": "NHC Chemistry", "header_images": [], "subquestions": [{"sub_id": "Exam-05-T9-1", "tags": ["2.2"], "text_open": "9.1 The benzoin condensation is familiar. Give the SMILES of the four key intermediates in the catalytic cycle shown.", "text_images": [{"file": "Exam-05-T9-1-1-prompt.png", "description": "Under NEt_3, 'CC[N+]1=CSC(CCO)=C1' gives 'CC[N+]1=[C-]SC(CCO)=C1'. Reaction with '[H]C(C1=CC=CC=C1CCC(C)=O)=O' gives A; proton transfer gives B; B gives C; proton transfer gives D; D gives 'O=C1C(O)(C)CCC2=CC=CC=C21' and regenerates 'CC[N+]1=[C-]SC(CCO)=C1'. SMILES/structure strings shown: 'CC[N+]1=[C-]SC(CCO)=C1'."}], "full_mark": 4}, {"sub_id": "Exam-05-T9-2", "tags": ["2.2"], "text_open": "9.3 Formaldehyde reacts poorly directly with NHCs. The reaction shown introduces a fragment equivalent to formaldehyde. Give the SMILES of the three key intermediates between the five-carbon aldose substrate and intermediate X.", "text_images": [{"file": "Exam-05-T9-2-1-prompt.png", "description": "'O[C@H]1[C@@H](O)[C@H](O)C(O)OC1' and 'O=C(C1=CC=CC=C1)/C=C/C2=CC=CC=C2' give 'C[N+]1=CSC(C)=C1C.[I-]' under 20 mol% 'O=C(C1=CC=CC=C1)CC(C([H])=O)C2=CC=CC=C2', 20 mol% K_2CO_3, MeCN, 130 °C, μW, 10 min. Intermediate X: 'O/C([H])=C1N(C)C(C)=C(C)S\\1'."}], "full_mark": 3}, {"sub_id": "Exam-05-T9-3", "tags": ["2.2"], "text_open": "9.4 When the reactant is not an aldehyde, NHCs can catalyze other reactions. The following reaction of an α-hydroxy ketone is NHC-catalyzed; the catalytic cycle is shown, and G carries one positive charge. 9.4.1 Compare the molecular formulas of reactants and products and identify the small-molecule product Y.", "text_images": [{"file": "Exam-05-T9-3-1-prompt.png", "description": "Reaction of 'O=C(C(C)(C)O)/C=C/C1=C(C)C=C(C)C=C1C' and 'CC1NCCCC1' under 10 mol% of the listed NHC catalyst, 10 mol% of the listed hydroxamic reagent, and 20 mol% DBU, DCM, room temperature gives 'C[C@@H]1N(C(CCC2=C(C)C=C(C)C=C2C)=O)CCCC1' and 'C[C@H]1NCCCC1'. SMILES/structure strings shown: 'CC1=CC(C)=CC(C)=C1[N+]2=CN3C(CCC3)=N2.[O-]Cl(=O)(=O)=O', 'O=C(CO1)N(O)[C@@H]([C@@H]1C2)C3=C2C=CC=C3'."}, {"file": "Exam-05-T9-3-2-prompt.png", "description": "Catalytic cycle: 'C[C@@H]1NCCCC1' reacts with H to give 'O=C(N1[C@@H](C)CCCC1)CCC2=C(C)C=C(C)C=C2C' and the cyclic hydroxamic product. The azolium catalyst and DBU form the NHC. The cyclic hydroxamic product and G give H and regenerate the NHC after loss of [H^+]. The NHC reacts with the α-hydroxy ketone to give E; E gives Y and F; protonation of F gives G. SMILES/structure strings shown: 'O=C1N(O)[C@H](C(C=CC=C2)=C2C3)[C@H]3OC1', 'O=C1N(O)[C@H](C(C=CC=C2)=C2C3)[C@H]3OC1', 'CC1=CC(C)=CC(C)=C1[N+]2=CN3C(CCC3)=N2', 'CC1=CC(C)=CC(C)=C1[N+]2=[C-]N3C(CCC3)=N2', 'CC1=CC(C)=CC(C)=C1[N+]2=[C-]N3C(CCC3)=N2', 'CC1=CC(C)=CC(C)=C1[N+]2=[C-]N3C(CCC3)=N2', 'O=C(C(C)(C)O)/C=C/C1=C(C)C=C(C)C=C1C'."}], "full_mark": 1}, {"sub_id": "Exam-05-T9-4", "tags": ["2.2", "2.3"], "text_open": "9.4.2 Give the SMILES of the four key intermediates E, F, G, and H.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-05-T9-5", "tags": ["2.2", "2.3"], "text_open": "9.5 With a vicinal diketone substrate, the reaction mode changes again. Give the SMILES of the four key intermediates. NEt_3 acts only as a base with the thiazolium ion; the resulting NHC first reacts with the vicinal diketone. Two addition-elimination events occur and each is treated as one step; their tetrahedral intermediates need not be shown. The NHC formed in the initial acid-base step also need not be shown.", "text_images": [{"file": "Exam-05-T9-5-1-prompt.png", "description": "'[Ar]C(C=C)=O' and 'O=C(C(C1=CC=CC=C1)=O)C2=CC=CC=C2' react under 'C[N+]1=CSC2=C1CCCCC2' and NEt_3 to give '[Ar]C(C(C(C1=CC=CC=C1)=O)CC(C2=CC=CC=C2)=O)=O'."}], "full_mark": 7}]} -{"problem_id": "Exam-05-T10", "header": "Problem 10 Sulfur-Nitrogen Chemistry", "header_images": [], "subquestions": [{"sub_id": "Exam-05-T10-1", "tags": ["1.5", "1.6"], "text_open": "10.1 The sulfilimine group, -NS(O)R_2, is an S,N substituent generally considered strongly electron-withdrawing. Calculations show that with a π acceptor attached, N-S contains a σ bond and a π bond strongly polarized toward N. 10.1.1 Give all resonance forms with single S-O bonds and identify the form closest to reality. Show exact connectivity.", "text_images": [{"file": "Exam-05-T10-1-1-prompt.png", "description": "'O=C(N=S(C)(C)=O)CC'."}], "full_mark": 4}, {"sub_id": "Exam-05-T10-2", "tags": ["2.3", "2.7"], "text_open": "10.1.2 Explain the regioselectivity and give the major-product SMILES with alkene geometry. No more than 100 words.", "text_images": [{"file": "Exam-05-T10-2-1-prompt.png", "description": "Reaction of 'O=S(C)(C)=NC#CC1=CC=CC=C1' with benzoic acid, 'OC(C1=CC=CC=C1)=O', gives the displayed product 'O=C(C1=CC=CC=C1)O/C(N=S(C)(C)=O)=C\\C2=CC=CC=C2'. A wavy bond between phenyl and the alkene means the actual E/Z configuration is unspecified in the question and must be inferred; geometry implied by the displayed SMILES is not necessarily correct."}], "full_mark": 2}, {"sub_id": "Exam-05-T10-3", "tags": ["2.2", "2.4"], "text_open": "10.1.3 Give the product and key intermediates in the actual mechanism. All key intermediates are neutral; the product contains a non-aromatic carbocycle.", "text_images": [{"file": "Exam-05-T10-3-1-prompt.png", "description": "Reaction of 'O=S(C)(C)=NC#CC1=CC=CC=C1' and 'ClC(C(C)C)=O' under NEt_3 to an unknown product."}], "full_mark": 3}, {"sub_id": "Exam-05-T10-4", "tags": ["2.2"], "text_open": "10.2 Sulfinamide derivatives enable asymmetric synthesis of α-amino imines and derivatives. Piv = pivaloyl, -COtBu. 10.2.1 Give the 3 key-intermediate SMILES.", "text_images": [{"file": "Exam-05-T10-4-1-prompt.png", "description": "Substrate A, 'CC/C(C1=CC=CC=C1)=N\\[S@@](C(C)(C)C)=N/C(C(C)(C)C)=O', reacts under 1) KOtBu, THF, -78 ℃, 0.5 h; 2) PFP, -78 ℃, 0.5 h, then -40 ℃, 1 h, to give B, 'CC(N(C(C(C)(C)C)=O)SC(C)(C)C)/C(C1=CC=CC=C1)=N/C2=C(F)C(F)=NC(F)=C2F', in 84% yield and 98.5:1.5 er. PFP is 'FC1=C(F)C(F)=C(F)N=C1F'."}], "full_mark": 3}, {"sub_id": "Exam-05-T10-5", "tags": ["2.3"], "text_open": "10.2.2 Classify the rearrangement in the final step.", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-05-T10-6", "tags": ["1.2", "2.3"], "text_open": "10.2.3 From the lowest-energy transition state, give the stereochemical SMILES of B.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-05-T10-7", "tags": ["1.2"], "text_open": "10.2.4 Assign the absolute configuration of B.", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-05-T10-8", "tags": ["1.5", "1.6"], "text_open": "10.3 Thiazole (1) is an aromatic S,N heterocycle whose C-H bond is more acidic than that of oxazole (2). 10.3.1 Which qualitative statement about C-S molecular orbitals is correct? (a) S is more electronegative, so its orbitals are lower and its coefficient in σ(C-S) is larger. (b) S is more electronegative, so its orbitals are higher and its coefficient in σ*(C-S) is larger. (c) S is larger, so its orbitals are lower and the C coefficient in σ(C-S) is larger. (d) S is in a higher period, so its orbitals are higher and its coefficient in σ*(C-S) is smaller.", "text_images": [{"file": "Exam-05-T10-8-1-prompt.png", "description": "Compound 1: '[H]C1=NC=CS1'; compound 2: '[H]C1=NC=CO1'; the right C-H bond is bold in each."}], "full_mark": 1}, {"sub_id": "Exam-05-T10-9", "tags": ["1.6", "1.7"], "text_open": "10.3.3 Based on 10.3.1, use orbital hyperconjugation to explain why thiazole is more C-H acidic than oxazole. No more than 100 words.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-05-T10-10", "tags": ["2.2", "1.7"], "text_open": "10.3.4 2-Bromothiazole reacts with LDA at -78 ℃ to give ring-opened F. Give two cyclic key intermediates and rank the acidities of thiazole hydrogens by position.", "text_images": [{"file": "Exam-05-T10-10-1-prompt.png", "description": "'BrC1=NC=CS1' gives 'Br/C(S[Li])=C/[N+]#[C-]' under LDA at -78 ℃."}, {"file": "Exam-05-T10-10-2-prompt.png", "description": "Thiazole 'C1=CSC=N1' labeled 2 at C between S and N, 4 at lower-left C, and 5 at upper-left C."}], "full_mark": 3}, {"sub_id": "Exam-05-T10-11", "tags": ["2.4"], "text_open": "10.4 Traditional thiazole syntheses often suffer unstable substrates and low yields; industrial methods use transition-metal coupling. In 2025, a metal-free synthesis from α-amino-acid derivatives gave substantial yields. 10.4.1 The reaction was discovered under Curtius-like conditions, but the expected rearrangement did not occur. Give the expected-product SMILES.", "text_images": [{"file": "Exam-05-T10-11-1-prompt.png", "description": "Reaction (a): the amino-acid substrate under SOCl_2, CH_2Cl_2, DBU, room temperature gives 'CC(C=C1)=CC=C1C2=NC=C(C3=CC=CC=C3)S2'. SMILES/structure strings shown: 'CC(C=C1)=CC=C1CN[C@@H](CC2=CC=CC=C2)C(O)=O'."}, {"file": "Exam-05-T10-11-2-prompt.png", "description": "Expected Curtius reaction of the same substrate under SOCl_2, DPPA, BnOH, K_2CO_3 to an unknown product. SMILES/structure strings shown: 'CC(C=C1)=CC=C1CN[C@@H](CC2=CC=CC=C2)C(O)=O', 'O=P(OC1=CC=CC=C1)(N=[N+]=[N-])OC2=CC=CC=C2'."}], "full_mark": 1}, {"sub_id": "Exam-05-T10-12", "tags": ["2.2", "2.4"], "text_open": "10.4.2 One substrate molecule first reacts with one equivalent SOCl_2 to give neutral X containing neither O nor S. Give X.", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-05-T10-13", "tags": ["2.2", "2.3"], "text_open": "10.4.3 Give three classes of sulfur-containing key intermediates from X to product.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-05-T10-14", "tags": ["2.1", "2.2"], "text_open": "10.4.4 Write the overall equation. DBU need not be included.", "text_images": [], "full_mark": 2}]} {"problem_id": "Exam-06-T1", "header": "Problem 1: The Immortal Red Dragon. The development of alchemy depended on the joint investigations of scholars in China and abroad. In Tiangong Kaiwu, Song Yingxing recorded a method for purifying the “Red Dragon.” Low-purity “Red Dragon” was first “calcined together for five double-hours” in a sealed iron vessel to obtain a silvery-white elemental metal that is liquid at room temperature. The metal was then reacted in the proportion “one jin of metal to two jin of shitingzhi (a nonmetallic element),” yielding “fourteen liang of vermilion and three liang five qian of secondary vermilion.” Both “vermilion” and “secondary vermilion” are high-purity forms of the “Red Dragon.”", "header_images": [], "subquestions": [{"sub_id": "Exam-06-T1-1", "tags": ["2.1"], "text_open": "1-1-2 Write the equation for the reaction that produces the silvery-white elemental metal.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-06-T1-2", "tags": ["2.1", "4.2"], "text_open": "1-2 Throughout the history of alchemy, many scholars explored the chemical properties of the “Red Dragon.” Because it is chemically stable, subduing the immortal “Red Dragon” was a longstanding goal of alchemists. (i) Boiling a mixture of “wood ash” and “lime” gives strongly basic “caustic potash.” Boiling this with excess sulfur produces a dark-red mixed solution called “liver of sulfur.” When the “Red Dragon” is placed in this solution, it “completely merges with the red holy water.” Adding acid generates foul-smelling “dragon's breath” and instantly turns the originally clear liquid turbid (reaction 1). After the reaction, a large amount of soft, pale yellow-brown powder remains at the bottom. (ii) The “Green Lion” is a salt with formula mass 278.02. When strongly heated in a furnace, it first releases clear water and then “roars” out pungent “white smoke,” which condenses into a highly corrosive viscous liquid called “oil of vitriol” (reaction 2). Heating “oil of vitriol” with “saltpeter,” a potassium salt, produces dense “red fumes” that condense into slightly yellow “strong water.” Adding “sal ammoniac,” an ammonium salt, to the “strong water” gives the “King's Sword,” meaning aqua regia. When the King's Sword pierces the “Red Dragon,” it dies amid “red fumes”; its body is completely consumed into clear water, leaving only “yellow grease” floating on the surface (reaction 3). The “yellow grease” is the same substance as shitingzhi. Write the equations for reactions 1-3. Note: reaction 1 comprises two reactions, and reaction 2 is the overall reaction by which the “Green Lion” forms “oil of vitriol.”", "text_images": [], "full_mark": 9}]} {"problem_id": "Exam-06-T3", "header": "Problem 3: Protection of Isocyanates (12 points, 5%)\nHMDI [OCN(CH_2)_6NCO] is a common monomer for polymerization reactions.", "header_images": [], "subquestions": [{"sub_id": "Exam-06-T3-1", "tags": ["1.7", "1.6"], "text_open": "3.3 Researchers recently developed a supramolecular method for protecting HMDI. The pentagonal-prismatic product P5A, obtained by Lewis-acid-catalyzed reaction of diethoxybenzene with paraformaldehyde, forms a 1:1 supramolecular complex with HMDI as shown below.\n3.3.1 Identify possible interactions between P5A and HMDI in the complex other than van der Waals forces.", "text_images": [{"file": "Exam-06-T3-1-1-prompt.png", "description": "Schematic 1:1 supramolecular complex of P5A and HMDI. The OCN-(CH_2)_6-NCO chain of HMDI threads through the cavity of pillar[5]arene P5A. The SMILES of P5A is 'CCOC1=CC(CC2=CC(OCC)=C(CC3=CC(OCC)=C(CC4=CC(OCC)=C(C5)C=C4OCC)C=C3OCC)C=C2OCC)=C(OCC)C=C1CC6=C(OCC)C=C5C(OCC)=C6'."}], "full_mark": 3}, {"sub_id": "Exam-06-T3-2", "tags": ["1.7", "2.8"], "text_open": "3.3.2 After synthesis, the complex can be washed with cyclohexane. When HMDI is needed for polymerization, the complex can be dissociated in DCM or chloroform. Explain why DCM or chloroform can dissociate the complex whereas cyclohexane cannot.", "text_images": [], "full_mark": 2}]} -{"problem_id": "Exam-06-T5", "header": "Problem 5: Layered Conductive Materials", "header_images": [], "subquestions": [{"sub_id": "Exam-06-T5-1", "tags": ["1.3"], "text_open": "5.1 CuSe crystal X has a layered structure. Its tetragonal unit cell is shown below, with dark spheres representing Cu and light spheres representing Se.\n5.1.1 Name the coordination polyhedron formed by the Se atoms around each Cu in crystal X.", "text_images": [{"file": "Exam-06-T5-1-1-prompt.png", "description": "A three-dimensional perspective view of a rectangular-prismatic unit cell drawn with thin dark-gray lines. Axes at the lower left show c vertically upward, b toward the right, and a toward the lower left. Ten larger blue spheres and four smaller green spheres are connected by cylindrical bonds. Blue spheres occupy all vertices of the prism and the face centers of the top and bottom faces. In the upper half, two green spheres are symmetrically positioned near the upper portions of the left and right faces. The left green sphere lies below several blue spheres at the upper left and is bonded to two blue spheres in the upper-left region and the blue sphere at the top center. The right green sphere similarly lies below the upper-right blue spheres and is bonded to two blue spheres in the upper-right region and the top-center blue sphere. In the lower half, two green spheres are symmetrically positioned near the lower portions of the front and back faces. The front green sphere is bonded to the lower-left and lower-right blue spheres of the front face and to the bottom-center blue sphere; the rear green sphere is bonded analogously to the corresponding rear-face blue spheres and bottom-center sphere. Overall, the green spheres lie between upper and lower sets of blue spheres, forming two layers in which green spheres bridge multiple blue spheres."}], "full_mark": 1}, {"sub_id": "Exam-06-T5-2", "tags": ["1.3"], "text_open": "5.1.2 Within each layer, these Cu-Se polyhedra are connected by ( ).\na. sharing vertices\nb. sharing edges\nc. sharing faces", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-06-T5-3", "tags": ["1.3"], "text_open": "5.1.3 State the lattice type of crystal X.", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-06-T5-4", "tags": ["1.3"], "text_open": "5.2 Translate the Cu-Se polyhedral layers in crystal X horizontally so that neighboring Se atoms in the upper and lower layers align along c, then fill all regular square-prismatic voids formed by Se with Tl to obtain tetragonal crystal Y.\n5.2.1 Write the chemical formula of crystal Y in the simplest integer ratio.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-06-T5-5", "tags": ["1.3"], "text_open": "5.2.4 State the lattice type of crystal Y.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-06-T5-6", "tags": ["1.3"], "text_open": "5.4 Replacing each single Cu-Se polyhedral layer in crystal Y with a double layer, in which the two component layers share one layer of Se atoms, gives crystal Z with a conduction mechanism similar to that of Y.\n5.4.1 Write the chemical formula of crystal Z.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-06-T5-7", "tags": ["1.3"], "text_open": "5.4.2 Approximate the 8 Se atoms around Tl in crystal Y as a perfect cube, and assume that the shapes and sizes of all coordination polyhedra remain unchanged in the transformation from Y to Z. State directly the value of unit-cell parameter c for the conventional unit cell of crystal Z.", "text_images": [], "full_mark": 2}]} -{"problem_id": "Exam-06-T8", "header": "Problem 8: Fundamental Organic Chemistry", "header_images": [], "subquestions": [{"sub_id": "Exam-06-T8-1", "tags": ["2.4", "2.7", "2.3"], "text_open": "8.2 Sulfur ylides are important nucleophilic carbon reagents that react with carbonyl compounds, α, β-unsaturated carbonyl compounds, and related substrates. Note: consider thermodynamic and kinetic products. 8.2.1 Compare the two reactions and write the SMILES representation of the major product of each.", "text_images": [{"file": "Exam-06-T8-1-1-prompt.png", "description": "'O=C1CCC(C(C)(C)C)CC1' reacts with '[CH2-][S+](C)C' in THF at 0 °C."}, {"file": "Exam-06-T8-1-2-prompt.png", "description": "'O=C1CCC(C(C)(C)C)CC1' reacts with 'C[S+](=O)(C)[CH2-]' in THF at 65 °C."}], "full_mark": 4}, {"sub_id": "Exam-06-T8-2", "tags": ["2.4", "2.7", "2.3"], "text_open": "8.2.2 Compare the two reactions and write the SMILES representation of the major product of each.", "text_images": [{"file": "Exam-06-T8-2-1-prompt.png", "description": "'O=C1CCC(C(=C)C)C=C1C' reacts with '[CH2-][S+](C)C' in THF at 0 °C."}, {"file": "Exam-06-T8-2-2-prompt.png", "description": "'CC(=C)C1CCC(=O)C(C)=C1' reacts with 'C[S+](=O)(C)[CH2-]' in THF at 50°C."}], "full_mark": 4}, {"sub_id": "Exam-06-T8-3", "tags": ["1.2", "1.7"], "text_open": "8.3 Molecular conformations. 8.3.1 Determine which side is favored in the following conformational equilibrium. 8.3.2 Determine which side is favored in the following conformational equilibrium. Note: combine the answers to 8.3.1-8.3.2 directly, for example, AA.", "text_images": [{"file": "Exam-06-T8-3-1-prompt.png", "description": "8.3.1 Interconversion equilibrium between molecule A and molecule B, whose SMILES are 'O=C1[C@@]2([H])[C@](N(C)CCC2)([H])CCC1' and 'O=C3[C@]4([H])[C@](N(C)CCC4)([H])CCC3', respectively."}, {"file": "Exam-06-T8-3-2-prompt.png", "description": "8.3.2 Interconversion equilibrium between molecule A and molecule B, whose SMILES are 'O=C1[C@@]2([H])[C@](N(C(C)=O)CCC2)([H])CCC1' and 'O=C3[C@]4([H])[C@](N(C(C)=O)CCC4)([H])CCC3', respectively."}], "full_mark": 2}, {"sub_id": "Exam-06-T8-4", "tags": ["4.3"], "text_open": "8.4.2 Examine the following four IR spectra and select from A-D the organic structure that best matches each spectrum. (A) phenol; (B) aniline; (C) N-methylaniline; (D) benzoic acid. Enter the capital letters in the order corresponding to (a)(b)(c)(d), for example, ABCD.", "text_images": [{"file": "Exam-06-T8-4-1-prompt.png", "description": "Four infrared spectra, each with a horizontal axis for vibrational absorption wavenumber in cm^{-1}; the four spectra are labeled (a), (b), (c), and (d) from left to right."}], "full_mark": 4}]} +{"problem_id": "Exam-06-T4", "header": "Problem 4 Determination of Tin by Potassium Iodate Titration. Tin is commonly determined by potassium iodate titration. The procedure includes sample decomposition, masking and separation, reductive fusion with zinc powder, reduction with aluminum powder, and titration of tin(II) ions with potassium iodate.", "header_images": [], "subquestions": [{"sub_id": "Exam-06-T4-1", "tags": ["2.1"], "text_open": "4.2 Sample pretreatment. Place 0.2026 g of the sample in a 250 mL beaker, add 2 g of potassium chlorate and 20 mL of concentrated hydrochloric acid, and digest at low temperature for 20 min. Add 60 mL of hydrochloric acid (1+1), 1 g of hydrazinium sulfate, and 0.3 g of potassium bromide, then heat and evaporate for 50 min. Remove from the heat and allow to cool slightly. Cover with a watch glass and add hydrogen peroxide dropwise until brownish-red bromine appears. Add 10 mL of EDTA solution (100 g/L) and 5 mL of beryllium sulfate solution (40 g/L), and neutralize with aqueous ammonia to pH = 9-10. Add a small amount of paper pulp and filter. Wash the precipitate with an aqueous ammonia-ammonium nitrate wash solution. Transfer the precipitate together with the filter paper to a crucible, dry, ash, and cool. 4.2.1 Write the ionic equation for the reduction of arsenic acid by hydrazine in hydrochloric acid medium, by which arsenic is removed through volatilization.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-06-T4-2", "tags": ["4.1", "4.7"], "text_open": "Reductive-fusion conditions: Add a zinc powder-borax-boric acid mixed flux to the crucible and cover it with additional mixed flux and NaCl. Place it in a furnace at 750 ℃, raise the temperature to 820 ℃, fuse for 30 min, then remove and cool.\n4.3.3 Explain how an excessively long fusion time or excessively high temperature affects the determination.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-06-T4-3", "tags": ["2.1", "4.5"], "text_open": "4.4.2 Write the ionic equations for the titration process.", "text_images": [], "full_mark": 2}]} +{"problem_id": "Exam-06-T7", "header": "Problem 7 Synthesis of Schiff Bases and Their Derived Complexes (25 points, 9%)\nOn 2023-4-28, a research team from Université de Bourgogne reported a class of Schiff-base ligands and their complexes that exhibit “fluxionality.” Here, “fluxionality” means that the “C=N” double bonds in the ligands can undergo “rotation” under specific conditions.", "header_images": [], "subquestions": [{"sub_id": "Exam-06-T7-1", "tags": ["2.4", "2.6"], "text_open": "7.1 The synthesis of Schiff-base ligands provides the basis for constructing their derived complexes. The following reactions synthesize the Schiff-base ligands L1H_{2} and L2H_{2}, respectively. The “C=N” bonds in both L1H_{2} and L2H_{2} are known to have the E configuration. Write the SMILES structures of L1H_{2} and L2H_{2}, respectively. Note: the figures are simplified representations of the synthetic routes and do not show the complete workup after each step.", "text_images": [{"file": "Exam-06-T7-1-1-prompt.png", "description": "Synthetic route to L1H_2: the molecule 'COC1=CC=CC=C1C(=O)N(CC)CC' is treated with (COCl_2) in Toluene at 333 K for 16h, then with NCCN 0.5 equiv. and NEt_3(5 equiv.) in DCM from 273 K to r.t. for 16h, subsequently with BBr_3, 3 equiv., in DCM from 273 K to r.t. for 16h, and finally with KHMDS (2 equiv) in THF to afford L1H_2."}, {"file": "Exam-06-T7-1-2-prompt.png", "description": "Synthetic route to L2H_2: the molecule 'OC1=CC=CC=C1C(=O)N(C)C' is treated under 1) SOCl_2, r.t., overnight, followed by 2) the molecule 'NC1=CC=CC=C1N', 0.45 equiv., and Et_3N (5 equiv.) in CH_2Cl_2 from 273 K to r.t. for 2.5h, to afford L2H_2."}], "full_mark": 4}, {"sub_id": "Exam-06-T7-2", "tags": ["1.4", "2.4"], "text_open": "7.2 The corresponding complexes can be constructed by reacting the above Schiff-base ligands with metal precursors. 7.2.1 L1H_{2} forms neutral complex A through the following reaction. L1H_{2} is known to exhibit almost no “fluxionality,” and Al in complex A is 4-coordinate. On this basis, write the SMILES structure of neutral complex A.", "text_images": [{"file": "Exam-06-T7-2-1-prompt.png", "description": "[L1H_4][Br_2] is treated under 1) KHMDS (2 equiv.), THF, r.t., 4h, followed by 2) AlMe_3, THF, r.t., 2h, to afford A."}], "full_mark": 3}, {"sub_id": "Exam-06-T7-3", "tags": ["1.4", "1.2", "2.4"], "text_open": "7.2.2 Because of the change in substituents, L2H_{2} exhibits some “fluxionality.” L2H_{2} forms neutral complex B_{1} through the following reaction. Given that Al in B_{1} is 4-coordinate, write the SMILES structure of B_{1}.", "text_images": [{"file": "Exam-06-T7-3-1-prompt.png", "description": "L2H_2 is treated with AlMe_3 in THF at 333K for 2h to afford B_1."}], "full_mark": 4}, {"sub_id": "Exam-06-T7-4", "tags": ["1.4", "2.2", "1.2"], "text_open": "7.2.3 The following reaction illustrates the above “fluxionality” of the ligand. Notably, complex B_{1} in 7.2.2 is chiral, and B_{1} can isomerize to B_{2} through the formation of a 5-coordinate intermediate C. Write the SMILES structure of intermediate C.", "text_images": [{"file": "Exam-06-T7-4-1-prompt.png", "description": "B1 ⇌ C ⇌ B2 isomerization scheme, emphasizing the conversion of B1 to B2 through 5-coordinate intermediate C. "}], "full_mark": 3}, {"sub_id": "Exam-06-T7-5", "tags": ["1.4", "1.2", "2.4"], "text_open": "7.2.4 The following reaction likewise demonstrates the “fluxionality” of the ligand. On this basis, write the SMILES structure of neutral complex D.", "text_images": [{"file": "Exam-06-T7-5-1-prompt.png", "description": "B_1 or B_2 is converted to D in the presence of 'CC(O)(C(F)(F)F)C(F)(F)F' in CD_2Cl_2 under reflux for 3 days."}], "full_mark": 3}, {"sub_id": "Exam-06-T7-6", "tags": ["1.4", "2.4"], "text_open": "7.3 Natural bond orbital calculations show that coordination of the second N atom in the ligand decreases the positive character of the Al center. Therefore, the more electron-deficient the Al center, the more readily it induces “rotation” of the ligand's C=N double bonds. On this basis, the research team further investigated how to prepare species with a more electron-deficient Al center.\n7.3.1 The following reactions provide further evidence for the theoretical prediction above. Given that both E and F are ionic complexes, write their SMILES structures.", "text_images": [{"file": "Exam-06-T7-6-1-prompt.png", "description": "A is converted to E with B(C_6F_5)_3 (1 equiv.) in CH_2Cl_2 at r.t. for 1h."}, {"file": "Exam-06-T7-6-2-prompt.png", "description": "B_1 or B_2 is converted to F with B(C_6F_5)_3 (1 equiv.) in CH_2Cl_2 at r.t. for 1h."}], "full_mark": 4}, {"sub_id": "Exam-06-T7-7", "tags": ["1.4", "2.4"], "text_open": "7.3.2 X-ray crystallography shows that E and F have dimeric structures in the solid state. Using E as an example, write the SMILES structure of ionic dimer G obtained by crystallization from pyridine, given that Al in G is 6-coordinate.", "text_images": [], "full_mark": 4}]} {"problem_id": "Exam-06-T9", "header": "Problem 9 Aziridines (30 points, 12%). Aziridine is an important electrophilic scaffold with diverse reactivity and broad applications in biomedicine and other fields.", "header_images": [], "subquestions": [{"sub_id": "Exam-06-T9-1", "tags": ["2.4", "2.2"], "text_open": "9.1 Nitrones react with alkynes to afford aziridine structures in high yields. However, when the nitrogen atom of the nitrone bears a tert-butyl group, aziridine intermediate B is converted into stable product C containing a five-membered ring. 9.1.1 Examine the reaction sequence and write the SMILES structures of A, B, and C. Stereochemistry is not required.", "text_images": [{"file": "Exam-06-T9-1-1-prompt.png", "description": "The molecule 'CC(C)(C)[N+]([O-])=C' reacts with the molecule 'COC(=O)C#CC(=O)OC' at 0 °C to form A. A then reacts at 80 °C to form intermediate B, which is ultimately converted to C."}], "full_mark": 4}, {"sub_id": "Exam-06-T9-2", "tags": ["2.3", "2.2"], "text_open": "9.1.2 Identify the type of rearrangement involved in the conversion of A to B.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-06-T9-3", "tags": ["2.3", "2.2"], "text_open": "9.1.3 The following reaction undergoes a process similar to that described above to form an aziridine intermediate. The aziridine moiety then fragments to form a new 1,3-dipole, followed by loss of an oxalate ester group to give the product. Based on this information, write the SMILES structures of the 4 key intermediates in the following reaction. Stereochemistry is not required.", "text_images": [{"file": "Exam-06-T9-3-1-prompt.png", "description": "The molecule 'CC1(C)N=C(C2=CC=CC=C2)C=[N+]1[O-]' reacts with the molecule 'COC(=O)C#CC(=O)OC' in CH_2Cl_2 at rt for 24 h to form the molecule 'CC(C)(N=C1C2=CC=CC=C2)N3C1=C(C(C(OC)=O)=C3C(OC)=O)C(OC)=O'."}], "full_mark": 6}, {"sub_id": "Exam-06-T9-4", "tags": ["2.4", "2.7", "1.2"], "text_open": "9.2 One method for synthesizing aziridines uses a Reformatsky reagent. The following reaction affords aziridine product D in relatively high conversion, together with a small amount of byproduct E, whose absolute configuration is not labeled. 9.2.1 Write the SMILES structure of aziridine product D, including stereochemistry.", "text_images": [{"file": "Exam-06-T9-4-1-prompt.png", "description": "'[H][C@@]1(N=C1C2=CC=CC=C2)C' reacts with 'BrC(C(OCC)=O)(C)C' in the presence of Zn in Et_2O/PhH (1:1) at 60°C for 2-5 h to form D and E. The SMILES of E is 'O=C(N3)C(C4(N(C4([H])C)C(C5=CC=CC=C5)=C3C)C6=CC=CC=C6)(C)C'. In the structure of E, the methyl-substituted carbon atom of the aziridine ring is labeled a."}], "full_mark": 2}, {"sub_id": "Exam-06-T9-5", "tags": ["2.2", "1.2"], "text_open": "9.2.2 The formation of E is known to proceed through an aziridine structure. Write the SMILES structures of the key intermediates in the conversion of the substrates to E, including stereochemistry.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-06-T9-6", "tags": ["1.2"], "text_open": "9.2.3 Use S/R notation to specify the absolute configuration of carbon atom a in product E.", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-06-T9-7", "tags": ["2.2", "1.5", "2.3"], "text_open": "9.3 Studies show that aziridine scaffolds can also be synthesized without adding an external dipole. Under photochemical conditions, an α-azidocinnamate forms a small-ring intermediate F whose structure resembles that of the nitrogen-containing substrate in the reaction in Problem 9.2. F can subsequently form dimeric and trimeric products. 9.3.1 During the oligomerization of F, photochemical ring opening first generates dipole G. Using the structure of the dimeric product, analyze G and write the SMILES structures of its reasonable resonance forms that satisfy an 8-electron configuration.", "text_images": [{"file": "Exam-06-T9-7-1-prompt.png", "description": "The substance whose structure in the figure has the SMILES '[N-]=[N+]=N/C(C(OC)=O)=C\\C1=CC=CC=C1' reacts under hν in acetone for 1 h at 30°C to form a product with the SMILES 'O=C(C2=N[C@H](N3[C@]2([C@H]3C4=CC=CC=C4)C(OC)=O)C5=CC=CC=C5)OC' and a trimer."}], "full_mark": 3}, {"sub_id": "Exam-06-T9-8", "tags": ["2.4", "2.2"], "text_open": "9.3.2 The trimer is formed when the photoexcited dimer undergoes addition with monomer F. On this basis, write the SMILES structure of the trimeric product. Stereochemistry is not required.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-06-T9-9", "tags": ["2.4", "2.3", "2.7"], "text_open": "9.3.3 The following azide is known to give different major products under thermal and photochemical conditions, although the two products have similar ring systems. Based on the information above, write the SMILES structures of the major products under thermal and photochemical conditions, respectively.", "text_images": [{"file": "Exam-06-T9-9-1-prompt.png", "description": "Structure of the azide substrate. The corresponding SMILES is 'O=C(/C(N=[N+]=[N-])=C/C1=CC=CC=N1)OCC'."}], "full_mark": 2}, {"sub_id": "Exam-06-T9-10", "tags": ["2.3", "2.4"], "text_open": "9.3.4 The following reaction synthesizes an aziridine scaffold through an ionic mechanism. Its design resembles that of the reaction above in that a dipolar intermediate is generated in situ, and the product contains a bridged ring system. On this basis, write the SMILES structures of the dipolar intermediate and the final product. Stereochemistry is not required.", "text_images": [{"file": "Exam-06-T9-10-1-prompt.png", "description": "The substance whose structure in the figure has the SMILES 'S=C(SC(C(O)=O)C1=CC=CC=C1)C2=CC=CC=C2' is treated sequentially under 1) Ac_2O/Et_3N and 2) 'CC3N=C3C4=CC=CC=C4'; the product is indicated by a question mark."}], "full_mark": 4}]} {"problem_id": "Exam-06-T10", "header": "Problem 10 Radical Cyclization and Photochemical Reactions (29 points, 12%). Compared with conventional ionic cyclization reactions, radical processes generally feature mild reaction conditions, broad functional-group compatibility, and, in some highly strained ring-forming processes, lower transition-state barriers. They have therefore attracted broad interest in natural-product synthesis and the construction of complex molecular scaffolds.", "header_images": [], "subquestions": [{"sub_id": "Exam-06-T10-1", "tags": ["2.2", "2.4", "1.2"], "text_open": "10.1 Radical cascade cyclization and ring expansion constitute an important strategy in modern organic synthesis for constructing medium-sized rings containing 8~11 members. The Dowd-Beckwith reaction, which efficiently converts a cyclic ketone scaffold into a medium-ring enone system, is a representative radical ring-expansion reaction. 10.1.1 Using the information above and the reaction shown below, write the SMILES structures of the key radical intermediates and the final product, paying attention to stereochemistry.", "text_images": [{"file": "Exam-06-T10-1-1-prompt.png", "description": "The molecule 'O=C(CCC[C@H]1[Sn](CCCC)(CCCC)CCCC)[C@]1(CCCCI)C' reacts under Bu_3SnH, AIBN, and PhH, reflux."}], "full_mark": 5}, {"sub_id": "Exam-06-T10-2", "tags": ["2.2", "2.4", "1.2"], "text_open": "10.1.2 Changing the configuration of one stereogenic center in the substrate causes a corresponding change in the product structure. Write the SMILES structure of the final product obtained from the substrate shown below under the same reaction conditions, and write the SMILES structure of the key intermediate conformation that accounts for the product stereochemistry.", "text_images": [{"file": "Exam-06-T10-2-1-prompt.png", "description": "Dowd-Beckwith substrate in which the configuration of a stereogenic center has been changed. The corresponding SMILES is 'O=C(CCC[C@@H]1[Sn](CCCC)(CCCC)CCCC)[C@]1(CCCCI)C'."}], "full_mark": 4}, {"sub_id": "Exam-06-T10-3", "tags": ["2.2", "2.4"], "text_open": "10.1.3 When the substrate is modified further, an unexpanded-ring product is observed in addition to the target medium-ring enone. Using the reaction below, write the SMILES structures of this unexpanded-ring product and the key intermediates through which it is formed.", "text_images": [{"file": "Exam-06-T10-3-1-prompt.png", "description": "The molecule 'O=C(CCC1)C([H])(CCCCI)[C@H]1[Sn](CCCC)(CCCC)CCCC' reacts under Bu_3SnH, AIBN, and PhH, reflux."}], "full_mark": 3}, {"sub_id": "Exam-06-T10-4", "tags": ["2.2", "2.4", "2.7"], "text_open": "10.2 Radicals can also be generated photochemically. For example, under irradiation, FeCl_{3} can undergo ligand-to-metal charge transfer, an LMCT process, to generate chlorine radicals capable of hydrogen-atom abstraction. 10.2.1 Under the standard reaction conditions, two products A and B are obtained in a ratio of 1: 1.4. When the temperature is raised to 60°C, the product ratio changes to 1: 4. Using this information, write the SMILES structures of the key intermediates leading to A and B, respectively. If the two pathways share a common intermediate, write that intermediate only once. Note: A and B are constitutional isomers.", "text_images": [{"file": "Exam-06-T10-4-1-prompt.png", "description": "The molecule 'CC(C(C)(C)C)=O' reacts with the molecule 'C=CC(OCC1=CC=CC=C1)=O' under FeCl_3 (25 mol%), a 390 nm LED, and MeCN, N_2, rt, 0.3 M, 36 h to form A and B."}], "full_mark": 7}, {"sub_id": "Exam-06-T10-5", "tags": ["2.7", "2.3", "1.7"], "text_open": "10.2.2 Keep the standard reaction conditions unchanged, but replace only the ketone substrate with the structure shown below. Determine whether the formation ratio of B relative to A increases or decreases, and explain why.", "text_images": [{"file": "Exam-06-T10-5-1-prompt.png", "description": "Structure of the replacement ketone substrate in 10.2.2. The corresponding SMILES is 'CC(C(C)C)=O'."}], "full_mark": 4}, {"sub_id": "Exam-06-T10-6", "tags": ["2.4", "2.3"], "text_open": "10.2.3 Keep the standard reaction conditions unchanged, but replace one of the alkene substrates with the compound shown below. Experimentally, the reaction does not produce the expected acyclic product but instead gives two cyclized products. Ignoring stereochemistry, write the SMILES structures of these two products.", "text_images": [{"file": "Exam-06-T10-6-1-prompt.png", "description": "Structure of the replacement alkene substrate in 10.2.3. The corresponding SMILES is 'N#C/C(C#N)=C/C1=CC=CC=C1'."}], "full_mark": 4}]} {"problem_id": "Exam-07-T3", "header": "Synthesis and Applications of Contrast Agents\nThe rapid expansion of medical magnetic resonance imaging has driven the development of contrast agents, most of which are complexes of element 64, Gd. Gd(III) is highly paramagnetic, making it an excellent candidate relaxation agent; among species that exist stably as hydrated ions, it is one of the most strongly magnetic.", "header_images": [], "subquestions": [{"sub_id": "Exam-07-T3-1", "tags": ["2.4", "1.4"], "text_open": "3.3 In one method for preparing a common Gd(III) contrast agent, aminoacetic acid is reacted with 1,2-dibromoethane in a 1:1 ratio in aqueous solution to give a tetrabasic acid X. Addition of X to $[Gd(H_2O)_8]^{3+}$ gives an anion A with a monocapped square-antiprismatic coordination geometry, one ligand of which is deprotonated X. Write the SMILES representations of X and A.", "text_images": [], "full_mark": 5}]} -{"problem_id": "Exam-07-T4", "header": "Determination of Sulfur Dioxide in Wine\nThe sulfur dioxide content of a wine sample is determined spectrophotometrically as follows. I. Construct a calibration curve. Accurately transfer 0.00 mL, 0.20 mL, 0.50 mL, 1.00 mL, 2.00 mL, and 3.00 mL of sulfur dioxide working standard solutions (containing 0.0 μg, 2.0 μg, 5.0 μg, 10.0 μg, 20.0 μg, and 30.0 μg sulfur dioxide, respectively) into 25 mL stoppered test tubes, and add formaldehyde-buffered absorbing solution to 10.00 mL. Then add, in sequence, 0.5 mL of 3 g/L ammonium sulfamate solution, 0.5 mL of 1.5 mol/L sodium hydroxide solution, and 1.0 mL of 0.5 g/L pararosaniline hydrochloride solution (structure shown below). Mix thoroughly and allow to stand for 20 min. Measure the absorbance of each standard solution at 579 nm using a UV–visible spectrophotometer. The following data are obtained:\n\n| $\\text{SO}_2$ content $m / \\mu\\text{g}$ | 0.0 | 2.0 | 5.0 | 10.0 | 20.0 | 30.0 |\n|---|---|---|---|---|---|---|\n| Absorbance $A$ | 0.003 | 0.062 | 0.159 | 0.312 | 0.618 | 0.928 |\n\nII. Accurately weigh 0.5000 g of the wine sample and add formaldehyde-buffered absorbing solution to 100 mL. After shaking for 2 h, take 10 mL of the solution and perform the spectrophotometric determination as in Step I. Repeat the measurement three times; the absorbances A’ are 0.637, 0.638, and 0.635. In China, the required sulfur dioxide content of alcoholic beverages is below $0.2500 \\ \\mathrm{g\\,dm^{-3}}$. Take the density of the wine to be $0.9 \\ \\mathrm{g\\,cm^{-3}}$. The structure of pararosaniline hydrochloride is 'NC1=CC=C(/C(C2=CC=C(N)C=C2)=C3C=CC(C=C\\3)=[NH2+])C=C1.[Cl-]'.", "header_images": [{"file": "Exam-07-T4-header-1.png", "description": "Structural diagram of pararosaniline hydrochloride; the SMILES representation is 'NC1=CC=C(/C(C2=CC=C(N)C=C2)=C3C=CC(C=C\\3)=[NH2+])C=C1.[Cl-]'."}], "subquestions": [{"sub_id": "Exam-07-T4-1", "tags": ["2.4"], "text_open": "4-1 The principle used in this titration is that A, formed by the addition reaction of sulfur dioxide with formaldehyde, forms a blue-violet complex with pararosaniline hydrochloride. Its absorbance is proportional to the sulfur dioxide concentration. Write the SMILES representation of A.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-07-T4-2", "tags": ["4.1", "2.8"], "text_open": "4-2 Why must 0.5 mL of 1.5 mol/L sodium hydroxide solution be added to adjust the pH before measurement? (No more than 50 words.)", "text_images": [], "full_mark": 2}]} +{"problem_id": "Exam-07-T5", "header": "Square-Planar Voids\nIn models of crystal packing and void filling, tetrahedral and octahedral voids are the most familiar. In fact, square-planar voids are also fairly common.", "header_images": [], "subquestions": [{"sub_id": "Exam-07-T5-1", "tags": ["1.3"], "text_open": "5.1 A binary compound AB belongs to the tetragonal crystal system. Its structure may be viewed as a simple cubic packing of B atoms in which A atoms occupy some of the square-planar voids. Among adjacent $AB_4$ quadrilaterals, the $AB_4$ quadrilaterals whose A atoms have the same c coordinate have identical spatial orientations and share edges, whereas the $AB_4$ quadrilaterals whose A atoms have different c coordinates share vertices. A projection of the crystal structure (containing more than one unit cell) along the c axis is shown below; the large spheres represent A atoms and the small spheres represent B atoms. The unit-cell parameters are a = 349.4 pm and c = 611.6 pm. All A atoms have the same chemical environment, and all B atoms likewise have the same chemical environment. Hint: Compare the unit-cell parameters with atomic radii when considering the unit-cell composition.\n5.1.1 Give the fraction of square-planar voids occupied by A atoms.", "text_images": [{"file": "Exam-07-T5-1-1-prompt.png", "description": "Orthographic top-view two-dimensional ball-and-stick model on a pure white background, with no text, axes, or labels. The main image contains 16 yellow spheres and 24 gray spheres: the 16 yellow spheres are arranged in 4 rows and 4 columns at the vertices of a square grid, while the 24 gray spheres lie exactly at the midpoints of the lines joining adjacent yellow spheres (12 in each of the horizontal and vertical directions). The spheres are connected by opaque, short, light-gray cylindrical rods. Each gray sphere is connected to two adjacent yellow spheres; depending on its position, each yellow sphere is connected to 2 to 4 adjacent gray spheres (coordination number four in the interior, three at an edge, and two at a corner). All spheres show a subtle 3D light-to-shadow gradient, and the rod ends are naturally occluded by the spheres."}], "full_mark": 2}, {"sub_id": "Exam-07-T5-2", "tags": ["1.3"], "text_open": "5.1.2 State the lattice type of the crystal.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-07-T5-3", "tags": ["1.3"], "text_open": "5.1.3 If the A atoms are regarded as forming the packing and the B atoms as filling the voids, what type of void do the B atoms occupy, and what fraction of these voids is occupied?", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-07-T5-4", "tags": ["1.3"], "text_open": "5.1.4 The theoretical density of the crystal is $10.10\\ \\mathrm{g\\,cm^{-3}}$. Determine the chemical formula of AB by calculation.", "text_images": [], "full_mark": 5}, {"sub_id": "Exam-07-T5-5", "tags": ["1.3"], "text_open": "5.2 NbO belongs to the cubic crystal system. In the unit cell of NbO, Nb atoms occupy all face centers and O atoms occupy all edge centers.\n5.2.2 For the crystal structure of NbO, determine whether each of the following statements is correct. Use T for true and F for false. 5.2.2.1 If octahedral voids are not counted as square-planar voids, do the Nb atoms occupy all square-planar voids formed by the O atoms?", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-07-T5-6", "tags": ["1.3"], "text_open": "5.2.2.3 The O atoms do not form a face-centered cubic packing, but in a projection of the crystal along a threefold axis, do the O atoms form a close-packed layer structure?", "text_images": [], "full_mark": 2}]} {"problem_id": "Exam-07-T7", "header": "The Chemistry of As, Sb, and Bi\nThe pnictogens are a familiar group of elements. N and P are well known for their wide range of applications in organic chemistry; As, Sb, and Bi can likewise bond to carbon and form many interesting compounds. ", "header_images": [], "subquestions": [{"sub_id": "Exam-07-T7-1", "tags": ["1.4"], "text_open": "7.1 The positive trivalent oxidation state is common among the pnictogens. Their lone pairs and vacant d orbitals give them Lewis basicity and acidity of varying strength. 7.1.1 In sharp contrast to $NR_3$, the inertness of the 6s electrons makes $BiR_3$ a rather weak base with poor coordination ability, although it can sometimes form complexes. The neutral complex A and the monocation B both contain a $BiPh_3$ ligand. In A, the mass fractions of Bi and the central metal Cr are 33.05 % and 8.22 %, respectively; in B, the mass fractions of Bi and the central metal Fe are 33.86 % and 9.04 %, respectively. The central metals in both A and B obey the 18-electron rule, and both complexes contain CO ligands. Write the chemical formulas of A and B.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-07-T7-2", "tags": ["1.4", "1.5"], "text_open": "7.1.2 Considering the structures of A and B, explain why the Bi-M bonds in them are stable. (No more than 50 words.)", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-07-T7-3", "tags": ["1.4", "2.4"], "text_open": "7.1.3 Compared with $BiR_3$, $SbR_3$ is slightly less acidic but markedly more basic, which enables a wider variety of reactions. Mixing equimolar amounts of $Me_2SbBr$, $Me_3Sb$, and $MeSbBr_2$ does not give mixed products from methyl-group exchange; instead, a 2:1 ionic compound is obtained. Both its cation and anion contain two Sb atoms. The cation contains no Br, and the anion has a center of symmetry. Write the SMILES representations of the cation and anion separately, and state the geometry at each metal center.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-07-T7-4", "tags": ["1.1", "1.4"], "text_open": "7.2 Reaction of $SbF_2OTf$ or $SbF(OTf)_2$ with the ligand $^{Dipp}BIAN$ (structure shown below) in $CH_2Cl_2$ gives the five-coordinate Sb complex $SbF_2(^{Dipp}BIAN)OTf$ or $SbF(^{Dipp}BIAN)(OTf)_2$, respectively.\n7.2.1 Use VSEPR theory to predict the molecular geometry of $SbF(^{Dipp}BIAN)(OTf)_2$.", "text_images": [{"file": "Exam-07-T7-4-1-prompt.png", "description": "Structure of $^{Dipp}BIAN$: 'CC(C)C(C=CC=C1C(C)C)=C1/N=C2C3=CC=CC4=C3C(C\\2=N/C5=C(C(C)C)C=CC=C5C(C)C)=CC=C4'."}], "full_mark": 2}, {"sub_id": "Exam-07-T7-5", "tags": ["1.4", "1.5"], "text_open": "7.2.2 The Sb-O bond in $SbF_2({}^{Dipp}BIAN)OTf$ is longer than that in $SbF({}^{Dipp}BIAN)(OTf)_2$. Briefly explain why.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-07-T7-6", "tags": ["2.4"], "text_open": "7.2.3 Replacing $F^-$ with $N_3^-$ greatly decreases the stability of the product. After $Sb(N_3)(^{Dipp}BIAN) (OTf)_2$ is left overnight in a protic solvent, its proton NMR spectrum shows that it has decomposed to form a new cation. This cation contains no Sb and contains a special five-membered ring in addition to the five-membered carbon ring in $^{Dipp}BIAN$. Write the SMILES representation of this cation; a textual description may be used to supplement the structure.", "text_images": [], "full_mark": 2}]} -{"problem_id": "Exam-07-T9", "header": "Bridged Polycyclic Heterocycles Involving Unexpected Rearrangements", "header_images": [], "subquestions": [{"sub_id": "Exam-07-T9-1", "tags": ["2.2", "2.4"], "text_open": "9.1 The reaction below is catalyzed by AgOTf.\n9.1.1 The reaction initially forms a nitrogen-containing heterocyclic intermediate. Write the SMILES representations of the 4 key intermediates and of the product formed after the 1,3-dipolar cycloaddition. Add a textual description if necessary.", "text_images": [{"file": "Exam-07-T9-1-1-prompt.png", "description": "'O/N=C/C1=CC=CC=C1C#CC2=CC=CC=C2' reacts with 'O=S(OC1=CC=CC=C1[Si](C)(C)C)(C(F)(F)F)=O' under AgOTf (10mol%), TBAF (3.0equiv), THF, 50°C conditions."}], "full_mark": 10}, {"sub_id": "Exam-07-T9-2", "tags": ["2.2"], "text_open": "9.1.2 However, the product described in 9.1.1 is not observed; the product formed after its further rearrangement is shown below. Write the SMILES representation of 1 key intermediate in the reaction.", "text_images": [{"file": "Exam-07-T9-2-1-prompt.png", "description": "'C12=C([C@H]3C(C=CC=C4)=C4O[C@@H]2C(C5=CC=CC=C5)=N3)C=CC=C1'"}], "full_mark": 2}, {"sub_id": "Exam-07-T9-3", "tags": ["2.2"], "text_open": "9.2 Some steps of the reaction below are very similar to those in 9.1. PhIO (iodosylbenzene) is a hypervalent iodine oxidant commonly used for the oxidative dearomatization of electron-rich arenes in the presence of a nucleophilic solvent. 9.2.1 Write the SMILES representation of the key OMe-containing electrophilic species formed by reaction of the starting material with PhIO in MeOH.", "text_images": [{"file": "Exam-07-T9-3-1-prompt.png", "description": "'CC1=CC=C(NS(C2=CC=C(C)C=C2)(=O)=O)C(C#CC3=CC=CC=C3)=C1' first reacts with iodosylbenzene, PhI=O, in methanol at room temperature for 5min, and then reacts with 'O/N=C/C1=CC=CC=C1C#CC2=CC=CC=C2' (1.1 equiv) under AgOTf (10 mol%) in dichloroethane at room temperature."}], "full_mark": 2}, {"sub_id": "Exam-07-T9-4", "tags": ["2.2", "2.4", "4.3"], "text_open": "9.2.2 The key electrophilic species then cyclizes to give a key intermediate, which undergoes a [3 + 3] cycloaddition with an intermediate from 9.1.1. Write the SMILES representations of the cyclized key intermediate and the cycloaddition product. Hint: The product contains a pyrrole ring system and has 1H NMR (400 MHz, $CDCl_3$) δ 7.95-7.93 (m, 2 H), 7.51-7.41 (m, 9 H), 7.31-7.27 (m, 2 H), 7.23 (d, J = 8.1 Hz, 2 H), 7.10 (d, J = 8.1 Hz, 2 H), 6.94 (d, J = 10.4 Hz, 1 H), 6.89-6.87 (m, 1 H), 5.94 (s, 1 H), 5.70 (s, 1 H), 5.68 (d, J = 10.3 Hz, 1 H), 4.66 (s, 1 H), 3.20 (s, 3 H), 2.31 (s, 3 H), 1.21 (s, 3 H).", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-07-T9-5", "tags": ["2.2", "2.4"], "text_open": "9.2.3 However, when the product above is stirred in 1,4-dioxane at 90 °C for 2 hours, it undergoes a reaction similar to that in 9.1. Write the SMILES representations of the 2 key intermediates and the final product.", "text_images": [], "full_mark": 6}, {"sub_id": "Exam-07-T9-6", "tags": ["2.2"], "text_open": "9.4 EWG is an acyclic electron-withdrawing group. By inference from 9.1 and 9.2, write the SMILES representation of 1 key tetracyclic intermediate in the reaction below.", "text_images": [{"file": "Exam-07-T9-6-1-prompt.png", "description": "'O/N=C/C1=CC=CC=C1C#CC2=CC=CC=C2' reacts with 'C=C=C-EWG' under 5 mol% AgOTf catalysis in DMSO, first at 60 °C for 20min and then at room temperature for 12h, to give 'O=C1C[C@@H]2C3=CC=CC=C3[C@@H](N=C2C4=CC=CC=C4)[C@@H]1[EWG]'."}], "full_mark": 2}]} +{"problem_id": "Exam-07-T8", "header": "Hyperconjugation and Its Applications to Structure and Reactivity", "header_images": [], "subquestions": [{"sub_id": "Exam-07-T8-1", "tags": ["1.5", "2.7", "2.4"], "text_open": "8.1 The molecule shown below can undergo intramolecular alkylation with the carbonyl oxygen acting as the nucleophile, forming 2 possible salt products. For this transformation, determine which reaction pathway is preferred and write the SMILES representation of the salt product corresponding to that pathway.", "text_images": [{"file": "Exam-07-T8-1-1-prompt.png", "description": "'BrCCNC(OCCBr)=O'"}], "full_mark": 2}, {"sub_id": "Exam-07-T8-2", "tags": ["1.2", "1.5"], "text_open": "8.2 Only one stereochemical configuration of the molecule shown below can react effectively with $O_3$. Write the SMILES representation of the reactive stereochemical configuration.", "text_images": [{"file": "Exam-07-T8-2-1-prompt.png", "description": "'[H][C@]12C(OCCC2)OCCC1'"}], "full_mark": 2}, {"sub_id": "Exam-07-T8-3", "tags": ["1.2", "1.5"], "text_open": "8.3 $N_2F_2$ has 2 stereoisomers. Determine which is thermodynamically more stable and which is kinetically more reactive, and explain the origin of the greater thermodynamic stability. (No more than 100 words.)", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-07-T8-4", "tags": ["1.2", "1.5"], "text_open": "8.4 Which of the following conformations is more stable?", "text_images": [{"file": "Exam-07-T8-4-1-prompt.png", "description": "Both molecules have the SMILES representation 'O=P(O[R]C)(O[R])[O-]' but differ in conformation. Structure A: the central P atom is bonded to =O directly below and to $-O^-$ at the upper left. Of the two -O-R groups bonded at the upper right, one P-O bond points into the plane of the page, with the R group on its oxygen extending vertically upward, while the other P-O bond points out of the plane, with the R group on its oxygen extending down and to the right. Structure B: the connectivity between P and the four oxygen atoms is the same as in structure A. For the -OR group at the upper right of P that points into the plane, the R group lies below and to the left of O and points into the plane; for the -OR group at the upper right of P that points into the plane, the R group lies below and to the left of O and points out of the plane. The overall difference from A is the positions of the R groups."}], "full_mark": 2}, {"sub_id": "Exam-07-T8-5", "tags": ["1.2", "1.5", "1.6"], "text_open": "8.5 The hyperconjugative stabilization of the C-S bond by $C-H_1$ and $C-H_2$ is 6.5 and 1.8 energy units, respectively. Determine which C-H bond provides the stronger hyperconjugative stabilization and explain why. (No more than 100 words.)", "text_images": [{"file": "Exam-07-T8-5-1-prompt.png", "description": "Chair conformation of 1,3-dithiane. Position 2 is labeled $H_1$ and position 5 is labeled $H_2$; both labeled H atoms occupy equatorial positions. '[H]C1CSC([H])SC1'"}], "full_mark": 4}]} {"problem_id": "Exam-08-T1", "header": "Problem 1 Peierls's Wondrous Gift\nIn the 18 century, Austrian mineralogist Müller von Reichenstein discovered a new substance while studying a gold ore. It had a silvery-white metallic luster but was brittle and a poor conductor; it burned with a blue flame and gave off an unusual odor resembling rotten radishes. German chemist Martin Klaproth later confirmed that it was a new element and named it after the Latin word for Earth. Interestingly, another member of the same group discovered decades later was named after the Moon to complement Earth. This element, denoted E, can be obtained by reduction of either its oxide A or its sodium oxoacid salt B; E has the same oxidation state in A and B. An early preparation heated B with carbon black. The simple hydride C of E is a foul-smelling, highly unstable gas whose aqueous solution is weakly acidic. Passing C into $\\mathrm{FeCl_3}$ solution gives a black precipitate. Another oxide D of E is an orange solid. When D dissolves in concentrated aqueous NaOH, the resulting anhydrous sodium salt contains 46.64% E by mass.", "header_images": [], "subquestions": [{"sub_id": "Exam-08-T1-1", "tags": ["2.1"], "text_open": "1-2-1 Write the equation for heating B with carbon black to obtain elemental E.", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-08-T1-2", "tags": ["2.1"], "text_open": "1-2-3 Write the equation for dissolution of D in concentrated aqueous NaOH.", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-08-T1-3", "tags": ["1.3", "1.6"], "text_open": "1-4-1 Structural studies of $\\alpha$-E show infinite one-dimensional zigzag chains in which each E atom bonds to its two neighboring E atoms, forming continuous $(E)_n$ chains with $n\\rightarrow\\infty$. Elemental halogen X is widely used for surgical disinfection. E and X form F, G, and H with E:X ratios of 1:1, 2:1, and 1:4, respectively. F retains the $(E)_n$ chain. Both E and X have two chemical environments, each in a 1:1 ratio. The coordination numbers of the two E atoms are in a 3:4 ratio, and one X atom forms only one covalent bond to E. G can be viewed as portions of two long F chains joined by additional E-E bonds to form an infinite six-membered-ring structure, and the E coordination numbers in G are the same as in F. H exists as a tetramer with two E environments, and every E atom is six-coordinate. Give SMILES descriptions of the structural motifs of F and G.", "text_images": [], "full_mark": 6}, {"sub_id": "Exam-08-T1-4", "tags": ["1.1", "1.6"], "text_open": "1-4-2 Give the SMILES structure of the H tetramer.", "text_images": [], "full_mark": 3}]} +{"problem_id": "Exam-08-T3", "header": "Problem 3 Phthalocyanine Complexes", "header_images": [], "subquestions": [{"sub_id": "Exam-08-T3-1", "tags": ["1.4", "2.4"], "text_open": "3.1 The diagram below shows simple syntheses of two mononuclear macrocyclic complexes.\nHint: A reducing substance is present in the reaction system.\n3.1.1 Give the SMILES structures of A and B.", "text_images": [{"file": "Exam-08-T3-1-1-prompt.png", "description": "Heating 4-iodophthalonitrile 'IC1=CC=C(C#N)C(C#N)=C1', 4,5-dibutoxyphthalonitrile 'N#CC1=CC(OCCCC)=C(OCCCC)C=C1C#N', and zinc acetate $\\mathrm{Zn(OAc)_2}$ in a 1:3:1 ratio gives A. Reaction of 4-iodophthalonitrile 'IC1=CC=C(C#N)C(C#N)=C1', 4,5-bis(propylsulfonyl)phthalonitrile 'N#CC1=CC(S(=O)(CCC)=O)=C(S(=O)(CCC)=O)C=C1C#N', and nickel chloride $\\mathrm{NiCl_2}$ in a 1:3:1 ratio gives B."}], "full_mark": 4}, {"sub_id": "Exam-08-T3-2", "tags": ["2.4"], "text_open": "3.2 Further reactions combine A and B into the dinuclear complex E, as shown in the route below. Give the SMILES structures of C, D, and E.", "text_images": [{"file": "Exam-08-T3-2-1-prompt.png", "description": "First, 1,4(1,4)-dibenzenacyclohexaphane (SMILES: 'C1(CC2)=CC=C(CCC3=CC=C2C=C3)C=C1') is treated with 2 equivalents of $\\mathrm{Br_2}$/$\\mathrm{CCl_4}$ and then reacted with tributyl(vinyl)tin (Bu_3SnCH=CH_2) under tetrakis(triphenylphosphine)palladium, Pd(PPh_3)_4, catalysis to give C. Next, C reacts with A in DMF in the presence of triethylamine ($\\mathrm{Pd(OAc)_2}$), tetrabutylammonium bromide (TBAB), and palladium acetate ($\\mathrm{NEt_3}$) to give D. Finally, D reacts with B under the same $\\mathrm{Pd(OAc)_2}$/DMF/$\\mathrm{NEt_3}$/TBAB conditions to give E."}], "full_mark": 6}]} +{"problem_id": "Exam-08-T5", "header": "Defective Cubic Perovskite\nA polymorph of $\\mathrm{CsFe}_x\\mathrm{Cl}_3$ has a defective cubic-perovskite structure. Relative to cubic perovskite, Cs occupies every Ca site, Cl every O site, and Fe occupies an ordered subset of the Ti sites. Equivalently, Cs and Cl jointly form close-packed layers in a 1:3 atomic ratio, and these layers are cubically stacked. Between every two adjacent Cs-Cl close-packed layers are octahedral voids, some formed only by Cl and others jointly by Cl and Cs. Fe fills some interlayer octahedral voids in the repeating sequence occupied, occupied, vacant, occupied, occupied, vacant, and so on. In each Fe-containing interlayer region, Fe occupies all and only the octahedral voids formed solely by Cl. This defect pattern changes the crystal to the hexagonal crystal family. The cell parameters are $a=736.2$ pm and $c=896.4$ pm. Two Fe atoms are located at $(2/3,1/3,0.18565)$ and $(1/3,2/3,0.81435)$, and their $\\mathrm{FeCl}_6$ octahedra are corner-sharing. In the projection down the crystallographic $c$ axis, the Fe atoms form a graphene-like pattern.", "header_images": [], "subquestions": [{"sub_id": "Exam-08-T5-1", "tags": ["1.3"], "text_open": "5.1 Determine the unit-cell composition using reasonable assumptions and approximations.", "text_images": [], "full_mark": 7}, {"sub_id": "Exam-08-T5-2", "tags": ["1.3"], "text_open": "5.4 Determine the crystal system.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-08-T5-3", "tags": ["1.3"], "text_open": "5.7 State the number of inversion centers in one primitive cell.", "text_images": [], "full_mark": 1}]} {"problem_id": "Exam-08-T6", "header": "Lithium or Sodium?\nBalancing fluctuations in renewable-electricity generation through electrochemical energy storage is essential to achieving carbon-peaking and carbon-neutrality goals. Lithium-ion batteries dominate the market, but constraints on lithium resources have stimulated research into sodium-ion batteries. Their operating principles are similar, yet their material structures and performance differ substantially. This problem considers two representative layered-oxide cathodes.", "header_images": [], "subquestions": [{"sub_id": "Exam-08-T6-1", "tags": ["3.2"], "text_open": "6.2 The following electrochemical data are given. NCM811 has an average discharge voltage of 3.8 V versus metallic lithium and an actual reversible specific capacity of $200\\,\\mathrm{mAh\\cdot g^{-1}}$. The P2 sodium material has an average discharge voltage of 3.2 V versus metallic sodium and an actual reversible specific capacity of $130\\,\\mathrm{mAh\\cdot g^{-1}}$. The volumetric energy density is $E_v=\\rho\\times C\\times U$, where $\\rho$ is density, $C$ specific capacity, and $U$ average voltage.\n6.2.1 Calculate the theoretical volumetric energy density of each cathode and their ratio $\\lambda$ (lithium-based/sodium-based).", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-08-T6-2", "tags": ["3.2"], "text_open": "The diffusion coefficient D of ions in a layered material and the temperature T obey the Arrhenius equation. For P2-type $Na_{0.67}Ni_{0.33}Mn_{0.67}O_2$, the experimentally measured pre-exponential factor is $D_0 = 5.0 × 10^{-4} cm^2·s^{-1}$ and the activation energy is E_a = 0.35 eV. The Arrhenius equation is $D = D_0 e^{-\\frac{E_a}{RT}}$.\n6.3.2 The lithium-ion diffusion activation energy in NCM811 is approximately \\(0.48\\,\\mathrm{eV}\\). If the pre-exponential factors have the same order of magnitude, compare sodium- and lithium-ion diffusion rates in such layered structures at the same temperature and briefly explain. (No more than 100 words.)", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-08-T6-3", "tags": ["1.7"], "text_open": "6.4 Sodium-ion batteries are viewed as a potential complement to lithium-ion batteries because sodium is abundant and the batteries offer good low-temperature performance and relatively high safety, but commercialization still faces challenges.\n6.4.1 Briefly explain, in terms of intrinsic material properties, why increasing the volumetric energy density of sodium-ion batteries may be difficult. (No more than 100 words.)", "text_images": [], "full_mark": 2}]} {"problem_id": "Exam-09-T1", "header": "Material-Balance Analysis of an Electrolytic Process\n\"If this silvery-white lightweight metal could be produced on a large scale at low cost, transportation and building structures would undergo a qualitative transformation.\" In the late 19th century, industry fiercely competed to develop a method for producing this lightweight metal. At almost the same time, chemists Charles M. Hall and Paul Héroult independently proposed, on opposite sides of the Atlantic, a high-temperature molten-salt electrolysis route: a reddish-brown ore A (whose principal component is A') is purified and then electrolytically reduced in a molten-salt bath to obtain the silvery-white metal B, thereby laying the foundation of the modern metallurgical industry for B. Natural ore A is usually reddish brown and often contains silicon dioxide impurities. During electrolysis, the mineral flux C is added to lower the melting point of the system, and a carbonaceous anode is used. The anode is continuously consumed, releasing gas D. Industrial electrolytic cells normally operate at atmospheric pressure and at approximately 955℃. Before Hall-Héroult electrolysis, ore A must be purified by the Bayer process. This process mainly comprises the following steps: first, crushed ore A is stirred with a 30 wt% NaOH solution at 473 K and 3.5 MPa for 2 h, converting it into the soluble complex E; the solution is then cooled to 320 K, seeded with particles approximately 75 μm in size (50 g·L^{-1}), and allowed to stand for 48 h, precipitating F with a relatively narrow particle-size distribution that facilitates subsequent calcination; finally, F is calcined in air at 1370 K for 6 h to regenerate high-purity A'. When mineral flux C is the principal component of the molten-salt bath, the bath usually contains two key substances: solid G, which must be replenished periodically, and compound H. Solid G is added periodically to maintain their molar ratio in the bath. If $Na_2CO_3$ impurity enters the bath, $Na_2CO_3$ reacts with G, releasing a small amount of gas D and producing compound H, which remains soluble in the molten-salt bath, together with another by-product. Define the bath ratio as BR = n(H) / n(G) = 2.50", "header_images": [], "subquestions": [{"sub_id": "Exam-09-T1-1", "tags": ["2.1", "3.6"], "text_open": "1.2 On a certain day, 15.0 kg of $Na_2CO_3$ impurity enters the molten-salt bath. Assume that all H produced dissolves in the bath and all D escapes. 1.2.1 Calculate the mass of A' additionally produced by the side reaction that day.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-09-T1-2", "tags": ["2.1"], "text_open": "1.2.2 Calculate the additional mass of G that should be added that day to restore BR to 2.50.", "text_images": [], "full_mark": 4}]} -{"problem_id": "Exam-09-T4", "header": "Kjeldahl Nitrogen Determination\nDetermination of nitrogen content has long been an important topic in analytical chemistry. In 1883, the Danish chemist Johan G.C.T. Kjeldahl invented the Kjeldahl method. The method mainly comprises digestion, alkalization, and titration: sulfuric-acid digestion converts the nitrogen in a sample into an ammonium salt; alkalization releases ammonia by distillation; and the absorbed ammonia is quantified indirectly by acid-base titration. The organic nitrogen content of a solid sample is to be determined. Answer the following questions based on the procedures described.", "header_images": [], "subquestions": [{"sub_id": "Exam-09-T4-1", "tags": ["4.1", "4.5"], "text_open": "4-1-1 First operation: digestion. Weigh $5.000 \\ \\mathrm{g}$ of the sample into a dry $100 \\ \\mathrm{mL}$ Kjeldahl flask. Add $0.4 \\ \\mathrm{g}$ of copper sulfate, $6 \\ \\mathrm{g}$ of potassium sulfate, and $20 \\ \\mathrm{mL}$ of anhydrous sulfuric acid. Swirl gently, place a small funnel in the flask mouth, and set the flask at an angle of 45° on an asbestos gauze with a small opening. Heat slowly. When the contents turn black and foaming has completely ceased, increase the heat to keep the liquid gently boiling. After the liquid becomes blue-green, clear, and transparent, continue heating for another $1 \\ \\mathrm{h}$. Remove the Kjeldahl flask and cool it to room temperature. Carefully add $20 \\ \\mathrm{mL}$ of water, quantitatively transfer the contents to a $100 \\ \\mathrm{mL}$ volumetric flask, rinse the inner wall of the Kjeldahl flask with a small amount of water, and combine the rinsings in the volumetric flask. Dilute to volume with water, mix, and reserve for use. Conduct a blank experiment in parallel. From the apparatus shown below, select the Kjeldahl flask suitable for digestion. The choices are labeled (a), (b), (c), and (d).", "text_images": [{"file": "Exam-09-T4-1-1-prompt.png", "description": "Image corresponding to option (a): a pear-shaped flask with a short neck and a ground-glass stopper."}, {"file": "Exam-09-T4-1-2-prompt.png", "description": "Image corresponding to option (b): a flask with a long neck, a pear-shaped bottom, and a ground-glass stopper."}, {"file": "Exam-09-T4-1-3-prompt.png", "description": "Image corresponding to option (c): an Erlenmeyer flask with a conical body, a flat bottom, and a ground-glass stopper."}, {"file": "Exam-09-T4-1-4-prompt.png", "description": "Image corresponding to option (d): a volumetric flask with a narrow neck, pear-shaped bottom, an annular calibration mark on the neck, and a ground-glass stopper."}], "full_mark": 1}, {"sub_id": "Exam-09-T4-2", "tags": ["4.1", "4.5"], "text_open": "4-2 Alkalization. Assemble the Kjeldahl distillation apparatus as shown below. Fill vessel 1 with water to 2/3 of its volume. Add several glass beads, several drops of methyl red, and several milliliters of sulfuric acid until the solution is red. Heat the water in the steam generator to boiling and maintain boiling. Add $10.0 \\ \\mathrm{mL}$ of $20 \\ \\mathrm{g/L}$ boric acid solution and 3 drops of mixed indicator A to the receiving flask, and immerse the lower end of apparatus 8 beneath the liquid surface in vessel 9. Using a pipette, accurately transfer $5.0 \\ \\mathrm{mL}$ of sample digest through apparatus 4 into reaction chamber 5. Rinse the small cup with $10 \\ \\mathrm{mL}$ of water and allow the rinsing to flow into reaction chamber 5, then tightly insert the rod-shaped glass stopper in apparatus 4. Slowly add $400 \\ \\mathrm{g/L}$ sodium hydroxide solution through apparatus 4 into reaction chamber 5 and immediately seal the system. After distilling for $15 \\ \\mathrm{min}$, move the distillate receiver so that the liquid surface is below the lower end of the condenser. Continue distilling for $1 \\ \\mathrm{min}$. Verify with pH paper that the distillate is neutral, rinse the outside of the lower end of the condenser with a small amount of water, and remove the receiving flask. Which of the following statements about the experimental operations and their purposes are correct?\n(a). Methyl red is added to apparatus 1 to generate acidic vapor that reacts with ammonia in apparatus 5 to form an ammonium salt.\n(b). After sodium hydroxide is added, the glass stopper must be inserted tightly and a water seal formed immediately to prevent loss of ammonia.\n(c). The principal purpose of apparatus 6 is to separate liquid water from the steam generated in apparatus 1.\n(d). After alkalization and collection for one digest have been completed, the next reaction may be started by opening screw clamp 7, closing screw clamp 3, and replacing the solution in vessel 9 with distilled water, in that order, to discharge the waste liquid and rinse the reaction chamber.", "text_images": [{"file": "Exam-09-T4-2-1-prompt.png", "description": "This black-and-white line drawing shows an assembly of laboratory glassware. Three main sections on the left, center, and right are connected by tubing, and the numbers 1 through 9 identify specific components. Left section: number 1 points to the liquid inside a long-neck round-bottom flask, represented by a horizontal dashed line. The bottom of the flask rests on a square base. Number 2 points to this base, which has a circular knob at the center of its front face and four feet underneath. The mouth of the round-bottom flask is sealed with a stopper. A very long straight vertical glass tube passes through the center of the stopper, extending below the liquid surface inside the flask and far above the flask mouth. A short right-angle glass tube also passes through the stopper and is connected at its end to U-shaped flexible tubing. Center section: the U-shaped flexible tubing from the left connects to a narrow inlet above the central apparatus. The main central apparatus is inclined from lower left to upper right and contains an outer jacket and an inner structure. Number 4 points to an upward-facing vertical inlet at the upper-right end of the inclined apparatus; it has a stopper through which a short straight tube is inserted. Number 5 points to an elongated cavity or inner-tube structure inside the inclined apparatus. Number 6 points to the enlarged spherical lower-left end of the inclined apparatus. Number 7 points to a narrow tube extending from the spherical end below number 6, connected to a short length of flexible tubing fitted with a clamp (pinchcock). Number 3 points to another narrow tube extending from the upper-middle part of the outer jacket, likewise connected to a short length of flexible tubing fitted with a clamp (pinchcock). The upper-right end of the inclined apparatus connects to the right section through an inverted U-shaped bent glass tube. Right section: number 8 points to a vertically oriented straight condenser. The end of the inverted U-shaped bent glass tube enters the top of the condenser. Two horizontal side-arm connections extend to the right from the outer jacket of condenser 8. A right-pointing arrow beside the upper side arm indicates outward flow, and a left-pointing arrow beside the lower side arm indicates inward flow. Number 9 points to an Erlenmeyer flask directly beneath the condenser. The flask contains liquid, represented by a horizontal dashed line. The inner straight tube of condenser 8 extends downward through the mouth of the flask and below the liquid surface inside it."}], "full_mark": 2}]} {"problem_id": "Exam-09-T6", "header": "Mg-Air Batteries\nWith their high specific energy density, stable operating voltage, simple device construction, and excellent safety, magnesium-air batteries have become candidates for emergency backup power in hospitals, emergency power for disaster relief, portable field energy-storage devices, and automotive power supplies.", "header_images": [], "subquestions": [{"sub_id": "Exam-09-T6-1", "tags": ["2.8", "3.4", "3.6"], "text_open": "6-2-2 An ideal anode film both suppresses hydrogen evolution at the anode and does not impede interfacial electrochemical reactions. A film that is both \"dense and porous\" has therefore become a research objective. The K_{sp} values of hydroxides of selected elements are shown below. From Cu, La, and Sr, select the metallic element that can be added to the Mg-alloy electrode, and explain your choice. Given K_{sp}[Mg(OH)_2] = 5.61 × 10^{-12}, K_{sp}[Cu(OH)_2] = 2.2 × 10^{-20}, K_{sp}[La(OH)_3] = 2.0 × 10^{-19}, and K_{sp}[Sr(OH)_2] = 3.2 × 10^{-4}.", "text_images": [{"file": "Exam-09-T6-1-1-prompt.png", "description": "This is a plot with logarithmic coordinates. Vertical axis: $\\boldsymbol{[M^{n+}]}$ (unit: mol/L), on a logarithmic scale ranging from $10^{-13}$ to $10^{1}$. Horizontal axis: $\\boldsymbol{pH}$, ranging from 0 to 14. Solubility-equilibrium curves for various metal hydroxides and metal oxides are plotted and labeled with different line styles, including solid and dashed lines. The substances include the following. Hydroxides: $Sn(OH)_2$, $Fe(OH)_2$, $Fe(OH)_3$, $Ga(OH)_3$, $Sn(OH)_4$, $Pb(OH)_4$, $Ce(OH)_4$, $Sc(OH)_3$, $In(OH)_3$, $Al(OH)_3$, $Dy(OH)_3$, $Y(OH)_3$, $Sm(OH)_3$, $Eu(OH)_3$, $Ce(OH)_3$, $Nd(OH)_3$, $Gd(OH)_3$, $La(OH)_3$, $Ni(OH)_2$, $Mn(OH)_2$, $Mg(OH)_2$, $Cd(OH)_2$, $Co(OH)_2$, $Zn(OH)_2$, $Pb(OH)_2$, and $Ca(OH)_2$. Oxides: $Fe_2O_3$, $FeO$, $MnO_2$, $Ga_2O_3$, $PbO_2$, $SnO_2$, $ZrO_2$, $In_2O_3$, $Al_2O_3$, $Sc_2O_3$, $Nd_2O_3$, $MgO$, $Dy_2O_3$, $SnO$, $ZnO$, $Sm_2O_3$, $Gd_2O_3$, $Y_2O_3$, $NiO$, $NiO_2$, $PbO$, $CoO_2$, and $MnO$. Overall trend of the curves: as pH increases, the corresponding metal-ion concentration $\\boldsymbol{[M^{n+}]}$ decreases."}], "full_mark": 4}, {"sub_id": "Exam-09-T6-2", "tags": ["2.8", "3.6"], "text_open": "6-2-3 In practical use, additional Ca is incorporated into the Mg alloy. Explain why. (K_{sp}[Ca(OH)_2] = 5.5 × 10^{-6})", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-09-T6-3", "tags": ["2.8", "3.6"], "text_open": "6-3 Investigation of second-phase effects\n\nAlloying is an effective means of modifying materials and can significantly improve the discharge performance of magnesium alloys. The type, morphology, and distribution of the second phase, i.e., another phase distributed discontinuously within the matrix phase, directly affect material performance. When a new metal is added to the base electrode material during alloy-electrode manufacture, a second phase forms. Microgalvanic couples between the second phase and matrix phase undergo galvanic corrosion. Galvanic corrosion occurs when two different metals are in contact while simultaneously exposed to an electrolyte, forming a spontaneous cell in which the more active metal acts as the anode and corrodes. The rate of galvanic corrosion depends on the contact area between the two metals.\n\n6-3-1 The anode efficiency is calculated as $P = \\frac{W_f}{\\Delta W}$, where $\\Delta W$ is the actual mass loss and $W_f$ is the theoretical mass loss, calculated from $W_f = \\frac{I \\times t}{F \\times \\sum\\limits_{i=1}^{n} \\frac{X_i \\times N_i}{M_i}}$. Here, $F$ is the Faraday constant, $X_i$ is the mass fraction of each element in the magnesium anode, $N_i$ is the corresponding number of electrons transferred by each element in the electrochemical process, and $M_i$ is the atomic mass of each element.\n\nAlloy materials A and B have the following compositions:\n\n| Alloy material | La | Ce | Ca | Al | Zn | Mg |\n| --- | --- | --- | --- | --- | --- | --- |\n| A | / | / | / | 6% | 3% | 91% |\n| B | 0.15% | 0.45% | 0.3% | 6% | 3% | 90.1% |\n\nFor $\\Delta W$, $\\Delta W_A = 36.27\\,\\text{mg}$ and $\\Delta W_B = 30.01\\,\\text{mg}$. Calculate the anode efficiencies of A and B at a current density of $0.5\\,\\text{mA/cm}^2$, duration of $6\\,\\text{h}$, and area of $14\\,\\text{cm}^2$.", "text_images": [], "full_mark": 6}, {"sub_id": "Exam-09-T6-4", "tags": ["2.8", "3.6"], "text_open": "6-3-2 Subsequent research proposed that the difference in efficiency between alloy materials A and B originates from a change in the second phase. After addition of the new metals, the second phase changes from the particulate and planar morphologies in A to the rod-like and relatively continuous β-Mg_{17}Al_{12}/τ-Mg_{32}(Al, Zn)_{49} and Al_{2}(La, Ce) structures in B. Explain the difference in efficiency on this basis.", "text_images": [], "full_mark": 2}]} -{"problem_id": "Exam-09-T7", "header": "Semiconductors and Electrochemistry\nSemiconductors are materials whose electrical conductivity at room temperature lies between that of conductors and insulators. Common semiconductor materials include silicon, germanium, and gallium arsenide. Based on band theory and Fermi statistics, doping with group V or group III elements, or with other materials that can donate electrons (donors) or accept electrons (acceptors), can regulate the Fermi level and carrier concentration of a semiconductor and thereby its various properties. Semiconductor carriers generally comprise electrons and holes. In this problem, involving an n-type semiconductor in which electrons are the majority carriers, only electrons need be considered. Nonuniform doping can create a carrier-concentration gradient in a semiconductor. The resulting diffusion causes charge separation and a built-in electric field; the built-in field opposes diffusion until equilibrium is reached. This is fundamentally similar to the electrochemical picture in which a chemical-potential difference drives charge transfer, which in turn produces a potential difference that opposes further charge transfer, ultimately establishing an electrode potential at equilibrium. By comparing a concentration-gradient-doped semiconductor with the electromotive force of an electrochemical concentration cell, this problem explores the universality of diffusion-electric-field equilibrium.\n7.1 Built-in electric field and diffusion equilibrium in a semiconductor\nConsider an n-type semiconductor at temperature T = 300 K with electron concentration varying with position x as $n(x)=N_{d0}\\exp(-x/L)$. Here $N_{d0}=1\\times10^{18}\\ \\mathrm{cm}^{-3}$, $L=0.1\\ \\mu\\mathrm{m}$, $D_n=225\\ \\mathrm{cm}^2\\mathrm{/s}$, $e=1.6\\times10^{-19}\\ \\mathrm{C}$, and $k=1.38\\times10^{-23}\\ \\mathrm{J/K}$. These quantities are, respectively, the donor concentration at x = 0 (which, under the approximation used here, may be regarded as the electron concentration at x = 0), characteristic length, electron diffusion coefficient, elementary charge, and Boltzmann constant.", "header_images": [], "subquestions": [{"sub_id": "Exam-09-T7-1", "tags": ["3.6"], "text_open": "7.2 Electrochemistry. 7.2.1 When an iron sheet is immersed in water, an electrical double layer forms near the iron electrode, as shown below. Briefly explain how the structure forms, and qualitatively sketch the absolute value of the potential as a function of distance from the metal surface, taking the potential infinitely far from the electrode to be zero.", "text_images": [{"file": "Exam-09-T7-1-1-prompt.png", "description": "This two-dimensional diagram, titled \"Figure 9.4 Schematic Structure of the Electrical Double Layer,\" shows the spatial distribution of an electrical double layer from left to right. At the far-left edge is a vertical row of minus signs (short horizontal lines). Immediately to their right is a vertical solid line shaded with diagonal hatching on its left side; the words \"Electrode surface\" are set vertically to the left of this line. A short distance to the right is a parallel vertical dashed line bearing a neat vertical series of small circles, each containing a \"+\" sign. The narrow interval between the solid line and this middle dashed line is marked at the bottom by a horizontal upward brace, beneath which the spacing symbol $d$ and the label \"Compact layer\" appear. Extending rightward from the middle dashed line to another parallel dashed boundary at the far right is a wider rectangular region. Several similar small circles containing \"+\" signs are scattered irregularly within it. This wider interval is likewise marked at the bottom by a horizontal brace, beneath which the spacing symbol $\\delta$ and the label \"Diffuse layer\" appear."}], "full_mark": 5}]} {"problem_id": "Exam-10-T3", "header": "Question 3 Anticoagulant Polymers (12 points, 5%). Anticoagulation is the prevention or treatment of thrombotic disease by inhibiting blood coagulation. Traditional natural anticoagulants, such as heparin, have limitations including structural complexity and substantial batch-to-batch variation. Developing synthetic anticoagulants with well-defined structures and controllable properties has therefore become an important research direction in medicinal chemistry and materials science.", "header_images": [], "subquestions": [{"sub_id": "Exam-10-T3-1", "tags": ["2.4", "2.5"], "text_open": "3.1.1 ACVA, with the condensed structural formula $\\mathrm{HOOCCH_2CH_2C(CN)(CH_3)N=NC(CN)(CH_3)CH_2CH_2COOH}$, is a water-soluble radical initiator and is used in a much smaller amount than reagent BM1429 in the synthesis. Write the SMILES strings of the major and minor polymer products, explicitly showing the end-group structures. The 2 products differ only in their end groups.", "text_images": [{"file": "Exam-10-T3-1-1-prompt.png", "description": "Polymerization of sodium 4-vinylbenzenesulfonate, whose SMILES may be written as 'C=CC1=CC=C(C=C1)S(=O)([O-])=O.[Na+]', in the presence of BM1429, with SMILES 'O=C(O)CCSC(=S)SC(C)C(=O)O', ACVA, and water at 80 °C for 4 h."}], "full_mark": 4}, {"sub_id": "Exam-10-T3-2", "tags": ["2.2", "2.3", "2.5"], "text_open": "3.1.2 Continue to use the reaction scheme from 3.1.1. Besides becoming an end group in the product, reagent BM1429 reversibly binds to the radical polymer chain, thereby lowering the radical concentration in the system and controlling the reaction. Write the SMILES string of the intermediate formed by this reversible binding. The end group remote from the reaction center need not be shown.", "text_images": [], "full_mark": 2}]} -{"problem_id": "Exam-10-T5", "header": "Structural Chemistry of Two-Dimensional Transition-Metal Dichalcogenides and Their Derivatives (26 points, 12%). Two-dimensional transition-metal dichalcogenides, or TMDs, generally have layered structures held together by van der Waals forces. Their symmetry and physical properties change substantially as the number of layers decreases.", "header_images": [], "subquestions": [{"sub_id": "Exam-10-T5-1", "tags": ["1.3"], "text_open": "Structural background: $WSe_2$ has a molybdenite-type hexagonal layered structure. The figure below is a projection along $a$; large spheres represent W and small spheres Se. One unit cell contains two layers along $c$.\n5-1-2 Explain how W and Se occupy the hexagonal plane lattice in a projection along the c axis, and give the projected relationship within one unit cell.", "text_images": [{"file": "Exam-10-T5-1-1-prompt.png", "description": "A two-dimensional projection of a vertical rectangular unit cell outlined by thin black lines. Axes at the lower left show c vertically upward and b horizontally to the left. The figure contains two larger gray spheres and eight smaller green spheres joined by slender cylindrical bonds. An upper gray sphere lies inside the upper half of the cell near its right boundary and is connected to four green spheres at the upper-left, lower-left, upper-right, and lower-right positions. The two green spheres on the left lie within the cell, whereas the two on the right lie outside the right boundary. A lower gray sphere lies inside the lower half near the left boundary and is likewise connected to four green spheres at the upper-left, lower-left, upper-right, and lower-right positions. The two green spheres on the left lie outside the left boundary, whereas the two on the right lie within the cell. The two gray spheres are not directly connected; together they form two separate four-coordinate units offset vertically and oriented in opposite horizontal directions."}], "full_mark": 2}]} {"problem_id": "Exam-10-T6", "header": "The 2025 Nobel Prize in Chemistry was awarded to Susumu Kitagawa, Richard Robson, and Omar M. Yaghi for their contributions to research on metal-organic frameworks, or MOFs. Chemists recently prepared a new Cr-MOF material, CPM-243, that resists strong acids and bases and remains stable over approximately 16 pH units under extreme conditions ranging from pH < 0 to pH > 14, redefining the feasibility of Cr-MOFs in the search for the most chemically stable MOFs.", "header_images": [], "subquestions": [{"sub_id": "Exam-10-T6-1", "tags": ["1.4", "2.4"], "text_open": "6-2 Cr-MOFs are currently known for MOFs based on $\\mathrm{Cr_3O}$ trimers, such as MIL-88(Cr), generally with dicarboxylate ligands such as terephthalate. Write the SMILES string of a single $[\\mathrm{Cr_3O(RCOO)_6}]$ metal-cluster node in this MOF. Use 'COC(C)OC' to represent $\\mathrm{RCOO^-}$ and disregard the charge on the node.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-10-T6-2", "tags": ["1.4", "2.1"], "text_open": "6-3-2 Known Cr-trimer MOFs have pendant ligands, such as F^{-}/OH^{-}/$\\mathrm{H_2O}$, and correspondingly low-connectivity frameworks. In CPM-243(Ac), chemists instead used the tripyridyl ligand tpt to provide three additional intercluster cross-links through Cr-N bonds, producing a 9-connected MOF framework. The synthesis was performed by adding chromium(III) acetate hydrate, $\\mathrm{Cr(Ac)_3 \\cdot 2H_2O}$, terephthalic acid, $\\mathrm{H_2bdc}$, 2,4,6-tris(4-pyridyl)-1,3,5-triazine, tpt, and glacial acetic acid, $\\mathrm{Hac}$, to a polytetrafluoroethylene liner. After adding 5.0 g of water and stirring for one hour, the vessel was sealed and heated for 3 days in an oven preheated to $220^\\circ\\mathrm{C}$. Given that CPM-243(Ac) has the formula $[(\\mathrm{Cr_3O})(\\mathrm{bdc})_x(\\mathrm{tpt})_y(\\mathrm{ac})]$, first determine $x$ and $y$ from the structure, and then write the reaction equation.", "text_images": [{"file": "Exam-10-T6-2-1-prompt.png", "description": "Structure of the organic compound tpt, with the SMILES 'C1(C2=CC=NC=C2)=NC(C3=CC=NC=C3)=NC(C4=CC=NC=C4)=N1'."}], "full_mark": 4}]} -{"problem_id": "Exam-10-T9", "header": "", "header_images": [], "subquestions": [{"sub_id": "Exam-10-T9-1", "tags": ["2.3", "2.4", "2.7"], "text_open": "9-1-1 In radical reactions of allylic compounds using NBS or HBr, substitution and addition compete, and different reagents show different selectivities. Write the major products of the following two reactions.", "text_images": [{"file": "Exam-10-T9-1-1-prompt.png", "description": "Two organic reaction schemes. In the first, 'CC1(C)C=CCCC1' reacts with NBS under irradiation. In the second, 'CC(C1CCCCC1)=C' reacts with HBr and $\\mathrm{(BzO)_2}$."}], "full_mark": 4}, {"sub_id": "Exam-10-T9-2", "tags": ["2.2", "2.3"], "text_open": "9-2-1 Radical reactions can also achieve efficient intramolecular N-C bond coupling. Write the key intermediates in the following reaction.", "text_images": [{"file": "Exam-10-T9-2-1-prompt.png", "description": "Organic reaction scheme in which 'C1(CN2CCCCC2)NCCCC1' reacts with NCS under irradiation to give 'N12CCCCC1CN3C2CCCC3'."}], "full_mark": 5}, {"sub_id": "Exam-10-T9-3", "tags": ["2.4"], "text_open": "9-3-2 Write the structure of the major product of the following reaction. Disregard stereochemistry.", "text_images": [{"file": "Exam-10-T9-3-1-prompt.png", "description": "Organic reaction scheme in which 'IC1C(OC/C=C/C)OCCC1' reacts with triethylborane and oxygen in water and hypophosphorous acid."}], "full_mark": 2}, {"sub_id": "Exam-10-T9-4", "tags": ["2.2", "2.3", "2.4"], "text_open": "9-3-4 The following reaction undergoes efficient triethylborane-catalyzed cyclization to give two isomers, A1 and A2, of which A2 is the major product. Write the SMILES string of A2 and those of the key intermediates leading to A2.", "text_images": [{"file": "Exam-10-T9-4-1-prompt.png", "description": "Organic reaction scheme in which 'CC1(C)C(O1)/C=C/CCCC#C' reacts with triethylborane and triphenyltin hydride in benzene to give A1, 'OC(C)(C)/C=C/C(CCC/1)C1=C[Sn](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4', and A2."}], "full_mark": 10}]} -{"problem_id": "Exam-11-T3", "header": "A photoresist is a polymeric material sensitive to light of specific wavelengths. Upon irradiation, it undergoes specific chemical reactions that substantially change its solubility in a developer, thereby forming the desired pattern on the surface of a semiconductor material. As a key patterning material in integrated-circuit fabrication, advances in photoresist chemistry directly determine the limits of process miniaturization and the industry's capacity for technological self-reliance.", "header_images": [], "subquestions": [{"sub_id": "Exam-11-T3-1", "tags": ["2.1", "2.5"], "text_open": "3.1 The novolak-diazonaphthoquinone (Novolak-DNQ) system is the most widely used photoresist for the g-line (436 nm) and i-line (365 nm). One component is a high-molecular-weight linear phenolic resin, and the other is diazonaphthoquinone, whose structure is shown below.\n3.1.1 The phenolic resin prepared under acidic conditions is a low-molecular-weight linear prepolymer. The curing agent hexamethylenetetramine may be added to an aqueous solution of the prepolymer to cure and cross-link it upon heating. Identify which substrate must be used in excess when preparing the acid-catalyzed phenolic resin, write the decomposition equation of the curing agent upon heating, and on this basis state the role of the curing agent.", "text_images": [{"file": "Exam-11-T3-1-1-prompt.png", "description": "'O=C1C2=CC=CC=C2C(S(=O)(O[Ar])=O)=CC1=[N+]=[N-]'"}], "full_mark": 5}, {"sub_id": "Exam-11-T3-2", "tags": ["1.6"], "text_open": "3.1.2 State the principal intermolecular forces between the two components of the Novolak-DNQ system.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-11-T3-3", "tags": ["2.3", "2.4"], "text_open": "3.1.3 In the Novolak-DNQ system, the photosensitive component undergoes rearrangement upon irradiation to form unstable structure A, which then forms B in the alkaline developer N(CH_3)_4^+OH^-, changing the solubility of the photoresist. Write the SMILES representations of A and B.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-11-T3-4", "tags": ["2.4", "2.5"], "text_open": "3.2 PVCi is a high-resolution photoresist. One method for synthesizing PVCi is shown below.\n3.2.1 The first step is a conventional free-radical polymerization of a double bond, and all subsequent steps are functional-group substitutions. Write the SMILES representations of X, Y, Z, and PVCi. The terminal-group structures need not be specified; use [*] to denote the connection sites at the two ends of the polymer backbone.", "text_images": [{"file": "Exam-11-T3-4-1-prompt.png", "description": "X (C_4H_6O_2) reacts under AIBN/CH_3OH conditions to give Y; Y reacts under CH_3OH/NaOH conditions to give Z; Z reacts with 'O=C(Cl)/C=C/C1=CC=CC=C1' to give PVCi."}], "full_mark": 6}, {"sub_id": "Exam-11-T3-5", "tags": ["2.3", "2.5"], "text_open": "3.2.2 Write the SMILES representation of the structure formed by the interchain reaction of PVCi upon irradiation at 360 nm. The terminal-group structures and degrees of polymerization need not be specified; use [*] and [**] to denote connection sites in the polymer chains.", "text_images": [], "full_mark": 2}]} -{"problem_id": "Exam-11-T5", "header": "Problem 5: Familiar Structures from an Unfamiliar Perspective (18 points, 7%)", "header_images": [], "subquestions": [{"sub_id": "Exam-11-T5-1", "tags": ["1.3"], "text_open": "5.2 Crystal Y has the chemical formula QF_y. Its projections along the a and c directions are shown below; black spheres represent Q ions and white spheres represent F^-. 5.2.1 Give the value of y.", "text_images": [{"file": "Exam-11-T5-1-1-prompt.png", "description": "The image contains two two-dimensional projections of a crystal structure. The left panel is a rectangular view of the b-c plane. Axes at the lower left show c vertically upward, b horizontally to the right, and a perpendicular to the page. A large black sphere is located at each of the four corners, with two-dimensional fractional coordinates (b, c) = (0.00, 0.00), (1.00, 0.00), (0.00, 1.00), and (1.00, 1.00). Two large black spheres lie inside the unit cell at approximately (0.33, 0.33) and (0.67, 0.67). The small white spheres in the unit cell include 4 single spheres at approximately (b, c) = (0.00, 0.50), (0.67, 0.17), (1.00, 0.50), and (0.33, 0.83), and 3 pairs of overlapping spheres at approximately (b, c) = (0.17, 0.17), (0.50, 0.50), and (0.83, 0.83). The right panel is a parallelogram view of the a-b plane. Axes at the lower left show a horizontally to the right, b diagonally toward the upper left, and c perpendicular to the page; the angle between a and b is approximately 120°. A large black sphere is located at each of the four corners, with two-dimensional fractional coordinates (a, b) = (0.00, 0.00), (1.00, 0.00), (0.00, 1.00), and (1.00, 1.00). Two large black spheres lie inside the unit cell at approximately (0.33, 0.67) and (0.67, 0.33). All the small white spheres in the unit cell are single spheres, 11 in total, at approximately (0.00, 0.50), (0.50, 0.00), (0.33, 0.17), (0.17, 0.33), (0.83, 0.17), (0.50, 0.50), (0.17, 0.83), (1.00, 0.50), (0.83, 0.67), (0.67, 0.83), and (0.50, 1.00)."}], "full_mark": 2}]} +{"problem_id": "Exam-11-T2", "header": "Problem 2: An Electrochemistry Medley", "header_images": [], "subquestions": [{"sub_id": "Exam-11-T2-1", "tags": ["3.6"], "text_open": "2.3.1 Shown below are the pH-potential diagram for the Cr-H_2O system at 298.15 K (the concentration of each soluble species is 1 mmol/L) and thermodynamic data for selected substances.\n\n| Substance | $\\Delta_f H^\\circ_m /(kJ \\cdot mol^{-1})$ | $\\Delta S^\\circ_m /(J \\cdot mol^{-1} \\cdot K^{-1})$ |\n| :--- | :--- | :--- |\n| H_2O(l) | -285.83 | 69.95 |\n| OH^-(aq) | -229.98 | -10.9 |\n| H_2(g) | - | 130.68 |\n| B | -985.09 | 95.4 |\n| F | -872.95 | 190.4 |\n| G | -890.89 | 5.4 |\n\nNote: the diagram below is a potential diagram for a very dilute aqueous solution (1 mmol/L), and each species contains only one Cr atom. In the pH-potential diagram, the slopes of lines cd, de, and ef are in the ratio 7:4:5. Using the information given and your chemical knowledge, identify the species represented by A-G.", "text_images": [{"file": "Exam-11-T2-1-1-prompt.png", "description": "This is a two-dimensional rectangular pH-potential diagram. The unlabeled horizontal axis ranges from 0 to 14, and the vertical axis, labeled “potential/V,” ranges from -1.5 to 1.5 V. The upper and lower boundaries are horizontal lines at 1.5 and -1.5, respectively; the left and right boundaries are vertical lines at 0 and 14, respectively. Thick black lines divide the rectangle into 8 regions: seven labeled A, B, C, D, E, F, and G, and one labeled $Cr^{2+}$. Region E occupies the upper-left corner and is a right trapezoid. Its left boundary is part of the left frame, its right boundary is a vertical line at about 0.4, its upper boundary is part of the top frame, and its lower boundary slopes downward from left to right. The upper-right vertex a is approximately (0.4, 1.5), and the lower-right vertex c is approximately (0.4, 1.3). Region F is pentagonal. Segment ac is its left boundary, segment ab on the top frame is its upper boundary, and segment be at about 6.6 is its right boundary. Its two lower boundaries are segments cd and de, both sloping downward from left to right. Point e is the lower-right vertex; d is approximately (3.2,0.9), and e is approximately (6.6,0.6). The vertical segment ac separates E from F. Segment cd separates A (lower left) from F (upper right), while segment de separates B (below) from F (above). At approximately 3.2, a vertical line extends downward from d to k, with h between them. Segment dh has A on the left and B on the right; segment hk has the $Cr^{2+}$ region on the left and C on the right. Point h is approximately (3.2,-0.4), and k is approximately (3.2,-0.9). At approximately 6.6, a vertical line extends from point b on the top frame down to e. Point b is approximately (6.6,1.5); segment be has F on the left and G on the right, and the vertical line does not continue below e. Region G lies above diagonal ef and below the top frame, extending from about 6.6 to the right boundary at 14; thus G lies above ef and B below it. Region A lies below cd and above horizontal gh, with part of the left frame as its left boundary and vertical dh as its right boundary. Point g is approximately (0.0,-0.4), and h is approximately (3.2,-0.4). Segment gh is horizontal, with A above and the $Cr^{2+}$ region below. Region B lies below the upper broken line def and above the lower diagonal hi. Its left boundary is vertical dh, and fi is part of the right frame. Thus B is to the right of dh and A to the left; B is below ef and G above; B is above hi and C below; B is to the left of vertical fi, with the area to its right outside the frame. Point f, approximately (14.0,-0.1), is the intersection of the upper diagonal with the right boundary. The $Cr^{2+}$ region lies below horizontal gh, above horizontal jk (about -0.9 V), and between 0 and 3.2. Part of the left frame is its left boundary, and hk is its right boundary. Thus $Cr^{2+}$ is below gh and A above; $Cr^{2+}$ is above jk and D below; $Cr^{2+}$ is left of hk and C right of it. Point j is approximately (0.0,-0.9), and k is approximately (3.2,-0.9). Region C lies below diagonal hi and above diagonal kl, extending from 3.2 to 14. Its left boundary is hk and its right boundary is il. Thus C is right of hk and $Cr^{2+}$ left of it; C is below hi and B above; C is above kl and D below; C is left of vertical il, with the area to its right outside the frame. Point i, approximately (14.0,-1.0), is the intersection of hi with the right boundary; point l, approximately (14.0,-1.5), is the intersection of kl with the right boundary. Region D lies below horizontal j-k and diagonal k-l. Its lower boundary is the bottom frame, and it spans 0 to 14. At horizontal coordinate 0 on the left boundary, D lies below j; D is below jk and the $Cr^{2+}$ region above; D is below kl and C above. Reading the coordinates from the axes to one decimal place gives approximately: a≈(0.4,1.5), b≈(6.6,1.5), c≈(0.4,1.3), d≈(3.2,0.9), e≈(6.6,0.6), f≈(14.0,-0.1), g≈(0.0,-0.4), h≈(3.2,-0.4), i≈(14.0,-1.0), j≈(0.0,-0.9), k≈(3.2,-0.9), and l≈(14.0,-1.5). Thus, from top to bottom, the three principal diagonal boundaries are an unlabeled lowercase point-c-d-e-f, h-i, and k-l. Above the first are regions E/F/G in sequence and below it are A/B: specifically, E lies above the unlabeled point-c segment, F above c-d-e, and G above e-f; A lies below the unlabeled point-c-d portion, and B below d-e-f. B is above h-i and C below; C is above k-l and D below. The principal horizontal boundaries are g-h (A above and $Cr^{2+}$ below) and j-k ($Cr^{2+}$ above and D below)."}], "full_mark": 7}, {"sub_id": "Exam-11-T2-2", "tags": ["3.4", "3.6"], "text_open": "2.3.3 Calculate the coordinates of point e from the thermodynamic data.", "text_images": [], "full_mark": 12}]} {"problem_id": "Exam-11-T7", "header": "Problem 7: Investigation and Isolation of Complexes of Group $\\mathrm{IVA}$ Elements. Transition-metal $\\pi$ complexes are key intermediates in many transition-metal-catalyzed reactions and have been widely used as catalysts and synthetic reagents. Recently, the first main-group-metal $\\pi$ complex was successfully isolated.", "header_images": [], "subquestions": [{"sub_id": "Exam-11-T7-1", "tags": ["1.4", "1.5"], "text_open": "7-1 Alkene complexes of main-group elements are difficult to synthesize. Bonding between a metal and an unsaturated hydrocarbon can be described by the Dewar-Chatt-Duncanson model, as shown below. Following the example, draw a diagram illustrating the orbital interactions in an alkene $\\pi$ complex of a heavier main-group element.", "text_images": [{"file": "Exam-11-T7-1-1-prompt.png", "description": "Left section: a red letter M appears at the top. Directly below M is a molecular structure containing a $\\mathrm{C=C}$ double bond, enclosed by a parallelogram to indicate that it lies in a plane. A vertical dashed line connects M to the center of the $\\mathrm{C=C}$ bond. Each carbon atom of the double bond is bonded to two R groups, whose three-dimensional orientations are indicated by solid and hashed wedges. Below is the label “M = Transition-metals.” The middle section shows two sets of orbitals, one above the other. Below is a $\\mathrm{C-C}$ framework with attached R groups. Each carbon bears an “8”-shaped orbital, and the two orbitals have the same phase, with both lower lobes unshaded. Above is a four-lobed metal orbital whose two horizontal lobes are shaded and whose two vertical lobes are unshaded. Its lower vertical lobe faces the carbon orbitals below. Two upward arrows run from the shaded upper lobes on the carbon atoms to the horizontal lobes of the metal orbital. The label below reads “$\\sigma$-donation.” 3. The right section likewise shows two sets of orbitals, one above the other. The orbitals on the two carbon atoms of the lower $\\mathrm{C-C}$ framework have opposite phases: the left orbital is unshaded below and shaded above, while the right orbital is shaded below and unshaded above. The metal orbital above has four lobes with diagonally distributed phases: the upper-right and lower-left lobes are shaded, while the upper-left and lower-right lobes are unshaded. Two downward arrows point from the two lower lobes of the metal orbital to the corresponding upper lobes of the carbon orbitals. The label below reads “$\\pi$-backdonation.”"}], "full_mark": 2}, {"sub_id": "Exam-11-T7-2", "tags": ["1.4", "1.5"], "text_open": "7-2 Zero-valent lead complex 1 undergoes a combination reaction with $\\mathrm{P_4}$ in toluene at room temperature to form lead bis(phosphene) $\\pi$ complex 2, in which $\\mathrm{Pb}$ is coordinated by two phosphenes containing $\\mathrm{P=P}$ bonds. Write the SMILES representation of 2; [R] may be used. The only atoms coordinated to $\\mathrm{Pb}$ are $\\mathrm{P}$ atoms.", "text_images": [{"file": "Exam-11-T7-2-1-prompt.png", "description": "Reactant SMILES: '[R]C1[Ge](C2=C(C=CC=C2C3=C(C=C(C=C3C(C)C)C(C)C)C(C)C)C4=C(C=C(C=C4C(C)C)C(C)C)C(C)C)[Pb][Ge](C5=C(C6=C(C=C(C=C6C(C)C)C(C)C)C(C)C)C=CC=C5C7=C(C=C(C=C7C(C)C)C(C)C)C(C)C)C1[R]'. The number “1” is shown beside it. A right-pointing arrow appears to its right, with 1.0 equiv $\\mathrm{P_4}$ above the arrow and toluene, RT, 1 h below it. The arrow points to the number 2. C[R]-C[R] denotes a C2B12 unit."}], "full_mark": 3}, {"sub_id": "Exam-11-T7-3", "tags": ["1.4"], "text_open": "7-4 A synthetic route to a complex containing a ligand of a Group $\\mathrm{IVA}$ element is as follows. At ambient temperature, $\\mathrm{AuCl}$ reacts with triethylphosphine in a 1:1 ratio to give complex A. At room temperature, ligand X (shown below) and $\\mathrm{KC_8}$ are added to a tetrahydrofuran solution of A, affording the desired highly symmetric precursor B, which contains three $\\mathrm{Au}$ atoms and has a planar framework. Addition of an equimolar amount of A to B gives molecule C, which contains one $\\mathrm{Au-Au}$ bond. Measurements indicate that C is aromatic. Note: this molecule has an extended conjugated system whose electrons originate from the $\\mathrm{d}$ electrons of $\\mathrm{Au}$. The structure of X is 'C[Si](C)(C1([Si](C)(C)C)CCC([Si](C)(C)C)([Si](C)(C)C)[Sn]1)C'. 7-4-1 Write the SMILES representations of B and C.", "text_images": [{"file": "Exam-11-T7-3-1-prompt.png", "description": "SMILES representation of X: 'C[Si](C)(C1([Si](C)(C)C)CCC([Si](C)(C)C)([Si](C)(C)C)[Sn]1)C'"}], "full_mark": 6}, {"sub_id": "Exam-11-T7-4", "tags": ["1.4", "1.5"], "text_open": "7-4-2 In fact, C itself is not aromatic, but polarization can generate an ionic structure whose anion is aromatic. Write the SMILES representations of the anion and cation in this structure, and state the number of $\\pi$ electrons in the extended conjugated system of the anion.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-11-T7-5", "tags": ["1.5", "1.6"], "text_open": "7-4-3 Researchers found that when the coordinating atom in X is $\\mathrm{Sn}$, C shows relatively weak aromaticity; replacing the coordinating atom in X with $\\mathrm{Ge}$ markedly enhances the aromaticity. Explain this observation.", "text_images": [], "full_mark": 2}]} {"problem_id": "Exam-11-T10", "header": "Chiral Boron Chemistry. Organoboron compounds occupy an important niche in asymmetric synthesis and have become versatile stereospecific building blocks. In general, the boron atom is an electron-deficient center; therefore, the most important intermediates in these reactions are boronate anions (ate complexes).", "header_images": [], "subquestions": [{"sub_id": "Exam-11-T10-1", "tags": ["1.5", "2.2"], "text_open": "10.1 Homologation is a common application of organoboron reagents in organic synthesis. A representative example is the Matteson homologation, whose general reaction pattern is shown below.\n10.1.1 An ate-complex intermediate is formed in the reaction. Write the SMILES representations of the 2 key boron-containing intermediates in the transformation, including stereochemistry.", "text_images": [{"file": "Exam-11-T10-1-1-prompt.png", "description": "The image shows a one-step Matteson homologation. The reactant is 'C[C@H]1[C@H](C)OB(C)O1', and the conditions are 1) CH_2Cl_2, LDA; 2) ZnCl_2; 3) RMgBr. The product is 'C[C@H]1[C@H](C)OB([C@@H](C)[R])O1'."}], "full_mark": 4}, {"sub_id": "Exam-11-T10-2", "tags": ["2.2", "2.7"], "text_open": "10.1.2 Studies show that the addition of ZnCl_2 is the key step controlling the stereospecificity of the product. ZnCl_2 acts as a chelating agent and forms a 5-membered chelate ring, and the principal factor affecting the stability of the migration transition state is 1,3-strain between bulky groups. On this basis, write the SMILES representation of the key intermediate leading to the major product, including stereochemistry.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-11-T10-3", "tags": ["2.7"], "text_open": "10.1.3 If CH_2Cl_2 in the first step is replaced with CH_3CHCl_2, the stereoselectivity of the product will: ( ).\n(a) increase (b) decrease (c) remain essentially unchanged", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-11-T10-4", "tags": ["1.2", "2.4", "2.7"], "text_open": "10.1.4 Based on the information above, write the SMILES representation of the product of the following reaction; the product contains neither silicon nor sulfur. Explain the stereoselectivity of the final step.", "text_images": [{"file": "Exam-11-T10-4-1-prompt.png", "description": "The image shows a three-step reaction. The reactant is 'CC1(C)C(C)(C)OB(CC2=CC=CC=C2)O1'. The first-step conditions are 'ClC([Li])[Si](C)(C)C', THF, -78°C then r.t., ZnCl_2. The second-step conditions are '[Li]C(S[Ar])O[R]', THF, -78°C, ZnCl_2. The third-step condition is H_2SO_4. The final product is unknown and indicated by a question mark."}], "full_mark": 3}, {"sub_id": "Exam-11-T10-5", "tags": ["1.5", "1.6"], "text_open": "10.2 A major advantage of organoboron reagents is the exceptional configurational stability of their chiral carbon atoms, which enables metal-free stereospecific coupling reactions.\n10.2.1 The reactivity of a boron reagent can be tuned by adding different nucleophiles. Among the ate complexes below, the bold C-B bond is most reactive in ( ) and least reactive in ( ).", "text_images": [{"file": "Exam-11-T10-5-1-prompt.png", "description": "The image shows four structures labeled (a), (b), (c), and (d) from left to right. (a) is 'CC1(C)O[B-](C)(OC)OC1(C)C'; (b) is 'CC1(C)O[B-](C)(N(C)C)OC1(C)C'; (c) is 'CC1(C)O[B-](C)(C)OC1(C)C'; and (d) is 'C[B-]1([C@@H]2CCC[C@H]1CCC2)C'. In every structure, a B-CH_3 bond is oriented axially with respect to the five-membered ring and is shown in bold."}], "full_mark": 2}, {"sub_id": "Exam-11-T10-6", "tags": ["1.2", "2.4"], "text_open": "10.2.2 Write the SMILES representations of products A and B of the following reactions, including stereochemistry.", "text_images": [{"file": "Exam-11-T10-6-1-prompt.png", "description": "The image shows two one-step reactions. The reactant in the first is '[H][C@@](C1=CC=CC=C1)(C)B2OC(C)(C)C(C)(C)O2', and the conditions are MeONH_2, nBuLi, then Boc_2O; the unknown product is labeled A. The reactant in the second is 'CC(C(C)(C)O1)(C)OB1[C@](C2=CC=CC=C2)(C)CC', and the conditions are CH_2Cl_2, nBuLi, ZnCl_2, then NaOH/H_2O_2; the unknown product is labeled B."}], "full_mark": 3}, {"sub_id": "Exam-11-T10-7", "tags": ["2.3", "2.7"], "text_open": "10.2.3 An organoboron reagent treated with an aryllithium reagent can trap an electrophile. Experiments show, however, that different electrophiles give different stereochemical outcomes: the ate complex reacts with I_2 with good stereoselectivity, whereas reaction with DIAD results in extensive racemization. The researchers propose that the two reactions proceed by different mechanisms. Explain these observations mechanistically. Note: carbocations are planar, carbon-centered radicals undergo rapid configurational inversion, and carbanions have a high barrier to configurational inversion.", "text_images": [{"file": "Exam-11-T10-7-1-prompt.png", "description": "The image shows a multistep reaction in which the substrate is the same but the conditions differ. The substrate is 'C[C@H](B1OC(C)C(C)(C)O1)CCC2=CC=CC=C2'. The first-step conditions are 4-MeOC_6H_4Li, THF, -78°C, and the first-step product is 'C[C@H]([B-]1(C2=CC=C(OC)C=C2)OC(C)(C)C(C)(C)O1)CCC3=CC=CC=C3.[Li+]'. This product follows two pathways. The upper-right pathway uses I_2 and gives 'C[C@@H](I)CCC1=CC=CC=C1' with 97% enantioselectivity. The lower-right pathway uses DIAD, whose SMILES representation is 'O=C(/N=N/C(OC(C)C)=O)OC(C)C', and gives 'C[C@@H](N(NC(OC(C)C)=O)C(OC(C)C)=O)CCC1=CC=CC=C1' with 13% enantioselectivity."}], "full_mark": 4}, {"sub_id": "Exam-11-T10-8", "tags": ["1.2", "2.2", "2.4"], "text_open": "10.3 Coupling reactions of organoboron reagents can also be conducted in tandem with other reactions, enabling a broader range of synthetic applications.\n10.3.1 The following reaction couples a boron reagent with a thiazole ring. Write the SMILES representations of the product and 3 key boron-containing intermediates, including stereochemistry.", "text_images": [{"file": "Exam-11-T10-8-1-prompt.png", "description": "The image shows a two-step reaction. The substrate is 'O=S1C=C(C)C2=CC=CC=C21'. The reaction conditions are 1) LDA, -78°C; 2) '[Bpin][C@@H](C)C1=CC=CC=C1', followed by Tf_2O, -78°C in the second step. The product is unknown and indicated by a question mark."}], "full_mark": 4}, {"sub_id": "Exam-11-T10-9", "tags": ["2.2", "2.4"], "text_open": "10.3.2 Examine the following reaction and write the SMILES representations of the product and the anionic ate-complex intermediate. Note: the purpose of adding acid is to promote the migration.", "text_images": [{"file": "Exam-11-T10-9-1-prompt.png", "description": "The image shows a multistep reaction. The substrate is 'C=[C]=COC'. The conditions over the first arrow are 1) nBuLi; 2) '[Bpin][C@@H](C)C1=CC=CC=C1'; 3) AcOH. The conditions over the second arrow are NaOH/H_2O_2, THF/H_2O. The product is unknown and indicated by a question mark."}], "full_mark": 4}]} +{"problem_id": "Exam-12-T1", "header": "Problem 1 Metallic Vitamins. Rare-earth elements are core strategic resources in modern materials and chemical industries and are known as \"industrial vitamins.\" Their unique electron configurations give them distinctive physicochemical properties, making them essential materials for emerging industries such as modern semiconductors and new energy. A rare-earth element M has density $7.563 \\ \\mathrm{g \\cdot cm^{-3}}$ and forms an alloy with Co in a fixed ratio for manufacturing strong magnets. It has the following chemical properties. (1) When elemental M is left in air, its metallic luster rapidly disappears and a composite thin layer whose principal component is oxide A forms on its surface. M reacts with hydrochloric acid to form blood-red chloride B. After M has dissolved completely, the blood-red color rapidly fades, producing pale-yellow chloride C and colorless gas D. (2) If M is placed in a tetrahydrofuran solution of an organic compound E containing $90.05\\%$ iodine, reaction at room temperature gives unstable green iodide F and releases gas G. Addition of a small amount of water turns the solution yellow. Gas G is a common plant-ripening hormone.", "header_images": [], "subquestions": [{"sub_id": "Exam-12-T1-1", "tags": ["1.4", "2.4"], "text_open": "1-4-1 Complexes of M also have interesting structures. Addition of 0.5 equivalents of $\\mathrm{P_2Ph_4}$ to $\\mathrm{MCp*_2(THF)_2}$ gives a chain polymer X. After polymer X is separated, the pentane mother liquor is left for several weeks, yielding a yellow-brown complex Y. Y contains two M atoms and a 14-membered macrocycle. The mass fraction of M is $22.19\\%$ in both X and Y. Write the SMILES structure of the repeat unit of X, using [R] for the site connected to the next repeat unit (for example, polypropylene may be written as C([R])C(C)[R]), and the SMILES structure of Y ($\\mathrm{Cp*}$ = pentamethylcyclopentadienyl anion).", "text_images": [], "full_mark": 4}]} {"problem_id": "Exam-12-T2", "header": "Problem 2 Schrödinger's Hydrogen-Atom Model\nThe year 2026 marks the 100th anniversary of the publication of Schrödinger's wave mechanics. In Bohr's model of the hydrogen atom, the electron is a particle moving in a fixed circular orbit. Schrödinger's wave mechanics instead treats the electron as having wave-particle duality and therefore as unable to possess simultaneously definite position and momentum. In wave mechanics, the state of motion of a particle is described by a wavefunction. In general, a wavefunction depends on time and position; if it is time-independent, it reduces to a function of position. The wavefunction itself does not directly give definite values of physical quantities such as position or momentum. Rather, it contains information about their probability distributions. Appropriate mathematical operations on the wavefunction yield the probability of finding the particle at a given location, the possible values of its momentum, and the energy of the system.\nHow is the wavefunction obtained? A fundamental postulate of quantum mechanics is that the wavefunction satisfies the Schrödinger equation. This equation is a partial differential equation, and its solutions are wavefunctions. Thus, one first writes the Schrödinger equation for an electron, solves it for a wavefunction, and then uses the resulting wavefunction to determine information about position, momentum, energy, and other quantities.\nIf the potential-energy distribution in space is time-independent, the wavefunction is a function only of position and the equation is called the time-independent Schrödinger equation. For a hydrogen atom, for example, the nucleus may be approximated as fixed; the equation describing the electron is then time-independent, and the electron wavefunction depends on its position coordinates. Position may be expressed in Cartesian coordinates $(x,y,z)$, but spherical polar coordinates $(r,\\theta,\\phi)$ are often used. The physical meanings of $r,\\theta,\\phi$ are shown below, with $r>0,0 \\leq \\theta \\leq \\pi,0 \\leq \\phi \\leq 2\\pi$.\nThe time-independent Schrödinger equation for the hydrogen atom is\n$$-\\frac{\\hbar^2}{2m}\\nabla^2\\psi(r,\\theta,\\phi)-\\frac{Ke^2}{r}\\psi(r,\\theta,\\phi)=E\\psi(r,\\theta,\\phi)$$\nwhere $\\nabla^2$ is the Laplace operator, which in spherical polar coordinates is\n$$\\nabla^2=\\frac{\\partial^2}{\\partial r^2}+\\frac{2}{r}\\frac{\\partial}{\\partial r}+\\frac{1}{r^2}\\left(\\frac{\\partial^2}{\\partial \\theta^2}+\\frac{1}{\\tan\\theta}\\frac{\\partial}{\\partial \\theta}+\\frac{1}{\\sin^2\\theta}\\frac{\\partial^2}{\\partial \\phi^2}\\right)$$\nHere $m$ is the electron mass, $K$ is the electrostatic-force constant, and $\\hbar$ is the reduced Planck constant. $E$ is called the energy eigenvalue, or simply the electron energy. Although the electron's momentum and position are uncertain, its energy is definite.\nNote: $\\partial$ denotes partial differentiation. If $f$ is a function of $x,y$, then $\\frac{\\partial f(x,y)}{\\partial x}$ is defined as the derivative of $f$ with respect to $x$ while the independent variable $y$ is held constant:\n$$\\frac{\\partial f(x,y)}{\\partial x}=\\lim_{\\Delta x\\to 0}\\frac{f(x+\\Delta x,y)-f(x,y)}{\\Delta x}$$\nWhen $f$ is independent of $y$, so that $\\frac{\\partial f}{\\partial y}=0$ identically, $f$ becomes a single-variable function of $x$ and $\\frac{\\partial f}{\\partial x}=\\frac{\\mathrm{d}f}{\\mathrm{d}x}$. A differentiation formula that may be useful is $(x^{\\alpha}e^{kx})'=(kx^{\\alpha}+\\alpha x^{\\alpha-1})e^{kx}$.\nThe time-independent Schrödinger equation for hydrogen has infinitely many solutions. Certain solutions correspond to the atomic orbitals of hydrogen; each orbital is described by a particular wavefunction. The wavefunctions for the orbitals in the first two shells are listed below. 1s: $\\psi_{1s}(r,\\theta,\\phi)=\\frac{1}{\\sqrt{\\pi a_0^3}}e^{-r/a_0}$; 2s: $\\psi_{2s}(r,\\theta,\\phi)=\\frac{1}{4\\sqrt{2\\pi a_0^3}}\\left(2-\\frac{r}{a_0}\\right)e^{-r/(2a_0)}$; $ 2p_x$: $\\psi_{2p_x}(r,\\theta,\\phi)=\\frac{1}{4\\sqrt{2\\pi a_0^3}}\\left(\\frac{r}{a_0}\\right)e^{-r/(2a_0)}\\sin\\theta\\cos\\phi$; $ 2p_y$: $\\psi_{2p_y}(r,\\theta,\\phi)=\\frac{1}{4\\sqrt{2\\pi a_0^3}}\\left(\\frac{r}{a_0}\\right)e^{-r/(2a_0)}\\sin\\theta\\sin\\phi$; $ 2p_z$: $\\psi_{2p_z}(r,\\theta,\\phi)=\\frac{1}{4\\sqrt{2\\pi a_0^3}}\\left(\\frac{r}{a_0}\\right)e^{-r/(2a_0)}\\cos\\theta$. Here $a_0=\\frac{\\hbar^2}{Ke^2m}=52.9\\text{pm}$ is the Bohr radius.", "header_images": [{"file": "Exam-12-T2-header-1.png", "description": "Schematic of spherical polar coordinates: the figure shows a three-dimensional Cartesian coordinate system with origin $O$ and right-handed $x, y, z$ axes. A sphere is centered at O, and point $P$ lies on the sphere in the first octant. The length OP is $r$, and the angle between ray OP and the positive z axis is \\theta. The projection of OP onto the xOy plane makes an angle \\phi with the positive x axis, measured from the positive x axis toward the projection of OP."}], "subquestions": [{"sub_id": "Exam-12-T2-1", "tags": ["3.9"], "text_open": "2.4 Which has the higher energy, the hydrogen 2s or $2p_z$ orbital? Given that the energy of the $2p_z$ orbital is $-3.40\\text{eV}$, calculate from the wavefunction and explain. Note: the energy eigenvalue in the Schrödinger equation represents the orbital energy.", "text_images": [], "full_mark": 7}]} {"problem_id": "Exam-12-T3", "header": "Problem 3 Pinene in Polymers\nPinene is extracted from pine oil and is an important class of natural products with broad applications in synthetic polymers.", "header_images": [], "subquestions": [{"sub_id": "Exam-12-T3-1", "tags": ["2.4", "2.6"], "text_open": "3.1 Cationic polymerization of \\beta-pinene can produce a high-molecular-weight polymer used in optoelectronic materials. A \\beta-pinene-based polymer synthesis is shown below. 3.1.1 Polymers A and B are known not to contain bridged rings. Write the SMILES structures of polymers A and B, explicitly including the end groups. For each polymer, provide three structures: (a) the repeat unit, using '[X]' and '[Y]' at connections to other repeat units or end groups; (b) the '[X]' end group, using * at the connection to the polymer backbone; and (c) the '[Y]' end group, using * at the connection to the polymer backbone.", "text_images": [{"file": "Exam-12-T3-1-1-prompt.png", "description": "Reaction scheme for 3.1: molecule 'CC1(C)[C@@H]2C(CC[C@H]1C2)=C', labeled \"\\beta-Pinene\" below, reacts with molecule 'CC(Cl)(C)C1=CC=CC=C1' under `EtAlCl_2` to form **A**. **A** then reacts further under `H_2, Pd/Al_2O_3` to give **B**."}], "full_mark": 4}, {"sub_id": "Exam-12-T3-2", "tags": ["2.3", "2.6"], "text_open": "3.1.2 If a polymer chain bearing pinene, or a derivative thereof, at both ends is desired, state how the preceding reaction should be modified.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-12-T3-3", "tags": ["2.3", "2.6", "2.7"], "text_open": "3.2 \\alpha-Pinene (structure shown below) is a biologically renewable feedstock. \\alpha-Pinene itself is difficult to polymerize into a polymer by radical polymerization, but the unsaturated ketone obtained by selective oxidation of \\alpha-pinene (structure shown below) can undergo conventional radical polymerization to give a high-molecular-weight polymer. 3.2.1 From the perspective of monomer structure, explain the difference in radical-polymerization reactivity between \\alpha-pinene and the unsaturated ketone.", "text_images": [{"file": "Exam-12-T3-3-1-prompt.png", "description": "The image shows two chemical structures. On the left is 'CC1(C)[C@@H]2C(C)=CC[C@H]1C2', labeled \"\\alpha-Pinene\" below. On the right is 'CC([C@H](C3)C4)(C)[C@@H]3C(C4=O)=C', labeled \"pinocarvone\" below."}], "full_mark": 2}, {"sub_id": "Exam-12-T3-4", "tags": ["2.4", "2.6"], "text_open": "3.2.2 Structural characterization shows that the backbone of the polymer obtained from the unsaturated ketone contains two different repeat-unit structures. Write the SMILES structures of both repeat units, using an asterisk (*) for the connection sites between repeat units.", "text_images": [], "full_mark": 4}]} -{"problem_id": "Exam-12-T5", "header": "Problem 5 Structure of a Barium Titanate\nA hexagonal-family form of $\\mathrm{BaTiO_3}$ has an interesting structure.\nIn this crystal, all Ba and O atoms together form a special stacking pattern. In a close-packed structure, designate a close-packed layer whose coordination environment above and below is the same as in hexagonal close packing as h, and one whose coordination environment is the same as in cubic close packing as c. For example, double hexagonal close packing (ABACABAC...) can be represented as hchchc.... The joint Ba/O stacking sequence is hcchcch...; considering only Ba gives the same hcchcch... sequence.\nEach Ba is coordinated by 12 O atoms to form a $\\mathrm{[BaO_{12}]}$ tetradecahedral unit. Ti occurs in two chemical environments. In one, two $\\mathrm{TiO_6}$ octahedra share a face to form a $\\mathrm{[Ti_2O_9]}$ unit; in the other, an isolated $\\mathrm{[TiO_6]}$ unit forms, meaning that it does not share a face with another $\\mathrm{TiO_6}$ octahedron, although corner sharing is possible. The ratio of the two types of Ti atoms is 2:1. In the conventional unit cell whose vertices are the Ti atoms of the isolated $\\mathrm{[TiO_6]}$ units, one Ba has coordinates (2/3, 1/3, 0.5967). The unit-cell parameters are a = b = 574.6 pm and c = 1399.9 pm.", "header_images": [], "subquestions": [{"sub_id": "Exam-12-T5-1", "tags": ["1.3"], "text_open": "5.1 Multiple selection: which of the following statements are correct?\n(a) The Ti arrangement resembles hexagonal diamond.\n(b) Each O layer forms a kagome-type \"six-pointed-star array\" like that shown in [Exam-12-T5-1-1-prompt.png].\n(c) If all Ba and O atoms are regarded as jointly forming the hcchcch... close-packed sequence and Ti atoms are treated as interstitial atoms, Ti occupies all octahedral voids formed solely by O atoms.\n(d) This $\\mathrm{BaTiO_3}$ contains 2 distinct Ti-O bonds and 2 distinct Ba-O bonds.", "text_images": [{"file": "Exam-12-T5-1-1-prompt.png", "description": "The image shows a translationally symmetric planar network made of black straight lines and circles and possessing additional symmetries. There are 3 sets of black lines in different directions; within each set, adjacent parallel lines are equally spaced. Together, the three sets partition the plane into closed regular hexagons and equilateral triangles of equal edge length, with a circle at every line intersection. Around each circle, the four polygons appear in the order triangle, hexagon, triangle, hexagon. In the overall connectivity, each of the three edges of every triangle is shared with a hexagonal region, and each of the 6 edges of every hexagon is shared with a hexagonal region."}], "full_mark": 4}, {"sub_id": "Exam-12-T5-2", "tags": ["1.3", "3.9"], "text_open": "5.4 Calculate the second-shortest Ba-Ba distance.", "text_images": [], "full_mark": 5}]} -{"problem_id": "Exam-12-T7", "header": "Problem 7 Pigment Chemistry\nColorful artistic treasures bear witness to human civilization, yet time has stripped some precious artifacts of their original colors. Light exposure is one important cause of these mysterious color changes.", "header_images": [], "subquestions": [{"sub_id": "Exam-12-T7-1", "tags": ["1.3", "1.7"], "text_open": "7.1 Compound A is a red pigment widely used since the Neolithic period, but the red pigment surfaces of many Roman murals are accompanied by gray or black deterioration. Analysis shows that under illumination pigment A absorbs environmental Cl^- and forms white-to-gray chloride B and black elemental substance C; gypsum may also form. Researchers simulated the color change by treating A with NaClO solution under basic conditions. Under light and oxygen, NaClO first forms neutral diatomic species D, a strongly oxidizing key intermediate in ozone-layer depletion. D rapidly oxidizes A to ternary compound E and then further oxidizes E to B. Under illumination, B also gradually forms C together with salt F. A projection of the crystal structure of E is shown below. 7.1.1 Write the chemical formulas of A, B, D, and E.", "text_images": [{"file": "Exam-12-T7-1-1-prompt.png", "description": "The figure contains two crystal-structure visualizations, with the right panel enlarging part of the left. The left panel is a two-dimensional framework. Blue spheres form an approximately close-packed, long-range-ordered pattern. Between some adjacent blue spheres, a red sphere lies near their midpoint and is joined to both by two-tone red-blue segments, red near the red sphere and blue near each blue sphere. Treating blue spheres as polygon vertices and each red sphere plus its two segments as an edge gives two polygon types: nearly equilateral triangles and nonagons with nearly equal edge lengths. Going clockwise, the approximate internal angles of a nonagon are 120°, 240°, 60°, 120°, 240°, 60°, 120°, 240°, and 60°. Every triangle has the same translational environment, as does every nonagon. The triangles and nonagons tile the plane without gaps. A green sphere lies near each of the three blue vertices of every triangle, but not near blue spheres that are not triangle vertices; the green spheres form an approximately kagome pattern. The right panel is a three-dimensional space-filling ball-and-stick model showing the bonding of blue, green, and red spheres in one triangle. Blue spheres occupy the three vertices; green spheres lie below the drawing plane at two blue vertices and above it at the third; red spheres lie approximately at the midpoints of the three sides."}], "full_mark": 7}, {"sub_id": "Exam-12-T7-2", "tags": ["2.1"], "text_open": "7.1.2 Write the ionic equations for oxidation of A to E and of E to B. Note: ignore gypsum formation; the reduction product in both reactions is $Cl^-$. ", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-12-T7-3", "tags": ["2.3"], "text_open": "7.2 Another common red pigment F (note that F in 7.2 is unrelated to the substance denoted F in 7.1) is also a semiconductor material. Its multivalent metal M can undergo interconversion between oxidation states under illumination. Studies show that on the surfaces of Van Gogh's oil paintings, F is photoexcited and its metal M is reduced to a single oxidation state. It then reacts with carbon dioxide released by oily binders such as palmitic acid to form intermediate H and ultimately converts completely into white I. On the surfaces of cave-temple and tomb murals, by contrast, F forms black oxide J under illumination.\n7.2.4 Based on the description in Section 7.2, under otherwise identical illumination, in which solution of equal concentration does F produce the larger photocurrent? ( ). (a) NaHCO_3 solution (b) NaCl solution", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-12-T7-4", "tags": ["2.4"], "text_open": "7.3 The most common red pigment used for dyeing textiles in antiquity was an organic pigment whose color arises from a hydroxyanthraquinone chromophore. The following structure is alizarin, a representative organic red dye: 'O=C(C1=C2C=CC(O)=C1O)C3=CC=CC=C3C2=O'. 7.3.1 Alizarin has two photoconversion pathways: excited-state proton transfer (ESPT) gives structure X_1, while electron transfer (ET) oxidizes it to ionic species X_2. Write the SMILES structures of X_1 and X_2. State in words the position of the radical in X_2.", "text_images": [{"file": "Exam-12-T7-4-1-prompt.png", "description": "The figure shows the structure of alizarin H_2Az: 'O=C(C1=C2C=CC(O)=C1O)C3=CC=CC=C3C2=O'."}], "full_mark": 2}]} +{"problem_id": "Exam-12-T4", "header": "Problem 4 Titrimetric Analysis of Etching Acids\nWet chemical etching is a major process in the semiconductor-chip industry. It uses acidic substances to corrode a substrate surface and thereby achieve etching. A mixture of hydrofluoric acid, fluorosilicic acid, and nitric acid is a common etchant. A rapid and accurate laboratory method for determining the contents of these three acids in an etchant is described below.", "header_images": [], "subquestions": [{"sub_id": "Exam-12-T4-1", "tags": ["4.2"], "text_open": "Procedure background: Accurately weigh 3.028 g of the sample. Place 1.502 g of it in a plastic beaker to prepare test solution A, and use the remaining sample to prepare test solution B by the same procedure. During preparation, add 10 mL of saturated $KNO_3$ solution and anhydrous ethanol, mix, dilute with anhydrous ethanol, and let stand for approximately 20 min until the precipitate has completely settled into a separate layer. Label this as test solution A. Prepare test solution B from the remaining sample by the same procedure.\n4.1.2 State the purpose of adding saturated KNO_3 solution.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-12-T4-2", "tags": ["2.1", "4.2"], "text_open": "4.2 Titrimetric analysis. Add 30 mL of water and 3 drops of phenolphthalein indicator to test solution A. Titrate with 0.5018 $\\mathrm{mol \\cdot L^{-1}}$ standard NaOH solution until the solution becomes faint pink; 24.39 mL of the standard NaOH solution is consumed. Add 30 mL of water and 2 drops of methyl-orange indicator to test solution B. Adjust the pH with NaOH solution until the solution just turns bright yellow. Using a fluoride-ion combination electrode as the measuring electrode, titrate with 0.08273 $\\mathrm{mol \\cdot L^{-1}}$ standard La(NO_3)_3 solution; precipitate forms continuously during the titration. At the endpoint, 25.11 mL of La(NO_3)_3 solution is consumed. Weigh 0.5003 g of the original sample into a beaker containing 15 mL of 0.5 $\\mathrm{mol \\cdot L^{-1}}$ NaOH solution, add water to 60 mL, and verify with pH paper that the pH is greater than 13. Heat slowly to boiling, maintain a gentle boil for 5 min, and cool to room temperature. Add 2 drops of methyl-red indicator, adjust with nitric acid until red, and then adjust with NaOH solution until the solution just turns bright yellow. Using a fluoride-ion combination electrode as the indicator electrode, titrate with 0.08273 $\\mathrm{mol \\cdot L^{-1}}$ standard La(NO_3)_3 solution. At the endpoint, 15.28 mL is consumed. 4.2.1 Write all reaction equations occurring in Sections 4.1 and 4.2; write repeated reactions only once.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-12-T4-3", "tags": ["4.7", "4.5"], "text_open": "4.2.2 Calculate the mass fractions of HF, H_2SiF_6, and HNO_3 in the sample. Note: the fluoride-ion electrode indicates the endpoint by the abrupt change in fluoride-ion concentration.", "text_images": [], "full_mark": 6}]} {"problem_id": "Exam-12-T8", "header": "Problem 8 Properties and Applications of Acetal Derivatives\nAn acetal is an organic compound formed when one molecule of an aldehyde or ketone undergoes acid-catalyzed nucleophilic addition with two molecules of alcohol followed by dehydration. Its core structure has two alkoxy groups bonded to the same carbon atom, giving acetals distinctive chemical properties and uses.", "header_images": [], "subquestions": [{"sub_id": "Exam-12-T8-1", "tags": ["1.5", "2.5", "2.7"], "text_open": "8-1 When benzaldehyde is used to protect mannitol (structure shown below; SMILES: 'OC[C@@H](O)[C@@H](O)[C@H](O)[C@H](O)CO'), the reaction ratio is 1:2, whereas with acetone the ratio becomes 1:3. This difference arises because benzaldehyde protects 1,3-diols to form six-membered cyclic acetals, while acetone protects 1,2-diols to form five-membered cyclic ketals. Explain the difference in selectivity.", "text_images": [{"file": "Exam-12-T8-1-1-prompt.png", "description": "The image shows the structure of mannitol, whose SMILES is 'OC[C@@H](O)[C@@H](O)[C@H](O)[C@H](O)CO'."}], "full_mark": 4}, {"sub_id": "Exam-12-T8-2", "tags": ["2.7", "2.8"], "text_open": "8-2-1 Replacing the oxygen atoms in an acetal with sulfur gives a thioacetal, which has different reactivity. Compare the stabilities of the following substances in $1 \\ \\mathrm{mol/L}$ hydrochloric acid and explain your answer.", "text_images": [{"file": "Exam-12-T8-2-1-prompt.png", "description": "Substance A on the left is a ketone with SMILES 'CC1CCCC1=O'; substance B on the right is a thioketone with SMILES 'CC1CCCC1=S'."}], "full_mark": 2}, {"sub_id": "Exam-12-T8-3", "tags": ["2.7", "2.8"], "text_open": "8-2-2 Compare the stabilities of the following substances in $1 \\ \\mathrm{mol/L}$ hydrochloric acid and explain your answer.", "text_images": [{"file": "Exam-12-T8-3-1-prompt.png", "description": "Substance A on the left is a thioketal with SMILES 'C1CCCC12SCCCS2'; substance B on the right is a ketal with SMILES 'C1CCCC12OCCCO2'."}], "full_mark": 2}, {"sub_id": "Exam-12-T8-4", "tags": ["2.4", "2.5"], "text_open": "8-3-1 A proposed synthetic route to M is shown below. Write the SMILES structures of N and M.", "text_images": [{"file": "Exam-12-T8-4-1-prompt.png", "description": "'CC(C)CC=O' reacts with SCCCS to form 'CC(C)CC1SCCCS1'. Subsequent reaction with $\\mathrm{n-BuLi}$ and 'C[Si](C1(SCCCS1)CC2CO2)(C)C' gives compound N. Under $\\mathrm{HgCl_2/MeOH}$ conditions, N is converted to compound M."}], "full_mark": 3}, {"sub_id": "Exam-12-T8-5", "tags": ["2.4", "2.5"], "text_open": "8-3-2 If the reagent in the initial step is replaced with ethylene dithiol, treatment of the thioacetal with butyllithium does not give the expected product. Instead, a side reaction occurs and produces a pungent odor. IR spectroscopy shows that the reaction mixture contains many products, one of which gives an absorption attributable to a carbon-carbon double bond in a hydrocarbon. Based on this information, write 2 products formed in the reaction.", "text_images": [], "full_mark": 3}]} -{"problem_id": "Exam-12-T9", "header": "Reactivity of Nitrogen-Containing Aromatic Rings", "header_images": [], "subquestions": [{"sub_id": "Exam-12-T9-1", "tags": ["1.6", "2.7"], "text_open": "9.1 Pyrrole (1) and pyridine (2) are two common nitrogen-containing aromatic rings with different reactivities.\n9.1.1 Using resonance structures, determine the nucleophilic sites of pyrrole in aromatic electrophilic substitution. Give the SMILES of the major resonance structures of pyrrole.", "text_images": [{"file": "Exam-12-T9-1-1-prompt.png", "description": "1: 'C1=CNC=C1'; 2: 'C1=NC=CC=C1'; 3: '[O-][N+]1=CC=CC=C1'"}], "full_mark": 3}, {"sub_id": "Exam-12-T9-2", "tags": ["1.6", "2.7"], "text_open": "9.1.2 Pyridine can be oxidized by H_2O_2 to **3**. Compare the nucleophilicities of **2** and **3**, and explain your answer.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-12-T9-3", "tags": ["2.2", "2.4"], "text_open": "9.2 A literature report describes a Lewis-acid-controlled electrophilic substitution between indole and triphenyl-1,3,5-triazinane.\n9.2.1 Without a Lewis acid, a short sequence gives final product C; the first step is a thermal pericyclic reaction. Write the SMILES structures of the two key intermediates and product C.", "text_images": [{"file": "Exam-12-T9-3-1-prompt.png", "description": "A ('C12=CC=CC=C1C=CN2') and B ('N1(C2=CC=CC=C2)CN(C3=CC=CC=C3)CN(C4=CC=CC=C4)C1') react in DCE at 80℃ for 24h to give C (C_15H_14N_2)."}], "full_mark": 3}, {"sub_id": "Exam-12-T9-4", "tags": ["2.4"], "text_open": "9.2.2 In the presence of a Lewis acid, C undergoes Lewis-acid-catalyzed fragmentation and rearrangement to give product D. D and C have the same molecular formula. Write the SMILES of D.", "text_images": [{"file": "Exam-12-T9-4-1-prompt.png", "description": "A ('C12=CC=CC=C1C=CN2') and B ('N1(C2=CC=CC=C2)CN(C3=CC=CC=C3)CN(C4=CC=CC=C4)C1') react under Eu(OTf)_3 in DCE at 60℃ for 1h to give D (88%) and E."}], "full_mark": 2}, {"sub_id": "Exam-12-T9-5", "tags": ["2.4"], "text_open": "9.2.3 In the presence of a Lewis acid, the reaction also produces a tetracyclic by-product E containing 2 five-membered rings and 2 six-membered rings. Write the SMILES of E.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-12-T9-6", "tags": ["2.2", "2.4"], "text_open": "9.4 A literature report describes an NHC-catalyzed synthesis of 2-pyrones. Interestingly, replacing the ketone substrate with an ester gives a fused-ring product with a different skeleton.\n9.4.1 The first step in both reactions is activation of substrate F by the NHC. Write the SMILES of the activated intermediate, including the complete N-heterocyclic carbene structure.", "text_images": [{"file": "Exam-12-T9-6-1-prompt.png", "description": "F ('O=C(OC(C)=O)C#CC1=C2C(C=CC=C2)=CC=C1OC') and G ('[R]C(CC1=NC2=CC=CC=C2S1)=O') react under 'CC(C)(C)[C@H]1COCC2=NN(C3=C(C)C=C(C)C=C3C)[C]N21' (10 mol%), base, and solvent. Here R = Ph or R = OEt."}], "full_mark": 1}, {"sub_id": "Exam-12-T9-7", "tags": ["2.2", "2.4"], "text_open": "9.4.3 For R = Ph, write the SMILES structures of the product and the other 2 intermediates besides the answer to 9.4.1. NHC substituents may be represented by R, and stereochemistry is not required. The product is known to contain 5 six-membered rings.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-12-T9-8", "tags": ["2.4"], "text_open": "9.4.4 Write the SMILES structure of the product when R = OEt; stereochemistry is not required. The product is known to contain 4 six-membered rings.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-12-T9-9", "tags": ["1.1"], "text_open": "9.4.5 Determine whether the products of the preceding reactions are chiral.", "text_images": [], "full_mark": 1}]} {"problem_id": "Exam-13-T1", "header": "Question 1 Vanadium Chemistry. The ionic radius of vanadium is close to those of Fe and Ti, so vanadium commonly occurs in ores associated with Ti and Fe. The principal extraction method first concentrates vanadium in slag, which is roasted to form $FeV_2O_4$. NaCl-$Na_2SO_4$ is then added and melted while oxygen is introduced, converting vanadium into a soluble 1:1 salt in reaction 1. Water leaching gives a vanadium-containing solution, after which an ammonium salt is added to separate the vanadium in reaction 2. Heating the resulting substance at 500 °C gives a simple orange-yellow vanadium oxide product in reaction 3. $VO_2$ has many interesting physical properties, including thermochromism. At atmospheric pressure, $VO_2$ has 2 phases, monoclinic M1 and rutile R, and a phase transition upon heating changes the color of $VO_2$. Thermodynamic data and unit-cell structures for the 2 phases of $VO_2$ near the transition temperature at $p^⊖$ are given below; the large spheres are V and the small spheres are O.\n\nThe first 2 columns of the table are thermodynamic data, and the final six columns are unit-cell parameters.\n\n| Phase | $\\Delta_{f}H_m^{\\ominus}/\\mathrm{kJ\\cdot mol^{-1}}$ | $S_m^{\\ominus}/\\mathrm{J\\cdot mol^{-1}\\cdot K^{-1}}$ | a/pm | b/pm | c/pm | α/° | β/° | γ/° |\n|---|---|---|---|---|---|---|---|---|\n| M1 phase | -714.4 | 61.62 | 575.2 | 452.5 | 538.2 | 90 | 122.6 | 90 |\n| R phase | -710.1 | 74.17 | 455.4 | 455.4 | 285.3 | 90 | 90 | 90 |", "header_images": [], "subquestions": [{"sub_id": "Exam-13-T1-1", "tags": ["4.2"], "text_open": "1.2 Briefly explain why $Na_2SO_4$ is added in the process represented by reaction 1.", "text_images": [], "full_mark": 2}]} {"problem_id": "Exam-13-T2", "header": "The history of first-generation glucose biosensors dates to 1962, when Clark and Lyons at Cincinnati Children's Hospital immobilized a thin layer of glucose oxidase, GOD, on an oxygen electrode behind a semipermeable membrane permeable only to $O_2$. GOD catalyzes oxidation of glucose by dissolved oxygen in blood, reducing the amount of $O_2$ passing through the membrane. Blood glucose is determined from the resulting change in oxygen-electrode potential.", "header_images": [], "subquestions": [{"sub_id": "Exam-13-T2-1", "tags": ["3.4"], "text_open": "2.2 First-generation glucose biosensors are strongly affected by fluctuations in blood oxygen partial pressure and are readily disturbed by environmental oxygen. Second-generation sensors address this limitation. After GOD oxidizes glucose and becomes reduced, a synthetic mediator M reoxidizes the reduced GOD. The reduced form of M is then oxidized at the measuring electrode to produce an electrical signal from which blood glucose is determined.\nOne glucose biosensor uses ferrocene as M. To test whether electrolyte pH affects the ferrocene redox rate, buffer solutions of different pH values must be prepared. Researchers prepare them by adding $HClO_4$ solution dropwise to $0.10\\ \\mathrm{mol\\cdot L^{-1}}$ K_2HPO_4 solution. The three pKa values of phosphoric acid are 2.15, 7.20, and 12.32, and the HClO_4 concentration is approximately $1.16\\ \\mathrm{mol\\cdot L^{-1}}$. 2.2.1 What volume of $1.16\\ \\mathrm{mol\\cdot L^{-1}}$ HClO_4 solution must be added to 100 mL of $0.10\\ \\mathrm{mol\\cdot L^{-1}}$ K_2HPO_4 solution to prepare a buffer at pH = 6.00?", "text_images": [], "full_mark": 5}]} -{"problem_id": "Exam-13-T5", "header": "Question 5 Crystal Distortions. Under ideal conditions, several classic crystal structures have high symmetry, including cubic perovskite, fluorite, and spinel structures. Under certain conditions, however, some crystal structures distort, breaking the original high symmetry and lowering it to tetragonal, trigonal, orthorhombic, monoclinic, or another crystal system.", "header_images": [{"file": "Exam-13-T5-header-1.png", "description": "A perspective ball-and-stick model of a three-dimensional crystal structure. Axes at the lower left show a in red pointing toward the lower left and into the view, b in green pointing horizontally right, and c in blue pointing vertically upward. The outer framework is a cube with a large light-blue sphere at each of its 8 vertices and thin light-blue rods joining adjacent vertices. A smaller red sphere is embedded near the midpoint of each of the 12 cube edges. The largest green sphere is suspended at the center of the cube and connects radially to the 12 red spheres through 12 thin green rods. The 8 red spheres in the top layer, c = 1, and bottom layer, c = 0, are accompanied by broad white arrows with black borders and bold black coordinates indicating displacement. In the top layer, the left sphere has a rightward arrow and (0.5,0.0098,1); the right sphere has a leftward arrow and (0.5,0.9902,1); the front sphere has a lower-right arrow and (1.0098,0.5,1); and the rear sphere has an upper-right arrow and (-0.0098,0.5,1). In the bottom layer, the left sphere has an outward left arrow and (0.5,-0.0098,0); the right sphere has an outward right arrow and (0.5,1.0098,0); the front sphere has a lower-left arrow and (0.9902,0.5,0); and the rear sphere, seen in perspective below and to the right of the central green sphere, has an upper-left arrow and (0.0098,0.5,0)."}], "subquestions": [{"sub_id": "Exam-13-T5-1", "tags": ["1.3"], "text_open": "5.2 At low temperature, $NaNiO_2$ distorts from a high-symmetry structure into a base-centered monoclinic crystal, as shown below; the large, medium, and small spheres are Na, Ni, and O, respectively. One O atom has coordinates (0.222, 0.5, 0.205), both Ni and Na occupy vertex or face-center positions, and the unit-cell parameters are a = 5.33 Å, b = 2.86 Å, c = 5.59 Å, and β = 110.5°. 5.2.1 Before distortion, the structure follows a close-packing model: O forms the closest-packed array, while alternating Na and Ni layers occupy octahedral holes in the O array. Use ABC for the close-packed O layers, abc for the Na layers, and a’b’c’ for the Ni layers. Represent the close-packing sequence before distortion using letters, giving one repeat unit, and state the Bravais lattice of the undistorted crystal.", "text_images": [{"file": "Exam-13-T5-1-1-prompt.png", "description": "A perspective three-dimensional ball-and-stick model enclosed by an inclined parallelepiped outlined with thin black lines. Axes at the lower left show a in red pointing roughly left, b in green pointing into the page, and c in blue pointing steeply toward the upper right. Three sphere sizes and colors represent the elements: the largest golden spheres are Na, medium silver-gray spheres are Ni, and the smallest red spheres are O. Thin cylindrical bonds are split between the colors of the connected atoms, such as yellow-red Na-O and silver-red Ni-O bonds. The structure is a highly ordered alternating layered arrangement along c. Silver-gray Ni layers lie near the upper and lower faces of the cell. Each Ni connects diagonally to 6 O atoms through 6 silver-red bonds, forming a typical $NiO_6$ octahedron. These edge-sharing octahedra extend laterally in planes perpendicular to c to form robust two-dimensional $NiO_2$ slabs. A layer of large golden Na spheres occupies the central plane in the open space between the 2 $NiO_2$ slabs. Each Na radiates 6 yellow-red bonds in an octahedral geometry, connecting upward to 3 O atoms at the bottom of the upper $NiO_2$ slab and downward to 3 O atoms at the top of the lower $NiO_2$ slab. Thus, rather than forming simple vertical linear units, the interlayer Na atoms act as structural anchors through a three-dimensional network of 6 contacts, 3 above and 3 below, joining the separated silver-red slabs along c into a continuous crystal network."}], "full_mark": 5}, {"sub_id": "Exam-13-T5-2", "tags": ["1.3"], "text_open": "5.2.2 Calculate all distinct Ni-O bond lengths after distortion.", "text_images": [], "full_mark": 4}]} +{"problem_id": "Exam-13-T4", "header": "Question 4 Determination of Iron in Iron Ore. The potassium dichromate method is a classic method for determining iron in iron ore, but its use of $HgCl_2$ creates toxicity and environmental concerns. Selected steps of an improved dichromate method, the $TiCl_3-K_2Cr_2O_7$ method, are as follows. ① Preparation of the $K_2Cr_2O_7$ standard solution: accurately weigh 0.9797 g of $K_2Cr_2O_7$ into a 100 mL beaker, dissolve it in water, quantitatively transfer it to a 250 mL volumetric flask, and dilute to volume to obtain the $K_2Cr_2O_7$ standard solution. ② Titration of the sample: accurately weigh 0.2026 g of powdered magnetite concentrate, add 20 mL of hydrochloric acid, approximately 6 mol/L, cover with a watch glass, and heat gently to near boiling. After the ore decomposes, the mixture is reddish brown; immediately add $SnCl_2$ solution dropwise until it becomes pale yellow. Once most of the ore has dissolved, keep the mixture at a gentle boil for 1-2 min until decomposition is complete and no black particles remain. If the yellow color is too deep, add a little more $SnCl_2$ solution to obtain a pale-yellow solution; if the solution is colorless, add a small amount of $KMnO_4$ until a pale-yellow color appears, and stop heating. Rinse the watch glass and beaker walls with a wash bottle, add 50 mL of water and 8 drops of $Na_2WO_4$ solution, and while swirling add $TiCl_3$ solution to endpoint 1, followed by 2 excess drops. Cool to room temperature and carefully add the $K_2Cr_2O_7$ standard solution to endpoint 2; V1 = 0.02 mL is consumed. Add another 50 mL of water, 10 mL of sulfuric-phosphoric mixed acid with concentrated sulfuric acid:concentrated phosphoric acid:water = 2:3:5 by volume, and 2 drops of sodium diphenylamine sulfonate. Immediately titrate with the $K_2Cr_2O_7$ standard solution to endpoint 3; V2 = 28.12 mL is consumed.", "header_images": [], "subquestions": [{"sub_id": "Exam-13-T4-1", "tags": ["4.7"], "text_open": "4.3 The following operations cause errors in the experiment. Explain the reason for each error. 4.3.1 The ore is heated at too high a temperature or for too long during decomposition with hydrochloric acid.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-13-T4-2", "tags": ["4.7"], "text_open": "4.3.2 The sulfuric-phosphoric mixed acid is not added before titration.", "text_images": [], "full_mark": 2}]} {"problem_id": "Exam-13-T8", "header": "Question 8 Fundamentals of Organic Chemistry. This problem concerns aldehyde and ketone hydration equilibria, asymmetric addition, neighboring-group participation, and pH-dependent hydrolysis mechanisms.", "header_images": [], "subquestions": [{"sub_id": "Exam-13-T8-1", "tags": ["1.2", "2.7", "2.4"], "text_open": "8.2 Addition reactions of unsymmetrical aldehydes and ketones generally exhibit some stereoselectivity. Select the major product formed in each of the following reactions. 8.2.3", "text_images": [{"file": "Exam-13-T8-1-1-prompt.png", "description": "Question 8.2.3: reduction of tert-butyl-substituted cyclohexanone by (iPrO)_3Al/iPrOH gives two possible cyclohexanol stereoisomers. The 4-tert-butylcyclohexanone reactant has a solid-wedge tert-butyl bond and SMILES 'CC(C)(C)C(CC1)CCC1=O'. A rightward arrow labeled '(iPrO)3Al' above, SMILES 'CC(O[Al](OC(C)C)OC(C)C)C', and iPrOH below, SMILES 'CC(O)C', gives an alcohol with a wavy OH bond and SMILES 'CC(C)(C)[C@H]2CCC(O)CC2'. Below, one diastereomer has a dashed OH bond and SMILES 'CC(C)(C)[C@H]4CC[C@@H](O)CC4'; '(b)' has a solid-wedge OH bond and SMILES 'CC(C)(C)[C@H]3CC[C@H](O)CC3'."}], "full_mark": 1}, {"sub_id": "Exam-13-T8-2", "tags": ["1.2", "2.7", "2.4"], "text_open": "8.2.4", "text_images": [{"file": "Exam-13-T8-2-1-prompt.png", "description": "A bicyclic ketone bearing a bridgehead methyl group has SMILES 'O=C1[C@@]2(C)CC[C@@H](C2)C1'. A rightward arrow labeled LiAlH_4 gives an alcohol with a wavy OH bond, indicating unspecified configuration at the new stereocenter, with SMILES 'OC3[C@@]4(C)CC[C@@H](C4)C3'. Below are two diastereomers: '(a)', with a dashed OH bond, has SMILES 'O[C@@H]5[C@@]6(C)CC[C@@H](C6)C5'; '(b)', with a solid-wedge OH bond, has SMILES 'O[C@H]7[C@@]8(C)CC[C@@H](C8)C7'."}], "full_mark": 1}, {"sub_id": "Exam-13-T8-3", "tags": ["2.3", "4.2"], "text_open": "8.4 Intramolecular catalysis sometimes occurs in aqueous hydrolysis of ketals, esters, and related compounds. This neighboring-group participation allows a group near the reaction center to provide an acidic, basic, or otherwise activating site, producing unusual pH dependence of the rate k. 8.4.1 Kinetic data show that the acetylsalicylate anion hydrolyzes much faster than neutral acetylsalicylic acid. Researchers proposed 2 intramolecular mechanisms, nucleophilic catalysis and general-base catalysis, as shown below. Isotope tracing shows that hydrolysis with H_{2}^{18}O does not produce salicylic acid containing ^{18}O. Which mechanism is correct?", "text_images": [{"file": "Exam-13-T8-3-1-prompt.png", "description": "Two catalytic mechanisms for hydrolysis of acetylsalicylate. In the English-localized upper half, the label is Nucleophilic catalysis:. Acetylsalicylate, SMILES 'CC(OC1=CC=CC=C1C([O-])=O)=O', forms an intermediate containing an ortho oxyanion and mixed-anhydride motif, SMILES 'CC(OC(C1=C([O-])C=CC=C1)=O)=O', which hydrolyzes under an arrow labeled H_2O to salicylic acid, SMILES 'O=C(C1=C(O)C=CC=C1)O'. In the lower half, the label is General-base catalysis:. Acetylsalicylate interacts with H-O-H; 2 curved arrows show the carboxylate oxyanion abstracting a water proton while the water oxygen attacks the acetyl carbonyl carbon. This gives a tetrahedral intermediate bearing methyl, hydroxyl, oxyanion, aryl ether, and ring carboxyl groups, SMILES 'CC(O)(OC1=CC=CC=C1C(O)=O)[O-]', which decomposes to the same salicylic acid, SMILES 'O=C(C1=C(O)C=CC=C1)O'."}], "full_mark": 2}, {"sub_id": "Exam-13-T8-4", "tags": ["1.6", "2.3", "1.7"], "text_open": "8.4.2 Introducing nitro groups at both positions meta to the acetyloxy group of acetylsalicylic acid reverses the hydrolysis mechanism in the resulting dinitro compound, so the other proposed mechanism operates. Give a brief reasonable explanation.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-13-T8-5", "tags": ["2.3", "3.7", "4.2"], "text_open": "8.4.3 The hydrolysis rate k of the cyclic ketal below varies with pH as shown. An anomalous plateau occurs over pH = 2-4. Propose the ketal-hydrolysis mechanism in this region.", "text_images": [{"file": "Exam-13-T8-5-1-prompt.png", "description": "k-pH curve for the cyclic ketal with SMILES 'O=C(C(C=CC=C1)=C1OC2OCCCC2)O'. A rectangular black coordinate frame has inward ticks on all sides and a decorative colored background at the left. The vertical axis is k/×10³ s⁻¹ with labels 0, 10, 20, 30, and 40 from bottom to top. The horizontal axis is pH with consecutive integers 1 through 7. A smooth black curve decreases overall with increasing pH and is relatively flat between pH 2 and 5. Along and near the curve are 12 hollow circular data markers with small center dots."}], "full_mark": 3}, {"sub_id": "Exam-13-T8-6", "tags": ["2.3", "3.7", "4.2"], "text_open": "8.4.4 The rate of ester hydrolysis of the substance below varies with pH as shown. Explain the k-pH curve mechanistically.", "text_images": [{"file": "Exam-13-T8-6-1-prompt.png", "description": "log k-pH curve for ester hydrolysis of the substrate with SMILES 'ClC1=C(OC(C)=O)C(CC2=C(Cl)C(Cl)=CC(Cl)=C2O)=C(Cl)C(Cl)=C1'. A black rectangular frame with inward ticks has vertical axis log k/s⁻¹ labeled -5, -4, -3, -2, and -1 from bottom to top and horizontal axis pH labeled 2, 4, 6, 8, 10, and 12. A continuous black curve with solid and hollow data markers first decreases, reaches a minimum near pH 5-6, then rises, forms a relatively flat plateau around pH 9-11, and bends upward again above pH 12."}], "full_mark": 6}]} -{"problem_id": "Exam-13-T10", "header": "Question 10 Dyotropic Rearrangements. A dyotropic rearrangement is a pericyclic isomerization in which two σ bonds migrate simultaneously within a molecule. Type I and type II rearrangements often show good stereoselectivity and can be used in natural-product synthesis.", "header_images": [], "subquestions": [{"sub_id": "Exam-13-T10-1", "tags": ["2.3", "1.2", "3.9"], "text_open": "10.1 Four-membered-ring lactones are common substrates for dyotropic rearrangements. The following rearrangement constructs a fused lactone containing 3 contiguous stereocenters in one step. 10.1.1 Draw a Newman projection of the reactive conformation of the substrate, viewed along its dashed-wedge bond.", "text_images": [{"file": "Exam-13-T10-1-1-prompt.png", "description": "Dynamic rearrangement equilibria labeled type I and type II. Type I shows an open-chain system in which groups A and B are attached to a carbon chain containing unsaturation; dashed lines indicate delocalization or partial double-bond character, and the equilibrium reversibly shifts the unsaturation or conjugated system along the chain while changing molecular configuration. Type II shows a three-dimensional bridged or cage-like polycycle. In the left isomer, A and B lie on a particular front-right C-C single bond and a C=C bond lies at the rear left. Rearrangement breaks the A/B-bearing bond and forms a new A/B-bearing bond at the front left while shifting the double bond concertedly."}, {"file": "Exam-13-T10-1-2-prompt.png", "description": "Question 10.1: four-membered-ring lactone substrate 'O=C1O[C@@H](C2CCCC2)[C@@H]1OC3=CC=CC=C3' undergoes an MgBr_2-promoted dyotropic rearrangement to give fused lactone '[H]C4(C(C(O5)=O)OC6=CC=CC=C6)C5([H])CCCC4' with 3 contiguous stereocenters."}], "full_mark": 2}, {"sub_id": "Exam-13-T10-2", "tags": ["2.4", "1.2", "2.7"], "text_open": "10.1.2 Write the product SMILES, including stereochemistry.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-13-T10-3", "tags": ["2.3", "2.4"], "text_open": "10.2 The natural product curcumanolide A can be synthesized concisely through the key steps of the following small-organic-molecule-catalyzed asymmetric intramolecular lactonization and dyotropic rearrangement. 10.2.1 Write the SMILES strings of at least 2 key intermediates in the first transformation.", "text_images": [{"file": "Exam-13-T10-3-1-prompt.png", "description": "A multistep synthesis. Reactant 'O=C(CCC1=O)C1(CCCC(O)=O)C' reacts under cat. and a catalyst bearing a dashed-wedge Ph group, SMILES '[C@H]7(CCN8C(SC9=C8C=CC=C9)=N7)C%10=CC=CC=C%10', with TsCl, LiCl, iPr_2NEt below the arrow, to give intermediate 'O=C(O2)[C@H]3CC[C@]4([C@]32CCC4=O)C'. mCPBA then gives A; cat. TMSOTf converts A to B; 2 parallel arrows convert B to curcumanolide A, SMILES 'O=C(/C(C5)=C(C)\\C)O[C@]65[C@H](CC[C@H]6C(OC)=O)C'."}], "full_mark": 4}, {"sub_id": "Exam-13-T10-4", "tags": ["2.8", "1.4", "2.7"], "text_open": "10.2.2 State the role of LiCl in the first transformation.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-13-T10-5", "tags": ["2.4", "1.2", "2.3"], "text_open": "10.2.3 In a later step, Lewis acid TMSOTf catalyzes dyotropic rearrangement of A to B. Write the SMILES strings of A and B, including stereochemistry.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-13-T10-6", "tags": ["1.4", "2.3", "2.4"], "text_open": "10.3 Dyotropic rearrangements can also occur at transition metals. Write the SMILES string of intermediate C in the following transformation.", "text_images": [{"file": "Exam-13-T10-6-1-prompt.png", "description": "A two-step reaction. A reactant bearing central Cp_2Zr and two furan rings, SMILES 'C1([ZrCp2]C2=CC=CO2)=CC=CO1', is heated at 185 °C to intermediate 'C'. C then reacts with MeO_2C-C≡C-CO_2Me, SMILES 'COC(=O)C#CC(=O)OC', at 105 °C to give a benzene product bearing two methoxycarbonyl groups, CO_2Me, and one furan on three consecutive carbons, SMILES 'COC(=O)C1=C(C(=O)OC)C(C2=CC=CO2)=CC=C1'."}], "full_mark": 2}, {"sub_id": "Exam-13-T10-7", "tags": ["2.3", "3.9"], "text_open": "10.4 Type II dyotropic rearrangements generally require aromatization as a driving force. The following transformation may involve a dyotropic mechanism, so researchers tested it using a deuterated substrate. If the reaction follows a dyotropic mechanism, which positions in the product are deuterated? Circle them on the answer sheet.", "text_images": [{"file": "Exam-13-T10-7-1-prompt.png", "description": "A consecutive 2-step reaction. Reactant 'O=C1OC(C2=C1CC3=C(C2)CC3)=O' and deuterated reactant 'C=C4C([2H])C=CC([2H])C4=C', which has 2 wavy bonds labeled D and the English-localized caption 1:1 mixture of cis/trans isomers, are joined by a plus sign and heated at 117 °C to intermediate '[2H]C5C=CC([2H])C6=C5CC(C7)(C(O8)=O)C(C6)(C8=O)CC(C7=C)=C', also bearing 2 wavy D bonds. Heating at 150 °C gives final product 'O=C(O9)C%10%11C(CC(C=CC=C%12)=C%12C%11)(C9=O)CC(C)=C(C)C%10'."}], "full_mark": 2}, {"sub_id": "Exam-13-T10-8", "tags": ["3.9", "1.2", "2.4"], "text_open": "10.5 The following reaction undergoes a dyotropic rearrangement. Write the product SMILES, including stereochemistry.", "text_images": [{"file": "Exam-13-T10-8-1-prompt.png", "description": "The molecule with SMILES 'ClC1=C(Cl)[C@@]2(Cl)[C@]3([H])[C@@]([C@]1(Cl)C2=O)([H])[C@H]4[C@@]5([H])[C@@]6(Cl)C(Cl)=C(Cl)[C@@]([R1]6)(Cl)[C@@]5([H])[C@@H]3[R2]4' is heated; the product is not shown. R1 represents CCl_2 and R2 represents CH_2."}], "full_mark": 3}]} +{"problem_id": "Exam-14-T1", "header": "Ancient Metallurgy and Transmutation Chemistry\nBecause natural gold and silver were rare and extremely valuable, people in antiquity devised many ingenious methods for preparing precious metals artificially. Ming-dynasty works such as Tiangong Kaiwu and Shuyuan Zaji recorded methods for converting pure copper into “imitation gold and silver” with certain minerals, while alchemists also performed “transmutation” experiments with arsenic-, mercury-, and lead-bearing minerals. Modern chemistry has shown that these remarkable phenomena involve complex principles of inorganic synthesis and thermodynamics.", "header_images": [], "subquestions": [{"sub_id": "Exam-14-T1-1", "tags": ["1.1", "2.1"], "text_open": "1.1.2 In the Jin dynasty, Ge Hong recorded in Baopuzi a method for transmuting pure copper into a silvery-white alloy using an arsenic-containing compound. In the actual reaction, heating this compound in a sealed system first causes it to sublime and form gaseous molecule A with an adamantane-type cage; A is then reduced by charcoal to gaseous molecule B with a regular tetrahedral geometry; B penetrates the surface of solid pure copper and ultimately forms a silvery-white copper-arsenic alloy (the atomic ratio of the two elements in the alloy is 1:3). Write the chemical formula of A and the overall equation for formation of the alloy (assume that CO is the only oxidation product of carbon).", "text_images": [], "full_mark": 3}]} {"problem_id": "Exam-14-T2", "header": "Electrochemistry of an Oxidoreductase\nLaccase is a copper-containing oxidoreductase that catalyzes the reduction of oxygen to water while oxidizing phenolic compounds and various other substrates. Its catalytic center contains a T1 copper center (electron-acceptor site) and a T2/T3 trinuclear copper cluster (redox site). Because of its high activity and selectivity, laccase is often used as a cathode catalyst in biofuel cells. This problem concerns a laccase from Trametes versicolor, which may be approximated as a spherical protein with diameter d of approximately 5.0 nm. The standard electrode potential of its T1 copper center is $E_{\\mathrm{T1}}^\\circ=+0.78\\ \\mathrm{V}$ (vs. SHE). The laccase is immobilized on different electrodes to investigate its direct electron transfer (DET) and mediated electron transfer (MET) behavior.", "header_images": [{"file": "Exam-14-T2-header-1.png", "description": "A set of colored ribbon and line structures comprising magenta, brown, and green regions.\n\nThe upper-central magenta region includes several ribbon arrows, folded ribbons, and thin lines. The green region on the right includes ribbon arrows, curved lines, and a helical coil near the bottom. The brown region on the left and at the bottom includes several ribbon arrows and thin lines.\n\nVisible labels are His 458, Asp 206, T1 Cu, and T2 & T3 Cu. His 458 is left of center, Asp 206 is upper-right of center, T1 Cu is in the middle-left region, and T2 & T3 Cu is below center.\n\nSeveral small spherical markers in blue, orange, red, purple, and other colors appear near the center. Black dashed circular or elliptical markers surround some spheres and nearby labels. There are no axes, table, legend, or title."}], "subquestions": [{"sub_id": "Exam-14-T2-1", "tags": ["3.7", "4.2"], "text_open": "2.1.2 For random orientations, the electron-transfer rate for the orientation with the T1 copper center nearest the electrode is approximately $1.1\\times10^{13}$ times that for the farthest orientation. Using the figure below, explain why immobilizing the enzyme on the electrode surface is necessary for the redox reaction. The small circle represents the T1 copper center (electron-acceptor site), and the triangle represents the T2/T3 trinuclear copper cluster (redox site). (No more than 50 words.)", "text_images": [{"file": "Exam-14-T2-1-1-prompt.png", "description": "The label Substrate appears at the upper left and Product at the upper right. A long horizontal light-gray rectangle with a dotted texture spans the bottom and is labeled Electrode at the lower right.\n\nA sequence of 5 blue circles is arranged from left to right above the gray rectangle. Each contains a red triangle and a small orange circle. The red triangles have different orientations: No. 1 points approximately upward, No. 2 upper-left, No. 3 left, No. 4 lower-left, and No. 5 downward.\n\nBelow blue circle No. 1 is a white rectangular box labeled e⁻, with a black arrow pointing upward from the box to the blue circle. Two curved arrows above circle No. 1 run from near Substrate toward Product.\n\nBlack dashed horizontal lines extend from several orange dots and connect to vertical double-headed arrows or scale lines on the right. The vertical double-headed arrow near circle No. 1 is labeled d. Similar vertical lines or double-headed arrows extend upward from the gray surface near the remaining circles, indicating progressively larger distances from the gray region to the orange dots."}], "full_mark": 2}]} -{"problem_id": "Exam-14-T5", "header": "Familiar Structures from an Unfamiliar Perspective (II)", "header_images": [], "subquestions": [{"sub_id": "Exam-14-T5-1", "tags": ["1.3"], "text_open": "Structural background: An orthorhombic crystal of $XO_n$ contains long chains of edge-sharing $[XO_6]$ octahedra whose projections coincide along $a$. Two long chains form a small unit, and these units repeat in the $bc$ plane. The figure below is the projection along $a$ (short side $b$, long side $c$).\n5.1.2 Give the value of n.", "text_images": [{"file": "Exam-14-T5-1-1-prompt.png", "description": "White background with a horizontally oriented thin dark-gray rectangular frame near the center. The frame lies behind colored spheres, rod-like connectors, and translucent purple polygonal faces, and some frame edges are obscured by foreground structures.\n\nThere are 24 visible spheres: 16 red and 8 purple or magenta. The red spheres comprise 1 at the far lower left, 1 at the upper left, 2 in the middle-left region, 1 at the lower left, 3 at the top, 2 at the center, 2 in the lower-central region, 2 in the middle-right region, and 2 at the far right. The purple or magenta spheres comprise 1 at the far left, 1 in the middle-left region, 1 upper-left of center, 1 upper-right of center, 1 in the middle-right region, 1 at the far right, 1 lower-left of center, and 1 lower-right of center.\n\nA local structure of 5 spheres appears on the left: 1 magenta sphere at the far left, 1 red sphere lower-left, 1 red sphere upper-left, 1 purple sphere in the middle-left region, and 1 red sphere left of center. Dark-purple or red rod-like connectors form triangular and rhomboidal outlines. A translucent purple face lies in this region, with the spheres and thick connectors drawn above it.\n\nThe upper region contains 6 principal spheres: red spheres at upper-left, upper-center, upper-right, and upper-center, plus purple or magenta spheres upper-left and upper-right of center. Dark-purple or red connectors and a large translucent purple polygonal face join them. The upper-left red sphere lies over the face edge; the upper-center and upper-right red spheres lie on the upper outline; the two purple or magenta spheres lie within or near the polygon boundary; and the red sphere above center obscures parts of the face and connectors.\n\nThe lower-central region contains 6 principal spheres: a middle-left red sphere, a lower-central red sphere, purple spheres lower-left and lower-right of center, another lower-central red sphere, and a lower-left red sphere. Rod-like connectors and a translucent purple polygonal face join them. A foreground lower-central red sphere obscures part of the face; the two purple spheres each connect to several surrounding red spheres; and the lower-left and lower-central red spheres lie along the lower boundary.\n\nA local structure of 6 spheres appears on the right: middle-right red and purple spheres, lower-right and upper-right red spheres, a far-right purple sphere, and a red sphere at the lower right. Red or dark-purple connectors join them. A translucent purple polygon lies between the middle-right red, upper-right red, lower-right red, and middle-right purple spheres. The far-right purple sphere is connected to the red spheres on the right and forms the rightmost point.\n\nThe dark-gray rectangular frame is farthest back; the translucent purple polygonal faces lie in front of it; and the spheres and thicker connectors are mostly in front of the faces. Some connectors are obscured by spheres, and parts of the translucent faces are obscured by red or purple spheres. The structure extends horizontally, with more dispersed ends and denser central, upper, and lower regions."}], "full_mark": 2}, {"sub_id": "Exam-14-T5-2", "tags": ["1.3"], "text_open": "5.1.3 State the lattice type of the crystal.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-14-T5-3", "tags": ["1.3"], "text_open": "5.2.1 Y is in the same group as X, and the structure of $\\mathrm{YO}_n$ is similar to that of $\\mathrm{XO}_n$. $\\mathrm{YO}_n$ is monoclinic, but its unit-cell parameter β = 119.97° is very close to 120°, and a ≈ c. It can therefore be described using a packing scheme analogous to that of a hexagonal crystal: O atoms form an approximate “hexagonal closest packing” along the b axis in an …ABAB… sequence; Y atoms regularly occupy half of the octahedral holes in each layer; and the projections of Y atoms in adjacent layers do not overlap along b. This can be denoted $\\ldots Ac_1Bc_2Ac_1Bc_2\\ldots$. The projection along b is shown below. (The Y atoms on the short diagonal are in the same layer and on the unit-cell surface; those on the long diagonal are in the same layer and inside the unit cell.) Give the number of Y atoms in one unit cell.", "text_images": [{"file": "Exam-14-T5-3-1-prompt.png", "description": "A colored ball-and-stick model on a white background. Behind the main structure is an inclined rectangular or parallelogram frame made of thin dark-gray lines. Portions of the frame are obscured by foreground spheres and connectors.\n\nA three-dimensional direction marker appears at the lower left: a red arrow points upper-left and is labeled a; a blue arrow points right and is labeled c; and a green dot or short arrow near their common origin is labeled b. The angle between the a and c directions is approximately 120°.\n\nThere are 20 visible spheres: 4 larger pale gray-green spheres and 16 red spheres. The 4 pale gray-green spheres lie at the upper-left, upper-right, lower-left, and lower-right positions, forming an approximate quadrilateral. The red spheres are distributed around them and along their connecting paths: 4 at the top, 3 on the left, 4 in the center, 3 at the bottom, and 2 on the right.\n\nThe upper-left pale gray-green sphere is connected to several red spheres: one above by an approximately vertical rod, one upper-left and one lower-left by diagonal rods, one upper-right by a diagonal rod, one middle red sphere lower-right, and one neighboring red sphere lower-left.\n\nThe upper-right pale gray-green sphere is connected to several red spheres: one above by an approximately vertical rod, one upper-center sphere to its upper-left, one central sphere to its lower-left, one below by an approximately vertical red segment, one upper-right, and one lower-right.\n\nThe lower-left pale gray-green sphere is connected to red spheres upper-left, above through an approximately vertical segment, upper-right, lower-right, below, and lower-left.\n\nThe lower-right pale gray-green sphere is connected to red spheres upper-left, upper-right, right, lower-right, below, and lower-left.\n\nThe connectors are mainly gray or pale gray-green, with red portions near some red spheres. Some connectors are obscured by spheres. Overall, the spheres and connectors lie in front of the gray frame, which remains visible through gaps behind the structure. One red sphere at the upper right is paler and appears translucent or highlighted; the remaining red spheres are mostly darker red with light-dark shading."}], "full_mark": 1}]} +{"problem_id": "Exam-14-T4", "header": "Determination of Iron in Tin Concentrate by Cerium(IV) Sulfate Titration\n4.1 Preparation of standard titrant solutions\nIron standard solution: Weigh 2.0000 g of pure iron wire (99.95%) into a beaker, add 50 mL of hydrochloric acid (1+1), cover with a watch glass, and heat on a low-temperature electric hot plate until completely dissolved (do not boil). Remove, cool, transfer to a 1000 mL volumetric flask, dilute to the mark with water, and mix thoroughly.\nCerium(IV) sulfate standard titrant: Weigh 18.1 g of cerium(IV) sulfate [Ce(SO4)2·4H2O] into a beaker, add a small amount of water, and add 80 mL of concentrated sulfuric acid with constant stirring. After complete dissolution, cool, dilute to 1000 mL with water, and mix.\nStandardization: Pipet 30.00 mL of the iron standard solution into a 300 mL conical flask, add 5 mL of concentrated hydrochloric acid, and carry out reduction and standardization according to Section 4.3. Conduct a blank test in parallel.\n4.2 Alkali fusion and leaching of the test portion\nPlace a 0.3126 g portion of tin concentrate in a 10 mL high-alumina crucible, add 2 g of sodium peroxide, and mix carefully with a thin glass rod. Wipe the rod with a small piece of filter paper and place the paper in the crucible. Place the crucible in a furnace preheated to 500 °C, continue heating to 650 °C–700 °C, and fuse until the melt is red and transparent. Remove the crucible and cool.\nPlace the crucible in a 250 mL beaker and add 20 mL of water for leaching. After the vigorous reaction ceases, neutralize to an acidic solution with sulfuric acid (1+1), add 5 mL of concentrated hydrochloric acid, wash the crucible with water, and heat the solution to boiling on a low-temperature electric hot plate.\n4.3 Reduction and titration\nWhile the solution is hot, add hydrochloric-acidified tin(II) chloride solution (100 g/L) dropwise with swirling until the solution is pale yellow. Rinse down the beaker wall, add 40 mL of water, add 5 drops of sodium tungstate solution (50 g/L) as indicator, and add hydrochloric-acidified titanium(III) chloride solution (50 g/L) dropwise until a blue color appears, followed by two excess drops. Cool to room temperature in a cold-water bath.\nAdd 50 mL of copper sulfate-ammonium sulfate solution (containing 0.02 g of copper sulfate and 80 g of ammonium sulfate per liter). After the blue color disappears, allow the solution to stand for 10 min. Add 25 mL of sulfuric acid-phosphoric acid solution (sulfuric acid:phosphoric acid:water = 1:1:4), add 7 drops of sodium diphenylamine sulfonate solution (5 g/L), and titrate with the cerium(IV) sulfate standard titrant to a stable purple-red endpoint.", "header_images": [], "subquestions": [{"sub_id": "Exam-14-T4-1", "tags": ["4.1", "4.2"], "text_open": "4.3.2 Explain why preliminary reduction with tin(II) chloride solution is required instead of adding sodium tungstate solution directly and reducing with titanium(III) chloride solution.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-14-T4-2", "tags": ["4.1", "4.2"], "text_open": "4.3.3 Explain why reduction with tin(II) chloride and titanium(III) chloride must be performed in a hot solution, whereas the subsequent operations require cooling to room temperature.", "text_images": [], "full_mark": 2}]} +{"problem_id": "Exam-14-T7", "header": "Bioinorganic Chemistry. 7.1 O2 is in fact highly toxic to many organisms. Oxygen carriers bound to metalloproteins enable the transport and storage of $O_2$ in organisms.", "header_images": [], "subquestions": [{"sub_id": "Exam-14-T7-1", "tags": ["1.4", "1.6", "2.4"], "text_open": "7.1.1 Arthropods transport oxygen using hemocyanin. Hemocyanin is an oligomeric protein containing a pair of closely spaced Cu atoms. Colorless deoxyhemocyanin A [Cu(I)] is represented below. A binds one molecule of $\\mathrm{O_2}$ to form bright-blue protein B. Given that Cu is in the +2 oxidation state in B, write the SMILES string of B (for example, A may be represented as '[N][Cu]([N])[N].[N][Cu]([N])[N]').", "text_images": [{"file": "Exam-14-T7-1-1-prompt.png", "description": "Simplified representation of the dicopper(I) center in deoxyhemocyanin A. The diagram is labeled with a bold uppercase A at the bottom center and is left-right symmetric. A central $Cu$ atom in the left half is connected by three straight lines to three $N$ atoms at the upper-left, directly left, and lower-left positions. These three $N$ atoms are joined in sequence by two arcs curving outward to the left: the upper-left $N$ is connected to the directly-left $N$, and the directly-left $N$ is connected to the lower-left $N$. A second central $Cu$ atom in the right half is similarly connected by three straight lines to three $N$ atoms at the upper-right, directly right, and lower-right positions. These three $N$ atoms are joined in sequence by two arcs curving outward to the right: the upper-right $N$ is connected to the directly-right $N$, and the directly-right $N$ is connected to the lower-right $N$. A long downward-curving arc at the bottom directly connects the lower-left $N$ and lower-right $N$ atoms."}], "full_mark": 3}, {"sub_id": "Exam-14-T7-2", "tags": ["1.4", "1.8", "2.4"], "text_open": "7.1.2 Hemerythrin is a member of a special family of proteins containing a diiron center. In addition to coordination by amino-acid side chains and two bridging carboxylates, the two iron atoms are bridged by a common diatomic anion. Hemerythrin C is represented below (the diatomic bridging ligand is not shown), and C binds one molecule of $\\mathrm{O_2}$ to form D. Given that $\\mathrm{O_2}$ coordinates to only one Fe atom and that the diatomic bridging ligand provides some noncovalent stabilization to $\\mathrm{O_2}$, write the SMILES string of D.", "text_images": [{"file": "Exam-14-T7-2-1-prompt.png", "description": "Simplified representation of the diiron center in hemerythrin C. C: '[R]/C1=O/[Fe]([*])([*])O/C([*])=O\\[Fe]([*])([*])O1', where * denotes a site at which Fe coordinates to the imidazole group of His and may be omitted here."}], "full_mark": 4}, {"sub_id": "Exam-14-T7-3", "tags": ["1.8", "1.4"], "text_open": "7.1.3 CO is well known to strongly inhibit the binding of $\\mathrm{O_2}$ to myoglobin, so suppressing CO coordination is essential. In fact, the structure of myoglobin weakens the binding of CO to the protein. Using the simplified myoglobin structure above, explain why CO binds less strongly to myoglobin than $\\mathrm{O_2}$, considering the coordination geometries of CO and $\\mathrm{O_2}$ and their noncovalent interactions. (No more than 100 words.)", "text_images": [{"file": "Exam-14-T7-3-1-prompt.png", "description": "Simplified structure of the myoglobin coordination site, drawn with black lines on a white background.\n\nA five-membered ring near the top bears the labels HN at the lower left and N at the upper right, with single- and double-bond lines in and around the ring. A downward-curving line from the lower-right side of the ring extends toward the upper-right region of an elliptical structure below.\n\nAn approximately elliptical ring of curved lines occupies the center. Four N labels appear around it at the upper-left, upper-right, lower-left, and lower-right positions, and Fe is labeled at the center.\n\nA vertical line extends downward from Fe through the lower part of the ellipse to the label His below.\n\nThe Chinese caption at the bottom reads “Simplified myoglobin structure” in the English-localized version. SMILES: '[Fe][*].[N].[N].[N].[N].C1=CN=CN1', where * denotes the site at which Fe coordinates to the imidazole group of His and may be omitted here."}], "full_mark": 4}]} {"problem_id": "Exam-14-T9", "header": "Synthesis of Weinreb Amides\nSince their discovery, N-methoxy-N-methylamides have been widely known as Weinreb amides (WAs). WAs exhibit reactivity distinct from that of ordinary amides: they can serve both as excellent acylating reagents and as aldehyde equivalents and are important building blocks in numerous organic syntheses.", "header_images": [], "subquestions": [{"sub_id": "Exam-14-T9-1", "tags": ["2.2", "2.4", "2.5"], "text_open": "9.1 The most common method for preparing Weinreb amides is condensation of a carboxylic acid with N,O-dimethylhydroxylamine. Because amide condensation is thermodynamically unfavorable, an additional coupling reagent is usually required. The carboxylic acid below is to be condensed with N,O-dimethylhydroxylamine to prepare an N-protected amino-acid WA. 9.1.1 DCC is selected as the coupling reagent. Write the SMILES strings of one key intermediate and the coproduct (the product other than the Weinreb amide).", "text_images": [{"file": "Exam-14-T9-1-1-prompt.png", "description": "Carboxylic-acid substrate: 'O=C(N[C@@H]([R])C(O)=O)OCC1=CC=CC=C1'; N,O-dimethylhydroxylamine: 'CNOC'"}, {"file": "Exam-14-T9-1-2-prompt.png", "description": "DCC coupling reagent: 'C1(N=C=NC2CCCCC2)CCCCC1'."}], "full_mark": 3}, {"sub_id": "Exam-14-T9-2", "tags": ["2.4", "2.3", "1.2"], "text_open": "9.1.2 Experimentally, using DCC alone gives a product racemized at the stereogenic carbon because the intermediate in Question 9.1.1 is further converted into a cyclic intermediate that undergoes isomerization and causes racemization. Write the SMILES string of the cyclic intermediate and draw the isomerization mechanism.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-14-T9-3", "tags": ["2.1"], "text_open": "9.2 A trichloromethyl carbinol can undergo a homologation-amination reaction to form a WA containing one additional carbon, as shown below. 9.2.1 Under these conditions, $\\mathrm{(PhSe)_2}$, $\\mathrm{NaBH_4}$, and EtOH generate a complex anion X that contains an Se-B bond but no B-H bond and may be regarded as an equivalent of $\\mathrm{PhSe^-}$. Write the equation for formation of anion X.", "text_images": [{"file": "Exam-14-T9-3-1-prompt.png", "description": "'[R]C(C(Cl)(Cl)Cl)O' reacts with 'CNOC' under $\\mathrm{(PhSe)_2}$, $\\mathrm{NaBH_4}$, NaOH, EtOH, 55°C to form '[R]CC(N(C)OC)=O'"}], "full_mark": 2}, {"sub_id": "Exam-14-T9-4", "tags": ["2.2", "2.4"], "text_open": "9.2.2 Detection of the intermediate below was used to infer the reaction mechanism. The trichloromethyl carbinol substrate first converts under basic conditions into a small-ring intermediate and then into the intermediate shown below. Using the information in Question 9.2.1, write the SMILES strings of 3 key intermediates leading to the final product, excluding the intermediate shown.", "text_images": [{"file": "Exam-14-T9-4-1-prompt.png", "description": "Structure of the intermediate detected in the reaction. SMILES: '[R]C(C(N(C)OC)=O)[Se]C1=CC=CC=C1'"}], "full_mark": 6}, {"sub_id": "Exam-14-T9-5", "tags": ["2.2"], "text_open": "9.2.3 Can (PhSe)_2 be regenerated in this reaction? ( )\n(a) Yes; (b) No", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-14-T9-6", "tags": ["2.4", "2.8", "2.5"], "text_open": "9.2.5 The experiment also produces a byproduct whose molecular mass is approximately 30 lower than that of the major product. Replacing the methyl group bonded to O in N,O-dimethylhydroxylamine with another group can markedly suppress formation of this byproduct, although the reaction time increases accordingly. Using the relevant information, write the SMILES string of one possible modified substrate.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-14-T9-7", "tags": ["2.4"], "text_open": "9.2.6 This experiment must be performed under oxygen-free conditions. If oxygen is present, another byproduct structurally similar to the main product is detected. Given that it contains multiple nitrogen atoms, write its SMILES string.", "text_images": [], "full_mark": 2}]} {"problem_id": "Exam-14-T10", "header": "Visible-Light Photocatalytic Reactions of Sulfonium Salts\nOn 2023-5-3, researchers from Qingdao University of Science and Technology published a study of visible-light photocatalytic reactions of sulfonium salts. Let the photosensitizer be denoted PC; in the photocatalytic pathway, light excites PC to PC*, which then participates in the reaction.", "header_images": [{"file": "Exam-14-T10-header-1.png", "description": "'CC1=CC=C([S+]2CCCC2)C=C1.F[P-](F)(F)(F)(F)F' reacts with 'C=CC1=CC=CC=C1' under hv, Photocatalyst (mol%), Base (3.0 equiv), DMSO, N2, 24 h, r.t."}], "subquestions": [{"sub_id": "Exam-14-T10-1", "tags": ["2.4", "2.2"], "text_open": "Standard conditions: The sulfonium salt shown reacts with styrene using PC7 (10 mol%) as the photosensitizer, $\\mathrm{Na_2CO_3}$ (3.0 equiv) as the base, and anhydrous DMSO as the solvent under irradiation with a 455 nm blue LED at room temperature for 24 hours.\n10.2.2 Researchers added 1,1-diphenylethylene under the standard conditions to trap a reaction intermediate, giving a product containing 3 rings. Write the SMILES string of this 3-ring product.", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-14-T10-2", "tags": ["2.4", "2.8"], "text_open": "10.2.3 Performing the reaction under the standard conditions but without DMSO gives a different product. Write its SMILES string.", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-14-T10-3", "tags": ["2.2", "2.3", "2.4"], "text_open": "10.3 Write the SMILES strings of the key intermediates and product of this reaction.", "text_images": [], "full_mark": 5}, {"sub_id": "Exam-14-T10-4", "tags": ["2.4", "2.5"], "text_open": "10.4 Treatment of the product from 10.3 with mCPBA (meta-chloroperoxybenzoic acid) gives two additional products. Write their SMILES strings.", "text_images": [], "full_mark": 2}]} -{"problem_id": "Exam-15-T3", "header": "Problem 3 Solid Electrolyte Interphase\nThe solid electrolyte interphase (SEI) is an important research area in modern electrochemistry. During the first charging of a lithium-ion battery, the electrode potential lies below the reduction potential of the electrolyte, so additives and lithium salts in the electrolyte undergo electrochemical reductive decomposition at the solid-liquid interface. Insoluble solid products deposit on the electrode surface and form a passivation film that is electronically insulating but ionically conducting, thereby slowing electrode corrosion. Vinylene carbonate (VC), a classic film-forming additive, reacts preferentially and promotes formation of a high-quality SEI.", "header_images": [{"file": "Exam-15-T3-header-1.png", "description": "The structures in the figure have the following SMILES: VC, 'O=C1OC=CO1'; EC, 'O=C2OCCO2'."}], "subquestions": [{"sub_id": "Exam-15-T3-1", "tags": ["2.2", "2.3", "2.4"], "text_open": "3.3 Studies show that VC preferentially undergoes cleavage of a C-O single bond and is reduced to radical anion A carrying 1 unit of negative charge. It then undergoes radical polymerization to form several polymeric products. 3.3.1 In one pathway, A acts as a chain initiator and reacts with VC to form acyclic polymer B. Write the SMILES of B.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-15-T3-2", "tags": ["2.1", "2.2", "3.6"], "text_open": "3.3.4 In the electrolyte, VC can also be electrochemically reduced to lithium oxalate and an enolate. Write the electrochemical reaction equation; EC participates in the reaction.", "text_images": [], "full_mark": 2}]} -{"problem_id": "Exam-15-T5", "header": "Problem 5 Versatile Rutile. Rutile ($TiO_2$) is a classic structure in the tetragonal crystal system. From a close-packing and interstitial-site perspective, oxygen atoms approximately form hexagonal close packing, and titanium atoms occupy half of the octahedral voids in an ordered manner. The $TiO_6$ octahedra are connected by shared vertices and shared edges. A rutile unit cell and its projection along the c axis are shown below.", "header_images": [{"file": "Exam-15-T5-header-1.png", "description": "The image is a three-dimensional ball-and-stick model of $TiO_2$ on a white background. All spheres have bright highlights and a smooth, glossy three-dimensional appearance. Thin black-gray lines outline a transparent rectangular unit-cell box in perspective. The 9 larger light-blue spheres represent Ti atoms, located at the 8 vertices and the exact center of the box. The 6 smaller bright-red spheres represent O atoms, all inside the cell around the central Ti. Atoms are connected by short two-color cylinders, light blue near Ti and red near O. The central Ti is connected to all 6 red spheres, forming an approximately octahedral $TiO_6$ environment. The left red sphere connects the central Ti and the upper-left-front and lower-left-front vertex Ti atoms; the right red sphere connects the central Ti and the upper-right-back and lower-right-back vertex Ti atoms. The other four red spheres cross around the center: the upper-left-back red sphere connects the upper-left-back vertex Ti, the upper-right-front red sphere connects the upper-right-front vertex Ti, the lower-left-back red sphere connects the lower-left-back vertex Ti, and the lower-right-front red sphere connects the lower-right-front vertex Ti. Natural perspective occlusion creates clear depth. There is no visible text, number, coordinate axis, arrow, legend, unit, or title."}, {"file": "Exam-15-T5-header-2.png", "description": "The image should be reconstructed as a flat two-dimensional vector diagram approximately 365×370 pixels, slightly taller than wide, on a pure-white borderless canvas. Taking the upper-left corner as origin (0,0), x to the right and y downward, first draw an unfilled thin solid gray square approximately 200 pixels per side, aligned horizontally and vertically, with corners near (85,84), (284,84), (284,284), and (85,284). Its line is about 1 pixel wide with slight antialiasing. No spheres or marks sit at the square corners; line midsections remain clear in white gaps, while corners are partly hidden by blue shapes. Place 5 nearly equal slanted rhombic parallelograms, each about 102 pixels per side with acute angles about 65° and obtuse angles about 115°. The 4 outer parallelograms at upper left, upper right, lower left, and lower right share an orientation elongated lower-left to upper-right, with uniform medium-light sky-blue fill, approximately RGB 117,182,224 or #75B6E0. Clockwise, their vertices are [(46,45),(146,23),(124,123),(24,145)], [(245,45),(345,23),(323,123),(223,145)], [(46,245),(146,223),(124,323),(24,345)], and [(245,245),(345,223),(323,323),(223,345)]. They extend well outside the gray square and approach the canvas edges but remain fully visible. The 5th parallelogram lies centrally, rotated approximately 90° relative to the outer 4 and elongated upper-left to lower-right, with clockwise vertices [(124,123),(223,145),(245,245),(146,223)]. Its fill is distinctly lighter, approximately RGB 144,196,230 or #90C4E6. All 5 have flat, texture-free, nearly opaque blue faces outlined by a very thin, approximately 1-pixel light-gray antialiased edge. A faint short gray diagonal divides each face visually into two triangular regions. For the 4 outer shapes the diagonals connect (46,45) to (124,123), (245,45) to (323,123), (46,245) to (124,323), and (245,245) to (323,323); for the center shape the diagonal connects (223,145) to (146,223). The 4 center-shape vertices are shared respectively with the nearest center vertices of the 4 outer shapes: (124,123), (223,145), (146,223), and (245,245). Except for these point contacts, the blue shapes do not overlap substantially. Draw 16 equal red three-dimensional spheres, diameter approximately 24–26 pixels, at all unique parallelogram vertices, centered at (146,23), (345,23), (46,45), (245,45), (124,123), (323,123), (24,145), (223,145), (146,223), (345,223), (46,245), (245,245), (124,323), (323,323), (24,345), and (223,345). The spheres lie above the blue faces and gray lines, hiding line ends. Each has a thin dark-red rim, a bright-red to deep-wine-red surface gradient, and a soft white or pale-pink highlight slightly upper-left of center, giving a smooth glass-or-plastic gloss without a cast shadow. The composition is approximately 180° rotationally symmetric. Outermost spheres leave only about 7–12 pixels of white margin. No text, number, letter, legend, coordinate axis, tick, arrow, symbol, watermark, background texture, or other decoration is present."}], "subquestions": [{"sub_id": "Exam-15-T5-1", "tags": ["1.1", "1.3"], "text_open": "5.2 α-$MnO_2$. α-$MnO_2$ has a rutile-derived structure. Rutile can be viewed as one-dimensional chains formed by edge-sharing $TiO_2$ octahedra, with chains of two different spatial orientations alternating and sharing vertices. α-$MnO_2$ can be viewed as two one-dimensional $MnO_2$ octahedral chains joined by shared edges to form a \"thick one-dimensional chain\"; thick chains of two different spatial orientations alternate and share vertices. The c-axis projection of the polyhedral connectivity is shown. Some thick chains of the same orientation cannot be translated into one another by vectors parallel to the ab plane because they occur at different c-axis heights. The lattice type of α-$MnO_2$ is body-centered tetragonal. 5.2.2 Which symmetry elements does α-$MnO_2$ possess? (a) 4-fold rotation axis; (b) 4-fold rotoinversion axis; (c) $4_1$ screw axis; (d) $4_2$ screw axis.", "text_images": [{"file": "Exam-15-T5-1-1-prompt.png", "description": "The image is a 427×424 pixel two-dimensional diagram on a pure-white background, showing a regular periodic network of purple polygons and red dots, with no axes, text, arrows, or labels. Approximately 72 red nodes are small spheres with a slight three-dimensional red gradient, brighter at the center and darker at the edge. They occupy vertices and connection points of the purple structure. The pattern repeats by approximately 192 pixels both horizontally and vertically, forming a 2×2 arrangement of four identical periodic units. The purple portions comprise approximately 16 large, separated structural units filled with light-to-deeper purple gradients, sometimes darker centrally and lighter at the edges, creating a translucent or three-dimensional effect. Red nodes serve as their vertices or junctions, forming combinations of rhombi, parallelograms, triangles, and irregular polygons. Two principal orientations occur: units extending upper-left to lower-right with roughly horizontal rhombic/parallelogram outlines, and units extending lower-left to upper-right with roughly vertical rhombic/triangular outlines. They alternate and connect through red nodes. In the upper-left region, nodes near (13,84), (51,58), (87,27), (125,51), (87,76), and (52,107) form a purple zigzag polygon extending upper-right. A translated neighboring group connects near (156,85), (182,124), and (205,84). The upper-right is the upper-left translated about 192 pixels right; the lower-left is translated about 192 pixels down; and the lower-right is translated both right and down. Typical red-node spacing is about 30–40 pixels. Connections are implied by polygon edges and vertices without black or other lines. Some nodes lie at junctions of multiple polygons, giving a periodic two-dimensional lattice or repeating coordination-framework appearance. The whole pattern is regular and approximately symmetric horizontally and vertically. Outermost nodes approach but do not touch the edges: top near (148,11) and (341,11), bottom near (85,412) and (278,411), left near (13,84) and (13,276), and right near (397,84) and (413,339). The purple-red structure occupies most of the white canvas and has obvious periodicity."}], "full_mark": 3}, {"sub_id": "Exam-15-T5-2", "tags": ["1.3"], "text_open": "5.2.3 How many neighboring $MnO_6$ octahedra share an edge with each $MnO_6$ octahedron?", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-15-T5-3", "tags": ["1.3"], "text_open": "5.3 Ca-Ti-O crystal. The figure shows the unit cell of a Ca-Ti-O crystal. Small spheres are oxygen, medium spheres titanium, and large spheres calcium. Ti and Ca each have only one chemical environment, while O has three. 5.3.1 Write the chemical formula of the crystal.", "text_images": [{"file": "Exam-15-T5-3-1-prompt.png", "description": "One unit cell contains 28 small red spheres: 8 on edges, 16 on faces, and 6 inside the cell. It contains 12 medium blue spheres: 4 on edges and 8 on faces. It contains 6 large white spheres: 2 on edges and 4 on faces."}], "full_mark": 2}]} +{"problem_id": "Exam-15-T1", "header": "Problem 1 The Immortal Mirror. In 1863, two chemists at the Freiberg Mining Academy in Germany unexpectedly observed a previously unknown spectral line while studying locally mined sphalerite. They assigned it to an undiscovered element M and successfully isolated elemental M from the mineral. At the subsequent Paris Exposition, elemental M was publicly displayed as \"the rarest metal,\" causing a sensation. At that time, M was used to make highly reflective mirrors; today it is more commonly an important semiconductor raw material used in conductive films.", "header_images": [], "subquestions": [{"sub_id": "Exam-15-T1-1", "tags": ["2.1"], "text_open": "1.1 Burning M in air gives a pale-yellow amorphous solid A. Heating A under hydrogen at 400~500°C reduces it to a black crystalline solid B (Reaction 1); dissolving B in dilute hydrochloric acid releases a colorless gas C. Above approximately 700°C, B decomposes to A and elemental M (Reaction 2). At high temperature, A reacts with ammonia to form a yellow semiconductor material D and releases a gas (Reaction 3).\n1.1.2 Write the chemical equations for Reactions 1~3.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-15-T1-2", "tags": ["1.4"], "text_open": "1.2 Chloride E of M is a white flaky solid that is extremely hygroscopic and dissolves in water with substantial heat release. In ether solution, E reacts with LiH to form complex F. F decomposes above 0°C and can be used as an in situ reducing agent in organic synthesis. E forms complex E·3THF with the electron-donor ligand THF. Single-crystal X-ray diffraction shows that the M center in this complex is six-coordinate with octahedral geometry. E has a layered crystal structure consisting of close-packed chloride ions with layers of octahedrally coordinated M centers. By contrast, chloride G of the congener whose period differs by only 1 commonly crystallizes as dimer H. 1.2.1 Write the chemical formulas of E~G, write the SMILES of H, and describe its three-dimensional structure in words.", "text_images": [], "full_mark": 5}, {"sub_id": "Exam-15-T1-3", "tags": ["1.5", "1.6"], "text_open": "1.3 In $E\\cdot3THF$, the $M$–Cl bond length is approximately 242 pm and the $M$–O bond length is approximately 225 pm. The ionic radius of $M$ is 80 pm, the ionic radius of $Cl^-$ is 181 pm, and the van der Waals radius of O in THF is approximately 152 pm.\n1.3.2 Use the bond-length data to compare the bonding character of the M-Cl and M-O bonds, including their relative ionic/covalent character, and explain.", "text_images": [], "full_mark": 3}]} +{"problem_id": "Exam-15-T4", "header": "Problem 4 Determination of Chloride by the Volhard Method. ① Pipette 20.00 mL of 0.1 mol/L standard silver nitrate titrant into a 250 mL conical flask. Add 30 mL water, 5 mL nitric acid solution, and 2 mL saturated ferric ammonium sulfate solution. While shaking, titrate with standard potassium thiocyanate titrant until a pale brown-red color appears and persists for 1 min. Record the volume of standard potassium thiocyanate consumed: V_{1} = 19.88 mL. ② Weigh 0.10 g sodium chloride that has been ignited at 500 ℃ ~ 600 ℃ to constant mass, accurate to 0.1 mg. Dissolve it in approximately 40 mL water in a beaker and transfer to a 100 mL volumetric flask. Add 5 mL nitric acid solution (1+3), then add 25.00 mL of 0.1 mol/L standard silver nitrate solution while shaking vigorously. Dilute to the mark with water and mix. Let stand in the dark for 5 min, filter through rapid filter paper, and discard the first 10 mL of filtrate. Accurately pipette 50.00 mL filtrate into a 250 mL conical flask, add 2 mL of 1 mol/L ferric ammonium sulfate solution, and titrate with standard potassium thiocyanate while shaking until a pale brown-red color appears and persists for 1 min. Record V_{2} = 4.01 mL. ③ Measure 50 mL test solution into a 100 mL colorimetric tube and otherwise follow the same procedure as in ②. Record V_{3} = 2.98 mL.", "header_images": [], "subquestions": [{"sub_id": "Exam-15-T4-1", "tags": ["4.5", "2.1"], "text_open": "4.1 Calculate the concentrations of the standard silver nitrate and potassium thiocyanate titrant solutions.", "text_images": [], "full_mark": 5}, {"sub_id": "Exam-15-T4-2", "tags": ["4.1", "4.7"], "text_open": "4.4.2 Explain why the first 10 mL of filtrate is discarded.", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-15-T4-3", "tags": ["4.2", "4.5"], "text_open": "4.5 If addition of standard silver nitrate produces a colloidal solution, what additional operation is required before filtration?", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-15-T4-4", "tags": ["4.5", "2.1", "4.7"], "text_open": "4.6 Calculate the $Cl^-$ concentration in the sample.", "text_images": [], "full_mark": 3}]} {"problem_id": "Exam-15-T6", "header": "Problem 6 Precipitation-Dissolution Equilibria. Constants at 25 ℃ may be used. Report all calculated results to 3 significant figures and solubilities in mol/L: dissociation constants of $H_2S$, Ka_1 = 1.3×10^{-7}, Ka_2 = 7.1×10^{-15}; Ksp(MnS)=3.2×10^{-11}; Ksp($Mn(OH)_2$)=1.6×10^{-13}.", "header_images": [], "subquestions": [{"sub_id": "Exam-15-T6-1", "tags": ["3.4"], "text_open": "6.3 Accounting for the pH change, calculate the solubility of MnS in pure water.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-15-T6-2", "tags": ["3.4"], "text_open": "6.4 Using the result from 6.3, demonstrate by calculation that $Mn(OH)_2$ precipitate actually forms in the system. Then calculate the solubility of MnS in pure water. Given Ksp($Mn(OH)_2$)=1.6×10^{-13}.", "text_images": [], "full_mark": 6}]} {"problem_id": "Exam-15-T7", "header": "Problem 7 Low-Valent Compounds of the Aluminum-Group Metals. Group 13 metals, the aluminum group (Al, Ga, In, Tl), can form chain, ring, and cluster compounds characterized by covalent metal-metal bonds and oxidation states below +III. Beginning with $E_2R_4$ compounds, this problem examines their bonding, stability, and reactivity. Chain and ring compounds have an E-E σ single bond (2c-2e) as their framework. After forming three covalent bonds, a metal has 6 valence-shell electrons and is Lewis acidic; the most common form is the $E_2R_4$ unit.", "header_images": [], "subquestions": [{"sub_id": "Exam-15-T7-1", "tags": ["1.4", "1.5"], "text_open": "7.1 Based on the electronic properties of the $E_2R_4$ unit, which of the following can stabilize the $E_2R_4$ unit? A: R is a strongly electron-withdrawing group. B: R is a sterically bulky group. C: ligand L is added to form $R_2(L)E{-}E(L)R_2$.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-15-T7-2", "tags": ["2.1", "2.2", "1.4"], "text_open": "7.2 \"Dihalides\" $E_2X_4$. Among the aluminum-group metals, Ga most readily forms $E_2X_4$, commonly as adduct $X_2(L)E{-}E(L)X_2$. If ligand L is also a halide anion, monoanionic or dianionic complexes form. 7.2.1 In one synthesis, Ga and $I_2$ react in toluene in a 2:1 molar ratio. After cooling to -78°C, 4 eq $NCy_2H$ is added. The mixture reacts for 12 h at room temperature, precipitate is removed by filtration, and the solution is concentrated and cooled to give colorless crystal A in 95% yield. Write the equations for the two steps forming A.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-15-T7-3", "tags": ["1.4", "4.4", "2.4"], "text_open": "7.2.2 Replacing the ligand with the bulky N-heterocyclic carbene IPr and changing the feed ratio to n(Ga) : n($I_2$) : n(IPr) = 2 : 1 : 2 gives ionic compound B. The Ga mass fraction in B is 8.98%, and the principal MS FAB fragment peaks are m/z 389 (100%, $[IPrH]^+$) and 909 (61%, $[Ga_2I_3(IPr)]^+$). Write the SMILES structure of B.", "text_images": [{"file": "Exam-15-T7-3-1-prompt.png", "description": "The IPr ligand in the prompt has SMILES 'CC(C1=C([N+]2=[C-]N(C=C2)C3=C(C=CC=C3C(C)C)C(C)C)C(C(C)C)=CC=C1)C'."}], "full_mark": 2}, {"sub_id": "Exam-15-T7-4", "tags": ["1.4", "2.4"], "text_open": "7.2.3 Passing $NH_3$ into a THF solution of GaBr gives a precipitate. Extraction with DMF and crystallization give crystal C, $C_48H_112Br_18Ga_8N_16O_16$. C is an ionic compound with an anion:cation ratio of 3 : 2, and its anion contains a $Ga_2$ core. Write a chemical formula representing the structure of C.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-15-T7-5", "tags": ["1.4", "2.4"], "text_open": "7.3 $E_2R_4$, where R is alkyl or amino. $E_2R_4$ compounds are usually prepared by reaction of dihalide $E_2X_4L_2$ or subhalide EX with alkyl or amino anions. 7.3.1 Monodentate ligand, silyl: $Ga^{I}[Ga^{III}Cl_4]$ reacts with $Li(THF)_3Si(SiMe_3)_3$ to give D, E, or F depending on feed ratio. n($Ga^{I}[Ga^{III}Cl_4]$) : n($Li(THF)_3Si(SiMe_3)_3$) | product | MS (m/z) | Ga oxidation state | Ga coordination number | Ga chemical environment | Cl element | ideal symmetry. 1 : 1 | tetranuclear D | 1053 (4), 667 (11), 316 (80), 73 (100) | +2 | 4 | only one environment | contains Cl and all Cl environments are identical | $D_{2d}$. 1 : 2 | dinuclear E | 563 (10), 73 (100) | +2 | 3 | | no Cl | $D_{2d}$. 1 : 3 | tetranuclear F | | +1 | 4 | | no Cl | $T_{d}$. Write the SMILES structures of D, E, and F.", "text_images": [], "full_mark": 6}, {"sub_id": "Exam-15-T7-6", "tags": ["1.4", "1.6", "4.3"], "text_open": "7.3.2 Bidentate ligand: the following dpp-bian ligand can coordinate as a monoanion or dianion. Its tripotassium salt $K_3[dpp-bian]$ reacts 1 : 1 with $GaCl_3$ in a reduction-coordination process to give dinuclear H_1. Its zinc salt $[(dpp-bian)ZnI]_2$ reacts with $GaCl_3$ and $K_4[dpp-bian]$ to give dinuclear H_2. Write SMILES structures reflecting their electronic properties, and explain why H_2 absorbs in the near-infrared at 950 nm whereas H_1 does not.", "text_images": [{"file": "Exam-15-T7-6-1-prompt.png", "description": "At far left, the figure shows ligand '[Ar]/N=C(/C2=N/[Ar])C3=CC=CC4=C3C2=CC=C4', labeled \"dpp-bian\", where Ar is 'CC(C)C1=CC=CC(C(C)C)=C1*', and two reaction pathways. In the upper path, `dpp-bian` reacts with `K (ex.)` in `THF` to form $K_4[dpp-bian]$, then with `+0.33 eq dpp-bian` to form $K_3[dpp-bian]$, and then under +$GaCl_3$ and `-3 KCl` gives H1, `[(dpp-bian)Ga-Ga(dpp-bian)]`. In the lower path, `dpp-bian` reacts with Zn (ex.), $I_2$ (0.5 eq), THF to form $[(dpp-bian)ZnI]_2$, then successively with $GaCl_3$ (2 eq) and $K_4[dpp-bian]$ (2 eq) to form H_2, `[(dpp-bian)Zn-Ga(dpp-bian)]`."}], "full_mark": 4}, {"sub_id": "Exam-15-T7-7", "tags": ["1.4", "2.4"], "text_open": "7.3.3 Tridentate ligand: in a suitable tridentate cage-ligand environment, the sixth-period element thallium (Tl) can rarely form a Tl-Tl metal bond. Tridentate ligand $H_3J$ = $(t{-}BuHN)_3SiMe$ can chelate one metal ion between every pair of nitrogen atoms and therefore bridge up to three metal atoms, forming diverse ring or cage structures. Reaction of $H_3J$ with excess n-BuLi in toluene gives dimeric lithium compound $(Li_3J)_2$, in which all Li atoms have identical environments and are three-coordinate to N. Reaction of $(Li_3J)_2$ with excess TlCl in dioxane gives dimeric thallium cluster $(Tl_3J)_2$, in which each Tl is two-coordinate to N and the cluster contains one relatively strong Tl-Tl bond. Write the SMILES structures of $(Li_3J)_2$ and $(Tl_3J)_2$.", "text_images": [], "full_mark": 4}]} {"problem_id": "Exam-16-T1", "header": "Problem 1: What Is That Smell?\nIn 1803, the British chemist Tennant dissolved crude platinum in dilute aqua regia and found a small amount of black, insoluble powder remaining at the bottom of the vessel. Although early researchers regarded it merely as residual graphite, he suspected that it might contain unknown substances awaiting discovery. Through systematic alternating acid and base treatments, Tennant successfully separated the black residue into two new metallic elements, one of which is metal M. M was named for its distinctive odor and also has an extremely high density and numerous oxidation states.", "header_images": [], "subquestions": [{"sub_id": "Exam-16-T1-1", "tags": ["1.4", "1.7", "2.4"], "text_open": "1.1 M forms several oxides, the best known of which is A. A can be prepared by heating powdered M in a stream of oxygen (Reaction 1). It is a pale-yellow crystal whose melting point is only slightly above body temperature and whose boiling point is only about 400 K. It is appreciably volatile even at room temperature and converts alkenes stereoselectively into cis vicinal diols through a concerted cycloaddition that forms the five-membered cyclic intermediate B. When heated with concentrated hydrochloric acid, A is reduced to complex C of M while gas D is released (Reaction 2). The anion of C has a regular octahedral geometry. M also forms complex ion E with 2,2′-bipyridine (bpy, C_{10}H_{8}N_{2}, molar mass 156.18 g·mol^{-1}). This complex has D_{3} symmetry and exists as Λ/Δ enantiomers; the oxidation state of its central ion is the same as that in C. The mass fraction of M in the anion of C is 47.21%. 1.1.1 Deduce and give the chemical formulas of B-E (give a generic structure for B).", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-16-T1-2", "tags": ["2.3", "2.7", "1.2"], "text_open": "1.3 The dihydroxylation of an alkene by A follows second-order kinetics and proceeds through a concerted [3+2] cycloaddition to form a five-membered cyclic intermediate. 1.3.1 Give the coordination geometry and oxidation state of M in the intermediate. Explain why the product has the cis rather than the trans configuration.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-16-T1-3", "tags": ["4.3", "3.4", "1.4"], "text_open": "1.4 Complex E exhibits a strong MLCT absorption band in the visible region and is deep red when dissolved in acetonitrile. To determine the coordination ratio of M to bpy in E, a student designed a Job plot (method of continuous variations). The total concentration of M and bpy was kept constant at 1.00 × 10^{-4} mol·L^{-1}, their mole ratio was varied, and the absorbance of each solution in acetonitrile was measured at the characteristic wavelength \\lambda_{max} = 480 nm. At this wavelength, neither free M ions nor bpy absorb, although complexes of M with bpy may absorb. A 1.00 cm cuvette was used. $$\\begin{array}{c|ccccccccc}\\text{Sample}&1&2&3&4&5&6&7&8&9\\\\ \\hline \\mathbf{M}\\text{ mole fraction }x(\\mathbf{M})&0.10&0.20&0.30&0.40&0.50&0.60&0.70&0.80&0.90\\\\ \\text{Absorbance }A&0.128&0.248&0.352&0.440&0.488&0.440&0.352&0.248&0.128\\end{array}$$ 1.4.1 Determine the coordination ratio n = [bpy] / [M] in complex ion E under these experimental conditions, show the reasoning, and explain why this coordination ratio is observed in the experiment.", "text_images": [], "full_mark": 3}]} {"problem_id": "Exam-16-T2", "header": "Problem 2: Thermogravimetric Analysis of a Complex Double Salt\nOn April 28, 2023, researchers from the Institute of General and Inorganic Chemistry published a thermogravimetric study of a double salt. A solution of 2 mL of 85% phosphoric acid diluted with 5 mL of deionized water was added in small portions to a suspension prepared beforehand from 320 mg of calcium carbonate and 5 mL of deionized water. During continuous stirring at room temperature, foaming caused by the release of carbon dioxide was first observed, after which the solution became clear. Another solution containing 560 mg of K_{2}HPO_{4} in 5 mL of deionized water was then added. The resulting solution remained completely clear, and product A was precipitated by adding 15 mL of acetone in five 3 mL portions while stirring continuously. The precipitate was filtered, washed twice with 2.5 mL portions of acetone, and dried under ambient conditions to give A. Thermogravimetric analysis of A gave the result shown below. A is stable at 95°C and completes thermal decomposition at 530 °C. The decomposition occurs in the ranges 95-143°C, 143-240°C, and 240-530°C, each accompanied by a corresponding mass loss. The theoretical mass losses for the three steps are 6.645%, 4.984%, and 6.645%, respectively.", "header_images": [], "subquestions": [{"sub_id": "Exam-16-T2-1", "tags": ["4.6", "4.7", "2.1"], "text_open": "2.1 Using the preparation and thermogravimetric experiments for A, determine the chemical formula of A by appropriate qualitative reasoning and quantitative calculation.", "text_images": [{"file": "Exam-16-T2-1-1-prompt.png", "description": "In this thermogravimetric-differential thermal analysis (TG-DTA) plot, the left vertical axis is mass percentage ($\\text{Mass / \\%}$), ranging from 80 to 100, and corresponds to the dark-red TG curve. As temperature increases, this curve displays several stepwise mass-loss stages and finally levels off at approximately $82\\%$. The right vertical axis is heat flow ($\\text{Heat flow / }\\mu\\text{V}$), ranging from -4 to 2; a downward arrow labeled \"$\\text{endo}$\" indicates that downward deflection is endothermic. The red DTA curve shows four prominent changes with increasing temperature: a sharp, strong downward endothermic peak at 120, a second distinct downward endothermic peak at 186 after recovery, a narrower third downward endothermic peak at 301 after another recovery, and then an upward rise to a broad local exothermic peak at 458, after which the curve trends downward overall."}], "full_mark": 6}, {"sub_id": "Exam-16-T2-2", "tags": ["2.1"], "text_open": "2.2 Write the chemical equations for the three mass-loss steps, expressing A and each intermediate product with formulas containing 2 Ca atoms.", "text_images": [], "full_mark": 3}]} {"problem_id": "Exam-16-T4", "header": "Problem 4 Determination of Metal Content by Disodium Ethylenediaminetetraacetate Titration. First prepare the standard solutions. Standard zinc titrant (0.0100 mol/L): Weigh 0.6538 g of pure zinc (99.99%) into a 300 mL conical flask fitted with a funnel. Add 15 mL hydrochloric acid. After the vigorous reaction has ceased, heat gently until the zinc has dissolved completely. Cool to room temperature, transfer to a 1000 mL volumetric flask, dilute to the mark with water, and mix. Standard disodium ethylenediaminetetraacetate titrant (0.01 mol/L): Weigh 3.7225 g of disodium ethylenediaminetetraacetate, dissolve it in 400 mL water, transfer to a 1000 mL volumetric flask, dilute to the mark with water, and mix. Standardization: Pipette 25.00 mL of the standard disodium ethylenediaminetetraacetate titrant into a 250 mL conical flask. Add 75 mL water, 10 mL ammoniacal buffer solution (pH = 10), and 3 drops of Eriochrome Black T solution (weigh 0.2 g Eriochrome Black T into a 50 mL beaker, add 15 mL triethanolamine and 5 mL ethanol (95%), and mix). Titrate with the standard zinc titrant to a red endpoint. Three parallel titrations consume 24.85 mL of the standard zinc titrant.", "header_images": [], "subquestions": [{"sub_id": "Exam-16-T4-1", "tags": ["4.1", "4.2"], "text_open": "4.2 Preparation and Pretreatment of the Sample Solution\nPlace the chromium-containing alloy sample (0.9985 g) in a 500 mL beaker, cover with a watch glass, add 10 mL of hydrochloric acid (1+1), and heat to decompose most of the sample. Add 40 mL of concentrated perchloric acid, heat until white fumes appear, and continue heating until the solution volume is 5 mL~10 mL. Remove from heat and cool. Add approximately 100 mL of water and 1 g of ammonium chloride. Neutralize with concentrated ammonia until a precipitate forms, then add hydrochloric acid (1+1) until the precipitate just dissolves, followed by an excess of 5~10 drops. Add hexamethylenetetramine solution (250 g/L) until a precipitate forms and boil. Keep at 80°C~90°C for 15 min~20 min, filter through quantitative filter paper, and wash the precipitate and filter paper with warm hexamethylenetetramine solution (5 g/L). Transfer the precipitate together with the filter paper back to the original beaker. Add 10 mL of concentrated nitric acid to dissolve the precipitate and destroy the filter paper, then add 20 mL of concentrated perchloric acid and heat until white fumes appear. Add concentrated hydrochloric acid in several portions to volatilize the remaining chromium, then continue fuming for approximately 10 min. Remove from heat and cool. 4.2.1 Explain why precipitation is performed with hexamethylenetetramine under heated conditions instead of directly with aqueous ammonia (no more than 100 words).", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-16-T4-2", "tags": ["4.1", "4.2"], "text_open": "4.2.2 Explain why the precipitate is washed with warm hexamethylenetetramine solution (5 g/L) rather than a cold solution (no more than 100 words).", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-16-T4-3", "tags": ["2.1", "4.5"], "text_open": "4.3 Titration of Iron Ions. Add approximately 100 mL of warm water to dissolve the salts, then add 2 mL of sulfosalicylic acid solution (50 g/L). Adjust the solution to pH 1.0–pH 2.2 with hydrochloric acid (1+2) or concentrated ammonia. Titrate with the standard disodium ethylenediaminetetraacetate titrant until the solution changes from red-violet to colorless or pale yellow; 21.64 mL of the standard disodium ethylenediaminetetraacetate titrant is consumed. 4.4 Titration of Aluminum Ions. Continue by adding an appropriate amount of the standard disodium ethylenediaminetetraacetate titrant. Add 20 mL of acetic acid-ammonium acetate buffer solution (pH = 5.5), heat to boiling for 2.5 min~3 min, and add 2~4 drops of xylenol orange solution (1 g/L) as the indicator. Titrate with the standard zinc titrant to a red endpoint (do not calculate the volume). Add 20 mL of ammonium fluoride solution (40 g/L) and boil for 2 min. Remove from heat and cool to room temperature. Add one further drop of xylenol orange solution as the indicator, then titrate with the standard zinc titrant to a red endpoint; 9.24 mL of the standard zinc titrant is consumed.\n4.4.2 Write the reaction equation for the step involving addition of ammonium fluoride (use Y^{4-} for the ethylenediaminetetraacetate tetraanion and HAc for acetic acid).", "text_images": [], "full_mark": 2}]} -{"problem_id": "Exam-16-T5", "header": "Problem 5: Fullerenes. In the 80s of the 20th century, the British astrophysicist Kroto and co-workers discovered a new class of carbon allotropes whose molecular shapes are polyhedra composed of pentagons and hexagons. The most representative example is C_{60}. Its structure resembles the dome of the United States pavilion at the World Exposition in Montreal, Canada, designed by the American architect Buckminster Fuller; these molecules (not only C_{60}) are therefore called fullerenes.", "header_images": [], "subquestions": [{"sub_id": "Exam-16-T5-1", "tags": ["1.3"], "text_open": "5.1 The C_{60} molecule is a 32-faced polyhedron composed of 12 regular pentagonal faces and 20 regular hexagonal faces, and it has high symmetry. Its structure is shown below. 5.1.4 Above 30℃, C_{60} crystals can precipitate from a benzene solution. Their density is 1.65 g/cm^{3}, and the C_{60} molecules adopt cubic close packing in the crystal. Use these data to estimate the van der Waals radius of the C_{60} molecule.", "text_images": [{"file": "Exam-16-T5-1-1-prompt.png", "description": "Structural diagram of C_60: a spherical cage fullerene composed of 12 pentagons and 20 hexagons, with vertices representing carbon atoms and edges representing C-C bonds."}], "full_mark": 4}, {"sub_id": "Exam-16-T5-2", "tags": ["1.1", "1.3", "1.5"], "text_open": "5.3 Smaller fullerenes. In a fullerene molecule, every carbon atom is bonded to three neighboring carbon atoms. From the Lewis-structure perspective, resonance structures can therefore be drawn in which each carbon atom participates in two single bonds and one double bond. In this sense, the smallest possible fullerene could be tetrahedral C_{4} (although it has not yet been synthesized), whereas icosahedral C_{12} is not a fullerene. In this problem (5.3), you need not consider the isolated-pentagon rule. 5.3.1 Which of the following molecule(s) could be fullerenes?\n(a) Octahedral C_6\n(b) Cubic C_8\n(c) C_{12} with the structure of the “truncated cube 3^84^6,” a polyhedron composed of regular triangles and squares whose vertices are the 12 edge midpoints of a cube\n(d) Dodecahedral C_{20}", "text_images": [{"file": "Exam-16-T5-2-1-prompt.png", "description": "The image shows three identical geometric wireframe diagrams side by side, labeled (a), (b), and (c) from left to right. Each diagram consists of a large outer equilateral triangle and four circular nodes: three nodes at the vertices of the large triangle and one node at its center. Three internal lines connect the central node to the three vertices, and three outer edges connect the vertices. In each diagram, two selected connections are drawn as double lines by adding a parallel thin line. In (a), the highlighted connections are the vertical internal line from the center to the top node and the bottom horizontal outer edge. In (b), they are the diagonal line from the center to the lower-right node and the diagonal internal line connecting the top node to the lower-left vertex. In (c), they are the diagonal internal line from the center to the lower-left node and the sloping outer edge from the top node to the lower-right node."}], "full_mark": 2}, {"sub_id": "Exam-16-T5-3", "tags": ["1.3"], "text_open": "5.4 Alkali-Metal Compounds of Fullerenes. Several compounds of C_{60} with alkali metals have been synthesized, some of which exhibit distinctive superconducting properties. We now examine the crystal structure of a compound A of C_{60} and potassium. A belongs to the cubic crystal system. In its cubic unit cell, a C_{60} anion occupies each vertex and the body center, while each pair of opposite faces contains 4 potassium atoms. These potassium atoms form the “cage structure” composed of rhombic and hexagonal faces shown below.\n5.4.2 Two potassium atoms in the crystal have fractional coordinates (1/2, 0.2795, 0) and (0.2205, 1/2, 0). Give the fractional coordinates of all remaining potassium atoms in the unit cell. Do not repeat the two coordinates already given.", "text_images": [{"file": "Exam-16-T5-3-1-prompt.png", "description": "The image shows a unit-cell diagram. Thin black lines outline a cubic unit cell, and purple spheres are located within it, with some pairs connected by purple line segments. At the lower left is a three-dimensional orthogonal coordinate system whose red, green, and blue arrows emerge from a gray central sphere: the red `a` axis points diagonally down-left, the green `b` axis points right, and the blue `c` axis points vertically upward. These axes are parallel to the three sets of unit-cell edges. There are 24 purple spheres distributed over the 6 faces of the cubic unit cell, 4 on each face. The 4 spheres on each face form a rhombus rather than a perfect square; both diagonals of the rhombus are perpendicular to the edges of the square unit-cell face, and the center of the rhombus coincides with the center of that face. Purple line segments occur in two places: along the edges of these rhombi and between the closest pair of purple spheres belonging to rhombi on adjacent unit-cell faces."}], "full_mark": 3}, {"sub_id": "Exam-16-T5-4", "tags": ["1.3"], "text_open": "5.4.3.1 Let the unit-cell edge length be a Å. Calculate the shortest K-K distance.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-16-T5-5", "tags": ["1.3", "1.1"], "text_open": "5.4.3.2 Determine whether the hexagonal faces of the cage-like structure are regular hexagons, and explain.", "text_images": [], "full_mark": 3}]} -{"problem_id": "Exam-16-T6", "header": "Problem 6: Polysulfide Anions\nA sodium-sulfur battery is a high-temperature molten secondary alkali-metal energy-storage battery that uses metallic sodium as the negative electrode, elemental sulfur as the positive electrode, and molten sodium beta-alumina as the solid electrolyte. It offers high energy density, inexpensive raw materials, and suitability for large-scale energy storage, and has become a research focus in electrochemical energy storage in recent years. Charging and discharging proceed through the formation of sodium polysulfides (Na_{2}S_{x}) with different chain lengths.", "header_images": [], "subquestions": [{"sub_id": "Exam-16-T6-1", "tags": ["2.1", "2.4"], "text_open": "6.2 Sodium polysulfides (Na_{2}S_{x}) are key electroactive intermediates in sodium-sulfur batteries, and their anions can be synthesized and observed in the laboratory using common reagents. Under an N_{2} atmosphere, excess sodium tert-butoxide is added to a DMF solution of S_{8}, reducing S_{8} to polysulfide anion A. A disproportionates into another polysulfide anion B containing fewer atoms. Electron paramagnetic resonance (EPR) measurements show that some B anions undergo homolysis in solution to form bent anion C. C is trapped in zeolite cages in lazurite crystals, [(Na,Ca)_{4-8}(AlSiO_{4})_{6}(SO_{4},S,Cl)_{2}], and appears blue in solution. The sodium salt corresponding to polysulfide anion B is a common component of sodium-sulfur batteries. If it is heated to 300°C and melted, anion C dimerizes again to give isomer D of polysulfide anion B; the sulfur atoms in D occupy only two distinct chemical environments. 6.2.1 Give the chemical formulas of A and B.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-16-T6-2", "tags": ["1.6", "2.4"], "text_open": "6.2.2 Give the SMILES structures of C and D, showing the Lewis connectivity, and indicate the formal charge on every sulfur atom.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-16-T6-3", "tags": ["1.6", "2.4"], "text_open": "6.3 Anion C can also be observed when S_{8} is dissolved in liquid ammonia, together with another hydrogen-free anion E. Early studies proposed the following mechanism: first, liquid ammonia undergoes autoionization to form NH_{2}^{-} and NH_{4}^{+}; second, NH_{2}^{-} attacks S_{8}, giving S_{8}NH_{2}^{-} by addition; third, S_{8}NH_{2}^{-} decomposes into cyclic anion F and a stable dibasic acid; fourth, F rearranges to form anion E and S_{8}, while the dibasic acid reacts with S_{8} and NH_{4}^{+} to form (NH_{4})_{2}S_{x}, which is subsequently converted into C. EPR measurements, however, show that anion F is not formed during the reaction. Moreover, the tendency of liquid ammonia to autoionize is extremely small and cannot explain the high solubility of S_{8} in liquid ammonia. Researchers therefore proposed a more plausible mechanism: first, S_{8}NH_{2}^{-} and NH_{3} participate in a proton-transfer equilibrium, forming S_{8}NH_{3} while generating NH_{2}^{-}, which promotes dissolution; second, NH_{3} can also react by nucleophilic substitution to form the small molecule S_{2}NH_{3}; third, S_{8}NH_{2}^{-} decomposes to anion B and intermediate G; fourth, G transfers all of its protons to NH_{3} to form anion H; fifth, H reacts with S_{2}NH_{3} to give anion E and NH_{3}, with all species in this reaction having a stoichiometric coefficient of 1. 6.3.1 Give the SMILES structures of E, F, G, and H, showing the Lewis connectivity. The Lewis structure of E is known to contain two S-S single bonds, and its sulfur atoms occupy two distinct chemical environments.", "text_images": [], "full_mark": 5}, {"sub_id": "Exam-16-T6-4", "tags": ["2.1", "2.2"], "text_open": "6.3.2 In fact, the fifth reaction proceeds in three steps, and both the first and second steps generate SNH_{3}. Write the chemical equations for the three steps in order.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-16-T6-5", "tags": ["3.2", "3.4"], "text_open": "6.4 Anion C is also relatively common in nature and occurs mainly in ores formed under high-temperature, high-pressure conditions. Calculations show that at 350℃ and 0.5 GPa, anion C can exist as a stable form of sulfur. Reaction 1 (unbalanced) is: H_2S(aq.) + SO_4^{2-}(aq.) + H^+(aq.) → C(aq.) + O_2(g.) + H_2O(l.), \\lg K_1 = -9.6 (when the stoichiometric coefficients of C and H^+ are both 1). The standard enthalpies of formation and standard entropies of several species are listed below (ignore the temperature dependence of enthalpy and entropy and ignore microscopic effects): H_2S(aq.): \\Delta_fH_m^⊖ = -39.0 kJ·mol^{-1}, S_m^⊖ = 122.0 J·mol^{-1}·K^{-1}; HS^-(aq.): -17.7, 61.1; HSO_4^-(aq.): -885.8, 126.9; SO_4^{2-}(aq.): -907.5, 17.0; H^+(aq.): 0, 0; O_2(g.): 0, 205.1; H_2O(l.): -285.8, 70.0.\n6.4.2 Derive the equation of the bold sloping line l_1 in the figure below. Hint: f_O2 denotes the fugacity of oxygen, which physically represents the effective pressure of a real gas. The total dissolved sulfur concentration in the figure is 1 wt% (that is, all dissolved sulfur-containing species in 1 kg of solution contain a total of 10 g of S atoms). Assume that the density of water under these conditions is ρ = 1.0 × 10^3 kg/m^3.", "text_images": [{"file": "Exam-16-T6-5-1-prompt.png", "description": "A two-dimensional predominance-diagram-style plot with a white background, black lines, and a rectangular frame. The top and right sides of the frame also have short tick marks. The horizontal axis is labeled “pH,” with major tick labels 1, 2, 3, 4, 5, 6, 7, 8, and 9; the axis extends beyond 9 without a clearly labeled 10. The vertical axis is labeled “$\\log_{10} f_{O_2}$,” with major ticks −22, −24, −26, −28, −30, −32, and −34 from top to bottom. Several solid black lines divide the plot into regions labeled with formulas or letters. The upper-left region is labeled “HSO_4^-,” the upper-right “SO_4^2-,” the lower-left “H_2S,” and the lower-right “HS^-.” Near the center-left is a light-gray irregular polygon labeled with a capital “C.” It has approximately five sides: an upper-left sloping boundary from about pH 1 and $\\log_{10} f_{O_2}$ about −24 to about pH 5.3 and $\\log_{10} f_{O_2}$ about −26; an upper-right sloping boundary from about pH 5.3 and $\\log_{10} f_{O_2}$ about −26 to about pH 7.6 and $\\log_{10} f_{O_2}$ about −28.4; a thinner lower-right boundary from about pH 7.6 and $\\log_{10} f_{O_2}$ about −28.4, extending down-left to about pH 6.0 and $\\log_{10} f_{O_2}$ about −31; a noticeably bold lower boundary from about pH 6.0 and $\\log_{10} f_{O_2}$ about −31, extending up-left to about pH 1 and $\\log_{10} f_{O_2}$ about −26.8; and a left boundary adjoining the vertical axis. A vertical solid line at about pH 5.3 runs from the top border to near the upper vertex of the gray region and separates HSO_4^- from SO_4^2-. Another vertical solid line at about pH 6.0 runs downward from the lower vertex of the gray region to the bottom border and separates H_2S from HS^-. A thinner sloping line also extends down-right from the upper-right vertex of the gray region to near the right border, separating SO_4^2- from HS^-."}], "full_mark": 4}, {"sub_id": "Exam-16-T6-6", "tags": ["2.2", "4.2"], "text_open": "6.5 Anion C also has many applications in organic synthesis. Use the results of the following control experiments to answer Questions 6.5.1-6.5.4. 6.5.1 According to the control experiments, from which species formed by conversion of elemental sulfur might the sulfur in the product originate?", "text_images": [{"file": "Exam-16-T6-6-1-prompt.png", "description": "Control experiments for 6.5.1, showing two parallel reactions. In the upper reaction, 'C1=CC=C(C#CC#CC2=CC=CC=C2)C=C1' (0.2 mmol) reacts with `S` (0.6 mmol) under 1.8 mmol NaOtBu in DMF, N_2, at 25°C to give 'C1=CC=C(C2=CC=C(C3=CC=CC=C3)S2)C=C1' in 65% yield. In the lower reaction, 'C1=CC=C(C#CC#CC2=CC=CC=C2)C=C1' (0.2 mmol) reacts with `S` (0.6 mmol) under 1.8 mmol NaO^tBu and 4.0 equiv TEMPO in ^tBuOH+DMF, N_2, at 25°C to give 'C1=CC=C(C2=CC=C(C3=CC=CC=C3)S2)C=C1' in 5% yield."}], "full_mark": 1}, {"sub_id": "Exam-16-T6-7", "tags": ["2.3", "2.7"], "text_open": "6.5.2 In the control experiments above, no radical intermediate was trapped in the first reaction, indicating that it does not follow a radical mechanism, whereas the third reaction proceeds through a radical mechanism. Explain, from the perspective of molecular structure, why the two reactions follow different mechanisms.", "text_images": [{"file": "Exam-16-T6-7-1-prompt.png", "description": "Control experiments for 6.5.2, showing four parallel reactions. In the first, 'C1=CC=C(C#CC#CC2=CC=CC=C2)C=C1' (0.2 mmol) reacts with `Na_2S · 9H_2O` (0.6 mmol) in DMF, air, at 25°C to give 'C1=CC=C(C2=CC=C(C3=CC=CC=C3)S2)C=C1' in 99% yield. In the second, 'C1=CC=C(C#CC#CC2=CC=CC=C2)C=C1' (0.2 mmol) reacts with `Na_2S · 9H_2O` (0.6 mmol) with 4.0 equiv TEMPO in DMF, air, at 25°C to give 'C1=CC=C(C2=CC=C(C3=CC=CC=C3)S2)C=C1' in 92% yield. In the third, 'C1=CC=C(C#CC2=CC=CC=C2)C=C1' (0.6 mmol) reacts with `K_2S` (0.36 mmol) in DMF, Ar, at 130°C to give 'C1(C2=CC=CC=C2)=C(C3=CC=CC=C3)C(C4=CC=CC=C4)=C(C5=CC=CC=C5)S1' in 65% yield. In the fourth, 'C1=CC=C(C#CC2=CC=CC=C2)C=C1' (0.6 mmol) reacts with `K_2S` (0.36 mmol) with 4.0 equiv TEMPO in DMF, Ar, at 130°C to give only a trace of 'C1(C2=CC=CC=C2)=C(C3=CC=CC=C3)C(C4=CC=CC=C4)=C(C5=CC=CC=C5)S1'."}], "full_mark": 2}, {"sub_id": "Exam-16-T6-8", "tags": ["2.2", "2.3"], "text_open": "6.5.3 Give the SMILES structures of two key intermediates in the third reaction of the control experiments in 6.5.2.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-16-T6-9", "tags": ["2.8", "2.3"], "text_open": "6.5.4 Explain separately why the yields of the following two reactions are substantially lower than that of the third reaction in 6.5.2.", "text_images": [{"file": "Exam-16-T6-9-1-prompt.png", "description": "Control experiments for 6.5.4, showing two parallel reactions. In the upper reaction, 'C1=CC=C(C#CC2=CC=CC=C2)C=C1' (0.6 mmol) reacts with `Na_2S · 9H_2O` (0.36 mmol) in DMF, Ar, at 130°C to give 'C1(C2=CC=CC=C2)=C(C3=CC=CC=C3)C(C4=CC=CC=C4)=C(C5=CC=CC=C5)S1' in 36% yield. In the lower reaction, 'C1=CC=C(C#CC2=CC=CC=C2)C=C1' (0.6 mmol) reacts with `S_8` (0.36 mmol) under NaO^tBu in DMF, Ar, at 130°C to give 'C1(C2=CC=CC=C2)=C(C3=CC=CC=C3)C(C4=CC=CC=C4)=C(C5=CC=CC=C5)S1' in less than 5% yield."}], "full_mark": 2}]} -{"problem_id": "Exam-16-T7", "header": "Problem 7: Baldwin’s Rules\nThis problem concerns the regioselectivity of intramolecular ring-closing reactions, Baldwin’s rules, and their exceptions.", "header_images": [], "subquestions": [{"sub_id": "Exam-16-T7-1", "tags": ["2.3", "2.7", "2.4"], "text_open": "7.1 Give the SMILES structure of the product formed by the intramolecular reaction of the molecule shown below under suitable conditions, and explain the selectivity.", "text_images": [{"file": "Exam-16-T7-1-1-prompt.png", "description": "Substrate in the prompt, with SMILES 'C=C(CC(N)C(OC)=O)C(OC)=O'."}], "full_mark": 2}, {"sub_id": "Exam-16-T7-2", "tags": ["2.3", "2.7", "2.4"], "text_open": "7.2 Baldwin’s rules are important empirical rules for explaining the regioselectivity of ring-closing reactions. Researchers used the reaction below in a classic mechanistic test of these rules. 7.2.1 Give the SMILES structures of the key intermediate in each of the two possible pathways and of the final product, and use Baldwin’s rules to identify the predominant pathway.", "text_images": [{"file": "Exam-16-T7-2-1-prompt.png", "description": "Reaction in the prompt: PhCH=CHCO_2Me reacts with NH_2NH_2 at 65°C."}], "full_mark": 5}, {"sub_id": "Exam-16-T7-3", "tags": ["2.3", "4.2"], "text_open": "7.2.2 Proof by contradiction is often used in scientific research to test reaction mechanisms. Using the substrate shown below as the main starting material, design an experiment to demonstrate that this reaction does not proceed through the pathway disfavored by Baldwin’s rules. State the experimental approach and the basis for the inference.", "text_images": [{"file": "Exam-16-T7-3-1-prompt.png", "description": "Substrate in the prompt, with SMILES 'O=C(/C=C/C1=CC=CC=C1)O[K]'."}], "full_mark": 2}, {"sub_id": "Exam-16-T7-4", "tags": ["2.3", "2.7", "2.4"], "text_open": "7.3 “Exceptions” to Baldwin’s rules. 7.3.1 Give the SMILES structure of the product formed by the intramolecular reaction of the molecule shown below under suitable basic conditions, and explain the selectivity.", "text_images": [{"file": "Exam-16-T7-4-1-prompt.png", "description": "Substrate in the prompt, with SMILES 'BrCCC(C)(C)C(C)=O'."}], "full_mark": 2}, {"sub_id": "Exam-16-T7-5", "tags": ["2.3", "1.6"], "text_open": "7.3.2 The molecule shown below readily undergoes ring closure under suitable basic conditions with the amide nitrogen as the nucleophilic center. Explain why.", "text_images": [{"file": "Exam-16-T7-5-1-prompt.png", "description": "Substrate in the prompt, containing a structure with SMILES 'O=C(N[Ar])CCCBr'."}], "full_mark": 2}, {"sub_id": "Exam-16-T7-6", "tags": ["2.3", "2.7", "2.4"], "text_open": "7.4 Give the SMILES structure of the major product formed by the intramolecular reaction of the molecule shown below under suitable basic conditions, and explain the selectivity.", "text_images": [{"file": "Exam-16-T7-6-1-prompt.png", "description": "Substrate in the prompt, with SMILES 'O=C(C)CC(C(C)=O)CC(C)=O'."}], "full_mark": 2}, {"sub_id": "Exam-16-T7-7", "tags": ["2.3", "2.4", "2.7"], "text_open": "7.5 Drawing on the discussion above, give the SMILES structure of the product formed by the intramolecular reaction of the molecule shown below under suitable conditions.", "text_images": [{"file": "Exam-16-T7-7-1-prompt.png", "description": "Substrate in the prompt, with SMILES 'O=C(CC(OCC)=O)CC(CCl)OC'."}], "full_mark": 2}]} -{"problem_id": "Exam-16-T9", "header": "Problem 9: Effects of Reaction Conditions and Mechanistic Investigation of Electrochemical Decarboxylative Alkylation\nElectrochemical anodic oxidation can use inexpensive and readily available quaternary-ammonium salts of alkyl carboxylic acids as alkyl sources, avoiding the need for additional oxidants and transition-metal catalysts during decarboxylation. The method therefore offers a simple system, mild conditions, and environmental advantages. Researchers used an unsaturated amide as a model substrate to investigate decarboxylative methylation/arylation. Tetrabutylammonium acetate served as both the methyl source and electrolyte, with carbon cloth as the anode, platinum as the cathode, and DMSO as the solvent. Under constant-current electrolysis at 3 mA with 6.0 equivalents of acetic acid as a proton source at 70°C for 9 h (5.0 F/mol), the target oxindole was obtained in 80% yield.", "header_images": [], "subquestions": [{"sub_id": "Exam-16-T9-1", "tags": ["3.5", "2.8", "2.3"], "text_open": "9.1 Further study showed that the acidity of the proton source strongly affects the reaction. When weaker proton sources, such as tert-butanol or hexafluoroisopropanol, or stronger proton sources, such as p-toluenesulfonic acid or trifluoroacetic acid, were used, the yield decreased substantially and the reaction was sometimes almost completely suppressed. Give a reasonable explanation for each observation.", "text_images": [{"file": "Exam-16-T9-1-1-prompt.png", "description": "Standard reaction in the prompt: 'FC(F)(C(C(N(C1=CC=CC=C1)C)=O)=C)F' reacts with `TBAOAc (1.5 equiv.)` using `C (+) / Pt (-)` under `HOAc (6.0 equiv.)`, `DMSO 4 mL, 70°C`, and `I = 3 mA, 5.0 F/mol`, producing 'CCC1(C(N(C2=CC=CC=C21)C)=O)C(F)(F)F'."}], "full_mark": 4}, {"sub_id": "Exam-16-T9-2", "tags": ["2.2", "2.3", "4.2"], "text_open": "9.3 The researchers performed the following experiments to investigate the reaction mechanism. Note: “standard condition” refers to the standard conditions described above, under which the yield is 80%. 9.3.1 To test for a free-radical intermediate, TEMPO was added as a radical trap. The target product did not form at all. Analysis of the reaction mixture together with high-resolution mass spectrometry revealed products formed by trapping with TEMPO. Give the SMILES structures of possible products.", "text_images": [{"file": "Exam-16-T9-2-1-prompt.png", "description": "Mechanistic experiment for 9.3.1: 'CN(C(C(C)=C)=O)C1=CC=CC=C1' reacts with the salt formed from anion 'CC([O-])=O' (shown with a circled minus sign) and cation 'CCCC[N+](CCCC)(CCCC)CCCC' (shown as `N^nBu_4` with a circled plus sign) under `standard condition` with *TEMPO (3.0 equiv.)*."}], "full_mark": 2}, {"sub_id": "Exam-16-T9-3", "tags": ["2.2", "2.3", "2.4"], "text_open": "9.3.3 On the basis of these mechanistic experiments, two pathways remain possible for the ring-closing step of the standard reaction. Give the SMILES structures of the key intermediates that may be involved in the standard reaction and identify the two possible pathways corresponding to the ring-closing step.", "text_images": [], "full_mark": 6}]} -{"problem_id": "Exam-17-T5", "header": "Problem 5 Dicarbon Anions", "header_images": [], "subquestions": [{"sub_id": "Exam-17-T5-1", "tags": ["1.3", "1.4"], "text_open": "5.2 Starting from crystalline $CaC_2$, inserting square Co layers into some of the interlayer spaces and replacing Ca with U gives a tetragonal $UCoC_2$ crystal. Its structure is shown below, where the large spheres represent U, the medium spheres represent Co, and the small spheres represent C. The lattice parameter c = 728 pm.\n5.2.1 Name the coordination polyhedron formed by the C atoms around Co.", "text_images": [{"file": "Exam-17-T5-1-1-prompt.png", "description": "An oblique three-dimensional ball-and-stick rendering of the tetragonal unit cell of $UCoC_2$ on a white background. Thin gray-black lines outline a rectangular-prismatic unit-cell frame elongated vertically along c. There are 10 large spheres, including spheres at face centers on the top and the upper and lower basal faces; 4 medium spheres lie on the side faces, one at an upper-middle position on each left/right face and one at a lower-middle position on each front/back face; and 8 small spheres form upper and lower groups, with two vertical pairs in each group along c joined by short brown sticks. The upper pair is on the front/back faces and the lower pair on the left/right faces. The small spheres are joined to adjacent large spheres by short sticks, forming a layered framework; the medium spheres have no sticks. At lower left, a blue arrow points upward and is labeled c, a green arrow points right and is labeled b, and a red arrow points lower-left and is labeled a. Bonding obscured by occlusion cannot be resolved. Apart from a, b, and c, the image contains no text or legend. According to the caption, the large, medium, and small spheres represent U, Co, and C, respectively. The coordinates of the two carbon atoms in the unit cell are (1/2, 0, 0.155) and (1/2, 0, 0.358)."}], "full_mark": 2}, {"sub_id": "Exam-17-T5-2", "tags": ["1.3"], "text_open": "5.3 A uranium carbide belongs to the cubic crystal system. Its unit-cell projections along the a, b, and c directions are all as shown below. The large spheres represent U and the small spheres represent C. All atoms of the same element are equivalent in the crystal, and the lattice parameter a = 808 pm.\n5.3.1 Write the chemical formula of this crystal.", "text_images": [{"file": "Exam-17-T5-2-1-prompt.png", "description": "Crystal-axis indicator on a white background: a solid red circle is at the lower left with a black letter a superimposed, indicating that the a axis points out of the page toward the viewer. A blue arrow extends upward from the circle and is labeled c above its tip; a green arrow extends rightward and is labeled b to the right of its tip. Apart from a, b, and c, the image contains no text or other graphics."}, {"file": "Exam-17-T5-2-2-prompt.png", "description": "Crystal-axis indicator on a white background: a solid green circle is at the lower left with a black letter b superimposed, indicating that the b axis points out of the page toward the viewer. A red arrow extends upward from the circle and is labeled a above its tip; a blue arrow extends rightward and is labeled c at its right end. Apart from a, b, and c, the image contains no text or other graphics."}, {"file": "Exam-17-T5-2-3-prompt.png", "description": "Crystal-axis indicator on a white background: a solid blue circle is at the lower left with a black letter c superimposed, indicating that the c axis points out of the page toward the viewer. A green arrow extends upward from the circle and is labeled b above its tip; a red arrow extends rightward and is labeled a at its right end. Apart from a, b, and c, the image contains no text or other graphics."}, {"file": "Exam-17-T5-2-4-prompt.png", "description": "Unit-cell projection of the uranium carbide in 5.3 (two-dimensional projection of a ball-and-stick model on a white background): thin black lines form a square unit-cell border, with some spheres lying on the border. The image contains only two sphere types: large light-blue highlighted spheres and small brown spheres. The large spheres occur at the four corners, at the middle of the top and bottom edges, and at several interior positions; some large spheres overlap in pairs. Some small spheres are joined by short brown sticks into paired or chain-like arrangements, while others are isolated. The image contains no text, letters, numbers, arrows, or legend. As specified in the prompt, the large spheres represent U and the small spheres represent C. The prompt states that the projections along the a, b, and c directions are all as shown; this image is that projection panel."}], "full_mark": 3}, {"sub_id": "Exam-17-T5-3", "tags": ["1.3"], "text_open": "5.3.4 One carbon atom in the unit cell has coordinates (1/2, 1/4, 0.536). Calculate the carbon-carbon bond length in this crystal.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-17-T5-4", "tags": ["1.3", "1.7"], "text_open": "5.4 The carbon-carbon bond lengths in ethane, ethene, and ethyne are approximately 154 pm, 134 pm, and 120 pm, respectively. On this basis, predict the negative charge on the dicarbon anion in $CaC_2$, $UCoC_2$, and the uranium carbide in 5.3, respectively. The carbon-carbon bond orders in the three compounds are distinct integers.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-17-T5-5", "tags": ["1.3", "1.7"], "text_open": "5.5 Crystalline $DyCoC_2$ has a structure entirely different from that of $UCoC_2$. Its three-dimensional structure and projection along the c direction are shown below, where the large spheres represent Dy, the medium spheres represent Co, and the small spheres represent C. The only common stable positive oxidation state of Dy is +3. A small sample of $DyCoC_2$ is hydrolyzed in 2 mol/L dilute hydrochloric acid. Gas-chromatographic analysis gives the following mass fractions of the gaseous hydrolysis products: 74% ethane, 13% propane, 12% methane, and 1% ethene; no ethyne is formed.\n5.5.1 Predict the carbon-carbon bond order in $DyCoC_2$.", "text_images": [{"file": "Exam-17-T5-5-1-prompt.png", "description": "Three-dimensional ball-and-stick model of crystalline $DyCoC_2$ (oblique rendering on a white background): thin black lines outline a parallelepiped unit cell. There are 10 large blue-violet highlighted spheres of varying shades: 8 at the 8 corners and 2 at the centers of the top and bottom faces. There are 3 medium dark-blue spheres in a horizontal layer through the middle of the cell, positioned left of center, right of center, and at the center. Approximately 9 small brown spheres surround the medium spheres and are connected to them by dark-blue or brown short sticks; some small spheres bridge two medium spheres, forming an in-layer zigzag network. There are no direct sticks between large spheres or between medium spheres. Some spheres are obscured, so the exact number of brown small spheres cannot be fully determined. A three-axis indicator at the lower left has a blue upward arrow labeled c, a green rightward arrow labeled b, and a red arrow pointing down-left labeled a. Apart from a, b, and c, the image contains no text or legend. As specified in the prompt, the large, medium, and small spheres represent Dy, Co, and C, respectively."}, {"file": "Exam-17-T5-5-2-prompt.png", "description": "Projection of crystalline $DyCoC_2$ along the c direction (two-dimensional projection of a ball-and-stick model on a pale gray-white background): thin black lines form a rectangular unit-cell border. Four large blue spheres lie at the corners; 1 slightly smaller blue sphere lies at the center of the rectangle without any bonds; and 3 medium blue spheres lie at the midpoints of the left and right edges and at the center of the lower edge. There are 10 small brown spheres: 2 terminal spheres outside the rectangle at the upper left and upper right, while the remaining 8 form 4 horizontal dumbbell pairs joined by two parallel short brown sticks. These pairs lie at the upper center, middle left, middle right, and outside the rectangle at the lower center. Each medium blue sphere is connected to 4 brown spheres by two-colored blue-brown short sticks, and adjacent blue spheres and brown pairs form triangular units. A three-axis indicator at the lower left has a red rightward arrow labeled a, a green upward arrow labeled b, and a blue circle bearing the letter c, indicating that the c axis points out of the page toward the viewer. Apart from a, b, and c, the image contains no text or legend. As specified in the prompt, the large, medium, and small spheres represent Dy, Co, and C, respectively."}], "full_mark": 1}, {"sub_id": "Exam-17-T5-6", "tags": ["1.6"], "text_open": "5.5.3 Briefly explain why the experimentally measured carbon-carbon bond length in crystalline $DyCoC_2$ is 138 pm.", "text_images": [], "full_mark": 2}]} -{"problem_id": "Exam-17-T6", "header": "Problem 6 Synthesis of Triphenylmethanol Using a Grignard Reagent\nGrignard reagents are common nucleophiles in organic synthesis. They must be used under anhydrous conditions free of oxygen and $CO_2$, and reactions are usually performed under nitrogen. In a teaching laboratory, however, the synthesis can be carried out with a simpler procedure. In this problem, bromobenzene is used to prepare phenylmagnesium bromide, which is then reacted with methyl benzoate to produce triphenylmethanol.", "header_images": [{"file": "Exam-17-T6-header-1.png", "description": "Experimental apparatus: a three-neck flask is centered, with its left neck closed by a ground-glass stopper. The middle neck holds a funnel-shaped dropping device with a sidearm and stopcock, labeled \"Apparatus 1\". The angled right neck is connected to a spherical condenser sloping upward to the right. Liquid is present in the lower part of the three-neck flask. The labels in the image are Apparatus 1, spherical condenser, three-neck flask, and to drying tube at the top of the condenser."}], "subquestions": [{"sub_id": "Exam-17-T6-1", "tags": ["2.8"], "text_open": "6.2 Synthesis of triphenylmethanol: cool the flask in a cold-water bath while slowly adding a solution of methyl benzoate in anhydrous diethyl ether with stirring. After addition is complete, heat the mixture under reflux for 1 h, cool the flask again in a cold-water bath, and add saturated ammonium chloride solution dropwise to decompose the addition product. Separate the layers and perform steam distillation to obtain the crude product.\n6.2.1 State the purpose of the saturated ammonium chloride workup, and explain why water or hydrochloric acid is not used directly for the workup.", "text_images": [], "full_mark": 3}]} -{"problem_id": "Exam-17-T8", "header": "Problem 8 Fundamentals of Organic Chemistry", "header_images": [], "subquestions": [{"sub_id": "Exam-17-T8-1", "tags": ["1.2"], "text_open": "8.2 Examine the reaction below and answer the question.\n8.2.1 Determine the configuration of the chiral carbon marked with * in the product, using S/R notation.", "text_images": [{"file": "Exam-17-T8-1-1-prompt.png", "description": "'O=C1C[C@H](O[Si](C)(C)C(C)(C)C)C=C1S(=O)(C2=CC=CC=C2)=O'>Me3Al (2.2 equiv.), Toluene, -78 ℃, 30 min>'O=C1C[C@H](O[Si](C)(C)C(C)(C)C)C(C)C1S(=O)(C2=CC=CC=C2)=O'. The chiral carbon marked with * in the product is the ring carbon bonded to the methyl group."}], "full_mark": 2}, {"sub_id": "Exam-17-T8-2", "tags": ["2.8"], "text_open": "8.2.3 When the Lewis acid is changed from $Me_3Al$ to $Et_3Al$ or $^iBu_3Al$, the yield decreases from 86% to 65% and 32%, respectively. Explain the decrease in yield.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-17-T8-3", "tags": ["1.2"], "text_open": "8.3 Consider the two conformations of $FCH_2C(=O)R$ shown below. They interconvert by rotation about the $C_\\alpha$-C(=O) single bond. The relative energy of the conformation on the right is defined as 0.0 kcal/mol.\n8.3.1 Define the energy of the left conformation relative to the right conformation as $\\Delta E = E_{\\text{left conformation}} - E_{\\text{right conformation}}$. Determine the sign of $\\Delta E$.", "text_images": [{"file": "Exam-17-T8-3-1-prompt.png", "description": "The molecular SMILES is '[R]C([CH2]F)=O'. The image shows two conformations connected by a reversible arrow. In the left conformation, F points downward in the plane of the page, COR points upward and to the right in the plane, and the two H atoms project in front of and behind the upper-left region. In the right conformation, F points upward and to the left in the plane, COR points upward and to the right in the plane, and the two H atoms project in front of and behind the lower region. The right conformation is labeled 0.0 kcal/mol; R = NH2, OMe, Me, H."}], "full_mark": 1}, {"sub_id": "Exam-17-T8-4", "tags": ["1.2", "1.6"], "text_open": "8.3.2 For $R = NH_2$, OMe, Me, and H, denote the corresponding values of $\\Delta E$ by A, B, C, and D, respectively. Arrange A, B, C, and D in increasing numerical order.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-17-T8-5", "tags": ["1.2"], "text_open": "8.4 The diagram below shows two conformations, A and B, of a difluoroalkane. Assume that they interconvert only by rotation about a carbon-carbon σ bond, without any change in configuration. Ignore intermolecular interactions and solvent effects, and compare only their intramolecular conformational energies. Determine which of A and B has the lower energy and state the principal reason.", "text_images": [{"file": "Exam-17-T8-5-1-prompt.png", "description": "Both A and B have the SMILES 'C[C@@H](F)C[C@H](C)F'. They are two conformations of the difluoroalkane and are connected by a reversible arrow. In A, the F at C2 points upward on a solid wedge, the F at C4 points downward and to the right on an ordinary single bond, and the methyl group at C4 points upward on a dashed wedge. In B, the F atoms at both C2 and C4 point upward on solid wedges, and the methyl group at C4 points downward and to the right on an ordinary single bond."}], "full_mark": 3}]} +{"problem_id": "Exam-17-T2", "header": "Problem 2 Determination of Paracetamol by Cyclic Voltammetry and Study of Its Electrochemistry\nCyclic voltammetry (CV) is an important electrochemical technique for investigating electrochemical reaction mechanisms and performing quantitative analysis. It uses a three-electrode system consisting of a working electrode, a reference electrode, and an auxiliary electrode. A triangular-wave potential that varies linearly with time is applied to the working electrode while the response current is recorded. One triangular-wave scan completes both an oxidation scan and a reduction scan, and the resulting current–potential curve is called a cyclic voltammogram. A substance that can undergo oxidation or reduction at an electrode within the scanned potential range is called an electroactive substance. A typical cyclic voltammogram exhibits an oxidation peak (anodic peak) and a reduction peak (cathodic peak). Under suitable conditions, the peak currents ($I_{pa}$ and $I_{pc}$) are proportional to the concentration of the electroactive substance and form the basis of quantitative analysis. The peak potentials ($E_{pa}$ and $E_{pc}$) and the peak-potential separation ($\\Delta E_p = E_{pa} - E_{pc}$) can be used to assess reaction reversibility and the number of electrons transferred. Changing the scan rate ($v$) can distinguish diffusion control from adsorption control and reveal the kinetics of subsequent chemical reactions.\nParacetamol (APAP, also known as acetaminophen) is a commonly used antipyretic analgesic. Studies show that APAP undergoes electrochemical oxidation at an electrode surface, and its oxidation product NAPQI undergoes subsequent chemical reactions in acidic media. A research group used cyclic voltammetry to investigate the electrochemical behavior of APAP systematically.\nAt room temperature, the peak current ($I_p$, in A) for a reversible electrochemical reaction in a quiescent solution at a planar electrode is related to the scan rate ($v$, in V/s), the concentration of the electroactive substance ($C$, in $\\mathrm{mol/cm^3}$), and other parameters by the Randles–Sevcik equation:\n$$I_p = 2.69 \\times 10^5 n^{3/2} A C D^{1/2} v^{1/2}$$\nn: number of electrons transferred in the reaction.\nA: effective area of the working electrode (unit: $\\mathrm{cm^2}$).\nD: diffusion coefficient of the electroactive substance (unit: $\\mathrm{cm^2/s}$), reflecting how rapidly the substance moves through solution.\nThe figure below is a typical cyclic voltammogram. Curve 1 is the background signal, and curve 2 is the normal response signal.", "header_images": [{"file": "Exam-17-T2-header-1.png", "description": "Typical cyclic voltammogram. The horizontal axis is potential, with tick marks of approximately 0.2, 0.1, 0, -0.1, and -0.2 from left to right; the vertical axis is current. The peak-current and peak-potential symbols $I_{pc}$, $I_{pa}$, $E_{pa}$, and $E_{pc}$ are labeled. Two closed curves are labeled 1 and 2. As stated in the question, curve 1 is the background signal and curve 2 is the normal response signal."}], "subquestions": [{"sub_id": "Exam-17-T2-1", "tags": ["4.7"], "text_open": "2.2 At a scan rate of $40\\ \\mathrm{mV\\cdot s^{-1}}$ in a pH = 2.2 buffer, the research group obtained the following cyclic-voltammetry data for standard APAP solutions:\n\n| APAP concentration/(mmol/L) | 0.10 | 0.40 | 1.00 | 3.00 | 5.00 |\n| ----------------- | ----- | ----- | ----- | ----- | ----- |\n| Anodic peak current $I_{pa}$/μA | 2.15 | 5.82 | 13.46 | 38.74 | 63.52 |\n\nUsing concentration as the horizontal coordinate and anodic peak current as the vertical coordinate, obtain the linear regression equation of $I_{pa}$ against APAP concentration and calculate the expected peak current for an APAP solution with a concentration of $3.50\\ \\mathrm{mmol\\cdot L^{-1}}$.", "text_images": [], "full_mark": 5}, {"sub_id": "Exam-17-T2-2", "tags": ["2.3"], "text_open": "2.3 To investigate the oxidation mechanism of APAP, the group performed CV scans at $40\\ \\mathrm{mV\\cdot s^{-1}}$ in media of different acidity (pH = 6.0, pH = 2.2, and $1.8\\ \\mathrm{mol\\cdot L^{-1}}\\ H_2SO_4$) and observed the following:\nAt pH = 6.0, both oxidation and reduction peaks appear, but $I_{pa}$ ≠ $I_{pc}$; at pH = 2.2, only the oxidation peak appears and the reduction peak disappears; under strongly acidic conditions ($1.8\\ \\mathrm{mol\\cdot L^{-1}}\\ H_2SO_4$), the reduction peak reappears.\n2.3.1 Explain why the reduction peak disappears at pH = 2.2.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-17-T2-3", "tags": ["2.3"], "text_open": "2.3.2 What chemical reaction is indicated by the reappearance of the reduction peak under strongly acidic conditions?", "text_images": [], "full_mark": 2}]} +{"problem_id": "Exam-17-T4", "header": "Problem 4 Gravimetric Determination of Tungsten (17 points, 8%)\nTungsten is an important strategic metal that has an irreplaceable strengthening role in cemented carbides, high-speed steel, and specialty steels. Accurate determination of tungsten content is essential for quality control of alloy materials. In the classical gravimetric method, tungsten is converted to a tungstic-acid precipitate, ignited to tungsten trioxide, and weighed. Accurate tungsten content is obtained by correcting for interference from coexisting elements such as molybdenum, chromium, and vanadium. This problem concerns precipitation and separation of tungsten, correction for impurities, and special treatment of niobium-containing samples, and tests understanding of the principles of gravimetric analysis and strategies for eliminating interference.\nThe hydrochloric acid used in this problem is diluted 1:1, the nitric acid is concentrated, and every reported mass has been corrected for the tare mass of the crucible.", "header_images": [], "subquestions": [{"sub_id": "Exam-17-T4-1", "tags": ["2.1"], "text_open": "4.1 Procedure for samples without niobium\nA sample of mass $m$ is weighed into a beaker. Hydrochloric acid and nitric acid are added, and the mixture is heated gently to dissolve the sample and evaporated to a syrup to remove nitrogen oxides. Hydrochloric acid is added to dissolve the salts, the solution is diluted with water and brought to the boil, and cinchonine hydrochloride solution is added slowly to precipitate all tungsten as tungstic acid. The precipitate is aged at 70-80℃ for 4 h and then filtered and washed with a cinchonine hydrochloride wash solution.\nThe precipitate and filter paper are placed in a platinum crucible, carbonized, and ignited at 750-800℃. The residue is moistened with water, hydrofluoric acid and sulfuric acid are added, and the mixture is heated to volatilize silicon. It is then ignited to constant mass, giving a mass $m_1$ of impure $WO_3$.\nSodium carbonate is added to the platinum crucible for fusion. After cooling, the melt is extracted with hot water and filtered, and the residue is ignited and weighed to give $m_2$. The filtrate is analyzed spectrophotometrically for molybdenum, chromium, and vanadium, which are converted to the corresponding oxide masses $m_3$, $m_4$, and $m_5$.\n4.1.1 Write the equation for oxidation of elemental tungsten to tungstic acid, $H_2WO_4$, in hydrochloric acid/nitric acid.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-17-T4-2", "tags": ["4.7", "2.1"], "text_open": "4.1.6 During aqueous extraction after sodium carbonate fusion, $WO_3$ is converted to $Na_2WO_4$ and enters solution, whereas $TiO_2$ and $Fe_2O_3$ remain insoluble. Analysis of a tungsten-steel sample gives the following data: $m$ = 0.5123 g, $m_1$ = 0.0842 g, $m_2$ = 0.0018 g, $m_3$ = 0.0023 g, $m_4$ = 0.0015 g, and $m_5$ = 0.0012 g. Calculate the mass fraction of tungsten in the sample.", "text_images": [], "full_mark": 2}]} +{"problem_id": "Exam-17-T7", "header": "Problem 7 Low-Valent Cluster Compounds of Group 13 Metals\nLow-valent compounds of Group 13 elements such as aluminum and gallium can form metal clusters with diverse structures. Their bonding modes and electronic structures are important for understanding the nature of metal-metal bonds. This problem concerns low-valent gallium and aluminum halides and their assembly reactions with ligands, focusing on structural analysis, oxidation-state calculations, and cluster reactivity.", "header_images": [], "subquestions": [{"sub_id": "Exam-17-T7-1", "tags": ["2.1"], "text_open": "Synthesis and structures of $E_3$ units: Ga-Ga bonds are relatively stable and can support extended compounds.\n7.1 Chain-like $E_3$ structure: reaction of freshly prepared LiM with 1.67 eq GaI affords the trigallane C = $Ga_3M_3I_2$.\n7.1.1 Write the reaction equation for the formation of C from LiM and GaI.", "text_images": [{"file": "Exam-17-T7-1-1-prompt.png", "description": "Two-step reaction: '[dipp]N=C=N[dipp]'>t-BuLi (1eq), n-hexane, 0 °C>intermediate Li+M-, where M denotes the ligand and the SMILES for M- is '[dipp]/N=C([N-][dipp])/C(C)(C)C'. The M- anion has an N,N'-bis(dipp) amidinate framework with t-Bu bonded to the central carbon. The second step is '[dipp]/N=C([N-][dipp])/C(C)(C)C'>GaI (1.67eq), Toluene, -78 °C>C, where C = $Ga_3M_3I_2$ and its SMILES is 'I[Ga]1([Ga]2([Ga]3(I)N([dipp])C(C(C)(C)C)=[N]3[dipp])N([dipp])C(C(C)(C)C)=[N]2[dipp])N([dipp])C(C(C)(C)C)=[N]1[dipp]'. Three Ga atoms form a chain through Ga-Ga single bonds, each terminal Ga bears one I, and each Ga is chelated by one M ligand. No wedge or dashed bonds are shown."}], "full_mark": 2}, {"sub_id": "Exam-17-T7-2", "tags": ["2.1"], "text_open": "7.1.2 Write the equation for the substitution reaction of trigallane C with $NaBEt_3H$ to form Et(M)Ga-Ga(M)-Ga(M)H.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-17-T7-3", "tags": ["2.4", "1.6"], "text_open": "7.2 Cyclic $E_3$ structure: cyclotrigallanes have intriguing electronic properties. Reaction of $(Mes_2C_6H_3)Li$ with $GaCl_3$ gives $(Mes_2C_6H_3)GaCl_2$, which is reduced with potassium metal in diethyl ether to afford the cyclic gallium compound D = $K_2[(Mes_2C_6H_3)Ga]_3$. This compound has a nearly equilateral triangular $Ga_3$ core, with an average Ga-Ga bond length of 2.426 Å that is markedly shorter than a typical single bond, indicating a higher bond order. The two potassium ions lie on opposite sides of the $Ga_3$ plane. Write the SMILES string for D and determine whether its anion is aromatic.", "text_images": [{"file": "Exam-17-T7-3-1-prompt.png", "description": "Two-step reaction: $GaCl_3$ reacts with LiAr (1.0 eq) in $Et_2O$ at -78 °C; $K^0$ (4.0 eq) is then added and the mixture is reacted at rt. for 5 days to give $K_2(ArGa)_3·(Et_2O)_2$. The structure defining Ar = $(Mes_2C_6H_3)$ appears at the right: the top carbon of the central benzene ring, drawn in an alternating-double-bond Kekulé form, bears a wavy bond marking the attachment point to Ga, and each of the two ortho carbons bears a 2,4,6-trimethylphenyl group (Mes). The encodable structure of Ar is represented by the SMILES '[*]C1=C(C2=C(C)C=C(C)C=C2C)C=CC=C1C3=C(C)C=C(C)C=C3C', where [*] is the attachment point corresponding to the wavy bond."}], "full_mark": 3}, {"sub_id": "Exam-17-T7-4", "tags": ["2.4", "1.4"], "text_open": "7.4 Eight-membered-ring structure: gaseous GaI is cocondensed with excess $PEt_3$ in toluene at -196 °C and then warmed slowly, giving F = $Ga_8I_8·6PEt_3$. This cluster has $C_{2h}$ symmetry and contains a nearly planar eight-membered $Ga_8$ ring. Two iodine atoms bridge across the ring to form a four-membered $Ga_2I_2$ ring. All eight Ga atoms are four-coordinate, and the average Ga-Ga bond length is 2.466 Å.\n7.4.1 Write the SMILES string for F.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-17-T7-5", "tags": ["1.7", "2.4"], "text_open": "7.4.2 The Ga/I/$PEt_3$ system forms the eight-membered-ring cluster F ($Ga_8I_8$), whereas the Al/Br/$NEt_3$ system forms the four-membered-ring cluster E ($Al_4Br_4$) under similar conditions. Which explanations are the most reasonable? Select all that apply.\n(A) Ga has a smaller atomic radius than Al and a shorter Ga-Ga bond; formation of a four-membered ring would therefore cause greater ring strain and bond-angle distortion, so an eight-membered ring is favored to relieve strain.\n(B) I has a larger atomic radius than Br and is well suited for bridging; transannular bridges stabilize the eight-membered ring.\n(C) $PEt_3$ is a stronger electron donor than $NEt_3$, increasing the electron density at Ga and shortening the Ga-Ga bonds, which favors formation of a larger ring system.\n(D) The Ga-Ga bonds in the eight-membered ring have a higher bond order than the Al-Al bonds in the four-membered ring, giving greater thermodynamic stability.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-17-T7-6", "tags": ["2.4", "1.2"], "text_open": "7.5 Fused-ring structure: attempts to synthesize amine-stabilized $Ga_8I_8·6NEt_3$ were unsuccessful, but partial oxidation unexpectedly gave G = $Ga_8I_8O_2·6NEt_3$. Its structure has $C_i$ symmetry and consists of two nearly planar $Ga_4O$ rings fused through a shared edge of a central four-membered $Ga_2O_2$ ring. All Ga atoms are four-coordinate. Write the SMILES string for G.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-17-T7-7", "tags": ["1.1"], "text_open": "7.6 Tetrahedral type: high-temperature gaseous GaBr is cocondensed with a toluene solution of $NHEt_2$. The resulting dark solution is concentrated and allowed to stand at 25 °C, yielding crystals of cluster H = $Ga_5Br_7(NHEt_2)_5$. H contains a $Ga_5$ core in which the central Ga atom is tetrahedrally surrounded by four Ga atoms.\n7.6.2 In the sterically congested H molecule, the 12 heteroatoms—the halogens and the O/N atoms of the ligands—adopt an arrangement that maximizes their mutual separation. What polyhedron is formed by these 12 vertices?", "text_images": [{"file": "Exam-17-T7-7-1-prompt.png", "description": "Structure of H = $Ga_5Br_7(NHEt_2)_5$: the central Ga is connected by single bonds to four peripheral Ga atoms in a tetrahedral arrangement. The encodable structure of H is represented by the SMILES '[L][Ga](Br)(Br)[Ga]([Ga]([L])(Br)Br)([Ga]([L])([L])Br)[Ga]([L])(Br)Br', where [L] denotes the $NHEt_2$ ligand labeled in the image. A total of 5 L ligands and 7 Br atoms are distributed among the peripheral Ga atoms. The lower right of the image defines L = $NHEt_2$, and all bonds are solid single bonds."}], "full_mark": 1}]} +{"problem_id": "Exam-17-T9", "header": "Problem 9 Stereochemistry\nCareful design of reactant structures and reaction conditions can provide highly selective stereochemical control.", "header_images": [], "subquestions": [{"sub_id": "Exam-17-T9-1", "tags": ["2.4", "1.2"], "text_open": "9.1 Garner's aldehyde is frequently used for stereochemical control in synthesis. The route below prepares it from the natural product S-serine.\n9.1.2 Garner's aldehyde can undergo further nucleophilic addition. Write the SMILES strings of the major products under the different reaction conditions below, representing them as anions and specifying stereochemistry.", "text_images": [{"file": "Exam-17-T9-1-1-prompt.png", "description": "Two-step reaction: 'OC[C@@H](C(O)=O)N'>>'OC[C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O'; 'OC[C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O'>DMSO, (COCl)2, DIPEA>'[H]C([C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O)=O'."}, {"file": "Exam-17-T9-1-2-prompt.png", "description": "Two parallel reactions: '[H]C([C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O)=O'>Nu^->[A]; '[H]C([C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O)=O'>Nu^-, [M]>[B]. The products are labeled only [A] and [B]; their structures are not shown."}], "full_mark": 4}, {"sub_id": "Exam-17-T9-2", "tags": ["2.7", "1.2"], "text_open": "9.1.3 In practice, the two reaction pathways compete. Determine which of the following conditions favor formation of product B, and briefly explain why.\n(1) Solvent: a. THF; b. toluene", "text_images": [{"file": "Exam-17-T9-2-1-prompt.png", "description": "'[H]C([C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O)=O'>nucleophile, solvent, additive>[A]+[B]. The products are labeled only [A] and [B]; their structures are not shown."}], "full_mark": 2}, {"sub_id": "Exam-17-T9-3", "tags": ["2.7", "1.2"], "text_open": "(3) Additive: a. DMPU; b. $ZnCl_2$", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-17-T9-4", "tags": ["1.2", "2.7"], "text_open": "9.1.4 Garner's aldehyde can also be synthesized from R-serine. In the reaction below, replacing Boc with Ts reverses the selectivity. Explain why.", "text_images": [{"file": "Exam-17-T9-4-1-prompt.png", "description": "'CC1(C)N([R])[C@@H](C=O)CO1'>Et-MgBr, THF>'CC1(C)N([R])[C@@H]([C@@H](O)CC)CO1' + 'CC1(C)N([R])[C@@H]([C@H](O)CC)CO1'. The image also states that the product ratio is 1:9 when R=Boc and 8:1 when R=Ts. R is the placeholder group used in the original diagram."}], "full_mark": 2}, {"sub_id": "Exam-17-T9-5", "tags": ["2.4", "1.2"], "text_open": "9.2 Garner's aldehyde can also participate in multicomponent reactions that construct new stereocenters. Write the SMILES string of the product of the reaction below.", "text_images": [{"file": "Exam-17-T9-5-1-prompt.png", "description": "Multicomponent reaction: 'C1(C=CC=C2)=C2NC=C1'.'CC(OC[C@H]1C=O)(C)N1C(OC(C)(C)C)=O'.'O=C1CC(OC(C)(C)O1)=O'>CH3CN, D,L-proline>. No product structure is shown to the right of the arrow."}], "full_mark": 2}]} {"problem_id": "Exam-18-T1", "header": "Problem 1 Structural Chemistry of Aluminosilicates\nThe fundamental structural unit of silicates is the $[SiO_4]$ tetrahedron. Sharing corner oxygen atoms between silicate tetrahedra can form finite polynuclear anions, one-dimensional chains, two-dimensional layers, and three-dimensional infinite frameworks. Zeolite molecular sieves are aluminosilicates with regular channels and cage structures; their frameworks can be regarded as three-dimensional networks of corner-sharing $[SiO_4]$ and $[AlO_4]$ tetrahedra.", "header_images": [], "subquestions": [{"sub_id": "Exam-18-T1-1", "tags": ["1.3"], "text_open": "1.1.4 Describe the long-chain silicate structure with general formula $[Si_4O_{11}]_n^{6n-}$, given that it contains 12-membered rings formed by alternating Si and O atoms.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-18-T1-2", "tags": ["1.3"], "text_open": "1.2 Sodalite belongs to the cubic crystal system. In its conventional unit cell, a sodalite cage (β cage) occupies each vertex and the body center, and the cages are connected by sharing hexagonal faces. In addition, each β cage is connected to other β cages by sharing square faces. Each β cage contains four sodium ions arranged as a regular tetrahedron, with a chloride ion at the center of the tetrahedron; each element has only one crystallographically distinct environment.\n1.2.4 In the β cage of sodalite, the number of Al atoms is the maximum allowed by Lowenstein's rule. State the lattice type of sodalite.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-18-T1-3", "tags": ["1.3"], "text_open": "1.3 In Type A zeolite (LTA), a larger α cage forms at the lattice center. An isolated α cage belongs to the Oh point group and has polyhedral notation [4126886]. The LTA framework may be regarded as a three-dimensional infinite structure in which α cages are connected by sharing regular octagonal faces in a “primitive cubic packing” arrangement. 1.3.1 If each LTA unit cell is regarded as containing 1 α cage, state the number of T atoms in one unit cell of Type A zeolite.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-18-T1-4", "tags": ["1.3", "2.1"], "text_open": "1.4 In the laboratory, 4Å molecular sieve A can be synthesized hydrothermally from sodium silicate, sodium aluminate, and sodium hydroxide.\n1.4.3 Write the overall equation for the synthesis of molecular sieve A.", "text_images": [], "full_mark": 2}]} {"problem_id": "Exam-18-T2", "header": "Problem 2 Fluoride-Ion-Selective Electrode\nThe core sensing element of a fluoride-ion-selective electrode is a LaF$_3$ single-crystal membrane doped with a small amount of Eu$^{2+}$ and sealed into the end of the electrode tube. The tube contains an internal reference solution, generally with fixed concentrations of NaF and NaCl, and an internal Ag/AgCl reference electrode. During doping, Eu$^{2+}$ replaces some La$^{3+}$ lattice sites, creating fluoride-ion vacancies to maintain electroneutrality. When the electrode is immersed in the sample solution, F$^{-}$ undergoes selective exchange at the membrane-solution interface and migrates through the crystal via these vacancies. Because the activities of F$^{-}$ differ across the membrane, the phase-boundary potentials on the two sides also differ; their difference is the membrane potential. Measuring this potential permits rapid and selective determination of fluoride in a sample. The structure is shown below.", "header_images": [{"file": "Exam-18-T2-header-1.png", "description": "Schematic of the apparatus and electrode structures for determining F⁻ in water with a fluoride-ion-selective electrode, comprising panels A, B, and C. Panel A, measurement apparatus: the negative and positive terminals of a high-input-impedance millivoltmeter (mV) are connected by wires to a fluoride-ion-selective electrode and a type 232 saturated calomel electrode with a single salt bridge, respectively. Both electrodes are immersed in liquid consisting of water sample + TISAB at pH≈5.5 in a plastic beaker. A magnetic stir bar with a rotation arrow is at the bottom, and the beaker is placed on a magnetic stirrer. Panel B, enlarged fluoride-ion-selective electrode: a top wire connects through an internal Ag/AgCl reference electrode into the internal reference solution of constant F⁻ activity inside the electrode body. A LaF₃ single-crystal sensing membrane doped with EuF₂ is at the bottom, with F⁻ in the sample solution outside the membrane. Panel C, enlarged type 232 saturated calomel electrode: a wire connects through a platinum wire in contact with Hg to Hg. A filling port, opened during measurement, is beside the platinum wire. Below the Hg are, in sequence, a Hg+Hg₂Cl₂ paste layer, saturated KCl filling solution, and KCl crystals. A porous ceramic liquid junction at the bottom extends into the sample solution. The electrode notation Pt|Hg(l)|Hg₂Cl₂(s)|KCl(saturated)‖sample solution appears at the bottom, with the note “type 232 has a single salt bridge; type 217 has a double salt bridge.”"}], "subquestions": [{"sub_id": "Exam-18-T2-1", "tags": ["3.6"], "text_open": "2.2 For the doped LaF$_3$ single-crystal membrane, the membrane potential is described by the following equation, where $a$ denotes activity and may be approximated by concentration at low F$^{-}$ concentration:\n$$E_{membrane}=\\frac{\\mu_{outside}^{0}-\\mu_{inside}^{0}}{nF}+\\frac{RT}{nF}\\ln\\frac{a_{outside}}{a_{inside}}=k+\\frac{RT}{nF}\\ln\\frac{a_{outside}}{a_{inside}}$$\n2.2.1 Calculate how the millivoltmeter reading changes when the concentration of the ion being measured decreases by a factor of 10.", "text_images": [], "full_mark": 2}]} +{"problem_id": "Exam-18-T3", "header": "Problem 3 Latent Monomers and Controlled Metathesis Polymerization\nControlling the temporal and spatial sequence of polymerization is an important topic in conventional polymer synthesis. Chemists have generally developed “latent catalysts” that switch between inactive and active states in response to external stimuli such as light or heat. Recent research, however, has introduced a new “latent monomer” strategy: isomerization of a photoswitchable molecule generates an active monomer in situ and on demand, enabling efficient control of polymerization.\nThis problem concerns a classic class of highly reactive ring-opening metathesis polymerization (ROMP) monomers: norbornadiene (NBD) derivatives. Under ultraviolet irradiation of a specific wavelength, they undergo intramolecular cycloaddition to form highly strained quadricyclane (QC) structures.", "header_images": [], "subquestions": [{"sub_id": "Exam-18-T3-1", "tags": ["2.4"], "text_open": "3.1 Synthesis and Photothermal Isomerization of the Latent Monomer\nMonomer A, an NBD derivative, can be prepared by reacting cyclopentadiene with diethyl acetylenedicarboxylate at room temperature. Under 313 nm ultraviolet irradiation, A is quantitatively converted into latent monomer B, a QC derivative. Adding silica-coated gold nanobipyramids (AuBPs) to B and irradiating at 850 nm with infrared light regenerates A.\n3.1.1 Write the SMILES strings for A and B.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-18-T3-2", "tags": ["2.4"], "text_open": "3.3 The complete inertness of latent monomers toward the catalyst at room temperature enables the synthesis of complex polymer topologies that are difficult to obtain by conventional metathesis polymerization. The researchers replaced the ethoxycarbonyl groups, $-COOC_2H_5$, in B with pent-4-enyloxycarbonyl groups, $-COO(CH_2)_3CH=CH_2$, to give monomer F. F was mixed with the Grubbs catalyst and reacted at 40℃, releasing gas X and giving a soluble linear polymer G with double bonds in its main chain. G was coated onto a microscope slide placed over a masked LED and then heated to 80℃, rapidly converting the film into an insoluble crosslinked network H. Write the SMILES strings for polymers G and H.", "text_images": [], "full_mark": 4}]} +{"problem_id": "Exam-18-T4", "header": "Problem 4 Complexometric Titration\nEthylenediaminetetraacetic acid (EDTA) is an effective chelating ligand widely used in chemical analysis. Approximately 20 g of a standard reagent is roughly weighed, 250mL of water is added, and the mixture is heated to dissolve the solid, cooled, and mixed thoroughly. The measured pH of the solution is 4.2. The stepwise ionization constants of EDTA are shown below.\n\n| | $K_{a1}$ | $K_{a2}$ | $K_{a3}$ | $K_{a4}$ | $K_{a5}$ | $K_{a6}$ |\n| --- | --- | --- | --- | --- | --- | --- |\n| | $10^{-0.9}$ | $10^{-1.6}$ | $10^{-2.0}$ | $10^{-2.67}$ | $10^{-6.16}$ | $10^{-10.26}$ |", "header_images": [], "subquestions": [{"sub_id": "Exam-18-T4-1", "tags": ["4.5", "3.4"], "text_open": "The standard solution is then standardized. Weigh 0.2500 g of working-reference ZnO that has been ignited to constant mass in a furnace at 800 ℃. Moisten it with a small amount of water, dissolve it in 2 mL of 20% hydrochloric acid solution, and add 100 mL of water. Adjust the solution to pH 8 with hexamethylenetetramine (HMTA) solution, add 10 mL of $(CH_2)_6N_4$-$(CH_2)_6N_4$·HCl buffer to adjust the pH to 10, and then add 5 drops of Eriochrome Black T indicator solution. Titrate with the prepared EDTA sodium-salt solution until the color changes from purple to pure blue. Conduct a blank determination in parallel. After blank correction, the volume of EDTA consumed is $V = 15.35\\ \\mathrm{mL}$.\n4.2.2 The buffering ability of a buffer solution is commonly measured by its buffer capacity, denoted β. It is defined as the amount db in mol of strong base required to raise the pH of 1L of buffer solution by dpH units, with the mathematical expression\n$\\beta = \\frac{db}{dpH} = -2.30[H^+] \\times \\frac{db}{d[H^+]}$\nGiven $pK_b((CH_2)_6N_4) = 4.9$, calculate the buffer capacity of a pH = 10 $(CH_2)_6N_4$-$(CH_2)_6N_4$·HCl buffer in which the analytical concentration of $(CH_2)_6N_4$ is 0.1000 mol/L. Hint: express $b$ as a function of $[H^+]$ and differentiate with respect to $[H^+]$ to obtain $\\frac{db}{d[H^+]}$. B may be used to represent $(CH_2)_6N_4$ in the calculation.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-18-T4-2", "tags": ["4.5", "3.4"], "text_open": "4.2.3 Neglecting volume changes during standardization, the change in solution pH over the entire process must not exceed 0.2 pH unit to ensure an accurate endpoint. Determine the minimum value of $[(CH_2)_6N_4]$.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-18-T4-3", "tags": ["4.1"], "text_open": "4.3 During standardization, why is the $(CH_2)_6N_4$-$(CH_2)_6N_4$·HCl buffer not added directly to adjust the solution to pH=10?", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-18-T4-4", "tags": ["4.5", "1.4"], "text_open": "4.4 The EDTA standard solution standardized in this problem can be used to determine the concentrations of Cu2+ and Mg2+ in a solution by either of the following methods. Method 1: Masking Cu2+ ions. Adjust the solution to pH 5.5 with an HMTA buffer and add excess thiosulfate to mask Cu2+. In this reaction, thiosulfate acts as both a reducing agent and a ligand. Method 2: Selective determination of $Cu^{2+}$. Adjust the solution to pH 5 with an HMTA buffer. Using 5-Br-PADAT as the indicator, selectively determine $Cu^{2+}$ with an EDTP standard solution whose principal component is disodium ethylenediaminetetrapropionate. 4.4.2 Based on the data below, explain why EDTP shows high selectivity for $Cu^{2+}$.\n\n| | $Cu^{2+}$ | $Zn^{2+}$ | $Cd^{2+}$ | $Mg^{2+}$ |\n| --- | --- | --- | --- | --- |\n| Ionic radius/pm | 73 | 74 | 95 | 72 |\n| $\\lg K_{M-EDTP}$ | 15.4 | 7.8 | 6.0 | 1.8 |\n| $\\lg K_{M-EDTA}$ | 18.8 | 16.50 | 16.46 | 8.7 |", "text_images": [], "full_mark": 2}]} +{"problem_id": "Exam-18-T5", "header": "Problem 5 Pharmacokinetics and Molecular Design\nPharmacokinetics quantitatively studies the absorption, distribution, metabolism, and excretion of drugs in living organisms and uses mathematical principles and methods to describe how drug concentration in blood varies with time. Pharmacokinetic parameters reveal pharmacological shortcomings of drug molecules and guide molecular modifications that improve specific properties.\nPropranolol (P), metoprolol (M), betaxolol (B), talinolol (T), and atenolol (A) are β-blockers. Propranolol (P) binds both β1 receptors, found mainly in the heart, and β2 receptors, found mainly in bronchial smooth muscle. At therapeutic doses, the other drugs bind β1 receptors with high selectivity. Although their therapeutic mechanisms are similar, their pharmacological properties differ substantially. Their structures are shown below.\nParameters that may be used in this problem are listed below.\n\n| Drug | Abbreviation | lg $P$ | Conjugate-acid $pK_a$ | $F_a$ | $E_{gi}$ | $E_H$ | $V_d$ / (L·kg$^{-1}$) |\n| --- | --- | --- | --- | --- | --- | --- | --- |\n| Propranolol | P | 3.20 | 9.45 | 0.95 | 0.10 | 0.64 | 3.8 |\n| Metoprolol | M | 2.90 | 9.68 | 0.90 | 0.02 | 0.60 | 5.6 |\n| Betaxolol | B | 2.42 | 9.40 | 0.95 | 0.00 | 0.19 | 1.6 |\n| Talinolol | T | 3.20 | 9.40 | 0.95 | 0.36 | 0.09 | 3.3 |\n| Atenolol | A | 0.25 | 9.60 | 0.50 | 0.00 | 0.01 | 0.70 |", "header_images": [{"file": "Exam-18-T5-header-1.png", "description": "Propranolol (P) has SMILES 'CC(NCC(COC1=CC=CC2=C1C=CC=C2)O)C'; metoprolol (M) has SMILES 'CC(NCC(COC1=CC=C(CCOC)C=C1)O)C'; betaxolol (B) has SMILES 'CC(NCC(COC1=CC=C(CCOCC2CC2)C=C1)O)C'; talinolol (T) has SMILES 'O=C(NC1=CC=C(OCC(CNC(C)(C)C)O)C=C1)NC2CCCCC2'; and atenolol (A) has SMILES 'NC(CC1=CC=C(OCC(CNC(C)(C)C)O)C=C1)=O'."}], "subquestions": [{"sub_id": "Exam-18-T5-1", "tags": ["1.7"], "text_open": "5.3 The second pharmacokinetic process is distribution. After a drug molecule enters the systemic circulation, differences in lipophilicity cause some drugs to enter tissues and even the brain rapidly, while others remain in the blood. To compare drug distribution in the human body, the apparent volume of distribution $V_d$ is defined as follows. When drug distribution reaches dynamic equilibrium, assume that the drug concentration in every tissue and organ equals that in plasma; $V_d$ is the total fluid volume that would be required to contain all the drug in the body at that concentration. The formula is:\n$V_d = \\dfrac{\\text{total amount of drug in the body}\\,(\\mathrm{mg/kg})}{\\text{plasma drug concentration}\\,C\\,(\\mathrm{mg/L})}$\n5.3.2 A standard adult weighing 70kg has an actual total body-fluid volume of only approximately 40 L, including approximately 3 L of plasma. Calculate the approximate $V_d$ of a drug confined entirely to the plasma.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-18-T5-2", "tags": ["2.9"], "text_open": "5.4 The third pharmacokinetic process is clearance, which is divided mainly into hepatic and renal clearance. Highly lipophilic drugs generally tend toward hepatic clearance, and hepatic clearance capacity may be regarded as proportional to $E_h$, whereas highly water-soluble drugs generally tend toward renal clearance.\nThe approximate half-lives of the five drugs are P 4 h, M 5 h, B 18 h, T 12 h, and A 6.5 h. In the absence of other side effects, a drug with a longer half-life is preferred to maintain smooth dosing.\nThe following patients require a β-blocker from the five drugs above. Select the most appropriate drug for each patient and give only its abbreviation.\n5.4.3 Patient C has mild liver disease and cannot use a drug with $E_H$ greater than 0.10.", "text_images": [], "full_mark": 1}]} {"problem_id": "Exam-18-T6", "header": "Problem 6 Low-Valent Compounds of Group 13 Metals · II\nLow-valent compounds of Group 13 elements such as aluminum, gallium, and indium (in the +1 oxidation state) have attracted widespread interest in synthetic chemistry and materials science because of their distinctive electronic structures and reactivities. These species often exist as oligomers, clusters, or coordination complexes. Their structural diversity arises from weak interactions between metal centers and their sensitivity to ligands. This problem focuses on the synthesis and structures of low-valent gallium and indium compounds and their coordination chemistry with Lewis acids, testing your understanding of cluster electron counting, aromaticity criteria, and coordination geometries.", "header_images": [], "subquestions": [{"sub_id": "Exam-18-T6-1", "tags": ["1.4", "1.6"], "text_open": "Part I: Core Frameworks of Gallium Clusters—Incomplete Cubes and Four-Membered Rings\n6.1 An electron-precise tetragallane anion: Ultrasonically prepared “GaI” reacts with $Li(THF)_3Si(SiMe_3)_3$ in toluene at -78°C. Warming affords compound $A = [Li(THF)_4]^+[R_4Ga_4I_3]^-$ ($R = Si(SiMe_3)_3$). The anion of A has an “incomplete-cube” framework: four Ga atoms and three I atoms occupy four and three vertices of a cube, respectively, forming a $Ga_4I_3$ core. Write the SMILES string for the anion of A. Each framework bond is known to be a 2c–2e bond, and every Ga atom obeys the octet rule.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-18-T6-2", "tags": ["1.4"], "text_open": "6.2 Cyclic tetragallane anions: planar and butterfly conformations\n6.2.1 $ArGaCl_2$ ($Ar$ = 2,6-bis(2,4,6-triisopropylphenyl)phenyl) reacts with excess K in diethyl ether to form $B = K_2[Ga_4Ar_2]$. The $Ga_4$ unit in the anion of B is an almost square four-membered ring. Each $K^+$ coordinates to the $Ga_4$ ring and two benzene rings. Write the SMILES string for B.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-18-T6-3", "tags": ["1.5", "1.7"], "text_open": "6.2.2 $RGaCl_2\\cdot THF$ ($R = Si(t-Bu)_3$) reacts with excess Na (or K) in heptane to form $C = Na_2Ga_4R_4(THF)_2$. The $Ga_4$ core of C adopts a butterfly (nonplanar) conformation, and the Ga–Ga bond lengths are nearly equal (average 2.43 Å) and appreciably shorter than those in the tetrahedrane $Ga_4R_4$ (2.57 Å). Two $Na^+(THF)$ units lie on opposite sides of the ring. Using the Hückel rule (Hückel aromaticity when the number of delocalized electrons $=4n+2$), determine whether the anion of C is aromatic and explain why it is nonplanar, in contrast to planar B.", "text_images": [{"file": "Exam-18-T6-3-1-prompt.png", "description": "The image shows the single-step reaction 'Cl[Ga-](Cl)([O+]1CCCC1)[Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C'>Na (ex.), n-C7H16, 100 °C>'CC([Si](C(C)(C)C)(C(C)(C)C)[Ga]12([Ga@]34[Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)[Ga@@]([Ga]35([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)[Na@@]14C6CCCO6)([Na@@]25C7CCCO7)[Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)(C)C'. In the reactant $RGaCl_2\\cdot THF$ ($R = Si(t-Bu)_3$), THF is connected to Ga by a coordinate bond, represented in the SMILES by charge separation as [O+]/[Ga-]. In the product $C = Na_2Ga_4R_4(THF)_2$, the Ga4 ring adopts a butterfly conformation (the @/@@ stereochemical markers are retained in the SMILES), and the two Na+(THF) units lie on opposite sides of the ring."}], "full_mark": 4}, {"sub_id": "Exam-18-T6-4", "tags": ["1.4", "2.4"], "text_open": "6.4 Propeller-shaped molecule: Lithium amide $LiN(SiMe_3)Dipp$ (2.2 eq) reacts with GaCl (1 eq) in toluene/diethyl ether (3:1) to give the cluster $D = [Li(Et_2O)_4][Ga_8(R)_2(R')_4(R'')_2OLi]$, where $R = N(SiMe_3)Dipp$, $R'$ = NHDipp, and $R''$ = NDipp. In the anion, the two central Ga atoms are bonded only to other Ga atoms. Each of the six peripheral Ga atoms bears one terminal N ligand and is also bonded to a bridging N or O atom. Write the SMILES string for the anion of D; $Li^+$ and solvent molecules need not be shown. The authors propose that this cluster is obtained by substitution of the subhalide shown below, with the principal framework retained during substitution.", "text_images": [{"file": "Exam-18-T6-4-1-prompt.png", "description": "The image shows a single-step reaction with the reactant 'Cl[Ga]12Cl[Ga](Cl)(Cl2)[Ga]3([Ga](Cl)(Cl4)Cl[Ga]54Cl)[Ga]51[Ga]6(Cl)Cl[Ga]3(Cl)Cl6', followed by >LiN(SiMe3)Dipp (2.2 eq), PhMe/Et2O>[D]. The SMILES representation of LiN(SiMe3)Dipp is '[Li]N([Si](C)(C)C)C1=C(C(C)C)C=CC=C1C(C)C'. The reactant on the left is a gallium subhalide cluster containing 8 Ga and 12 Cl atoms: 6 terminal Cl atoms, 4 bridging Cl atoms, and 2 Cl atoms shown as short dashed bridges. The lithium amide $LiN(SiMe_3)Dipp$ and (2.2 eq) are written above the arrow, and PhMe/Et2O below it. The product on the right is identified only by the formula $D = [Li(Et_2O)_4][Ga_8(N(SiMe_3)Dipp)_2(NHDipp)_4(NDipp)_2OLi]$; its structure is not drawn."}], "full_mark": 2}, {"sub_id": "Exam-18-T6-5", "tags": ["1.4", "2.4"], "text_open": "Part III: E→E Coordinate Bonds. Some E–E bonds can be viewed as coordinate bonds formed when a lone pair on a low-valent metal(I) center coordinates to a trivalent metal(III) Lewis acid. Special ligands are required to obtain stable monomeric low-valent metal(I) species. The following questions examine two ligand systems—a tridentate borate ligand and crown ethers—and their stabilization of monovalent gallium and indium.\n6.5 Metastable “GaI” reacts with $[Tp(t-Bu)_2]Na$ to give the monomer $E = [Tp(t-Bu)_2]Ga$. E can react further with $GaI_3$ to form the adduct $F = [Tp(t-Bu)_2]Ga\\rightarrow GaI_3$.\n6.5.1 Write the SMILES strings for E and F.", "text_images": [{"file": "Exam-18-T6-5-1-prompt.png", "description": "The image shows the two-step reaction '[H]B1(N2[N]3=C(C(C)(C)C)C=C2C(C)(C)C)N4[N]([Na]3[N]5=C(C(C)(C)C)C=C(C(C)(C)C)N51)=C(C(C)(C)C)C=C4C(C)(C)C'>GaI>[E]; [E]>GaI3, NEt3>[F]'. On the left is the structure of $[Tp(t-Bu)_2]Na$: B is bonded to one H and one N atom from each of three pyrazole rings. Each five-membered pyrazole ring contains two adjacent N atoms, one C=C bond and one C=N bond within the ring, and a t-Bu group on each of two ring carbon atoms. The other N atom of each ring coordinates to the central Na, with two of the N–Na bonds shown as dashed lines. E and F are indicated only by their letter labels; their structures are not drawn."}], "full_mark": 3}, {"sub_id": "Exam-18-T6-6", "tags": ["1.4", "1.1"], "text_open": "6.6 In addition to $[Tp(t-Bu)_2]$, crown ethers such as 18-crown-6 can also stabilize monovalent aluminum-group metals.\n6.6.1 Soluble $InOTf$ reacts with 18-crown-6 in toluene to give adduct G. Write the SMILES string for G and indicate the coordination geometry of In.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-18-T6-7", "tags": ["2.4", "1.7"], "text_open": "6.6.2 Unlike the parent $InOTf$, G rapidly undergoes oxidative addition with $CHCl_3$ to form the $In(III)$ compound H, whereas the parent $InOTf$ does not react under the same conditions. Write the SMILES string for H and explain why G is more reactive than $InOTf$.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-18-T6-8", "tags": ["2.4", "1.4"], "text_open": "6.6.3 Under different conditions, $[In_2Cl_4]$ can form valence isomers:\nIn toluene, it exists as the ion pair J, in which one of the cation and anion has a regular tetrahedral geometry.\nIn THF, it exists as the symmetric molecule K containing an In–In bond, with each In atom five-coordinate.\nIn the presence of dibenzo-18-crown-6, it exists as the $In(I)\\rightarrow In(III)$ donor–acceptor adduct L.\nDibenzo-18-crown-6:\nWrite the SMILES strings for J, K, and L, and indicate the oxidation state of each In atom.", "text_images": [{"file": "Exam-18-T6-8-1-prompt.png", "description": "The image shows the structure of dibenzo-18-crown-6, whose SMILES representation is 'C12=CC=CC=C1OCCOCCOC(C=CC=C3)=C3OCCOCCO2'. The 6 O atoms and 8 methylene bridges alternate within an 18-membered ring. Each of the 2 benzene rings, drawn in Kekulé form with alternating double bonds, is fused into the crown ring through 2 adjacent carbon atoms. All bonds other than those within the benzene rings are single bonds."}], "full_mark": 9}]} -{"problem_id": "Exam-18-T7", "header": "Problem 7 Rearrangement Reactions", "header_images": [], "subquestions": [{"sub_id": "Exam-18-T7-1", "tags": ["2.2", "2.3"], "text_open": "7.1 The following reaction involves a pericyclic reaction. Write the SMILES strings for 3 key intermediates in the reaction pathway.", "text_images": [{"file": "Exam-18-T7-1-1-prompt.png", "description": "The figure shows a one-step reaction: 'ClC(/C(C)=C/CC)=O'.'CC1=CCCC1'>AlCl3>'O=C1[C@@H](C)[C@@](CC)(C)C2=C1CCC2'. In the product, two carbon atoms bear a methyl and an ethyl group, respectively, on wedged bonds. The @/@@ stereochemical descriptors are retained in the SMILES."}], "full_mark": 6}, {"sub_id": "Exam-18-T7-2", "tags": ["2.2", "2.4"], "text_open": "7.2 In different solvents, a carbene can give different rearrangement products. Under irradiation, the following diazo compound forms a carbene and rearranges to ketene A.\n7.2.1 Write the SMILES string for A.", "text_images": [{"file": "Exam-18-T7-2-1-prompt.png", "description": "The figure shows a one-step reaction: 'O=C1OC2=CC=CC=C2C1=[N+]=[N-]'>hν>[A]. The product is labeled only as [A], and its structure is not shown."}], "full_mark": 2}, {"sub_id": "Exam-18-T7-3", "tags": ["2.4"], "text_open": "7.2.2 In an aqueous $10^{-3}$ mol/L $HClO_4$ solution, A undergoes ring closure to form B. Write the SMILES string for B.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-18-T7-4", "tags": ["2.4", "1.2"], "text_open": "7.2.3 In hexane, A undergoes dimerization to give several products. Write the SMILES strings for C, which contains 4 six-membered rings, and D, which contains 2 five-membered rings and 2 six-membered rings. Both products are centrosymmetric.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-18-T7-5", "tags": ["2.2", "2.4"], "text_open": "7.3 Purpurogallin is a natural pigment formed by oxidation of pyrogallol, followed by dimerization and loss of a small molecule.\n7.3.1 Write the SMILES strings for 2 key intermediates in the second transformation.", "text_images": [{"file": "Exam-18-T7-5-1-prompt.png", "description": "The figure shows a three-step reaction: 'OC1=C(O)C(O)=CC=C1'>[O]>'O=C(C=CC=C1O)C1=O'; 'O=C(C=CC=C1O)C1=O'>H2O>'OC1=CC(CC=CC(O)=C2O)=C2C(O)=C1O'; 'OC1=CC(CC=CC(O)=C2O)=C2C(O)=C1O'>[O]>Purpurogallin. The final product is labeled only by name, and its structure is not shown."}], "full_mark": 2}, {"sub_id": "Exam-18-T7-6", "tags": ["2.4", "1.7"], "text_open": "7.3.2 Purpurogallin is known to be an aromatic compound. Write its SMILES string.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-18-T7-7", "tags": ["2.2", "2.7", "2.3"], "text_open": "7.4 Ozonolysis of allylic alcohols shows unusual selectivity.\n7.4.1 Write the SMILES strings for 2 key intermediates in the first step and 1 key intermediate in the second step.", "text_images": [{"file": "Exam-18-T7-7-1-prompt.png", "description": "The figure shows a two-step reaction: 'OC1(C)C=CC(C)(C)CC1'>O3>'CC1(C(O)OO2)OC2C(C)(C)CC1'; 'CC1(C(O)OO2)OC2C(C)(C)CC1'>NaOH>'O=C1C=CC(C)(C)CC1'."}], "full_mark": 6}, {"sub_id": "Exam-18-T7-8", "tags": ["2.4", "2.5", "2.7"], "text_open": "7.4.2 Write the SMILES string for the product of the following reaction. It contains no cyclic structure.", "text_images": [{"file": "Exam-18-T7-8-1-prompt.png", "description": "The figure shows a one-step reaction: 'C/C(C)=C1[C@@](C)(C[C@H](C)CC/1)O'>1) O3, 2) NaOH>[product not shown]. In the starting material, OH is shown on a hashed wedge, one methyl group on a solid wedge, and the opposite methyl group on a dashed bond. The @/@@ stereochemical descriptors are retained in the SMILES."}], "full_mark": 2}]} -{"problem_id": "Exam-18-T8", "header": "Problem 8 Reactions of Amino-Acid Side Chains\nAmino-acid side chains can undergo chemical reactions when they contain functional groups. These groups include hydroxy, phenolic, thiol, indolyl, imidazolyl, guanidino, methylthio, non-α−amino, and non-α−carboxy groups. Each functional group can react with many reagents, and some of these reactions form the basis of chemical modification of proteins. Stereochemistry is not considered in this problem.", "header_images": [], "subquestions": [{"sub_id": "Exam-18-T8-1", "tags": ["2.4", "2.5"], "text_open": "8.2 Toi reaction: In sodium borate buffer at pH 8~9 and 25~35℃, the guanidino side chain of arginine reacts with 1,2−cyclohexanedione to form condensation product E$^+$. E$^+$ contains a newly formed ring. Its chemical formula is $C_{12}H_{23}N_4O_4^+$, and if only hydrogens attached to carbon are considered, there are 6 distinct chemical environments. Write the SMILES string for E$^+$.", "text_images": [{"file": "Exam-18-T8-1-1-prompt.png", "description": "'O=C(O)C(N)CCCNC(N)=N'.'O=C1CCCCC1=O'>>[E+]. No conditions are written above the arrow, and the product structure is not drawn but is labeled E$^+$. The arginine structure is labeled Arginine."}], "full_mark": 2}, {"sub_id": "Exam-18-T8-2", "tags": ["2.2", "2.5"], "text_open": "8.3 Patchornik reaction: Under mild conditions, the indole ring of a tryptophan residue in a protein can be oxidized by NCS and undergo rearrangement, causing cleavage of the peptide bond. Write the SMILES strings for the two key intermediates between the tryptophan residue and intermediate X.", "text_images": [{"file": "Exam-18-T8-2-1-prompt.png", "description": "The figure shows a two-step reaction: 'O=C(N[R2])C(N[R1])CC1=CNC2=CC=CC=C21'>NCS, H2O>'O=C1C(N[R1])CC2(O1)C3=CC=CC=C3NC2Cl'; 'O=C1C(N[R1])CC2(O1)C3=CC=CC=C3NC2Cl'>NCS, H2O>'O=C1C(N[R1])CC2(O1)C3=CC=CC=C3NC2=O'. The intermediate structure containing Cl is labeled X, while the product on the right has no letter label. R1 and R2 are the original placeholders in the figure."}], "full_mark": 4}, {"sub_id": "Exam-18-T8-3", "tags": ["2.2", "2.5"], "text_open": "8.4.1 Hopkins-Cole reaction: A protein solution is mixed with Hopkins-Cole reagent, whose principal active component is glyoxylic acid. Concentrated sulfuric acid is then carefully added down the wall of the tube so that it forms a separate lower layer. If the protein contains tryptophan, a violet ring appears at the interface between the two liquid layers. However, this reaction is not simple or reliable for quantitative determination of tryptophan because product formation is slow, various color changes occur concurrently, and the products have low stability. Studies have shown that tryptophan can form two differently colored products with glyoxylic acid. When a small amount of aldehyde is used and the reaction temperature is below 15°C, the first colored product is carmine. Write the SMILES strings for three key intermediates between the tryptophan residues and intermediate Y.", "text_images": [{"file": "Exam-18-T8-3-1-prompt.png", "description": "The figure shows a two-step reaction: '[R]C1=CNC2=CC=CC=C21'.'[R]C1=CNC2=CC=CC=C21'.'O=C([H])C(O)=O'>'[R]C1=CNC2=CC=C(C(C(O)=O)C3=CC=C(NC=C4[R])C4=C3)C=C21'; '[R]C1=CNC2=CC=C(C(C(O)=O)C3=CC=C(NC=C4[R])C4=C3)C=C21'>[O], -CO2, -H2O>'[R]C1=CNC2=CC=C(/C=C(C=C3)/C=C4C3=NC=C4[R])C=C21'. The intermediate structure after the first arrow is labeled Y. R is the original placeholder in the figure."}], "full_mark": 6}, {"sub_id": "Exam-18-T8-4", "tags": ["2.2", "2.5"], "text_open": "8.4.2 Raising the temperature to approximately 50°C provides sufficient activation energy to promote a further condensation reaction, forming a second blue-violet product. Write the SMILES strings for two key intermediates between the reactant in the scheme and intermediate Z.", "text_images": [{"file": "Exam-18-T8-4-1-prompt.png", "description": "The figure shows a two-step reaction: 'NC(C(O)=O)CC1=CN=C2C1=C/C(C=C2)=C/C3=CC=C(NC=C4CC(N)C(O)=O)C4=C3'>glyoxylic acid, -H2O>'O=C(C(NC1C(O)=O)CC2=C1N=C3C2=C/C(C=C3)=C/C4=CC=C(NC5=C6CC(NC5C(O)=O)C(O)=O)C6=C4)O'; 'O=C(C(NC1C(O)=O)CC2=C1N=C3C2=C/C(C=C3)=C/C4=CC=C(NC5=C6CC(NC5C(O)=O)C(O)=O)C6=C4)O'>-CO2>'O=C(C(NC1)CC2=C1N=C3C2=C/C(C=C3)=C/C4=CC=C(NC5=C6CC(NC5)C(O)=O)C6=C4)O'. The upper-right structure is labeled Z, and the structure of glyoxylic acid is shown above the first arrow."}], "full_mark": 4}]} +{"problem_id": "Exam-18-T9", "header": "Problem 9 Skeletal Reconstruction of Alkaloids\nLate-stage skeletal modification and precise remodeling of natural products provide an ideal shortcut for rapidly expanding compound libraries and discovering breakthrough drugs. In 2026, Amir H. Hoveyda's team reported an isolable, general-purpose platform generated by Hofmann elimination (HE), followed by a general strategy for divergent construction of new nonnatural frameworks through rearrangement, ring-closing metathesis, or intramolecular hydroamination.", "header_images": [], "subquestions": [{"sub_id": "Exam-18-T9-1", "tags": ["2.2", "2.4"], "text_open": "9.1.1 Under specific conditions, anhydrovinblastine reacts with a chlorine source to introduce a chlorine atom and form electrically neutral intermediate A1. In the presence of $AgBF_4$ and solvent, intramolecular reorganization occurs through intermediates A2 and A3, ultimately giving vinorelbine, a widely used clinical anticancer drug. Write the SMILES strings for the key intermediates A1, A2, and A3.", "text_images": [{"file": "Exam-18-T9-1-1-prompt.png", "description": "Organic reaction scheme: '[H][C@]12CN(CCC3=C([C@]([R])(C(OC)=O)C2)NC4=CC=CC=C43)CC(CC)=C1'>AgBF4, THF/H_2O (1/1)>'[H][C@]1(C[C@@]([R])(C(OC)=O)C2=C(C3)C4=CC=CC=C4N2)CN3CC(CC)=C1'. The chlorine source is also drawn above the arrow and has SMILES 'CN1C2=CC=CC=C2N=N1.[Cl]'. R is the original placeholder in the figure."}], "full_mark": 3}, {"sub_id": "Exam-18-T9-2", "tags": ["2.2", "2.4"], "text_open": "9.1.2 Under further basic conditions, nucleophilic conjugate addition introduces an alkynyl aldehyde fragment into vinorelbine to give A4. C-N bond cleavage then forms A5, followed by cyclization and desilylation to afford a two-carbon ring-expanded analogue. Write the SMILES strings for the two key intermediates A4 and A5.", "text_images": [{"file": "Exam-18-T9-2-1-prompt.png", "description": "Organic reaction scheme: '[H][C@]1(C[C@@]([R])(C(OC)=O)C2=C(C3)C4=CC=CC=C4N2)CN3CC(CC)=C1'>'C[Si](C#CC=O)(C)C', Me3Si removal on workup>'[H][C@]1(C[C@]([R])(C(OC)=O)C2=C(C/C(C=O)=C\\3)C4=CC=CC=C4N2)CN3CC(CC)=C1'. R is the original placeholder in the figure."}], "full_mark": 2}, {"sub_id": "Exam-18-T9-3", "tags": ["2.2", "2.3"], "text_open": "9.2 While studying skeletal deconstruction of vinpocetine, the researchers encountered an unexpected product. Reaction of vinpocetine with 5.0 equivalents of allyl iodide efficiently formed a quaternary ammonium salt. Subsequent treatment in potassium ethoxide/ethanol (KOEt/EtOH) at 80°C for 2 h gave not only the expected ring-opened product in 58% yield but also a deallylated secondary-amine byproduct in 32% yield, as shown. Isotopic-labeling experiments demonstrated that formation of the byproduct did not arise from direct $S_N2$ allylic nucleophilic substitution, but proceeded through a rapid and reversible acid-base proton-exchange process.\n9.2.1 Write the SMILES string for the key enammonium cation intermediate B formed after deprotonation of the quaternary ammonium salt and used to explain the deallylated byproduct.", "text_images": [{"file": "Exam-18-T9-3-1-prompt.png", "description": "The figure shows a two-step reaction: '[R]C1=C[C@]2(CC)CCCN([C@@H]23)CCC4=C3N1C5=CC=CC=C45'>5.0 equiv. allyl iodide, MeCN, 3 h, 80 °C>'[R]C1=C[C@@]2(CCC[N@@+]3(CCC4=C([C@H]23)N1C5=CC=CC=C45)C(C=C)[H])CC.[I-]'; '[R]C1=C[C@@]2(CCC[N@@+]3(CCC4=C([C@H]23)N1C5=CC=CC=C45)C(C=C)[H])CC.[I-]'>KOEt, EtOH, + aq. workup>'[R]C(N1C2=C(C(C=C)=C1[C@H]3NCCC4)C=CC=C2)=C[C@@]34CC'. R is the original placeholder in the figure."}], "full_mark": 1}, {"sub_id": "Exam-18-T9-4", "tags": ["2.3", "2.2"], "text_open": "9.2.2 If the KOEt/EtOH system above is replaced by the deuterated reagent system KOEt/EtOD and the vincane-derived quaternary ammonium salt is treated at 80°C, partially deuterated products are obtained. Quantitative nuclear magnetic resonance ($^1H$ NMR) analysis shows unequal deuteration at the allylic positions of the recovered diene substrate: 33% at $H^b$ and 16% at $H^c$. At the most acidic benzylic proton $H^a$, no appreciable deuteration is observed because elimination is too rapid.\nWrite the SMILES string for intermediate C in this kinetic proton-exchange process.", "text_images": [{"file": "Exam-18-T9-4-1-prompt.png", "description": "Organic reaction scheme: '[R]C1=C[C@@]2(CCC[N@@+]3(CCC4=C([C@H]23)N1C5=CC=CC=C45)[C@@H](/C=C/[Hc])[Hb])CC.[I-]'>KOEt, EtOD>'[R]C1=C[C@@]2(CCC[N@@+]3(CCC4=C([C@H]23)N1C5=CC=CC=C45)[C@@H](/C=C/[Dc])[Db])CC.[I-]'. R is the original placeholder, and [Hb], [Hc], [Db], and [Dc] are the original hydrogen/deuterium labels in the figure."}], "full_mark": 1}, {"sub_id": "Exam-18-T9-5", "tags": ["2.7", "1.2"], "text_open": "9.3 When the proton environment of the reaction is changed, a reactive carbanion intermediate can be directed onto an entirely different rearrangement pathway.\n9.3.1 Vincane is reacted with allyl iodide to form intermediate D, which is then treated with KOEt in anhydrous acetonitrile. Rather than undergoing elimination, it gives a C-allylated product in 65% yield with exceptionally high asymmetric stereoselectivity of >98:2 d.r.\nGive the specific name of the concerted rearrangement in this step, and write the SMILES string for intermediate D with the stereochemistry of its stereogenic centers shown explicitly.", "text_images": [{"file": "Exam-18-T9-5-1-prompt.png", "description": "The figure shows a two-step reaction: 'CC[C@]12CCCN3CCC4=C([C@@H]23)N(CC1)C5=CC=CC=C45'>2.0 equiv. allyl iodide, MeCN>[D]; [D]>3.0 equiv. KOEt, (no EtOH), 1 h, 22 °C>'C=CC[C@]12C(N(CC3)C4=CC=CC=C54)=C5CCN1CCC[C@@]32CC'."}], "full_mark": 2}, {"sub_id": "Exam-18-T9-6", "tags": ["2.2", "2.4"], "text_open": "9.3.2 The product from the previous step reacts with MeOTf to form quaternary ammonium salt E, which is again treated with strong base for elimination. DFT calculations show that, to relieve steric compression, a particular C-N bond in E is substantially elongated to 1.61 Å and is therefore highly susceptible to cleavage. Elimination ultimately occurs selectively at the less accessible but more acidic C-H bond, giving the ten-membered-ring exo-bridged diene F. Write the SMILES strings for quaternary ammonium salt E and elimination product F.", "text_images": [{"file": "Exam-18-T9-6-1-prompt.png", "description": "The figure shows a two-step reaction: 'C=CC[C@]12C(N(CC3)C4=CC=CC=C54)=C5CCN1CCC[C@@]32CC'>3.0 equiv. MeOTf, DCE, 16 h, 70 °C>[E]; [E]>3.0 equiv. KOEt, EtOH, 3 h, 80 °C>[F]. E and F are shown only as labels; their structures are not drawn."}], "full_mark": 2}, {"sub_id": "Exam-18-T9-7", "tags": ["2.2", "2.4"], "text_open": "9.4 While attempting ring-closing metathesis (RCM) of an oxidized tetrahydropalmatine derivative to construct a medium-sized oxygen-containing ring, the research team encountered an unusual reverse process. When the Lewis acid $BPh_3$ was added to the W catalyst, the reaction did not give the expected ten-membered-ring metathesis product. Instead, it afforded a ring-contracted isomer bearing a stereogenic vinyl substituent in 61% yield.\nA Lewis acid-base adduct forms first, followed by an extremely rare reverse Meisenheimer rearrangement (RMR) to give boron-complex intermediate G. Write the SMILES strings for G and the boron-containing organic byproduct formed with the product.", "text_images": [{"file": "Exam-18-T9-7-1-prompt.png", "description": "The figure shows a three-step reaction: 'C=CC(C=C1OC)=C(C=C1OC)[C@@H]2CC3=CC=C(C(OC)=C3CN2OCC=C)OC'>1.2 equiv. BPh_3, Benzene, 15 h, 60 °C, W10%>'COC1=CC2=C(C=C1OC)/C=C\\C[O+]([B-](C3=CC=CC=C3)(C4=CC=CC=C4)C5=CC=CC=C5)N(CC6=C7OC)[C@H]2CC6=CC=C7OC'; 'COC1=CC2=C(C=C1OC)/C=C\\C[O+]([B-](C3=CC=CC=C3)(C4=CC=CC=C4)C5=CC=CC=C5)N(CC6=C7OC)[C@H]2CC6=CC=C7OC'>[G]; [G]>'C=C[C@@H](C1=C2C=C(OC)C(OC)=C1)N3CC4=C(C[C@@H]23)C=CC(OC)=C4OC', with a branch arrow pointing to byproduct ?."}], "full_mark": 2}, {"sub_id": "Exam-18-T9-8", "tags": ["2.3", "2.4"], "text_open": "9.5 While attempting to isomerize the double bonds of the substrate above, the researchers found that treatment in a $KO^tBu$/DMSO system at room temperature (22°C) gave quantitative conversion in only 30 min, with >98% yield, to an aryl-substituted dehydroisoquinoline derivative. In subsequent biological-activity tests, this molecule showed remarkably potent anticancer activity at nanomolar concentrations.\nTheoretical calculations (DFT) and deuterium-control experiments showed that rapid elimination first forms the $(DMSO)_3K^+$-associated π-allyl-type conjugated complex intermediate $H_1$. With assistance from the potassium ion, this intermediate then undergoes a thermodynamically highly favorable anionic [1,6]-hydrogen proton-transfer process, an 8-electron pericyclic reaction with an activation barrier of only 15.9 kcal/mol, to form $H_2$. Write the SMILES strings for intermediates $H_1$ and $H_2$.", "text_images": [{"file": "Exam-18-T9-8-1-prompt.png", "description": "The figure shows a three-step reaction: 'C=CC(C=C1OC)=C(C=C1OC)[C@@H]2CC3=CC=C(C(OC)=C3CN2OCC=C)OC'>1.0 equiv. KOt-Bu, DMSO, 0.5 h, 22 °C>[H1]; [H1]>[H2]; [H2]>'[H]CCC(C=C1OC)=C(C=C1OC)C2=CC3=CC=C(C(OC)=C3C=N2)OC'."}], "full_mark": 2}]} +{"problem_id": "Exam-19-T1", "header": "Question 1 Chemistry in Language\nMāori is the language of the indigenous Māori people of New Zealand and belongs to the Malayo-Polynesian branch of the Austronesian language family. Whereas element names in many other languages are borrowed from English or Latin, a number of Māori element names are constructed from native morphemes. The Chinese and English names of several elements and their Māori equivalents are given below. Examine these element names, infer the word-formation rules, and use those rules to answer the questions that follow.\n\n| Chinese name (English translation) | English | Māori | Chinese name (English translation) | English | Māori |\n| --- | --- | --- | --- | --- | --- |\n| hydrogen | hydrogen | hauwai | helium | helium | haumāmā |\n| lithium | lithium | konukōhatu | carbon | carbon | waro |\n| sodium | sodium | konutai | oxygen | oxygen | hāora |\n| chlorine | chlorine | haumāota | calcium | calcium | konupūmā |\n| chromium | chromium | konukita | nickel | nickel | konukōreko |\n\nFor reference, English translations of several Māori words are provided below:\n\n| | | | |\n| --- | --- | --- | --- |\n| hā — breath | ora — alive | wai — water | tai — sea |\n| kōhatu — stone | kita — bright | uku — clay | kōreko — shiny |\n| mohe — malleable | pūmā — grayish white | māota — fresh green, vitality | |", "header_images": [], "subquestions": [{"sub_id": "Exam-19-T1-1", "tags": ["1.7"], "text_open": "1.2 One day, a Māori friend tells you about several elements and their compounds. Infer their meanings from the context.\nHygroscopic solid A contains only the elements hautawa and konumohe. It fumes in air and sublimes under vacuum. Thorough heating of A in air produces compound B together with colored fumes (Reaction 1). B is also an important industrial raw material for producing konumohe, but New Zealand relies mainly on imports of this material.\nElemental konumohe has many uses. Dissolve it in a caustic soda solution, adjust the pH until precipitate C forms, and collect the precipitate by filtration. Adding soda ash and C to an aqueous solution of the simple monoprotic acid corresponding to haukōwhai, followed by recrystallization, gives a 3:1 crystal D (Reaction 2). D has a lower melting point than B and can be used in the metallurgical industry.\nThe Māori names haukōwhai and hautawa are both derived from their colors. The two elements belong to the same group of the periodic table, but their elemental substances have different physical states at room temperature.\nElemental haukōwhai reacts with 2% caustic soda solution to form mainly colorless gas E (Reaction 3). The molecule of E has a structure similar to that of water and reacts under acidic conditions with a simple potassium salt of hautawa to form elemental hautawa (Reaction 4).\n1.2.1 From the analysis, directly give the element names corresponding to hautawa, konumohe, and haukōwhai, and the chemical formulas of A–E.", "text_images": [], "full_mark": 8}, {"sub_id": "Exam-19-T1-2", "tags": ["2.1"], "text_open": "1.2.2 Write the equations for Reactions 1–4.", "text_images": [], "full_mark": 6}, {"sub_id": "Exam-19-T1-3", "tags": ["1.7"], "text_open": "1.3.2 Some people argue that the prefix hau in hautawa is inconsistent with the properties of the element. Do you agree? Explain your reasoning.", "text_images": [], "full_mark": 2}]} +{"problem_id": "Exam-19-T2", "header": "Question 2 Photochromic Organic Transistor\nOrganic field-effect transistors (OFETs) offer advantages such as flexibility, printability, and low-temperature fabrication, and are widely used in flexible displays, sensors, and large-area electronic circuits. Organic molecules with distinctive photochemical properties can therefore be used to make light-controlled switches or transistors. Researchers fabricated a photochromic organic transistor using a DAE film as the transistor channel layer. The device supports reversible writing and erasing. The molecular structure of DAE, the organic transistor structure, and the structure of a conventional bipolar transistor are shown below.\nHere, Channel means the conduction channel. From a purely functional perspective, the Source (S terminal) of the field-effect transistor is the source electrode and corresponds to emitter E of the bipolar transistor; Gate (G terminal) is the gate electrode and corresponds to base B; Drain (D terminal) is the drain electrode and corresponds to collector C.", "header_images": [{"file": "Exam-19-T2-header-1.png", "description": "The SMILES representation of DAE is 'FC1(F)C(F)(F)C(C2=C(C)SC(C3=CC=C(C4=CC=CC=C4)C=C3)=C2)=C(C5=C(C)SC(C6=CC=C(C7=CC=CC=C7)C=C6)=C5)C1(F)F'."}, {"file": "Exam-19-T2-header-2.png", "description": "The figure is a structural schematic of an organic field-effect transistor (OFET). From bottom to top it shows a p+-Si substrate (with a lead from the bottom connected to a variable DC supply and labeled Gate), a SiO₂ layer, Au electrodes on the left and right, and a 1D channel spanning the two electrodes. The left electrode is grounded by a lead and labeled Source; the right electrode is connected by a lead to a variable DC supply and labeled Drain. Above the device, the tip of a probe (nanomanipulator/AFM probe) points vertically toward the 1D channel. A translucent cover layer at upper right is labeled DAE and PMMA. The lower right is labeled XY-axes stage and includes arrows showing three-dimensional directions of motion."}, {"file": "Exam-19-T2-header-3.png", "description": "The figure shows the circuit symbol of a conventional bipolar transistor. Inside a circle is an inverted trapezoidal arrangement. A vertical line at the top connects to B (base), a horizontal line at the right connects to C (collector), and a horizontal line at the left connects to E (emitter). The end of the E line carries an arrow pointing toward the lower diagonal line inside the circle."}], "subquestions": [{"sub_id": "Exam-19-T2-1", "tags": ["3.5", "1.7"], "text_open": "The channel contains a mixture of DAE and its isomer X. Let α denote the fraction of X in the channel. Experiments show that after sufficiently prolonged irradiation at wavelength λ1, the maximum isomer fraction α_{max} = 0.75. If a further 40s scan at wavelength λ2 is then applied, α decreases by 0.10. After sufficiently prolonged irradiation at wavelength λ2, the minimum isomer fraction α_{max} = 0.10. If a further 40s scan at wavelength λ1 is then applied, α increases by 0.20. For a reversible first-order isomerization, \\alpha(t) = \\alpha_{\\infty} + (\\alpha_0 - \\alpha_{\\infty})e^{-kt}. Here k is the apparent rate constant, \\alpha_{\\infty} is the isomer fraction at the photostationary state, and \\alpha_ is the isomer fraction at t = 0.\n2.3 In the saturation region, the magnitude of the drain current through a single channel satisfies\n$$|I_D|=\\frac{μC_iW}{2L}(V_G-V_T)^2$$\nwhere W is the channel width; L is the channel length (the length of semiconductor traversed by the current from S to D), taken as 40 μm in this part; μ is the carrier mobility in the saturation region, which describes the overall rate of motion of electrons and holes through the material; $C_i$ is the capacitance per unit area of the insulating layer; $V_G$ is the gate voltage; and $V_T$ is the threshold voltage, which may be approximated as a constant independent of $V_G$ and μ. For n identical channels, approximately\n$$|I_D|=\\frac{μnC_iW}{2L}(V_G-V_T)^2$$\nIn this problem, the saturation-region carrier mobility may be taken as proportional to α.\n2.3.1 The source is grounded and the drain voltage $V_D$ is fixed so that the transistor remains in the saturation region. After irradiation with ultraviolet and visible light of specified wavelengths, respectively, the gate voltage $V_G$ is scanned while the drain current is monitored. The results are shown in the left-hand figure below. The region 0 > $V_G$ > -20V is outside the saturation region and is irrelevant to this question. The transistor properties differ markedly after the two irradiations. Determine whether the transistor after ultraviolet irradiation corresponds to curve group A or B, and briefly explain why. (Figure labels: Current; Voltage; Drawing Cycle.)", "text_images": [{"file": "Exam-19-T2-1-1-prompt.png", "description": "The figure is a coordinate plot (left-hand figure for 2.3.1). The vertical axis is labeled 'Drain Current (pA)', with tick labels -8, -6, -4, -2, 0, and 2 from top to bottom. The horizontal axis is labeled 'Gate Voltage (V)', with tick labels 0, -20, -40, and -60 from left to right. Several dark solid curves nearly overlap from 0 to -20 V and separate into two groups beyond -20 V. Group A (labeled 1 and 3,5) has currents of approximately -6 pA and -4 pA at -60 V. Group B (labeled 6 and 2,4,0) has currents between 0 and 1 pA at -60 V. The upper left is labeled 'UV: 25 mW/cm²' and 'VIS: 500 mW/cm²'."}, {"file": "Exam-19-T2-1-2-prompt.png", "description": "The figure is a coordinate plot (right-hand figure for 2.3.1). The vertical axis is labeled 'Drain Current (pA)', with tick labels 2, 0, -2, -4, -6, and -8 from bottom to top. The horizontal axis is labeled 'Drawing Cycle', with ticks at 0, 5, 10, 15, and 20. A black solid polyline with filled black square markers oscillates in a sawtooth pattern between approximately -0.5 and -6 pA. The absolute magnitudes of the peaks and troughs decrease gradually as the cycle number increases. Gray arrows above each peak and below each trough indicate application of light. No legend is present."}], "full_mark": 2}, {"sub_id": "Exam-19-T2-2", "tags": ["3.6", "3.8"], "text_open": "Starting from the $\\lambda_2$ photostationary state, two consecutive $\\lambda_1$ scans under the stated conditions give an isomer-X fraction $\\alpha=0.44$.\nProlonged irradiation at $\\lambda_1$ makes the isomer-X fraction $\\alpha$ reach 0.75, whereas prolonged irradiation at $\\lambda_2$ makes $\\alpha$ reach 0.10.\n2.3.3 For a transistor with n = 1, after prolonged scanning at wavelength $λ_1$, two data points in the saturation region of its $I_D$–$V_G$ curve are measured: (-40 V, -2.0 pA) and (-50 V, -3.9 pA). A transistor with n = 3 is irradiated for a long time at wavelength $λ_2$ and then subjected to two consecutive 40 s scans at wavelength $λ_1$. Neglecting losses, calculate $I_D$ at $V_G$ = -45 V.", "text_images": [], "full_mark": 5}, {"sub_id": "Exam-19-T2-3", "tags": ["3.6"], "text_open": "2.4 This transistor can be used to construct a light-controlled analog adder. Its structure is shown below. The three branches (1, 2, and 3) forming the Y share one gate G, with $V_G$ = -40 V. The distance between S and D is 60 μm. Branch 1 contains 3 channels and branch 2 contains 2 channels. Every channel operates in the saturation region and has the same width. The section connecting the three channels (the section parallel to S and D) lies midway between S and D and may be treated as an ideal conductor. Initially, every channel has been irradiated for a long time at wavelength $λ_1$. (Hint: note the definition of L. Neglect losses.)\n2.4.1 Calculate the initial magnitude of the drain current.", "text_images": [{"file": "Exam-19-T2-3-1-prompt.png", "description": "The figure is a schematic of the Y-shaped light-controlled analog adder, shown as a three-dimensional rectangular trough-like device. The left side is labeled S and connected by a bent lead to a ground symbol. The right side is labeled D and connected by a bent lead to a DC-source symbol. The front of the bottom is labeled G and connected by a vertical lead to a ground symbol. Internal stepped layered regions labeled 1, 2, and 3 correspond to the three branches forming the Y; branches 1 and 2 converge into branch 3."}], "full_mark": 3}, {"sub_id": "Exam-19-T2-4", "tags": ["3.6", "3.8"], "text_open": "2.4.2 For a channel in which μ varies, the equivalent mobility of the channel can be obtained by averaging μ as follows:\n$$μ_{eff}=\\frac{L_1+L_2}{\\frac{L_1}{μ_1}+\\frac{L_2}{μ_2}}$$\nEach channel in branch 1 is now subjected to one 40 s scan at wavelength $λ_2$. Calculate the change in the magnitude of the drain current.", "text_images": [], "full_mark": 3}]} +{"problem_id": "Exam-19-T3", "header": "Question 3 Degradable Thermosetting Resin\nConventional thermosetting polymers form covalently cross-linked networks that provide exceptional mechanical strength and heat resistance, but also prevent melt processing and make recycling and degradation difficult. In recent years, orthogonal polymerization of different functional groups within a single monomer has emerged as a leading strategy for synthesizing polymers that can be degraded at predetermined sites. 2,3-Dihydrofuran (DHF) is a monomer that can be prepared from bio-based feedstocks. Recently, chemists used DHF in a one-pot process involving two successive polymerizations with entirely different mechanisms to produce a thermosetting resin that can be recycled through multiple pathways.", "header_images": [], "subquestions": [{"sub_id": "Exam-19-T3-1", "tags": ["2.4"], "text_open": "3.1 In the presence of a trace amount of second-generation Grubbs catalyst, DHF slowly undergoes ring-opening metathesis polymerization (ROMP) in the dark to give rubbery linear polymer A. In the presence of a photoacid generator (PAG, which produces the superacid HSbF6 upon irradiation), it rapidly undergoes cationic polymerization to give highly rigid linear polymer B.\n3.1.2 Write the SMILES strings for linear polymers A and B.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-19-T3-2", "tags": ["2.4"], "text_open": "3.2 To prepare a thermosetting resin, the researchers mixed DHF, second-generation Grubbs catalyst, and PAG. The mixture was first allowed to react in the dark for 10 minutes to form an appropriate amount of polymer A, then irradiated with 456 nm blue light for 6 minutes to initiate cationic polymerization. This produced cross-linked thermosetting resin C, which is insoluble and infusible. Write the SMILES string for C when the two polymer chains are fully cross-linked, with no unreacted cross-linking sites. Hint: to simplify the structure, assume that both linear polymers A and B participating in the reaction have degree of polymerization n.", "text_images": [], "full_mark": 2}]} {"problem_id": "Exam-19-T4", "header": "Question 4 Determination of Iron by Titanium(III) Chloride–Potassium Dichromate Titration\nFerromanganese-silicon alloy is widely used as a deoxidizer and alloying additive in steelmaking, and its iron content strongly affects its quality. Potassium dichromate titration is a classical method for determining iron. After acid decomposition of the sample, titanium(III) chloride reduces $Fe^{3+}$ to $Fe^{2+}$; the solution is then titrated with standard potassium dichromate using sodium diphenylamine sulfonate as the indicator. This problem covers sample decomposition, prereduction, removal of interferences, and blank correction.", "header_images": [], "subquestions": [{"sub_id": "Exam-19-T4-1", "tags": ["4.1"], "text_open": "4.1 Sample decomposition\nPlace a 0.5026 g ferromanganese-silicon alloy sample in a platinum dish. Add 10 mL of concentrated nitric acid, then add approximately 2 mL of concentrated hydrofluoric acid dropwise in portions. Heat gently until the sample is completely decomposed. Add 5 mL of concentrated perchloric acid and heat until faint perchloric acid fumes appear, then remove from heat. Transfer the solution to a 250 mL conical flask and continue heating until nearly dry (syrupy), then remove from heat. Add 20 mL of hydrochloric acid (1+1), 1–2 drops of hydrogen peroxide (27%), and approximately 40 mL of water. Dissolve the salts until the test solution is clear, heat to boiling until large bubbles evolve, remove from heat, and cool to room temperature. Hint: high-valent manganese oxides are strong oxidants under acidic conditions and, being insoluble in water, adsorb iron ions and interfere with the determination of iron.\n4.1.2 State the purpose of fuming with perchloric acid and of subsequently continuing to heat the solution until nearly dry.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-19-T4-2", "tags": ["2.1", "4.1"], "text_open": "4.2 Reduction and titration\nDilute the test solution with water to approximately 120 mL and add four drops of sodium tungstate solution (250 g/L). Add titanium(III) chloride solution (1.5%–2%) dropwise until the solution turns blue, then add a further 1–2 drops. With continuous swirling, titrate with potassium dichromate solution (1.250 mmol/L) until the solution becomes colorless, adding slowly near the endpoint; do not record this reading. Immediately add 10 mL of sulfuric acid–phosphoric acid mixture (sulfuric acid:phosphoric acid:water = 2:3:5), then add 3 drops of sodium diphenylamine sulfonate indicator solution (2 g/L). Titrate with standard potassium dichromate solution (2.500 mmol/L) to a violet-blue endpoint and record the consumed volume C.\n4.2.2 Write the reaction equations for the addition of titanium(III) chloride solution and explain why a further 1–2 drops of titanium(III) chloride must be added.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-19-T4-3", "tags": ["4.1", "4.2"], "text_open": "4.2.4 Which statements about the functions of adding the sulfuric acid–phosphoric acid mixture are correct?\nA. It provides the acidity required for titration and prevents hydrolysis of metal ions.\nB. Phosphoric acid forms a colorless complex with iron ions, eliminating interference from their yellow color.\nC. It lowers the potential of the $Fe^{3+}$/$Fe^{2+}$ redox couple, making the titration jump more pronounced.\nD. It decolorizes sodium tungstate, preventing interference with the subsequent titration.", "text_images": [], "full_mark": 3}]} {"problem_id": "Exam-19-T8", "header": "Question 8 Remarkable Phosphorus Heterocycles\nIntroductory heterocyclic chemistry most often focuses on N, O, and S heteroatoms. Phosphorus, however, has a large valence-shell space and a rich range of coordination numbers and oxidation states. It can form low-coordinate λ species as well as hypercoordinate $\\lambda^5$ species, where λ denotes the number of bonds to P. Phosphorus heterocycles exhibit remarkable structures and properties and are uniquely valuable in studies of low-valent metal ligands and tunable fluorophores. (Stereochemistry need not be considered in this problem.)", "header_images": [], "subquestions": [{"sub_id": "Exam-19-T8-1", "tags": ["1.6", "4.1"], "text_open": "8.1.2 Which of the following measures are commonly used to evaluate the aromaticity of a compound?\n(a) C–X–C bond angle (b) Degree of bond-length equalization (c) Nucleus-independent chemical shift calculation (d) Standard molar enthalpy of formation", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-19-T8-2", "tags": ["2.4"], "text_open": "8.1.4 P–C bonds are highly reactive. Write the SMILES string for product A from the following 1:1 addition reaction using the specified SMILES.", "text_images": [{"file": "Exam-19-T8-2-1-prompt.png", "description": "'C1(C2=CC=CC=C2)=CC(C3=CC=CC=C3)=CC(C4=NC=CC=C4)=P1'.'FC(C#CC(F)(F)F)(F)F'>100°C>[A]. The structure of product A is not drawn; it is labeled only as A."}], "full_mark": 2}, {"sub_id": "Exam-19-T8-3", "tags": ["2.2", "2.4"], "text_open": "8.1.5 The following reaction ultimately gives product B in which the phosphabenzene ring has been disrupted. Write the SMILES strings for Int and B. B has no mirror plane.", "text_images": [{"file": "Exam-19-T8-3-1-prompt.png", "description": "The figure shows a two-step reaction: 'CC(C=C1)=CC=C1C2=CC(C3=CC=CC=C3)=PC(C4=CC=CC=C4)=C2'>PhLi, THF, -78°C to r.t.>[Int]; [Int]>H2O, -LiOH>[B]. The structures of Int and B are not drawn; they are labeled only as Int and B."}], "full_mark": 3}, {"sub_id": "Exam-19-T8-4", "tags": ["2.2", "2.4"], "text_open": "8.2.1 Several routes are available for synthesizing $\\lambda^3$-phosphabenzenes and their derivatives. Early chemists began with phosphole and used relatively harsh conditions. Write the SMILES strings for intermediates C and D in the following synthesis.", "text_images": [{"file": "Exam-19-T8-4-1-prompt.png", "description": "The figure shows a three-step reaction: 'C1(C2=CC=CC=C2)=CC=C(C3=CC=CC=C3)P1C4=CC=CC=C4'>230°C, [1,5]-σ>[C]; [C].'[R]C#C[R]'>>[D]; [D]>230°C, -CPh2>'[R]C1=PC(C2=CC=CC=C2)=CC=C1[R]'. The structures of intermediates C and D are not drawn; they are labeled only as C and D."}], "full_mark": 3}, {"sub_id": "Exam-19-T8-5", "tags": ["2.2", "2.4"], "text_open": "8.2.2 Milder synthetic routes were later developed, beginning with preparation of $\\lambda^5$ phosphorus intermediate a.\n8.2.2.1 Given that F is an ionic compound, write the SMILES strings for intermediates E and F and identify, in order, the types of reactions occurring from F to a.", "text_images": [{"file": "Exam-19-T8-5-1-prompt.png", "description": "The figure shows four steps: 'ClP(/C=C/OCC)N(C)C'>1) 'C=CC[Mg]Cl', THF, -35°C; 2) MeOH, r.t.>[E]; [E]>1) H2O2·CO(NH2)2; 2) MeOTf, DCM, -70°C>[F]; [F]>tBuOK, -70°C>[a]; [a]>[H]>'C1=CP=CC=C1'. The structures of E and F are not drawn and are labeled only as E and F; a has structure 'COP1(OC)=CC=CC=C1'."}], "full_mark": 6}, {"sub_id": "Exam-19-T8-6", "tags": ["2.2", "2.4"], "text_open": "8.3 Phosphenium ions are often regarded as cationic heavy-element analogs of carbenes, but extending classical carbocation rearrangements such as the Wagner–Meerwein rearrangement directly to phosphorus chemistry remains challenging. Researchers proposed that a carbocation generated adjacent to P in a phosphirane might rearrange to a phosphenium ion, opening a new ring-forming pathway. In this problem, $W(CO)_5$ coordinated to P does not participate directly in the transformations.\n8.3.1 The following reaction gives different major products at different temperatures. Write the SMILES strings for G and H, knowing that neither product contains a C=P double bond.", "text_images": [{"file": "Exam-19-T8-6-1-prompt.png", "description": "The figure shows two parallel reactions of the same reactant: 'O=C(C1=CC=CC=C1)[P+]2([W-])CC2.[C-]#[O+].[C-]#[O+].[C-]#[O+].[C-]#[O+].[C-]#[O+]'>AlCl3, toluene, 100°C, 1h>[G]; the same reactant>AlCl3, toluene, 0°C, 1h>[H]. The structures of G and H are not drawn; they are labeled G and H with molecular formulas C14H8ClO5PW and C21H16ClO5PW."}], "full_mark": 4}, {"sub_id": "Exam-19-T8-7", "tags": ["2.4"], "text_open": "8.3.3 Write the SMILES string for the major product of the following reaction, whose mechanism resembles that forming G.", "text_images": [{"file": "Exam-19-T8-7-1-prompt.png", "description": "The figure shows the one-step reaction 'O=C(C1=CC=C(OC)C=C1)[P+]2([W-])C(C)C2.[C-]#[O+].[C-]#[O+].[C-]#[O+].[C-]#[O+].[C-]#[O+]'>AlCl3, toluene, r.t.>product. The product structure is not drawn and no letter label is assigned."}], "full_mark": 2}, {"sub_id": "Exam-19-T8-8", "tags": ["2.4"], "text_open": "8.3.4 To verify the existence of a phosphenium-ion intermediate, the researchers carried out the following reaction. Write the SMILES string for the major product, knowing that it contains four rings.", "text_images": [{"file": "Exam-19-T8-8-1-prompt.png", "description": "The figure shows the one-step reaction 'O=C(C1=C(OC)C=CC=C1)[P+]2([W-])CC2.[C-]#[O+].[C-]#[O+].[C-]#[O+].[C-]#[O+].[C-]#[O+]'>AlCl3, toluene, 60°C, 1h>product. The product structure is not drawn; molecular formula C21H15O6PW is shown."}], "full_mark": 3}]} -{"problem_id": "Exam-19-T9", "header": "Question 9 The Appeal of Self-Immolative Reactions\nFragmentation is an important and elegant class of skeletal-reorganization processes in organic chemistry. Classical fragmentations are usually controlled by specific electronic effects and cleave otherwise inert chemical bonds through concerted electron transfer.", "header_images": [], "subquestions": [{"sub_id": "Exam-19-T9-1", "tags": ["2.2", "2.3"], "text_open": "9.1 Write the SMILES strings for at least 2 key intermediates in the following Eschenmoser–Tanabe fragmentation.", "text_images": [{"file": "Exam-19-T9-1-1-prompt.png", "description": "The figure shows the one-step sequence 'CC1CC2(OC2C(C1)=O)C'>1. TsNHNH2, rt; 2. 55°C, 2h, 50%>'CC(CC(CC#C)C)=O'."}], "full_mark": 4}, {"sub_id": "Exam-19-T9-2", "tags": ["2.4", "2.2"], "text_open": "In recent years, chemists have introduced the concept of fragmentation driven by a cascade of electron transfers into drug delivery and smart sensing, developing self-immolative reactions. Such systems usually consist of a trigger group, a conjugated linker, and a leaving payload. When a specific stimulus activates the trigger, long-range concerted fragmentation begins within the molecule, causing spontaneous disassembly and release of the target molecule.\n9.2 In the following molecule, reduction serves as the specific stimulus. After neutral intermediate A forms, self-immolation occurs immediately. Write the SMILES string for intermediate A.", "text_images": [{"file": "Exam-19-T9-2-1-prompt.png", "description": "The figure shows two steps: 'O=C(NC1=CC=C(CC(NCCN(C)C)=O)C=C1)OCC2=CC=C([N+]([O-])=O)C=C2'>60Co γ-ray, iPrOH/H2O>[A]; [A]>H2O>'O=C(NC1=CC=C(CC(NCCN(C)C)=O)C=C1)OCC2=CC=C(NO)C=C2'.'CO2'.'OCC1=CC=C(NO)C=C1'. Intermediate A is represented only by a bold boxed A and its structure is not shown; CO2 appears as a text label."}], "full_mark": 2}, {"sub_id": "Exam-19-T9-3", "tags": ["2.4", "2.1"], "text_open": "9.3 A slight modification of the substrate in 9.2 allows more complex products to be obtained through a cascade reaction in $H_2O$. The reaction produces 2 inorganic and 2 organic products. Write the SMILES strings for the 2 organic products and the formulas of the 2 inorganic products.", "text_images": [{"file": "Exam-19-T9-3-1-prompt.png", "description": "The figure shows the one-step reaction 'O=C(NC1=CC=C([N+]([O-])=O)C=C1)OCC2=CC=C(B(O)O)C=C2'>H2O2>[products not shown]."}], "full_mark": 6}, {"sub_id": "Exam-19-T9-4", "tags": ["2.4", "1.2"], "text_open": "9.4 Ring formation is fundamental to organic reactions. Certain ring-forming reactions can also serve as specific stimuli that initiate self-immolation.\n9.4.1 In the reaction above, the substrate undergoes Michael addition to give B. Self-immolation of B forms seven-membered-ring compound C. Write the SMILES string for C and the chemical formula of the small inorganic molecule produced.", "text_images": [{"file": "Exam-19-T9-4-1-prompt.png", "description": "The figure shows the one-step reaction 'C=CC(OC1=CC=C(COC(NC2=CC=CC=C2)=S)C=C1)=O'.'SC[C@H](N)C(O)=O'>PBS (PH=7.4, 10mM)>[B]. The cysteine thiol is shown with a dashed wedge to indicate stereochemistry, retained by the @ marker in the SMILES. The product is labeled only B and its structure is not shown."}], "full_mark": 3}, {"sub_id": "Exam-19-T9-5", "tags": ["2.4", "2.3"], "text_open": "9.4.2 Ring formation is not the endpoint; a pericyclic reaction follows. This strategy uses the inverse-electron-demand Diels–Alder reaction (IEDDA) between a trans-cyclooctene derivative (TCO) and a tetrazine derivative to trigger self-immolation. In the reaction below, IEDDA between TCO and tetrazine releases a gas and forms a dihydropyridazine intermediate. After the pericyclic reaction, an N atom of the dihydropyridazine acts as the initiating group for self-immolation, producing 2 aromatic organic compounds. Write the SMILES strings for both products.", "text_images": [{"file": "Exam-19-T9-5-1-prompt.png", "description": "The figure shows the one-step reaction 'CC1=NN=C(C)N=N1'.'O=C(NC1=CC=CC=C1)OC2CCCCC/C=C/2'>>[products not shown]."}], "full_mark": 4}, {"sub_id": "Exam-19-T9-6", "tags": ["2.3", "2.8"], "text_open": "9.4.3 The dihydropyridazine intermediate must undergo a formal σ-migration rearrangement before it can initiate 1,4-elimination, the subsequent self-immolation. Name this rearrangement and explain why it is faster in water than in an organic phase.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-19-T9-7", "tags": ["2.2", "2.4"], "text_open": "9.5 The following substance can undergo a self-immolative reaction through rearrangement.\nWrite the SMILES strings for 1 key rearrangement intermediate containing 2 five-membered rings and the 2 organic molecules produced by self-immolation.", "text_images": [{"file": "Exam-19-T9-7-1-prompt.png", "description": "The figure shows the one-step reaction 'CC(C)(C(C1=CC=CC=C1)O)S(C2=NC3=CC=CC=C3S2)(=O)=O'>10% DMSO/PBS, pH = 7.4, 37°C>[products not shown]."}], "full_mark": 4}, {"sub_id": "Exam-19-T9-8", "tags": ["2.1", "2.4"], "text_open": "9.6 The following polymer readily depolymerizes through self-immolation under acidic conditions. Write the reaction equation.", "text_images": [{"file": "Exam-19-T9-8-1-prompt.png", "description": "The figure shows a cyclic polymer in which three benzene-ring units alternate with O–CH units to form a ring. The visible cyclic skeleton has SMILES 'C1(C2OC3OC(OC4OC(OC5O2)C6=C5C=CC=C6)C7=C4C=CC=C7)=C3C=CC=C1'. A curved bracket marks the repeating portion with subscript n-2, so the full structure is an n-mer of monomer '[*]OC1OC([*])C2=C1C=CC=C2'. H+ is written above the arrow and the product is not shown."}], "full_mark": 3}]} {"problem_id": "Exam-20-T1", "header": "The Ingenious Green Shoot\nThe heavy element M was first discovered by the British chemist William Crookes (1832-1919) while he was analyzing selenium-containing flue dust sent from a German sulfuric-acid factory. On observing through a spectroscope a bright emerald-green spectral line never seen before, he concluded that he had discovered a new element. Elemental M is silvery white and extremely soft, leaves a dark-gray streak on paper, and rapidly tarnishes when a freshly cut surface is exposed to air. M has two common oxidation states: the lower oxidation state is stable in aqueous solution, whereas compounds in the higher oxidation state readily decompose on heating and are strong oxidants.\nWhen heated in air to approximately 300 °C, M burns to form black hygroscopic solid A. A dissolves in water to give a clear colorless solution containing the strong base B, whose aqueous basicity is close to that of KOH. Passing excess $CO_2$ through a solution of B precipitates colorless crystals C. Mixing C with nitric acid and potassium chromate solution gives yellow precipitate D (reaction 1). Adding hydrogen peroxide to a solution of B produces dark-brown powder E (reaction 2). Heating E with a KI solution acidified with hydrochloric acid gives yellow precipitate F and purple-black solid G (reaction 3). The mass fraction of M in F is 61.69 %.", "header_images": [], "subquestions": [{"sub_id": "Exam-20-T1-1", "tags": ["2.1"], "text_open": "1.1.1 Identify element M and write the chemical formulas of A-G.", "text_images": [], "full_mark": 8}, {"sub_id": "Exam-20-T1-2", "tags": ["2.1", "4.2"], "text_open": "1.2 The chlorides of M have unusual properties. Its lower-valent chloride H is a white solid but gradually turns purple in sunlight. Its higher-valent chloride J can be prepared from H by the combined action of concentrated hydrochloric acid and chlorine. Dissolving J in concentrated hydrochloric acid gives the tetrahedral complex anion K.\n1.2.1 Write the equation for the gradual photochemical conversion of H to the purple material, and explain the composition and origin of the color of the purple product.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-20-T1-3", "tags": ["1.7"], "text_open": "1.2.4 Experimentally, H is much more stable than J. Explain the key atomic-structure effect responsible for this behavior and why it is especially pronounced for M within its group.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-20-T1-4", "tags": ["2.1", "3.4"], "text_open": "1.3 The following experiment is performed to determine the cumulative stability constant $\\beta_4$ of complex anion K formed when J dissolves in hydrochloric acid.\nStep 1: Accurately weigh a quantity of J, dissolve it in 0.500 mol·L$^{-1}$ HCl, and dilute to 50.0 mL. Place a portion of the solution in a 1.00 cm quartz cuvette and measure an absorbance of $A_1 = 0.680$ at 230 nm. At this wavelength, the molar absorptivity of K is $\\varepsilon = 8.50 \\times 10^3\\ \\mathrm{L\\cdot mol^{-1}\\cdot cm^{-1}}$. M ions and $Cl^-$ do not absorb at this wavelength, and at this HCl concentration almost all M is present as K.\nStep 2: Transfer 10.00 mL of the solution above, add 10.00 mL of 0.100 mol·L$^{-1}$ $AgNO_3$ solution, stir thoroughly to complete precipitation, and filter. Measure the absorbance of the filtrate under the same conditions, obtaining $A_2 = 0.306$. After addition of $AgNO_3$, the amount of excess $Ag^+$ is negligible. Ignore dissolution of $AgCl$ and activity-coefficient effects.\n1.3.1 Calculate the equilibrium concentration of free $Cl^-$ in the filtrate after reaction with the $AgNO_3$ solution.", "text_images": [], "full_mark": 5}]} {"problem_id": "Exam-20-T2", "header": "A Solvent-Responsive Color-Changing Complex\nThe pentacyanoferrate(II) complex ion X is a solvatochromic complex that is highly sensitive to changes in the dielectric constant and hydrogen-bond-donor ability of its solvent environment. Studies show that the color of X changes in opposite directions with solvent composition in the two binary solvent systems methanol/water and formamide (FA)/water.", "header_images": [], "subquestions": [{"sub_id": "Exam-20-T2-1", "tags": ["1.4", "1.6"], "text_open": "2.1 The color of complex ion X arises primarily from metal-to-ligand charge transfer, MLCT.\n2.1.2 In the MLCT mechanism of X, the excited state $X^{*}$ can be represented by a limiting resonance form of the ground-state ion in which one nitrogen atom of the organic ligand is a radical cation. Write the SMILES string of this limiting resonance form and correctly label the oxidation state of Fe.", "text_images": [{"file": "Exam-20-T2-1-1-prompt.png", "description": "X has the SMILES 'COC(C=C1)=CC=C1[N+](C=C2)=CC=C2C3=CC=[N]([Fe+2]([C-]#N)([C-]#N)([C-]#N)([C-]#N)[C-]#N)C=C3'. Complex ion X has an overall charge of two negative units."}], "full_mark": 3}, {"sub_id": "Exam-20-T2-2", "tags": ["1.7"], "text_open": "2.1.3 State which has the larger dipole moment: ground-state X or its excited state $X^{*}$.", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-20-T2-3", "tags": ["3.2", "1.7"], "text_open": "2.2 To quantify the effects of solvent dielectric constant and hydrogen-bond-donor ability on the color of the complex ion, the following linear relationship is used:\n$E_{CT} = E_0 + s\\cdot f(ε) + k\\cdot α$\nwhere ε is the dielectric constant of the solvent, α is the hydrogen-bond-donor acidity of the solvent, the Onsager polarity function is $f(ε)=\\frac{2(ε-1)}{2ε+1}$, and s and k are the corresponding linear coefficients.\n2.2.3 For a binary methanol/water solvent, the fitted linear coefficients are\n$E_0$ = −28.112 kcal·mol^{-1}; s = 123.758 kcal·mol^{-1}; k = −61.985 kcal·mol^{-1}.\nThe dielectric constant ε of the binary methanol/water solvent is approximately linearly related to the mole fraction of methanol $x_{MeOH}$:\n$ε = 79.2 - 43.6\\cdot x_{MeOH}$\nThe hydrogen-bond-donor acidity α of the mixed solvent is approximately independent of solvent composition. Calculate the slope of $E_{CT}$ with respect to $x_{MeOH}$ for the complex in the mixed solvent at $x_{MeOH}$ = 0.50. Hint: for $u = u(v(x))$, $du/dx = (du/dv) \\cdot (dv/dx)$.", "text_images": [], "full_mark": 6}, {"sub_id": "Exam-20-T2-4", "tags": ["1.7", "3.9"], "text_open": "2.3 In a binary formamide (FA)/water solvent, FA is the more polar component. Preferential solvation makes the local solvent mole fraction y around the solute differ from the bulk solvent mole fraction x. The Suppan dielectric-enrichment model is used:\n$\\frac{y_n}{y_p}=\\frac{x_n}{x_p}e^{-Z_{ps}}$\nwhere the subscript n denotes the less polar solvent and p the more polar solvent. In an FA/water mixture, the solvation index is $Z_{ps}=0.412$.\n2.3.2 Define the local enrichment factor of the more polar solvent p as $η_p = y_p / x_p$. Derive an expression for $η_p$ in terms of $x_p$ and $Z_{ps}$, and calculate $η_p$ at $x_p$ = 0.600.", "text_images": [], "full_mark": 5}, {"sub_id": "Exam-20-T2-5", "tags": ["1.6", "1.7"], "text_open": "Experimental observation: in methanol/water, the absorption peak shifts to the blue as the mole fraction of the more polar component, water, increases. In FA/water, the absorption peak shifts to the red as the mole fraction of the more polar component, FA, increases, and preferential solvation occurs.\n2.4 Based on the information and model analysis above, in a methanol/water solvent the ( ) between solute and solvent determines the direction of the solute's absorption-peak shift, and the solvent predominantly stabilizes ( ). In an FA/water solvent, the ( ) between solute and solvent determines the direction of the absorption-peak shift, and the solvent predominantly stabilizes ( ).\n(a) Dipole/electrostatic interactions; (b) hydrogen-bonding interactions; (c) ground-state X; (d) excited-state $X^{*}$", "text_images": [], "full_mark": 2}]} +{"problem_id": "Exam-20-T3", "header": "Halogen-Bond-Donor-Catalyzed Synthesis of Poly(ortho esters)\nPolymers have transformed modern life because of their low cost and practicality. Polyesters are high-performance, versatile engineering plastics and can also be made into polyester fibers and films; the family includes polyester resins and elastomers.", "header_images": [], "subquestions": [{"sub_id": "Exam-20-T3-1", "tags": ["2.4", "2.3"], "text_open": "3.1 Cyclic ketene acetals undergo radical polymerization in the presence of peroxides to generate poly(regular lactone) chains. Write the SMILES strings of A-D in the reaction scheme below. Polymer end groups need not be shown. Only D is a polymer.", "text_images": [{"file": "Exam-20-T3-1-1-prompt.png", "description": "Four-step reaction: 'BrCC1OCCO1'>tBuOK>A; A>ROOR, heat>[B]; [B]>isomerization>[C]; [C]>D. R is the placeholder used in the original diagram, and [B] and [C] denote radical intermediates."}], "full_mark": 5}, {"sub_id": "Exam-20-T3-2", "tags": ["2.4", "2.6"], "text_open": "3.2 Poly(ortho esters) (POEs) are renewable, biodegradable polymers whose degradation kinetics and surface-erosion behavior can be tuned and whose degradation products are harmless. Earlier synthetic methods, however, gave POEs with molecular weights that were too low and properties that could not be well controlled, making them unsuitable as general-purpose materials such as plastics and rubbers. The reaction below uses renewable feedstocks and an innovative initiator to achieve controlled ring-opening polymerization of the monomer and produce POEs with controllable molecular weights.\n3.2.1 In this reaction, two renewable starting materials first form the spiro-ortho-ester monomer E, MSOE, under $BF_3$ catalysis. The target polymer P1 is then synthesized using the initiator MOMBr and the catalyst ion XBD. After monomer E reacts with the active center of the polymer chain, it can isomerize and polymerize to give the acyclic polymer P2. Based on this information, write the SMILES strings of E, P1, and P2. Polymer end groups need not be shown.", "text_images": [{"file": "Exam-20-T3-2-1-prompt.png", "description": "Two-step reaction: 'CC1CO1' + 'O=C1CCCCCO1' > $BF_3\\cdot OEt_2$ (5.0 mol%) > E; E + 'BrCOC' (MOMBr) > XBD > P1 (major) + P2 (minor)."}], "full_mark": 6}, {"sub_id": "Exam-20-T3-3", "tags": ["2.4"], "text_open": "3.2.2 At pH = 3, P1 can react with water and degrade, allowing recovery of small-molecule ester monomers. Write the SMILES strings of the possible recovered small molecules.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-20-T3-4", "tags": ["1.6", "1.1"], "text_open": "3.3 The catalyst ion XBD is a trinuclear halogen-bond-donor catalyst whose mechanism involves a special secondary interaction: the halogen bond. A halogen bond is fundamentally an electrostatic interaction arising from a region of accumulated positive charge at the extension opposite a covalent bond to a halogen atom, especially Cl, Br, or I. This region is called a \"σ hole\" in the literature and is illustrated below.\n3.3.1 Based on this information, write the SMILES strings of the halogen-bonded aggregates $(Cl_2)_2$ and $(RBr)_3$. The latter has a rotational axis of order higher than two; treat R as a spherical ligand. Specify the halogen-bond angles rigorously.", "text_images": [{"file": "Exam-20-T3-4-1-prompt.png", "description": "Schematic of a σ hole on a halogen atom: a blue gradient sphere at the center bears the capital letter E. The letter R is connected to the left of the sphere by a short horizontal line. To the right is a concentric region with an outer green ring, a middle yellow ring, and a red central point. A red δ+ at the upper right has a curved arrow pointing to the concentric region, while a blue δ- at the lower left has a curved arrow pointing to the bottom of the sphere."}], "full_mark": 4}, {"sub_id": "Exam-20-T3-5", "tags": ["2.3", "2.1"], "text_open": "3.3.2 The structure of the catalyst ion XBD used in 3.2 is shown below. Write the reaction equations for chain initiation by the initiating system, MOMBr and the catalyst ion XBD. XBD may be abbreviated; there is no need to write a SMILES string containing XBD. On this basis, determine whether the reaction is cationic or anionic polymerization. Hint: chain initiation consists of two elementary steps.", "text_images": [{"file": "Exam-20-T3-5-1-prompt.png", "description": "The SMILES for XBD is '[R]N(N=[N+]1C)C(I)=C1C2=CC(C3=C(I)N([R])N=[N+]3C)=CC(C4=C(I)N([R])N=[N+]4C)=C2'. The 3 [R] groups are placeholders used in the original diagram, and the 3 [N+] atoms carry positive-charge labels."}], "full_mark": 5}]} +{"problem_id": "Exam-20-T4", "header": "Determination of Silicon by the Silicomolybdenum Blue Spectrophotometric Method\nSilicon is an important element in iron ores and steel materials, and its content significantly affects steel properties. In the commonly used silicomolybdenum blue spectrophotometric method, silicic acid reacts with ammonium molybdate in acidic medium to form silicomolybdenum yellow, which is then reduced to silicomolybdenum blue for absorbance measurement. This problem concerns sample fusion, removal of interferences, control of color-development conditions, and quantification by a calibration curve.", "header_images": [], "subquestions": [{"sub_id": "Exam-20-T4-1", "tags": ["4.1", "4.7"], "text_open": "4.1 Sample pretreatment\nWeigh 0.1284 g of sample into a 50 mL silver crucible, add 2 g of sodium peroxide and mix, then cover with another 0.5 g of sodium peroxide. Cover the crucible, place it in a furnace at 300 °C, slowly raise the temperature to 750 °C-800 °C, and fuse for 20 min. Remove and allow to cool slightly. Add 100 mL of boiling water to leach the melt completely and wash out the crucible. While swirling, slowly add 30 mL of concentrated hydrochloric acid. After cooling, transfer the solution to a 200 mL volumetric flask, dilute to the mark with water, mix, filter, and discard the initial portion of the filtrate.\n4.1.6 The acidified solution must be dry-filtered through medium-speed filter paper, and the first few milliliters of filtrate must be discarded. Explain the purposes of this procedure. If the initial filtrate were not discarded, would the measured result be too high or too low?", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-20-T4-2", "tags": ["4.2"], "text_open": "4.2 Color development for the sample\nTransfer 10.00 mL of filtrate to a 100 mL volumetric flask and add blank-test solution to bring the volume to 20 mL. Add 10 mL of hydrochloric acid (1 + 3), add potassium permanganate solution (20 g/L) dropwise until the solution is pale red, add 10 mL of water, and mix.\nAdd 1 mL of anhydrous ethanol and mix. While continuously swirling, add 5 mL of ammonium molybdate solution (100 g/L), mix, and allow to stand at 15 °C-25 °C for 15 min.\nAdd 15 mL of hydrochloric acid (1 + 3), followed by 5 mL of oxalic acid-tartaric acid mixture containing 50 g/L oxalic acid and 40 g/L tartaric acid. Mix and immediately add 5 mL of ascorbic acid solution (20 g/L), dilute to the mark with water, mix, and allow to stand at 25 °C-35 °C for 15 min.\nTransfer a portion of the colored solution to a 3 cm cuvette. Using the colored blank-test solution carried through with the sample as reference, measure an absorbance of 0.425 at 810 nm.\n4.2.1 Explain why adding potassium permanganate to a pale-red endpoint eliminates interference from lower-valent iron and organic matter, and state the effect of excess potassium permanganate.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-20-T4-3", "tags": ["4.2"], "text_open": "4.2.4 State which interfering ions are masked by oxalic acid and tartaric acid, respectively, in the color-development system, and explain why ascorbic acid must not be added before the oxalic acid-tartaric acid mixture.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-20-T4-4", "tags": ["4.7"], "text_open": "4.3 Construction of the calibration curve\nTransfer 0 mL, 1.00 mL, 2.00 mL, 4.00 mL, 6.00 mL, 8.00 mL, and 10.00 mL portions of silicon dioxide standard solution (20.0 μg/mL) to a series of 100 mL volumetric flasks. Add water to 20 mL and develop the color according to the procedure in 4.2. Transfer each colored solution to a 3 cm cuvette and measure the absorbance at 810 nm against a reagent blank. Construct a calibration curve using the amount of silicon dioxide in μg as the horizontal coordinate and absorbance as the vertical coordinate.\n\n| Volume of silicon dioxide standard solution ($V$ / mL) | 0.00 | 1.00 | 2.00 | 4.00 | 6.00 | 8.00 | 10.00 |\n| --- | --- | --- | --- | --- | --- | --- | --- |\n| Amount of silicon dioxide ($m$ / μg) | 0 | 20.00 | 40.00 | 80.00 | 120.0 | 160.0 | 200.0 |\n| Absorbance ($A$) | 0.000 | 0.082 | 0.163 | 0.326 | 0.488 | 0.651 | 0.812 |\n\nCalculate the silicon dioxide content of the sample.", "text_images": [], "full_mark": 5}]} {"problem_id": "Exam-20-T5", "header": "The Vanishing Sixfold Axis\nCrystals in the hexagonal crystal system are characterized by a sixfold axis along the c direction. In addition to the simplest sixfold rotation axis, for which rotation by 60° about the axis brings the crystal into coincidence with itself, a crystal may contain either of the following two types of sixfold axes:\nFirst, a $6_n$ screw axis, for which rotation by 60° followed by translation along c by n / 6 of the unit-cell length brings the crystal into coincidence with itself.\nSecond, a sixfold rotoinversion axis, for which rotation by 60° followed by inversion through a point on the axis brings the crystal into coincidence with itself.", "header_images": [], "subquestions": [{"sub_id": "Exam-20-T5-1", "tags": ["1.3"], "text_open": "5.3 The symmetry of an individual close-packed layer does not necessarily represent the symmetry of the whole crystal. Consider a crystal described as follows. Its close-packed layers are analogous to those in perovskite: metal M and vacancies occupy the $Ca^{2+}$ and $O^{2-}$ sites, respectively, in a 1:3 ratio. These close-packed layers undergo simple hexagonal stacking along c with a four-layer repeat, and the projections of M along c do not overlap within a repeat. X atoms occupy all trigonal-prismatic voids jointly formed by M atoms and vacancies.\n5.3.1 Write the chemical formula of this crystal.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-20-T5-2", "tags": ["1.3"], "text_open": "5.3.3 State the Bravais lattice type of this crystal.", "text_images": [], "full_mark": 3}]} -{"problem_id": "Exam-20-T6", "header": "Co(Salen) Oxygen Carrier\nMany proteins containing transition-metal ions can bind molecular oxygen to form complexes and reversibly take up and release oxygen for biological processes. Examples include the iron-containing proteins myoglobin and hemoglobin and the copper-containing protein hemocyanin. The Co-Schiff-base complex [Co(II)(Salen)], cobalt N,N'-ethylenebis(salicylideneiminate), is an important representative synthetic model oxygen carrier.", "header_images": [{"file": "Exam-20-T6-header-1.png", "description": "[Co(II)(Salen)] has the SMILES 'C12=C(O[Co]34OC5=CC=CC=C5C=[N]3CC[N]4=C2)C=CC=C1'. The image shows the structure of [Co(II)(Salen)]. The Salen ligand coordinates Co tetradentately through 2 phenoxide O atoms and 2 imine N atoms. The benzene rings are drawn in alternating-single-and-double-bond Kekulé form, both C=N bonds are double bonds, and the central Co atom is represented as four-coordinate by the ring-closure labels [Co]34. No charges or stereochemical markers are shown. The caption [Co(II)(Salen)] appears below the structure."}], "subquestions": [{"sub_id": "Exam-20-T6-1", "tags": ["1.4"], "text_open": "6.1 Depending on the preparation method, [Co(II)(Salen)] can exist in two solid forms: dark-red, gelatinous, active form A and dark-red, crystalline, inactive form B. A rapidly absorbs oxygen at room temperature, whereas B is stable at room temperature and absorbs almost no oxygen. Both are bimolecular dimers.\n6.1.1 A and B dimerize differently: A through Co-Co bonding and B through Co-O bonding. Based on this description, write the actual SMILES strings of A and B, using a textual description if necessary. The Salen ligand may be abbreviated.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-20-T6-2", "tags": ["1.4", "2.4"], "text_open": "6.1.2 Inactive B dissolves in certain solvents, and the resulting solution acquires the ability to absorb oxygen. [Co(II)(Salen)] dissolved in solvent S can react with oxygen by different pathways to form product C with $Co:O_2$ = 1:1 or product D with $Co:O_2$ = 2:1. The Co centers in both C and D are octahedral. Write the SMILES strings of C and D.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-20-T6-3", "tags": ["1.4", "2.8"], "text_open": "6.2 Inactive [Co(II)(Salen)] can be prepared as follows. Add salicylaldehyde and ethylenediamine in stoichiometric amounts successively to a three-neck flask, followed by ethanol. Yellow plate-like crystals form immediately. Assemble the apparatus shown at right, maintain the water bath at approximately 80 °C, and reflux the ethanol until the yellow crystals dissolve completely. After passing nitrogen through the two-way stopcock on the right for some time, use the syringe on the left to inject a solution of $Co(CH_3COO)_2$ slowly through the septum into the three-neck flask. Continue refluxing to obtain a dark-red crystalline precipitate. Stop heating and cool the flask in an ice-water bath while maintaining nitrogen flow and stirring. Collect the crystals on a sintered-glass funnel and dry them by suction. Wash the precipitate successively with deionized water and anhydrous ethanol, then rinse once with acetone and dry by suction to obtain dry red-brown [Co(II)(Salen)].\n6.2.1 What consequences would result from adding excess ethylenediamine? State 2 consequences.", "text_images": [{"file": "Exam-20-T6-3-1-prompt.png", "description": "Apparatus for preparing [Co(II)(Salen)] in 6.2: a three-neck round-bottom flask rests on a base, with a visible liquid level inside. A syringe with plunger is inserted obliquely upward into the left neck. The middle neck is connected to a vertical condenser whose top connects through a bent tube to a U-tube containing liquid at the bottom; the top of the right arm is open. The right neck is connected to an addition component with a stopcock, the two-way stopcock, and a bulb-shaped upper section. The image contains no text labels."}], "full_mark": 2}, {"sub_id": "Exam-20-T6-4", "tags": ["4.1"], "text_open": "6.2.2 What are the purposes of adding deionized water to the U-tube apparatus connected above the condenser? State 2 purposes.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-20-T6-5", "tags": ["4.1"], "text_open": "6.2.3 What problem may occur while cooling the flask in an ice-water bath, and what measure should be taken?", "text_images": [], "full_mark": 3}]} -{"problem_id": "Exam-20-T8", "header": "Fundamentals of Organic Chemistry\nThe rates and selectivities of organic reactions depend not only on electronic structure but also strongly on molecular conformation and spatial arrangement in the transition state.", "header_images": [], "subquestions": [{"sub_id": "Exam-20-T8-1", "tags": ["1.6"], "text_open": "8.1 First consider peroxyacid epoxidation of alkenes from the perspective of frontier molecular orbitals. Examine the general reaction below.\n8.1.2 Under identical reaction conditions, substrate structure affects the rate of epoxidation. Arrange the following compounds in decreasing order of reaction rate.", "text_images": [{"file": "Exam-20-T8-1-1-prompt.png", "description": "General reaction for peroxyacid epoxidation of an alkene: '[R]/C([R])=C([R])/[R]'.'[R]C(OO)=O'>>'[R]C1([R])C([R])([R])O1'. No reaction conditions are labeled on the arrow; R is the placeholder used in the original diagram."}, {"file": "Exam-20-T8-1-2-prompt.png", "description": "The SMILES for A is 'C1CC=CCC1'; the SMILES for B is 'OC1C=CCCC1'; the SMILES for C is 'O=C(OC1C=CCCC1)C'; and the SMILES for D is 'OC1CC=CCC1'."}], "full_mark": 3}, {"sub_id": "Exam-20-T8-2", "tags": ["1.2"], "text_open": "8.2 When the oxygen-transfer reagent is changed, the spatial course of the reaction still strongly affects the epoxidation outcome. Consider the alkene epoxidation involving the dioxirane below.\n8.2.1 For a more structurally complex substrate, its preferred conformation further affects the reaction. Write the SMILES string of the preferred conformation of the actual reactive species derived from the molecule shown under these reaction conditions.", "text_images": [{"file": "Exam-20-T8-2-1-prompt.png", "description": "The molecule is represented by the SMILES 'O=C1[C@H]2[C@H](OC(C)(C)O2)CO[C@@]13OC(C)(C)OC3'. It contains a C=O group and two O—C(CH₃)₂—O ketal fragments; stereochemistry is indicated by solid and dashed wedge bonds."}], "full_mark": 2}, {"sub_id": "Exam-20-T8-3", "tags": ["2.4", "1.2"], "text_open": "8.3 The epoxidation above illustrates how molecular conformation and spatial arrangement in the transition state affect selectivity. Similar stereochemical control can also be used to form carbon-carbon bonds. The enolate alkylation below gives its product with very high stereoselectivity.\n8.3.1 Write the SMILES string of the major alkylation product, explicitly specifying its stereochemistry.", "text_images": [{"file": "Exam-20-T8-3-1-prompt.png", "description": "The substrate is represented by the SMILES 'O=C1OC[C@@H](C(C)C)N1C(CC)=O', with stereochemistry indicated by a solid wedge bond. The arrow is labeled 'LDA, THF, -78°C' above and 'then PhCH2Br' below; no product structure is drawn to the right."}], "full_mark": 2}]} -{"problem_id": "Exam-20-T9", "header": "Problem 9 Prins Cyclization (25 points, 12%)\nThe Prins cyclization is a powerful synthetic transformation that rapidly constructs tetrahydropyran rings. Tetrahydropyran is a common structural unit in many bioactive molecules, including spongistatins and bryostatins. Researchers have used the Prins cyclization in numerous elegant strategies for the total synthesis of natural-product targets.", "header_images": [], "subquestions": [{"sub_id": "Exam-20-T9-1", "tags": ["2.4", "2.2"], "text_open": "9.1 The following is a classic Prins cyclization that gives the final product with high stereoselectivity, where R = alkyl.\n9.1.1 Write the SMILES strings of A-D in the scheme, including stereochemistry. B is a highly strained intermediate.", "text_images": [{"file": "Exam-20-T9-1-1-prompt.png", "description": "Prins cyclization sequence: '[R]C1=CCCCC1'>H2C=O, H2SO4>[A⇌B]; [B]>H2O>[C]; [C]>H2C=O, H2SO4>[D]. In the image, the last arrow points leftward from C to D. A-D are shown only as letter labels and no structures are provided; R is the placeholder used in the original diagram."}], "full_mark": 5}, {"sub_id": "Exam-20-T9-2", "tags": ["2.2", "2.4"], "text_open": "9.1.2 The following cascade comprises a Prins cyclization and a rearrangement and ultimately gives a product containing a five-membered ring. Write the SMILES string of the product and of at least 2 charged key intermediates, including stereochemistry.", "text_images": [{"file": "Exam-20-T9-2-1-prompt.png", "description": "Single-step reaction: 'C=C(C)[C@H]([C@H](C(C)=C)O)O'>Me2C=O, H2SO4>[product not shown]. Stereochemistry of the substrate's two OH groups is shown by wedge bonds and retained by the @ markers in the SMILES."}], "full_mark": 4}, {"sub_id": "Exam-20-T9-3", "tags": ["1.2", "2.7"], "text_open": "9.2 A key advantage of the Prins cyclization is that, starting from a single stereocenter and a defined alkene geometry, it can construct up to four stereocenters on a tetrahydropyran ring with high stereoselectivity. In some cases, however, enantiomeric excess (ee) is lost.\n9.2.1 In the Prins cyclization below, the ee of the major product is lower than that of the starting material, and small amounts of two byproducts are detected.\nPredict the configurations of the chiral carbons in the product by assigning the absolute configurations of chiral carbons a and b in the major product using S/R notation.", "text_images": [{"file": "Exam-20-T9-3-1-prompt.png", "description": "Single-step reaction: 'O[C@H](CC=C)C1=CC=CC=C1' reacts with PhCH2CH2CHO under BF3·OEt2 and AcOH conditions to give the major product, 'O=C(O[C@H]1C[C@@H](CCC2=CC=CC=C2)O[C@@H](C3=CC=CC=C3)C1)C', byproduct 1, 'O=C(OC1CC(C2=CC=CC=C2)OC(C3=CC=CC=C3)C1)C', and byproduct 2, 'O=C(OC1CC(CCC2=CC=CC=C2)OC(CCC3=CC=CC=C3)C1)C'. Starting material M is labeled 87% ee and the major product 67% ee. Chiral carbons a and b are labeled on the ring of the major product; OAc at a is shown on a solid wedge, while CH2CH2Ph at b is shown on a dashed bond."}], "full_mark": 2}, {"sub_id": "Exam-20-T9-4", "tags": ["2.3", "2.2"], "text_open": "9.2.2 The pathway to the major product can undergo racemization through a rearrangement. Experimentally, the enantiomer M’ of substrate M is detected. If M and $PhCH_2CH_2CHO$ are replaced by structure N below, racemization through this rearrangement cannot occur under the same conditions. In light of the byproduct structures and the analysis above, write the SMILES strings of the reactants and products in each step of the mechanism by which M forms M’, and label the reaction conditions. Electron-flow arrows need not be shown.", "text_images": [{"file": "Exam-20-T9-4-1-prompt.png", "description": "Substrate N has the SMILES 'C=CC[C@@H](OC(OC(C)=O)CC1=CC=CC=C1)C2=CC=CC=C2'. N is the structure label in the image."}], "full_mark": 4}, {"sub_id": "Exam-20-T9-5", "tags": ["2.3", "2.7"], "text_open": "9.2.3 Explain why a racemization process analogous to the one above cannot occur when substrate N is used.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-20-T9-6", "tags": ["2.2", "2.4"], "text_open": "9.3 Researchers have developed many broadly useful Prins reactions using unconventional substrates such as alkynols, allenes, and small-ring compounds.\n9.3.1 The reaction below is based on a Prins-type cyclization of a 3,5-diynol with an aldehyde and proceeds through a domino sequence to synthesize a 2,3-dialkylidenetetrahydrofuran framework. The compound shown at the right was detected in the reaction mixture. This compound is not necessarily itself an intermediate and may arise from an intermediate. Based on this information, write the SMILES strings of the key intermediates. Stereochemistry is not required.", "text_images": [{"file": "Exam-20-T9-6-1-prompt.png", "description": "Single-step reaction: 'OCCC#CC#CC1=CC=CC=C1'.'O=CC1=CC=CC=C1'>HBF4·OEt2 (3 equiv), MeOH (3 equiv), DCM, rt, 8 h>'O=C(/C=C1OCCC\\1=C/C2=CC=CC=C2)C3=CC=CC=C3'."}, {"file": "Exam-20-T9-6-2-prompt.png", "description": "The image is labeled 'Detected in the reaction mixture:' and the detected compound has the SMILES 'OCC/C(C(C#CC1=CC=CC=C1)=O)=C\\C2=CC=CC=C2'."}], "full_mark": 4}, {"sub_id": "Exam-20-T9-7", "tags": ["2.4", "2.7"], "text_open": "9.3.2 The following reaction proceeds through a three-membered-ring intermediate and gives the product with high stereoselectivity, governed primarily by stereoelectronic effects. Write the SMILES string of the product, including stereochemistry, and briefly explain the stereoselectivity.", "text_images": [{"file": "Exam-20-T9-7-1-prompt.png", "description": "Single-step reaction: 'C=C([Si](C)(C1=CC=CC=C1)C)C[C@@H](O)C'.'[R]C=O'>TMSOTf>[?]. The product structure is not shown and is represented by '?'. R is the placeholder used in the original diagram and was retained after manual confirmation."}], "full_mark": 4}]} {"problem_id": "Exam-20-T10", "header": "Photocatalytic Reactions", "header_images": [], "subquestions": [{"sub_id": "Exam-20-T10-1", "tags": ["2.2", "2.4"], "text_open": "10.1 Researchers studied the visible-light photocatalytic difluoroallylation of a silane shown below. The photosensitizer is 4-CzIPN, denoted PC. Under irradiation, PC is excited to PC* and can undergo one-electron reduction to the reduced radical anion PC$^{·-}$. PC$^{·-}$ is unstable and is readily oxidized back to PC. The detailed structure of the photosensitizer need not be considered in any subquestion.\n10.1.1 Based on the reaction mechanism, write the SMILES strings of 1 nitrogen-containing key intermediate and 2 fluorine-containing key intermediates other than PC and its various forms. Quinuclidine is 1-azabicyclo[2.2.2]octane.", "text_images": [{"file": "Exam-20-T10-1-1-prompt.png", "description": "Single-step reaction: 'CC1=CC=C(C(C(F)(F)F)=C)C=C1'.Et3SiH>4-CzIPN, Blue LED; quinuclidine, KHCO3, MeCN>'CC1=CC=C(C(C[Si](CC)(CC)CC)=C(F)F)C=C1'. At the left of the arrow, a p-methylphenyl-C(CF3)=CH2 substrate is joined by a plus sign to Et3SiH; the product is p-methylphenyl-C(=CF2)-CH2SiEt3."}], "full_mark": 6}, {"sub_id": "Exam-20-T10-2", "tags": ["2.4"], "text_open": "10.1.2 When the reaction solvent is changed from the original solvent to THF, a different product is formed. This product still contains 2 fluorine atoms. Write the SMILES string of this byproduct.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-20-T10-3", "tags": ["2.2", "2.4"], "text_open": "10.1.3 Keep the standard reaction conditions unchanged, but replace the para methyl group on the substrate's aryl ring with a para acetyl group (-COCH$_3$). The expected difluoroallylsilane is not obtained; instead, a different major product is formed. Carefully examine the mechanism in 10.1.1, write the SMILES string of this major product, and briefly explain why an analogue of the 10.1.1 product is not formed.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-20-T10-4", "tags": ["2.4"], "text_open": "10.2 The researchers further found that the α-trifluoromethyl alkene derivative above can react with 1,3-dioxolane in a similar visible-light photocatalytic system. Product A is formed predominantly in MeCN, whereas product B is formed predominantly in DMSO. B contains only 1 fluorine atom. Write the SMILES string of B.", "text_images": [{"file": "Exam-20-T10-4-1-prompt.png", "description": "Single-step reaction: 'C=C(C(F)(F)F)C1=CC=C(C2=CC=CC=C2)C=C1'.'O1CCOC1'>4-CzIPN(1 mol%), quinuclidine(20 mol%); NaHCO3(1.5 equiv), Blue LED, Solv., rt, 12 h>. At the left of the arrow, a 4-biphenyl-C(CF3)=CH2 substrate is joined by a plus sign to 1,3-dioxolane; no product structure is drawn to the right."}], "full_mark": 2}]} {"problem_id": "Exam-21-T1", "header": "Chasing the Rainbow\nA silvery-gray, high-melting metal M has no independent mineral in nature and occurs mainly in molybdenite. Industrially, molybdenite is roasted to concentrate the oxides of M.", "header_images": [], "subquestions": [{"sub_id": "Exam-21-T1-1", "tags": ["3.2"], "text_open": "M forms numerous oxides in a wide range of oxidation states and colors, earning them the name \"chemical rainbow.\" Oxide A contains M in its highest oxidation state. It is a yellow, highly deliquescent crystalline solid and dissolves in water to form a strong acid in which the mass fraction of M is 74.12%. Heating a mixture of A and elemental M in a sealed tube gives red crystals B with a metallic luster. Further heating to about 600 °C converts B into the blue solid M2O5; at still higher temperatures, M2O5 decomposes into the gray solid C. M has a different oxidation state in each of A, B, and C, and the oxidation state of M in C is the one most commonly found in oxides of its group. In the gas phase, the M atoms in A have tetrahedral coordination, whereas in the solid state A consists of corner-sharing octahedra and tetrahedra. Passing $Cl_2$ through a heated mixture of M powder and $KCl$ gives the red trimeric chloride D. 1.4 Thermodynamic data show that the reaction of C with $O_2$ to form B is spontaneous at 400 °C; experimentally, however, B decomposes at 400 °C by the reaction $3B(s)\\rightarrow A(s)+C(s)$. Explain this apparent contradiction.", "text_images": [], "full_mark": 4}]} {"problem_id": "Exam-21-T2", "header": "Molecular Qubit Frameworks\nMolecular qubit frameworks (MQFs) are among the leading candidate materials for quantum-information chips. Molecular qubits are functional materials that use electron-spin states within molecules to store and process quantum information. Under an applied magnetic field, two controllable quantum states can be selected from the low-energy levels of a magnetic center to serve as the qubit states \"0\" and \"1\". Long-term information storage requires minimizing environmental perturbations to the electron spins and isolating the magnetic centers from one another to obtain a long coherence time. A research team successfully constructed a diluted array of molecular spin qubits by doping trace amounts of $Nd^{3+}$ magnetic spin centers into a rigid yttrium-based MOF.", "header_images": [], "subquestions": [{"sub_id": "Exam-21-T2-1", "tags": ["1.4", "1.1"], "text_open": "2.1 The yttrium-based MOF has the formula $[C_5N_2H_{11}][Y(C_2O_4)_2]$. The cation does not coordinate to $Y^{3+}$. The MOF crystallizes in the monoclinic system with $\\beta$ = 130.6°. Each $C_2O_4^{2-}$ bridges two $Y^{3+}$ ions in a $\\mu_2$-$\\kappa^2$O,O′;$\\kappa^2$O″,O‴ mode, chelating each $Y^{3+}$ ion through two O atoms. Answer the following questions about this compound.\n2.1.1 The cation of this compound contains a six-membered ring, its two nitrogen atoms have identical chemical environments, and it contains no tertiary carbon atom. Write the SMILES string for the cation.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-21-T2-2", "tags": ["1.4"], "text_open": "2.1.2 Give the coordination number of $Y^{3+}$.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-21-T2-3", "tags": ["3.8", "4.7"], "text_open": "2.3 In quantum-information storage, environmental lattice vibrations (phonon scattering) strongly perturb spin qubits and shorten their lifetimes. Theoretical studies show that, in this porous oxalate framework, the spin–lattice relaxation rate $1/T_1$ of $Nd^{3+}$ in the higher-temperature range 6.5K–8.5K is driven primarily by the combined Raman process and Orbach process and obeys the following kinetic equation:\n$\\frac{1}{T_1}=A_{Raman}T^7+A_{Orbach}\\exp\\left(-\\frac{U_{eff}}{T}\\right)$\nHere, $A_{Raman}$ and $A_{Orbach}$ are the prefactors for the respective mechanisms, and $U_{eff}$ is the effective energy barrier of the low-lying excited state arising from strong spin–orbit coupling. The experimentally measured $U_{eff}$ for this material is 32.6 K.\nThe table below gives experimental spin–lattice relaxation times $T_1$ measured for the extremely dilute sample at different temperatures $T$ using a pulsed EPR spectrometer:\n\n| Measurement temperature $T$ / K | Measured spin-relaxation time $T_1$ / μs |\n| --- | --- |\n| 6.8 | 6.359 |\n| 7.2 | 4.623 |\n| 7.6 | 3.252 |\n| 8.0 | 2.347 |\n| 8.5 | 1.612 |\n\n2.3.1 Using the data above, calculate $A_{Raman}$ and $A_{Orbach}$ by linear regression.", "text_images": [], "full_mark": 6}]} +{"problem_id": "Exam-21-T3", "header": "Preparation and Properties of a Thermochromic Copper(II) Complex\nMaterials whose compounds exhibit thermochromism through structural phase transitions are widely used in anti-counterfeiting applications. To synthesize the thermochromic complex $[Cu(deen)_2](ClO_4)_2$ (deen denotes N,N-diethylethylenediamine), hydrated copper perchlorate must first be prepared, followed by introduction of the deen ligand.", "header_images": [], "subquestions": [{"sub_id": "Exam-21-T3-1", "tags": ["1.4", "4.3"], "text_open": "Hydrated copper perchlorate can be prepared by a precipitation–dissolution method. Dissolve $CuSO_4\\cdot 5H_2O$ in water, add $NaOH$ solution with stirring, and heat slowly on a hot plate until the precipitate turns black. Maintain gentle boiling for 5–10 min, allow the mixture to cool slightly, and collect the precipitate by suction filtration. Rinse it with a small amount of deionized water, then use a glass rod to transfer the precipitate into an evaporating dish containing 10 mL of water. Add perchloric acid, gently guide the precipitate into the solution with the glass rod, and stir until it dissolves completely to form a copper perchlorate solution. Concentrate the solution by gentle heating to about 2 mL. On cooling, crystals of hydrated copper perchlorate are obtained. The preparation of $[Cu(deen)_2](ClO_4)_2$ begins with the hydrated copper perchlorate crystals obtained above. Adding absolute ethanol to these crystals gives a clear blue solution. In a dry beaker, dissolve N,N-diethylethylenediamine in ethanol to obtain a clear pale-yellow solution. Slowly add the ethanolic deen solution to the ethanolic copper perchlorate solution with stirring; the mixture immediately turns deep blue-violet. Stir for 2–3min, then allow it to stand for about 10 min. A large amount of pink precipitate forms at the bottom of the vessel while the supernatant remains deep blue-violet. Collect the product by suction filtration and dry it to obtain $[Cu(deen)_2](ClO_4)_2$. 3.2.2 Dissolving the red $[Cu(deen)_2](ClO_4)_2$ in water again gives a deep-blue solution. Explain why. (No more than 50 words)", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-21-T3-2", "tags": ["1.4"], "text_open": "3.3 In a thermochromism experiment on $[Cu(deen)_2](ClO_4)_2$, a small amount of the $[Cu(deen)_2](ClO_4)_2$ complex is held between pieces of adhesive tape and heated with a hair dryer. The sample changes from pink to blue-violet and returns to pink on cooling.\n3.3.1 Based on crystal field theory, describe in words as necessary the structural change of the $[Cu(deen)_2]^{2+}$ unit upon heating and cooling. The cooled unit may be approximated as centrosymmetric, whereas after heating it is no longer centrosymmetric.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-21-T3-3", "tags": ["4.6"], "text_open": "3.4 To investigate the thermochromism of $[Cu(deen)_2]X_2$ complexes formed with monovalent counteranion $X^-$, differential scanning calorimetry is performed on $[Cu(deen)_2](ClO_4)_2$ and $[Cu(deen)_2](NO_3)_2$. Multiple interactions occur between the cations and anions in both complexes. The results are shown below. For both complexes, the color change is associated with a phase-transition temperature, and the chemical composition remains unchanged across the color change. The temperature at the peak maximum is defined as the transition temperature.\n3.4.1 From the graph, directly give the thermochromic transition temperatures of $[Cu(deen)_2](ClO_4)_2$ and $[Cu(deen)_2](NO_3)_2$.", "text_images": [{"file": "Exam-21-T3-3-1-prompt.png", "description": "Untitled coordinate plot. The horizontal axis is labeled \"Temperature/°C\" with major ticks at 50, 100, 150, and 200 and minor ticks at 25, 75, 125, and 175. The vertical axis is labeled \"Thermogravimetry\" vertically and has no numerical ticks. Two solid black curves are shown. The legend at lower left uses short line segments for \"$ClO_4$(a)\" and \"$NO_3$(b),\" and the curves are labeled (a) and (b). Curve (a), corresponding to $[Cu(deen)_2](ClO_4)_2$, has a lower baseline and a sharp upward peak at about 45–50°C, remains flat thereafter, and drops steeply after about 200°C. Curve (b), corresponding to $[Cu(deen)_2](NO_3)_2$, has a higher baseline, a broad upward peak at about 110°C, a sharp upward peak at about 145°C, another smaller upward peak at about 180°C, and a steep drop after about 200°C. No arrows are shown."}], "full_mark": 4}, {"sub_id": "Exam-21-T3-4", "tags": ["1.4", "4.6"], "text_open": "3.4.2 Using the information above, explain the origin of the difference between the transition temperatures of $[Cu(deen)_2](ClO_4)_2$ and $[Cu(deen)_2](NO_3)_2$. (No more than 100 words)", "text_images": [], "full_mark": 4}]} +{"problem_id": "Exam-21-T4", "header": "Determination of Manganese by Ammonium Nitrate Oxidation Titrimetry\nManganese is an important element in steels and alloys, and its content directly affects their strength, hardness, and wear resistance. Ammonium nitrate oxidation titrimetry is a classical method for determining manganese: ammonium nitrate oxidizes Mn(II) to Mn(III) in a phosphoric acid medium, and the product is then titrated with a standard ammonium iron(II) sulfate solution. This problem addresses standardization of the titrant, control of the oxidation conditions, and calculation of the result.", "header_images": [], "subquestions": [{"sub_id": "Exam-21-T4-1", "tags": ["1.4"], "text_open": "4.2 Sample treatment and titration\nWeigh 0.2026 g of sample into a 300 mL conical flask. Add 15 mL of concentrated phosphoric acid and 5 mL of concentrated nitric acid, and heat until the sample dissolves completely. Continue heating and evaporating until the surface becomes still and faint fumes just begin to appear (about 220–240 °C), then remove the flask from the heat. Immediately add 3.0 g ± 0.2 g of ammonium nitrate. While swirling the flask, use a rubber-bulb pipette to blow tangentially across the mouth of the flask to remove nitrogen oxides until the yellow fumes disappear completely. Allow the mixture to cool slightly, add 60 mL of distilled water, mix thoroughly, and cool to room temperature. Titrate with the standard ammonium iron(II) sulfate solution until the solution is pale red. Add 2 drops of N-phenylanthranilic acid indicator solution (2 g/L) and continue titrating to a bright-yellow endpoint. The standard ammonium iron(II) sulfate solution consumed is 16.48 mL, and the blank consumes 0.84 mL.\n4.2.2 Give the predominant form of manganese after oxidation by ammonium nitrate. (Hint: the ligand is dihydrogen phosphate.)", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-21-T4-2", "tags": ["2.1"], "text_open": "4.2.4 Write the net ionic equation for the titration. In the products, the metal ions are present as free ions rather than complexes.", "text_images": [], "full_mark": 2}]} +{"problem_id": "Exam-21-T5", "header": "Structures of Silicides", "header_images": [], "subquestions": [{"sub_id": "Exam-21-T5-1", "tags": ["1.3"], "text_open": "5.1 Consider a hypothetical crystal X. Si atoms form graphite-like layers A stacked along the c direction. The projections of all layers A along c coincide, producing a series of layers of hexagonal-prismatic voids between them. Metal M regularly occupies 2/3 of the hexagonal-prismatic voids in each such layer to form layers B, and the projections of all layers B along c also coincide.\n5.2 Crystal Y is obtained by replacing some Si atoms in layers A of crystal X with M atoms. In crystal Y, no two Si atoms and no two M atoms are adjacent within the same layer A. Likewise, between two neighboring layers A along c, no Si atom is adjacent to another Si atom and no M atom is adjacent to another M atom.\n5.2.1 Give the chemical formula of crystal Y in terms of M and Si, reduced approximately to the simplest ratio, and the number of formula units in one conventional unit cell.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-21-T5-2", "tags": ["1.3"], "text_open": "5.3 Further distortion of layers A in crystal Y gives hexagonal crystal Z, a real silicide. The projection of crystal Z along c is shown below, where the small dark spheres represent Si atoms and the large light-colored spheres represent M atoms. After distortion, all Si atoms still have identical chemical environments, and all M atoms within the same type of layer likewise have identical chemical environments.\n5.3.1 In crystal Z, one Si atom has coordinates (0.6000, 0, 0.2500), and two M atoms have coordinates (0.2300, 0, 0.2500) and (0.3333, 0.6667, 0.0000), respectively. Give the coordinates of all remaining Si and M atoms.", "text_images": [{"file": "Exam-21-T5-2-1-prompt.png", "description": "Projection of crystal Z along the c direction on a white background. Thin gray lines outline a right-leaning parallelogram unit cell, with approximately horizontal top and bottom edges and side edges slanting upward to the right. The diagram contains 22 shaded, highlighted spheres of two types: 12 large light-blue spheres and 10 small dark-blue spheres. There are 16 spheres on the unit-cell boundary, 4 on each edge; every edge has 2 large light-blue spheres and 2 small dark-blue spheres arranged in the order large light-blue, small dark-blue, small dark-blue, large light-blue. The interior contains 6 additional spheres: 4 large light-blue spheres at upper left, upper right, lower left, and lower right, and 2 small dark-blue spheres at center-right and lower-center-left, approximately along the shorter diagonal. At lower left, a red solid arrow points horizontally right and is labeled a; a green solid arrow points upper left and is labeled b. Their common origin is a blue circle containing a white c. As specified in the prompt, the dark spheres are Si atoms and the light-colored spheres are M atoms."}], "full_mark": 6}, {"sub_id": "Exam-21-T5-3", "tags": ["1.3"], "text_open": "5.3.2 During the distortion, the coordination polyhedron around an M atom in layer A, formed by Si atoms in the same and adjacent layers, is also distorted. State how many distinct Si–M distances occur in the distorted coordination polyhedron.", "text_images": [], "full_mark": 2}]} +{"problem_id": "Exam-21-T6", "header": "Pharmacokinetics and Molecular Design (II)\nPharmacokinetics quantitatively studies the absorption, distribution, metabolism, and excretion of drugs in living organisms and uses mathematical principles and methods to describe how plasma drug concentration changes with time. Chemical-kinetic modeling provides a systematic understanding of drug metabolism in vivo. This problem concerns the oxidative metabolism of diclofenac in the body.", "header_images": [], "subquestions": [{"sub_id": "Exam-21-T6-1", "tags": ["3.7"], "text_open": "The Michaelis equation is $r=r_{max}[S]/(K_m+[S])$.\nExperiments show that $[S]$ and $r$ for the 4′-hydroxylation of diclofenac catalyzed by human liver microsomes obey the following relationship:\n| Group | $[S]_1$ / $\\mu\\mathrm{M}$ | $r_1$ / $(\\mathrm{pmol \\cdot min^{-1} \\cdot mg^{-1}})$ | $[S]_2$ / $\\mu\\mathrm{M}$ | $r_2$ / $(\\mathrm{pmol \\cdot min^{-1} \\cdot mg^{-1}})$ |\n| --- | --- | --- | --- | --- |\n| Low concentration | 0.90 | 39.27 | 1.80 | 72.00 |\n| High concentration | 90.0 | 392.73 | 180.0 | 411.43 |\n\n6.2.4 Under the same conditions, the 5-hydroxylation has $K_m = 43\\ \\mu\\mathrm{M}$ and $r_{max} = 15.4\\ \\mathrm{pmol \\cdot min^{-1} \\cdot mg^{-1}}$. Intrinsic clearance $CL_{int}$ is defined by $CL_{int} = r_{max}/K_m$. Calculate $CL_{int}$ for the 4′- and 5-hydroxylation pathways and briefly explain the kinetic meaning of $CL_{int}$. Hint: the drug concentration in the liver is much lower than $K_m$.", "text_images": [], "full_mark": 5}, {"sub_id": "Exam-21-T6-2", "tags": ["2.9", "3.7"], "text_open": "6.2.5 The kinetic parameters for the individual steps in the pathways above are shown below:\n| Pathway | Substrate → product | Enzyme | $CL_{int}$ / $(\\mu\\mathrm{L}\\cdot\\mathrm{min}^{-1}\\cdot\\mathrm{mg}^{-1})$ |\n| --- | --- | --- | --- |\n| 4′-hydroxylation pathway | DF → 4′-OH-DF | CYP2C9 | / |\n| 4′-hydroxylation pathway | 4′-OH-DF → A | CYP2C9 | 358 |\n| 5-hydroxylation pathway | DF → 5-OH-DF | CYP3A4 | / |\n| 5-hydroxylation pathway | 5-OH-DF → B | CYP2C9 | 103 |\n\nThe researchers also compared detoxification of the quinone products from the two hydroxylation pathways. The measured curves are shown below, where \"conjugate\" denotes a detoxification product.\nExplain why both steps of the 5-hydroxylation pathway have lower $CL_{int}$ values than those of the 4′-hydroxylation pathway, yet the 5-hydroxylation pathway is more toxic. (No more than 100 words)", "text_images": [{"file": "Exam-21-T6-2-1-prompt.png", "description": "Coordinate plot containing two side-by-side panels labeled A and B at upper left. Both horizontal axes are labeled 'GSH concentration (mM)' with ticks at 0, 2, 4, and 6; both vertical axes are labeled 'Conjugate formation (%)' with ticks at 0, 50, and 100. Each panel contains one solid black curve with filled circular data markers, some with vertical error bars. In panel A, the curve rises rapidly from near the origin as GSH concentration increases, approaches 100% at about 2 mM, and remains on a plateau. In panel B, the curve rises more gradually from near the origin and approaches 100% at about 5–6 mM, without a clear plateau. There is no legend or title."}], "full_mark": 2}, {"sub_id": "Exam-21-T6-3", "tags": ["1.7", "2.9"], "text_open": "6.3 Diclofenac has a half-life of about 1.0h in humans and is cleared mainly through CYP2C9-catalyzed 4′-hydroxylation of an aromatic ring. The structurally similar fenclorac is more resistant to aromatic hydroxylation and has a half-life exceeding 20h. Its structure is shown below.\nFrom a structural perspective, explain why fenclorac has a longer duration of action. (No more than 50 words)", "text_images": [{"file": "Exam-21-T6-3-1-prompt.png", "description": "The SMILES representation of fenclorac is 'OC(CC1=C(OC2=CC=C(Cl)C=C2Cl)C=CC=C1)=O'."}], "full_mark": 2}]} {"problem_id": "Exam-21-T7", "header": "Silver in the Making\nIn the worldview of alchemists, metals were not inert matter but substances that \"grew\" and \"evolved\" within Earth's crust. They believed that all base metals, such as lead and iron, were evolving toward the perfect noble metals, gold and silver. Metal M has the largest atomic number among elements with no definite year or person of discovery. As early as about 1500 CE, the Inca used M, Cu, and Sn to make alloys for the tumi knife.", "header_images": [], "subquestions": [{"sub_id": "Exam-21-T7-1", "tags": ["2.1", "1.1"], "text_open": "7.17.1 When miners and alchemists unearthed M near silver veins, glinting with a faint pink metallic luster, they found it harder than lead, more brittle than tin, and remarkably low-melting. They called M \"silver in the making\" (Tectum Argenti). Adding oxide A of M to excess hydrochloric acid gives a solution of B. Solid B and M are mixed and heated to 325℃, then cooled at 270℃ for 1–2 weeks. Excess B is removed by extraction with benzene, yielding a black solid chloride C containing 16.52% chlorine. C consists of a nine-atom cluster cation D, a mononuclear dianion E, and a dinuclear dianion F; M has the same oxidation state in B, E, and F (reaction 1). Adding excess concentrated sodium hydroxide to a nitrate solution of M may be regarded as forming the simple hydroxide precipitate G. Passing chlorine through this suspension gives the brown, strongly oxidizing powder H (reaction 2).\n7.1.2 Write equations for reactions 1 and 2. Explain why $NaBiO_3$ is a strong oxidizing agent.", "text_images": [], "full_mark": 4}]} -{"problem_id": "Exam-21-T8", "header": "Lithiation Reactions\nFrom the late 30s of the 20th century to the early 50s, Gilmann and Wittig successively discovered lithium–halogen exchange in aryl halides and directed ortho-lithiation of aromatic compounds by substituent heteroatoms. These reactions have become important methods for forming carbon–lithium bonds and constructing complex molecular structures.", "header_images": [], "subquestions": [{"sub_id": "Exam-21-T8-1", "tags": ["2.4"], "text_open": "8.1 Write the SMILES strings for the major products A and B in the reactions below.", "text_images": [{"file": "Exam-21-T8-1-1-prompt.png", "description": "Two parallel reactions share the same substrate: 'ClC1=CC(Br)=CC=C1O[Si](C(C)C)(C(C)C)C(C)C'>1) LDA 1.5eq, THF, -78°C; 2) MeI>[A]; 'ClC1=CC(Br)=CC=C1O[Si](C(C)C)(C(C)C)C(C)C'>1) nBuLi 1.1eq, THF, -78°C; 2) MeI>[B]. The structures of products A and B are not drawn; they are shown only as A and B."}], "full_mark": 2}, {"sub_id": "Exam-21-T8-2", "tags": ["1.4"], "text_open": "8.2 Ortho-lithiation can be viewed as being controlled by two limiting mechanisms: a coordination mechanism and an acid–base mechanism. In most cases, both mechanisms influence the reaction to different extents. A schematic of the coordination mechanism is shown below.\n8.2.1 TMEDA (tetramethylethylenediamine) and butyllithium can form a dinuclear complex with the formula $Li_2Bu_2(TMEDA)_2$, in which both Li atoms are 4-coordinate. Write its SMILES string.", "text_images": [{"file": "Exam-21-T8-2-1-prompt.png", "description": "Schematic of the coordination mechanism: substrate '[H]C1=C(C=CC=C1)CN([R])[R]' forms intermediate '[H]C1=C(C=CC=C1)C[N]([R])([R])[Li]CCCC' through a reversible arrow labeled nBuLi, and the intermediate gives product '[R][N]1([R])CC2=C(C=CC=C2)[Li]1' through a single arrow. A curved arrow runs from the nBu-Li bond to the ortho H atom of the benzene ring. In the product, Li and N are joined by a dashed coordination bond to form a five-membered chelate ring. R2 is the original placeholder in the figure for the two R substituents on N; Li coordination in the intermediate and product is shown by dashed lines."}], "full_mark": 2}, {"sub_id": "Exam-21-T8-3", "tags": ["2.3", "2.7"], "text_open": "The following lithiation gives different products with different additives. No workup has been performed, and each product contains only 1 lithium atom.\n8.2.2 Write the SMILES strings for the major products C and D, and identify the limiting mechanism that may be regarded as controlling each reaction.", "text_images": [{"file": "Exam-21-T8-3-1-prompt.png", "description": "Two parallel reactions share the same substrate: 'COC1=CC=C(CN(C)C)C=C1'>nBuLi>[C]; 'COC1=CC=C(CN(C)C)C=C1'>nBuLi, TMEDA>[D]. The structures of products C and D are not drawn; they are shown only as C and D."}], "full_mark": 4}, {"sub_id": "Exam-21-T8-4", "tags": ["2.4"], "text_open": "8.3 Lithiation reactions are also prone to side reactions, making control of the reaction conditions especially important.\n8.3.1 The following reaction did not give the expected major product. Instead, E, which has the same molecular formula as the expected product, was obtained in 70% yield. Write the SMILES string for E.", "text_images": [{"file": "Exam-21-T8-4-1-prompt.png", "description": "The substrate on the left is 'CC1=C(OC(N(CC)CC)=O)C=CC=C1'. A reaction arrow labeled sec-BuLi, TMEDA, THF above and -78°C-r.t. below points to product E. A box on the right labeled 'Expected product' gives the SMILES representation 'CC1=C(OC(N(CC)CC)=O)C([Li])=CC=C1'. The structure of E is not drawn and is shown only as E."}], "full_mark": 2}, {"sub_id": "Exam-21-T8-5", "tags": ["2.4", "2.2"], "text_open": "8.3.3 The following reaction causes fragmentation of the substrate. Write the SMILES strings for all products.", "text_images": [{"file": "Exam-21-T8-5-1-prompt.png", "description": "The substrate on the left is 'C1(C2=CC=CC=C2)=NOC=C1C3=CC=CC=C3'. A reaction arrow labeled RLi points to '? + ?'; the product structures are not drawn."}], "full_mark": 2}, {"sub_id": "Exam-21-T8-6", "tags": ["2.4"], "text_open": "8.4 Under base-induced conditions, a halogen substituent in a halogenated aromatic compound may migrate to a new substitution site. Such reactions are vividly known as halogen-dance reactions. 8.4.2 Some studies suggest that increasing solvent polarity accelerates the halogen-dance reaction. Write the SMILES string for the major product of the reaction below, knowing that it contains only two halogen atoms.", "text_images": [{"file": "Exam-21-T8-6-1-prompt.png", "description": "The substrate on the left is 'IC1=C(Br)C(F)=CC=C1'. A reaction arrow labeled 1. BuLi, toluene, -100°C above and 2. THF, -75°C; 3. CO2, then HCl below points to the right; the product structure is not drawn."}], "full_mark": 2}]} {"problem_id": "Exam-21-T9", "header": "Problem 9 Reactions of Sulfur Ylides\nSulfur ylides are important intermediates in organic synthesis with distinctive reactivity. Through the Corey–Chaykovsky reaction, they enable transformations such as cyclopropanation, epoxidation, and aziridination. Recent advances in organocatalytic sulfur-ylide reactions have opened new routes to the enantioselective synthesis of complex molecules.", "header_images": [], "subquestions": [{"sub_id": "Exam-21-T9-1", "tags": ["2.2"], "text_open": "9.1 A sulfur ylide and an α,β-unsaturated ketone undergo the asymmetric cyclopropanation shown below under catalysis by a chiral amine organocatalyst.\n9.1.1 The chiral amine catalyst first undergoes condensation with one substrate. Write the SMILES string for key condensation intermediate A.", "text_images": [{"file": "Exam-21-T9-1-1-prompt.png", "description": "Single-step reaction: 'O=C(/C=S(C)/C)C1=CC=CC=C1'.'O=C(C)/C=C/C1=CC=CC=C1' gives 'O=C([C@@H]1[C@@H](C2=CC=CC=C2)[C@@H]1C(C)=O)C3=CC=CC=C3' under amine-cat (20 mol%), PhCO2H (20 mol%), CHCl3, MS, 30°C, 72h. The two substrates are separated by a plus sign. The sulfur-ylide SMILES is shown in a neutral form without formal charges. The cyclopropane product contains 3 stereocenters, with @/@@ retained in the SMILES; PhC(O)- is shown by a solid wedge and Ph by a dashed wedge. Below the reaction, the chiral amine catalyst amine-cat is shown as 'N[C@@H](C1=CC=CC=C1)[C@H](C2=CC=CC=C2)NC3CCCCC3', with C-NH_2 on a solid wedge and C-NH on a dashed wedge."}], "full_mark": 2}, {"sub_id": "Exam-21-T9-2", "tags": ["2.2", "2.4"], "text_open": "9.2 In addition to conventional cyclopropanation, some sulfur ylides conjugated with double bonds can generate distinctive fused-ring products.\n9.2.1 Write the SMILES strings for the three key intermediates in the following reaction, ignoring stereochemistry.", "text_images": [{"file": "Exam-21-T9-2-1-prompt.png", "description": "Single-step reaction: 'O=C1N(C2=CC=CC=C2)C3=CC=CC=C3C1(OC(OC(C)(C)C)=O)C(C(OCC)=O)=C'.'C[S+](C)C/C=C/C(OCC)=O.[Br-]'>NaOH, CH_3CN>'O=C1N(C2=CC=CC=C2)C3=CC=CC=C3[C@@]14[C@](C5)(C(OCC)=O)[C@@]4([H])C=C5C(OCC)=O'.'O=C1N(C2=CC=CC=C2)C3=CC=CC=C3[C@@]14[C@@](C5)(C(OCC)=O)[C@]4([H])C=C5C(OCC)=O'. The two products are separated by a plus sign and are stereoisomers; @/@@ stereochemical markers are retained in the SMILES."}], "full_mark": 6}, {"sub_id": "Exam-21-T9-3", "tags": ["2.4"], "text_open": "9.2.2 Using the information from the preceding question, write the SMILES string for the product of the following reaction.", "text_images": [{"file": "Exam-21-T9-3-1-prompt.png", "description": "Single-step reaction: 'O=S1(C2=CC=CC=C2C(/C=C/C3=CC=CC=C3)=N1)=O'.'C[S+](C)C/C=C/C(OC)=O.[Br-]'>CsCO3 (3 eq), CH3CN, rt>[product not shown; only ? appears]. Reactant 2 is labeled (2.5 eq)."}], "full_mark": 2}, {"sub_id": "Exam-21-T9-4", "tags": ["2.3", "2.4"], "text_open": "9.2.3 A similar substrate gives a product different from those in the preceding two reactions. Explain why.", "text_images": [{"file": "Exam-21-T9-4-1-prompt.png", "description": "Single-step reaction: 'O=C(C1=CC(Br)=CC=C1S/2)C2=C\\C3=CC=CC=C3'.'C[S+](C)C/C=C/C(OC)=O.[Br-]'>NaH, CHCl_3, 1min>'O=C(OC)C1=CCOC2=C(SC3=CC=C(Br)C=C32)C1C4=CC=CC=C4'."}], "full_mark": 3}, {"sub_id": "Exam-21-T9-5", "tags": ["2.2", "2.4"], "text_open": "9.3.1 Researchers have also developed a DABCO-catalyzed (3+3) cyclization between a 2-(acetoxymethyl)buta-2,3-dienoate and a sulfur ylide, providing a convenient route to 4H-pyrans bearing a vinyl sulfide group. Write the SMILES strings for the four key intermediates in the following reaction, using $NR_3$ in place of DABCO.", "text_images": [{"file": "Exam-21-T9-5-1-prompt.png", "description": "Single-step reaction: 'C=C=C(C(OCC1=CC=CC=C1)=O)COC(C)=O'.'O=C(C[S+](C)C)C1=CC=CC=C1.[Br-]'>20 mol% DABCO, 1.2 eq K2CO3, acetone>'CC1=C(C(OCC2=CC=CC=C2)=O)CC(SC)=C(C3=CC=CC=C3)O1'."}], "full_mark": 8}, {"sub_id": "Exam-21-T9-6", "tags": ["2.3"], "text_open": "9.3.2 Replacing the carbonyl-stabilized sulfur ylide with an ester-stabilized ylide gives a different product. Explain why.", "text_images": [{"file": "Exam-21-T9-6-1-prompt.png", "description": "Single-step reaction: 'C=C=C(C(OCC1=CC=CC=C1)=O)COC(C)=O'.'O=C(C[S+](C)C)OCC.[Br-]' gives 'C#CC1(C(OCC2=CC=CC=C2)=O)CC1C(OCC)=O' under 20 mol% DABCO, 1.2 eq K_2CO_3, acetone."}], "full_mark": 2}, {"sub_id": "Exam-21-T9-7", "tags": ["2.4"], "text_open": "9.3.3 If the allene substrate is replaced by a less reactive alkene, cyclization cannot be completed. Write the SMILES string for the product of the following reaction.", "text_images": [{"file": "Exam-21-T9-7-1-prompt.png", "description": "Single-step reaction: 'C=C(C(OCC)=O)COC(C)=O'.'O=C(C[S+](C)C)C1=CC=CC=C1.[Br-]'>20 mol% DABCO, 1.2 eq K_2CO_3, acetone>[product not shown; only ? appears]."}], "full_mark": 2}]} +{"problem_id": "Exam-21-T10", "header": "Problem 10 Building a Very Familiar Aromatic Ring—the Simple Shikimic Acid Pathway\nAmino acids are essential substances that our bodies require continuously. Some can be synthesized in the body, whereas a few cannot and must be obtained from food, including Phe (phenylalanine), Tyr (tyrosine), and Trp (tryptophan), whose structures are shown below.\nHow do plants, microorganisms, and other organisms synthesize these amino acids? A widespread natural route produces aromatic compounds and many related six-membered cyclic compounds. It is named after a key intermediate: the shikimic acid pathway. Note: stereochemistry is required throughout this problem unless otherwise stated.", "header_images": [{"file": "Exam-21-T10-header-1.png", "description": "Phe has the SMILES representation '[H][C@@](N)(C(O)=O)CC1=CC=CC=C1'; Tyr, 'N[C@]([H])(C(O)=O)CC1=CC=C(O)C=C1'; and Trp, 'N[C@]([H])(C(O)=O)CC1=CNC2=C1C=CC=C2'. The three structures are shown side by side, with -NH2 at each stereogenic carbon drawn on a solid wedge."}], "subquestions": [{"sub_id": "Exam-21-T10-1", "tags": ["2.4", "1.2"], "text_open": "10.1 The shikimic acid pathway begins with condensation of phosphoenolpyruvate (PEP) and erythrose 4-phosphate (E4P).\n10.1.1 Write the SMILES string for the major organic product A. A carries one negative charge, contains only one phosphorus atom, and the reaction creates one new stereocenter of absolute configuration R.", "text_images": [{"file": "Exam-21-T10-1-1-prompt.png", "description": "Single-step reaction: 'O=P(OC(C(O)=O)=C)([O-])O' (PEP, with one negative charge on a phosphate oxygen).'O=P(O)([O-])OC[C@@H](O)[C@@H](O)C([H])=O' (E4P, with one negative charge on a phosphate oxygen)>>[A]. No reaction conditions are written, and the product is shown only as A without an explicit structure."}], "full_mark": 3}, {"sub_id": "Exam-21-T10-2", "tags": ["2.4", "1.2"], "text_open": "10.1.2 A readily isomerizes, so the reaction actually gives a mixture of A and its isomers. Write the SMILES string for the principal component M other than A.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-21-T10-3", "tags": ["2.2", "2.3"], "text_open": "10.2 A cyclizes to form 3-dehydroquinate in a process mediated by $NAD^+$ and NADH. Earlier studies proposed the following mechanism. First, A is oxidized by 1 molecule of $NAD^+$ to give Int1, with loss of one stereocenter. Int1 undergoes β-elimination of the phosphate group and tautomerization to give Int2, which contains an α, β-diketone unit. Int2 is then reduced by 1 molecule of NADH to give Int3, restoring the lost stereocenter with its original relative configuration. Int3 undergoes intramolecular condensation to form 3-dehydroquinate ($C_7H_{10}O_6$). Write the SMILES strings for Int1, Int2, Int3, and 3-dehydroquinate.", "text_images": [{"file": "Exam-21-T10-3-1-prompt.png", "description": "NAD+ has the SMILES representation 'NC(C1=CC=C[N+]([R])=C1)=O'; NADH has the SMILES representation 'NC(C1=CN([R])C=CC1)=O'. R is the original placeholder in the figure for the substituent on the pyridinium/1,4-dihydropyridine ring nitrogen."}], "full_mark": 9}]} +{"problem_id": "Exam-22-T1", "header": "Problem 1 Chemistry of the Rare-Earth Elements\nRare-earth elements are indispensable strategic resources for modern high technology, new energy, advanced materials, defense, and related fields. Often called the “vitamins of industry” and a “treasure house of new materials,” they comprise scandium, yttrium, and all the lanthanides. For convenience, the symbol Ln is commonly used in coordination and inorganic chemistry to denote the lanthanides or the entire rare-earth series. This problem examines rare-earth chemistry in detail.\nNote: Different letters or alphanumeric combinations in this problem may represent the same element or chemical formula.", "header_images": [], "subquestions": [{"sub_id": "Exam-22-T1-1", "tags": ["1.3"], "text_open": "1.2 Consider a rare-earth element M1 as an example of rare-earth chemistry. M1 has two common oxidation states in its compounds. Elemental M1 is a silvery-white metal that slowly tarnishes in air. M1 dissolves in dilute hydrochloric acid, releasing colorless gas A and giving a pale blue-green solution (Reaction 1). Adding dilute sulfuric acid dropwise to this solution immediately produces white precipitate B, which is insoluble in dilute nitric acid. After B is filtered, washed, and dried, it is ignited to constant mass at 800 °C under Ar; the white precipitate becomes dark-brown powder C (Reaction 2), an oxide of M1. If the mass of B is 2.480 g, 1.760 g of C is obtained after ignition. A separate portion of B is heated with dilute hydrochloric acid and bromine water until it dissolves (Reaction 3), during which the solution changes from pale blue-green to colorless (or very pale pink). The solution is divided into two portions. Excess aqueous ammonia and oxalic acid are added separately to give white gelatinous precipitate D (with aqueous ammonia) and white precipitate E (with oxalic acid), respectively. Filtering and igniting either D or E at 900 °C likewise gives C. C is mixed with Y2O3 in a certain ratio and calcined at high temperature; product F emits bright red fluorescence under ultraviolet irradiation. 1.2.3 Write ionic equations for Reactions 1 and 3 and a chemical equation for Reaction 2.", "text_images": [], "full_mark": 6}, {"sub_id": "Exam-22-T1-2", "tags": ["1.4"], "text_open": "1.4 Coordination Compounds of Rare-Earth Elements\n1 equivalent of elemental rare-earth metal M5 and 2.8 equivalents of ethylene glycol monomethyl ether are added together to a toluene solution to form X. Addition of 2,6-dimethylphenol then precipitates orange crystals Y. Y is filtered and washed; upon slight heating it becomes a different crystal Z. Further characterization provides the following information:\n① The effective magnetic moment of X is 7.9 $\\mu_B$.\n② Crystal diffraction shows that Y belongs to the orthorhombic crystal system, with $a$ = 12.748 Å, $b$ = 12.900 Å, and $c$ = 29.220 Å. The unit cell contains 2 formula units, and the theoretical density of Y is 1.428 $\\mathrm{g/cm^3}$.\n③ Elemental analysis of Z gives the following experimental mass fractions: $\\omega(C)$ = 39.60%, $\\omega(O)$ = 18.86%, and $\\omega(H)$ = 5.74%. Crystal diffraction shows that Z belongs to the tetragonal crystal system, with $a$ = 14.837 Å and $c$ = 15.117 Å. The crystal contains 2 formula units, and the density of Z is 1.1854 times the theoretical density of Y.\n④ The anion of Z contains an $[M5_4O_4]$ cluster framework and exactly two types of ligands. One ligand adopts two different coordination modes. All M5 atoms are 7-coordinate, and a particular conformation has 1 twofold rotation axis.\n⑤ The oxidation state of M5 is the same in X, Y, and Z.\n1.4.2 Determine the chemical formulas of X, Y, and Z, showing your calculations. In your answer, the ethylene glycol monomethyl ether anion may be abbreviated as $mee^-$, and the 2,6-dimethylphenoxide anion as $dmp^-$.\n", "text_images": [], "full_mark": 9}, {"sub_id": "Exam-22-T1-3", "tags": ["1.4"], "text_open": "1.4.3 Write the SMILES strings for the cation and anion of Z. You may treat all $\\sigma$ coordinate bonds as ordinary single bonds and represent the structures using SMILES.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-22-T1-4", "tags": ["2.1"], "text_open": "1.5 Electrochemistry of the Rare-Earth Elements\nThe following standard electrode potentials are given for acidic medium at 25 °C:\n$O_2(g) + 4H^+ + 4e^- \\rightarrow 2H_2O$ $E° = +1.229\\ \\mathrm{V}$\n$M1^{n+} + e^- \\rightarrow M1^{(n-1)+}$ $E° = -0.35\\ \\mathrm{V}$\n1.5.1 At pH = 0 and 25 °C, calculate the common logarithm $\\lg K$ of the standard equilibrium constant for oxidation of $M1^{(n-1)+}$ to $M1^{n+}$ by $O_2$.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-22-T1-5", "tags": ["2.1"], "text_open": "1.5.2 The partial pressure of oxygen in air is $p(O_2)$ = 0.21 bar. Use a calculation to determine whether $M1^{(n-1)+}$ will be oxidized by oxygen in air when pH = 4.00 and $[M1^{(n-1)+}]$ = $[M1^{n+}]$ = 1.00 mol·L-1.", "text_images": [], "full_mark": 2}]} {"problem_id": "Exam-22-T2", "header": "Problem 2 Determination of Nickel by EDTA Titration\nNickel is an important strategic metal widely used in stainless steel, high-temperature alloys, and battery materials. Accurate determination of its content is essential for quality control. EDTA complexometric titration is commonly used to determine nickel, but the reaction between $Ni^{2+}$ and EDTA is slow. Back titration (adding excess EDTA first and then back-titrating with a standard $Cu^{2+}$ solution) is therefore often used to improve endpoint sensitivity. This problem concerns the preparation of standard solutions, the principle of back titration, and the elimination of interferences.", "header_images": [], "subquestions": [{"sub_id": "Exam-22-T2-1", "tags": ["2.1"], "text_open": "2.2 Sample Decomposition\nWeigh $m$ = 0.3000 g of the sample into a 250 mL polytetrafluoroethylene beaker. Add 15 mL of concentrated nitric acid and 10 mL of concentrated hydrochloric acid, and heat at low temperature on a hot plate until no obvious reaction remains. If the silicon content of the sample exceeds 1%, add 10–15 drops of concentrated hydrofluoric acid to assist dissolution. Add 10 mL of perchloric acid and continue heating at low temperature until the sample is completely decomposed. Fume with perchloric acid until nearly dry, then remove from the heat. Add 10 mL of hydrochloric acid (1+1), rinse down the beaker wall with water, add water to approximately 50 mL, and gently boil to dissolve the salts. Cool to room temperature under running water, transfer into a 200 mL volumetric flask, dilute to the mark with water, and mix thoroughly.\nIf glassware is used during decomposition, additional Si will be introduced. Write the reaction equation involving $SiO_2$ in the glass.", "text_images": [], "full_mark": 2}]} {"problem_id": "Exam-22-T3", "header": "Problem 3 Body-Centered Unit Cells and Interstitial Sites", "header_images": [], "subquestions": [{"sub_id": "Exam-22-T3-1", "tags": ["1.3"], "text_open": "3.3 Changing the metal species in the tetragonal crystal above and introducing a slight distortion gives a monoclinic crystal. Its projection along the $b$ direction is shown below, with black spheres representing chromium atoms, gray spheres representing manganese atoms, and white spheres representing phosphorus atoms.\n3.3.1 Give the chemical formula of this crystal.", "text_images": [{"file": "Exam-22-T3-1-1-prompt.png", "description": "Projection of the monoclinic crystal along the $b$ direction. The main drawing is enclosed by a thin solid black, inclined parallelepiped wireframe representing the unit-cell boundary. Its bottom edge is horizontal, and its inclined edges run from upper left to lower right with a slope of approximately -1. The bottom edge is approximately equal in length to its corresponding height. The positive $c$ direction runs from left to right, while the positive $a$ direction runs from lower right to upper left along an inclined edge. The drawing contains 12 solid black spheres representing chromium atoms, 12 gray spheres representing manganese atoms, and 8 white spheres representing phosphorus atoms. Short line segments connect the spheres into a network. The centers of the gray spheres have the following $(a,c)$ coordinates: (0.66936, 0.94746), (0.33064, 0.05254), (0.53460, 0.39020), (0.46540, 0.60980), (0.35230, 0.32464), (0.64770, 0.67536), (0.16936, 0.94746), (0.83064, 0.05254), (0.03460, 0.39020), (0.96540, 0.60980), (0.85230, 0.32464), and (0.14771, 0.67536). The centers of the black spheres have the following $(a,c)$ coordinates: (0.10287, 0.18012), (0.89713, 0.81988), (0.57754, 0.97218), (0.42246, 0.02781), (0.22212, 0.39334), (0.77788, 0.60666), (0.60287, 0.18012), (0.39713, 0.81988), (0.07754, 0.97218), (0.92246, 0.02781), (0.72212, 0.39334), and (0.27788, 0.60666). The centers of the white spheres have the following $(a,c)$ coordinates: (0.53944, 0.33443), (0.46056, 0.66557), (0.29436, 0.24658), (0.70564, 0.75341), (0.03944, 0.33443), (0.96056, 0.66557), (0.79436, 0.24658), and (0.20564, 0.75341). A coordinate legend appears at the lower left: a small solid green circle bears a white lowercase $b$; a red arrow pointing toward the upper left is labeled with a lowercase $a$; and a blue arrow pointing right is labeled with a lowercase $c$. Apart from the legend, the figure contains no text."}], "full_mark": 3}]} -{"problem_id": "Exam-22-T6", "header": "Problem 6 Heteroatoms (23 points, 9%)\nIntroducing heteroatoms often changes chemical reactivity. The following questions examine aromatic substitution, ylide reactions, and amide reduction.", "header_images": [], "subquestions": [{"sub_id": "Exam-22-T6-1", "tags": ["2.7"], "text_open": "6.1 First consider substitution reactions of aromatic systems. Compounds A and B are separately treated with persulfate under basic conditions.\n6.1.1 Is the major substitution product of each of A and B the ortho or the para product?", "text_images": [{"file": "Exam-22-T6-1-1-prompt.png", "description": "The SMILES representation of A is '[O-]C1=CC=CC=C1', and that of B is 'NC1=CC=CC=C1'. A boxed letter A or B appears below the corresponding structure. To the right of each structure is a single-headed arrow pointing right, with $S_2O_8^{2-}$ written above it; the product is not shown. All bonds are solid lines, and no stereochemistry is indicated."}], "full_mark": 2}, {"sub_id": "Exam-22-T6-2", "tags": ["2.2"], "text_open": "6.1.2 Write the SMILES string for 1 key intermediate in the formation of the ortho product.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-22-T6-3", "tags": ["2.4", "2.2"], "text_open": "6.2.2 After comparing the ylide reactions above, consider two related transformations. E reacts separately with a carbonyl compound (an aldehyde or ketone) and with an imine, giving an alkene in each case.\nWrite the SMILES strings for F and at least 3 key intermediates in its formation. Note: some intermediates contain a five-membered ring, and sulfur is present in the products only as elemental sulfur.", "text_images": [{"file": "Exam-22-T6-3-1-prompt.png", "description": "The figure shows two parallel reactions: 'CN(C)/C(SCC(OC)=O)=[N+](C)/C'.'[H]C(C1=CC=CC=C1)=O'>BTMG, THF, -78 ℃, 2h>[F]; and 'CN(C)/C(SCC(OC)=O)=[N+](C)/C'.'[H]/C(C1=CC=CC=C1)=N\\S(C2=CC=C(C)C=C2)(=O)=O'>BTMG, CHCl3, -78 ℃ to rt, 2h>[G]. The letter E is shown below each 'CN(C)/C(SCC(OC)=O)=[N+](C)/C' structure. F and G are represented only by boxed letters, and their structures are not shown. In both reactions, the counterion of E is NTf_2^-, whose structure is not shown."}], "full_mark": 9}]} -{"problem_id": "Exam-22-T7", "header": "Problem 7 Skeletal Rearrangements of Maltol Derivatives\nThe three-dimensional construction of molecular frameworks is crucial in biochemistry and pharmaceutical research. Complex frameworks generally require changing catalysts or reagents, or even carrying out several separate reaction steps. In the reactions below, researchers investigated cascade skeletal rearrangements of maltol derivatives in depth. Simple modification of the substrate affords products with entirely different frameworks, ranging from planar to three-dimensional structures.", "header_images": [], "subquestions": [{"sub_id": "Exam-22-T7-1", "tags": ["2.2", "2.4"], "text_open": "7.1 On heating, maltol ether A undergoes rearrangement to give the key intermediate X. Trapping X with a nucleophile affords product B. When no nucleophile is added, the dimerization product C of X is also detected.\n7.1.1 Formation of the key intermediate X is known to involve only pericyclic reactions and simple tautomerization. Write the SMILES strings for 3 key intermediates in the formation of B. Stereochemistry is not required.", "text_images": [{"file": "Exam-22-T7-1-1-prompt.png", "description": "The figure shows a one-step reaction: 'O=C1C(OCC#C[H])=C(C)OC=C1'>BnSH, DCE, 150℃, 13 h>'OC1=C(O)C(C)=CC(CSCC2=CC=CC=C2)=C1C=O'. The structure on the left is labeled A, and that on the right is labeled B."}, {"file": "Exam-22-T7-1-2-prompt.png", "description": "The figure shows a one-step reaction: 'O=C1C(OCC#C[H])=C(C)OC=C1'>PhCl, 150℃, 13 h>'CC1=C2C(CCC3=C2C(C)=C(O)C(O)=C3C=O)=C(C=O)C(O)=C1O'. The structure on the left is labeled A, and the product on the right is labeled C."}], "full_mark": 4}, {"sub_id": "Exam-22-T7-2", "tags": ["2.2", "2.4"], "text_open": "7.1.2 Write the SMILES string for one key intermediate in the formation of C from X. Stereochemistry is not required.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-22-T7-3", "tags": ["2.4"], "text_open": "7.2 Replacing the nucleophile with styrene gives a reaction similar to that above and affords product D, which contains a five-membered ring. Write the SMILES string for D. Stereochemistry is not required.", "text_images": [{"file": "Exam-22-T7-3-1-prompt.png", "description": "The figure shows a one-step reaction: 'O=C1C(OCC#C[H])=C(C)OC=C1'>'C=CC1=CC=CC=C1' (styrene), 150℃, 13 h>[D]. The structure on the left is labeled A. On the right, only the molecular formula C17H16O3 of product D is given; its structure is not shown."}], "full_mark": 2}, {"sub_id": "Exam-22-T7-4", "tags": ["1.2", "2.3"], "text_open": "7.3 Introducing an allyl group into the maltol substrate above gives substrate E. Under similar conditions, E affords product F and the product G with a new framework. (PMB denotes the p-methoxybenzyl group.)\n7.3.1 The formation of G is known to proceed through an intermediate similar to X. Analyze the three-dimensional conformation of the favored transition state for the step that forms G, and use S/R notation to assign the configuration of the carbon atom in G whose stereochemistry is not labeled.", "text_images": [{"file": "Exam-22-T7-4-1-prompt.png", "description": "The figure shows a one-step reaction: 'O=C1C(OCC#CCC(C)=C)=C(C)OC=C1'>DCE, PMBOH, 150℃, 13 h>'OC1=C(O)C(C)=C(CC(C)=C)C(COCC2=CC=C(OC)C=C2)=C1C=O'+'O=C1C2(O)[C@@](C(CC(C2)=C)=C=C)(C)OC=C1'. The structure on the left is labeled E, and the two products on the right are labeled F and G, respectively."}], "full_mark": 2}, {"sub_id": "Exam-22-T7-5", "tags": ["2.4"], "text_open": "7.3.2 If substrate E in the reaction above is replaced by E’ shown below, determine whether the yield of G’, which has the same framework as G, will increase or decrease. Briefly explain your answer.", "text_images": [{"file": "Exam-22-T7-5-1-prompt.png", "description": "The SMILES representation of E’ is 'O=C1C(OCC#CCC(C)=C)=C(C(C)C)OC=C1'. The structure is labeled E’."}], "full_mark": 3}, {"sub_id": "Exam-22-T7-6", "tags": ["2.3", "2.4"], "text_open": "7.4 Introducing a new group into the substrate allows further cascade cyclization based on the fundamental reaction mode above.\n7.4.1 The following reaction undergoes a Nazarov-type cyclization to form the aromatic product I, which contains multiple fused rings. Write the SMILES string for I and explain why no external Lewis acid catalyst is required.", "text_images": [{"file": "Exam-22-T7-6-1-prompt.png", "description": "The figure shows a one-step reaction: 'O=C1C(OCC#CC2=CSC=C2)=C(C)OC=C1'>DCE, 150℃, 13 h>[I]. The structure on the left is labeled S. On the right, only the molecular formula C13H10O3S of product I is given; its structure is not shown."}], "full_mark": 4}, {"sub_id": "Exam-22-T7-7", "tags": ["2.2", "2.3"], "text_open": "7.4.2 When the substrate shown below is reacted under similar conditions, no product analogous to that in 7.4.1 is observed. Instead, two products, L and O, with complex three-dimensional frameworks are obtained.\nIntermediate J forms by the same process as X. J then undergoes two consecutive concerted reactions, both through six-membered cyclic transition states, to give product L. Alternatively, J can first form M, which undergoes an electrocyclic reaction to give the charge-separated intermediate N. Group migration in N then affords the bridged product O. Based on this information, write the SMILES strings for K, N, and O in the scheme.", "text_images": [{"file": "Exam-22-T7-7-1-prompt.png", "description": "The figure shows a multistep reaction sequence: 'O=C1C(OCC#CC2=C(C=C)C=CC=C2)=C(C)OC=C1'>PhCl, 150℃, 13 h>[J]; [J]>[K]; [K]>'O=C1[C@]([C@@]2(C)OC=C1)(O)CC3=C2C4=CC=CC=C4C=C3' (L); [J]>'C=C1CC(O/C2=C/C=O)C(C=CC=C3)=C3/C1=C(C)/C2=O' (M); [M]>[N]; [N]>[O]. J, K, N, and O are shown only as letter labels, whereas the structures of L and M are drawn. The [C@]/[C@@] descriptors in the SMILES representation of L correspond to the stereochemical orientations shown by wedged bonds in the figure."}], "full_mark": 6}]} +{"problem_id": "Exam-22-T5", "header": "Unlike most covalent compounds, whose structures can be described using 2c-2e bonds, polyhedral clusters are often electron-deficient systems and require multicenter bonding models.\nWade's rules, more generally known as the polyhedral skeletal electron pair theory (PSEPT), are an important tool for understanding cluster structures. Let $n$ denote the number of atoms or isolobal fragments that form the cluster framework. Subtracting the electrons used in terminal two-center two-electron bonds, nonskeletal lone pairs, and the internal bonding of ligands from the total valence-electron count gives the number of skeletal electrons. Skeletal electron-pair counts of $n+1$, $n+2$, $n+3$, and $n+4$ correspond, respectively, to closo, nido, arachno, and hypho structures. A closo cluster is a complete triangulated polyhedron with $n$ vertices, whereas nido, arachno, and hypho clusters may be regarded as the corresponding closo parent polyhedron with one, two, or three vertices removed, respectively. For example, in $B_5H_9$, 5 BH fragments contribute 10 electrons and 4 additional hydrogen atoms contribute another 4, giving 14 skeletal electrons or 7 skeletal electron pairs. Since the skeletal electron-pair count is $n+2$, the cluster is nido.\nThis problem uses low-valent clusters of Group 13 metals (Al, Ga, In, and Tl) to examine cluster synthesis, structure, bonding, and reactivity.", "header_images": [], "subquestions": [{"sub_id": "Exam-22-T5-1", "tags": ["1.6"], "text_open": "5.1 $B_4Cl_4$ and $B_4(t\\text{-}Bu)_4$ have tetrahedral cores, whereas the acetamido-substituted boron compound $B_4(NHAc)_4$ adopts a butterfly-shaped cyclic structure. Explain why.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-22-T5-2", "tags": ["2.1"], "text_open": "5.2 Write equations for the following reactions.\n5.2.1 $AlCl\\cdot x\\,Et_2O$ ($x>1$) reacts with $MgCp^*_2$ at low temperature in toluene/diethyl ether to form $Al_4Cp^*_4$.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-22-T5-3", "tags": ["2.1"], "text_open": "5.2.2 $GaI$ reacts with $Li(THF)_3Ge(SiMe_3)_3$ in toluene to form $[GaGe(SiMe_3)_3]_4$.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-22-T5-4", "tags": ["2.4"], "text_open": "5.4 Tetrahedral clusters not only have unusual structures but also show diverse oxidation chemistry, giving different products under different conditions. Draw the structures of products A, B, D, and E below. 5.4.1 Oxidation of [GaSi(t-Bu)_3]_4 with oxygen gives cubic cluster A = [OGaSi(t-Bu)_3]_4.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-22-T5-5", "tags": ["2.4"], "text_open": "5.4.2 Further hydrolysis of A gives product B, which contains only one ring and has only one chemical environment for its Ga atoms. Write the SMILES string for B. If necessary, describe its detailed structure in no more than 50 words. Multiple reasonable structures may be possible; writing any one reasonable SMILES string is sufficient.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-22-T5-6", "tags": ["2.4"], "text_open": "5.4.3 Oxidation of $C = [InC(SiMe_3)_3]_4$ with o-nitrosotoluene gives a cubic cluster and product D, which contains two benzene rings. Write the SMILES string for aromatic product D and, if necessary, describe its detailed structure in no more than 50 words.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-22-T5-7", "tags": ["2.4"], "text_open": "5.4.4 Oxidation of C with 1,2-thiirane gives a trigonal-bipyramidal cluster E. Write the SMILES string for E and, if necessary, describe its detailed structure in no more than 50 words.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-22-T5-8", "tags": ["2.2"], "text_open": "5.6 Reaction of $GaI$ with $Li(THF)_3R$ ($R = SiMe(SiMe_3)_2$) in toluene gives the octahedral cluster $G = [GaR]_6$. In contrast, reaction of $GaI$ with $Na(THF)_2R'$ ($R' = Si(t\\text{-}Bu)_3$) in toluene forms the octahedral cluster $H = [Ga_6R'_4Bn_2]^{2-}$, where Bn denotes benzyl.\n5.6.1 State the source of the benzyl groups in H.", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-22-T5-9", "tags": ["1.7"], "text_open": "5.6.3 Explain why G is a highly distorted octahedron whereas H is nearly a regular octahedron.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-22-T5-10", "tags": ["1.7"], "text_open": "5.8 Electron-rich cluster compounds: When the number of ligands in a cluster is much smaller than the number of metal atoms, its structure approaches that of a bare metal-atom cluster; such species are called electron-rich cluster compounds. A representative example is the metalloid cluster $K = Ga_{84}[N(SiMe_3)_2]_{20}^{4-}$. This cluster contains 84 Ga atoms, 64 of which are “bare.” Its structure has three layers: a central $Ga_2$ dumbbell with a bond length of 235 pm, approaching a $Ga\\equiv Ga$ triple bond; a middle $Ga_{32}$ football-shaped tube; and an outer assembly of three $Ga_{10}$ rings surrounded by 20 $N(SiMe_3)_2$ ligands.\nCompare K with the fullerene $Sc_2@C_{84}$, and identify which structural units in K correspond to the encapsulated $Sc_2$ and to the $C_{84}$ carbon cage, respectively.", "text_images": [{"file": "Exam-22-T5-10-1-prompt.png", "description": "Unlabeled three-dimensional ball-and-stick molecular model. The overall structure is a vertically symmetric multilayer assembly. The upper and lower layers each contain a polyhedron made of dark-blue spheres, approximately 12 in total. A larger central polyhedron is formed by approximately 12 purple spheres, and the outermost framework contains approximately 24 orange-yellow spheres. Spheres of the same color are connected by solid lines of the same color, with radial connections between layers. The drawing represents the layered structure of the metalloid cluster $K = Ga_{84}[N(SiMe_3)_2]_{20}^{4-}$ described in the question."}], "full_mark": 2}]} {"problem_id": "Exam-22-T8", "header": "Problem 8 Cascade Cyclizations Involving \\alpha-Oxoketenes (25 points, 11%)\n\\alpha-Oxoketenes are highly reactive intermediates that readily react with various dienes, dipoles, and nucleophiles. Unlike ordinary ketenes, \\alpha-oxoketenes favor [4+2] rather than [2+2] cyclization reactions.", "header_images": [{"file": "Exam-22-T8-header-1.png", "description": "General reaction scheme for an \\alpha-oxoketene with nucleophiles and unsaturated components, using R1, R2, Nu, Y, and Z as placeholders. The \\alpha-oxoketene has SMILES '[R2]C(C([R1])=C=O)=O'. Three arrows originate from this structure. The first points to a molecule with SMILES '[R2]C(C([R1])C([Nu])=O)=O', with NuH written above the arrow. The second points to a molecule with SMILES '[R2]C(C(CC=[Z][Y]1)([R1])C1=O)=O', with a molecule having SMILES 'C=C[Z]=[Y]' shown above the arrow. The third points to a molecule with SMILES '[R2]C(O[Y][Z]1)=C([R1])C1=O', with a molecule having SMILES '[Z]=[Y]' shown above the arrow."}], "subquestions": [{"sub_id": "Exam-22-T8-1", "tags": ["2.2", "2.4"], "text_open": "8.1.1 The Wolff rearrangement is one of the best methods for generating \\alpha-oxoketenes in situ. The resulting intermediate can be trapped immediately by various nucleophiles or participate in cycloaddition reactions, rapidly increasing molecular complexity. For example, dimedone A gives diazoketone B as shown, which rearranges to the corresponding \\alpha-oxoketene C. Write the SMILES string for one key electrically neutral intermediate M in the conversion of A to B, and write the SMILES string for C.", "text_images": [{"file": "Exam-22-T8-1-1-prompt.png", "description": "The figure shows a two-step reaction: 'O=C1CC(C)(C)CC(C1)=O'>TsN3, NEt3, MeCN, rt, 13 h>'O=C1CC(C)(C)CC(C1=[N+]=[N-])=O'. Only “heat” is written above the second arrow, and the product is labeled “\\alpha-oxo ketene C” without a drawn structure."}], "full_mark": 2}, {"sub_id": "Exam-22-T8-2", "tags": ["2.4"], "text_open": "8.1.4 Diazoketone B generally undergoes thermolysis to form a highly reactive \\alpha-oxoketene, which then reacts with nucleophiles. However, B can also add to nucleophiles at room temperature. For example, reaction of A and B in aqueous ammonia gives the tetracyclic compound G, which contains two unusual six-membered rings. The $^{1}$H NMR spectrum of G shows four sets of resonances at $\\delta$ 12.4, 2.6, 2.5, and 1.1. Write the SMILES string for G.", "text_images": [{"file": "Exam-22-T8-2-1-prompt.png", "description": "The figure shows a one-step reaction: 'O=C1CC(C)(C)CC(C1)=O'.'O=C1CC(C)(C)CC(C1=[N+]=[N-])=O'>NH3·H2O>[G]. G is shown only as a label; its structure is not drawn."}], "full_mark": 2}, {"sub_id": "Exam-22-T8-3", "tags": ["2.4"], "text_open": "8.2 Reactions of \\alpha-Oxoketenes (I)\nThe 1,5-benzodiazepin-2-one, or diazepine-type, framework is an important pharmacologically active structure. The following method constructs this framework in a single step.\n8.2.1 An \\alpha-oxoketene contains two electrophilic carbon atoms in a 1,3 relationship and can react with a 1,4-dinucleophile to give the seven-membered diazepine-type compound H1. The reaction is performed under microwave irradiation without additives, proceeds within minutes, and gives moderate to good yields. Write the SMILES string for H1.", "text_images": [{"file": "Exam-22-T8-3-1-prompt.png", "description": "The figure shows a one-step reaction: 'NC1=CC=CC=C1N'.'O=C1CC(C)(C)CC(C1=[N+]=[N-])=O' (B, 1 eq)>Toluene, microwave, 180 °C, 6 min>[H1]. H1 is shown only as a label; its structure is not drawn."}], "full_mark": 2}, {"sub_id": "Exam-22-T8-4", "tags": ["2.4", "2.3"], "text_open": "8.2.2 When the reaction is performed with a seven-membered cyclic ketone diazo compound, the expected product is not obtained; instead, the rearranged imidazolone product H2 forms. Write the SMILES string for H2. Note: lowering the temperature to 130℃ gives reaction product H3, whose structure is similar to H1, and H2 and H3 are constitutional isomers.", "text_images": [{"file": "Exam-22-T8-4-1-prompt.png", "description": "The figure shows a one-step reaction: 'NC1=CC=CC=C1N'.'O=C1CCCCC(C1=[N+]=[N-])=O' (seven-membered cyclic ketone diazo compound, 1 eq)>Toluene, microwave, 180 °C, 3 min>[H2]. H2 is shown only as a label; its structure is not drawn."}], "full_mark": 2}, {"sub_id": "Exam-22-T8-5", "tags": ["2.4", "4.3"], "text_open": "8.3 Reactions of \\alpha-Oxoketenes (II)\n8.3.1 Diels-Alder reaction product: In the following reaction, two reactive species undergo a Diels-Alder reaction to give product J. The IR spectrum of J shows the following peak wavenumbers, IR (neat): 2956, 2866, 1759, 1692, 1409, 1342, 1245, 1144, and 697 cm$^{-1}$. Write the SMILES string for product J.", "text_images": [{"file": "Exam-22-T8-5-1-prompt.png", "description": "The figure shows a one-step reaction: 'O=C1CC(C)(C)CC(C1=[N+]=[N-])=O'.'O=C(N=[N+]=[N-])C1=CC=CC=C1' (5 eq)>Toluene, 130 °C>[J]. J is shown only as a label; its structure is not drawn."}], "full_mark": 2}, {"sub_id": "Exam-22-T8-6", "tags": ["2.4"], "text_open": "8.3.2 Cascade ring closure involving Diels-Alder reactions. The following transformation undergoes one intermolecular Diels-Alder reaction followed by one intramolecular Diels-Alder reaction, constructing the hexacyclic compound K in a single operation. Write the SMILES string for K. Stereochemistry is not required.", "text_images": [{"file": "Exam-22-T8-6-1-prompt.png", "description": "The figure shows a multicomponent reaction: 'CC(CC(C1=[N+]=[N-])=O)(C)CC1=O'.'[H]C(C1=NC(C2=CC=CC=C2)=CO1)=O'.'NCC=C'>Toluene, microwave, 140 °C, 5 min>[K]. K is shown only as a label; its structure is not drawn."}], "full_mark": 2}, {"sub_id": "Exam-22-T8-7", "tags": ["2.2", "2.3", "2.4"], "text_open": "8.3.3 Spirocyclic ring systems can be constructed by reactions of ketenes with imines. The potential-energy diagram for the following reaction is shown below.\nIn the formation of M from C and L, 3 key intermediates, INT1, INT2, and INT3, are involved. They are constitutional isomers, and conversion of INT1 to INT2 involves an s-trans to s-cis conformational change of the diene. Write the SMILES strings for INT1, INT2, and INT3, and identify the type of pericyclic reaction by which INT2 forms product M.", "text_images": [{"file": "Exam-22-T8-7-1-prompt.png", "description": "Reaction potential-energy diagram. The vertical axis is labeled “Relative energy” in kJ/mol, with tick marks at -75, -50, -25, 0, 25, and 50. The horizontal axis is labeled “Reaction coordinate.” In the legend, a solid line denotes pathway I and a dashed line denotes pathway II. Solid pathway I starts at 0 kJ/mol, crosses a 44.8 kJ/mol barrier to intermediate INT1 at 7.6 kJ/mol, crosses a 49.2 kJ/mol barrier to INT2 at 24.2 kJ/mol, and then crosses a 56.9 kJ/mol barrier to the product at -73.4 kJ/mol. Dashed pathway II starts from INT1, crosses a 27.9 kJ/mol barrier to INT3 at -17.3 kJ/mol, and then joins the product at -73.4 kJ/mol."}, {"file": "Exam-22-T8-7-2-prompt.png", "description": "The figure shows a one-step reaction: 'CC(CC(C1=[N+]=[N-])=O)(C)CC1=O'.'C/N=C/C=C/C1=CC=CC=C1'>Toluene, 160 °C, 15 min>'O=C(CC(C)(C)C1)[C@@]21[C@@H](C3=CC=CC=C3)C=CN(C)C2=O'."}], "full_mark": 7}]} {"problem_id": "Exam-23-T1", "header": "Question 1 Chemistry of Element X (28 points, 12%)\nX was first discovered in 1898 and is widely used in the electronics and electric-light-source industries. It is relatively unreactive and exhibits several oxidation states in its compounds. Common fluorides of X include A, B, and C, all colorless crystals at room temperature, with the mass fraction of fluorine increasing in that order.", "header_images": [], "subquestions": [{"sub_id": "Exam-23-T1-1", "tags": ["2.1"], "text_open": "1.1 A, B, and C can all be prepared by reacting elemental X with fluorine under suitable conditions. A is soluble in water and reacts slowly with it. At 275 ℃, A reacts with CrO_2F_2 to form D, which contains 41.6% Cr. B hydrolyzes rapidly in water and undergoes disproportionation to produce two gases; at 273.15 K and 100 kPa, the gas mixture has a density of 3.91 g·L^-1. Complete hydrolysis of C gives an explosive oxide of X.\n1.1.2 Write the equations for the reaction of A with $CrO_2F_2$ to form D and for the reaction of B with water.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-23-T1-2", "tags": ["2.1"], "text_open": "1.1.3 Dry A and B can both be stored in glass containers, but placing C in a glass or quartz container, whose principal component may be taken as $SiO_2$, is highly dangerous. Use a chemical equation to explain why C cannot be stored in a quartz container.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-23-T1-3", "tags": ["1.4", "1.1"], "text_open": "1.2 In addition to forming covalent compounds with elements such as fluorine and oxygen, X forms a rich variety of ionic compounds. The earliest reported X-containing anion can be prepared as follows: C hydrolyzes and disproportionates under basic conditions to give two gases and octahedral anion E. In addition, C reacts with an equimolar amount of CsF to give F; heating F converts it to G, in which the mass fraction of X is 23.91%.\n1.2.2 In the X-containing anions above, the X–O and X–F distances are close to the sums of the corresponding covalent radii, indicating covalent bonding. In some X-containing anions discovered later, however, the X–halogen distance is substantially longer than the sum of the covalent radii but shorter than the sum of the van der Waals radii. To explain this observation, scientists proposed the concept of an X bond (AeB). When X is bonded to a highly electronegative element or electron-withdrawing group, its electron distribution changes markedly, creating a region of high electrostatic potential along the extension of the $\\sigma$-bond axis, a $\\sigma$ hole. X thereby acts as an electrophilic center that can interact with nucleophiles such as halide ions.\n1.2.2.1 F hydrolyzes slowly in moist air to give H, whose formula may be regarded as an adduct of an oxide of X with $CsF$. In the crystal, the anion of H forms an infinite one-dimensional chain. Write its SMILES string.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-23-T1-4", "tags": ["1.4", "1.1"], "text_open": "1.2.2.2 In the presence of $CsF$, partial hydrolysis of C gives I. I contains 47.93% X by mass, and all X atoms have the same chemical environment. Write the SMILES string for the anion of I.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-23-T1-5", "tags": ["3.2"], "text_open": "1.3 The physical properties of A, B, and C have also been studied extensively. Researchers measured the standard equilibrium constants of the following reactions at several temperatures; all species are gaseous and treated as ideal gases.\n\n| Temperature (℃) / Reaction | 250 | 300 | 350 | 400 |\n| --- | --- | --- | --- | --- |\n| $X + F_2 \\rightarrow A$ | $8.79\\times10^{4}$ | $1.02\\times10^{4}$ | $1.67\\times10^{3}$ | $3.59\\times10^{2}$ |\n| $A + F_2 \\rightarrow B$ | $1.43\\times10^{3}$ | $1.55\\times10^{2}$ | 27.2 | 4.86 |\n| $B + F_2 \\rightarrow C$ | 0.944 | 0.211 | 0.0558 | 0.0182 |\n\n1.3.1 Calculate $\\Delta_r H_m^θ$ and $\\Delta_r S_m^θ$ for each reaction, assuming $\\Delta_r H_m^θ$ and $\\Delta_r S_m^θ$ are temperature-independent.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-23-T1-6", "tags": ["3.4"], "text_open": "1.3.3 At 300 ℃, 1 mol of X and 5 mol of $F_2$ are introduced into a 1 L vessel. Calculate the mole fraction of C in the equilibrium gas mixture.", "text_images": [], "full_mark": 6}]} -{"problem_id": "Exam-23-T3", "header": "Question 3 Reversible-Deactivation Anionic Polymerization (8 points, 4%)\nConventional anionic polymerization of methacrylates, such as ethyl methacrylate shown below, requires highly hazardous organolithium initiators. Because the chain end is extremely reactive, the reaction must be conducted at very low temperature, such as -78 ℃, under rigorously anhydrous and oxygen-free conditions; otherwise severe side reactions terminate the chains.", "header_images": [{"file": "Exam-23-T3-header-1.png", "description": "The SMILES representation of ethyl methacrylate (EMA) is 'O=C(OCC)C(C)=C'."}], "subquestions": [{"sub_id": "Exam-23-T3-1", "tags": ["2.2"], "text_open": "3.1 Chemists recently developed a $CO_2$-mediated reversible-deactivation anionic polymerization (CMAP). The method operates at 90 ℃, uses a stable and safe solid potassium salt as initiator, and produces polymers with a very narrow molecular-weight distribution, or very low dispersity $Đ$, under a $CO_2$ atmosphere. The key conditions are shown below.\n3.1.1 Anionic polymerization initiator A first decarboxylates to form an enolate that initiates polymerization. Write the SMILES string for A.", "text_images": [{"file": "Exam-23-T3-1-1-prompt.png", "description": "'O=C(C(C)=C)OCC'>1.0 equiv. A; 18-crown-6 (1.2 equiv.); TEMPO (0.1 equiv.); 1 atm $CO_2$; DMF, 90°C, 6 h>'O=C(C(CC(C(OCC)=O)(C)C(O[K])=O)(C)C)OCC'. The product is poly(ethyl methacrylate), poly-EMA. Its repeating unit is enclosed in brackets with subscript n at the lower right. The listed product SMILES schematically represents an EtO-C(=O)-C(CH3)2- initiating end connected through -CH2- to a dormant-end unit -C(CH3)(CO2Et)(CO2K)."}], "full_mark": 1}, {"sub_id": "Exam-23-T3-2", "tags": ["2.8"], "text_open": "3.1.2 State the principal chemical role of 18-crown-6 (shown as 18-crown-6 in the reaction scheme) in the system.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-23-T3-3", "tags": ["2.4", "2.3"], "text_open": "3.2 Control of side reactions is also an important topic in anionic polymerization.\n3.2.1 In conventional high-temperature anionic polymerization without $CO_2$ protection, the active chain end of a polymethacrylate readily attacks an upstream carbonyl group in the polymer backbone. This intramolecular backbiting completely terminates the chain. Write the SMILES string for the final cyclic end group formed after backbiting termination of active poly(ethyl methacrylate), poly-EMA; only one chain end need be shown.", "text_images": [], "full_mark": 2}]} {"problem_id": "Exam-23-T4", "header": "Question 4 Determination of Aluminum by Titration (25 points, 10%)\nAluminum is an important component of light alloys, and its content directly affects strength, corrosion resistance, and processability. EDTA displacement titration is a classical method for determining aluminum. Excess EDTA is first added to complex aluminum ions, and the excess is back-titrated with standard zinc solution. Fluoride is then added to displace EDTA from Al–EDTA, and the released EDTA is titrated with standard zinc solution to determine the aluminum content.", "header_images": [], "subquestions": [{"sub_id": "Exam-23-T4-1", "tags": ["2.1"], "text_open": "4.1 Preparation of Standard Solutions. Disodium ethylenediaminetetraacetate (EDTA) solution (0.02 mol/L): Weigh 7.4448 g EDTA into a 400 mL beaker, dissolve in water, transfer to a 1000 mL volumetric flask, dilute to the mark, and mix. Standard zinc solution (0.02000 mol/L): Weigh 1.6276 g of zinc oxide (primary standard reagent) that has been dried to constant mass at 110 ℃ into a 400 mL beaker. Add 20 mL hydrochloric acid (1 + 1) and heat to dissolve. Add approximately 200 mL water and then add aqueous ammonia (1 + 1) dropwise until Congo red paper changes from blue to violet. Cool, transfer to a 1000 mL volumetric flask, dilute to the mark with water, and mix. 4.2 Dissolution of the Sample. Accurately weigh 1.0008 g of the alloy sample into a 400 mL glass beaker. Add 20 mL~30 mL hydrochloric acid (1 + 1) and heat to dissolve; for samples that are difficult to dissolve, an appropriate amount of concentrated hydrochloric acid or concentrated nitric acid may be added. Add 5 mL concentrated nitric acid dropwise for oxidation, then add 10 mL~30 mL sulfuric acid (1 + 1). Evaporate until sulfuric acid fumes have evolved for 1 min. Remove from heat and cool slightly, add 50 mL water, and heat to dissolve the salts. 4.3 Separation of the Precipitate. Acidity adjustment: Add 100 mL of complexing-agent solution (containing 200 g/L ammonium citrate and 25 g/L ammonium oxalate) to the test solution, and adjust the pH to 4~5 with aqueous ammonia (1 + 1) and sulfuric acid (1 + 8). If a precipitate forms in a tungsten-containing sample, first add concentrated aqueous ammonia until basic and heat to dissolve the tungstic acid precipitate. After cooling, adjust the pH to 4~5 with sulfuric acid (1 + 1) and aqueous ammonia (1 + 1). If the test solution is turbid, heat it to boiling until clear, cool, and readjust the pH to 4~5. Precipitation and dissolution: Transfer the test solution to a 400 mL plastic beaker, add a small amount of paper pulp, and add 100 mL sodium fluoride solution (35 g/L) with stirring. Stir thoroughly for 5 min and let stand for 40 min. Filter through slow filter paper containing a small amount of paper pulp, wash 7~8 times with sodium fluoride solution (5 g/L), and discard the filtrate. Return the precipitate and filter paper to the original glass beaker and immediately add 20 mL hydrochloric acid-boric acid mixed acid (boric acid:hydrochloric acid:water = 15:25:60) to dissolve the precipitate; it must be added immediately, or complete dissolution will be difficult. Add 20 mL~25 mL concentrated nitric acid and boil without covering with a watch glass to destroy the filter paper. Cool slightly, add 7 mL concentrated perchloric acid, and heat to near dryness with fuming. Cool slightly and add 50 mL water to dissolve the salts.\n4.3.3 Write the ionic equation for addition of sodium fluoride in the “Separation of the Precipitate” step. The purpose of adding paper pulp during filtration is (select one).\nA. Increase precipitate purity\nB. Loosen the gelatinous precipitate and prevent passage through the filter\nC. Act as an indicator\nD. Adsorb impurity ions", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-23-T4-2", "tags": ["2.1"], "text_open": "4.3.4 Write the ionic equation for dissolution of the precipitate by the hydrochloric acid–boric acid mixture.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-23-T4-3", "tags": ["4.5"], "text_open": "4.4 Titration\nAccurately add 20.00 mL EDTA solution to the test solution and boil for 2–3 min. Add 2 drops of 2,4-dinitrophenol solution (2 g/L), then add ammonia (1 + 1) dropwise until the solution is pale yellow. Continue boiling for 1–2 min and cool; if the yellow color fades, add ammonia dropwise until pale yellow again. Add 20 mL acetate–ammonium acetate buffer (pH 5.0–5.5) and 4 drops of xylenol orange solution (2 g/L). Titrate with standard zinc solution until violet-red, without recording the reading. Add 1–2 g sodium fluoride, boil for 2–3 min, and cool to room temperature. If the solution is violet-red, add hydrochloric acid (1 + 1) dropwise until the color just disappears. Add another 10 mL acetate–ammonium acetate buffer and 1–2 drops of xylenol orange solution. Titrate with standard zinc solution to a violet-red endpoint; 12.66 mL is consumed.\n4.4.1 Does the EDTA solution require standardization in this experiment? Briefly explain.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-23-T4-4", "tags": ["2.1"], "text_open": "4.4.2 Write the ionic equation for addition of sodium fluoride followed by boiling, using $H_2Y^{2-}$ for the doubly charged ethylenediaminetetraacetate ion.", "text_images": [], "full_mark": 2}]} {"problem_id": "Exam-23-T5", "header": "Problem 5 Water-Molecule Cages (24 points, 10%)\nThe cubic unit-cell structure of the alloy $Zr_3Au$ is shown below. Au atoms occupy the corners and body center of the cell. Each face contains 2 Zr atoms arranged symmetrically along a direction parallel to a coordinate axis (their three coordinates are 0, 1/2, and either 1/4 or 3/4).", "header_images": [{"file": "Exam-23-T5-header-1.png", "description": "Schematic cubic unit cell of the alloy $Zr_3Au$, drawn as a thin gray wireframe cube. Two kinds of glossy spheres are shown: dark, nearly black spheres near the cube corners and body center, and light gray-white spheres arranged symmetrically near the faces. Thin gray lines connect the spheres. The spheres carry no labels. At lower left is a three-colored coordinate triad: a red arrow points down-left and is labeled with a lowercase a, a green arrow points horizontally right and is labeled b, and a blue arrow points vertically upward and is labeled c. No other text appears in the image."}], "subquestions": [{"sub_id": "Exam-23-T5-1", "tags": ["1.3"], "text_open": "5.1 State the lattice type of crystalline $Zr_3Au$.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-23-T5-2", "tags": ["1.3"], "text_open": "5.2 State the number of nearest, equidistant Zr atoms around each Au atom.", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-23-T5-3", "tags": ["1.3"], "text_open": "5.3 For each Zr atom, state separately the number of nearest, equidistant Au atoms, the number of nearest, equidistant Zr atoms, and the number of next-nearest, equidistant Zr atoms.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-23-T5-4", "tags": ["1.3"], "text_open": "5.4 Treat the Zr atoms described in 5.2 as the coordinating atoms of Au, and the Zr and Au atoms described in 5.3 as the coordinating atoms of Zr. For each coordination polyhedron in the crystal, connecting its center to all of its vertices produces a number of small tetrahedra. State:\n(1) the number of faces of the coordination polyhedron around an Au atom;\n(2) the number of faces of the coordination polyhedron around a Zr atom;\n(3) the number of coordination polyhedra that contain each small tetrahedron;\n(4) the number of small tetrahedra in one unit cell.", "text_images": [], "full_mark": 7}, {"sub_id": "Exam-23-T5-5", "tags": ["1.3"], "text_open": "5.5 Replace every atomic position in $Zr_3Au$ with a methane molecule and place one water molecule at the center of each small tetrahedron described above. The resulting structure is that of methane hydrate. Write the formula of methane hydrate in the form $xCH_4\\cdot yH_2O$.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-23-T5-6", "tags": ["1.3", "2.1"], "text_open": "5.6 The unit-cell parameter of methane hydrate is $a = 1.20\\ \\mathrm{nm}$. Calculate the maximum theoretical volume of methane gas that can be released from $1\\ \\mathrm{m^3}$ of methane hydrate at 273 K and 1 atm.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-23-T5-7", "tags": ["1.3"], "text_open": "5.7 The cubic crystal of another similar hydrate can be described as follows. Hexadecahedra $[5^{12}6^4]$ whose vertices are water molecules (the notation denotes 12 pentagonal faces and 4 hexagonal faces; the polyhedron has the same symmetry as a regular tetrahedron and is shown below) are joined pairwise by sharing hexagonal faces, forming an infinite diamond-like structure. These hexadecahedra clearly do not fill all space: the four tetrahedral-void positions that are unoccupied in diamond remain. In fact, exactly four pentagonal dodecahedra, also with water molecules at their vertices, fit into each such void. The dodecahedra share pentagonal faces with one another and with the hexadecahedra, and each water molecule in the final structure is shared by four polyhedra. Place one $CCl_4$ molecule at the center of every hexadecahedron and one Xe atom at the center of every dodecahedron.\nStructure of the hexadecahedron $[5^{12}6^4]$\n5.7.1 Write the formula of this crystal in the form $xCCl_4\\cdot yXe\\cdot zH_2O$.", "text_images": [{"file": "Exam-23-T5-7-1-prompt.png", "description": "Wireframe representation of the hexadecahedron $[5^{12}6^4]$. Its vertices are shown as small hollow white circles with thin black outlines, and its edges as thin black lines. Edges facing the viewer are solid, while hidden rear edges are dashed. Alternating hexagonal and pentagonal faces form an approximately spherical cage. The image contains no text, numbers, arrows, or coordinate-axis labels."}], "full_mark": 3}]} {"problem_id": "Exam-23-T6", "header": "Problem 6 Determination of the Oxygen-Uptake Ratio of an Oxygen Carrier (23 points, 10%)\nA laboratory has prepared inactive Co(Salen) and wishes to determine the oxygen-uptake ratio $n_{O_2}:n_{[Co(II)(Salen)]}$ after the carrier has fully absorbed $O_2$ from its surroundings under specified conditions. Inactive Co(Salen) itself has almost no oxygen-uptake capacity, but it dissolves in coordinating solvents such as DMF or DMSO, and the resulting solution can absorb oxygen. The main procedure is described below:", "header_images": [{"file": "Exam-23-T6-header-1.png", "description": "Schematic apparatus for measuring the oxygen-uptake ratio, with no text, letters, numbers, or arrow labels. At left is a U-shaped gas burette. Its left arm is vertical and bears evenly spaced short, unnumbered graduations. Both arms contain liquid, with the level in the left arm higher than that in the right. In the center is a three-neck round-bottom flask containing liquid in its lower portion and resting above a rectangular base. The left neck connects to a downward-sloping bent addition tube that also contains liquid; its liquid level lies below that in the flask, and its downward-facing open end is not submerged. The middle neck connects to a long vertical tube that bends left at the top and enters the right arm of the gas burette; this connecting tube contains no liquid. The right neck connects to a two-way stopcock assembly with a spherical bulb above it, a short tube at the top, and two symmetric thin rods; this part contains no liquid."}], "subquestions": [{"sub_id": "Exam-23-T6-1", "tags": ["4.1"], "text_open": "6.1 Assembly of the Measurement Apparatus. Accurately weigh 0.1136 g Co(Salen) and transfer it through a dry glass funnel into a clean, dry three-neck flask. Assemble the apparatus as shown below. Apply a layer of petroleum jelly, as thin as possible, to the ground-glass joints of the bent tube, gas inlet tube, and two-way stopcock. Place 5 mL DMF (or DMSO) in the bent tube and connect the charging bent tube to the flask. Use a narrow-tip glass dropper to add 1 mL deionized water to the gas burette, whose graduations increase upward. Inflate the oxygen bag and connect it to the two-way stopcock on the right. Apply pressure to the oxygen bag so that air is expelled through the graduated gas burette. When the oxygen in the bag has been almost completely discharged, close the stopcock. Add approximately 5 mL more deionized water to the graduated gas burette so that the difference between the liquid levels exceeds 2 cm. Observe the liquid levels; the system may be considered to have reached thermal equilibrium when the change in liquid level is less than 0.03 mL over three consecutive minutes.\n6.2 Determination of the Oxygen-Uptake Ratio\nAfter assembling the apparatus, determine the oxygen-uptake ratio of the Co(Salen) sample as follows.\nRecord the laboratory temperature and atmospheric pressure. Remove the oxygen bag, open the stopcock to the atmosphere, and keep the gas burette vertical. When the liquid levels in the two arms are equal (difference less than 0.03 mL), close the stopcock and record the position of the liquid level in the gas burette. Then introduce DMF (or DMSO) into the flask, stir rapidly, and observe the changes in the reactants. Promptly add deionized water to the gas burette with a narrow-tip glass dropper to keep the liquid levels in its two arms equal. Record the liquid-level position every 20~40 seconds until ________. Calculate the actual volume of oxygen absorbed.\n6.2.1 State how the DMF (or DMSO) should be introduced into the flask.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-23-T6-2", "tags": ["4.2"], "text_open": "6.2.2 State the observation that should fill the blank in the procedure above.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-23-T6-3", "tags": ["4.7"], "text_open": "6.2.4 Suppose the preceding group cleaned all of the apparatus with a volatile organic solvent S, which has no coordinating ability, after performing the same experiment. Before starting, a student finds a small amount of liquid solvent S remaining in the three-neck flask. If the apparatus is used directly, will the measured oxygen-uptake ratio be higher or lower than the true value? Briefly explain.", "text_images": [], "full_mark": 3}]} {"problem_id": "Exam-23-T7", "header": "Problem 7 Reactivity, Preparation, and Applications of Silanes (32 points, 10%)\nSilanes ($Si_nH_{2n+2}$) and their halogenated derivatives are important precursors for high-purity silicon films, semiconductor devices, and solar cells. Unlike hydrocarbons, silanes contain highly reactive Si–Si and Si–H bonds, creating distinctive challenges in their synthesis and purification.", "header_images": [], "subquestions": [{"sub_id": "Exam-23-T7-1", "tags": ["1.2"], "text_open": "7.4.3 Mixing the tridentate ligand pedeta (pentaethyldiethylenetriamine, $Et_2NCH_2CH_2NEtCH_2CH_2NEt_2$) with trichlorosilane in DCM at room temperature gives A = $[(pedeta)H_2SiCl^+]_2[Si_6Cl_{14}^{2-}]$.\n7.4.3.1 How many stereoisomers can the cation of A theoretically have?", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-23-T7-2", "tags": ["1.4"], "text_open": "7.4.3.2 The anion of A has an approximate sixfold rotation axis. Write its SMILES string.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-23-T7-3", "tags": ["2.3"], "text_open": "7.4.3.3 Mechanistic experiments show that adding 1,3-butadiene to this reaction system does not produce detectable 1,1-dichlorosilacyclopent-3-ene. Which of the following is therefore the most likely key one-silicon intermediate?\nA. $:SiCl_2$\nB. $SiCl_3$\nC. $SiCl_3^-$\nD. $SiCl_4^-$", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-23-T7-4", "tags": ["2.3"], "text_open": "7.5 Wurtz coupling. More complex silane molecules can be obtained through Wurtz coupling.\n7.5.1 Mechanistically, which class does the first reaction belong to?\nA. Nucleophilic aromatic substitution\nB. Electrophilic aromatic substitution\nC. Radical substitution\nD. Transition-metal-catalyzed reaction", "text_images": [{"file": "Exam-23-T7-4-1-prompt.png", "description": "The image shows two reaction sequences. The upper row contains three steps: 'C1([Si]2(C3=CC=CC=C3)[Si](C4=CC=CC=C4)(C5=CC=CC=C5)[Si](C6=CC=CC=C6)(C7=CC=CC=C7)[Si](C8=CC=CC=C8)(C9=CC=CC=C9)[Si]2(C%10=CC=CC=C%10)C%11=CC=CC=C%11)=CC=CC=C1'>CF3SO3H>'O=S(O[Si]1(C2=CC=CC=C2)[Si](C3=CC=CC=C3)(C4=CC=CC=C4)[Si](C5=CC=CC=C5)(C6=CC=CC=C6)[Si](C7=CC=CC=C7)(C8=CC=CC=C8)[Si]1(C9=CC=CC=C9)C%10=CC=CC=C%10)(C(F)(F)F)=O'; 'O=S(O[Si]1(C2=CC=CC=C2)[Si](C3=CC=CC=C3)(C4=CC=CC=C4)[Si](C5=CC=CC=C5)(C6=CC=CC=C6)[Si](C7=CC=CC=C7)(C8=CC=CC=C8)[Si]1(C9=CC=CC=C9)C%10=CC=CC=C%10)(C(F)(F)F)=O'>NaBr (3eq), Et2O>[B]; [B]>naphthalene (3eq), Li (6eq), DME, -30 °C>'C1([Si]2(C3=CC=CC=C3)[Si](C4=CC=CC=C4)(C5=CC=CC=C5)[Si]([Si]6(C7=CC=CC=C7)[Si](C8=CC=CC=C8)(C9=CC=CC=C9)[Si](C%10=CC=CC=C%10)(C%11=CC=CC=C%11)[Si](C%12=CC=CC=C%12)(C%13=CC=CC=C%13)[Si]6(C%14=CC=CC=C%14)C%15=CC=CC=C%15)(C%16=CC=CC=C%16)[Si](C%17=CC=CC=C%17)(C%18=CC=CC=C%18)[Si]2(C%19=CC=CC=C%19)C%20=CC=CC=C%20)=CC=CC=C1'. In the lower row, the starting material at left is omitted. An arrow labeled 'HCl (g), AlCl3' above and 'PhH, 50 °C' below leads to an intermediate labeled only C. C is the product obtained by treating the upper-row product (Si10Ph18) with HCl/AlCl3; the connection is omitted because it does not fit on one line. [C]>LiAlH4 (6eq), C6H14/Et2O, -40 °C>'[H][Si]1([H])[Si]([H])([H])[Si]([H])([Si]2([H])[Si]([H])([H])[Si]([H])([H])[Si]([H])([H])[Si]2([H])[H])[Si]([H])([H])[Si]1([H])[H]'."}], "full_mark": 1}, {"sub_id": "Exam-23-T7-5", "tags": ["2.4"], "text_open": "7.5.2 Write the SMILES strings for intermediates B and C.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-23-T7-6", "tags": ["4.3"], "text_open": "7.5.3 Identify which silicon atom in the product (fused cyclopentasilane) appears farthest upfield in the NMR spectrum, that is, at the smallest chemical shift δ.", "text_images": [], "full_mark": 1}]} -{"problem_id": "Exam-23-T10", "header": "Problem 10 Dearomatization of Benzene Rings (27 points, 13%)\nAromaticity makes the benzene ring structurally stable and provides substantial resonance stabilization. Directly disrupting its aromatic conjugation or cleaving a C–C bond in the ring framework normally requires considerable energy. Dearomatization or ring-opening cleavage of a benzene ring is therefore highly challenging.", "header_images": [], "subquestions": [{"sub_id": "Exam-23-T10-1", "tags": ["2.2"], "text_open": "10.1 In 2024, researchers developed a novel benzene-ring-opening strategy that directly converts a range of aryl azide derivatives into valuable multiply conjugated enynes.\n10.1.1 Write the SMILES string for intermediate A formed before $^tBuONa$ is added.", "text_images": [{"file": "Exam-23-T10-1-1-prompt.png", "description": "Single-step reaction: 'O=C(C1=CC=CC=C1N=[N+]=[N-])C2=CC=CC=C2'>TosNHNH2 (1.2 equiv), TFA (20 mol%), PhCl (0.2 M); then $^tBuONa$ (1.5 equiv), PhCl (0.1 M), 130°C>'N#C/C=C\\C=C/C#CC1=CC=CC=C1'."}], "full_mark": 1}, {"sub_id": "Exam-23-T10-2", "tags": ["2.2", "2.3"], "text_open": "10.1.2 To investigate the mechanism, the researchers performed the following experiments.\nCalculations indicate that after $^tBuONa$ is added, loss of one molecule of gas first produces highly reactive intermediate B. B then reacts with the azide group to form charge-separated tricyclic intermediate C. A concerted reaction of C expels another gas molecule to give the product. Write the SMILES strings for B and C.", "text_images": [{"file": "Exam-23-T10-2-1-prompt.png", "description": "Two parallel mechanistic experiments. Upper reaction: the starting material is labeled only A and its structure is not drawn; conditions are TEMPO (2.0 equiv), $^tBuONa$ (1.5 equiv), PhCl, heat, giving 'N#C/C=C\\C=C/C#CC1=CC=CC=C1' in 70% yield. Lower reaction: 'C1(C(C2=CC=CC=C2)=NN=N3)=C3C=CC=C1'>$^tBuONa$ (1.5 equiv), PhCl, heat>'N#C/C=C\\C=C/C#CC1=CC=CC=C1', 0% yield."}], "full_mark": 3}, {"sub_id": "Exam-23-T10-3", "tags": ["2.4"], "text_open": "10.1.3 Considering the reaction sequence above, write the SMILES strings for the products of the following reactions. (Hint: each reaction proceeds through a high-energy intermediate similar to B.)", "text_images": [{"file": "Exam-23-T10-3-1-prompt.png", "description": "Two parallel reactions, arranged vertically, with neither product drawn. Upper reaction: 'O=C(C1=CC=CC=C1N=[N+]=[N-])C2=CC=C(C)C=C2'>TosNHNH2 (1.2 equiv), TFA (20 mol%), PhCl; then $^tBuONa$ (1.5 equiv), PhCl, heat>[product not shown]. Lower reaction: 'O=C(C)/C(C(C)=O)=C/C#CC1=CC=CC=C1N=[N+]=[N-]'>PhCl (0.05 M), heat>[product not shown]."}], "full_mark": 3}, {"sub_id": "Exam-23-T10-4", "tags": ["2.2"], "text_open": "10.2 In 2025, the group of Professor Daniele Leonori reported oxidative cleavage of electron-rich aromatic rings using photoexcited nitroarenes. (λ = 390 nm denotes irradiation with 390 nm light.)\n10.2.1 Under irradiation, the two substrates react to form a five-membered-ring intermediate Int, whose fragmentation gives the product. Write the SMILES string for Int; the aromatic ring attached to the nitro group may be abbreviated as -Ar.", "text_images": [{"file": "Exam-23-T10-4-1-prompt.png", "description": "Photooxidative-cleavage reaction: 'COC1=CC=CC2=CC=CC=C21'.'O=[N+](C1=CC=CC=C1)[O-].[R]C'>CH2Cl2 (0.1 M), -65°C, 5h, λ = 390 nm>'O=C(OC)C1=CC=CC=C1/C=C\\C=O'. [R] in substrate 2 is the original R-label placeholder."}], "full_mark": 1}, {"sub_id": "Exam-23-T10-5", "tags": ["3.9", "1.6"], "text_open": "10.2.2 The researchers reacted differently substituted nitroarenes with a substrate containing both an aromatic ring and an alkene and found that the substituents affect chemoselectivity. For example, nitroarenes N1 and N2 react with D to give selective oxidation of either the alkene or the aromatic ring.\nStudies of nitroarene excited states show that their lowest triplet state may have either a $^3(n,\\pi^*)$ or $^3(\\pi,\\pi^*)$ electronic configuration. These correspond, respectively, to an n to $\\pi^*$ transition, representable as a dioxygen-diradical species, and a $\\pi, \\pi^*$ transition, involving intramolecular electron transfer from the benzene unit to the nitro group with the unpaired electron delocalized over the entire $\\pi$ system. Based on the electronic properties of the substituents, which state is lower in energy for N1, and which is lower for N2?\n(a) n, $\\pi^*$ state; (b) $\\pi, \\pi^*$ state.", "text_images": [{"file": "Exam-23-T10-5-1-prompt.png", "description": "Three compounds, labeled D, N1, and N2 from left to right. The SMILES for N1 is 'O=[N+](C1=CC=C(OC)C(OC)=C1)[O-]'; for N2, 'O=[N+](C1=CC(C(F)(F)F)=CC(C(F)(F)F)=C1)[O-]'; and for D, 'C/C(C)=C\\CCOC1=CC=CC2=C1C=CC=C2'."}], "full_mark": 2}, {"sub_id": "Exam-23-T10-6", "tags": ["2.7", "3.9"], "text_open": "10.2.3 A triplet excited state with an n, $\\pi^*$ configuration can undergo cycloaddition directly. A triplet excited state with a $\\pi, \\pi^*$ configuration cannot undergo direct concerted cycloaddition; calculations indicate the following pathway. It first forms a complex with the other substrate, in which dispersion is the principal intermolecular attraction. Partial electron transfer between the substrates then gives the nitro group a partial negative charge and the other substrate a partial positive charge, promoting formation of the first C–O bond. Finally, intersystem crossing and ring closure give the five-membered intermediate. Accordingly, a nitroarene excited state with an n, $\\pi^*$ configuration reacts more readily with ( ), whereas one with a $\\pi, \\pi^*$ configuration reacts more readily with ( ). Explain each choice.\n(a) An electron-rich aromatic ring; (b) an isolated double bond.", "text_images": [], "full_mark": 5}, {"sub_id": "Exam-23-T10-7", "tags": ["2.4"], "text_open": "10.2.4 Write the SMILES string for the product formed when D is oxidized by N2 under irradiation.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-23-T10-8", "tags": ["2.2", "2.3"], "text_open": "10.3 In 2024, Professor Youliang Wang's group achieved thermochemical dearomatization of a benzene ring, constructing a complex ring system in one step.\nExperiments show that when TFA is replaced by deuterated TFA and the reaction is run for 10 h, 99% of the circled hydrogen atom in the product is replaced by deuterium.\n10.3.1 The reaction includes a pericyclic step. Write the SMILES strings for 3–4 key intermediates and name the pericyclic reaction.", "text_images": [{"file": "Exam-23-T10-8-1-prompt.png", "description": "Single-step reaction: 'C=CCN(C1=CC=CC=C1)C(C#CC2=CC=CC=C2)=O'>TFA, 50°C, 12h>'O=C(N1C2)C([H])=C(C31C4C5C=CC3C2C4)C6=C5C=CC=C6'. The hydrogen at C([H]) in the product is circled, corresponding to the deuterium-labeling experiment in the prompt."}], "full_mark": 7}, {"sub_id": "Exam-23-T10-9", "tags": ["2.4"], "text_open": "10.3.2 The reaction also forms two byproducts. E has a ring system similar to that of the major product, whereas F contains only three rings and no bridged ring. Write the SMILES strings for E and F.", "text_images": [], "full_mark": 3}]} diff --git a/test/instance_ids.json b/test/instance_ids.json index 3f3bdac0ff86df394ada090fffe58db3fa721c95..bb58bba456e620457bd757f4dc61e16d36f49f16 100644 --- a/test/instance_ids.json +++ b/test/instance_ids.json @@ -1,107 +1,107 @@ [ + "Exam-01-T1", "Exam-01-T2", - "Exam-01-T3", "Exam-01-T4", - "Exam-01-T8", - "Exam-01-T9", + "Exam-01-T6", + "Exam-01-T7", "Exam-01-T10", + "Exam-02-T1", + "Exam-02-T3", "Exam-02-T8", - "Exam-02-T10", + "Exam-02-T9", "Exam-03-T1", - "Exam-03-T3", - "Exam-03-T5", - "Exam-03-T6", + "Exam-03-T7", + "Exam-03-T10", "Exam-04-T1", + "Exam-04-T2", "Exam-04-T5", "Exam-04-T7", - "Exam-04-T10", + "Exam-04-T8", "Exam-05-T1", - "Exam-05-T5", + "Exam-05-T4", "Exam-05-T9", - "Exam-05-T10", "Exam-06-T1", "Exam-06-T3", - "Exam-06-T5", - "Exam-06-T8", + "Exam-06-T4", + "Exam-06-T7", "Exam-06-T9", "Exam-06-T10", "Exam-07-T3", - "Exam-07-T4", + "Exam-07-T5", "Exam-07-T7", - "Exam-07-T9", + "Exam-07-T8", "Exam-08-T1", + "Exam-08-T3", + "Exam-08-T5", "Exam-08-T6", "Exam-09-T1", - "Exam-09-T4", "Exam-09-T6", - "Exam-09-T7", "Exam-10-T3", - "Exam-10-T5", "Exam-10-T6", - "Exam-10-T9", - "Exam-11-T3", - "Exam-11-T5", + "Exam-11-T2", "Exam-11-T7", "Exam-11-T10", + "Exam-12-T1", "Exam-12-T2", "Exam-12-T3", - "Exam-12-T5", - "Exam-12-T7", + "Exam-12-T4", "Exam-12-T8", - "Exam-12-T9", "Exam-13-T1", "Exam-13-T2", - "Exam-13-T5", + "Exam-13-T4", "Exam-13-T8", - "Exam-13-T10", + "Exam-14-T1", "Exam-14-T2", - "Exam-14-T5", + "Exam-14-T4", + "Exam-14-T7", "Exam-14-T9", "Exam-14-T10", - "Exam-15-T3", - "Exam-15-T5", + "Exam-15-T1", + "Exam-15-T4", "Exam-15-T6", "Exam-15-T7", "Exam-16-T1", "Exam-16-T2", "Exam-16-T4", - "Exam-16-T5", - "Exam-16-T6", - "Exam-16-T7", - "Exam-16-T9", - "Exam-17-T5", - "Exam-17-T6", - "Exam-17-T8", + "Exam-17-T2", + "Exam-17-T4", + "Exam-17-T7", + "Exam-17-T9", "Exam-18-T1", "Exam-18-T2", + "Exam-18-T3", + "Exam-18-T4", + "Exam-18-T5", "Exam-18-T6", - "Exam-18-T7", - "Exam-18-T8", + "Exam-18-T9", + "Exam-19-T1", + "Exam-19-T2", + "Exam-19-T3", "Exam-19-T4", "Exam-19-T8", - "Exam-19-T9", "Exam-20-T1", "Exam-20-T2", + "Exam-20-T3", + "Exam-20-T4", "Exam-20-T5", - "Exam-20-T6", - "Exam-20-T8", - "Exam-20-T9", "Exam-20-T10", "Exam-21-T1", "Exam-21-T2", + "Exam-21-T3", + "Exam-21-T4", + "Exam-21-T5", + "Exam-21-T6", "Exam-21-T7", - "Exam-21-T8", "Exam-21-T9", + "Exam-21-T10", + "Exam-22-T1", "Exam-22-T2", "Exam-22-T3", - "Exam-22-T6", - "Exam-22-T7", + "Exam-22-T5", "Exam-22-T8", "Exam-23-T1", - "Exam-23-T3", "Exam-23-T4", "Exam-23-T5", "Exam-23-T6", - "Exam-23-T7", - "Exam-23-T10" + "Exam-23-T7" ] diff --git a/test/tasks/Exam-01-T1/problem.json b/test/tasks/Exam-01-T1/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..a2cc3b34c5a5030aba2d1f8cd8a405975153ce37 --- /dev/null +++ b/test/tasks/Exam-01-T1/problem.json @@ -0,0 +1,28 @@ +{ + "problem_id": "Exam-01-T1", + "header": "Divine Gold!", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-01-T1-1", + "tags": [ + "2.1", + "1.4", + "2.4" + ], + "text_open": "1-2-2 Anhydrous AuCl_3 reacts with a small amount of CO to produce “AuCl_2,” which exists as a tetramer. The molecule possesses only a C_2 axis and a mirror plane perpendicular to the C_2 axis. Write the chemical equation for the reaction and the SMILES structure of tetrameric AuCl_2.", + "text_images": [], + "full_mark": 4 + }, + { + "sub_id": "Exam-01-T1-2", + "tags": [ + "1.4", + "2.4" + ], + "text_open": "1-3 Au can form cluster compounds with unusual frameworks. Based on the descriptions, write reasonable SMILES structures for the following complexes. 1-3-1 Complex Au_4(dppm)_3I_2: the numbers of Au atoms in two chemical environments are in a ratio of 1:3; every Au is coordinated by one I atom; and all dppm ligands have the same chemical environment. (dppm=Ph_2PCH_2PPh_2)", + "text_images": [], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-01-T6/problem.json b/test/tasks/Exam-01-T6/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..b9a24b47a219353ec0da6f8a53fe8be1b727b70d --- /dev/null +++ b/test/tasks/Exam-01-T6/problem.json @@ -0,0 +1,45 @@ +{ + "problem_id": "Exam-01-T6", + "header": "Crystal Structures of Layered Alloys", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-01-T6-1", + "tags": [ + "1.3", + "1.4" + ], + "text_open": "Structural background: Crystal X is a layered Al-Ni-U alloy in which Ni and U form layer M and Al forms layer N, with an infinite $\\cdots MN\\cdots$ sequence. In M, Ni has a graphene-like arrangement and U occupies the center of every Ni hexagon. Along $c$, all M-layer projections coincide and all N-layer projections coincide. In N, Al forms regular hexagons and equilateral triangles; Ni lies at every Al-triangle center and U at every Al-hexagon center.\nConsidering the coordination of Ni by Al and U, name the coordination polyhedron of Ni and state how many coordinating atoms are U and how many are Al.", + "text_images": [], + "full_mark": 3 + }, + { + "sub_id": "Exam-01-T6-2", + "tags": [ + "1.3" + ], + "text_open": "State the crystal system of crystal X and write its chemical formula.", + "text_images": [], + "full_mark": 3 + }, + { + "sub_id": "Exam-01-T6-3", + "tags": [ + "1.1", + "1.3" + ], + "text_open": "Crystal Y is an Al-Rh-U alloy in which all U atoms and some Rh atoms form layer P, while all Al atoms and the remaining Rh atoms form layer Q. The crystal extends infinitely in a …PQ… sequence to form a three-dimensional structure. The arrangement of U in layer P resembles that of Al in layer N of crystal X, while Rh occupies the center of every hexagon. Of the six U atoms in each hexagon, three are closer to the central Rh atom, causing distortion of layer P; correspondingly, the other three U atoms are farther from Rh, such that all triangles in layer P remain equilateral. Each U atom is shared by two hexagons, is closer to the central Rh atom of one and farther from that of the other, and lies on the line connecting those two Rh atoms. In layer Q, all Al atoms are arranged as equilateral Al_3 triangles. In projection along the c axis, the projections of all P layers coincide, as do those of all Q layers. The centers of the Al_3 triangles coincide with the Rh atoms in layer P, while the centers of the U_3 triangles coincide with the Rh atoms in layer Q. The projection of each Al atom lies on the line connecting a Rh atom in layer P and the more distant U atom in its hexagon. Define a unit cell with the Rh atoms in layer P as vertices. In this unit cell, one U atom has coordinates (0.5806, 0.0000, 0.5000). Give the coordinates of all other U atoms in the unit cell, to four decimal places, and the two types of U-U-U bond angles in a U_6 hexagon, to one decimal place.", + "text_images": [], + "full_mark": 6 + }, + { + "sub_id": "Exam-01-T6-4", + "tags": [ + "1.3" + ], + "text_open": "State the crystal system of crystal Y and write its chemical formula.", + "text_images": [], + "full_mark": 3 + } + ] +} diff --git a/test/tasks/Exam-01-T7/problem.json b/test/tasks/Exam-01-T7/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..1bf49724f6592f258a5ed9acc5d4ccaa75427241 --- /dev/null +++ b/test/tasks/Exam-01-T7/problem.json @@ -0,0 +1,16 @@ +{ + "problem_id": "Exam-01-T7", + "header": "Efficiency of Oxidative Phosphorylation", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-01-T7-1", + "tags": [ + "3.2" + ], + "text_open": "A molecular-level hypothesis has been proposed: for every $2e^-$ transferred through the electron-transfer chain, Complex III transports $2HSuc^-$ across the membrane. The opposing concentration gradients that must be overcome in the two transport events are approximately $1:7$ and $1:8$, respectively. Energy losses from other ion-transport processes at Complexes I and IV may be neglected. The inner-side condition is $pH=8$, the outer-side condition is $pH=5$, succinic acid has $pK_{a1,2}=4.2,5.6$, and the temperature is $310\\,\\mathrm{K}$.\nFor this redox reaction, $\\Delta_rG_m^{\\ominus *}\\approx-2.20\\times10^2\\ \\mathrm{kJ\\,mol^{-1}}$; take the magnitude of the available energy as $220\\ \\mathrm{kJ\\,mol^{-1}}$.\nATP synthase can transport only $HSuc^-$ ions on the outer side of the membrane; neither $H_2Suc$ nor $Suc^{2-}$ outside the membrane can pass through ATP synthase. Thus, of the energy stored in $H_2Suc$, $HSuc^-$, and $Suc^{2-}$ during the redox stage, only the fraction associated with external $HSuc^-$ can be utilized. On this basis, calculate $\\eta_{redox}$.", + "text_images": [], + "full_mark": 3 + } + ] +} diff --git a/test/tasks/Exam-02-T1/problem.json b/test/tasks/Exam-02-T1/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..7bb2fa061756e3fdd5dbf1446cc0ef2980950d11 --- /dev/null +++ b/test/tasks/Exam-02-T1/problem.json @@ -0,0 +1,38 @@ +{ + "problem_id": "Exam-02-T1", + "header": "Problem 1: Carbon Coordination! Since Frankland reported Et_2Zn in 1849, molecules containing carbon-metal bonds have been known and have continually attracted intensive study. Carbon atoms have been found to exhibit a wide variety of electron-donation numbers and coordination numbers when coordinated to metals.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-02-T1-1", + "tags": [ + "1.4", + "1.6" + ], + "text_open": "1-1 For the ligands Ph_3PCPPh_3 (denoted L) and RCR (where R is an alkyl group), state the coordination mode by which each preferentially coordinates to a transition-metal atom.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-02-T1-2", + "tags": [ + "1.1", + "1.4" + ], + "text_open": "1-2 L reacts with ZnI_2 or CdI_2 in a 1:1 ratio, but the products have different structures. The ZnI_2 complex has 2 mutually perpendicular mirror planes and 1 twofold rotation axis but no inversion center. The CdI_2 complex has only 1 mirror plane and 1 twofold rotation axis but does possess an inversion center. Draw both structures.", + "text_images": [], + "full_mark": 4 + }, + { + "sub_id": "Exam-02-T1-3", + "tags": [ + "2.1", + "2.4", + "1.4" + ], + "text_open": "1-5 As stated in the introduction, carbon atoms can donate different numbers of electrons in complexes. Reaction of WCl_6 with (CH_3)_3CCH_2Li and Ph_2PCH_2CH_2PPh_2 gives a mononuclear W complex containing 3 carbon-tungsten bonds whose lengths are all distinctly different. Write the equation for its preparation.", + "text_images": [], + "full_mark": 3 + } + ] +} diff --git a/test/tasks/Exam-02-T3/images/Exam-02-T3-header-1.png b/test/tasks/Exam-02-T3/images/Exam-02-T3-header-1.png new file mode 100644 index 0000000000000000000000000000000000000000..374631955e8b9a423c4df807f06dc639d286fae6 --- /dev/null +++ b/test/tasks/Exam-02-T3/images/Exam-02-T3-header-1.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:e7ff2d77065feea0be832e538059888fcb0e0410bb72a65ed394904b79811b8b +size 847567 diff --git a/test/tasks/Exam-02-T3/problem.json b/test/tasks/Exam-02-T3/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..7428a983b370c287a3475b9348c01b2298855eff --- /dev/null +++ b/test/tasks/Exam-02-T3/problem.json @@ -0,0 +1,41 @@ +{ + "problem_id": "Exam-02-T3", + "header": "Problem 3: Bi-Ion Removal Battery. A bi-ion removal battery (BDI) can intercalate cations and anions from seawater into crystalline phases analogous to those in lithium-ion batteries and store them through conversion reactions in the cathode and anode, enabling efficient seawater desalination with high ion-removal capacity and high charge efficiency. Studies show that a $\\mathrm{NaBi_3O_4Cl_2}$ electrode can simultaneously remove $\\mathrm{Na^+}$ and $\\mathrm{Cl^-}$ from seawater.", + "header_images": [ + { + "file": "Exam-02-T3-header-1.png", + "description": "On the left is a stacked structure of red blocks. Following the process indicated by the blue arrow “$\\mathrm{-Na^+, -Cl^-}$,” it becomes the central stacked structure of red and orange blocks. The process indicated by the red arrow “$\\mathrm{+Na^+, +Cl^-}$” then converts it to the structure on the right, composed of red and blue blocks. A further process indicated by the blue arrow “$\\mathrm{-Na^+, -Cl^-}$” returns it to the central structure. The legend identifies red blocks as $\\mathrm{NaBi_3O_4Cl_2}$, blue blocks as substance X, and orange blocks as $\\mathrm{Bi}$." + } + ], + "subquestions": [ + { + "sub_id": "Exam-02-T3-1", + "tags": [ + "2.1", + "3.5" + ], + "text_open": "Substance X is $BiOCl$.\n3-2-3 Write the electrode-reaction equations occurring at the $\\mathrm{NaBi_3O_4Cl_2}$ electrode during charging.", + "text_images": [], + "full_mark": 4 + }, + { + "sub_id": "Exam-02-T3-2", + "tags": [ + "3.5" + ], + "text_open": "3-3 To construct a complete battery, $\\mathrm{NaTi_2(PO_4)_3@C}$ is used as the electrode on the other side. 3-3-1 State the type of the $\\mathrm{NaTi_2(PO_4)_3@C}$ electrode during seawater desalination.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-02-T3-3", + "tags": [ + "2.1", + "3.5" + ], + "text_open": "3-3-2 Write the electrode-reaction equation occurring at the $\\mathrm{NaTi_2(PO_4)_3@C}$ electrode during seawater desalination.", + "text_images": [], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-02-T9/images/Exam-02-T9-1-1-prompt.png b/test/tasks/Exam-02-T9/images/Exam-02-T9-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..9ff5c96e0ff31312240d69f308491d7e80c170e5 --- /dev/null +++ b/test/tasks/Exam-02-T9/images/Exam-02-T9-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:c37fcd43b53c668e4366808de2697d1cb5707f8ed8a996d0cf4e62c78023317a +size 2511 diff --git a/test/tasks/Exam-02-T9/images/Exam-02-T9-2-1-prompt.png b/test/tasks/Exam-02-T9/images/Exam-02-T9-2-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..5476348ae61b7f53b9087c88f3dceb7dbda2f15d --- /dev/null +++ b/test/tasks/Exam-02-T9/images/Exam-02-T9-2-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:af606053ad73a7534b7315d8d15c2093b828bc91311fea6411fd3db0ce760a0c +size 15937 diff --git a/test/tasks/Exam-02-T9/images/Exam-02-T9-4-1-prompt.png b/test/tasks/Exam-02-T9/images/Exam-02-T9-4-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..637bb279fea25b35087a03bc556f03516f604b57 --- /dev/null +++ b/test/tasks/Exam-02-T9/images/Exam-02-T9-4-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:871c3fe0bead09eeb07cbe88695c49fa019974a4f096b773ac4e9481da426b93 +size 4919 diff --git a/test/tasks/Exam-02-T9/images/Exam-02-T9-4-2-prompt.png b/test/tasks/Exam-02-T9/images/Exam-02-T9-4-2-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..0e1eb64dde35b0bf902124c9c6420f1069fccf54 --- /dev/null +++ b/test/tasks/Exam-02-T9/images/Exam-02-T9-4-2-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:2d809f7739e0a26059fbcc204da3b6d13210978aee547affd687a52082f26cda +size 6509 diff --git a/test/tasks/Exam-02-T9/images/Exam-02-T9-4-3-prompt.png b/test/tasks/Exam-02-T9/images/Exam-02-T9-4-3-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..a17f16cf0799c72618ac7bebc269b9ad41b20bd0 --- /dev/null +++ b/test/tasks/Exam-02-T9/images/Exam-02-T9-4-3-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:520efdaf691d939f214e24ca39efbe39af22c47be2e8094185c0c42f6db5f25b +size 754094 diff --git a/test/tasks/Exam-02-T9/problem.json b/test/tasks/Exam-02-T9/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..dc30877814097f0872e8542a2ff3379e7b33adca --- /dev/null +++ b/test/tasks/Exam-02-T9/problem.json @@ -0,0 +1,82 @@ +{ + "problem_id": "Exam-02-T9", + "header": "Problem 9: Selectivity in Chemistry", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-02-T9-1", + "tags": [ + "2.7", + "1.2", + "2.4" + ], + "text_open": "9-1 Reduction of the following compound with different reducing agents gives different stereochemical outcomes. Write the SMILES representation of major product 9A1 obtained with LiAlH_{4} and that of major product 9A2 obtained with ^{i}Bu_{2}AlH.", + "text_images": [ + { + "file": "Exam-02-T9-1-1-prompt.png", + "description": "The structure in the image has SMILES 'O=C(C1)C[C@H]2N([C@@H]1CC2)C'." + } + ], + "full_mark": 4 + }, + { + "sub_id": "Exam-02-T9-2", + "tags": [ + "2.8" + ], + "text_open": "$Zn(OTf)_2$ is used in the selective esterification shown below. It acts as a Lewis acid to activate the ester group and coordinates separately to the two substrates to control the reaction site.\n9-2-2 What amount of Zn(OTf)_{2} should be used in this reaction?", + "text_images": [ + { + "file": "Exam-02-T9-2-1-prompt.png", + "description": "The image shows a reaction proceeding from left to right. At the far left is 'O=C(O/N=C/c2ncccc2)CCc3ccccc3', labeled **9B** below, joined by a plus sign to 'O[C@@H]1[C@@H](O)[C@H](CO)O[C@@H](OCCCCCCCC)[C@H]1NC(CO)=O' in the center. Above the reaction arrow is “Zn(OTf)_2”; below it are “DMF/THF/DME,” “(1:5:9),” and “30°C, 18h,” in sequence." + } + ], + "full_mark": 2 + }, + { + "sub_id": "Exam-02-T9-3", + "tags": [ + "2.4", + "2.7", + "1.2" + ], + "text_open": "9-2-3 Write the SMILES representation of the major product, given that only one hydroxy group is esterified.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-02-T9-4", + "tags": [ + "2.2", + "2.3" + ], + "text_open": "9-3-2 Compound 9D is the addition product of acetonitrile and hydrogen peroxide, and 9G contains a peroxide bond. Write the SMILES representations of small molecules 9D and 9F in the catalytic cycle below.", + "text_images": [ + { + "file": "Exam-02-T9-4-1-prompt.png", + "description": "Structure of compound 9C, with SMILES 'CC1(OC[C@]2(OC[C@H]3OC(C)(O[C@H]3C2=O)C)O1)C'." + }, + { + "file": "Exam-02-T9-4-2-prompt.png", + "description": "The image shows a reaction segment in which the far-left compound 'C/C=C/c1ccccc1' reacts to the right under an arrow labeled “9C (15 mol%), H_2O_2 (4 equiv.)” above and “K_2CO_3, CH_3CN” and “0°C, 12h” below." + }, + { + "file": "Exam-02-T9-4-3-prompt.png", + "description": "The image is a catalytic cycle. At the top is 'CC1(OC[C@]2(OC[C@H]3OC(C)(O[C@H]3C2=O)C)O1)C', labeled “9C” below. On the right, `CH_3CN + H_2O_2` points left by a straight arrow to “9D.” A curved arrow from “9D” runs down and left into the main cycle, merging with the downward path from “9C,” and both lead to “9E.” Two curved arrows descend from “9E,” one down-right to “9F” and the other down-left to “9G.” A curved arrow from “9G” runs up and left, meeting an upward curved arrow from 'C/C=C/c1ccccc1' at the lower left; together they lead to `[transition state]^‡`. Finally, a curved arrow from `[transition state]^‡` runs up and right to close the cycle at “9C,” with a curved branch along this upward path pointing up-left to the word “product.”" + } + ], + "full_mark": 4 + }, + { + "sub_id": "Exam-02-T9-5", + "tags": [ + "2.7", + "1.2", + "2.4" + ], + "text_open": "9-3-3 Predict the stereochemistry and write the SMILES representation of the product.", + "text_images": [], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-03-T10/images/Exam-03-T10-1-1-prompt.png b/test/tasks/Exam-03-T10/images/Exam-03-T10-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..ecebc2c6132edd6cfc3d65b4e2a63914fda78d43 --- /dev/null +++ b/test/tasks/Exam-03-T10/images/Exam-03-T10-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:ac50027644194d22d93b6a48a700fc24a05deea4745057bfb36aaf4dd3b1c28f +size 21144 diff --git a/test/tasks/Exam-03-T10/images/Exam-03-T10-3-1-prompt.png b/test/tasks/Exam-03-T10/images/Exam-03-T10-3-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..8a109e314598c26b62f006d1ec4356a28c7cc6c9 --- /dev/null +++ b/test/tasks/Exam-03-T10/images/Exam-03-T10-3-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:e806771d810e6cdfd250be76b5b75cff932067e74d3afa32ac723dd66f839485 +size 11452 diff --git a/test/tasks/Exam-03-T10/images/Exam-03-T10-4-1-prompt.png b/test/tasks/Exam-03-T10/images/Exam-03-T10-4-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..acc250b4aa9095ef52486cc75698192bd8edabe7 --- /dev/null +++ b/test/tasks/Exam-03-T10/images/Exam-03-T10-4-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:968eb104eb67c7edd85a6ecd875b8e6069125de877235f72c56076f8dcdf7408 +size 14061 diff --git a/test/tasks/Exam-03-T10/images/Exam-03-T10-5-1-prompt.png b/test/tasks/Exam-03-T10/images/Exam-03-T10-5-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..ec0da1c0428a5d1455afc6a6f20c344323cdbb36 --- /dev/null +++ b/test/tasks/Exam-03-T10/images/Exam-03-T10-5-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:43d8e1c872a2c1ea474c2e9afab66ca7d88bb00306756866580be9025d16e546 +size 11246 diff --git a/test/tasks/Exam-03-T10/problem.json b/test/tasks/Exam-03-T10/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..c67303e3d8a7b30bc876d0179c53224daff73c76 --- /dev/null +++ b/test/tasks/Exam-03-T10/problem.json @@ -0,0 +1,97 @@ +{ + "problem_id": "Exam-03-T10", + "header": "Problem 10 Synthesis of Indigoid Compounds (25 points)", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-03-T10-1", + "tags": [ + "2.4", + "2.2" + ], + "text_open": "10-1 In 1909, H. Friedländer noted that an excellent purple pigment mentioned in the Bible—the pigment that the Romans called Tyrian purple after the Phoenician port of Tyre in present-day Lebanon—was in fact 6,6'-dibromoindigo, F. This precious early dye was extracted from small purple sea snails, the murex; as many as 12000 snails were required to prepare 1.5 g of dye. After nearly a century of continuing research into its synthesis, Pinkney and Chammers proposed a new method, which Iming and coworkers improved into an efficient route. 10-1-1 Give the SMILES of intermediate E.", + "text_images": [ + { + "file": "Exam-03-T10-1-1-prompt.png", + "description": "10-1 Synthetic route. This multistep scheme begins at the upper left with 'CC1=CC=C(N)C=C1', which reacts under HNO_3 and H_2SO_4 to give intermediate B. B reacts under 1) NaNO_2, HBr and 2) CuBr to give C. C then reacts under CrO_3 and (CH_3CO)_2O to give D. In the second row, D reacts under H_2SO_4 and H_2O to give E. Finally, E reacts under NaOH and H_2O, (CH_3)_2CO to give the product F at the lower right, with SMILES 'O=C1NC2=C(C=C(Br)C=C2)C1=C3NC4=C(C=C(Br)C=C4)C3=O'." + } + ], + "full_mark": 2 + }, + { + "sub_id": "Exam-03-T10-2", + "tags": [ + "2.7", + "1.6" + ], + "text_open": "10-1-2 Explain the regioselectivity of the conversion of A into B. (No more than 50 words.)", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-03-T10-3", + "tags": [ + "2.1", + "2.2" + ], + "text_open": "10-2 In the route above, conversion of E into F is clearly the key step. In fact, as early as 1882, Baeyer used this reaction to synthesize indigo in one step from o-nitrobenzaldehyde. 10-2-1 Write the chemical equation for this reaction, representing o-nitrobenzaldehyde and indigo by their SMILES.", + "text_images": [ + { + "file": "Exam-03-T10-3-1-prompt.png", + "description": "One-step synthesis. At the far left is 'O=CC1=CC=CC=C1[N+]([O-])=O', labeled G below. Under NaOH and H_2O, (CH_3)_2CO, it gives the product at the far right, 'O=C1NC(C=CC=C2)=C2/C1=C3NC4=C(C=CC=C4)C/3=O', labeled H below." + } + ], + "full_mark": 2 + }, + { + "sub_id": "Exam-03-T10-4", + "tags": [ + "2.2", + "2.3" + ], + "text_open": "10-2-2 The final step of the reaction is shown below. Considering the byproducts formed in the equation in 10-2-1, give the SMILES of the four key intermediates in the conversion of G into M. For intermediates related by proton gain, loss, or transfer, only one need be given.", + "text_images": [ + { + "file": "Exam-03-T10-4-1-prompt.png", + "description": "10-2-2 Question figure. A reaction proceeds from left to right. At the far left is 'O=C1C=NC2=C1C=CC=C2', labeled M below, followed by a plus sign and the middle molecule 'OC1=C(NC2=C1C=CC=C2)O', labeled N below. A one-way right-pointing reaction arrow leads to the product at the far right, 'O=C5NC(C=CC=C6)=C6/C5=C7NC8=C(C=CC=C8)C/7=O', labeled H below, indicating that M and N react to form H." + } + ], + "full_mark": 6 + }, + { + "sub_id": "Exam-03-T10-5", + "tags": [ + "2.2", + "2.5" + ], + "text_open": "10-3 Nitrogen plays a key role in the reaction above, but analogous substrates and methods are unsuitable for obtaining the sulfur-substituted product. After the nineteen-eighties, Scottish chemists synthesized thioindigo, K, using the reaction below. The reaction still proceeds through an intermediate containing a fused six- and five-membered ring system. It is known that contraction of the six-membered ring into the five-membered ring does not occur by ring opening followed by ring closure. Answer the following questions. 10-3-1 In what form does the oxygen atom of the sulfoxide depart?", + "text_images": [ + { + "file": "Exam-03-T10-5-1-prompt.png", + "description": "Structure/reaction description for synthesis of thioindigo K. Sulfoxide J gives thioindigo K under NaOAc/HOAc. The SMILES of J is 'BrC(C1=O)=C([S+](C2=C1C=CC=C2)[O-])Br'; thioindigo K may be represented as 'O=C1C2=C(S/C1=C3SC4=C(C\\3=O)C=CC=C4)C=CC=C2'." + } + ], + "full_mark": 1 + }, + { + "sub_id": "Exam-03-T10-6", + "tags": [ + "2.3", + "2.7" + ], + "text_open": "10-3-2 During ring contraction, which atom is lost? ( ) (a) the carbon atom nearer the carbonyl group (b) the carbon atom nearer the sulfoxide group", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-03-T10-7", + "tags": [ + "2.2", + "2.3" + ], + "text_open": "10-3-4 Give the SMILES of the six key intermediates in the reaction. For intermediates related by proton gain, loss, or transfer, only one need be given. A substitution proceeding by addition-elimination or by an S_{N1} process counts as only one step.", + "text_images": [], + "full_mark": 9 + } + ] +} diff --git a/test/tasks/Exam-03-T7/problem.json b/test/tasks/Exam-03-T7/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..00e1b446ea6f74b89c590b445d1608c2776c7218 --- /dev/null +++ b/test/tasks/Exam-03-T7/problem.json @@ -0,0 +1,36 @@ +{ + "problem_id": "Exam-03-T7", + "header": "Problem 7 Li-CO_2 Batteries Based on Dimerization with a Solid Redox Mediator. 7-1 As energy demand and global greenhouse-gas emissions increase, efficiently storing energy while reducing CO_2 emissions has become a major scientific and technological challenge. Li-CO_2 batteries are a focus of next-generation energy-storage research because of their high theoretical energy density, approximately 1876 Wh/kg, and their ability to convert the greenhouse gas CO_2 into electrical energy. Conventional Li-CO_2 batteries, however, face three bottlenecks: (1) low discharge voltage because direct electrochemical reduction of CO_2 is kinetically slow; (2) high charging overpotential because the main discharge product is difficult to decompose and the charging voltage is typically above 4.3 V; and (3) short cycle life because conventional catalysts cannot stabilize the discharge product, so cycle life is usually fewer than 100 cycles.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-03-T7-1", + "tags": [ + "2.1", + "3.5" + ], + "text_open": "7-1-1 Write the typical cathodic discharge reaction of a Li-CO_2 battery. The products are Li_2CO_3 and an elemental substance in a 2:1 ratio.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-03-T7-2", + "tags": [ + "1.7", + "3.5" + ], + "text_open": "Earlier soluble redox mediators had three problems: they diffused between the cathode and anode and underwent side reactions; their transport through the electrolyte was slow at high current density; and they decomposed during cycling to form irreversible byproducts.\n7-3 To address these problems, researchers developed a Li-CO_2 battery based on a solid redox mediator. The mediator, RM(II)-BTC, is a microporous MOF formed by coordination of Cu(II) with trimesic acid (BTC). It is combined with carbon nanotubes to form a three-dimensional conductive network and has a high specific surface area and a pore diameter of 9 Å. The electrochemical performance of the battery is greatly improved: (1) the discharge voltage increases to 2.8 V and the charging voltage decreases to 3.7 V; (2) cycle life exceeds 400 cycles; and (3) excellent kinetics are obtained at high current density. 7-3-1 Based on the information in 7-3, state at least two advantages of the proposed Li-CO_2 battery over earlier Li-CO_2 batteries. Use no more than 20 words for each point.", + "text_images": [], + "full_mark": 4 + }, + { + "sub_id": "Exam-03-T7-3", + "tags": [ + "3.8" + ], + "text_open": "The DFT pathway is defined as follows. A is the initial state. A reacts with $Li^+$, $e^-$, and $CO_2$ to form B, with $\\Delta E_1=-1.50\\ \\mathrm{eV}$. B dimerizes to C, with $\\Delta E_2=+0.80\\ \\mathrm{eV}$. C binds Cu(I) with an oxidation-state change to give D, the final state containing stably adsorbed $Li_2C_2O_4$, with $\\Delta E_3=-2.72\\ \\mathrm{eV}$. $E(A)=-300.00\\ \\mathrm{eV}$; $E(TS_2)-E(C)=+0.45\\ \\mathrm{eV}$ and $E(TS_3)-E(D)=+3.62\\ \\mathrm{eV}$.\n7-4-2 Calculate and compare the maximum energy barriers of Steps 2 and 3, defined as the difference between the transition state and the corresponding precursor intermediate, and identify the rate-controlling step of the pathway.", + "text_images": [], + "full_mark": 3 + } + ] +} diff --git a/test/tasks/Exam-04-T2/problem.json b/test/tasks/Exam-04-T2/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..160452037e9c61ac6df104d5dceec6e6de71dacb --- /dev/null +++ b/test/tasks/Exam-04-T2/problem.json @@ -0,0 +1,45 @@ +{ + "problem_id": "Exam-04-T2", + "header": "Problem 2 Fundamentals of Iodometric Titration", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-04-T2-1", + "tags": [ + "3.6", + "3.4" + ], + "text_open": "2-1-2 Calculate the conditional reduction potential of $\\mathrm{Fe(CN)_6^{3-}}$ to $\\mathrm{Fe(CN)_6^{4-}}$ in $\\mathrm{mol}\\cdot\\mathrm{L}^{-1}$ HCl. Given $\\varphi^\\Theta_{\\mathrm{Fe(CN)_6^{3-}/Fe(CN)_6^{4-}}}=0.36\\ \\mathrm{V}$; $\\mathrm{H_3Fe(CN)_6}$ is a strong acid; $\\mathrm{H_4Fe(CN)_6}$ has $\\mathrm{p}K_{\\mathrm{a3}}=2.2$ and $\\mathrm{p}K_{\\mathrm{a4}}=4.2$; neglect ionic effects.", + "text_images": [], + "full_mark": 5 + }, + { + "sub_id": "Exam-04-T2-2", + "tags": [ + "2.1", + "4.1" + ], + "text_open": "2-2 Dissolve a 1.000 g sample containing sodium iodate and sodium periodate in a 250 mL volumetric flask and dilute to the mark. Transfer 50.00 mL to an Erlenmeyer flask, adjust to weakly basic pH, add excess potassium iodide, and titrate with 0.04000 mol·L^{-1} sodium thiosulfate; 10.00 mL is consumed. Transfer another 20.00 mL, acidify with hydrochloric acid, add excess potassium iodide, and titrate with sodium thiosulfate of the same concentration; 30.00 mL is consumed. 2-2-1 What reaction occurs in the first titration? Write its ionic equation and explain why the solution must be adjusted to weakly basic conditions.", + "text_images": [], + "full_mark": 4 + }, + { + "sub_id": "Exam-04-T2-3", + "tags": [ + "4.5" + ], + "text_open": "2-2-2 Calculate the mass percentages of sodium iodate and sodium periodate.", + "text_images": [], + "full_mark": 9 + }, + { + "sub_id": "Exam-04-T2-4", + "tags": [ + "4.5" + ], + "text_open": "2-2-3 In theory, this method can measure the mass percentages of sodium iodate and sodium periodate fairly accurately. In a student laboratory, however, it is difficult to obtain an accurate result, defined as an error within 0.1%. Explain the principal reason in terms of experimental readings. (No more than 20 words.)", + "text_images": [], + "full_mark": 3 + } + ] +} diff --git a/test/tasks/Exam-04-T8/images/Exam-04-T8-1-1-prompt.png b/test/tasks/Exam-04-T8/images/Exam-04-T8-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..ec3ded3725baa13b2486b506a75293d96bca643f --- /dev/null +++ b/test/tasks/Exam-04-T8/images/Exam-04-T8-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:e27253cc5ab0c2c2908bd45962ca07cc33e78ed23ef20bb2ef8c74ae63a76990 +size 3932 diff --git a/test/tasks/Exam-04-T8/images/Exam-04-T8-1-2-prompt.png b/test/tasks/Exam-04-T8/images/Exam-04-T8-1-2-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..0b703f599b88a15dac9d27e4ab8bd5aa4ac37150 --- /dev/null +++ b/test/tasks/Exam-04-T8/images/Exam-04-T8-1-2-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:ea0a09fc5c9bf7fa6bc334de0adbc5ae890446cfd8a9c51836ea2e21ad2237f9 +size 3292 diff --git a/test/tasks/Exam-04-T8/images/Exam-04-T8-1-3-prompt.png b/test/tasks/Exam-04-T8/images/Exam-04-T8-1-3-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..6bae47403849e1f2abf5f9526ece3dc3b1d8a764 --- /dev/null +++ b/test/tasks/Exam-04-T8/images/Exam-04-T8-1-3-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:235cc227e235e25e07afc316427d45b200136c9dbec4e2cbd16dc70667e11893 +size 2368 diff --git a/test/tasks/Exam-04-T8/images/Exam-04-T8-1-4-prompt.png b/test/tasks/Exam-04-T8/images/Exam-04-T8-1-4-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..a2dd461a567fe3599141314309402293f2284561 --- /dev/null +++ b/test/tasks/Exam-04-T8/images/Exam-04-T8-1-4-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:27652f19a3e7dfc25d077c74794fc34351d73628f123762d72a93885177cc8e2 +size 2838 diff --git a/test/tasks/Exam-04-T8/images/Exam-04-T8-2-1-prompt.png b/test/tasks/Exam-04-T8/images/Exam-04-T8-2-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..437355916571c8fe5af530f73df0f9c5cb511c27 --- /dev/null +++ b/test/tasks/Exam-04-T8/images/Exam-04-T8-2-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:f9c8ef321973845a28beac3239ab0798c5c559a58b3fbb753b56533eb12258e3 +size 1686 diff --git a/test/tasks/Exam-04-T8/images/Exam-04-T8-2-2-prompt.png b/test/tasks/Exam-04-T8/images/Exam-04-T8-2-2-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..3959ed9738735ba459822eb22c784e0f9c395b30 --- /dev/null +++ b/test/tasks/Exam-04-T8/images/Exam-04-T8-2-2-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:4359d1d7c44fba93319629f96b55b93be81ef716565d7f2bd163f327b2291b56 +size 1869 diff --git a/test/tasks/Exam-04-T8/images/Exam-04-T8-2-3-prompt.png b/test/tasks/Exam-04-T8/images/Exam-04-T8-2-3-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..2eef5f43974d26915f99a5781c10f97e35f0b3e2 --- /dev/null +++ b/test/tasks/Exam-04-T8/images/Exam-04-T8-2-3-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:91bd93df27509f2ea553cad1ef9644679fe07e80d987dcc8fe66a19366b38499 +size 1695 diff --git a/test/tasks/Exam-04-T8/images/Exam-04-T8-3-1-prompt.png b/test/tasks/Exam-04-T8/images/Exam-04-T8-3-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..2a5648851af167a95731cdb496ee86819d0ed934 --- /dev/null +++ b/test/tasks/Exam-04-T8/images/Exam-04-T8-3-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:2e76694d5f4e90b94863181ec08bc1df63c6c83619734f1d94fbaa9818694e99 +size 5763 diff --git a/test/tasks/Exam-04-T8/problem.json b/test/tasks/Exam-04-T8/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..64796f81f60691d505fcd6c0ef08709fd3bf6a26 --- /dev/null +++ b/test/tasks/Exam-04-T8/problem.json @@ -0,0 +1,69 @@ +{ + "problem_id": "Exam-04-T8", + "header": "Problem 8 Basic Organic Chemistry", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-04-T8-1", + "tags": [ + "1.7" + ], + "text_open": "8-2 Arrange the following radicals in decreasing order of stability.", + "text_images": [ + { + "file": "Exam-04-T8-1-1-prompt.png", + "description": "Option a, SMILES '[CH]1(C2)CC[C]2CC1'." + }, + { + "file": "Exam-04-T8-1-2-prompt.png", + "description": "Option b, SMILES '[C]1(CC2)CCC2CC1'." + }, + { + "file": "Exam-04-T8-1-3-prompt.png", + "description": "Option c, SMILES '[CH]1(C[CH](C2)C3)C[C]2C[CH]3C1'." + }, + { + "file": "Exam-04-T8-1-4-prompt.png", + "description": "Option d, SMILES 'C12CCCC[C]1CCCC2'." + } + ], + "full_mark": 2 + }, + { + "sub_id": "Exam-04-T8-2", + "tags": [ + "4.1" + ], + "text_open": "8-3 Identify the three compounds shown by chemical tests. Available reagents are: (a) acetone (b) NaI (c) I_2 (d) concentrated hydrochloric acid (e) CuSO4 solution (f) NaOH solution.", + "text_images": [ + { + "file": "Exam-04-T8-2-1-prompt.png", + "description": "Compound to be identified, SMILES 'CC(C)CCCl'." + }, + { + "file": "Exam-04-T8-2-2-prompt.png", + "description": "Compound to be identified, SMILES 'CC(C)CCO'." + }, + { + "file": "Exam-04-T8-2-3-prompt.png", + "description": "Compound to be identified, SMILES 'CC(C)CC=O'." + } + ], + "full_mark": 4 + }, + { + "sub_id": "Exam-04-T8-3", + "tags": [ + "1.7" + ], + "text_open": "8-4 At room temperature, X-ray diffraction shows that D-glucose crystallized from water contains only the α anomer, whereas its specific rotation after dissolution in water shows that the solution contains a mixture of the α and β anomers. 8-4-2 State which anomer is more abundant in aqueous solution and explain why.", + "text_images": [ + { + "file": "Exam-04-T8-3-1-prompt.png", + "description": "Structure of the α anomer, SMILES 'O[C@H]1[C@H](O)[C@@H](CO)O[C@H](O)[C@@H]1O'." + } + ], + "full_mark": 3 + } + ] +} diff --git a/test/tasks/Exam-05-T4/problem.json b/test/tasks/Exam-05-T4/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..eae240c54509c15fea783f89a7c6ab0c2bc628a9 --- /dev/null +++ b/test/tasks/Exam-05-T4/problem.json @@ -0,0 +1,54 @@ +{ + "problem_id": "Exam-05-T4", + "header": "Problem 4 Phosphoric Acid and Phosphate Chemistry. Phosphates are important in industrial production, food additives, and other fields, so measuring their content and understanding their properties are significant.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-05-T4-1", + "tags": [ + "2.1" + ], + "text_open": "4-1 Aqueous sodium phosphate dodecahydrate, Na_3PO_4·12H_2O, is strongly basic and is used in washing powders and fat-saponification agents. It is prepared by reacting calcium dihydrogen phosphate with sodium carbonate and then adding sodium hydroxide solution. Write the preparation equation.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-05-T4-2", + "tags": [ + "4.5" + ], + "text_open": "Titration notation: Add a volume $V_0$ of standard HCl of concentration $c_1$ to the sample, then titrate with standard NaOH of concentration $c_2$. The first jump near pH 4.0 consumes a NaOH volume $V_1$, and continuing to the second jump near pH 8.8 consumes an additional volume $V_2$.\n4-2-2 What relation among V_0, V_1, V_2, c_1, and c_2 indicates that the sample contains Na_2HPO_4? Express it symbolically.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-05-T4-3", + "tags": [ + "4.5", + "4.7" + ], + "text_open": "4-2-3 Calculate the mass fraction ω(%) of Na_3PO_4 in the sample. Significant figures are required.", + "text_images": [], + "full_mark": 4 + }, + { + "sub_id": "Exam-05-T4-4", + "tags": [ + "1.7", + "2.1" + ], + "text_open": "4-3-2 Phosphorus atoms in different oxidation states can also be linked in catenated phosphoric acids. H_4P_2O_5 contains linked P(Ⅱ) and P(Ⅳ). It has two hydrolysis pathways under basic conditions. Only the first generates gas, and each gives only one phosphorus-containing product. Write the ionic equations for both pathways.", + "text_images": [], + "full_mark": 4 + }, + { + "sub_id": "Exam-05-T4-5", + "tags": [ + "2.4" + ], + "text_open": "4-3-3 Hydrolysis of cyclic H_6P_6O_12 is complex. First, one P-P bond cleaves to form a linear catenated hexaphosphoric acid by disproportionation; then terminal P-P bonds cleave stepwise. How many ways can the linear hexaphosphoric acid hydrolyze to catenated pentaphosphoric acid? Give the products of each. Use P for a phosphorus-containing group and a superscript to denote oxidation state; for example, catenated diphosphoric acid is P^4-P^4.", + "text_images": [], + "full_mark": 4 + } + ] +} diff --git a/test/tasks/Exam-06-T4/problem.json b/test/tasks/Exam-06-T4/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..0a9be2947e2d85ca591a743d55c5790870c7ba34 --- /dev/null +++ b/test/tasks/Exam-06-T4/problem.json @@ -0,0 +1,36 @@ +{ + "problem_id": "Exam-06-T4", + "header": "Problem 4 Determination of Tin by Potassium Iodate Titration. Tin is commonly determined by potassium iodate titration. The procedure includes sample decomposition, masking and separation, reductive fusion with zinc powder, reduction with aluminum powder, and titration of tin(II) ions with potassium iodate.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-06-T4-1", + "tags": [ + "2.1" + ], + "text_open": "4.2 Sample pretreatment. Place 0.2026 g of the sample in a 250 mL beaker, add 2 g of potassium chlorate and 20 mL of concentrated hydrochloric acid, and digest at low temperature for 20 min. Add 60 mL of hydrochloric acid (1+1), 1 g of hydrazinium sulfate, and 0.3 g of potassium bromide, then heat and evaporate for 50 min. Remove from the heat and allow to cool slightly. Cover with a watch glass and add hydrogen peroxide dropwise until brownish-red bromine appears. Add 10 mL of EDTA solution (100 g/L) and 5 mL of beryllium sulfate solution (40 g/L), and neutralize with aqueous ammonia to pH = 9-10. Add a small amount of paper pulp and filter. Wash the precipitate with an aqueous ammonia-ammonium nitrate wash solution. Transfer the precipitate together with the filter paper to a crucible, dry, ash, and cool. 4.2.1 Write the ionic equation for the reduction of arsenic acid by hydrazine in hydrochloric acid medium, by which arsenic is removed through volatilization.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-06-T4-2", + "tags": [ + "4.1", + "4.7" + ], + "text_open": "Reductive-fusion conditions: Add a zinc powder-borax-boric acid mixed flux to the crucible and cover it with additional mixed flux and NaCl. Place it in a furnace at 750 ℃, raise the temperature to 820 ℃, fuse for 30 min, then remove and cool.\n4.3.3 Explain how an excessively long fusion time or excessively high temperature affects the determination.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-06-T4-3", + "tags": [ + "2.1", + "4.5" + ], + "text_open": "4.4.2 Write the ionic equations for the titration process.", + "text_images": [], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-06-T7/images/Exam-06-T7-1-1-prompt.png b/test/tasks/Exam-06-T7/images/Exam-06-T7-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..9c958768a06ae689dbdf481b36d744c57ef19f48 --- /dev/null +++ b/test/tasks/Exam-06-T7/images/Exam-06-T7-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:cae3d1c0e3aee5bdb8838335552eec415ef04b17f454be93bc8fb46504ed5e2a +size 12021 diff --git a/test/tasks/Exam-06-T7/images/Exam-06-T7-1-2-prompt.png 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+1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:7ecd99ab323b8f0e4ab1d5c1df3d8b780725f2d244b7d90a2598c551370a2768 +size 2847 diff --git a/test/tasks/Exam-06-T7/problem.json b/test/tasks/Exam-06-T7/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..d9c6aefee7c73f8952a6e9c9c14b3b81e9c12f6e --- /dev/null +++ b/test/tasks/Exam-06-T7/problem.json @@ -0,0 +1,118 @@ +{ + "problem_id": "Exam-06-T7", + "header": "Problem 7 Synthesis of Schiff Bases and Their Derived Complexes (25 points, 9%)\nOn 2023-4-28, a research team from Université de Bourgogne reported a class of Schiff-base ligands and their complexes that exhibit “fluxionality.” Here, “fluxionality” means that the “C=N” double bonds in the ligands can undergo “rotation” under specific conditions.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-06-T7-1", + "tags": [ + "2.4", + "2.6" + ], + "text_open": "7.1 The synthesis of Schiff-base ligands provides the basis for constructing their derived complexes. The following reactions synthesize the Schiff-base ligands L1H_{2} and L2H_{2}, respectively. The “C=N” bonds in both L1H_{2} and L2H_{2} are known to have the E configuration. Write the SMILES structures of L1H_{2} and L2H_{2}, respectively. Note: the figures are simplified representations of the synthetic routes and do not show the complete workup after each step.", + "text_images": [ + { + "file": "Exam-06-T7-1-1-prompt.png", + "description": "Synthetic route to L1H_2: the molecule 'COC1=CC=CC=C1C(=O)N(CC)CC' is treated with (COCl_2) in Toluene at 333 K for 16h, then with NCCN 0.5 equiv. and NEt_3(5 equiv.) in DCM from 273 K to r.t. for 16h, subsequently with BBr_3, 3 equiv., in DCM from 273 K to r.t. for 16h, and finally with KHMDS (2 equiv) in THF to afford L1H_2." + }, + { + "file": "Exam-06-T7-1-2-prompt.png", + "description": "Synthetic route to L2H_2: the molecule 'OC1=CC=CC=C1C(=O)N(C)C' is treated under 1) SOCl_2, r.t., overnight, followed by 2) the molecule 'NC1=CC=CC=C1N', 0.45 equiv., and Et_3N (5 equiv.) in CH_2Cl_2 from 273 K to r.t. for 2.5h, to afford L2H_2." + } + ], + "full_mark": 4 + }, + { + "sub_id": "Exam-06-T7-2", + "tags": [ + "1.4", + "2.4" + ], + "text_open": "7.2 The corresponding complexes can be constructed by reacting the above Schiff-base ligands with metal precursors. 7.2.1 L1H_{2} forms neutral complex A through the following reaction. L1H_{2} is known to exhibit almost no “fluxionality,” and Al in complex A is 4-coordinate. On this basis, write the SMILES structure of neutral complex A.", + "text_images": [ + { + "file": "Exam-06-T7-2-1-prompt.png", + "description": "[L1H_4][Br_2] is treated under 1) KHMDS (2 equiv.), THF, r.t., 4h, followed by 2) AlMe_3, THF, r.t., 2h, to afford A." + } + ], + "full_mark": 3 + }, + { + "sub_id": "Exam-06-T7-3", + "tags": [ + "1.4", + "1.2", + "2.4" + ], + "text_open": "7.2.2 Because of the change in substituents, L2H_{2} exhibits some “fluxionality.” L2H_{2} forms neutral complex B_{1} through the following reaction. Given that Al in B_{1} is 4-coordinate, write the SMILES structure of B_{1}.", + "text_images": [ + { + "file": "Exam-06-T7-3-1-prompt.png", + "description": "L2H_2 is treated with AlMe_3 in THF at 333K for 2h to afford B_1." + } + ], + "full_mark": 4 + }, + { + "sub_id": "Exam-06-T7-4", + "tags": [ + "1.4", + "2.2", + "1.2" + ], + "text_open": "7.2.3 The following reaction illustrates the above “fluxionality” of the ligand. Notably, complex B_{1} in 7.2.2 is chiral, and B_{1} can isomerize to B_{2} through the formation of a 5-coordinate intermediate C. Write the SMILES structure of intermediate C.", + "text_images": [ + { + "file": "Exam-06-T7-4-1-prompt.png", + "description": "B1 ⇌ C ⇌ B2 isomerization scheme, emphasizing the conversion of B1 to B2 through 5-coordinate intermediate C. " + } + ], + "full_mark": 3 + }, + { + "sub_id": "Exam-06-T7-5", + "tags": [ + "1.4", + "1.2", + "2.4" + ], + "text_open": "7.2.4 The following reaction likewise demonstrates the “fluxionality” of the ligand. On this basis, write the SMILES structure of neutral complex D.", + "text_images": [ + { + "file": "Exam-06-T7-5-1-prompt.png", + "description": "B_1 or B_2 is converted to D in the presence of 'CC(O)(C(F)(F)F)C(F)(F)F' in CD_2Cl_2 under reflux for 3 days." + } + ], + "full_mark": 3 + }, + { + "sub_id": "Exam-06-T7-6", + "tags": [ + "1.4", + "2.4" + ], + "text_open": "7.3 Natural bond orbital calculations show that coordination of the second N atom in the ligand decreases the positive character of the Al center. Therefore, the more electron-deficient the Al center, the more readily it induces “rotation” of the ligand's C=N double bonds. On this basis, the research team further investigated how to prepare species with a more electron-deficient Al center.\n7.3.1 The following reactions provide further evidence for the theoretical prediction above. Given that both E and F are ionic complexes, write their SMILES structures.", + "text_images": [ + { + "file": "Exam-06-T7-6-1-prompt.png", + "description": "A is converted to E with B(C_6F_5)_3 (1 equiv.) in CH_2Cl_2 at r.t. for 1h." + }, + { + "file": "Exam-06-T7-6-2-prompt.png", + "description": "B_1 or B_2 is converted to F with B(C_6F_5)_3 (1 equiv.) in CH_2Cl_2 at r.t. for 1h." + } + ], + "full_mark": 4 + }, + { + "sub_id": "Exam-06-T7-7", + "tags": [ + "1.4", + "2.4" + ], + "text_open": "7.3.2 X-ray crystallography shows that E and F have dimeric structures in the solid state. Using E as an example, write the SMILES structure of ionic dimer G obtained by crystallization from pyridine, given that Al in G is 6-coordinate.", + "text_images": [], + "full_mark": 4 + } + ] +} diff --git a/test/tasks/Exam-07-T5/images/Exam-07-T5-1-1-prompt.png b/test/tasks/Exam-07-T5/images/Exam-07-T5-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..7abca96644e2b8b9766231edc2c33ec8ccf6c366 --- /dev/null +++ b/test/tasks/Exam-07-T5/images/Exam-07-T5-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:e539f58daad9933d00faded44b1edec9c7c764d998124489cb909f1eecb06c36 +size 78788 diff --git a/test/tasks/Exam-07-T5/problem.json b/test/tasks/Exam-07-T5/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..27f6c69d21ac6e89768a31c524089c2a4c593245 --- /dev/null +++ b/test/tasks/Exam-07-T5/problem.json @@ -0,0 +1,66 @@ +{ + "problem_id": "Exam-07-T5", + "header": "Square-Planar Voids\nIn models of crystal packing and void filling, tetrahedral and octahedral voids are the most familiar. In fact, square-planar voids are also fairly common.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-07-T5-1", + "tags": [ + "1.3" + ], + "text_open": "5.1 A binary compound AB belongs to the tetragonal crystal system. Its structure may be viewed as a simple cubic packing of B atoms in which A atoms occupy some of the square-planar voids. Among adjacent $AB_4$ quadrilaterals, the $AB_4$ quadrilaterals whose A atoms have the same c coordinate have identical spatial orientations and share edges, whereas the $AB_4$ quadrilaterals whose A atoms have different c coordinates share vertices. A projection of the crystal structure (containing more than one unit cell) along the c axis is shown below; the large spheres represent A atoms and the small spheres represent B atoms. The unit-cell parameters are a = 349.4 pm and c = 611.6 pm. All A atoms have the same chemical environment, and all B atoms likewise have the same chemical environment. Hint: Compare the unit-cell parameters with atomic radii when considering the unit-cell composition.\n5.1.1 Give the fraction of square-planar voids occupied by A atoms.", + "text_images": [ + { + "file": "Exam-07-T5-1-1-prompt.png", + "description": "Orthographic top-view two-dimensional ball-and-stick model on a pure white background, with no text, axes, or labels. The main image contains 16 yellow spheres and 24 gray spheres: the 16 yellow spheres are arranged in 4 rows and 4 columns at the vertices of a square grid, while the 24 gray spheres lie exactly at the midpoints of the lines joining adjacent yellow spheres (12 in each of the horizontal and vertical directions). The spheres are connected by opaque, short, light-gray cylindrical rods. Each gray sphere is connected to two adjacent yellow spheres; depending on its position, each yellow sphere is connected to 2 to 4 adjacent gray spheres (coordination number four in the interior, three at an edge, and two at a corner). All spheres show a subtle 3D light-to-shadow gradient, and the rod ends are naturally occluded by the spheres." + } + ], + "full_mark": 2 + }, + { + "sub_id": "Exam-07-T5-2", + "tags": [ + "1.3" + ], + "text_open": "5.1.2 State the lattice type of the crystal.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-07-T5-3", + "tags": [ + "1.3" + ], + "text_open": "5.1.3 If the A atoms are regarded as forming the packing and the B atoms as filling the voids, what type of void do the B atoms occupy, and what fraction of these voids is occupied?", + "text_images": [], + "full_mark": 3 + }, + { + "sub_id": "Exam-07-T5-4", + "tags": [ + "1.3" + ], + "text_open": "5.1.4 The theoretical density of the crystal is $10.10\\ \\mathrm{g\\,cm^{-3}}$. Determine the chemical formula of AB by calculation.", + "text_images": [], + "full_mark": 5 + }, + { + "sub_id": "Exam-07-T5-5", + "tags": [ + "1.3" + ], + "text_open": "5.2 NbO belongs to the cubic crystal system. In the unit cell of NbO, Nb atoms occupy all face centers and O atoms occupy all edge centers.\n5.2.2 For the crystal structure of NbO, determine whether each of the following statements is correct. Use T for true and F for false. 5.2.2.1 If octahedral voids are not counted as square-planar voids, do the Nb atoms occupy all square-planar voids formed by the O atoms?", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-07-T5-6", + "tags": [ + "1.3" + ], + "text_open": "5.2.2.3 The O atoms do not form a face-centered cubic packing, but in a projection of the crystal along a threefold axis, do the O atoms form a close-packed layer structure?", + "text_images": [], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-07-T8/images/Exam-07-T8-1-1-prompt.png b/test/tasks/Exam-07-T8/images/Exam-07-T8-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..95be584ff691330fc0038abd7def9865b52688f6 --- /dev/null +++ b/test/tasks/Exam-07-T8/images/Exam-07-T8-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:9e4a03312ce5fb3063e51347d1836cf061d1c320b61920c2f9c5dcea71bd8cbf +size 5556 diff --git a/test/tasks/Exam-07-T8/images/Exam-07-T8-2-1-prompt.png b/test/tasks/Exam-07-T8/images/Exam-07-T8-2-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..10473bec693659084b399607abbec758951e3cac --- /dev/null +++ b/test/tasks/Exam-07-T8/images/Exam-07-T8-2-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:e67a7b59dffa8bd51fba95817099e140e244ebf7b90f66ae75df4ed08b4b8205 +size 3588 diff --git a/test/tasks/Exam-07-T8/images/Exam-07-T8-4-1-prompt.png b/test/tasks/Exam-07-T8/images/Exam-07-T8-4-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..3511d0169b5e06f7cd84f3eb7ef838bbc999a97f --- /dev/null +++ b/test/tasks/Exam-07-T8/images/Exam-07-T8-4-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:b66a20bbe7cef484978ca57daff9f96c7dbedc0553a6093f47c9ceaf4e5c6af3 +size 12293 diff --git a/test/tasks/Exam-07-T8/images/Exam-07-T8-5-1-prompt.png b/test/tasks/Exam-07-T8/images/Exam-07-T8-5-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..59c9b943d3de6329bba3a2a706f27ba94c44a944 --- /dev/null +++ b/test/tasks/Exam-07-T8/images/Exam-07-T8-5-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:50ed37ec5876a05c9f7728ad9e3cf878d1901377b4e4fdea29c9afc1eba693db +size 5020 diff --git a/test/tasks/Exam-07-T8/problem.json b/test/tasks/Exam-07-T8/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..695cca272f2980a5f4db0a97690ab5d43be46c7c --- /dev/null +++ b/test/tasks/Exam-07-T8/problem.json @@ -0,0 +1,79 @@ +{ + "problem_id": "Exam-07-T8", + "header": "Hyperconjugation and Its Applications to Structure and Reactivity", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-07-T8-1", + "tags": [ + "1.5", + "2.7", + "2.4" + ], + "text_open": "8.1 The molecule shown below can undergo intramolecular alkylation with the carbonyl oxygen acting as the nucleophile, forming 2 possible salt products. For this transformation, determine which reaction pathway is preferred and write the SMILES representation of the salt product corresponding to that pathway.", + "text_images": [ + { + "file": "Exam-07-T8-1-1-prompt.png", + "description": "'BrCCNC(OCCBr)=O'" + } + ], + "full_mark": 2 + }, + { + "sub_id": "Exam-07-T8-2", + "tags": [ + "1.2", + "1.5" + ], + "text_open": "8.2 Only one stereochemical configuration of the molecule shown below can react effectively with $O_3$. Write the SMILES representation of the reactive stereochemical configuration.", + "text_images": [ + { + "file": "Exam-07-T8-2-1-prompt.png", + "description": "'[H][C@]12C(OCCC2)OCCC1'" + } + ], + "full_mark": 2 + }, + { + "sub_id": "Exam-07-T8-3", + "tags": [ + "1.2", + "1.5" + ], + "text_open": "8.3 $N_2F_2$ has 2 stereoisomers. Determine which is thermodynamically more stable and which is kinetically more reactive, and explain the origin of the greater thermodynamic stability. (No more than 100 words.)", + "text_images": [], + "full_mark": 4 + }, + { + "sub_id": "Exam-07-T8-4", + "tags": [ + "1.2", + "1.5" + ], + "text_open": "8.4 Which of the following conformations is more stable?", + "text_images": [ + { + "file": "Exam-07-T8-4-1-prompt.png", + "description": "Both molecules have the SMILES representation 'O=P(O[R]C)(O[R])[O-]' but differ in conformation. Structure A: the central P atom is bonded to =O directly below and to $-O^-$ at the upper left. Of the two -O-R groups bonded at the upper right, one P-O bond points into the plane of the page, with the R group on its oxygen extending vertically upward, while the other P-O bond points out of the plane, with the R group on its oxygen extending down and to the right. Structure B: the connectivity between P and the four oxygen atoms is the same as in structure A. For the -OR group at the upper right of P that points into the plane, the R group lies below and to the left of O and points into the plane; for the -OR group at the upper right of P that points into the plane, the R group lies below and to the left of O and points out of the plane. The overall difference from A is the positions of the R groups." + } + ], + "full_mark": 2 + }, + { + "sub_id": "Exam-07-T8-5", + "tags": [ + "1.2", + "1.5", + "1.6" + ], + "text_open": "8.5 The hyperconjugative stabilization of the C-S bond by $C-H_1$ and $C-H_2$ is 6.5 and 1.8 energy units, respectively. Determine which C-H bond provides the stronger hyperconjugative stabilization and explain why. (No more than 100 words.)", + "text_images": [ + { + "file": "Exam-07-T8-5-1-prompt.png", + "description": "Chair conformation of 1,3-dithiane. Position 2 is labeled $H_1$ and position 5 is labeled $H_2$; both labeled H atoms occupy equatorial positions. '[H]C1CSC([H])SC1'" + } + ], + "full_mark": 4 + } + ] +} diff --git a/test/tasks/Exam-08-T3/images/Exam-08-T3-1-1-prompt.png b/test/tasks/Exam-08-T3/images/Exam-08-T3-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..3c0bd9bc6b0b4c6e4ac1c7d377149d33e04f8144 --- /dev/null +++ b/test/tasks/Exam-08-T3/images/Exam-08-T3-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:617888ba08f049a8a1702d50d37cf37b9056ceabc0c8531221b02543f81e16b9 +size 47197 diff --git a/test/tasks/Exam-08-T3/images/Exam-08-T3-2-1-prompt.png b/test/tasks/Exam-08-T3/images/Exam-08-T3-2-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..d883df96bb14e2ae51f29f9a9e8c5b93f40ab68e --- /dev/null +++ b/test/tasks/Exam-08-T3/images/Exam-08-T3-2-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:c955ebc1451d5637ec1067f00e04d372237484d07d8287e545cc243ffecf4b30 +size 40280 diff --git a/test/tasks/Exam-08-T3/problem.json b/test/tasks/Exam-08-T3/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..5ed697e8f6ee8004dff9377c332cd25cd59037bf --- /dev/null +++ b/test/tasks/Exam-08-T3/problem.json @@ -0,0 +1,36 @@ +{ + "problem_id": "Exam-08-T3", + "header": "Problem 3 Phthalocyanine Complexes", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-08-T3-1", + "tags": [ + "1.4", + "2.4" + ], + "text_open": "3.1 The diagram below shows simple syntheses of two mononuclear macrocyclic complexes.\nHint: A reducing substance is present in the reaction system.\n3.1.1 Give the SMILES structures of A and B.", + "text_images": [ + { + "file": "Exam-08-T3-1-1-prompt.png", + "description": "Heating 4-iodophthalonitrile 'IC1=CC=C(C#N)C(C#N)=C1', 4,5-dibutoxyphthalonitrile 'N#CC1=CC(OCCCC)=C(OCCCC)C=C1C#N', and zinc acetate $\\mathrm{Zn(OAc)_2}$ in a 1:3:1 ratio gives A. Reaction of 4-iodophthalonitrile 'IC1=CC=C(C#N)C(C#N)=C1', 4,5-bis(propylsulfonyl)phthalonitrile 'N#CC1=CC(S(=O)(CCC)=O)=C(S(=O)(CCC)=O)C=C1C#N', and nickel chloride $\\mathrm{NiCl_2}$ in a 1:3:1 ratio gives B." + } + ], + "full_mark": 4 + }, + { + "sub_id": "Exam-08-T3-2", + "tags": [ + "2.4" + ], + "text_open": "3.2 Further reactions combine A and B into the dinuclear complex E, as shown in the route below. Give the SMILES structures of C, D, and E.", + "text_images": [ + { + "file": "Exam-08-T3-2-1-prompt.png", + "description": "First, 1,4(1,4)-dibenzenacyclohexaphane (SMILES: 'C1(CC2)=CC=C(CCC3=CC=C2C=C3)C=C1') is treated with 2 equivalents of $\\mathrm{Br_2}$/$\\mathrm{CCl_4}$ and then reacted with tributyl(vinyl)tin (Bu_3SnCH=CH_2) under tetrakis(triphenylphosphine)palladium, Pd(PPh_3)_4, catalysis to give C. Next, C reacts with A in DMF in the presence of triethylamine ($\\mathrm{Pd(OAc)_2}$), tetrabutylammonium bromide (TBAB), and palladium acetate ($\\mathrm{NEt_3}$) to give D. Finally, D reacts with B under the same $\\mathrm{Pd(OAc)_2}$/DMF/$\\mathrm{NEt_3}$/TBAB conditions to give E." + } + ], + "full_mark": 6 + } + ] +} diff --git a/test/tasks/Exam-08-T5/problem.json b/test/tasks/Exam-08-T5/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..f0a3769e966acfb6de2579bd46a2243a2aee03ad --- /dev/null +++ b/test/tasks/Exam-08-T5/problem.json @@ -0,0 +1,34 @@ +{ + "problem_id": "Exam-08-T5", + "header": "Defective Cubic Perovskite\nA polymorph of $\\mathrm{CsFe}_x\\mathrm{Cl}_3$ has a defective cubic-perovskite structure. Relative to cubic perovskite, Cs occupies every Ca site, Cl every O site, and Fe occupies an ordered subset of the Ti sites. Equivalently, Cs and Cl jointly form close-packed layers in a 1:3 atomic ratio, and these layers are cubically stacked. Between every two adjacent Cs-Cl close-packed layers are octahedral voids, some formed only by Cl and others jointly by Cl and Cs. Fe fills some interlayer octahedral voids in the repeating sequence occupied, occupied, vacant, occupied, occupied, vacant, and so on. In each Fe-containing interlayer region, Fe occupies all and only the octahedral voids formed solely by Cl. This defect pattern changes the crystal to the hexagonal crystal family. The cell parameters are $a=736.2$ pm and $c=896.4$ pm. Two Fe atoms are located at $(2/3,1/3,0.18565)$ and $(1/3,2/3,0.81435)$, and their $\\mathrm{FeCl}_6$ octahedra are corner-sharing. In the projection down the crystallographic $c$ axis, the Fe atoms form a graphene-like pattern.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-08-T5-1", + "tags": [ + "1.3" + ], + "text_open": "5.1 Determine the unit-cell composition using reasonable assumptions and approximations.", + "text_images": [], + "full_mark": 7 + }, + { + "sub_id": "Exam-08-T5-2", + "tags": [ + "1.3" + ], + "text_open": "5.4 Determine the crystal system.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-08-T5-3", + "tags": [ + "1.3" + ], + "text_open": "5.7 State the number of inversion centers in one primitive cell.", + "text_images": [], + "full_mark": 1 + } + ] +} diff --git a/test/tasks/Exam-11-T2/images/Exam-11-T2-1-1-prompt.png b/test/tasks/Exam-11-T2/images/Exam-11-T2-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..d694ca86e1ff7a6b734fd64114351e6d7828398a --- /dev/null +++ b/test/tasks/Exam-11-T2/images/Exam-11-T2-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:2d8636e9664746bb57b5519df9ae16b74b6e54a8ab2f09b7e0f30e89fd5d1134 +size 1051769 diff --git a/test/tasks/Exam-11-T2/problem.json b/test/tasks/Exam-11-T2/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..9f9db1d3d731763ae6365397e013663ecd33f73a --- /dev/null +++ b/test/tasks/Exam-11-T2/problem.json @@ -0,0 +1,31 @@ +{ + "problem_id": "Exam-11-T2", + "header": "Problem 2: An Electrochemistry Medley", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-11-T2-1", + "tags": [ + "3.6" + ], + "text_open": "2.3.1 Shown below are the pH-potential diagram for the Cr-H_2O system at 298.15 K (the concentration of each soluble species is 1 mmol/L) and thermodynamic data for selected substances.\n\n| Substance | $\\Delta_f H^\\circ_m /(kJ \\cdot mol^{-1})$ | $\\Delta S^\\circ_m /(J \\cdot mol^{-1} \\cdot K^{-1})$ |\n| :--- | :--- | :--- |\n| H_2O(l) | -285.83 | 69.95 |\n| OH^-(aq) | -229.98 | -10.9 |\n| H_2(g) | - | 130.68 |\n| B | -985.09 | 95.4 |\n| F | -872.95 | 190.4 |\n| G | -890.89 | 5.4 |\n\nNote: the diagram below is a potential diagram for a very dilute aqueous solution (1 mmol/L), and each species contains only one Cr atom. In the pH-potential diagram, the slopes of lines cd, de, and ef are in the ratio 7:4:5. Using the information given and your chemical knowledge, identify the species represented by A-G.", + "text_images": [ + { + "file": "Exam-11-T2-1-1-prompt.png", + "description": "This is a two-dimensional rectangular pH-potential diagram. The unlabeled horizontal axis ranges from 0 to 14, and the vertical axis, labeled “potential/V,” ranges from -1.5 to 1.5 V. The upper and lower boundaries are horizontal lines at 1.5 and -1.5, respectively; the left and right boundaries are vertical lines at 0 and 14, respectively. Thick black lines divide the rectangle into 8 regions: seven labeled A, B, C, D, E, F, and G, and one labeled $Cr^{2+}$. Region E occupies the upper-left corner and is a right trapezoid. Its left boundary is part of the left frame, its right boundary is a vertical line at about 0.4, its upper boundary is part of the top frame, and its lower boundary slopes downward from left to right. The upper-right vertex a is approximately (0.4, 1.5), and the lower-right vertex c is approximately (0.4, 1.3). Region F is pentagonal. Segment ac is its left boundary, segment ab on the top frame is its upper boundary, and segment be at about 6.6 is its right boundary. Its two lower boundaries are segments cd and de, both sloping downward from left to right. Point e is the lower-right vertex; d is approximately (3.2,0.9), and e is approximately (6.6,0.6). The vertical segment ac separates E from F. Segment cd separates A (lower left) from F (upper right), while segment de separates B (below) from F (above). At approximately 3.2, a vertical line extends downward from d to k, with h between them. Segment dh has A on the left and B on the right; segment hk has the $Cr^{2+}$ region on the left and C on the right. Point h is approximately (3.2,-0.4), and k is approximately (3.2,-0.9). At approximately 6.6, a vertical line extends from point b on the top frame down to e. Point b is approximately (6.6,1.5); segment be has F on the left and G on the right, and the vertical line does not continue below e. Region G lies above diagonal ef and below the top frame, extending from about 6.6 to the right boundary at 14; thus G lies above ef and B below it. Region A lies below cd and above horizontal gh, with part of the left frame as its left boundary and vertical dh as its right boundary. Point g is approximately (0.0,-0.4), and h is approximately (3.2,-0.4). Segment gh is horizontal, with A above and the $Cr^{2+}$ region below. Region B lies below the upper broken line def and above the lower diagonal hi. Its left boundary is vertical dh, and fi is part of the right frame. Thus B is to the right of dh and A to the left; B is below ef and G above; B is above hi and C below; B is to the left of vertical fi, with the area to its right outside the frame. Point f, approximately (14.0,-0.1), is the intersection of the upper diagonal with the right boundary. The $Cr^{2+}$ region lies below horizontal gh, above horizontal jk (about -0.9 V), and between 0 and 3.2. Part of the left frame is its left boundary, and hk is its right boundary. Thus $Cr^{2+}$ is below gh and A above; $Cr^{2+}$ is above jk and D below; $Cr^{2+}$ is left of hk and C right of it. Point j is approximately (0.0,-0.9), and k is approximately (3.2,-0.9). Region C lies below diagonal hi and above diagonal kl, extending from 3.2 to 14. Its left boundary is hk and its right boundary is il. Thus C is right of hk and $Cr^{2+}$ left of it; C is below hi and B above; C is above kl and D below; C is left of vertical il, with the area to its right outside the frame. Point i, approximately (14.0,-1.0), is the intersection of hi with the right boundary; point l, approximately (14.0,-1.5), is the intersection of kl with the right boundary. Region D lies below horizontal j-k and diagonal k-l. Its lower boundary is the bottom frame, and it spans 0 to 14. At horizontal coordinate 0 on the left boundary, D lies below j; D is below jk and the $Cr^{2+}$ region above; D is below kl and C above. Reading the coordinates from the axes to one decimal place gives approximately: a≈(0.4,1.5), b≈(6.6,1.5), c≈(0.4,1.3), d≈(3.2,0.9), e≈(6.6,0.6), f≈(14.0,-0.1), g≈(0.0,-0.4), h≈(3.2,-0.4), i≈(14.0,-1.0), j≈(0.0,-0.9), k≈(3.2,-0.9), and l≈(14.0,-1.5). Thus, from top to bottom, the three principal diagonal boundaries are an unlabeled lowercase point-c-d-e-f, h-i, and k-l. Above the first are regions E/F/G in sequence and below it are A/B: specifically, E lies above the unlabeled point-c segment, F above c-d-e, and G above e-f; A lies below the unlabeled point-c-d portion, and B below d-e-f. B is above h-i and C below; C is above k-l and D below. The principal horizontal boundaries are g-h (A above and $Cr^{2+}$ below) and j-k ($Cr^{2+}$ above and D below)." + } + ], + "full_mark": 7 + }, + { + "sub_id": "Exam-11-T2-2", + "tags": [ + "3.4", + "3.6" + ], + "text_open": "2.3.3 Calculate the coordinates of point e from the thermodynamic data.", + "text_images": [], + "full_mark": 12 + } + ] +} diff --git a/test/tasks/Exam-12-T1/problem.json b/test/tasks/Exam-12-T1/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..7d262d3bc7e8f36e829e14f37821d2092f78a35d --- /dev/null +++ b/test/tasks/Exam-12-T1/problem.json @@ -0,0 +1,17 @@ +{ + "problem_id": "Exam-12-T1", + "header": "Problem 1 Metallic Vitamins. Rare-earth elements are core strategic resources in modern materials and chemical industries and are known as \"industrial vitamins.\" Their unique electron configurations give them distinctive physicochemical properties, making them essential materials for emerging industries such as modern semiconductors and new energy. A rare-earth element M has density $7.563 \\ \\mathrm{g \\cdot cm^{-3}}$ and forms an alloy with Co in a fixed ratio for manufacturing strong magnets. It has the following chemical properties. (1) When elemental M is left in air, its metallic luster rapidly disappears and a composite thin layer whose principal component is oxide A forms on its surface. M reacts with hydrochloric acid to form blood-red chloride B. After M has dissolved completely, the blood-red color rapidly fades, producing pale-yellow chloride C and colorless gas D. (2) If M is placed in a tetrahydrofuran solution of an organic compound E containing $90.05\\%$ iodine, reaction at room temperature gives unstable green iodide F and releases gas G. Addition of a small amount of water turns the solution yellow. Gas G is a common plant-ripening hormone.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-12-T1-1", + "tags": [ + "1.4", + "2.4" + ], + "text_open": "1-4-1 Complexes of M also have interesting structures. Addition of 0.5 equivalents of $\\mathrm{P_2Ph_4}$ to $\\mathrm{MCp*_2(THF)_2}$ gives a chain polymer X. After polymer X is separated, the pentane mother liquor is left for several weeks, yielding a yellow-brown complex Y. Y contains two M atoms and a 14-membered macrocycle. The mass fraction of M is $22.19\\%$ in both X and Y. Write the SMILES structure of the repeat unit of X, using [R] for the site connected to the next repeat unit (for example, polypropylene may be written as C([R])C(C)[R]), and the SMILES structure of Y ($\\mathrm{Cp*}$ = pentamethylcyclopentadienyl anion).", + "text_images": [], + "full_mark": 4 + } + ] +} diff --git a/test/tasks/Exam-12-T4/problem.json b/test/tasks/Exam-12-T4/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..bbd590e6562db33ba8de34ccb6cc425c5ebcd622 --- /dev/null +++ b/test/tasks/Exam-12-T4/problem.json @@ -0,0 +1,36 @@ +{ + "problem_id": "Exam-12-T4", + "header": "Problem 4 Titrimetric Analysis of Etching Acids\nWet chemical etching is a major process in the semiconductor-chip industry. It uses acidic substances to corrode a substrate surface and thereby achieve etching. A mixture of hydrofluoric acid, fluorosilicic acid, and nitric acid is a common etchant. A rapid and accurate laboratory method for determining the contents of these three acids in an etchant is described below.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-12-T4-1", + "tags": [ + "4.2" + ], + "text_open": "Procedure background: Accurately weigh 3.028 g of the sample. Place 1.502 g of it in a plastic beaker to prepare test solution A, and use the remaining sample to prepare test solution B by the same procedure. During preparation, add 10 mL of saturated $KNO_3$ solution and anhydrous ethanol, mix, dilute with anhydrous ethanol, and let stand for approximately 20 min until the precipitate has completely settled into a separate layer. Label this as test solution A. Prepare test solution B from the remaining sample by the same procedure.\n4.1.2 State the purpose of adding saturated KNO_3 solution.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-12-T4-2", + "tags": [ + "2.1", + "4.2" + ], + "text_open": "4.2 Titrimetric analysis. Add 30 mL of water and 3 drops of phenolphthalein indicator to test solution A. Titrate with 0.5018 $\\mathrm{mol \\cdot L^{-1}}$ standard NaOH solution until the solution becomes faint pink; 24.39 mL of the standard NaOH solution is consumed. Add 30 mL of water and 2 drops of methyl-orange indicator to test solution B. Adjust the pH with NaOH solution until the solution just turns bright yellow. Using a fluoride-ion combination electrode as the measuring electrode, titrate with 0.08273 $\\mathrm{mol \\cdot L^{-1}}$ standard La(NO_3)_3 solution; precipitate forms continuously during the titration. At the endpoint, 25.11 mL of La(NO_3)_3 solution is consumed. Weigh 0.5003 g of the original sample into a beaker containing 15 mL of 0.5 $\\mathrm{mol \\cdot L^{-1}}$ NaOH solution, add water to 60 mL, and verify with pH paper that the pH is greater than 13. Heat slowly to boiling, maintain a gentle boil for 5 min, and cool to room temperature. Add 2 drops of methyl-red indicator, adjust with nitric acid until red, and then adjust with NaOH solution until the solution just turns bright yellow. Using a fluoride-ion combination electrode as the indicator electrode, titrate with 0.08273 $\\mathrm{mol \\cdot L^{-1}}$ standard La(NO_3)_3 solution. At the endpoint, 15.28 mL is consumed. 4.2.1 Write all reaction equations occurring in Sections 4.1 and 4.2; write repeated reactions only once.", + "text_images": [], + "full_mark": 4 + }, + { + "sub_id": "Exam-12-T4-3", + "tags": [ + "4.7", + "4.5" + ], + "text_open": "4.2.2 Calculate the mass fractions of HF, H_2SiF_6, and HNO_3 in the sample. Note: the fluoride-ion electrode indicates the endpoint by the abrupt change in fluoride-ion concentration.", + "text_images": [], + "full_mark": 6 + } + ] +} diff --git a/test/tasks/Exam-13-T4/problem.json b/test/tasks/Exam-13-T4/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..0f4149d4f1837a41dc20c6695df2e01d95a94cef --- /dev/null +++ b/test/tasks/Exam-13-T4/problem.json @@ -0,0 +1,25 @@ +{ + "problem_id": "Exam-13-T4", + "header": "Question 4 Determination of Iron in Iron Ore. The potassium dichromate method is a classic method for determining iron in iron ore, but its use of $HgCl_2$ creates toxicity and environmental concerns. Selected steps of an improved dichromate method, the $TiCl_3-K_2Cr_2O_7$ method, are as follows. ① Preparation of the $K_2Cr_2O_7$ standard solution: accurately weigh 0.9797 g of $K_2Cr_2O_7$ into a 100 mL beaker, dissolve it in water, quantitatively transfer it to a 250 mL volumetric flask, and dilute to volume to obtain the $K_2Cr_2O_7$ standard solution. ② Titration of the sample: accurately weigh 0.2026 g of powdered magnetite concentrate, add 20 mL of hydrochloric acid, approximately 6 mol/L, cover with a watch glass, and heat gently to near boiling. After the ore decomposes, the mixture is reddish brown; immediately add $SnCl_2$ solution dropwise until it becomes pale yellow. Once most of the ore has dissolved, keep the mixture at a gentle boil for 1-2 min until decomposition is complete and no black particles remain. If the yellow color is too deep, add a little more $SnCl_2$ solution to obtain a pale-yellow solution; if the solution is colorless, add a small amount of $KMnO_4$ until a pale-yellow color appears, and stop heating. Rinse the watch glass and beaker walls with a wash bottle, add 50 mL of water and 8 drops of $Na_2WO_4$ solution, and while swirling add $TiCl_3$ solution to endpoint 1, followed by 2 excess drops. Cool to room temperature and carefully add the $K_2Cr_2O_7$ standard solution to endpoint 2; V1 = 0.02 mL is consumed. Add another 50 mL of water, 10 mL of sulfuric-phosphoric mixed acid with concentrated sulfuric acid:concentrated phosphoric acid:water = 2:3:5 by volume, and 2 drops of sodium diphenylamine sulfonate. Immediately titrate with the $K_2Cr_2O_7$ standard solution to endpoint 3; V2 = 28.12 mL is consumed.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-13-T4-1", + "tags": [ + "4.7" + ], + "text_open": "4.3 The following operations cause errors in the experiment. Explain the reason for each error. 4.3.1 The ore is heated at too high a temperature or for too long during decomposition with hydrochloric acid.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-13-T4-2", + "tags": [ + "4.7" + ], + "text_open": "4.3.2 The sulfuric-phosphoric mixed acid is not added before titration.", + "text_images": [], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-14-T1/problem.json b/test/tasks/Exam-14-T1/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..41cbddf4217d489888993aff9b38ba591f3d5f59 --- /dev/null +++ b/test/tasks/Exam-14-T1/problem.json @@ -0,0 +1,17 @@ +{ + "problem_id": "Exam-14-T1", + "header": "Ancient Metallurgy and Transmutation Chemistry\nBecause natural gold and silver were rare and extremely valuable, people in antiquity devised many ingenious methods for preparing precious metals artificially. Ming-dynasty works such as Tiangong Kaiwu and Shuyuan Zaji recorded methods for converting pure copper into “imitation gold and silver” with certain minerals, while alchemists also performed “transmutation” experiments with arsenic-, mercury-, and lead-bearing minerals. Modern chemistry has shown that these remarkable phenomena involve complex principles of inorganic synthesis and thermodynamics.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-14-T1-1", + "tags": [ + "1.1", + "2.1" + ], + "text_open": "1.1.2 In the Jin dynasty, Ge Hong recorded in Baopuzi a method for transmuting pure copper into a silvery-white alloy using an arsenic-containing compound. In the actual reaction, heating this compound in a sealed system first causes it to sublime and form gaseous molecule A with an adamantane-type cage; A is then reduced by charcoal to gaseous molecule B with a regular tetrahedral geometry; B penetrates the surface of solid pure copper and ultimately forms a silvery-white copper-arsenic alloy (the atomic ratio of the two elements in the alloy is 1:3). Write the chemical formula of A and the overall equation for formation of the alloy (assume that CO is the only oxidation product of carbon).", + "text_images": [], + "full_mark": 3 + } + ] +} diff --git a/test/tasks/Exam-14-T4/problem.json b/test/tasks/Exam-14-T4/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..d6e7437b821bc0ffe57f1c3778f41e25cf3ef93e --- /dev/null +++ b/test/tasks/Exam-14-T4/problem.json @@ -0,0 +1,27 @@ +{ + "problem_id": "Exam-14-T4", + "header": "Determination of Iron in Tin Concentrate by Cerium(IV) Sulfate Titration\n4.1 Preparation of standard titrant solutions\nIron standard solution: Weigh 2.0000 g of pure iron wire (99.95%) into a beaker, add 50 mL of hydrochloric acid (1+1), cover with a watch glass, and heat on a low-temperature electric hot plate until completely dissolved (do not boil). Remove, cool, transfer to a 1000 mL volumetric flask, dilute to the mark with water, and mix thoroughly.\nCerium(IV) sulfate standard titrant: Weigh 18.1 g of cerium(IV) sulfate [Ce(SO4)2·4H2O] into a beaker, add a small amount of water, and add 80 mL of concentrated sulfuric acid with constant stirring. After complete dissolution, cool, dilute to 1000 mL with water, and mix.\nStandardization: Pipet 30.00 mL of the iron standard solution into a 300 mL conical flask, add 5 mL of concentrated hydrochloric acid, and carry out reduction and standardization according to Section 4.3. Conduct a blank test in parallel.\n4.2 Alkali fusion and leaching of the test portion\nPlace a 0.3126 g portion of tin concentrate in a 10 mL high-alumina crucible, add 2 g of sodium peroxide, and mix carefully with a thin glass rod. Wipe the rod with a small piece of filter paper and place the paper in the crucible. Place the crucible in a furnace preheated to 500 °C, continue heating to 650 °C–700 °C, and fuse until the melt is red and transparent. Remove the crucible and cool.\nPlace the crucible in a 250 mL beaker and add 20 mL of water for leaching. After the vigorous reaction ceases, neutralize to an acidic solution with sulfuric acid (1+1), add 5 mL of concentrated hydrochloric acid, wash the crucible with water, and heat the solution to boiling on a low-temperature electric hot plate.\n4.3 Reduction and titration\nWhile the solution is hot, add hydrochloric-acidified tin(II) chloride solution (100 g/L) dropwise with swirling until the solution is pale yellow. Rinse down the beaker wall, add 40 mL of water, add 5 drops of sodium tungstate solution (50 g/L) as indicator, and add hydrochloric-acidified titanium(III) chloride solution (50 g/L) dropwise until a blue color appears, followed by two excess drops. Cool to room temperature in a cold-water bath.\nAdd 50 mL of copper sulfate-ammonium sulfate solution (containing 0.02 g of copper sulfate and 80 g of ammonium sulfate per liter). After the blue color disappears, allow the solution to stand for 10 min. Add 25 mL of sulfuric acid-phosphoric acid solution (sulfuric acid:phosphoric acid:water = 1:1:4), add 7 drops of sodium diphenylamine sulfonate solution (5 g/L), and titrate with the cerium(IV) sulfate standard titrant to a stable purple-red endpoint.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-14-T4-1", + "tags": [ + "4.1", + "4.2" + ], + "text_open": "4.3.2 Explain why preliminary reduction with tin(II) chloride solution is required instead of adding sodium tungstate solution directly and reducing with titanium(III) chloride solution.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-14-T4-2", + "tags": [ + "4.1", + "4.2" + ], + "text_open": "4.3.3 Explain why reduction with tin(II) chloride and titanium(III) chloride must be performed in a hot solution, whereas the subsequent operations require cooling to room temperature.", + "text_images": [], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-14-T7/images/Exam-14-T7-1-1-prompt.png b/test/tasks/Exam-14-T7/images/Exam-14-T7-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..d90321936efc90c9ea288babc8540703d1787f2a --- /dev/null +++ b/test/tasks/Exam-14-T7/images/Exam-14-T7-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:7390eafb7a9a4137f120a66f2b350e91bbdddac7718ea67bf5da17dccdc5a67d +size 3938 diff --git a/test/tasks/Exam-14-T7/images/Exam-14-T7-2-1-prompt.png b/test/tasks/Exam-14-T7/images/Exam-14-T7-2-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..2707025a2feff7db7e10084346a92b478e488dfb --- /dev/null +++ b/test/tasks/Exam-14-T7/images/Exam-14-T7-2-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:f1c09c0c7a7fdb915f7751cf31e31588ec5edd70f41bbacfeabaae71beafcee6 +size 5007 diff --git a/test/tasks/Exam-14-T7/images/Exam-14-T7-3-1-prompt.png b/test/tasks/Exam-14-T7/images/Exam-14-T7-3-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..e1bbbc10d7706814479b4f4d28d1c4984f6d4db3 --- /dev/null +++ b/test/tasks/Exam-14-T7/images/Exam-14-T7-3-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:3ace9314ac6c83bb645c2166461cf68dd3be45da5fa83c3d13ffaf52aaea3799 +size 67066 diff --git a/test/tasks/Exam-14-T7/problem.json b/test/tasks/Exam-14-T7/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..0c358303db999c38e23e04ed812dd3e9b3e3854c --- /dev/null +++ b/test/tasks/Exam-14-T7/problem.json @@ -0,0 +1,54 @@ +{ + "problem_id": "Exam-14-T7", + "header": "Bioinorganic Chemistry. 7.1 O2 is in fact highly toxic to many organisms. Oxygen carriers bound to metalloproteins enable the transport and storage of $O_2$ in organisms.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-14-T7-1", + "tags": [ + "1.4", + "1.6", + "2.4" + ], + "text_open": "7.1.1 Arthropods transport oxygen using hemocyanin. Hemocyanin is an oligomeric protein containing a pair of closely spaced Cu atoms. Colorless deoxyhemocyanin A [Cu(I)] is represented below. A binds one molecule of $\\mathrm{O_2}$ to form bright-blue protein B. Given that Cu is in the +2 oxidation state in B, write the SMILES string of B (for example, A may be represented as '[N][Cu]([N])[N].[N][Cu]([N])[N]').", + "text_images": [ + { + "file": "Exam-14-T7-1-1-prompt.png", + "description": "Simplified representation of the dicopper(I) center in deoxyhemocyanin A. The diagram is labeled with a bold uppercase A at the bottom center and is left-right symmetric. A central $Cu$ atom in the left half is connected by three straight lines to three $N$ atoms at the upper-left, directly left, and lower-left positions. These three $N$ atoms are joined in sequence by two arcs curving outward to the left: the upper-left $N$ is connected to the directly-left $N$, and the directly-left $N$ is connected to the lower-left $N$. A second central $Cu$ atom in the right half is similarly connected by three straight lines to three $N$ atoms at the upper-right, directly right, and lower-right positions. These three $N$ atoms are joined in sequence by two arcs curving outward to the right: the upper-right $N$ is connected to the directly-right $N$, and the directly-right $N$ is connected to the lower-right $N$. A long downward-curving arc at the bottom directly connects the lower-left $N$ and lower-right $N$ atoms." + } + ], + "full_mark": 3 + }, + { + "sub_id": "Exam-14-T7-2", + "tags": [ + "1.4", + "1.8", + "2.4" + ], + "text_open": "7.1.2 Hemerythrin is a member of a special family of proteins containing a diiron center. In addition to coordination by amino-acid side chains and two bridging carboxylates, the two iron atoms are bridged by a common diatomic anion. Hemerythrin C is represented below (the diatomic bridging ligand is not shown), and C binds one molecule of $\\mathrm{O_2}$ to form D. Given that $\\mathrm{O_2}$ coordinates to only one Fe atom and that the diatomic bridging ligand provides some noncovalent stabilization to $\\mathrm{O_2}$, write the SMILES string of D.", + "text_images": [ + { + "file": "Exam-14-T7-2-1-prompt.png", + "description": "Simplified representation of the diiron center in hemerythrin C. C: '[R]/C1=O/[Fe]([*])([*])O/C([*])=O\\[Fe]([*])([*])O1', where * denotes a site at which Fe coordinates to the imidazole group of His and may be omitted here." + } + ], + "full_mark": 4 + }, + { + "sub_id": "Exam-14-T7-3", + "tags": [ + "1.8", + "1.4" + ], + "text_open": "7.1.3 CO is well known to strongly inhibit the binding of $\\mathrm{O_2}$ to myoglobin, so suppressing CO coordination is essential. In fact, the structure of myoglobin weakens the binding of CO to the protein. Using the simplified myoglobin structure above, explain why CO binds less strongly to myoglobin than $\\mathrm{O_2}$, considering the coordination geometries of CO and $\\mathrm{O_2}$ and their noncovalent interactions. (No more than 100 words.)", + "text_images": [ + { + "file": "Exam-14-T7-3-1-prompt.png", + "description": "Simplified structure of the myoglobin coordination site, drawn with black lines on a white background.\n\nA five-membered ring near the top bears the labels HN at the lower left and N at the upper right, with single- and double-bond lines in and around the ring. A downward-curving line from the lower-right side of the ring extends toward the upper-right region of an elliptical structure below.\n\nAn approximately elliptical ring of curved lines occupies the center. Four N labels appear around it at the upper-left, upper-right, lower-left, and lower-right positions, and Fe is labeled at the center.\n\nA vertical line extends downward from Fe through the lower part of the ellipse to the label His below.\n\nThe Chinese caption at the bottom reads “Simplified myoglobin structure” in the English-localized version. SMILES: '[Fe][*].[N].[N].[N].[N].C1=CN=CN1', where * denotes the site at which Fe coordinates to the imidazole group of His and may be omitted here." + } + ], + "full_mark": 4 + } + ] +} diff --git a/test/tasks/Exam-15-T1/problem.json b/test/tasks/Exam-15-T1/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..cbb39d7965a3aa6b5570e21155b1e06e26f5552d --- /dev/null +++ b/test/tasks/Exam-15-T1/problem.json @@ -0,0 +1,35 @@ +{ + "problem_id": "Exam-15-T1", + "header": "Problem 1 The Immortal Mirror. In 1863, two chemists at the Freiberg Mining Academy in Germany unexpectedly observed a previously unknown spectral line while studying locally mined sphalerite. They assigned it to an undiscovered element M and successfully isolated elemental M from the mineral. At the subsequent Paris Exposition, elemental M was publicly displayed as \"the rarest metal,\" causing a sensation. At that time, M was used to make highly reflective mirrors; today it is more commonly an important semiconductor raw material used in conductive films.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-15-T1-1", + "tags": [ + "2.1" + ], + "text_open": "1.1 Burning M in air gives a pale-yellow amorphous solid A. Heating A under hydrogen at 400~500°C reduces it to a black crystalline solid B (Reaction 1); dissolving B in dilute hydrochloric acid releases a colorless gas C. Above approximately 700°C, B decomposes to A and elemental M (Reaction 2). At high temperature, A reacts with ammonia to form a yellow semiconductor material D and releases a gas (Reaction 3).\n1.1.2 Write the chemical equations for Reactions 1~3.", + "text_images": [], + "full_mark": 3 + }, + { + "sub_id": "Exam-15-T1-2", + "tags": [ + "1.4" + ], + "text_open": "1.2 Chloride E of M is a white flaky solid that is extremely hygroscopic and dissolves in water with substantial heat release. In ether solution, E reacts with LiH to form complex F. F decomposes above 0°C and can be used as an in situ reducing agent in organic synthesis. E forms complex E·3THF with the electron-donor ligand THF. Single-crystal X-ray diffraction shows that the M center in this complex is six-coordinate with octahedral geometry. E has a layered crystal structure consisting of close-packed chloride ions with layers of octahedrally coordinated M centers. By contrast, chloride G of the congener whose period differs by only 1 commonly crystallizes as dimer H. 1.2.1 Write the chemical formulas of E~G, write the SMILES of H, and describe its three-dimensional structure in words.", + "text_images": [], + "full_mark": 5 + }, + { + "sub_id": "Exam-15-T1-3", + "tags": [ + "1.5", + "1.6" + ], + "text_open": "1.3 In $E\\cdot3THF$, the $M$–Cl bond length is approximately 242 pm and the $M$–O bond length is approximately 225 pm. The ionic radius of $M$ is 80 pm, the ionic radius of $Cl^-$ is 181 pm, and the van der Waals radius of O in THF is approximately 152 pm.\n1.3.2 Use the bond-length data to compare the bonding character of the M-Cl and M-O bonds, including their relative ionic/covalent character, and explain.", + "text_images": [], + "full_mark": 3 + } + ] +} diff --git a/test/tasks/Exam-15-T4/problem.json b/test/tasks/Exam-15-T4/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..7c8fbc1b7adf7add2944dd0d69a64e3f7a64b82d --- /dev/null +++ b/test/tasks/Exam-15-T4/problem.json @@ -0,0 +1,48 @@ +{ + "problem_id": "Exam-15-T4", + "header": "Problem 4 Determination of Chloride by the Volhard Method. ① Pipette 20.00 mL of 0.1 mol/L standard silver nitrate titrant into a 250 mL conical flask. Add 30 mL water, 5 mL nitric acid solution, and 2 mL saturated ferric ammonium sulfate solution. While shaking, titrate with standard potassium thiocyanate titrant until a pale brown-red color appears and persists for 1 min. Record the volume of standard potassium thiocyanate consumed: V_{1} = 19.88 mL. ② Weigh 0.10 g sodium chloride that has been ignited at 500 ℃ ~ 600 ℃ to constant mass, accurate to 0.1 mg. Dissolve it in approximately 40 mL water in a beaker and transfer to a 100 mL volumetric flask. Add 5 mL nitric acid solution (1+3), then add 25.00 mL of 0.1 mol/L standard silver nitrate solution while shaking vigorously. Dilute to the mark with water and mix. Let stand in the dark for 5 min, filter through rapid filter paper, and discard the first 10 mL of filtrate. Accurately pipette 50.00 mL filtrate into a 250 mL conical flask, add 2 mL of 1 mol/L ferric ammonium sulfate solution, and titrate with standard potassium thiocyanate while shaking until a pale brown-red color appears and persists for 1 min. Record V_{2} = 4.01 mL. ③ Measure 50 mL test solution into a 100 mL colorimetric tube and otherwise follow the same procedure as in ②. Record V_{3} = 2.98 mL.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-15-T4-1", + "tags": [ + "4.5", + "2.1" + ], + "text_open": "4.1 Calculate the concentrations of the standard silver nitrate and potassium thiocyanate titrant solutions.", + "text_images": [], + "full_mark": 5 + }, + { + "sub_id": "Exam-15-T4-2", + "tags": [ + "4.1", + "4.7" + ], + "text_open": "4.4.2 Explain why the first 10 mL of filtrate is discarded.", + "text_images": [], + "full_mark": 1 + }, + { + "sub_id": "Exam-15-T4-3", + "tags": [ + "4.2", + "4.5" + ], + "text_open": "4.5 If addition of standard silver nitrate produces a colloidal solution, what additional operation is required before filtration?", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-15-T4-4", + "tags": [ + "4.5", + "2.1", + "4.7" + ], + "text_open": "4.6 Calculate the $Cl^-$ concentration in the sample.", + "text_images": [], + "full_mark": 3 + } + ] +} diff --git a/test/tasks/Exam-17-T2/images/Exam-17-T2-header-1.png b/test/tasks/Exam-17-T2/images/Exam-17-T2-header-1.png new file mode 100644 index 0000000000000000000000000000000000000000..b2d2fe54c4e09cf8b2aea76822e1e241950005d3 --- /dev/null +++ b/test/tasks/Exam-17-T2/images/Exam-17-T2-header-1.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:daf5975087ee04d545688808446e0b7a7839dd9955c786702679b8e18b0c6abc +size 226189 diff --git a/test/tasks/Exam-17-T2/problem.json b/test/tasks/Exam-17-T2/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..c79dccffabeffd71f20138d0d11d67239e0e3797 --- /dev/null +++ b/test/tasks/Exam-17-T2/problem.json @@ -0,0 +1,39 @@ +{ + "problem_id": "Exam-17-T2", + "header": "Problem 2 Determination of Paracetamol by Cyclic Voltammetry and Study of Its Electrochemistry\nCyclic voltammetry (CV) is an important electrochemical technique for investigating electrochemical reaction mechanisms and performing quantitative analysis. It uses a three-electrode system consisting of a working electrode, a reference electrode, and an auxiliary electrode. A triangular-wave potential that varies linearly with time is applied to the working electrode while the response current is recorded. One triangular-wave scan completes both an oxidation scan and a reduction scan, and the resulting current–potential curve is called a cyclic voltammogram. A substance that can undergo oxidation or reduction at an electrode within the scanned potential range is called an electroactive substance. A typical cyclic voltammogram exhibits an oxidation peak (anodic peak) and a reduction peak (cathodic peak). Under suitable conditions, the peak currents ($I_{pa}$ and $I_{pc}$) are proportional to the concentration of the electroactive substance and form the basis of quantitative analysis. The peak potentials ($E_{pa}$ and $E_{pc}$) and the peak-potential separation ($\\Delta E_p = E_{pa} - E_{pc}$) can be used to assess reaction reversibility and the number of electrons transferred. Changing the scan rate ($v$) can distinguish diffusion control from adsorption control and reveal the kinetics of subsequent chemical reactions.\nParacetamol (APAP, also known as acetaminophen) is a commonly used antipyretic analgesic. Studies show that APAP undergoes electrochemical oxidation at an electrode surface, and its oxidation product NAPQI undergoes subsequent chemical reactions in acidic media. A research group used cyclic voltammetry to investigate the electrochemical behavior of APAP systematically.\nAt room temperature, the peak current ($I_p$, in A) for a reversible electrochemical reaction in a quiescent solution at a planar electrode is related to the scan rate ($v$, in V/s), the concentration of the electroactive substance ($C$, in $\\mathrm{mol/cm^3}$), and other parameters by the Randles–Sevcik equation:\n$$I_p = 2.69 \\times 10^5 n^{3/2} A C D^{1/2} v^{1/2}$$\nn: number of electrons transferred in the reaction.\nA: effective area of the working electrode (unit: $\\mathrm{cm^2}$).\nD: diffusion coefficient of the electroactive substance (unit: $\\mathrm{cm^2/s}$), reflecting how rapidly the substance moves through solution.\nThe figure below is a typical cyclic voltammogram. Curve 1 is the background signal, and curve 2 is the normal response signal.", + "header_images": [ + { + "file": "Exam-17-T2-header-1.png", + "description": "Typical cyclic voltammogram. The horizontal axis is potential, with tick marks of approximately 0.2, 0.1, 0, -0.1, and -0.2 from left to right; the vertical axis is current. The peak-current and peak-potential symbols $I_{pc}$, $I_{pa}$, $E_{pa}$, and $E_{pc}$ are labeled. Two closed curves are labeled 1 and 2. As stated in the question, curve 1 is the background signal and curve 2 is the normal response signal." + } + ], + "subquestions": [ + { + "sub_id": "Exam-17-T2-1", + "tags": [ + "4.7" + ], + "text_open": "2.2 At a scan rate of $40\\ \\mathrm{mV\\cdot s^{-1}}$ in a pH = 2.2 buffer, the research group obtained the following cyclic-voltammetry data for standard APAP solutions:\n\n| APAP concentration/(mmol/L) | 0.10 | 0.40 | 1.00 | 3.00 | 5.00 |\n| ----------------- | ----- | ----- | ----- | ----- | ----- |\n| Anodic peak current $I_{pa}$/μA | 2.15 | 5.82 | 13.46 | 38.74 | 63.52 |\n\nUsing concentration as the horizontal coordinate and anodic peak current as the vertical coordinate, obtain the linear regression equation of $I_{pa}$ against APAP concentration and calculate the expected peak current for an APAP solution with a concentration of $3.50\\ \\mathrm{mmol\\cdot L^{-1}}$.", + "text_images": [], + "full_mark": 5 + }, + { + "sub_id": "Exam-17-T2-2", + "tags": [ + "2.3" + ], + "text_open": "2.3 To investigate the oxidation mechanism of APAP, the group performed CV scans at $40\\ \\mathrm{mV\\cdot s^{-1}}$ in media of different acidity (pH = 6.0, pH = 2.2, and $1.8\\ \\mathrm{mol\\cdot L^{-1}}\\ H_2SO_4$) and observed the following:\nAt pH = 6.0, both oxidation and reduction peaks appear, but $I_{pa}$ ≠ $I_{pc}$; at pH = 2.2, only the oxidation peak appears and the reduction peak disappears; under strongly acidic conditions ($1.8\\ \\mathrm{mol\\cdot L^{-1}}\\ H_2SO_4$), the reduction peak reappears.\n2.3.1 Explain why the reduction peak disappears at pH = 2.2.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-17-T2-3", + "tags": [ + "2.3" + ], + "text_open": "2.3.2 What chemical reaction is indicated by the reappearance of the reduction peak under strongly acidic conditions?", + "text_images": [], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-17-T4/problem.json b/test/tasks/Exam-17-T4/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..c8327e7dbf062c8cefa4637caf808744644ae4f1 --- /dev/null +++ b/test/tasks/Exam-17-T4/problem.json @@ -0,0 +1,26 @@ +{ + "problem_id": "Exam-17-T4", + "header": "Problem 4 Gravimetric Determination of Tungsten (17 points, 8%)\nTungsten is an important strategic metal that has an irreplaceable strengthening role in cemented carbides, high-speed steel, and specialty steels. Accurate determination of tungsten content is essential for quality control of alloy materials. In the classical gravimetric method, tungsten is converted to a tungstic-acid precipitate, ignited to tungsten trioxide, and weighed. Accurate tungsten content is obtained by correcting for interference from coexisting elements such as molybdenum, chromium, and vanadium. This problem concerns precipitation and separation of tungsten, correction for impurities, and special treatment of niobium-containing samples, and tests understanding of the principles of gravimetric analysis and strategies for eliminating interference.\nThe hydrochloric acid used in this problem is diluted 1:1, the nitric acid is concentrated, and every reported mass has been corrected for the tare mass of the crucible.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-17-T4-1", + "tags": [ + "2.1" + ], + "text_open": "4.1 Procedure for samples without niobium\nA sample of mass $m$ is weighed into a beaker. Hydrochloric acid and nitric acid are added, and the mixture is heated gently to dissolve the sample and evaporated to a syrup to remove nitrogen oxides. Hydrochloric acid is added to dissolve the salts, the solution is diluted with water and brought to the boil, and cinchonine hydrochloride solution is added slowly to precipitate all tungsten as tungstic acid. The precipitate is aged at 70-80℃ for 4 h and then filtered and washed with a cinchonine hydrochloride wash solution.\nThe precipitate and filter paper are placed in a platinum crucible, carbonized, and ignited at 750-800℃. The residue is moistened with water, hydrofluoric acid and sulfuric acid are added, and the mixture is heated to volatilize silicon. It is then ignited to constant mass, giving a mass $m_1$ of impure $WO_3$.\nSodium carbonate is added to the platinum crucible for fusion. After cooling, the melt is extracted with hot water and filtered, and the residue is ignited and weighed to give $m_2$. The filtrate is analyzed spectrophotometrically for molybdenum, chromium, and vanadium, which are converted to the corresponding oxide masses $m_3$, $m_4$, and $m_5$.\n4.1.1 Write the equation for oxidation of elemental tungsten to tungstic acid, $H_2WO_4$, in hydrochloric acid/nitric acid.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-17-T4-2", + "tags": [ + "4.7", + "2.1" + ], + "text_open": "4.1.6 During aqueous extraction after sodium carbonate fusion, $WO_3$ is converted to $Na_2WO_4$ and enters solution, whereas $TiO_2$ and $Fe_2O_3$ remain insoluble. Analysis of a tungsten-steel sample gives the following data: $m$ = 0.5123 g, $m_1$ = 0.0842 g, $m_2$ = 0.0018 g, $m_3$ = 0.0023 g, $m_4$ = 0.0015 g, and $m_5$ = 0.0012 g. Calculate the mass fraction of tungsten in the sample.", + "text_images": [], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-17-T7/images/Exam-17-T7-1-1-prompt.png b/test/tasks/Exam-17-T7/images/Exam-17-T7-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..7ddc101c455472d248debc0f71d11ea55e8e27af --- /dev/null +++ b/test/tasks/Exam-17-T7/images/Exam-17-T7-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:428534b76acb199c62a5fbb1b02b3f5d6cc1449bad3b72639f158dc91960a5af +size 80032 diff --git a/test/tasks/Exam-17-T7/images/Exam-17-T7-3-1-prompt.png b/test/tasks/Exam-17-T7/images/Exam-17-T7-3-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..190735f830ca0fdc11bc09d8359aaadadfaab63c --- /dev/null +++ b/test/tasks/Exam-17-T7/images/Exam-17-T7-3-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:b6f672b778fe431cb7ec63869662d2f52c62a109825292fb006c371defe87b4c +size 56376 diff --git a/test/tasks/Exam-17-T7/images/Exam-17-T7-7-1-prompt.png b/test/tasks/Exam-17-T7/images/Exam-17-T7-7-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..dfc2da2808f719b939ac9e8d334576c4307bb5ad --- /dev/null +++ b/test/tasks/Exam-17-T7/images/Exam-17-T7-7-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:22c75c472291aebfa4e0195687f123b987e990e5e325ae5e00c018902c7e6552 +size 57525 diff --git a/test/tasks/Exam-17-T7/problem.json b/test/tasks/Exam-17-T7/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..3463440fb19a6f0c80f517b20fcd2c50c72441cd --- /dev/null +++ b/test/tasks/Exam-17-T7/problem.json @@ -0,0 +1,89 @@ +{ + "problem_id": "Exam-17-T7", + "header": "Problem 7 Low-Valent Cluster Compounds of Group 13 Metals\nLow-valent compounds of Group 13 elements such as aluminum and gallium can form metal clusters with diverse structures. Their bonding modes and electronic structures are important for understanding the nature of metal-metal bonds. This problem concerns low-valent gallium and aluminum halides and their assembly reactions with ligands, focusing on structural analysis, oxidation-state calculations, and cluster reactivity.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-17-T7-1", + "tags": [ + "2.1" + ], + "text_open": "Synthesis and structures of $E_3$ units: Ga-Ga bonds are relatively stable and can support extended compounds.\n7.1 Chain-like $E_3$ structure: reaction of freshly prepared LiM with 1.67 eq GaI affords the trigallane C = $Ga_3M_3I_2$.\n7.1.1 Write the reaction equation for the formation of C from LiM and GaI.", + "text_images": [ + { + "file": "Exam-17-T7-1-1-prompt.png", + "description": "Two-step reaction: '[dipp]N=C=N[dipp]'>t-BuLi (1eq), n-hexane, 0 °C>intermediate Li+M-, where M denotes the ligand and the SMILES for M- is '[dipp]/N=C([N-][dipp])/C(C)(C)C'. The M- anion has an N,N'-bis(dipp) amidinate framework with t-Bu bonded to the central carbon. The second step is '[dipp]/N=C([N-][dipp])/C(C)(C)C'>GaI (1.67eq), Toluene, -78 °C>C, where C = $Ga_3M_3I_2$ and its SMILES is 'I[Ga]1([Ga]2([Ga]3(I)N([dipp])C(C(C)(C)C)=[N]3[dipp])N([dipp])C(C(C)(C)C)=[N]2[dipp])N([dipp])C(C(C)(C)C)=[N]1[dipp]'. Three Ga atoms form a chain through Ga-Ga single bonds, each terminal Ga bears one I, and each Ga is chelated by one M ligand. No wedge or dashed bonds are shown." + } + ], + "full_mark": 2 + }, + { + "sub_id": "Exam-17-T7-2", + "tags": [ + "2.1" + ], + "text_open": "7.1.2 Write the equation for the substitution reaction of trigallane C with $NaBEt_3H$ to form Et(M)Ga-Ga(M)-Ga(M)H.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-17-T7-3", + "tags": [ + "2.4", + "1.6" + ], + "text_open": "7.2 Cyclic $E_3$ structure: cyclotrigallanes have intriguing electronic properties. Reaction of $(Mes_2C_6H_3)Li$ with $GaCl_3$ gives $(Mes_2C_6H_3)GaCl_2$, which is reduced with potassium metal in diethyl ether to afford the cyclic gallium compound D = $K_2[(Mes_2C_6H_3)Ga]_3$. This compound has a nearly equilateral triangular $Ga_3$ core, with an average Ga-Ga bond length of 2.426 Å that is markedly shorter than a typical single bond, indicating a higher bond order. The two potassium ions lie on opposite sides of the $Ga_3$ plane. Write the SMILES string for D and determine whether its anion is aromatic.", + "text_images": [ + { + "file": "Exam-17-T7-3-1-prompt.png", + "description": "Two-step reaction: $GaCl_3$ reacts with LiAr (1.0 eq) in $Et_2O$ at -78 °C; $K^0$ (4.0 eq) is then added and the mixture is reacted at rt. for 5 days to give $K_2(ArGa)_3·(Et_2O)_2$. The structure defining Ar = $(Mes_2C_6H_3)$ appears at the right: the top carbon of the central benzene ring, drawn in an alternating-double-bond Kekulé form, bears a wavy bond marking the attachment point to Ga, and each of the two ortho carbons bears a 2,4,6-trimethylphenyl group (Mes). The encodable structure of Ar is represented by the SMILES '[*]C1=C(C2=C(C)C=C(C)C=C2C)C=CC=C1C3=C(C)C=C(C)C=C3C', where [*] is the attachment point corresponding to the wavy bond." + } + ], + "full_mark": 3 + }, + { + "sub_id": "Exam-17-T7-4", + "tags": [ + "2.4", + "1.4" + ], + "text_open": "7.4 Eight-membered-ring structure: gaseous GaI is cocondensed with excess $PEt_3$ in toluene at -196 °C and then warmed slowly, giving F = $Ga_8I_8·6PEt_3$. This cluster has $C_{2h}$ symmetry and contains a nearly planar eight-membered $Ga_8$ ring. Two iodine atoms bridge across the ring to form a four-membered $Ga_2I_2$ ring. All eight Ga atoms are four-coordinate, and the average Ga-Ga bond length is 2.466 Å.\n7.4.1 Write the SMILES string for F.", + "text_images": [], + "full_mark": 4 + }, + { + "sub_id": "Exam-17-T7-5", + "tags": [ + "1.7", + "2.4" + ], + "text_open": "7.4.2 The Ga/I/$PEt_3$ system forms the eight-membered-ring cluster F ($Ga_8I_8$), whereas the Al/Br/$NEt_3$ system forms the four-membered-ring cluster E ($Al_4Br_4$) under similar conditions. Which explanations are the most reasonable? Select all that apply.\n(A) Ga has a smaller atomic radius than Al and a shorter Ga-Ga bond; formation of a four-membered ring would therefore cause greater ring strain and bond-angle distortion, so an eight-membered ring is favored to relieve strain.\n(B) I has a larger atomic radius than Br and is well suited for bridging; transannular bridges stabilize the eight-membered ring.\n(C) $PEt_3$ is a stronger electron donor than $NEt_3$, increasing the electron density at Ga and shortening the Ga-Ga bonds, which favors formation of a larger ring system.\n(D) The Ga-Ga bonds in the eight-membered ring have a higher bond order than the Al-Al bonds in the four-membered ring, giving greater thermodynamic stability.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-17-T7-6", + "tags": [ + "2.4", + "1.2" + ], + "text_open": "7.5 Fused-ring structure: attempts to synthesize amine-stabilized $Ga_8I_8·6NEt_3$ were unsuccessful, but partial oxidation unexpectedly gave G = $Ga_8I_8O_2·6NEt_3$. Its structure has $C_i$ symmetry and consists of two nearly planar $Ga_4O$ rings fused through a shared edge of a central four-membered $Ga_2O_2$ ring. All Ga atoms are four-coordinate. Write the SMILES string for G.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-17-T7-7", + "tags": [ + "1.1" + ], + "text_open": "7.6 Tetrahedral type: high-temperature gaseous GaBr is cocondensed with a toluene solution of $NHEt_2$. The resulting dark solution is concentrated and allowed to stand at 25 °C, yielding crystals of cluster H = $Ga_5Br_7(NHEt_2)_5$. H contains a $Ga_5$ core in which the central Ga atom is tetrahedrally surrounded by four Ga atoms.\n7.6.2 In the sterically congested H molecule, the 12 heteroatoms—the halogens and the O/N atoms of the ligands—adopt an arrangement that maximizes their mutual separation. What polyhedron is formed by these 12 vertices?", + "text_images": [ + { + "file": "Exam-17-T7-7-1-prompt.png", + "description": "Structure of H = $Ga_5Br_7(NHEt_2)_5$: the central Ga is connected by single bonds to four peripheral Ga atoms in a tetrahedral arrangement. The encodable structure of H is represented by the SMILES '[L][Ga](Br)(Br)[Ga]([Ga]([L])(Br)Br)([Ga]([L])([L])Br)[Ga]([L])(Br)Br', where [L] denotes the $NHEt_2$ ligand labeled in the image. A total of 5 L ligands and 7 Br atoms are distributed among the peripheral Ga atoms. The lower right of the image defines L = $NHEt_2$, and all bonds are solid single bonds." + } + ], + "full_mark": 1 + } + ] +} diff --git a/test/tasks/Exam-17-T9/images/Exam-17-T9-1-1-prompt.png b/test/tasks/Exam-17-T9/images/Exam-17-T9-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..4378a6671430a026a482a8896b2e51fc9702f7a9 --- /dev/null +++ b/test/tasks/Exam-17-T9/images/Exam-17-T9-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:4de7ab1879e1fec710df4685e95d7be74d374fd8b161ff806f2b90b4abdbccad +size 51169 diff --git a/test/tasks/Exam-17-T9/images/Exam-17-T9-1-2-prompt.png b/test/tasks/Exam-17-T9/images/Exam-17-T9-1-2-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..797469dc3d73515c171cfe55512a7961273a3bf6 --- /dev/null +++ b/test/tasks/Exam-17-T9/images/Exam-17-T9-1-2-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:15c8756f20448077a06a1d1598863078ef1cfe75abd5844e393a107cf142bbe7 +size 43629 diff --git a/test/tasks/Exam-17-T9/images/Exam-17-T9-2-1-prompt.png b/test/tasks/Exam-17-T9/images/Exam-17-T9-2-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..246d1fadb4086cdf9b9714bd8c8907570f6e7b14 --- /dev/null +++ b/test/tasks/Exam-17-T9/images/Exam-17-T9-2-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:65e8da18a04e831d248dc2761061644f695a6873b4152de4029e676c337c30b2 +size 925448 diff --git a/test/tasks/Exam-17-T9/images/Exam-17-T9-4-1-prompt.png b/test/tasks/Exam-17-T9/images/Exam-17-T9-4-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..4058d4217a8ecc9c50632bf80e8e4b1c4ae10b3b --- /dev/null +++ b/test/tasks/Exam-17-T9/images/Exam-17-T9-4-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:d8b885ec10917daccb26b2746a3051483e5fafe27c515e1815ec834700d9095e +size 58032 diff --git a/test/tasks/Exam-17-T9/images/Exam-17-T9-5-1-prompt.png b/test/tasks/Exam-17-T9/images/Exam-17-T9-5-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..b8c7942ddb256f19ddca15cc53747d1a09c267bd --- /dev/null +++ b/test/tasks/Exam-17-T9/images/Exam-17-T9-5-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:64deca9309bd9b50c22bab4ab1eb7d7465356b198435ea8bdfd90ac703368df4 +size 43369 diff --git a/test/tasks/Exam-17-T9/problem.json b/test/tasks/Exam-17-T9/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..950c9e512e757bbfc5e61ff26617a2ada2c97166 --- /dev/null +++ b/test/tasks/Exam-17-T9/problem.json @@ -0,0 +1,81 @@ +{ + "problem_id": "Exam-17-T9", + "header": "Problem 9 Stereochemistry\nCareful design of reactant structures and reaction conditions can provide highly selective stereochemical control.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-17-T9-1", + "tags": [ + "2.4", + "1.2" + ], + "text_open": "9.1 Garner's aldehyde is frequently used for stereochemical control in synthesis. The route below prepares it from the natural product S-serine.\n9.1.2 Garner's aldehyde can undergo further nucleophilic addition. Write the SMILES strings of the major products under the different reaction conditions below, representing them as anions and specifying stereochemistry.", + "text_images": [ + { + "file": "Exam-17-T9-1-1-prompt.png", + "description": "Two-step reaction: 'OC[C@@H](C(O)=O)N'>>'OC[C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O'; 'OC[C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O'>DMSO, (COCl)2, DIPEA>'[H]C([C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O)=O'." + }, + { + "file": "Exam-17-T9-1-2-prompt.png", + "description": "Two parallel reactions: '[H]C([C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O)=O'>Nu^->[A]; '[H]C([C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O)=O'>Nu^-, [M]>[B]. The products are labeled only [A] and [B]; their structures are not shown." + } + ], + "full_mark": 4 + }, + { + "sub_id": "Exam-17-T9-2", + "tags": [ + "2.7", + "1.2" + ], + "text_open": "9.1.3 In practice, the two reaction pathways compete. Determine which of the following conditions favor formation of product B, and briefly explain why.\n(1) Solvent: a. THF; b. toluene", + "text_images": [ + { + "file": "Exam-17-T9-2-1-prompt.png", + "description": "'[H]C([C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O)=O'>nucleophile, solvent, additive>[A]+[B]. The products are labeled only [A] and [B]; their structures are not shown." + } + ], + "full_mark": 2 + }, + { + "sub_id": "Exam-17-T9-3", + "tags": [ + "2.7", + "1.2" + ], + "text_open": "(3) Additive: a. DMPU; b. $ZnCl_2$", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-17-T9-4", + "tags": [ + "1.2", + "2.7" + ], + "text_open": "9.1.4 Garner's aldehyde can also be synthesized from R-serine. In the reaction below, replacing Boc with Ts reverses the selectivity. Explain why.", + "text_images": [ + { + "file": "Exam-17-T9-4-1-prompt.png", + "description": "'CC1(C)N([R])[C@@H](C=O)CO1'>Et-MgBr, THF>'CC1(C)N([R])[C@@H]([C@@H](O)CC)CO1' + 'CC1(C)N([R])[C@@H]([C@H](O)CC)CO1'. The image also states that the product ratio is 1:9 when R=Boc and 8:1 when R=Ts. R is the placeholder group used in the original diagram." + } + ], + "full_mark": 2 + }, + { + "sub_id": "Exam-17-T9-5", + "tags": [ + "2.4", + "1.2" + ], + "text_open": "9.2 Garner's aldehyde can also participate in multicomponent reactions that construct new stereocenters. Write the SMILES string of the product of the reaction below.", + "text_images": [ + { + "file": "Exam-17-T9-5-1-prompt.png", + "description": "Multicomponent reaction: 'C1(C=CC=C2)=C2NC=C1'.'CC(OC[C@H]1C=O)(C)N1C(OC(C)(C)C)=O'.'O=C1CC(OC(C)(C)O1)=O'>CH3CN, D,L-proline>. No product structure is shown to the right of the arrow." + } + ], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-18-T3/problem.json b/test/tasks/Exam-18-T3/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..20149e18c8541785e8504a630fd8f3c97a080c70 --- /dev/null +++ b/test/tasks/Exam-18-T3/problem.json @@ -0,0 +1,25 @@ +{ + "problem_id": "Exam-18-T3", + "header": "Problem 3 Latent Monomers and Controlled Metathesis Polymerization\nControlling the temporal and spatial sequence of polymerization is an important topic in conventional polymer synthesis. Chemists have generally developed “latent catalysts” that switch between inactive and active states in response to external stimuli such as light or heat. Recent research, however, has introduced a new “latent monomer” strategy: isomerization of a photoswitchable molecule generates an active monomer in situ and on demand, enabling efficient control of polymerization.\nThis problem concerns a classic class of highly reactive ring-opening metathesis polymerization (ROMP) monomers: norbornadiene (NBD) derivatives. Under ultraviolet irradiation of a specific wavelength, they undergo intramolecular cycloaddition to form highly strained quadricyclane (QC) structures.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-18-T3-1", + "tags": [ + "2.4" + ], + "text_open": "3.1 Synthesis and Photothermal Isomerization of the Latent Monomer\nMonomer A, an NBD derivative, can be prepared by reacting cyclopentadiene with diethyl acetylenedicarboxylate at room temperature. Under 313 nm ultraviolet irradiation, A is quantitatively converted into latent monomer B, a QC derivative. Adding silica-coated gold nanobipyramids (AuBPs) to B and irradiating at 850 nm with infrared light regenerates A.\n3.1.1 Write the SMILES strings for A and B.", + "text_images": [], + "full_mark": 3 + }, + { + "sub_id": "Exam-18-T3-2", + "tags": [ + "2.4" + ], + "text_open": "3.3 The complete inertness of latent monomers toward the catalyst at room temperature enables the synthesis of complex polymer topologies that are difficult to obtain by conventional metathesis polymerization. The researchers replaced the ethoxycarbonyl groups, $-COOC_2H_5$, in B with pent-4-enyloxycarbonyl groups, $-COO(CH_2)_3CH=CH_2$, to give monomer F. F was mixed with the Grubbs catalyst and reacted at 40℃, releasing gas X and giving a soluble linear polymer G with double bonds in its main chain. G was coated onto a microscope slide placed over a masked LED and then heated to 80℃, rapidly converting the film into an insoluble crosslinked network H. Write the SMILES strings for polymers G and H.", + "text_images": [], + "full_mark": 4 + } + ] +} diff --git a/test/tasks/Exam-18-T4/problem.json b/test/tasks/Exam-18-T4/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..4a477af0b4e0d291f621adde68807c14840ac236 --- /dev/null +++ b/test/tasks/Exam-18-T4/problem.json @@ -0,0 +1,46 @@ +{ + "problem_id": "Exam-18-T4", + "header": "Problem 4 Complexometric Titration\nEthylenediaminetetraacetic acid (EDTA) is an effective chelating ligand widely used in chemical analysis. Approximately 20 g of a standard reagent is roughly weighed, 250mL of water is added, and the mixture is heated to dissolve the solid, cooled, and mixed thoroughly. The measured pH of the solution is 4.2. The stepwise ionization constants of EDTA are shown below.\n\n| | $K_{a1}$ | $K_{a2}$ | $K_{a3}$ | $K_{a4}$ | $K_{a5}$ | $K_{a6}$ |\n| --- | --- | --- | --- | --- | --- | --- |\n| | $10^{-0.9}$ | $10^{-1.6}$ | $10^{-2.0}$ | $10^{-2.67}$ | $10^{-6.16}$ | $10^{-10.26}$ |", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-18-T4-1", + "tags": [ + "4.5", + "3.4" + ], + "text_open": "The standard solution is then standardized. Weigh 0.2500 g of working-reference ZnO that has been ignited to constant mass in a furnace at 800 ℃. Moisten it with a small amount of water, dissolve it in 2 mL of 20% hydrochloric acid solution, and add 100 mL of water. Adjust the solution to pH 8 with hexamethylenetetramine (HMTA) solution, add 10 mL of $(CH_2)_6N_4$-$(CH_2)_6N_4$·HCl buffer to adjust the pH to 10, and then add 5 drops of Eriochrome Black T indicator solution. Titrate with the prepared EDTA sodium-salt solution until the color changes from purple to pure blue. Conduct a blank determination in parallel. After blank correction, the volume of EDTA consumed is $V = 15.35\\ \\mathrm{mL}$.\n4.2.2 The buffering ability of a buffer solution is commonly measured by its buffer capacity, denoted β. It is defined as the amount db in mol of strong base required to raise the pH of 1L of buffer solution by dpH units, with the mathematical expression\n$\\beta = \\frac{db}{dpH} = -2.30[H^+] \\times \\frac{db}{d[H^+]}$\nGiven $pK_b((CH_2)_6N_4) = 4.9$, calculate the buffer capacity of a pH = 10 $(CH_2)_6N_4$-$(CH_2)_6N_4$·HCl buffer in which the analytical concentration of $(CH_2)_6N_4$ is 0.1000 mol/L. Hint: express $b$ as a function of $[H^+]$ and differentiate with respect to $[H^+]$ to obtain $\\frac{db}{d[H^+]}$. B may be used to represent $(CH_2)_6N_4$ in the calculation.", + "text_images": [], + "full_mark": 4 + }, + { + "sub_id": "Exam-18-T4-2", + "tags": [ + "4.5", + "3.4" + ], + "text_open": "4.2.3 Neglecting volume changes during standardization, the change in solution pH over the entire process must not exceed 0.2 pH unit to ensure an accurate endpoint. Determine the minimum value of $[(CH_2)_6N_4]$.", + "text_images": [], + "full_mark": 4 + }, + { + "sub_id": "Exam-18-T4-3", + "tags": [ + "4.1" + ], + "text_open": "4.3 During standardization, why is the $(CH_2)_6N_4$-$(CH_2)_6N_4$·HCl buffer not added directly to adjust the solution to pH=10?", + "text_images": [], + "full_mark": 1 + }, + { + "sub_id": "Exam-18-T4-4", + "tags": [ + "4.5", + "1.4" + ], + "text_open": "4.4 The EDTA standard solution standardized in this problem can be used to determine the concentrations of Cu2+ and Mg2+ in a solution by either of the following methods. Method 1: Masking Cu2+ ions. Adjust the solution to pH 5.5 with an HMTA buffer and add excess thiosulfate to mask Cu2+. In this reaction, thiosulfate acts as both a reducing agent and a ligand. Method 2: Selective determination of $Cu^{2+}$. Adjust the solution to pH 5 with an HMTA buffer. Using 5-Br-PADAT as the indicator, selectively determine $Cu^{2+}$ with an EDTP standard solution whose principal component is disodium ethylenediaminetetrapropionate. 4.4.2 Based on the data below, explain why EDTP shows high selectivity for $Cu^{2+}$.\n\n| | $Cu^{2+}$ | $Zn^{2+}$ | $Cd^{2+}$ | $Mg^{2+}$ |\n| --- | --- | --- | --- | --- |\n| Ionic radius/pm | 73 | 74 | 95 | 72 |\n| $\\lg K_{M-EDTP}$ | 15.4 | 7.8 | 6.0 | 1.8 |\n| $\\lg K_{M-EDTA}$ | 18.8 | 16.50 | 16.46 | 8.7 |", + "text_images": [], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-18-T5/images/Exam-18-T5-header-1.png b/test/tasks/Exam-18-T5/images/Exam-18-T5-header-1.png new file mode 100644 index 0000000000000000000000000000000000000000..73f3c0b4c311a027a03f151b23fd0064490c5c0a --- /dev/null +++ b/test/tasks/Exam-18-T5/images/Exam-18-T5-header-1.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:986302b14d7f80500ac4611f639c31a47c22d1f92ee366592f6a77b3a7bba163 +size 1339205 diff --git a/test/tasks/Exam-18-T5/problem.json b/test/tasks/Exam-18-T5/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..6d09ad362bb2c3d62f744ba3bd3c782307c91733 --- /dev/null +++ b/test/tasks/Exam-18-T5/problem.json @@ -0,0 +1,30 @@ +{ + "problem_id": "Exam-18-T5", + "header": "Problem 5 Pharmacokinetics and Molecular Design\nPharmacokinetics quantitatively studies the absorption, distribution, metabolism, and excretion of drugs in living organisms and uses mathematical principles and methods to describe how drug concentration in blood varies with time. Pharmacokinetic parameters reveal pharmacological shortcomings of drug molecules and guide molecular modifications that improve specific properties.\nPropranolol (P), metoprolol (M), betaxolol (B), talinolol (T), and atenolol (A) are β-blockers. Propranolol (P) binds both β1 receptors, found mainly in the heart, and β2 receptors, found mainly in bronchial smooth muscle. At therapeutic doses, the other drugs bind β1 receptors with high selectivity. Although their therapeutic mechanisms are similar, their pharmacological properties differ substantially. Their structures are shown below.\nParameters that may be used in this problem are listed below.\n\n| Drug | Abbreviation | lg $P$ | Conjugate-acid $pK_a$ | $F_a$ | $E_{gi}$ | $E_H$ | $V_d$ / (L·kg$^{-1}$) |\n| --- | --- | --- | --- | --- | --- | --- | --- |\n| Propranolol | P | 3.20 | 9.45 | 0.95 | 0.10 | 0.64 | 3.8 |\n| Metoprolol | M | 2.90 | 9.68 | 0.90 | 0.02 | 0.60 | 5.6 |\n| Betaxolol | B | 2.42 | 9.40 | 0.95 | 0.00 | 0.19 | 1.6 |\n| Talinolol | T | 3.20 | 9.40 | 0.95 | 0.36 | 0.09 | 3.3 |\n| Atenolol | A | 0.25 | 9.60 | 0.50 | 0.00 | 0.01 | 0.70 |", + "header_images": [ + { + "file": "Exam-18-T5-header-1.png", + "description": "Propranolol (P) has SMILES 'CC(NCC(COC1=CC=CC2=C1C=CC=C2)O)C'; metoprolol (M) has SMILES 'CC(NCC(COC1=CC=C(CCOC)C=C1)O)C'; betaxolol (B) has SMILES 'CC(NCC(COC1=CC=C(CCOCC2CC2)C=C1)O)C'; talinolol (T) has SMILES 'O=C(NC1=CC=C(OCC(CNC(C)(C)C)O)C=C1)NC2CCCCC2'; and atenolol (A) has SMILES 'NC(CC1=CC=C(OCC(CNC(C)(C)C)O)C=C1)=O'." + } + ], + "subquestions": [ + { + "sub_id": "Exam-18-T5-1", + "tags": [ + "1.7" + ], + "text_open": "5.3 The second pharmacokinetic process is distribution. After a drug molecule enters the systemic circulation, differences in lipophilicity cause some drugs to enter tissues and even the brain rapidly, while others remain in the blood. To compare drug distribution in the human body, the apparent volume of distribution $V_d$ is defined as follows. When drug distribution reaches dynamic equilibrium, assume that the drug concentration in every tissue and organ equals that in plasma; $V_d$ is the total fluid volume that would be required to contain all the drug in the body at that concentration. The formula is:\n$V_d = \\dfrac{\\text{total amount of drug in the body}\\,(\\mathrm{mg/kg})}{\\text{plasma drug concentration}\\,C\\,(\\mathrm{mg/L})}$\n5.3.2 A standard adult weighing 70kg has an actual total body-fluid volume of only approximately 40 L, including approximately 3 L of plasma. Calculate the approximate $V_d$ of a drug confined entirely to the plasma.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-18-T5-2", + "tags": [ + "2.9" + ], + "text_open": "5.4 The third pharmacokinetic process is clearance, which is divided mainly into hepatic and renal clearance. Highly lipophilic drugs generally tend toward hepatic clearance, and hepatic clearance capacity may be regarded as proportional to $E_h$, whereas highly water-soluble drugs generally tend toward renal clearance.\nThe approximate half-lives of the five drugs are P 4 h, M 5 h, B 18 h, T 12 h, and A 6.5 h. In the absence of other side effects, a drug with a longer half-life is preferred to maintain smooth dosing.\nThe following patients require a β-blocker from the five drugs above. Select the most appropriate drug for each patient and give only its abbreviation.\n5.4.3 Patient C has mild liver disease and cannot use a drug with $E_H$ greater than 0.10.", + "text_images": [], + "full_mark": 1 + } + ] +} diff --git a/test/tasks/Exam-18-T9/images/Exam-18-T9-1-1-prompt.png b/test/tasks/Exam-18-T9/images/Exam-18-T9-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..54e271e51a105a8a5013d5735c0571ad01d7688b --- /dev/null +++ b/test/tasks/Exam-18-T9/images/Exam-18-T9-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:7a464b22a03bd0e098e3f7ad345b8118fbcb7d751c09820359753613dab4855b +size 78094 diff --git a/test/tasks/Exam-18-T9/images/Exam-18-T9-2-1-prompt.png b/test/tasks/Exam-18-T9/images/Exam-18-T9-2-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..e0df638428925e1535e429546c4210a7a6765604 --- /dev/null +++ b/test/tasks/Exam-18-T9/images/Exam-18-T9-2-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:f949cefce115fff422c2c12e4ed92d8815fbd1fcdfb03ab2a4aa6f57c75df12b +size 80232 diff --git a/test/tasks/Exam-18-T9/images/Exam-18-T9-3-1-prompt.png b/test/tasks/Exam-18-T9/images/Exam-18-T9-3-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..462d0b25eb317041f6a34967493d0260c6253d75 --- /dev/null +++ b/test/tasks/Exam-18-T9/images/Exam-18-T9-3-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:946d84b60387ae95a989a8bdeeb4e0c2be426a2fcb4be0285851f17d1563e8cf +size 105345 diff --git a/test/tasks/Exam-18-T9/images/Exam-18-T9-4-1-prompt.png b/test/tasks/Exam-18-T9/images/Exam-18-T9-4-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..0b58247af816c95d2d4c16eb230c17b7fbdd5f14 --- /dev/null +++ b/test/tasks/Exam-18-T9/images/Exam-18-T9-4-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:fff56cf77d465a4f4a47cf3bf9b24f2beffa295b94e5550bbbbef4cb744a092b +size 61878 diff --git a/test/tasks/Exam-18-T9/images/Exam-18-T9-5-1-prompt.png b/test/tasks/Exam-18-T9/images/Exam-18-T9-5-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..bd2fcdfb576693ea25653621df5255faf127352b --- /dev/null +++ b/test/tasks/Exam-18-T9/images/Exam-18-T9-5-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:07664fe7da79385081d0e90b428127633ad18a0e6b968a3c896a6c44e21f5b02 +size 75371 diff --git a/test/tasks/Exam-18-T9/images/Exam-18-T9-6-1-prompt.png b/test/tasks/Exam-18-T9/images/Exam-18-T9-6-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..285783ceb60b5700e86c9254e17592878228e1d7 --- /dev/null +++ b/test/tasks/Exam-18-T9/images/Exam-18-T9-6-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:438bb20d8e0e82004f40c1f10a51df71e32a16a70fe57cb632c41725aa7474a6 +size 53814 diff --git a/test/tasks/Exam-18-T9/images/Exam-18-T9-7-1-prompt.png b/test/tasks/Exam-18-T9/images/Exam-18-T9-7-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..85ec8505753b04da5bf2ddf2a66378db1843d65f --- /dev/null +++ b/test/tasks/Exam-18-T9/images/Exam-18-T9-7-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:af84d47aaca31fd727d58fd8f71d7537503bd101188276e7dc9eb4a7ff963a3e +size 131099 diff --git a/test/tasks/Exam-18-T9/images/Exam-18-T9-8-1-prompt.png b/test/tasks/Exam-18-T9/images/Exam-18-T9-8-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..cd2cb3408ac8bf282c4a9168473b9607ffa0f72b --- /dev/null +++ b/test/tasks/Exam-18-T9/images/Exam-18-T9-8-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:9ac8815c0aa08a497529c1db935fc5586c5f66bd39e29bba0ef7b1d9738ddc72 +size 83715 diff --git a/test/tasks/Exam-18-T9/problem.json b/test/tasks/Exam-18-T9/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..d267d3c1fc2317eb37e34010c09726eb074d9440 --- /dev/null +++ b/test/tasks/Exam-18-T9/problem.json @@ -0,0 +1,127 @@ +{ + "problem_id": "Exam-18-T9", + "header": "Problem 9 Skeletal Reconstruction of Alkaloids\nLate-stage skeletal modification and precise remodeling of natural products provide an ideal shortcut for rapidly expanding compound libraries and discovering breakthrough drugs. In 2026, Amir H. Hoveyda's team reported an isolable, general-purpose platform generated by Hofmann elimination (HE), followed by a general strategy for divergent construction of new nonnatural frameworks through rearrangement, ring-closing metathesis, or intramolecular hydroamination.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-18-T9-1", + "tags": [ + "2.2", + "2.4" + ], + "text_open": "9.1.1 Under specific conditions, anhydrovinblastine reacts with a chlorine source to introduce a chlorine atom and form electrically neutral intermediate A1. In the presence of $AgBF_4$ and solvent, intramolecular reorganization occurs through intermediates A2 and A3, ultimately giving vinorelbine, a widely used clinical anticancer drug. Write the SMILES strings for the key intermediates A1, A2, and A3.", + "text_images": [ + { + "file": "Exam-18-T9-1-1-prompt.png", + "description": "Organic reaction scheme: '[H][C@]12CN(CCC3=C([C@]([R])(C(OC)=O)C2)NC4=CC=CC=C43)CC(CC)=C1'>AgBF4, THF/H_2O (1/1)>'[H][C@]1(C[C@@]([R])(C(OC)=O)C2=C(C3)C4=CC=CC=C4N2)CN3CC(CC)=C1'. The chlorine source is also drawn above the arrow and has SMILES 'CN1C2=CC=CC=C2N=N1.[Cl]'. R is the original placeholder in the figure." + } + ], + "full_mark": 3 + }, + { + "sub_id": "Exam-18-T9-2", + "tags": [ + "2.2", + "2.4" + ], + "text_open": "9.1.2 Under further basic conditions, nucleophilic conjugate addition introduces an alkynyl aldehyde fragment into vinorelbine to give A4. C-N bond cleavage then forms A5, followed by cyclization and desilylation to afford a two-carbon ring-expanded analogue. Write the SMILES strings for the two key intermediates A4 and A5.", + "text_images": [ + { + "file": "Exam-18-T9-2-1-prompt.png", + "description": "Organic reaction scheme: '[H][C@]1(C[C@@]([R])(C(OC)=O)C2=C(C3)C4=CC=CC=C4N2)CN3CC(CC)=C1'>'C[Si](C#CC=O)(C)C', Me3Si removal on workup>'[H][C@]1(C[C@]([R])(C(OC)=O)C2=C(C/C(C=O)=C\\3)C4=CC=CC=C4N2)CN3CC(CC)=C1'. R is the original placeholder in the figure." + } + ], + "full_mark": 2 + }, + { + "sub_id": "Exam-18-T9-3", + "tags": [ + "2.2", + "2.3" + ], + "text_open": "9.2 While studying skeletal deconstruction of vinpocetine, the researchers encountered an unexpected product. Reaction of vinpocetine with 5.0 equivalents of allyl iodide efficiently formed a quaternary ammonium salt. Subsequent treatment in potassium ethoxide/ethanol (KOEt/EtOH) at 80°C for 2 h gave not only the expected ring-opened product in 58% yield but also a deallylated secondary-amine byproduct in 32% yield, as shown. Isotopic-labeling experiments demonstrated that formation of the byproduct did not arise from direct $S_N2$ allylic nucleophilic substitution, but proceeded through a rapid and reversible acid-base proton-exchange process.\n9.2.1 Write the SMILES string for the key enammonium cation intermediate B formed after deprotonation of the quaternary ammonium salt and used to explain the deallylated byproduct.", + "text_images": [ + { + "file": "Exam-18-T9-3-1-prompt.png", + "description": "The figure shows a two-step reaction: '[R]C1=C[C@]2(CC)CCCN([C@@H]23)CCC4=C3N1C5=CC=CC=C45'>5.0 equiv. allyl iodide, MeCN, 3 h, 80 °C>'[R]C1=C[C@@]2(CCC[N@@+]3(CCC4=C([C@H]23)N1C5=CC=CC=C45)C(C=C)[H])CC.[I-]'; '[R]C1=C[C@@]2(CCC[N@@+]3(CCC4=C([C@H]23)N1C5=CC=CC=C45)C(C=C)[H])CC.[I-]'>KOEt, EtOH, + aq. workup>'[R]C(N1C2=C(C(C=C)=C1[C@H]3NCCC4)C=CC=C2)=C[C@@]34CC'. R is the original placeholder in the figure." + } + ], + "full_mark": 1 + }, + { + "sub_id": "Exam-18-T9-4", + "tags": [ + "2.3", + "2.2" + ], + "text_open": "9.2.2 If the KOEt/EtOH system above is replaced by the deuterated reagent system KOEt/EtOD and the vincane-derived quaternary ammonium salt is treated at 80°C, partially deuterated products are obtained. Quantitative nuclear magnetic resonance ($^1H$ NMR) analysis shows unequal deuteration at the allylic positions of the recovered diene substrate: 33% at $H^b$ and 16% at $H^c$. At the most acidic benzylic proton $H^a$, no appreciable deuteration is observed because elimination is too rapid.\nWrite the SMILES string for intermediate C in this kinetic proton-exchange process.", + "text_images": [ + { + "file": "Exam-18-T9-4-1-prompt.png", + "description": "Organic reaction scheme: '[R]C1=C[C@@]2(CCC[N@@+]3(CCC4=C([C@H]23)N1C5=CC=CC=C45)[C@@H](/C=C/[Hc])[Hb])CC.[I-]'>KOEt, EtOD>'[R]C1=C[C@@]2(CCC[N@@+]3(CCC4=C([C@H]23)N1C5=CC=CC=C45)[C@@H](/C=C/[Dc])[Db])CC.[I-]'. R is the original placeholder, and [Hb], [Hc], [Db], and [Dc] are the original hydrogen/deuterium labels in the figure." + } + ], + "full_mark": 1 + }, + { + "sub_id": "Exam-18-T9-5", + "tags": [ + "2.7", + "1.2" + ], + "text_open": "9.3 When the proton environment of the reaction is changed, a reactive carbanion intermediate can be directed onto an entirely different rearrangement pathway.\n9.3.1 Vincane is reacted with allyl iodide to form intermediate D, which is then treated with KOEt in anhydrous acetonitrile. Rather than undergoing elimination, it gives a C-allylated product in 65% yield with exceptionally high asymmetric stereoselectivity of >98:2 d.r.\nGive the specific name of the concerted rearrangement in this step, and write the SMILES string for intermediate D with the stereochemistry of its stereogenic centers shown explicitly.", + "text_images": [ + { + "file": "Exam-18-T9-5-1-prompt.png", + "description": "The figure shows a two-step reaction: 'CC[C@]12CCCN3CCC4=C([C@@H]23)N(CC1)C5=CC=CC=C45'>2.0 equiv. allyl iodide, MeCN>[D]; [D]>3.0 equiv. KOEt, (no EtOH), 1 h, 22 °C>'C=CC[C@]12C(N(CC3)C4=CC=CC=C54)=C5CCN1CCC[C@@]32CC'." + } + ], + "full_mark": 2 + }, + { + "sub_id": "Exam-18-T9-6", + "tags": [ + "2.2", + "2.4" + ], + "text_open": "9.3.2 The product from the previous step reacts with MeOTf to form quaternary ammonium salt E, which is again treated with strong base for elimination. DFT calculations show that, to relieve steric compression, a particular C-N bond in E is substantially elongated to 1.61 Å and is therefore highly susceptible to cleavage. Elimination ultimately occurs selectively at the less accessible but more acidic C-H bond, giving the ten-membered-ring exo-bridged diene F. Write the SMILES strings for quaternary ammonium salt E and elimination product F.", + "text_images": [ + { + "file": "Exam-18-T9-6-1-prompt.png", + "description": "The figure shows a two-step reaction: 'C=CC[C@]12C(N(CC3)C4=CC=CC=C54)=C5CCN1CCC[C@@]32CC'>3.0 equiv. MeOTf, DCE, 16 h, 70 °C>[E]; [E]>3.0 equiv. KOEt, EtOH, 3 h, 80 °C>[F]. E and F are shown only as labels; their structures are not drawn." + } + ], + "full_mark": 2 + }, + { + "sub_id": "Exam-18-T9-7", + "tags": [ + "2.2", + "2.4" + ], + "text_open": "9.4 While attempting ring-closing metathesis (RCM) of an oxidized tetrahydropalmatine derivative to construct a medium-sized oxygen-containing ring, the research team encountered an unusual reverse process. When the Lewis acid $BPh_3$ was added to the W catalyst, the reaction did not give the expected ten-membered-ring metathesis product. Instead, it afforded a ring-contracted isomer bearing a stereogenic vinyl substituent in 61% yield.\nA Lewis acid-base adduct forms first, followed by an extremely rare reverse Meisenheimer rearrangement (RMR) to give boron-complex intermediate G. Write the SMILES strings for G and the boron-containing organic byproduct formed with the product.", + "text_images": [ + { + "file": "Exam-18-T9-7-1-prompt.png", + "description": "The figure shows a three-step reaction: 'C=CC(C=C1OC)=C(C=C1OC)[C@@H]2CC3=CC=C(C(OC)=C3CN2OCC=C)OC'>1.2 equiv. BPh_3, Benzene, 15 h, 60 °C, W10%>'COC1=CC2=C(C=C1OC)/C=C\\C[O+]([B-](C3=CC=CC=C3)(C4=CC=CC=C4)C5=CC=CC=C5)N(CC6=C7OC)[C@H]2CC6=CC=C7OC'; 'COC1=CC2=C(C=C1OC)/C=C\\C[O+]([B-](C3=CC=CC=C3)(C4=CC=CC=C4)C5=CC=CC=C5)N(CC6=C7OC)[C@H]2CC6=CC=C7OC'>[G]; [G]>'C=C[C@@H](C1=C2C=C(OC)C(OC)=C1)N3CC4=C(C[C@@H]23)C=CC(OC)=C4OC', with a branch arrow pointing to byproduct ?." + } + ], + "full_mark": 2 + }, + { + "sub_id": "Exam-18-T9-8", + "tags": [ + "2.3", + "2.4" + ], + "text_open": "9.5 While attempting to isomerize the double bonds of the substrate above, the researchers found that treatment in a $KO^tBu$/DMSO system at room temperature (22°C) gave quantitative conversion in only 30 min, with >98% yield, to an aryl-substituted dehydroisoquinoline derivative. In subsequent biological-activity tests, this molecule showed remarkably potent anticancer activity at nanomolar concentrations.\nTheoretical calculations (DFT) and deuterium-control experiments showed that rapid elimination first forms the $(DMSO)_3K^+$-associated π-allyl-type conjugated complex intermediate $H_1$. With assistance from the potassium ion, this intermediate then undergoes a thermodynamically highly favorable anionic [1,6]-hydrogen proton-transfer process, an 8-electron pericyclic reaction with an activation barrier of only 15.9 kcal/mol, to form $H_2$. Write the SMILES strings for intermediates $H_1$ and $H_2$.", + "text_images": [ + { + "file": "Exam-18-T9-8-1-prompt.png", + "description": "The figure shows a three-step reaction: 'C=CC(C=C1OC)=C(C=C1OC)[C@@H]2CC3=CC=C(C(OC)=C3CN2OCC=C)OC'>1.0 equiv. KOt-Bu, DMSO, 0.5 h, 22 °C>[H1]; [H1]>[H2]; [H2]>'[H]CCC(C=C1OC)=C(C=C1OC)C2=CC3=CC=C(C(OC)=C3C=N2)OC'." + } + ], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-19-T1/problem.json b/test/tasks/Exam-19-T1/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..6378a58b0664480e956fb0c5b2671208c323d6cd --- /dev/null +++ b/test/tasks/Exam-19-T1/problem.json @@ -0,0 +1,34 @@ +{ + "problem_id": "Exam-19-T1", + "header": "Question 1 Chemistry in Language\nMāori is the language of the indigenous Māori people of New Zealand and belongs to the Malayo-Polynesian branch of the Austronesian language family. Whereas element names in many other languages are borrowed from English or Latin, a number of Māori element names are constructed from native morphemes. The Chinese and English names of several elements and their Māori equivalents are given below. Examine these element names, infer the word-formation rules, and use those rules to answer the questions that follow.\n\n| Chinese name (English translation) | English | Māori | Chinese name (English translation) | English | Māori |\n| --- | --- | --- | --- | --- | --- |\n| hydrogen | hydrogen | hauwai | helium | helium | haumāmā |\n| lithium | lithium | konukōhatu | carbon | carbon | waro |\n| sodium | sodium | konutai | oxygen | oxygen | hāora |\n| chlorine | chlorine | haumāota | calcium | calcium | konupūmā |\n| chromium | chromium | konukita | nickel | nickel | konukōreko |\n\nFor reference, English translations of several Māori words are provided below:\n\n| | | | |\n| --- | --- | --- | --- |\n| hā — breath | ora — alive | wai — water | tai — sea |\n| kōhatu — stone | kita — bright | uku — clay | kōreko — shiny |\n| mohe — malleable | pūmā — grayish white | māota — fresh green, vitality | |", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-19-T1-1", + "tags": [ + "1.7" + ], + "text_open": "1.2 One day, a Māori friend tells you about several elements and their compounds. Infer their meanings from the context.\nHygroscopic solid A contains only the elements hautawa and konumohe. It fumes in air and sublimes under vacuum. Thorough heating of A in air produces compound B together with colored fumes (Reaction 1). B is also an important industrial raw material for producing konumohe, but New Zealand relies mainly on imports of this material.\nElemental konumohe has many uses. Dissolve it in a caustic soda solution, adjust the pH until precipitate C forms, and collect the precipitate by filtration. Adding soda ash and C to an aqueous solution of the simple monoprotic acid corresponding to haukōwhai, followed by recrystallization, gives a 3:1 crystal D (Reaction 2). D has a lower melting point than B and can be used in the metallurgical industry.\nThe Māori names haukōwhai and hautawa are both derived from their colors. The two elements belong to the same group of the periodic table, but their elemental substances have different physical states at room temperature.\nElemental haukōwhai reacts with 2% caustic soda solution to form mainly colorless gas E (Reaction 3). The molecule of E has a structure similar to that of water and reacts under acidic conditions with a simple potassium salt of hautawa to form elemental hautawa (Reaction 4).\n1.2.1 From the analysis, directly give the element names corresponding to hautawa, konumohe, and haukōwhai, and the chemical formulas of A–E.", + "text_images": [], + "full_mark": 8 + }, + { + "sub_id": "Exam-19-T1-2", + "tags": [ + "2.1" + ], + "text_open": "1.2.2 Write the equations for Reactions 1–4.", + "text_images": [], + "full_mark": 6 + }, + { + "sub_id": "Exam-19-T1-3", + "tags": [ + "1.7" + ], + "text_open": "1.3.2 Some people argue that the prefix hau in hautawa is inconsistent with the properties of the element. Do you agree? Explain your reasoning.", + "text_images": [], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-19-T2/images/Exam-19-T2-1-1-prompt.png b/test/tasks/Exam-19-T2/images/Exam-19-T2-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..40970b9c454d047ddaa06ce5a8810e1317674cb8 --- /dev/null +++ b/test/tasks/Exam-19-T2/images/Exam-19-T2-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:d985b3685526804a252b79f3db46cbd7201627050b759d8e0cab993744d755e6 +size 71881 diff --git a/test/tasks/Exam-19-T2/images/Exam-19-T2-1-2-prompt.png b/test/tasks/Exam-19-T2/images/Exam-19-T2-1-2-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..4c47d76b972a2fe82d291f009b8e0b350fac26e3 --- /dev/null +++ b/test/tasks/Exam-19-T2/images/Exam-19-T2-1-2-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:e137a957af2bcce1d424b2bfc2b0aabc0243cdd43643c87aed21bf064493193c +size 74130 diff --git a/test/tasks/Exam-19-T2/images/Exam-19-T2-3-1-prompt.png b/test/tasks/Exam-19-T2/images/Exam-19-T2-3-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..2ed1cbb5a88ebca66ff4352f15581f0082caff21 --- /dev/null +++ b/test/tasks/Exam-19-T2/images/Exam-19-T2-3-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:013b1009cbbf0802c9c0c81999378a32a419c2421855c1c6c7130c18171d4e4e +size 27744 diff --git a/test/tasks/Exam-19-T2/images/Exam-19-T2-header-1.png b/test/tasks/Exam-19-T2/images/Exam-19-T2-header-1.png new file mode 100644 index 0000000000000000000000000000000000000000..595035509fc56b7338d9f75cab748f282fc0e82b --- /dev/null +++ b/test/tasks/Exam-19-T2/images/Exam-19-T2-header-1.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:e75d53c8612779f2aa64d775dcd91a96b68ab68b099a3e1b4a504adbe0af31d6 +size 147329 diff --git a/test/tasks/Exam-19-T2/images/Exam-19-T2-header-2.png b/test/tasks/Exam-19-T2/images/Exam-19-T2-header-2.png new file mode 100644 index 0000000000000000000000000000000000000000..e7f1d538e756f55e068c2f885786b8637791edbf --- /dev/null +++ b/test/tasks/Exam-19-T2/images/Exam-19-T2-header-2.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:388d5c5fae2cdb823604c0c05c49a430ddc713a54458048b11028c94df60eb98 +size 76883 diff --git a/test/tasks/Exam-19-T2/images/Exam-19-T2-header-3.png b/test/tasks/Exam-19-T2/images/Exam-19-T2-header-3.png new file mode 100644 index 0000000000000000000000000000000000000000..287b7a3839307114af5bdab1bdd22818f67a120f --- /dev/null +++ b/test/tasks/Exam-19-T2/images/Exam-19-T2-header-3.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:09b48af4d8b20c3e573bb2c5c1b2ffc5ca2aba9ca7b2cfdac5796392b5562b98 +size 6312 diff --git a/test/tasks/Exam-19-T2/problem.json b/test/tasks/Exam-19-T2/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..e7c68a723bb6127fa11d39cf9f6e1ff57dee99b5 --- /dev/null +++ b/test/tasks/Exam-19-T2/problem.json @@ -0,0 +1,73 @@ +{ + "problem_id": "Exam-19-T2", + "header": "Question 2 Photochromic Organic Transistor\nOrganic field-effect transistors (OFETs) offer advantages such as flexibility, printability, and low-temperature fabrication, and are widely used in flexible displays, sensors, and large-area electronic circuits. Organic molecules with distinctive photochemical properties can therefore be used to make light-controlled switches or transistors. Researchers fabricated a photochromic organic transistor using a DAE film as the transistor channel layer. The device supports reversible writing and erasing. The molecular structure of DAE, the organic transistor structure, and the structure of a conventional bipolar transistor are shown below.\nHere, Channel means the conduction channel. From a purely functional perspective, the Source (S terminal) of the field-effect transistor is the source electrode and corresponds to emitter E of the bipolar transistor; Gate (G terminal) is the gate electrode and corresponds to base B; Drain (D terminal) is the drain electrode and corresponds to collector C.", + "header_images": [ + { + "file": "Exam-19-T2-header-1.png", + "description": "The SMILES representation of DAE is 'FC1(F)C(F)(F)C(C2=C(C)SC(C3=CC=C(C4=CC=CC=C4)C=C3)=C2)=C(C5=C(C)SC(C6=CC=C(C7=CC=CC=C7)C=C6)=C5)C1(F)F'." + }, + { + "file": "Exam-19-T2-header-2.png", + "description": "The figure is a structural schematic of an organic field-effect transistor (OFET). From bottom to top it shows a p+-Si substrate (with a lead from the bottom connected to a variable DC supply and labeled Gate), a SiO₂ layer, Au electrodes on the left and right, and a 1D channel spanning the two electrodes. The left electrode is grounded by a lead and labeled Source; the right electrode is connected by a lead to a variable DC supply and labeled Drain. Above the device, the tip of a probe (nanomanipulator/AFM probe) points vertically toward the 1D channel. A translucent cover layer at upper right is labeled DAE and PMMA. The lower right is labeled XY-axes stage and includes arrows showing three-dimensional directions of motion." + }, + { + "file": "Exam-19-T2-header-3.png", + "description": "The figure shows the circuit symbol of a conventional bipolar transistor. Inside a circle is an inverted trapezoidal arrangement. A vertical line at the top connects to B (base), a horizontal line at the right connects to C (collector), and a horizontal line at the left connects to E (emitter). The end of the E line carries an arrow pointing toward the lower diagonal line inside the circle." + } + ], + "subquestions": [ + { + "sub_id": "Exam-19-T2-1", + "tags": [ + "3.5", + "1.7" + ], + "text_open": "The channel contains a mixture of DAE and its isomer X. Let α denote the fraction of X in the channel. Experiments show that after sufficiently prolonged irradiation at wavelength λ1, the maximum isomer fraction α_{max} = 0.75. If a further 40s scan at wavelength λ2 is then applied, α decreases by 0.10. After sufficiently prolonged irradiation at wavelength λ2, the minimum isomer fraction α_{max} = 0.10. If a further 40s scan at wavelength λ1 is then applied, α increases by 0.20. For a reversible first-order isomerization, \\alpha(t) = \\alpha_{\\infty} + (\\alpha_0 - \\alpha_{\\infty})e^{-kt}. Here k is the apparent rate constant, \\alpha_{\\infty} is the isomer fraction at the photostationary state, and \\alpha_ is the isomer fraction at t = 0.\n2.3 In the saturation region, the magnitude of the drain current through a single channel satisfies\n$$|I_D|=\\frac{μC_iW}{2L}(V_G-V_T)^2$$\nwhere W is the channel width; L is the channel length (the length of semiconductor traversed by the current from S to D), taken as 40 μm in this part; μ is the carrier mobility in the saturation region, which describes the overall rate of motion of electrons and holes through the material; $C_i$ is the capacitance per unit area of the insulating layer; $V_G$ is the gate voltage; and $V_T$ is the threshold voltage, which may be approximated as a constant independent of $V_G$ and μ. For n identical channels, approximately\n$$|I_D|=\\frac{μnC_iW}{2L}(V_G-V_T)^2$$\nIn this problem, the saturation-region carrier mobility may be taken as proportional to α.\n2.3.1 The source is grounded and the drain voltage $V_D$ is fixed so that the transistor remains in the saturation region. After irradiation with ultraviolet and visible light of specified wavelengths, respectively, the gate voltage $V_G$ is scanned while the drain current is monitored. The results are shown in the left-hand figure below. The region 0 > $V_G$ > -20V is outside the saturation region and is irrelevant to this question. The transistor properties differ markedly after the two irradiations. Determine whether the transistor after ultraviolet irradiation corresponds to curve group A or B, and briefly explain why. (Figure labels: Current; Voltage; Drawing Cycle.)", + "text_images": [ + { + "file": "Exam-19-T2-1-1-prompt.png", + "description": "The figure is a coordinate plot (left-hand figure for 2.3.1). The vertical axis is labeled 'Drain Current (pA)', with tick labels -8, -6, -4, -2, 0, and 2 from top to bottom. The horizontal axis is labeled 'Gate Voltage (V)', with tick labels 0, -20, -40, and -60 from left to right. Several dark solid curves nearly overlap from 0 to -20 V and separate into two groups beyond -20 V. Group A (labeled 1 and 3,5) has currents of approximately -6 pA and -4 pA at -60 V. Group B (labeled 6 and 2,4,0) has currents between 0 and 1 pA at -60 V. The upper left is labeled 'UV: 25 mW/cm²' and 'VIS: 500 mW/cm²'." + }, + { + "file": "Exam-19-T2-1-2-prompt.png", + "description": "The figure is a coordinate plot (right-hand figure for 2.3.1). The vertical axis is labeled 'Drain Current (pA)', with tick labels 2, 0, -2, -4, -6, and -8 from bottom to top. The horizontal axis is labeled 'Drawing Cycle', with ticks at 0, 5, 10, 15, and 20. A black solid polyline with filled black square markers oscillates in a sawtooth pattern between approximately -0.5 and -6 pA. The absolute magnitudes of the peaks and troughs decrease gradually as the cycle number increases. Gray arrows above each peak and below each trough indicate application of light. No legend is present." + } + ], + "full_mark": 2 + }, + { + "sub_id": "Exam-19-T2-2", + "tags": [ + "3.6", + "3.8" + ], + "text_open": "Starting from the $\\lambda_2$ photostationary state, two consecutive $\\lambda_1$ scans under the stated conditions give an isomer-X fraction $\\alpha=0.44$.\nProlonged irradiation at $\\lambda_1$ makes the isomer-X fraction $\\alpha$ reach 0.75, whereas prolonged irradiation at $\\lambda_2$ makes $\\alpha$ reach 0.10.\n2.3.3 For a transistor with n = 1, after prolonged scanning at wavelength $λ_1$, two data points in the saturation region of its $I_D$–$V_G$ curve are measured: (-40 V, -2.0 pA) and (-50 V, -3.9 pA). A transistor with n = 3 is irradiated for a long time at wavelength $λ_2$ and then subjected to two consecutive 40 s scans at wavelength $λ_1$. Neglecting losses, calculate $I_D$ at $V_G$ = -45 V.", + "text_images": [], + "full_mark": 5 + }, + { + "sub_id": "Exam-19-T2-3", + "tags": [ + "3.6" + ], + "text_open": "2.4 This transistor can be used to construct a light-controlled analog adder. Its structure is shown below. The three branches (1, 2, and 3) forming the Y share one gate G, with $V_G$ = -40 V. The distance between S and D is 60 μm. Branch 1 contains 3 channels and branch 2 contains 2 channels. Every channel operates in the saturation region and has the same width. The section connecting the three channels (the section parallel to S and D) lies midway between S and D and may be treated as an ideal conductor. Initially, every channel has been irradiated for a long time at wavelength $λ_1$. (Hint: note the definition of L. Neglect losses.)\n2.4.1 Calculate the initial magnitude of the drain current.", + "text_images": [ + { + "file": "Exam-19-T2-3-1-prompt.png", + "description": "The figure is a schematic of the Y-shaped light-controlled analog adder, shown as a three-dimensional rectangular trough-like device. The left side is labeled S and connected by a bent lead to a ground symbol. The right side is labeled D and connected by a bent lead to a DC-source symbol. The front of the bottom is labeled G and connected by a vertical lead to a ground symbol. Internal stepped layered regions labeled 1, 2, and 3 correspond to the three branches forming the Y; branches 1 and 2 converge into branch 3." + } + ], + "full_mark": 3 + }, + { + "sub_id": "Exam-19-T2-4", + "tags": [ + "3.6", + "3.8" + ], + "text_open": "2.4.2 For a channel in which μ varies, the equivalent mobility of the channel can be obtained by averaging μ as follows:\n$$μ_{eff}=\\frac{L_1+L_2}{\\frac{L_1}{μ_1}+\\frac{L_2}{μ_2}}$$\nEach channel in branch 1 is now subjected to one 40 s scan at wavelength $λ_2$. Calculate the change in the magnitude of the drain current.", + "text_images": [], + "full_mark": 3 + } + ] +} diff --git a/test/tasks/Exam-19-T3/problem.json b/test/tasks/Exam-19-T3/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..7d7747c79b951c164409930daf647088ee4ab835 --- /dev/null +++ b/test/tasks/Exam-19-T3/problem.json @@ -0,0 +1,25 @@ +{ + "problem_id": "Exam-19-T3", + "header": "Question 3 Degradable Thermosetting Resin\nConventional thermosetting polymers form covalently cross-linked networks that provide exceptional mechanical strength and heat resistance, but also prevent melt processing and make recycling and degradation difficult. In recent years, orthogonal polymerization of different functional groups within a single monomer has emerged as a leading strategy for synthesizing polymers that can be degraded at predetermined sites. 2,3-Dihydrofuran (DHF) is a monomer that can be prepared from bio-based feedstocks. Recently, chemists used DHF in a one-pot process involving two successive polymerizations with entirely different mechanisms to produce a thermosetting resin that can be recycled through multiple pathways.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-19-T3-1", + "tags": [ + "2.4" + ], + "text_open": "3.1 In the presence of a trace amount of second-generation Grubbs catalyst, DHF slowly undergoes ring-opening metathesis polymerization (ROMP) in the dark to give rubbery linear polymer A. In the presence of a photoacid generator (PAG, which produces the superacid HSbF6 upon irradiation), it rapidly undergoes cationic polymerization to give highly rigid linear polymer B.\n3.1.2 Write the SMILES strings for linear polymers A and B.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-19-T3-2", + "tags": [ + "2.4" + ], + "text_open": "3.2 To prepare a thermosetting resin, the researchers mixed DHF, second-generation Grubbs catalyst, and PAG. The mixture was first allowed to react in the dark for 10 minutes to form an appropriate amount of polymer A, then irradiated with 456 nm blue light for 6 minutes to initiate cationic polymerization. This produced cross-linked thermosetting resin C, which is insoluble and infusible. Write the SMILES string for C when the two polymer chains are fully cross-linked, with no unreacted cross-linking sites. Hint: to simplify the structure, assume that both linear polymers A and B participating in the reaction have degree of polymerization n.", + "text_images": [], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-20-T3/images/Exam-20-T3-1-1-prompt.png b/test/tasks/Exam-20-T3/images/Exam-20-T3-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..5a5ce35aea6f2f58c2aa72da267a3ea800dc294a --- /dev/null +++ b/test/tasks/Exam-20-T3/images/Exam-20-T3-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:3ecad2b388ae301b85d9c7c49aca1e231309ba842124a5c5ba8ac5266004fdcd +size 419777 diff --git a/test/tasks/Exam-20-T3/images/Exam-20-T3-2-1-prompt.png b/test/tasks/Exam-20-T3/images/Exam-20-T3-2-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..e1de77dbf7a40cf467cf148fe8696a4eced08222 --- /dev/null +++ b/test/tasks/Exam-20-T3/images/Exam-20-T3-2-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:199dcd0240957126e23443457a3c00867ec79661a1d662d56bc9b86cb7165176 +size 180090 diff --git a/test/tasks/Exam-20-T3/images/Exam-20-T3-4-1-prompt.png b/test/tasks/Exam-20-T3/images/Exam-20-T3-4-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..fe0f2058658eb9e6c57d05c9ab59166855f330a6 --- /dev/null +++ b/test/tasks/Exam-20-T3/images/Exam-20-T3-4-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:bfef5665604e31c7cea834b993e37fffe4a114a4975a6bc780c767107e894832 +size 13317 diff --git a/test/tasks/Exam-20-T3/images/Exam-20-T3-5-1-prompt.png b/test/tasks/Exam-20-T3/images/Exam-20-T3-5-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..5519af0727f894d6008ac273c4c102b6e69b6d83 --- /dev/null +++ b/test/tasks/Exam-20-T3/images/Exam-20-T3-5-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:d7758ef643f5e18c79c79bfeb2f42490f31b00a2475cc81531f8c4f81eceb8a6 +size 126316 diff --git a/test/tasks/Exam-20-T3/problem.json b/test/tasks/Exam-20-T3/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..47cdfa496217caed5a9c28ec493dc1f4869c49fc --- /dev/null +++ b/test/tasks/Exam-20-T3/problem.json @@ -0,0 +1,76 @@ +{ + "problem_id": "Exam-20-T3", + "header": "Halogen-Bond-Donor-Catalyzed Synthesis of Poly(ortho esters)\nPolymers have transformed modern life because of their low cost and practicality. Polyesters are high-performance, versatile engineering plastics and can also be made into polyester fibers and films; the family includes polyester resins and elastomers.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-20-T3-1", + "tags": [ + "2.4", + "2.3" + ], + "text_open": "3.1 Cyclic ketene acetals undergo radical polymerization in the presence of peroxides to generate poly(regular lactone) chains. Write the SMILES strings of A-D in the reaction scheme below. Polymer end groups need not be shown. Only D is a polymer.", + "text_images": [ + { + "file": "Exam-20-T3-1-1-prompt.png", + "description": "Four-step reaction: 'BrCC1OCCO1'>tBuOK>A; A>ROOR, heat>[B]; [B]>isomerization>[C]; [C]>D. R is the placeholder used in the original diagram, and [B] and [C] denote radical intermediates." + } + ], + "full_mark": 5 + }, + { + "sub_id": "Exam-20-T3-2", + "tags": [ + "2.4", + "2.6" + ], + "text_open": "3.2 Poly(ortho esters) (POEs) are renewable, biodegradable polymers whose degradation kinetics and surface-erosion behavior can be tuned and whose degradation products are harmless. Earlier synthetic methods, however, gave POEs with molecular weights that were too low and properties that could not be well controlled, making them unsuitable as general-purpose materials such as plastics and rubbers. The reaction below uses renewable feedstocks and an innovative initiator to achieve controlled ring-opening polymerization of the monomer and produce POEs with controllable molecular weights.\n3.2.1 In this reaction, two renewable starting materials first form the spiro-ortho-ester monomer E, MSOE, under $BF_3$ catalysis. The target polymer P1 is then synthesized using the initiator MOMBr and the catalyst ion XBD. After monomer E reacts with the active center of the polymer chain, it can isomerize and polymerize to give the acyclic polymer P2. Based on this information, write the SMILES strings of E, P1, and P2. Polymer end groups need not be shown.", + "text_images": [ + { + "file": "Exam-20-T3-2-1-prompt.png", + "description": "Two-step reaction: 'CC1CO1' + 'O=C1CCCCCO1' > $BF_3\\cdot OEt_2$ (5.0 mol%) > E; E + 'BrCOC' (MOMBr) > XBD > P1 (major) + P2 (minor)." + } + ], + "full_mark": 6 + }, + { + "sub_id": "Exam-20-T3-3", + "tags": [ + "2.4" + ], + "text_open": "3.2.2 At pH = 3, P1 can react with water and degrade, allowing recovery of small-molecule ester monomers. Write the SMILES strings of the possible recovered small molecules.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-20-T3-4", + "tags": [ + "1.6", + "1.1" + ], + "text_open": "3.3 The catalyst ion XBD is a trinuclear halogen-bond-donor catalyst whose mechanism involves a special secondary interaction: the halogen bond. A halogen bond is fundamentally an electrostatic interaction arising from a region of accumulated positive charge at the extension opposite a covalent bond to a halogen atom, especially Cl, Br, or I. This region is called a \"σ hole\" in the literature and is illustrated below.\n3.3.1 Based on this information, write the SMILES strings of the halogen-bonded aggregates $(Cl_2)_2$ and $(RBr)_3$. The latter has a rotational axis of order higher than two; treat R as a spherical ligand. Specify the halogen-bond angles rigorously.", + "text_images": [ + { + "file": "Exam-20-T3-4-1-prompt.png", + "description": "Schematic of a σ hole on a halogen atom: a blue gradient sphere at the center bears the capital letter E. The letter R is connected to the left of the sphere by a short horizontal line. To the right is a concentric region with an outer green ring, a middle yellow ring, and a red central point. A red δ+ at the upper right has a curved arrow pointing to the concentric region, while a blue δ- at the lower left has a curved arrow pointing to the bottom of the sphere." + } + ], + "full_mark": 4 + }, + { + "sub_id": "Exam-20-T3-5", + "tags": [ + "2.3", + "2.1" + ], + "text_open": "3.3.2 The structure of the catalyst ion XBD used in 3.2 is shown below. Write the reaction equations for chain initiation by the initiating system, MOMBr and the catalyst ion XBD. XBD may be abbreviated; there is no need to write a SMILES string containing XBD. On this basis, determine whether the reaction is cationic or anionic polymerization. Hint: chain initiation consists of two elementary steps.", + "text_images": [ + { + "file": "Exam-20-T3-5-1-prompt.png", + "description": "The SMILES for XBD is '[R]N(N=[N+]1C)C(I)=C1C2=CC(C3=C(I)N([R])N=[N+]3C)=CC(C4=C(I)N([R])N=[N+]4C)=C2'. The 3 [R] groups are placeholders used in the original diagram, and the 3 [N+] atoms carry positive-charge labels." + } + ], + "full_mark": 5 + } + ] +} diff --git a/test/tasks/Exam-20-T4/problem.json b/test/tasks/Exam-20-T4/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..0266cbeaacb537fcba9d8150a5ba5a9dad34ce76 --- /dev/null +++ b/test/tasks/Exam-20-T4/problem.json @@ -0,0 +1,44 @@ +{ + "problem_id": "Exam-20-T4", + "header": "Determination of Silicon by the Silicomolybdenum Blue Spectrophotometric Method\nSilicon is an important element in iron ores and steel materials, and its content significantly affects steel properties. In the commonly used silicomolybdenum blue spectrophotometric method, silicic acid reacts with ammonium molybdate in acidic medium to form silicomolybdenum yellow, which is then reduced to silicomolybdenum blue for absorbance measurement. This problem concerns sample fusion, removal of interferences, control of color-development conditions, and quantification by a calibration curve.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-20-T4-1", + "tags": [ + "4.1", + "4.7" + ], + "text_open": "4.1 Sample pretreatment\nWeigh 0.1284 g of sample into a 50 mL silver crucible, add 2 g of sodium peroxide and mix, then cover with another 0.5 g of sodium peroxide. Cover the crucible, place it in a furnace at 300 °C, slowly raise the temperature to 750 °C-800 °C, and fuse for 20 min. Remove and allow to cool slightly. Add 100 mL of boiling water to leach the melt completely and wash out the crucible. While swirling, slowly add 30 mL of concentrated hydrochloric acid. After cooling, transfer the solution to a 200 mL volumetric flask, dilute to the mark with water, mix, filter, and discard the initial portion of the filtrate.\n4.1.6 The acidified solution must be dry-filtered through medium-speed filter paper, and the first few milliliters of filtrate must be discarded. Explain the purposes of this procedure. If the initial filtrate were not discarded, would the measured result be too high or too low?", + "text_images": [], + "full_mark": 3 + }, + { + "sub_id": "Exam-20-T4-2", + "tags": [ + "4.2" + ], + "text_open": "4.2 Color development for the sample\nTransfer 10.00 mL of filtrate to a 100 mL volumetric flask and add blank-test solution to bring the volume to 20 mL. Add 10 mL of hydrochloric acid (1 + 3), add potassium permanganate solution (20 g/L) dropwise until the solution is pale red, add 10 mL of water, and mix.\nAdd 1 mL of anhydrous ethanol and mix. While continuously swirling, add 5 mL of ammonium molybdate solution (100 g/L), mix, and allow to stand at 15 °C-25 °C for 15 min.\nAdd 15 mL of hydrochloric acid (1 + 3), followed by 5 mL of oxalic acid-tartaric acid mixture containing 50 g/L oxalic acid and 40 g/L tartaric acid. Mix and immediately add 5 mL of ascorbic acid solution (20 g/L), dilute to the mark with water, mix, and allow to stand at 25 °C-35 °C for 15 min.\nTransfer a portion of the colored solution to a 3 cm cuvette. Using the colored blank-test solution carried through with the sample as reference, measure an absorbance of 0.425 at 810 nm.\n4.2.1 Explain why adding potassium permanganate to a pale-red endpoint eliminates interference from lower-valent iron and organic matter, and state the effect of excess potassium permanganate.", + "text_images": [], + "full_mark": 3 + }, + { + "sub_id": "Exam-20-T4-3", + "tags": [ + "4.2" + ], + "text_open": "4.2.4 State which interfering ions are masked by oxalic acid and tartaric acid, respectively, in the color-development system, and explain why ascorbic acid must not be added before the oxalic acid-tartaric acid mixture.", + "text_images": [], + "full_mark": 3 + }, + { + "sub_id": "Exam-20-T4-4", + "tags": [ + "4.7" + ], + "text_open": "4.3 Construction of the calibration curve\nTransfer 0 mL, 1.00 mL, 2.00 mL, 4.00 mL, 6.00 mL, 8.00 mL, and 10.00 mL portions of silicon dioxide standard solution (20.0 μg/mL) to a series of 100 mL volumetric flasks. Add water to 20 mL and develop the color according to the procedure in 4.2. Transfer each colored solution to a 3 cm cuvette and measure the absorbance at 810 nm against a reagent blank. Construct a calibration curve using the amount of silicon dioxide in μg as the horizontal coordinate and absorbance as the vertical coordinate.\n\n| Volume of silicon dioxide standard solution ($V$ / mL) | 0.00 | 1.00 | 2.00 | 4.00 | 6.00 | 8.00 | 10.00 |\n| --- | --- | --- | --- | --- | --- | --- | --- |\n| Amount of silicon dioxide ($m$ / μg) | 0 | 20.00 | 40.00 | 80.00 | 120.0 | 160.0 | 200.0 |\n| Absorbance ($A$) | 0.000 | 0.082 | 0.163 | 0.326 | 0.488 | 0.651 | 0.812 |\n\nCalculate the silicon dioxide content of the sample.", + "text_images": [], + "full_mark": 5 + } + ] +} diff --git a/test/tasks/Exam-21-T10/images/Exam-21-T10-1-1-prompt.png b/test/tasks/Exam-21-T10/images/Exam-21-T10-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..0ca2554562547e584d555011636f72f59a4e07f2 --- /dev/null +++ b/test/tasks/Exam-21-T10/images/Exam-21-T10-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:439a942469ec91ed4bf26e5bb1be17986f56cd45cdbde8e7bd7b8fff92f64fe1 +size 79629 diff --git a/test/tasks/Exam-21-T10/images/Exam-21-T10-3-1-prompt.png b/test/tasks/Exam-21-T10/images/Exam-21-T10-3-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..eaff1bc45d8b0f0e45a2d0014ff873d91291066f --- /dev/null +++ b/test/tasks/Exam-21-T10/images/Exam-21-T10-3-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:28d0f68052cd21f6c1861279b951623b2c9c11f72ce0fad16a3c94b8caf1c43a +size 79767 diff --git a/test/tasks/Exam-21-T10/images/Exam-21-T10-header-1.png b/test/tasks/Exam-21-T10/images/Exam-21-T10-header-1.png new file mode 100644 index 0000000000000000000000000000000000000000..66f367d9f2138cee1c0850e631e8a5ddb9417894 --- /dev/null +++ b/test/tasks/Exam-21-T10/images/Exam-21-T10-header-1.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:ea60b3b8f101104b9c2495f6572ab483ef3e3c175fd491c893877697a431b974 +size 191664 diff --git a/test/tasks/Exam-21-T10/problem.json b/test/tasks/Exam-21-T10/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..67c41254f3da581289f97ded7d0dc059bc7a9519 --- /dev/null +++ b/test/tasks/Exam-21-T10/problem.json @@ -0,0 +1,52 @@ +{ + "problem_id": "Exam-21-T10", + "header": "Problem 10 Building a Very Familiar Aromatic Ring—the Simple Shikimic Acid Pathway\nAmino acids are essential substances that our bodies require continuously. Some can be synthesized in the body, whereas a few cannot and must be obtained from food, including Phe (phenylalanine), Tyr (tyrosine), and Trp (tryptophan), whose structures are shown below.\nHow do plants, microorganisms, and other organisms synthesize these amino acids? A widespread natural route produces aromatic compounds and many related six-membered cyclic compounds. It is named after a key intermediate: the shikimic acid pathway. Note: stereochemistry is required throughout this problem unless otherwise stated.", + "header_images": [ + { + "file": "Exam-21-T10-header-1.png", + "description": "Phe has the SMILES representation '[H][C@@](N)(C(O)=O)CC1=CC=CC=C1'; Tyr, 'N[C@]([H])(C(O)=O)CC1=CC=C(O)C=C1'; and Trp, 'N[C@]([H])(C(O)=O)CC1=CNC2=C1C=CC=C2'. The three structures are shown side by side, with -NH2 at each stereogenic carbon drawn on a solid wedge." + } + ], + "subquestions": [ + { + "sub_id": "Exam-21-T10-1", + "tags": [ + "2.4", + "1.2" + ], + "text_open": "10.1 The shikimic acid pathway begins with condensation of phosphoenolpyruvate (PEP) and erythrose 4-phosphate (E4P).\n10.1.1 Write the SMILES string for the major organic product A. A carries one negative charge, contains only one phosphorus atom, and the reaction creates one new stereocenter of absolute configuration R.", + "text_images": [ + { + "file": "Exam-21-T10-1-1-prompt.png", + "description": "Single-step reaction: 'O=P(OC(C(O)=O)=C)([O-])O' (PEP, with one negative charge on a phosphate oxygen).'O=P(O)([O-])OC[C@@H](O)[C@@H](O)C([H])=O' (E4P, with one negative charge on a phosphate oxygen)>>[A]. No reaction conditions are written, and the product is shown only as A without an explicit structure." + } + ], + "full_mark": 3 + }, + { + "sub_id": "Exam-21-T10-2", + "tags": [ + "2.4", + "1.2" + ], + "text_open": "10.1.2 A readily isomerizes, so the reaction actually gives a mixture of A and its isomers. Write the SMILES string for the principal component M other than A.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-21-T10-3", + "tags": [ + "2.2", + "2.3" + ], + "text_open": "10.2 A cyclizes to form 3-dehydroquinate in a process mediated by $NAD^+$ and NADH. Earlier studies proposed the following mechanism. First, A is oxidized by 1 molecule of $NAD^+$ to give Int1, with loss of one stereocenter. Int1 undergoes β-elimination of the phosphate group and tautomerization to give Int2, which contains an α, β-diketone unit. Int2 is then reduced by 1 molecule of NADH to give Int3, restoring the lost stereocenter with its original relative configuration. Int3 undergoes intramolecular condensation to form 3-dehydroquinate ($C_7H_{10}O_6$). Write the SMILES strings for Int1, Int2, Int3, and 3-dehydroquinate.", + "text_images": [ + { + "file": "Exam-21-T10-3-1-prompt.png", + "description": "NAD+ has the SMILES representation 'NC(C1=CC=C[N+]([R])=C1)=O'; NADH has the SMILES representation 'NC(C1=CN([R])C=CC1)=O'. R is the original placeholder in the figure for the substituent on the pyridinium/1,4-dihydropyridine ring nitrogen." + } + ], + "full_mark": 9 + } + ] +} diff --git a/test/tasks/Exam-21-T3/images/Exam-21-T3-3-1-prompt.png b/test/tasks/Exam-21-T3/images/Exam-21-T3-3-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..dd6470088e3c889433dbbcefb0c97b559b5bef63 --- /dev/null +++ b/test/tasks/Exam-21-T3/images/Exam-21-T3-3-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:e860267bb2b12e00750a07d1c5a3ebfb9a3f4d1dbe7a6796d0e03586a56b5290 +size 829850 diff --git a/test/tasks/Exam-21-T3/problem.json b/test/tasks/Exam-21-T3/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..316c256d3f29bc9b1c73c79c60d7c635d9b186c1 --- /dev/null +++ b/test/tasks/Exam-21-T3/problem.json @@ -0,0 +1,50 @@ +{ + "problem_id": "Exam-21-T3", + "header": "Preparation and Properties of a Thermochromic Copper(II) Complex\nMaterials whose compounds exhibit thermochromism through structural phase transitions are widely used in anti-counterfeiting applications. To synthesize the thermochromic complex $[Cu(deen)_2](ClO_4)_2$ (deen denotes N,N-diethylethylenediamine), hydrated copper perchlorate must first be prepared, followed by introduction of the deen ligand.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-21-T3-1", + "tags": [ + "1.4", + "4.3" + ], + "text_open": "Hydrated copper perchlorate can be prepared by a precipitation–dissolution method. Dissolve $CuSO_4\\cdot 5H_2O$ in water, add $NaOH$ solution with stirring, and heat slowly on a hot plate until the precipitate turns black. Maintain gentle boiling for 5–10 min, allow the mixture to cool slightly, and collect the precipitate by suction filtration. Rinse it with a small amount of deionized water, then use a glass rod to transfer the precipitate into an evaporating dish containing 10 mL of water. Add perchloric acid, gently guide the precipitate into the solution with the glass rod, and stir until it dissolves completely to form a copper perchlorate solution. Concentrate the solution by gentle heating to about 2 mL. On cooling, crystals of hydrated copper perchlorate are obtained. The preparation of $[Cu(deen)_2](ClO_4)_2$ begins with the hydrated copper perchlorate crystals obtained above. Adding absolute ethanol to these crystals gives a clear blue solution. In a dry beaker, dissolve N,N-diethylethylenediamine in ethanol to obtain a clear pale-yellow solution. Slowly add the ethanolic deen solution to the ethanolic copper perchlorate solution with stirring; the mixture immediately turns deep blue-violet. Stir for 2–3min, then allow it to stand for about 10 min. A large amount of pink precipitate forms at the bottom of the vessel while the supernatant remains deep blue-violet. Collect the product by suction filtration and dry it to obtain $[Cu(deen)_2](ClO_4)_2$. 3.2.2 Dissolving the red $[Cu(deen)_2](ClO_4)_2$ in water again gives a deep-blue solution. Explain why. (No more than 50 words)", + "text_images": [], + "full_mark": 4 + }, + { + "sub_id": "Exam-21-T3-2", + "tags": [ + "1.4" + ], + "text_open": "3.3 In a thermochromism experiment on $[Cu(deen)_2](ClO_4)_2$, a small amount of the $[Cu(deen)_2](ClO_4)_2$ complex is held between pieces of adhesive tape and heated with a hair dryer. The sample changes from pink to blue-violet and returns to pink on cooling.\n3.3.1 Based on crystal field theory, describe in words as necessary the structural change of the $[Cu(deen)_2]^{2+}$ unit upon heating and cooling. The cooled unit may be approximated as centrosymmetric, whereas after heating it is no longer centrosymmetric.", + "text_images": [], + "full_mark": 3 + }, + { + "sub_id": "Exam-21-T3-3", + "tags": [ + "4.6" + ], + "text_open": "3.4 To investigate the thermochromism of $[Cu(deen)_2]X_2$ complexes formed with monovalent counteranion $X^-$, differential scanning calorimetry is performed on $[Cu(deen)_2](ClO_4)_2$ and $[Cu(deen)_2](NO_3)_2$. Multiple interactions occur between the cations and anions in both complexes. The results are shown below. For both complexes, the color change is associated with a phase-transition temperature, and the chemical composition remains unchanged across the color change. The temperature at the peak maximum is defined as the transition temperature.\n3.4.1 From the graph, directly give the thermochromic transition temperatures of $[Cu(deen)_2](ClO_4)_2$ and $[Cu(deen)_2](NO_3)_2$.", + "text_images": [ + { + "file": "Exam-21-T3-3-1-prompt.png", + "description": "Untitled coordinate plot. The horizontal axis is labeled \"Temperature/°C\" with major ticks at 50, 100, 150, and 200 and minor ticks at 25, 75, 125, and 175. The vertical axis is labeled \"Thermogravimetry\" vertically and has no numerical ticks. Two solid black curves are shown. The legend at lower left uses short line segments for \"$ClO_4$(a)\" and \"$NO_3$(b),\" and the curves are labeled (a) and (b). Curve (a), corresponding to $[Cu(deen)_2](ClO_4)_2$, has a lower baseline and a sharp upward peak at about 45–50°C, remains flat thereafter, and drops steeply after about 200°C. Curve (b), corresponding to $[Cu(deen)_2](NO_3)_2$, has a higher baseline, a broad upward peak at about 110°C, a sharp upward peak at about 145°C, another smaller upward peak at about 180°C, and a steep drop after about 200°C. No arrows are shown." + } + ], + "full_mark": 4 + }, + { + "sub_id": "Exam-21-T3-4", + "tags": [ + "1.4", + "4.6" + ], + "text_open": "3.4.2 Using the information above, explain the origin of the difference between the transition temperatures of $[Cu(deen)_2](ClO_4)_2$ and $[Cu(deen)_2](NO_3)_2$. (No more than 100 words)", + "text_images": [], + "full_mark": 4 + } + ] +} diff --git a/test/tasks/Exam-21-T4/problem.json b/test/tasks/Exam-21-T4/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..fc2a9923d6d16d8a7dbeda469668e87ffd0710b9 --- /dev/null +++ b/test/tasks/Exam-21-T4/problem.json @@ -0,0 +1,25 @@ +{ + "problem_id": "Exam-21-T4", + "header": "Determination of Manganese by Ammonium Nitrate Oxidation Titrimetry\nManganese is an important element in steels and alloys, and its content directly affects their strength, hardness, and wear resistance. Ammonium nitrate oxidation titrimetry is a classical method for determining manganese: ammonium nitrate oxidizes Mn(II) to Mn(III) in a phosphoric acid medium, and the product is then titrated with a standard ammonium iron(II) sulfate solution. This problem addresses standardization of the titrant, control of the oxidation conditions, and calculation of the result.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-21-T4-1", + "tags": [ + "1.4" + ], + "text_open": "4.2 Sample treatment and titration\nWeigh 0.2026 g of sample into a 300 mL conical flask. Add 15 mL of concentrated phosphoric acid and 5 mL of concentrated nitric acid, and heat until the sample dissolves completely. Continue heating and evaporating until the surface becomes still and faint fumes just begin to appear (about 220–240 °C), then remove the flask from the heat. Immediately add 3.0 g ± 0.2 g of ammonium nitrate. While swirling the flask, use a rubber-bulb pipette to blow tangentially across the mouth of the flask to remove nitrogen oxides until the yellow fumes disappear completely. Allow the mixture to cool slightly, add 60 mL of distilled water, mix thoroughly, and cool to room temperature. Titrate with the standard ammonium iron(II) sulfate solution until the solution is pale red. Add 2 drops of N-phenylanthranilic acid indicator solution (2 g/L) and continue titrating to a bright-yellow endpoint. The standard ammonium iron(II) sulfate solution consumed is 16.48 mL, and the blank consumes 0.84 mL.\n4.2.2 Give the predominant form of manganese after oxidation by ammonium nitrate. (Hint: the ligand is dihydrogen phosphate.)", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-21-T4-2", + "tags": [ + "2.1" + ], + "text_open": "4.2.4 Write the net ionic equation for the titration. In the products, the metal ions are present as free ions rather than complexes.", + "text_images": [], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-21-T5/images/Exam-21-T5-2-1-prompt.png b/test/tasks/Exam-21-T5/images/Exam-21-T5-2-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..d194a80721e25a0d2ab44cb6c054d14eefe53cf9 --- /dev/null +++ b/test/tasks/Exam-21-T5/images/Exam-21-T5-2-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:3d7b28e7eeb2d1d381ff4b5b6ff85d5495cf09adf2e5f9321fc64a2e92bfb309 +size 109587 diff --git a/test/tasks/Exam-21-T5/problem.json b/test/tasks/Exam-21-T5/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..d3a3a978e354123a0a35082a3b2b7305e6cb36c7 --- /dev/null +++ b/test/tasks/Exam-21-T5/problem.json @@ -0,0 +1,39 @@ +{ + "problem_id": "Exam-21-T5", + "header": "Structures of Silicides", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-21-T5-1", + "tags": [ + "1.3" + ], + "text_open": "5.1 Consider a hypothetical crystal X. Si atoms form graphite-like layers A stacked along the c direction. The projections of all layers A along c coincide, producing a series of layers of hexagonal-prismatic voids between them. Metal M regularly occupies 2/3 of the hexagonal-prismatic voids in each such layer to form layers B, and the projections of all layers B along c also coincide.\n5.2 Crystal Y is obtained by replacing some Si atoms in layers A of crystal X with M atoms. In crystal Y, no two Si atoms and no two M atoms are adjacent within the same layer A. Likewise, between two neighboring layers A along c, no Si atom is adjacent to another Si atom and no M atom is adjacent to another M atom.\n5.2.1 Give the chemical formula of crystal Y in terms of M and Si, reduced approximately to the simplest ratio, and the number of formula units in one conventional unit cell.", + "text_images": [], + "full_mark": 3 + }, + { + "sub_id": "Exam-21-T5-2", + "tags": [ + "1.3" + ], + "text_open": "5.3 Further distortion of layers A in crystal Y gives hexagonal crystal Z, a real silicide. The projection of crystal Z along c is shown below, where the small dark spheres represent Si atoms and the large light-colored spheres represent M atoms. After distortion, all Si atoms still have identical chemical environments, and all M atoms within the same type of layer likewise have identical chemical environments.\n5.3.1 In crystal Z, one Si atom has coordinates (0.6000, 0, 0.2500), and two M atoms have coordinates (0.2300, 0, 0.2500) and (0.3333, 0.6667, 0.0000), respectively. Give the coordinates of all remaining Si and M atoms.", + "text_images": [ + { + "file": "Exam-21-T5-2-1-prompt.png", + "description": "Projection of crystal Z along the c direction on a white background. Thin gray lines outline a right-leaning parallelogram unit cell, with approximately horizontal top and bottom edges and side edges slanting upward to the right. The diagram contains 22 shaded, highlighted spheres of two types: 12 large light-blue spheres and 10 small dark-blue spheres. There are 16 spheres on the unit-cell boundary, 4 on each edge; every edge has 2 large light-blue spheres and 2 small dark-blue spheres arranged in the order large light-blue, small dark-blue, small dark-blue, large light-blue. The interior contains 6 additional spheres: 4 large light-blue spheres at upper left, upper right, lower left, and lower right, and 2 small dark-blue spheres at center-right and lower-center-left, approximately along the shorter diagonal. At lower left, a red solid arrow points horizontally right and is labeled a; a green solid arrow points upper left and is labeled b. Their common origin is a blue circle containing a white c. As specified in the prompt, the dark spheres are Si atoms and the light-colored spheres are M atoms." + } + ], + "full_mark": 6 + }, + { + "sub_id": "Exam-21-T5-3", + "tags": [ + "1.3" + ], + "text_open": "5.3.2 During the distortion, the coordination polyhedron around an M atom in layer A, formed by Si atoms in the same and adjacent layers, is also distorted. State how many distinct Si–M distances occur in the distorted coordination polyhedron.", + "text_images": [], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-21-T6/images/Exam-21-T6-2-1-prompt.png b/test/tasks/Exam-21-T6/images/Exam-21-T6-2-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..695f003e67354225370a191c6546fec6f81d3a94 --- /dev/null +++ b/test/tasks/Exam-21-T6/images/Exam-21-T6-2-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:ecdb03a7b10587870a7a4a61a278c75e2779e242b49e18c17b7adef7893c223f +size 670609 diff --git a/test/tasks/Exam-21-T6/images/Exam-21-T6-3-1-prompt.png b/test/tasks/Exam-21-T6/images/Exam-21-T6-3-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..c75e230451c1fb4aa220f7c24ba10f29205e381d --- /dev/null +++ b/test/tasks/Exam-21-T6/images/Exam-21-T6-3-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:1221df2b58f97d026c251ee2500a562053a637de2a1b540f8c4e4c31d79ede3f +size 812829 diff --git a/test/tasks/Exam-21-T6/problem.json b/test/tasks/Exam-21-T6/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..a58141cc97a466e77a6314f3367c8598b28ca95e --- /dev/null +++ b/test/tasks/Exam-21-T6/problem.json @@ -0,0 +1,46 @@ +{ + "problem_id": "Exam-21-T6", + "header": "Pharmacokinetics and Molecular Design (II)\nPharmacokinetics quantitatively studies the absorption, distribution, metabolism, and excretion of drugs in living organisms and uses mathematical principles and methods to describe how plasma drug concentration changes with time. Chemical-kinetic modeling provides a systematic understanding of drug metabolism in vivo. This problem concerns the oxidative metabolism of diclofenac in the body.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-21-T6-1", + "tags": [ + "3.7" + ], + "text_open": "The Michaelis equation is $r=r_{max}[S]/(K_m+[S])$.\nExperiments show that $[S]$ and $r$ for the 4′-hydroxylation of diclofenac catalyzed by human liver microsomes obey the following relationship:\n| Group | $[S]_1$ / $\\mu\\mathrm{M}$ | $r_1$ / $(\\mathrm{pmol \\cdot min^{-1} \\cdot mg^{-1}})$ | $[S]_2$ / $\\mu\\mathrm{M}$ | $r_2$ / $(\\mathrm{pmol \\cdot min^{-1} \\cdot mg^{-1}})$ |\n| --- | --- | --- | --- | --- |\n| Low concentration | 0.90 | 39.27 | 1.80 | 72.00 |\n| High concentration | 90.0 | 392.73 | 180.0 | 411.43 |\n\n6.2.4 Under the same conditions, the 5-hydroxylation has $K_m = 43\\ \\mu\\mathrm{M}$ and $r_{max} = 15.4\\ \\mathrm{pmol \\cdot min^{-1} \\cdot mg^{-1}}$. Intrinsic clearance $CL_{int}$ is defined by $CL_{int} = r_{max}/K_m$. Calculate $CL_{int}$ for the 4′- and 5-hydroxylation pathways and briefly explain the kinetic meaning of $CL_{int}$. Hint: the drug concentration in the liver is much lower than $K_m$.", + "text_images": [], + "full_mark": 5 + }, + { + "sub_id": "Exam-21-T6-2", + "tags": [ + "2.9", + "3.7" + ], + "text_open": "6.2.5 The kinetic parameters for the individual steps in the pathways above are shown below:\n| Pathway | Substrate → product | Enzyme | $CL_{int}$ / $(\\mu\\mathrm{L}\\cdot\\mathrm{min}^{-1}\\cdot\\mathrm{mg}^{-1})$ |\n| --- | --- | --- | --- |\n| 4′-hydroxylation pathway | DF → 4′-OH-DF | CYP2C9 | / |\n| 4′-hydroxylation pathway | 4′-OH-DF → A | CYP2C9 | 358 |\n| 5-hydroxylation pathway | DF → 5-OH-DF | CYP3A4 | / |\n| 5-hydroxylation pathway | 5-OH-DF → B | CYP2C9 | 103 |\n\nThe researchers also compared detoxification of the quinone products from the two hydroxylation pathways. The measured curves are shown below, where \"conjugate\" denotes a detoxification product.\nExplain why both steps of the 5-hydroxylation pathway have lower $CL_{int}$ values than those of the 4′-hydroxylation pathway, yet the 5-hydroxylation pathway is more toxic. (No more than 100 words)", + "text_images": [ + { + "file": "Exam-21-T6-2-1-prompt.png", + "description": "Coordinate plot containing two side-by-side panels labeled A and B at upper left. Both horizontal axes are labeled 'GSH concentration (mM)' with ticks at 0, 2, 4, and 6; both vertical axes are labeled 'Conjugate formation (%)' with ticks at 0, 50, and 100. Each panel contains one solid black curve with filled circular data markers, some with vertical error bars. In panel A, the curve rises rapidly from near the origin as GSH concentration increases, approaches 100% at about 2 mM, and remains on a plateau. In panel B, the curve rises more gradually from near the origin and approaches 100% at about 5–6 mM, without a clear plateau. There is no legend or title." + } + ], + "full_mark": 2 + }, + { + "sub_id": "Exam-21-T6-3", + "tags": [ + "1.7", + "2.9" + ], + "text_open": "6.3 Diclofenac has a half-life of about 1.0h in humans and is cleared mainly through CYP2C9-catalyzed 4′-hydroxylation of an aromatic ring. The structurally similar fenclorac is more resistant to aromatic hydroxylation and has a half-life exceeding 20h. Its structure is shown below.\nFrom a structural perspective, explain why fenclorac has a longer duration of action. (No more than 50 words)", + "text_images": [ + { + "file": "Exam-21-T6-3-1-prompt.png", + "description": "The SMILES representation of fenclorac is 'OC(CC1=C(OC2=CC=C(Cl)C=C2Cl)C=CC=C1)=O'." + } + ], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-22-T1/problem.json b/test/tasks/Exam-22-T1/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..9580da63a33c86b1e99c4dbb3f37079a755f156b --- /dev/null +++ b/test/tasks/Exam-22-T1/problem.json @@ -0,0 +1,52 @@ +{ + "problem_id": "Exam-22-T1", + "header": "Problem 1 Chemistry of the Rare-Earth Elements\nRare-earth elements are indispensable strategic resources for modern high technology, new energy, advanced materials, defense, and related fields. Often called the “vitamins of industry” and a “treasure house of new materials,” they comprise scandium, yttrium, and all the lanthanides. For convenience, the symbol Ln is commonly used in coordination and inorganic chemistry to denote the lanthanides or the entire rare-earth series. This problem examines rare-earth chemistry in detail.\nNote: Different letters or alphanumeric combinations in this problem may represent the same element or chemical formula.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-22-T1-1", + "tags": [ + "1.3" + ], + "text_open": "1.2 Consider a rare-earth element M1 as an example of rare-earth chemistry. M1 has two common oxidation states in its compounds. Elemental M1 is a silvery-white metal that slowly tarnishes in air. M1 dissolves in dilute hydrochloric acid, releasing colorless gas A and giving a pale blue-green solution (Reaction 1). Adding dilute sulfuric acid dropwise to this solution immediately produces white precipitate B, which is insoluble in dilute nitric acid. After B is filtered, washed, and dried, it is ignited to constant mass at 800 °C under Ar; the white precipitate becomes dark-brown powder C (Reaction 2), an oxide of M1. If the mass of B is 2.480 g, 1.760 g of C is obtained after ignition. A separate portion of B is heated with dilute hydrochloric acid and bromine water until it dissolves (Reaction 3), during which the solution changes from pale blue-green to colorless (or very pale pink). The solution is divided into two portions. Excess aqueous ammonia and oxalic acid are added separately to give white gelatinous precipitate D (with aqueous ammonia) and white precipitate E (with oxalic acid), respectively. Filtering and igniting either D or E at 900 °C likewise gives C. C is mixed with Y2O3 in a certain ratio and calcined at high temperature; product F emits bright red fluorescence under ultraviolet irradiation. 1.2.3 Write ionic equations for Reactions 1 and 3 and a chemical equation for Reaction 2.", + "text_images": [], + "full_mark": 6 + }, + { + "sub_id": "Exam-22-T1-2", + "tags": [ + "1.4" + ], + "text_open": "1.4 Coordination Compounds of Rare-Earth Elements\n1 equivalent of elemental rare-earth metal M5 and 2.8 equivalents of ethylene glycol monomethyl ether are added together to a toluene solution to form X. Addition of 2,6-dimethylphenol then precipitates orange crystals Y. Y is filtered and washed; upon slight heating it becomes a different crystal Z. Further characterization provides the following information:\n① The effective magnetic moment of X is 7.9 $\\mu_B$.\n② Crystal diffraction shows that Y belongs to the orthorhombic crystal system, with $a$ = 12.748 Å, $b$ = 12.900 Å, and $c$ = 29.220 Å. The unit cell contains 2 formula units, and the theoretical density of Y is 1.428 $\\mathrm{g/cm^3}$.\n③ Elemental analysis of Z gives the following experimental mass fractions: $\\omega(C)$ = 39.60%, $\\omega(O)$ = 18.86%, and $\\omega(H)$ = 5.74%. Crystal diffraction shows that Z belongs to the tetragonal crystal system, with $a$ = 14.837 Å and $c$ = 15.117 Å. The crystal contains 2 formula units, and the density of Z is 1.1854 times the theoretical density of Y.\n④ The anion of Z contains an $[M5_4O_4]$ cluster framework and exactly two types of ligands. One ligand adopts two different coordination modes. All M5 atoms are 7-coordinate, and a particular conformation has 1 twofold rotation axis.\n⑤ The oxidation state of M5 is the same in X, Y, and Z.\n1.4.2 Determine the chemical formulas of X, Y, and Z, showing your calculations. In your answer, the ethylene glycol monomethyl ether anion may be abbreviated as $mee^-$, and the 2,6-dimethylphenoxide anion as $dmp^-$.\n", + "text_images": [], + "full_mark": 9 + }, + { + "sub_id": "Exam-22-T1-3", + "tags": [ + "1.4" + ], + "text_open": "1.4.3 Write the SMILES strings for the cation and anion of Z. You may treat all $\\sigma$ coordinate bonds as ordinary single bonds and represent the structures using SMILES.", + "text_images": [], + "full_mark": 4 + }, + { + "sub_id": "Exam-22-T1-4", + "tags": [ + "2.1" + ], + "text_open": "1.5 Electrochemistry of the Rare-Earth Elements\nThe following standard electrode potentials are given for acidic medium at 25 °C:\n$O_2(g) + 4H^+ + 4e^- \\rightarrow 2H_2O$ $E° = +1.229\\ \\mathrm{V}$\n$M1^{n+} + e^- \\rightarrow M1^{(n-1)+}$ $E° = -0.35\\ \\mathrm{V}$\n1.5.1 At pH = 0 and 25 °C, calculate the common logarithm $\\lg K$ of the standard equilibrium constant for oxidation of $M1^{(n-1)+}$ to $M1^{n+}$ by $O_2$.", + "text_images": [], + "full_mark": 3 + }, + { + "sub_id": "Exam-22-T1-5", + "tags": [ + "2.1" + ], + "text_open": "1.5.2 The partial pressure of oxygen in air is $p(O_2)$ = 0.21 bar. Use a calculation to determine whether $M1^{(n-1)+}$ will be oxidized by oxygen in air when pH = 4.00 and $[M1^{(n-1)+}]$ = $[M1^{n+}]$ = 1.00 mol·L-1.", + "text_images": [], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-22-T5/images/Exam-22-T5-10-1-prompt.png b/test/tasks/Exam-22-T5/images/Exam-22-T5-10-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..494cb877903d57d534a2c62a78a6cbc7d8a56631 --- /dev/null +++ b/test/tasks/Exam-22-T5/images/Exam-22-T5-10-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:33c94df44d296f58900fa7567cd30332f506a33ae89282a7b419fce502e73f14 +size 127661 diff --git a/test/tasks/Exam-22-T5/problem.json b/test/tasks/Exam-22-T5/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..801bfdf066f6f086f19b59d245a314890601a64c --- /dev/null +++ b/test/tasks/Exam-22-T5/problem.json @@ -0,0 +1,102 @@ +{ + "problem_id": "Exam-22-T5", + "header": "Unlike most covalent compounds, whose structures can be described using 2c-2e bonds, polyhedral clusters are often electron-deficient systems and require multicenter bonding models.\nWade's rules, more generally known as the polyhedral skeletal electron pair theory (PSEPT), are an important tool for understanding cluster structures. Let $n$ denote the number of atoms or isolobal fragments that form the cluster framework. Subtracting the electrons used in terminal two-center two-electron bonds, nonskeletal lone pairs, and the internal bonding of ligands from the total valence-electron count gives the number of skeletal electrons. Skeletal electron-pair counts of $n+1$, $n+2$, $n+3$, and $n+4$ correspond, respectively, to closo, nido, arachno, and hypho structures. A closo cluster is a complete triangulated polyhedron with $n$ vertices, whereas nido, arachno, and hypho clusters may be regarded as the corresponding closo parent polyhedron with one, two, or three vertices removed, respectively. For example, in $B_5H_9$, 5 BH fragments contribute 10 electrons and 4 additional hydrogen atoms contribute another 4, giving 14 skeletal electrons or 7 skeletal electron pairs. Since the skeletal electron-pair count is $n+2$, the cluster is nido.\nThis problem uses low-valent clusters of Group 13 metals (Al, Ga, In, and Tl) to examine cluster synthesis, structure, bonding, and reactivity.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-22-T5-1", + "tags": [ + "1.6" + ], + "text_open": "5.1 $B_4Cl_4$ and $B_4(t\\text{-}Bu)_4$ have tetrahedral cores, whereas the acetamido-substituted boron compound $B_4(NHAc)_4$ adopts a butterfly-shaped cyclic structure. Explain why.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-22-T5-2", + "tags": [ + "2.1" + ], + "text_open": "5.2 Write equations for the following reactions.\n5.2.1 $AlCl\\cdot x\\,Et_2O$ ($x>1$) reacts with $MgCp^*_2$ at low temperature in toluene/diethyl ether to form $Al_4Cp^*_4$.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-22-T5-3", + "tags": [ + "2.1" + ], + "text_open": "5.2.2 $GaI$ reacts with $Li(THF)_3Ge(SiMe_3)_3$ in toluene to form $[GaGe(SiMe_3)_3]_4$.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-22-T5-4", + "tags": [ + "2.4" + ], + "text_open": "5.4 Tetrahedral clusters not only have unusual structures but also show diverse oxidation chemistry, giving different products under different conditions. Draw the structures of products A, B, D, and E below. 5.4.1 Oxidation of [GaSi(t-Bu)_3]_4 with oxygen gives cubic cluster A = [OGaSi(t-Bu)_3]_4.", + "text_images": [], + "full_mark": 3 + }, + { + "sub_id": "Exam-22-T5-5", + "tags": [ + "2.4" + ], + "text_open": "5.4.2 Further hydrolysis of A gives product B, which contains only one ring and has only one chemical environment for its Ga atoms. Write the SMILES string for B. If necessary, describe its detailed structure in no more than 50 words. Multiple reasonable structures may be possible; writing any one reasonable SMILES string is sufficient.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-22-T5-6", + "tags": [ + "2.4" + ], + "text_open": "5.4.3 Oxidation of $C = [InC(SiMe_3)_3]_4$ with o-nitrosotoluene gives a cubic cluster and product D, which contains two benzene rings. Write the SMILES string for aromatic product D and, if necessary, describe its detailed structure in no more than 50 words.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-22-T5-7", + "tags": [ + "2.4" + ], + "text_open": "5.4.4 Oxidation of C with 1,2-thiirane gives a trigonal-bipyramidal cluster E. Write the SMILES string for E and, if necessary, describe its detailed structure in no more than 50 words.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-22-T5-8", + "tags": [ + "2.2" + ], + "text_open": "5.6 Reaction of $GaI$ with $Li(THF)_3R$ ($R = SiMe(SiMe_3)_2$) in toluene gives the octahedral cluster $G = [GaR]_6$. In contrast, reaction of $GaI$ with $Na(THF)_2R'$ ($R' = Si(t\\text{-}Bu)_3$) in toluene forms the octahedral cluster $H = [Ga_6R'_4Bn_2]^{2-}$, where Bn denotes benzyl.\n5.6.1 State the source of the benzyl groups in H.", + "text_images": [], + "full_mark": 1 + }, + { + "sub_id": "Exam-22-T5-9", + "tags": [ + "1.7" + ], + "text_open": "5.6.3 Explain why G is a highly distorted octahedron whereas H is nearly a regular octahedron.", + "text_images": [], + "full_mark": 3 + }, + { + "sub_id": "Exam-22-T5-10", + "tags": [ + "1.7" + ], + "text_open": "5.8 Electron-rich cluster compounds: When the number of ligands in a cluster is much smaller than the number of metal atoms, its structure approaches that of a bare metal-atom cluster; such species are called electron-rich cluster compounds. A representative example is the metalloid cluster $K = Ga_{84}[N(SiMe_3)_2]_{20}^{4-}$. This cluster contains 84 Ga atoms, 64 of which are “bare.” Its structure has three layers: a central $Ga_2$ dumbbell with a bond length of 235 pm, approaching a $Ga\\equiv Ga$ triple bond; a middle $Ga_{32}$ football-shaped tube; and an outer assembly of three $Ga_{10}$ rings surrounded by 20 $N(SiMe_3)_2$ ligands.\nCompare K with the fullerene $Sc_2@C_{84}$, and identify which structural units in K correspond to the encapsulated $Sc_2$ and to the $C_{84}$ carbon cage, respectively.", + "text_images": [ + { + "file": "Exam-22-T5-10-1-prompt.png", + "description": "Unlabeled three-dimensional ball-and-stick molecular model. The overall structure is a vertically symmetric multilayer assembly. The upper and lower layers each contain a polyhedron made of dark-blue spheres, approximately 12 in total. A larger central polyhedron is formed by approximately 12 purple spheres, and the outermost framework contains approximately 24 orange-yellow spheres. Spheres of the same color are connected by solid lines of the same color, with radial connections between layers. The drawing represents the layered structure of the metalloid cluster $K = Ga_{84}[N(SiMe_3)_2]_{20}^{4-}$ described in the question." + } + ], + "full_mark": 2 + } + ] +}