diff --git "a/dev/dataset.jsonl" "b/dev/dataset.jsonl"
--- "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 $V_0c_1<c_2(V_1+2V_2)$. (2 points)"], "rubric_images": []}, {"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, "rubric": ["First determine that $30 × 1.057 - 9.60 × 1.032 < 2 × 10.75 × 1.032$, showing that the sample contains Na_2HPO_4. (2 points)", "$\\omega =\\frac{30.00\\times 1.057-9.60\\times 1.032-10.75\\times 1.032\\times {{10}^{-3}}\\times 163.94}{2.013}=87.22\\%$. (2 points; directly obtaining n(Na_3PO_4) = 0.01071 mol without the determination also earns 2 points. Deduct 1 point for incorrect significant figures.)"], "rubric_images": []}, {"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, "rubric": ["Pathway one: (HP_2O_5)^(3-) + OH^- = (P_2O_6)^(4-) + H_2. (2 points)", "Pathway 2: (HP_2O_5)^(3-) + OH^- = 2(HPO_3)^(2-). (2 points)", "Writing reactants or products in acid form earns 1 point per equation. Answers containing (P_2O_5)^(4-) or (PO_3)^(3-) receive no credit."], "rubric_images": []}, {"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, "rubric": ["Three. (1 point)", "P^1 + P^4-P^3-P^3-P^3-P^4. (1 point) P^3 + P^2-P^3-P^3-P^3-P^4. (1 point) P^5 + P^2-P^3-P^3-P^3-P^2. (1 point)", "Deduct 1 point for an incorrect hydrolysis pathway and 1 point if the monophosphorus product is omitted."], "rubric_images": []}]}
+{"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, "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": []}]}
+{"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": []}]}
 {"problem_id": "Exam-05-T7", "header": "Problem 7 Explosive Chemistry. An explosion rapidly converts a system's energy into mechanical work through gas expansion, usually with intense heat, light, and sound. Explosive chemistry appears both as a laboratory hazard and in high-density energy-storage materials.", "header_images": [], "subquestions": [{"sub_id": "Exam-05-T7-1", "tags": ["2.4"], "text_open": "7-1-2 Isopropanol deterioration is modeled by acetone reacting in acid with equal numbers of H_2O_2 molecules: two acetone molecules give DADP and 3 give TATP. Both are highly symmetric peroxides. Give their SMILES.", "text_images": [], "full_mark": 2, "rubric": ["7-1-2 DADP: 'CC1(C)OOC(C)(C)OO1'; TATP: 'CC1(C)OOC(C)(C)OOC(C)(C)OO1'. (1 point each)"], "rubric_images": [{"file": "Exam-05-T7-1-1-rubric.png", "description": "DADP and TATP structures as listed. SMILES/structure strings shown: 'CC1(C)OOC(C)(C)OO1', 'CC1(C)OOC(C)(C)OOC(C)(C)OO1'."}]}, {"sub_id": "Exam-05-T7-2", "tags": ["2.2", "2.3"], "text_open": "7-1-3 One TATP first gives cyclic A plus acetone; A then gives acetone, dipolar B, and gas. Write 2 equations with A and B as SMILES.", "text_images": [], "full_mark": 2, "rubric": ["7-1-3 Reaction 1: 'CC1(C)OOC(C)(C)OOC(C)(C)OO1>>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'."}]}]}