| {"problem_id": "Exam-01-T2", "header": "Determination of Ionic Constituents\nDuring the preparation of K_3[Al(C_2O_4)_3]·3H_2O crystals, some Cr^3+ ions were introduced to replace Al^3+. The following titration experiments were designed to determine the Al and Cr contents of the crystal sample. Titration experiment I: dissolve a 0.5062 g crystal sample in water and transfer it to a 100 mL conical flask. Add a buffer to adjust the pH, then add a solution containing 2.024×10^-3 mol of EDTA. Heat at a gentle boil for 10 min, cool, and add indicator A. Titrate to the endpoint with 0.09142 mol/L standard Zn^2+ solution; 10.40 mL of the standard Zn^2+ solution is consumed. Titration experiment II: dissolve a separate 2.5015 g crystal sample in water and transfer it to a 100 mL conical flask. Acidify with sulfuric acid, add excess (NH_4)_2S_2O_8 and a catalytic amount of Ag^+, then heat at boiling for 10 min. Cool, add indicator B, and titrate with 0.1515 mol/L standard Fe^2+ solution; 17.94 mL of the standard Fe^2+ solution is consumed. Note: assume that the two samples have the same composition and that impurities do not participate in the titration reactions. Report calculated results to 3 significant figures.", "header_images": [], "subquestions": [{"sub_id": "Exam-01-T2-1", "tags": ["2.1", "4.5", "4.2"], "text_open": "Write the ionic equations for the principal reactions involved after acidification with sulfuric acid in titration experiment II.", "text_images": [], "full_mark": 8}, {"sub_id": "Exam-01-T2-2", "tags": ["4.5", "4.7", "2.1"], "text_open": "Calculate the mass fractions of Al and Cr in the sample.", "text_images": [], "full_mark": 6}]} | |
| {"problem_id": "Exam-01-T3", "header": "Lithium Redox-Flow Battery\nA lithium redox-flow battery based on redox-targeting reactions is shown below. Redox targeting links an active material to an electrode by introducing into the electrolyte a highly soluble redox mediator paired with that active material. This approach both circumvents the solubility limitation of the active material and allows low-solubility active materials to be stored in tanks.", "header_images": [{"file": "Exam-01-T3-header-1.png", "description": "Schematic of an electrochemical flow apparatus. At the center is a battery stack composed of multiple plate-like layers, with blue outer plates and black and green layers between them. The left blue plate is labeled “+” and the right blue plate “-”. A wire at the top connects to a glowing light bulb; \\(e\\) is labeled above the wire, and an arrow shows electron transfer from the blue plate labeled “-” to the blue plate labeled “+”. To the left of the stack is a cylindrical vessel containing a pale-blue liquid and a cylindrical material labeled \\(\\mathrm{LiFePO_4}\\). \\(\\mathrm{Fe(CN)_6^{4-}}\\) is labeled above it and \\(\\mathrm{Fe(CN)_6^{3-}}\\) below it. Blue tubing connects the left vessel to the central stack and forms a circulation loop through a bottom pump labeled \\(\\mathrm{Pump}\\). The circulation direction is: blue plate labeled “+” → lower blue tubing → \\(\\mathrm{Pump}\\) → bottom of the left vessel (\\(\\mathrm{Fe(CN)_6^{3-}}\\)) → \\(\\mathrm{LiFePO_4}\\) → upper part of the left vessel (\\(\\mathrm{Fe(CN)_6^{4-}}\\)) → upper blue tubing → blue plate on the side labeled “+”. To the right of the stack is a cylindrical vessel containing a pale-yellow liquid and a cylindrical material labeled \\(\\mathrm{LiTi_2(PO_4)_3}\\). \\(\\mathrm{S_2^{2-}}\\) is labeled above it and \\(\\mathrm{S^{2-}}\\) below it. Yellow tubing connects the right vessel to the central stack and forms a circulation loop through a bottom pump labeled \\(\\mathrm{Pump}\\). The circulation direction is: blue plate labeled “-” → lower yellow tubing → \\(\\mathrm{Pump}\\) → bottom of the right vessel (\\(\\mathrm{S^{2-}}\\)) → \\(\\mathrm{LiTi_2(PO_4)_3}\\) → upper part of the right vessel (\\(\\mathrm{S_2^{2-}}\\)) → upper yellow tubing → blue plate labeled “-”."}], "subquestions": [{"sub_id": "Exam-01-T3-1", "tags": ["2.1", "3.5"], "text_open": "3-2 Write the electrode reactions at the cathode and anode during charging.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-01-T3-2", "tags": ["2.1", "3.5"], "text_open": "3-3 Write the reactions occurring in the catholyte and anolyte tanks during charging.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-01-T3-3", "tags": ["3.5", "3.6"], "text_open": "3-4 This lithium redox-flow battery operates at a constant current of 2.0 A for 24 minutes, during which the mass of solid material in the catholyte tank changes by 1.52 g. Calculate the efficiency with which the battery converts chemical energy into electrical energy.", "text_images": [], "full_mark": 4}]} | |
| {"problem_id": "Exam-01-T4", "header": "Preparation of a Dinuclear Ti Complex\nIn benzene solution, 10.5 mmol of TiBr_4 is mixed with 5 mmol of B_2(NMe_2)_4. A dark-red solution forms immediately, followed by a brown precipitate. The reaction mixture is stirred vigorously at room temperature for 5 h, after which the volatile components are removed under vacuum and passed successively through cold traps at -22 ℃ and -196 ℃. The -196 ℃ trap contains the benzene solvent, whereas the substance collected in the -22 ℃ trap is identified as B(NMe_2)_3. The remaining liquid is removed by distillation, and the insoluble solid is washed with DCM to give a sample whose principal component is X. A 219.7 mg portion of the sample is treated with 25 % sulfuric acid solution, immediately producing a dark-purple solution with no gas evolution at room temperature. Addition of sufficient NaOH solution, from which dissolved air has been removed, to this dark-purple solution produces a blue-black precipitate that gradually evolves gas and is ultimately converted into a white precipitate. On the basis of IR, 1H-NMR, magnetic-moment, and other studies, a structure is proposed for X. In this structure, only one type of coordinating atom and two types of ligands surround Ti. The B-containing ligand, denoted L, is bidentate, and the three atoms bonded to each B atom are all different. The relative molecular mass of the molecule is 1114.3 g/mol.", "header_images": [], "subquestions": [{"sub_id": "Exam-01-T4-1", "tags": ["1.4", "2.1"], "text_open": "4-1 Determine the molecular formula of X by calculation and analysis.", "text_images": [], "full_mark": 6}, {"sub_id": "Exam-01-T4-2", "tags": ["1.4", "2.4"], "text_open": "4-2 Using the information above, write the SMILES structure of X.", "text_images": [], "full_mark": 2}]} | |
| {"problem_id": "Exam-01-T8", "header": "Fundamental Concepts and Reactions of Organic Compounds\nFor Problems 8-1 through 8-5, make the requested judgment and provide a reasonable explanation. For Problems 8-6 and 8-7, answer according to the stated conditions and requirements.", "header_images": [], "subquestions": [{"sub_id": "Exam-01-T8-1", "tags": ["2.3"], "text_open": "Does Dewar benzene, whose structure is shown below, convert to benzene under irradiation or heating? Limit your answer to no more than 50 words.", "text_images": [{"file": "Exam-01-T8-1-1-prompt.png", "description": "Structure of Dewar benzene: 'C12C=CC1C=C2'"}], "full_mark": 3}, {"sub_id": "Exam-01-T8-2", "tags": ["1.2", "2.7"], "text_open": "Which compound is oxidized more rapidly by periodic acid, cis-1,2-cyclohexanediol or trans-1,2-cyclohexanediol? Limit your answer to no more than 50 words.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-01-T8-3", "tags": ["1.5", "1.6"], "text_open": "For the carbene shown below, is the singlet or triplet state more stable? Limit your answer to no more than 100 words.", "text_images": [{"file": "Exam-01-T8-3-1-prompt.png", "description": "Structure of the carbene in the question: '[C@@H]12CCC([C]2)C=C1'"}], "full_mark": 3}, {"sub_id": "Exam-01-T8-4", "tags": ["2.3", "2.4"], "text_open": "The compound shown below has an intense floral fragrance and is a useful perfume ingredient, but under basic conditions it is converted into an odorless isomer. Write the SMILES structure of this isomer.", "text_images": [{"file": "Exam-01-T8-4-1-prompt.png", "description": "Structure of the fragrance compound in the question: 'CC(C)(O)[C@H]1CC[C@@H](C=O)CC1'"}], "full_mark": 2}]} | |
| {"problem_id": "Exam-01-T9", "header": "How Well Do You Really Know Carbon-Phosphorus Double Bonds?\nIn studying organic chemistry, we seem to restrict our understanding of the reactivity of carbon-phosphorus double bonds to their “metathesis” reactions with carbon-oxygen double bonds. With the assistance of transition metals, however, double bonds between carbon and trivalent phosphorus display distinctive modes of formation and reactivity. In this problem, whenever a subquestion asks for the SMILES of a complex, supplement it with a verbal description of the specific coordination mode.", "header_images": [], "subquestions": [{"sub_id": "Exam-01-T9-1", "tags": ["1.4", "2.2", "2.4"], "text_open": "9-1 Carbon-phosphorus double bonds can be constructed in many ways and can coordinate to transition metals in more than one mode. 9-1-1 In the reaction below, [Mn] denotes (η^{5}-C_{5}H_{4}CH_{3})Mn(CO)_{2}, and the reaction produces 4 molecules of THF. Given that the major product contains 2 phosphorus-oxygen single bonds, write the SMILES structure of the major product. 4 [Mn]·THF + 2 ^{t}BuCP + H_{2}O →", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-01-T9-2", "tags": ["1.4", "2.2", "2.3"], "text_open": "9-1-2 The following example provides an initial illustration of the role played by transition metals in forming carbon-phosphorus double bonds. Use SMILES structures and any necessary verbal descriptions to represent the three key electrically neutral intermediates in the reaction. Coordination of double bonds to Mo need not be shown in any of them.", "text_images": [{"file": "Exam-01-T9-2-1-prompt.png", "description": "The reaction in the figure shows the anionic dinuclear molybdenum complex 'O=C[Mo](C=O)[Mo](PC1=CC=CC=C1)(C=O)C=O' reacting with acryloyl chloride, 'ClC(C=C)=O', to form a product in which the phosphaalkene ligand 'C/C=P\\C2=CC=CC=C2' is bound to the dinuclear molybdenum framework 'O=C[Mo](C=O)[Mo](C=O)C=O'. In the reactant, a phenylphosphido group bridges two Mo atoms, each bearing two carbonyl ligands; on one side, a coordinate bond shown by an arrow points to the left Mo atom. After reaction, the newly formed phosphaalkene ligand coordinates in a bidentate bridging mode: the lone pair on phosphorus forms a coordinate bond, shown by an arrow, to the left Mo atom, while the $P=C$ double bond coordinates through a $\\pi$ bond, shown by an arrow, to the right Mo atom."}], "full_mark": 3}, {"sub_id": "Exam-01-T9-3", "tags": ["1.4", "2.2", "2.6"], "text_open": "9-1-3 Write the SMILES structures of the six key intermediates in the reaction shown below. Note: the step involving addition of tert-butyllithium corresponds to three key intermediates; each other step corresponds to one key intermediate.", "text_images": [{"file": "Exam-01-T9-3-1-prompt.png", "description": "Description of the structures/reaction in the figure: '[H]C#CC1=CC=CC=C1' undergoes steps involving NEt_3/PCl_3, ArLi, tBuLi, ML_n, and irradiation/O_2 to form a species whose ligand has the SMILES 'O=C(C(/C(C(C1=CC=CC=C1)=O)=P/C)=P\\C)C2=CC=CC=C2'. The ligand coordinates to [ML_n] through its two P atoms. Ar denotes an aryl substituent, and [ML_n] denotes a metal coordination fragment."}], "full_mark": 6}, {"sub_id": "Exam-01-T9-4", "tags": ["2.2", "2.3", "1.4"], "text_open": "9-2 An important class of reactions of carbon-phosphorus double bonds involves metal carbenes. 9-2-1 In the following reaction, addition between the two substrates gives A_{1}, and A_{1} then converts into its isomer A_{2}. A_{2} undergoes a hydrogen migration to give A_{3}; A_{3} eliminates HNMe_{2} to give A_{4}; and a further elimination gives A_{5} together with byproduct A_{6}. A_{5} undergoes a simple reaction to afford the final product. W remains six-coordinate throughout the process. In A_{2}, W is coordinated by the carbon-phosphorus double bond, whereas in A_{3}, A_{4}, and A_{6}, W is coordinated by P. On this basis, write the SMILES structures of A_{1}~A_{6}.", "text_images": [{"file": "Exam-01-T9-4-1-prompt.png", "description": "Description of the structures/reaction in the figure: 'CN(C)C(=P)N(C)C' reacts with 'C1CC1C(O[Si](C)(C)C)=[W]([C]#O)([C]#O)([C]#O)([C]#O)[C]#O' to form 'O=C1C2(CC2)C(=O)C13CC3'."}], "full_mark": 6}, {"sub_id": "Exam-01-T9-5", "tags": ["2.2", "2.3", "1.4"], "text_open": "9-2-2 When the phosphorus atom no longer bears a hydrogen atom, a hydrogen migration analogous to the process above cannot occur. In the following reaction, the two substrates undergo a process similar to that in the preceding subquestion, proceeding through B_{1} to B_{2}. Write the SMILES structure of one key electrically neutral intermediate in the conversion of B_{2} to the products.", "text_images": [{"file": "Exam-01-T9-5-1-prompt.png", "description": "Description of the structures/reaction in the figure: 'CN(C)C(=P[Fe])N(C)C' + 'CCOC(=[M])C' -> 'CN(C)/C=C/C(=[M])OCC' + 'CCOC(C)=P([Fe])[M]. Here M denotes a metal. The ligands on Fe and M are omitted in this question.'"}], "full_mark": 3}]} | |
| {"problem_id": "Exam-01-T10", "header": "Light", "header_images": [], "subquestions": [{"sub_id": "Exam-01-T10-1", "tags": ["2.2", "2.4"], "text_open": "10-1 Light can excite certain chemical bonds. Write the SMILES structures of [C] and D. Stereochemistry is not required.", "text_images": [{"file": "Exam-01-T10-1-1-prompt.png", "description": "In the figure, 'O=C(C1=CC=CC=C1C)C2=CC=CC=C2' forms [A] under hv; [A] is converted to [B]; [B] is converted to [C]; and [C] reacts with 'O=C(/C=C/C(OCC)=O)OCC' to form D. A, B, and C are enclosed in square brackets to indicate highly reactive intermediates."}], "full_mark": 8}, {"sub_id": "Exam-01-T10-2", "tags": ["2.2", "2.3", "2.4"], "text_open": "10-3 The figure above is a simplified diagram of a photoredox catalytic mechanism. Such reactions use a photocatalyst capable of both single-electron oxidation and reduction under visible-light irradiation to initiate radical reactions and, after a series of complex steps, quench the radical process. 10-3-1 Based on the information above, write the SMILES structures of the 3 key intermediates in the reaction shown and of the other product in addition to the product depicted. PMP denotes p-methoxyphenyl.", "text_images": [{"file": "Exam-01-T10-2-1-prompt.png", "description": "Schematic of a photocatalyst quenching cycle. “P-Cat” is at the bottom center and “P-Cat*” at the top, with an upward vertical arrow between them labeled “hν.” Blue circular arrows connect “P-Cat*,” “P-Cat^+•,” “P-Cat^-•,” and “P-Cat”: the left blue arrow points downward from P-Cat* to P-Cat^+•, and the lower-left blue arrow points down and right from P-Cat^+• to P-Cat; the right blue arrow points downward from P-Cat* to P-Cat^-•, and the lower-right blue arrow points down and left from P-Cat^-• to P-Cat. “Oxidative Quenching” is labeled at the upper left, near “A” and “A^-•,” with a curved red arrow from A to A^-•. At the lower left are “D” and “D^+•,” with a curved red arrow from D to D^+•. “Reductive Quenching” is labeled at the upper right, near “D” and “D^+•,” with a curved red arrow from D to D^+•. At the lower right are “A” and “A^-•,” with a curved red arrow from A to A^-•."}, {"file": "Exam-01-T10-2-2-prompt.png", "description": "'CC1=CC=C(S(N=CC2=CC=CC=C2)(=O)=O)C=C1' and 'C[Si](COC1=CC=C(OC)C=C1)(C)C' react under 10 W blue LEDs, P-Cat (0.5 mol %), MeOH (Solv.), rt to form 'COC(C=C1)=CC=C1OCC(NS(C2=CC=C(C)C=C2)(=O)=O)C3=CC=CC=C3'."}], "full_mark": 6}, {"sub_id": "Exam-01-T10-3", "tags": ["2.2", "2.4", "2.5"], "text_open": "10-3-2 Under these conditions, the product can react further to remove the PMP protecting group and reveal an alcoholic hydroxy group. Write the SMILES structures of the 2 key intermediates in this reaction.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-01-T10-4", "tags": ["2.2", "2.3"], "text_open": "10-3-3 The photosensitizer P-Cat can capture O_2 from air. Write all species that may be formed.", "text_images": [], "full_mark": 6}]} | |
| {"problem_id": "Exam-02-T8", "header": "Problem 8: Familiar Named Reactions (14 points)", "header_images": [], "subquestions": [{"sub_id": "Exam-02-T8-1", "tags": ["2.7", "1.2"], "text_open": "8-1 The Baeyer-Villiger oxidation should be familiar, and migrating groups are known to follow an order of migratory aptitude. Explain why the following reaction gives a single product in very high yield.", "text_images": [{"file": "Exam-02-T8-1-1-prompt.png", "description": "A reaction proceeds from left to right. The molecule 'O=C1[C@@H]2C3C4CC2[C@H]5C4CC3[C@H]51' on the left reacts under a one-way arrow labeled “m-CPBA” above and “DCM,93%” below to give 'O=C1[C@@H]2C3C4CC2[C@H]5C4CC3[C@H]5O1' on the far right."}], "full_mark": 2}, {"sub_id": "Exam-02-T8-2", "tags": ["2.3", "2.7", "2.2"], "text_open": "8-3 The Schmidt rearrangement should be familiar. The following two substrates give different products under identical conditions. Explain the outcome for the second substrate.", "text_images": [{"file": "Exam-02-T8-2-1-prompt.png", "description": "The image shows two parallel reaction equations. In the upper reaction, the reactant on the left has SMILES 'O=C1CCCC1CCC(c2ccccc2)N=[N+]=[N-]' and reacts under an arrow labeled “CF_3SO_2OH” above and “0°C” below to give 'O=C3CCCC4N3C(c5ccccc5)CC4' on the right. In the lower reaction, the reactant on the left has SMILES 'O=C6CCCC6CCCC(c7ccccc7)N=[N+]=[N-]' and reacts under an arrow labeled “CF_3SO_2OH” above and “0°C” below to give 'O=C8CCCC89C(CCC9)Nc%10ccccc%10' on the right."}], "full_mark": 2}, {"sub_id": "Exam-02-T8-3", "tags": ["2.2", "2.3", "1.2"], "text_open": "8-4 At first glance, the following reaction appears simple and seemingly requires only elimination, hydride migration, and addition. Closer examination shows that this interpretation is inconsistent with the stereochemical information. Write the SMILES representations of all key intermediates actually involved, including their stereochemistry.", "text_images": [{"file": "Exam-02-T8-3-1-prompt.png", "description": "'C[C@@H]1CO[C@@]2([H])N1[C@](CC(OC)=C)(C#N)CCC2' and 'Cl[Mg]CCCC1OCCO1' react under tBuMe2SiOTf, DCM/Et2O/THF, then NH4Cl/H2O conditions to give 'C[C@@H]1CO[C@]2(CCCC3OCCO3)N1[C@]([H])(CC(OC)=C)CCC2'."}], "full_mark": 8}]} | |
| {"problem_id": "Exam-02-T10", "header": "Problem 10: Forever Unnamed", "header_images": [], "subquestions": [{"sub_id": "Exam-02-T10-1", "tags": ["2.2"], "text_open": "10-1 A neutral reaction intermediate can be obtained by the process shown below. Write its SMILES representation.", "text_images": [{"file": "Exam-02-T10-1-1-prompt.png", "description": "A chemical reaction equation: the reactants on the left are [R1]CC(=O)Cl and R_2NC, joined by a plus sign `+`; a horizontal right-pointing arrow in the center leads to bold X on the right."}], "full_mark": 2}, {"sub_id": "Exam-02-T10-2", "tags": ["2.4", "2.2", "1.2"], "text_open": "10-2-1 Further reactions can proceed from intermediates of type X. Write the SMILES representations of products A and B in the reactions below, given that B contains a five-membered aromatic ring. 10-2-2 Label the absolute configurations of all chiral carbon atoms in the structure of A from the preceding question.", "text_images": [{"file": "Exam-02-T10-2-1-prompt.png", "description": "The image shows two parallel reactions. In the upper reaction, the starting material is 'CC1=C(O[Si](C)(C)C(C)(C)C)CC[C@@H]1C[N+]#[C-]'. Above the arrow is “(a)tBuCOCl,DCM,r.t.” and below it “(b)AgBF_4,DCM,-78∘C”; the arrow leads to product A. In the lower reaction, '[R]C(Cl)=O' + '[Ar]C[N+]#[C-]' reacts under “base” to give product B."}], "full_mark": 4}, {"sub_id": "Exam-02-T10-3", "tags": ["2.2", "2.3"], "text_open": "10-2-3 Write the SMILES representation of another electrically neutral key intermediate Y, other than the type-X intermediate, involved in formation of B.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-02-T10-4", "tags": ["2.4", "2.3", "2.7"], "text_open": "10-2-4 If an appropriate amount of CH_{2}=CHCO_{2}Me is added to the reaction system that forms B, Y is trapped to form two products, C_{1} and C_{2}. Write their SMILES representations.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-02-T10-5", "tags": ["2.2"], "text_open": "10-3-1 Addition of other reagents changes the reaction pathway. Write the small-molecule by-product formed in the reaction below and the SMILES representation of another electrically neutral key intermediate, other than the type-X intermediate, involved in the reaction.", "text_images": [{"file": "Exam-02-T10-5-1-prompt.png", "description": "A continuous multistep reaction sequence. The substrate is `R_1COCl`; above the first arrow is R_2NC; above the second is `P(OR_3)_3`; and above the third is '[O-]/C([Ar])=C\\[N+]1=CC=CC=C1'. The final product is 'O=C([Ar])C1=C(C([R1])=C2C=CC=CN12)N[R2]'."}], "full_mark": 3}, {"sub_id": "Exam-02-T10-6", "tags": ["2.4", "2.2"], "text_open": "10-3-2 In the presence of Et_{3}N, EtO_{2}CCOCl, CyNC, P(Oi-Pr)_{3}, NCCH_{2}CN, and CS_{2} undergo a multicomponent reaction in MeCN. Write the SMILES representation of product D.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-02-T10-7", "tags": ["2.2", "2.4"], "text_open": "10-4 Under certain conditions, an α-carbonyl diazo compound can replace an acyl chloride in a similar reaction with an isocyanide. Write the SMILES representations of all intermediates and products in the scheme below. W and all three products contain a five-membered ring, but only the five-membered ring in E is aromatic; F also contains a six-membered ring.", "text_images": [{"file": "Exam-02-T10-7-1-prompt.png", "description": "A synthetic reaction scheme: '[R1]C(C(C([R1])=O)=[N+]=[N-])=O' is the starting material and is converted by an arrow to Z. Intermediate Z then reacts with `R_2NC` to form W. Intermediate **W** branches into three transformations: the upper branch reacts with `MeNHNH_2` to form E; the middle branch reacts with '[R1]C(C([R1])=O)=C=O' to form F; and the lower branch forms H under H^+ conditions."}], "full_mark": 10}]} | |
| {"problem_id": "Exam-03-T1", "header": "", "header_images": [], "subquestions": [{"sub_id": "Exam-03-T1-1", "tags": ["1.3"], "text_open": "1-3 The following structure is an As-S polyanion. Write its chemical formula, including the charge.", "text_images": [{"file": "Exam-03-T1-1-1-prompt.png", "description": "This schematic of an As-S polyanion shows a two-dimensional network of solid red circles (As atoms) and hollow white circles (S atoms) connected by black line segments. Its repeating motifs are cyclic clusters, each typically an eight-membered ring formed by alternating 4 red and four white nodes. Visually, each eight-membered ring contains a special terminal white node that is connected by only one line to a red node on the ring and points toward the center of the ring. The rings are not isolated; adjacent cyclic clusters are linked by individual bridging white nodes. Red nodes on adjacent clusters connect to a shared bridging white node between them. Every red node not connected to an inward-pointing white node participates in bridging, whereas a red node connected to an inward-pointing white node does not. Repetition in all directions forms a flexible, wavy-looking infinite two-dimensional network. At the image boundaries, some nodes have short black lines extending outward without terminating at another circle. Near the upper part of the image, the labels \"As\" and \"S\" point respectively to a specific red node and an adjacent white node."}], "full_mark": 2}, {"sub_id": "Exam-03-T1-2", "tags": ["2.4", "1.4"], "text_open": "1-4 Under illumination, Fe(CO)_5 reacts with (C_6F_5As)_4 to form complex X with ω(Fe) = 8.57%. 1-4-1 Write the chemical formula of complex X.", "text_images": [], "full_mark": 2}]} | |
| {"problem_id": "Exam-03-T3", "header": "Problem 3 Polymers. Global plastics production reached 3.907 hundred million tonnes in 2021, yet only 8.3% of plastics were recycled, resulting in severe accumulation of plastic waste and environmental problems. Developing next-generation chemically recyclable polymers that can be efficiently and selectively depolymerized back to their original monomers is therefore a viable route toward a closed-loop, sustainable plastics economy and a solution to end-of-life problems. Owing to their distinctive biorenewability and biodegradability, polyesters are among the most promising recyclable polymers.", "header_images": [], "subquestions": [{"sub_id": "Exam-03-T3-1", "tags": ["2.2", "2.3"], "text_open": "3-2 In 2024, researchers developed a polymer containing cyclobutane rings in its main chain. The polymer is thermally stable, resistant to hydrolysis, has tunable mechanical properties, and exhibits excellent chemical recyclability. It is prepared from monomer 4R-BL (R = Bu or Ph) using catalyst Y_1. Denote the resulting polymer as P(4^R-BL). 3-2-1 Draw and give the SMILES of the key intermediates in the initiation and chain-propagation steps of Y_1-catalyzed polymerization of 4R-BL. Ligands on the central Y atom that do not participate in the transformation may be omitted.", "text_images": [{"file": "Exam-03-T3-1-1-prompt.png", "description": "Structure of 4^R-BL: '[R]C(CC1O2)(C1)C2=O'."}, {"file": "Exam-03-T3-1-2-prompt.png", "description": "Structure of Y1: 'C[Si](C)(C[Y]123(O4CCCC4)OC5=C(C[N]1(CCO3C)CC6=C(O2)C(C(C)(C)C)=CC(C(C)(C)C)=C6)C=C(C(C)(C)C)C=C5C(C)(C)C)C'."}], "full_mark": 4}, {"sub_id": "Exam-03-T3-2", "tags": ["2.3", "2.8"], "text_open": "3-2-3 Adding one equivalent of isopropanol relative to Y_1 increases the initiation rate of polymerization. Explain why. (No more than 100 words.)", "text_images": [], "full_mark": 4}]} | |
| {"problem_id": "Exam-03-T5", "header": "Problem 5 Simple Substances, Complex Crystals", "header_images": [], "subquestions": [{"sub_id": "Exam-03-T5-1", "tags": ["1.1", "1.3"], "text_open": "5.1 Projections of a crystal along the b and c axes are shown below. Large black spheres represent A, white spheres represent B, and small black spheres represent O. 5.1.1 Name the coordination polyhedron of A and that of B.", "text_images": [{"file": "Exam-03-T5-1-1-prompt.png", "description": "Crystallographic projection of the unit cell viewed approximately along the b axis. The horizontal direction is the c axis and the vertical direction is the a axis. A black rectangle represents one unit cell, with a varying from 0 to 1 vertically and c from 0 to 1 horizontally. Blue, yellow, and red atoms are shown. Their approximate projected fractional coordinates (a,c), relative to the unit-cell rectangle, are as follows. Blue atoms: (0.85,0.50), (0.65,0.00), (0.30,0.00), and (0.15,0.50), together with periodic copies at the corresponding unit-cell boundaries. Yellow atoms: (0.85,0.00), (0.65,0.50), (0.30,0.50), and (0.15,0.00), together with periodic copies at c=1. Red atoms: approximately (1.20,0.50), (1.02,-0.17), (1.02,0.17), (1.02,0.83), (1.02,1.17), (0.70,0.00), (0.70,1.00), (0.48,-0.33), (0.48,1.33), (0.48,0.33), (0.48,0.67), (0.30,0.00), (0.30,1.00), (0.15,-0.17), (0.15,0.17), (0.15,0.83), and (0.15,1.17). Some atoms are partially obscured or overlap larger blue atoms. Coordinates outside the [0,1]×[0,1] rectangle represent visible periodic images. Red bonds connect neighboring red, blue, and yellow atoms."}, {"file": "Exam-03-T5-1-2-prompt.png", "description": "Crystallographic projection of the unit cell viewed approximately along the c axis. The horizontal direction is the a axis and the vertical direction is the b axis. A black rectangle represents one unit cell, with a varying from 0 to 1 horizontally and b from 0 to 1 vertically. Blue, yellow, and red atoms are shown. Their approximate projected fractional coordinates (a,b), relative to the unit-cell rectangle, are as follows. Blue atoms: (0.35,0.75), (0.85,0.75), (0.15,0.25), and (0.65,0.25). Yellow atoms: (0.85,0.75) and (0.15,0.25). Red atoms: approximately (0.30,1.08), (0.80,1.08), (0.05,0.92), (0.55,0.92), (1.05,0.92), (0.30,0.75), (0.80,0.75), (0.05,0.50), (0.55,0.50), (1.05,0.50), (0.30,0.25), (0.80,0.25), (0.05,0.08), (0.55,0.08), and (1.05,0.08), together with visible periodic images outside the unit-cell boundaries. Several red atoms visually overlap blue or yellow atoms. Red bonds connect neighboring red, blue, and yellow atoms."}], "full_mark": 4}, {"sub_id": "Exam-03-T5-2", "tags": ["1.3"], "text_open": "5.1.2 Using the coordination numbers obtained above, write the chemical formula of the crystal.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-03-T5-3", "tags": ["1.3"], "text_open": "5.1.3 State the lattice type of the crystal.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-03-T5-4", "tags": ["1.3"], "text_open": "5.2 Consider an ideal hexagonal crystal composed of X and O. The coordinates of three X atoms in the unit cell are (1/2, 1/2, 0), (0, 1/2, 1/3), and (1/2, 0, 2/3). The unit cell may be viewed as containing helical X-O-X-O chains extending along c. For example, when viewed along c, rotate the X atom at (1/2, 1/2, 0) clockwise by 120° about the line a = 1/3, b = 2/3 and then translate it upward by 1/3 of the c-axis length; it reaches (0, 1/2, 1/3). Applying the same clockwise rotation and upward translation to the X atom at (0, 1/2, 1/3) gives (1/2, 1, 2/3). Applying them once more gives (1/2, 1/2, 1), equivalent to (1/2, 1/2, 0). This is illustrated below; the numbers in the figure are the c-axis coordinates of the X atoms. 5.2.1 Call the helical sense described above left-handed. Correspondingly, a helix that rotates counterclockwise while advancing along c is called right-handed. The unit cell contains another helical chain whose axis lies at a = 2/3, b = 1/3. State whether this chain is left- or right-handed.", "text_images": [{"file": "Exam-03-T5-4-1-prompt.png", "description": "Schematic of the helical X-O-X-O chain, projected along c. The main outline is a solid parallelogram. Slightly left of center inside it is a solid black point surrounded by a circular path bearing three clockwise arrows. Three light-gray circles with black outlines are evenly spaced on this path. Their labels increase clockwise: \"0\" at the lower right, \"1/3\" at the lower left, and \"2/3\" at the top. At the lower left of the image is an axis legend: a horizontal right-pointing arrow labeled a, an arrow slanting up and left labeled b, and the letter c at their common origin."}], "full_mark": 2}, {"sub_id": "Exam-03-T5-5", "tags": ["1.1", "1.3"], "text_open": "Structural background: A crystal composed of C, D, and O has a body-centered lattice and formula $CDO_x$, with $a=b=433\\ \\mathrm{pm}$ and $c=664\\ \\mathrm{pm}$. C occupies the cell corners and body center, and one D atom has fractional coordinates $(0.50,0,0.75)$.\n5.3.2 Considering only C and D atoms, identify the polyhedron formed by D atoms around C and the polyhedron formed by C atoms around D.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-03-T5-6", "tags": ["1.3"], "text_open": "5.3.3 One O atom in the unit cell has fractional coordinates (0.138, 0.260, 0.131). Calculate its distance to the nearest C atom, d_{O-C}, and to the nearest D atom, d_{O-D}.", "text_images": [], "full_mark": 4}]} | |
| {"problem_id": "Exam-03-T6", "header": "Problem 6 Exploring Our Little Broken Planet. Rocks are generally mixtures of one or more minerals and can be classified as igneous, metamorphic, or sedimentary. Geological processes can transform each type into the others. Igneous rocks form when magma cools. Magma is a complex mixture without a fixed melting point; melting occurs over a temperature range, beginning at the solidus and becoming complete at the liquidus. Most rocks undergo partial melting. During magma cooling and crystallization, changes in physical and chemical conditions cause minerals of different compositions to crystallize in sequence. Consequently, the solid and residual melt often have different chemical compositions. This phenomenon is called chemical differentiation. Temperature, chemical composition, and gravity are important factors affecting chemical differentiation. Metamorphic rocks form when pre-existing rocks in the crust—igneous, sedimentary, or metamorphic—undergo changes in mineral composition, texture, structure, and chemical composition under the influence of temperature, pressure, and chemically active fluids. Metamorphism generally occurs below the solidus temperature of the rock (600-1000 ℃), so melting does not occur in the great majority of cases; pressure, however, has no strict upper limit.", "header_images": [{"file": "Exam-03-T6-header-1.png", "description": "Pressure-temperature (P-T) phase diagram labeled \"A\". The horizontal axis is temperature $T$ in $^\\circ C$, ranging from $1000\\sim2000^\\circ C$. The vertical axis is pressure $P$ in $GPa$, ranging from $0\\sim10GPa$. The diagram contains a curve labeled solidus, which is the thickest curve, a curve labeled liquidus to its right, and several nearly parallel thin curves between them. All curves trend downward as temperature and pressure increase and converge in the high-pressure, high-temperature region near $9GPa$ and $2000^\\circ C$."}], "subquestions": [{"sub_id": "Exam-03-T6-1", "tags": ["4.2"], "text_open": "6-1 As magma passes through the crust, some cools and solidifies within the crust to form intrusive rocks, while some reaches the surface to form extrusive rocks. Crystallinity mainly reflects the extent to which a substance has transformed from a disordered amorphous state into an ordered crystalline state. A rock has high crystallinity if its mineral grains are large and well crystallized, with regular and complete particle arrangements and few defects or disorders; otherwise, it has low crystallinity. State whether intrusive or extrusive rocks have higher crystallinity and briefly explain why. (No more than 50 words.)", "text_images": [], "full_mark": 2}]} | |
| {"problem_id": "Exam-04-T1", "header": "Problem 1 Minerals and Chemistry. In the 18 century, Austrian chemist F. J. Müller discovered a substance X resembling antimony in a gold-bearing ore. It was not until 16 years later that German mineralogist M. H. Klaproth formally discovered and named it. Roasting an X-containing mineral A in air gives B and elemental silver. Theoretically, roasting each 1.000 g of A produces 0.628 g of silver.", "header_images": [], "subquestions": [{"sub_id": "Exam-04-T1-1", "tags": ["2.1", "2.2"], "text_open": "It is known that element $X=Te$, mineral $A=Ag_2Te$, and product $B=TeO_2$.\n1-2 B can be further converted into other compounds of X. (i) B dissolves in a concentrated solution of C to give a solution containing D, while base G is also formed. D reacts with E in a 1:2 ratio, precipitating elemental X and producing C and F. (ii) B reacts with base G to form H. A solution in glacial acetic acid of the acid corresponding to the anion of excess H reacts with F in a 1:4 ratio to give the X-containing compound I and the X-free compound J. Reaction of I with G precipitates elemental X and also forms F and H. (iii) Under acidic conditions, B reacts with hydrogen peroxide to give K. K is reduced by a common monoprotic strong acid L, producing B, X, and elemental M. Interestingly, M reacts with F to form J, a reaction commonly used to titrate M in analytical chemistry. C is a binary compound, and C, D, E, F, H, I, and J are all 2:1 salts. Determine the formulas of compounds D, E, I, and J.", "text_images": [], "full_mark": 10}, {"sub_id": "Exam-04-T1-2", "tags": ["3.4", "3.6"], "text_open": "1-3 The standard electrode potential for conversion of K to B is $+1.02\\mathrm{V}$, that of $\\mathrm{Au^{3+}/Au}$ is $+1.50\\mathrm{V}$, and the logarithmic cumulative stability constant for $\\mathrm{Au^{3+}+4Cl^- = AuCl_4^-}$ is $\\lg \\beta_4 = 25.6$. By a brief calculation, determine whether a mixture of K and concentrated hydrochloric acid can dissolve gold at room temperature.", "text_images": [], "full_mark": 5}]} | |
| {"problem_id": "Exam-04-T5", "header": "Rutile is a typical tetragonal crystal structure. One unit cell is shown, with large light-blue spheres representing Ti and small red spheres representing O. This structure occurs widely among transition-metal compounds and gives rise to many functional materials.", "header_images": [{"file": "Exam-04-T5-header-1.png", "description": "Three-dimensional ball-and-stick model of a TiO_2 unit cell. Red a, green b, and blue c axes are labeled at the lower left. Large light-blue spheres and small red spheres are connected by light-blue bonds inside a tetragonal unit-cell frame outlined in black. The large light-blue spheres occupy the corners and center at (0, 0, 0) and (0.5, 0.5, 0.5). Nearby red spheres surround them in coordination environments at (0.305, 0.305, 0), (0.695, 0.695, 0), (0.805, 0.195, 0.5), and (0.195, 0.805, 0.5)."}], "subquestions": [{"sub_id": "Exam-04-T5-1", "tags": ["1.3"], "text_open": "5.2 Potassium tetrachlorocuprate dihydrate, $\\mathrm{K_2CuCl_4·2H_2O}$, is a bright-blue deliquescent crystal used mainly as a copper-ion source and catalyst in laboratories and industry. Its crystal symmetry resembles that of rutile and may be compared with the structure above. $\\mathrm{[Cu(H_2O)Cl_4]^(2-)}$ units occupy the corners and body center, while $\\mathrm{K^+}$ ions lie on the c = 0.250 and c = 0.750 planes. Cu-O bonds are parallel to c, and each Cu-Cl bond is parallel to a corresponding basal-plane diagonal. All identical atoms except Cl have equivalent chemical environments. The cell parameters are a = b = 745 pm, c = 788 pm, and α = β = γ = 90°. 5-2-1 Two Cl atoms have coordinates (0.720, 0.280, 0) and (0.720, 0.280, 0.500). Give the coordinates of the remaining 6 Cl atoms.", "text_images": [{"file": "Exam-04-T5-1-1-prompt.png", "description": "Unit-cell model of $\\mathrm{K_2CuCl_4·2H_2O}$ containing blue, purple, green, red, and white spheres. Blue spheres occupy the corners and center. Purple spheres lie on the four side faces, two vertically separated on each face. Green spheres lie on the top, bottom, and middle planes, four evenly arranged at the upper-left, lower-left, upper-right, and lower-right positions of each plane. Red spheres lie above and below each blue sphere, approximately at the same height as the purple spheres. White spheres are bonded beside the red spheres, two per red sphere, forming V shapes."}], "full_mark": 6}, {"sub_id": "Exam-04-T5-2", "tags": ["1.3"], "text_open": "5.3 See the color figure for this part. A tetragonal crystal composed of Na, H, and Ru also has a rutile-like structure. Its projection along c is shown. Yellow spheres are Na, blue spheres Ru, and pink and white spheres H. Light-yellow Na lies in the c = 0 plane and dark-yellow Na in the c = 1/2 plane; light-blue Ru lies in c = 0 and dark-blue Ru in c = 1/2. White H lies in c = 0, while pink H is distributed within the unit cell. All Ru atoms have equivalent environments. H occurs with coordination numbers 3, 4, and 5, with more four-coordinate H than three-coordinate H; Na has only one coordination number. All atoms shown lie partly or wholly inside the unit cell. Hint: mirror planes occur at c = 0 and c = 1/2. 5-3-1 Give the chemical formula of the crystal.", "text_images": [{"file": "Exam-04-T5-2-1-prompt.png", "description": "Color ball-and-stick projection with a thin dark-gray square marking the unit-cell or projection boundary and no axes or text labels. 40 distinguishable spheres are visible: 16 yellow or gold, 14 light pink, 6 white, and 4 dark blue. Some lie partly outside the square. The yellow/gold spheres are distributed around the top, sides, center, and bottom; the light-pink spheres surround and connect the blue and yellow spheres; the six white spheres occur mainly near the upper right, center right, lower left, and bottom; and the four dark-blue spheres lie at left center, upper right, lower left, and lower right-center. Short rods connect the spheres, with colors generally transitioning between adjacent atoms, such as yellow-pink, yellow-white, pink-blue, and white-blue. Together they form a continuous 3-dimensional network passing through the square boundary, with some spheres and bonds extending outside it. The numbered sphere positions are explicitly enumerated as follows: yellow/gold 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, and 16; light pink 1 through 14; white 1 through 6; and dark blue 1 through 4."}], "full_mark": 2}, {"sub_id": "Exam-04-T5-3", "tags": ["1.3"], "text_open": "5-3-2 Name the coordination polyhedron around Ru in the crystal.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-04-T5-4", "tags": ["1.3"], "text_open": "5-3-3 Give the number ratio of three-, 4-, and five-coordinate H atoms in the crystal.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-04-T5-5", "tags": ["1.3"], "text_open": "5-3-4 Give the coordination number of Na in the crystal.", "text_images": [], "full_mark": 2}]} | |
| {"problem_id": "Exam-04-T7", "header": "Problem 7 Microbial Fuel Cells. Data that may be used in this problem, all at 25 ℃ unless otherwise stated: $\\varphi^\\Theta_{\\mathrm{HCO_3^-}/\\mathrm{CH_3COO^-}} = 0.187\\ \\mathrm{V}$; $\\varphi^\\Theta_{\\mathrm{K_3Fe(CN)_6/K_4Fe(CN)_6}} = 0.361\\ \\mathrm{V}$; $\\mathrm{p}K_{\\mathrm{a1},\\mathrm{a2},\\mathrm{a3}}(\\mathrm{H_3PO_4}) = 2.16,\\ 7.21,\\ 12.3$; $\\mathrm{p}K_\\mathrm{a}(\\mathrm{CH_3COOH}) = 4.76$; $\\mathrm{p}K_{\\mathrm{a1},\\mathrm{a2}}(\\mathrm{H_2CO_3}) = 6.37,\\ 10.33$; $\\Delta_\\mathrm{r}H_\\mathrm{c}(\\mathrm{CH_3COOH}) = -875\\ \\mathrm{kJ\\cdot mol^{-1}}$; $\\Delta_\\mathrm{r}H_\\mathrm{c}(n-\\mathrm{C_3H_7COOH}) = -2184\\ \\mathrm{kJ\\cdot mol^{-1}}$. A microbial fuel cell (MFC) uses microorganisms as catalysts to convert the chemical energy of organic and inorganic substances directly into electrical energy. MFCs oxidatively decompose organic matter in wastewater, such as glucose and acetic acid, while generating electricity. This treatment both reduces the chemical oxygen demand of the wastewater and converts organic matter into useful energy.", "header_images": [], "subquestions": [{"sub_id": "Exam-04-T7-1", "tags": ["3.4", "3.6"], "text_open": "The electrode-solution compositions, as analytical concentrations, are: electrode A, 0.0223 M $Na_2HPO_4$, 0.0359 M $NaH_2PO_4$, 0.0504 M $K_3[Fe(CN)_6]$, and 0.0167 M $K_4[Fe(CN)_6]$; electrode B, 0.0223 M $Na_2HPO_4$, 0.0359 M $NaH_2PO_4$, 0.0132 M $CH_3COONa$, and 0.0075 M $NaHCO_3$. The pH at A is 7.00 and B may be treated as having the same pH.\n7-1-2 Assuming the pH values at electrodes A and B are approximately equal, calculate the electrode potentials of A and B under these conditions. Give 3 significant figures.", "text_images": [], "full_mark": 4}]} | |
| {"problem_id": "Exam-04-T10", "header": "Problem 10 Hydrazones and Their Salts: Our Advantage", "header_images": [], "subquestions": [{"sub_id": "Exam-04-T10-1", "tags": ["2.2", "2.3"], "text_open": "Reaction background: The reaction shown below first converts the ketone into a hydrazone intermediate and then effects elimination at the carbonyl $\\alpha$ carbon. The “second step” is conversion of the hydrazone intermediate to the final product under $H^+/H_2O$.\n10-1-2 Explain why the hydrazone is introduced, and give the SMILES of one key intermediate in the second step.", "text_images": [{"file": "Exam-04-T10-1-1-prompt.png", "description": "Two-step reaction. The leftmost structure is '[*]C1C([*])CC[C@]2([H])[C@@H](Br)C(CC[C@@]21C)=O'. Each * is attached to a wavy line denoting an additional ring system that does not participate and need not be considered. Reaction with ethoxycarbonyl hydrazine 'NNC(OCC)=O' gives '[*]C1C(CC[C@@]2([C@H](/C(CC[C@@]21C)=N\\NC(OCC)=O)Br)[H])[*]', with the same meaning for * and the upper-right wavy line. H^+/H_2O then gives '[*]C1C(CCC2=CC(CC[C@@]21C)=O)[*]', again with the same meaning for *."}], "full_mark": 3}, {"sub_id": "Exam-04-T10-2", "tags": ["2.3"], "text_open": "10-2 A hydrazone can be reduced under basic conditions to a C-C double bond. 10-2-1 A dianion is known to appear during the reaction. Which statements are correct? (a) The first proton removed is on N. (b) The first proton removed is on C. (c) The reaction rate is controlled by departure of Ts^-. (d) The reaction rate is controlled by the second deprotonation. (e) It cannot be determined.", "text_images": [{"file": "Exam-04-T10-2-1-prompt.png", "description": "Reaction of hydrazone '[R1]/C(C[R2])=N/NS(C1=CC=C(C)C=C1)(=O)=O' under 1) Base and 2) TMSCl gives the C-C double-bond product '[R1]/C([Si](C)(C)C)=C/[R2]'."}], "full_mark": 1}, {"sub_id": "Exam-04-T10-3", "tags": ["2.7", "2.8"], "text_open": "Consider the reaction shown; Quench means quenching. With potassium carbonate as base, product A forms. With LDA, A and a small amount of B form. The substituent R also affects the outcome. 10-2-2 With LDA, arrange the following R groups in decreasing B yield/A yield: (a) R = H; (b) R = Me; (c) R = F.", "text_images": [{"file": "Exam-04-T10-3-1-prompt.png", "description": "Multiple pathways from common substrate 'O=C(CC(C)(C)C1)C([R])C1=O'. Downward reaction with TsNHNH_2, 1 eq, gives bracketed intermediate M. M gives A, 'O=C1CC(C)(C)CC=C1[R]', under 1) K_2CO_3, heat; 2) Quench. Under 1) LDA; 2) Quench, M gives both A and B, where B is 'O=C1CC(C)(C)C=CC1[R]'. SMILES/structure strings shown: 'O=C1CC(C)(C)CC=C1[R]'."}], "full_mark": 1}, {"sub_id": "Exam-04-T10-4", "tags": ["2.2", "2.4"], "text_open": "10-3 Similarly, the N-N bond of a hydrazone can be converted into a diazo group, which undergoes many interesting reactions. 10-3-1 Give the SMILES of intermediate C.", "text_images": [{"file": "Exam-04-T10-4-1-prompt.png", "description": "Two-step reaction. The left substrate is 'BrC1=C(OC)C(OC)=C(C=NNS(C2=CC=C(C)C=C2)(=O)=O)C(COC)=C1'. It reacts with 'BrC1=C(OC)C=CC(B(O)O)=C1' under K_2CO_3, 1,4-dioxane, 110 ℃ to give unknown C. C then gives 'BrC1=C(OC)C(OC)=C(CC2=CC=C(OC)C(Br)=C2)C(CO)=C1' under BBr_3, CH_2Cl, then THF/H_2O."}], "full_mark": 2}, {"sub_id": "Exam-04-T10-5", "tags": ["1.2", "2.4"], "text_open": "10-3-2 Based on the reaction above, give the SMILES of products D, E, and F. Include stereochemistry.", "text_images": [{"file": "Exam-04-T10-5-1-prompt.png", "description": "Reaction producing D. Substrate '[R2][C@@]([R3])([BPin])[C@](C/1)([R1])CC1=N\\NC2=C(C=C(C=C2C)C)C' gives unknown D under Cs_2CO_3, 3.0 equiv, in 1,4-dioxane at 100 ℃."}, {"file": "Exam-04-T10-5-2-prompt.png", "description": "Reaction producing E. 'CC1=CC=C(S(NN=C2[C@@H](CCN=[N+]=[N-])CCCC2)(=O)=O)C=C1' plus '[R1]B(O)O' gives unknown E under K_2CO_3, 3.0 equiv, in 1,4-dioxane at 150 ℃."}, {"file": "Exam-04-T10-5-3-prompt.png", "description": "Reaction producing F. 'CC1=CC=C(S(NN=C2CCN(CC#N)CC2)(=O)=O)C=C1' plus '[R2]/C=C/B(O)O' gives unknown F under K_2CO_3, 2.0 equiv, in 1,4-dioxane with MW at 120 ℃."}], "full_mark": 6}, {"sub_id": "Exam-04-T10-6", "tags": ["2.2"], "text_open": "10-4 Under suitable conditions, a hydrazone can generate a highly reactive carbene. When first proposed, the reaction shown was called an \"unusual carbon-skeleton rearrangement\". After spectroscopic structure determination and isotope labeling, a mechanism was proposed that includes an intramolecular ring closure and a \"questionable\" addition to an sp-hybridized carbon. 10-4-1 Give the SMILES of 2 neutral key intermediates.", "text_images": [{"file": "Exam-04-T10-6-1-prompt.png", "description": "One-step reaction. Reactant 'CC1=CC=C(C=C1)S(N/N=C(C2C(/C3=N/NS(C4=CC=C(C=C4)C)(=O)=O)C2)\\C5C3C5)(=O)=O' under excess NaH in diglyme at 190 ℃ gives 'C12=CC=NN1C=CC3C2C3'."}], "full_mark": 4}, {"sub_id": "Exam-04-T10-7", "tags": ["2.3", "3.7"], "text_open": "10-4-2 When the two three-membered rings of the reactant are on the same side, the yield is 60%; when on opposite sides, it is 40%. Explain this in terms of activation energy.", "text_images": [], "full_mark": 1}]} | |
| {"problem_id": "Exam-05-T1", "header": "Problem 1 Chemistry of Element M. Historically, the mining and use of minerals largely determined the progress of elemental research. Element M and another element are named after the same country. M is about as abundant as nitrogen in Earth's crust but is extremely dispersed. Minerals containing M alone are uncommon, but binary compound A of M and element N often occurs with sphalerite. A contains 48.20% M by mass and is widely used in optoelectronics.", "header_images": [], "subquestions": [{"sub_id": "Exam-05-T1-1", "tags": ["2.1"], "text_open": "Given $M=Ga$, $B=GaCl$, $C=GaCl_3$, and $N=As$.\n1-1-2 Write the equation for the reaction of B with N.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-05-T1-2", "tags": ["1.4", "1.5"], "text_open": "1-1-3 D can also be prepared in benzene without heating. Explain why D forms readily in benzene by giving the SMILES of the relevant structure. No more than 100 words.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-05-T1-3", "tags": ["1.4", "2.4"], "text_open": "1-2 At 250 K, chloride C reacts with $\\mathrm{Me_3SiH}$ to form E. E commonly exists as a dimer in which all chlorine atoms have equivalent environments. 1-2-1 Give the SMILES of the actual structure of E.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-05-T1-4", "tags": ["1.4", "2.4", "2.7"], "text_open": "1-2-2 At low temperature, the different reactions of E with ammonia and trimethylamine attracted attention. IR and other data show that the products have similar compositions but different coordination numbers of M; the coordination number in the ammonia product equals that in the E dimer. Neither product is dimeric or polymeric. Give the product SMILES for reaction with trimethylamine and ammonia, and explain the difference in no more than 20 words.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-05-T1-5", "tags": ["3.2"], "text_open": "1-3 The physical properties of element M are equally distinctive. At standard atmospheric pressure, the thermodynamic data for M and related compounds are given below. Assume that all $\\Delta H$ and $\\Delta S$ values are independent of temperature. Throughout the temperature range considered, the most stable oxide of M remains in the solid state.\n| Substance | $\\Delta_\\mathrm{f} H_m^\\circ \\, (\\mathrm{kJ\\cdot mol^{-1}})$ | $S_m^\\circ \\, (\\mathrm{J\\cdot mol^{-1}\\cdot K^{-1}})$ |\n|---|---:|---:|\n| $M(s)$ | 0 | 40.9 |\n| $M(l)$ | 5.6 | 56.0 |\n| $M(g)$ | 277.0 | 169.0 |\n| $O_2(g)$ | 0 | 205.2 |\n| Most stable oxide of M $(s)$ | -1089.0 | 88.0 |\n1-3-2 Determine the temperature range in which M reacts spontaneously with oxygen at standard atmospheric pressure.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-05-T1-6", "tags": ["3.2", "3.4"], "text_open": "1-3-3 Calculate the ambient pressure when the boiling point of M is 2200 K.", "text_images": [], "full_mark": 3}]} | |
| {"problem_id": "Exam-05-T5", "header": "Problem 5 Structural Chemistry of Metal M. Formulas must use M and should be in approximately simplest ratios.", "header_images": [], "subquestions": [{"sub_id": "Exam-05-T5-1", "tags": ["1.3"], "text_open": "5-1 The crystal structure of chloride A of M and its projection along c are shown; large spheres are M and small spheres Cl. 5-1-1 Name the coordination polyhedron formed by Cl around one M.", "text_images": [{"file": "Exam-05-T5-1-1-prompt.png", "description": "Perspective view of crystal A inside a gray dashed hexagonal frame. Large dark-gray spheres are M and small light-gray spheres Cl. 2 M atoms at upper left and lower right share two bridging Cl atoms; each M also connects radially to 7 terminal Cl atoms. Around each M are three upper, three lower, and three equatorial Cl atoms. An axis triad has c upward, b lower right, and a left."}, {"file": "Exam-05-T5-1-2-prompt.png", "description": "Projection of A along c within a dashed parallelogram. Large spheres are M and small spheres Cl. Two M atoms, one at (1/3,2/3), share two bridging Cl atoms. Each M also has 4 radial terminal Cl atoms in projection, with three nearer and three farther Cl neighbors overall. The axis marker has a rightward, b upper-left, and c out of the page."}], "full_mark": 2}, {"sub_id": "Exam-05-T5-2", "tags": ["1.3"], "text_open": "5-1-2 Give the number of distinct Cl environments and the coordination number of Cl.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-05-T5-3", "tags": ["1.3"], "text_open": "5-1-3 Give the crystal system and formula of A.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-05-T5-4", "tags": ["1.3"], "text_open": "5-2 M forms complex compounds with Ni and O. Projections of a mixed oxide X along b and c are shown; large spheres are M, small spheres O, and medium spheres Ni. Give its formula.", "text_images": [{"file": "Exam-05-T5-4-1-prompt.png", "description": "Crystallographic projection of the unit cell of composite oxide X viewed along b. The vertical a-axis and horizontal c-axis each range from 0 to 1. Large spheres are M, medium spheres Ni, and small spheres O. Approximate fractional projection coordinates (a,c) are: M, (1.00,0.25), (0.00,0.75), (0.50,0.25), (0.50,0.75); Ni, the four vertices (0.00,0.00), (1.00,0.00), (0.00,1.00), (1.00,1.00), the edge midpoints (1.00,0.50), (0.00,0.50), (0.50,0.00), (0.50,1.00), and the face center (0.50,0.50); O, approximately (0.76,0.04), (0.68,0.04), (0.42,0.25), (0.12,0.25), (0.76,0.50), (0.68,0.50), (0.30,0.50), (0.24,0.50), (0.88,0.75), (0.58,0.75), (0.30,0.96), and (0.24,0.96). The lower-left direction marker points a upward and c rightward."}, {"file": "Exam-05-T5-4-2-prompt.png", "description": "Crystallographic projection of the unit cell of composite oxide X viewed along c. The horizontal a-axis and vertical b-axis each range from 0 to 1. Large spheres are M, medium spheres Ni, and small spheres O. Approximate fractional projection coordinates (a,b) are: M, upper-left (0.02,0.90), lower-right (0.98,0.10), and the two vertically overlapping central sites (0.50,0.55) and (0.50,0.45); Ni, near the edge centers (0.00,0.47), (1.00,0.47), (0.53,1.00), and (0.53,0.00); O, next to boundary Ni sites at approximately (0.08,0.47), (0.92,0.47), (0.45,0.98), and (0.61,0.02), plus interior sites at (0.30,0.70), (0.78,0.78), (0.22,0.22), and (0.70,0.30). The lower-left direction marker points b upward and a rightward."}], "full_mark": 2}, {"sub_id": "Exam-05-T5-5", "tags": ["1.3"], "text_open": "5-3 Compound Y, prepared from an oxide of M, NiO, and B_2O_3, contains M, Ni, O, and B and has a primitive monoclinic lattice. It consists of M cations, Ni^2+, and a network borate built from spiro units sharing sp^3-hybridized B atoms. The network contains [B_3O_3] six-membered and [B_4O_4] 8-membered rings. In every [B_4O_4], all four B atoms are four-coordinate and shared with neighboring [B_3O_3] rings; two ring O atoms coordinate one metal each and two coordinate two metals each. In every [B_3O_3], two B atoms are four-coordinate and shared with adjacent rings, while one is three-coordinate. The three ring O atoms each coordinate one metal; half the terminal O atoms coordinate two metals and half three. A monoclinic unit cell has [B_4O_4] at the corners and body center, each connected to four of eight neighboring [B_4O_4] rings through two [B_3O_3] rings. 5-3-1 Determine the number of B atoms per unit cell.", "text_images": [{"file": "Exam-05-T5-5-1-prompt.png", "description": "Schematic of two [B_4O_4] rings connected through two vertex-sharing [B_3O_3] rings, topologically like 'C1CCCC2(CCCC3(CC4(CCCCCCC4)CCC3)C2)CCC1'."}], "full_mark": 2}, {"sub_id": "Exam-05-T5-6", "tags": ["1.3"], "text_open": "5-3-2 Give the simplest formula of the borate anion.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-05-T5-7", "tags": ["1.4", "2.1"], "text_open": "5-3-3 M is 9-coordinate, Ni^2+ five-coordinate, and M:Ni = 1:1. Give the formula of Y.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-05-T5-8", "tags": ["1.4"], "text_open": "5-3-4 State the oxidation state of M.", "text_images": [], "full_mark": 1}]} | |
| {"problem_id": "Exam-05-T9", "header": "NHC Chemistry", "header_images": [], "subquestions": [{"sub_id": "Exam-05-T9-1", "tags": ["2.2"], "text_open": "9.1 The benzoin condensation is familiar. Give the SMILES of the four key intermediates in the catalytic cycle shown.", "text_images": [{"file": "Exam-05-T9-1-1-prompt.png", "description": "Under NEt_3, 'CC[N+]1=CSC(CCO)=C1' gives 'CC[N+]1=[C-]SC(CCO)=C1'. Reaction with '[H]C(C1=CC=CC=C1CCC(C)=O)=O' gives A; proton transfer gives B; B gives C; proton transfer gives D; D gives 'O=C1C(O)(C)CCC2=CC=CC=C21' and regenerates 'CC[N+]1=[C-]SC(CCO)=C1'. SMILES/structure strings shown: 'CC[N+]1=[C-]SC(CCO)=C1'."}], "full_mark": 4}, {"sub_id": "Exam-05-T9-2", "tags": ["2.2"], "text_open": "9.3 Formaldehyde reacts poorly directly with NHCs. The reaction shown introduces a fragment equivalent to formaldehyde. Give the SMILES of the three key intermediates between the five-carbon aldose substrate and intermediate X.", "text_images": [{"file": "Exam-05-T9-2-1-prompt.png", "description": "'O[C@H]1[C@@H](O)[C@H](O)C(O)OC1' and 'O=C(C1=CC=CC=C1)/C=C/C2=CC=CC=C2' give 'C[N+]1=CSC(C)=C1C.[I-]' under 20 mol% 'O=C(C1=CC=CC=C1)CC(C([H])=O)C2=CC=CC=C2', 20 mol% K_2CO_3, MeCN, 130 °C, μW, 10 min. Intermediate X: 'O/C([H])=C1N(C)C(C)=C(C)S\\1'."}], "full_mark": 3}, {"sub_id": "Exam-05-T9-3", "tags": ["2.2"], "text_open": "9.4 When the reactant is not an aldehyde, NHCs can catalyze other reactions. The following reaction of an α-hydroxy ketone is NHC-catalyzed; the catalytic cycle is shown, and G carries one positive charge. 9.4.1 Compare the molecular formulas of reactants and products and identify the small-molecule product Y.", "text_images": [{"file": "Exam-05-T9-3-1-prompt.png", "description": "Reaction of 'O=C(C(C)(C)O)/C=C/C1=C(C)C=C(C)C=C1C' and 'CC1NCCCC1' under 10 mol% of the listed NHC catalyst, 10 mol% of the listed hydroxamic reagent, and 20 mol% DBU, DCM, room temperature gives 'C[C@@H]1N(C(CCC2=C(C)C=C(C)C=C2C)=O)CCCC1' and 'C[C@H]1NCCCC1'. SMILES/structure strings shown: 'CC1=CC(C)=CC(C)=C1[N+]2=CN3C(CCC3)=N2.[O-]Cl(=O)(=O)=O', 'O=C(CO1)N(O)[C@@H]([C@@H]1C2)C3=C2C=CC=C3'."}, {"file": "Exam-05-T9-3-2-prompt.png", "description": "Catalytic cycle: 'C[C@@H]1NCCCC1' reacts with H to give 'O=C(N1[C@@H](C)CCCC1)CCC2=C(C)C=C(C)C=C2C' and the cyclic hydroxamic product. The azolium catalyst and DBU form the NHC. The cyclic hydroxamic product and G give H and regenerate the NHC after loss of [H^+]. The NHC reacts with the α-hydroxy ketone to give E; E gives Y and F; protonation of F gives G. SMILES/structure strings shown: 'O=C1N(O)[C@H](C(C=CC=C2)=C2C3)[C@H]3OC1', 'O=C1N(O)[C@H](C(C=CC=C2)=C2C3)[C@H]3OC1', 'CC1=CC(C)=CC(C)=C1[N+]2=CN3C(CCC3)=N2', 'CC1=CC(C)=CC(C)=C1[N+]2=[C-]N3C(CCC3)=N2', 'CC1=CC(C)=CC(C)=C1[N+]2=[C-]N3C(CCC3)=N2', 'CC1=CC(C)=CC(C)=C1[N+]2=[C-]N3C(CCC3)=N2', 'O=C(C(C)(C)O)/C=C/C1=C(C)C=C(C)C=C1C'."}], "full_mark": 1}, {"sub_id": "Exam-05-T9-4", "tags": ["2.2", "2.3"], "text_open": "9.4.2 Give the SMILES of the four key intermediates E, F, G, and H.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-05-T9-5", "tags": ["2.2", "2.3"], "text_open": "9.5 With a vicinal diketone substrate, the reaction mode changes again. Give the SMILES of the four key intermediates. NEt_3 acts only as a base with the thiazolium ion; the resulting NHC first reacts with the vicinal diketone. Two addition-elimination events occur and each is treated as one step; their tetrahedral intermediates need not be shown. The NHC formed in the initial acid-base step also need not be shown.", "text_images": [{"file": "Exam-05-T9-5-1-prompt.png", "description": "'[Ar]C(C=C)=O' and 'O=C(C(C1=CC=CC=C1)=O)C2=CC=CC=C2' react under 'C[N+]1=CSC2=C1CCCCC2' and NEt_3 to give '[Ar]C(C(C(C1=CC=CC=C1)=O)CC(C2=CC=CC=C2)=O)=O'."}], "full_mark": 7}]} | |
| {"problem_id": "Exam-05-T10", "header": "Problem 10 Sulfur-Nitrogen Chemistry", "header_images": [], "subquestions": [{"sub_id": "Exam-05-T10-1", "tags": ["1.5", "1.6"], "text_open": "10.1 The sulfilimine group, -NS(O)R_2, is an S,N substituent generally considered strongly electron-withdrawing. Calculations show that with a π acceptor attached, N-S contains a σ bond and a π bond strongly polarized toward N. 10.1.1 Give all resonance forms with single S-O bonds and identify the form closest to reality. Show exact connectivity.", "text_images": [{"file": "Exam-05-T10-1-1-prompt.png", "description": "'O=C(N=S(C)(C)=O)CC'."}], "full_mark": 4}, {"sub_id": "Exam-05-T10-2", "tags": ["2.3", "2.7"], "text_open": "10.1.2 Explain the regioselectivity and give the major-product SMILES with alkene geometry. No more than 100 words.", "text_images": [{"file": "Exam-05-T10-2-1-prompt.png", "description": "Reaction of 'O=S(C)(C)=NC#CC1=CC=CC=C1' with benzoic acid, 'OC(C1=CC=CC=C1)=O', gives the displayed product 'O=C(C1=CC=CC=C1)O/C(N=S(C)(C)=O)=C\\C2=CC=CC=C2'. A wavy bond between phenyl and the alkene means the actual E/Z configuration is unspecified in the question and must be inferred; geometry implied by the displayed SMILES is not necessarily correct."}], "full_mark": 2}, {"sub_id": "Exam-05-T10-3", "tags": ["2.2", "2.4"], "text_open": "10.1.3 Give the product and key intermediates in the actual mechanism. All key intermediates are neutral; the product contains a non-aromatic carbocycle.", "text_images": [{"file": "Exam-05-T10-3-1-prompt.png", "description": "Reaction of 'O=S(C)(C)=NC#CC1=CC=CC=C1' and 'ClC(C(C)C)=O' under NEt_3 to an unknown product."}], "full_mark": 3}, {"sub_id": "Exam-05-T10-4", "tags": ["2.2"], "text_open": "10.2 Sulfinamide derivatives enable asymmetric synthesis of α-amino imines and derivatives. Piv = pivaloyl, -COtBu. 10.2.1 Give the 3 key-intermediate SMILES.", "text_images": [{"file": "Exam-05-T10-4-1-prompt.png", "description": "Substrate A, 'CC/C(C1=CC=CC=C1)=N\\[S@@](C(C)(C)C)=N/C(C(C)(C)C)=O', reacts under 1) KOtBu, THF, -78 ℃, 0.5 h; 2) PFP, -78 ℃, 0.5 h, then -40 ℃, 1 h, to give B, 'CC(N(C(C(C)(C)C)=O)SC(C)(C)C)/C(C1=CC=CC=C1)=N/C2=C(F)C(F)=NC(F)=C2F', in 84% yield and 98.5:1.5 er. PFP is 'FC1=C(F)C(F)=C(F)N=C1F'."}], "full_mark": 3}, {"sub_id": "Exam-05-T10-5", "tags": ["2.3"], "text_open": "10.2.2 Classify the rearrangement in the final step.", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-05-T10-6", "tags": ["1.2", "2.3"], "text_open": "10.2.3 From the lowest-energy transition state, give the stereochemical SMILES of B.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-05-T10-7", "tags": ["1.2"], "text_open": "10.2.4 Assign the absolute configuration of B.", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-05-T10-8", "tags": ["1.5", "1.6"], "text_open": "10.3 Thiazole (1) is an aromatic S,N heterocycle whose C-H bond is more acidic than that of oxazole (2). 10.3.1 Which qualitative statement about C-S molecular orbitals is correct? (a) S is more electronegative, so its orbitals are lower and its coefficient in σ(C-S) is larger. (b) S is more electronegative, so its orbitals are higher and its coefficient in σ*(C-S) is larger. (c) S is larger, so its orbitals are lower and the C coefficient in σ(C-S) is larger. (d) S is in a higher period, so its orbitals are higher and its coefficient in σ*(C-S) is smaller.", "text_images": [{"file": "Exam-05-T10-8-1-prompt.png", "description": "Compound 1: '[H]C1=NC=CS1'; compound 2: '[H]C1=NC=CO1'; the right C-H bond is bold in each."}], "full_mark": 1}, {"sub_id": "Exam-05-T10-9", "tags": ["1.6", "1.7"], "text_open": "10.3.3 Based on 10.3.1, use orbital hyperconjugation to explain why thiazole is more C-H acidic than oxazole. No more than 100 words.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-05-T10-10", "tags": ["2.2", "1.7"], "text_open": "10.3.4 2-Bromothiazole reacts with LDA at -78 ℃ to give ring-opened F. Give two cyclic key intermediates and rank the acidities of thiazole hydrogens by position.", "text_images": [{"file": "Exam-05-T10-10-1-prompt.png", "description": "'BrC1=NC=CS1' gives 'Br/C(S[Li])=C/[N+]#[C-]' under LDA at -78 ℃."}, {"file": "Exam-05-T10-10-2-prompt.png", "description": "Thiazole 'C1=CSC=N1' labeled 2 at C between S and N, 4 at lower-left C, and 5 at upper-left C."}], "full_mark": 3}, {"sub_id": "Exam-05-T10-11", "tags": ["2.4"], "text_open": "10.4 Traditional thiazole syntheses often suffer unstable substrates and low yields; industrial methods use transition-metal coupling. In 2025, a metal-free synthesis from α-amino-acid derivatives gave substantial yields. 10.4.1 The reaction was discovered under Curtius-like conditions, but the expected rearrangement did not occur. Give the expected-product SMILES.", "text_images": [{"file": "Exam-05-T10-11-1-prompt.png", "description": "Reaction (a): the amino-acid substrate under SOCl_2, CH_2Cl_2, DBU, room temperature gives 'CC(C=C1)=CC=C1C2=NC=C(C3=CC=CC=C3)S2'. SMILES/structure strings shown: 'CC(C=C1)=CC=C1CN[C@@H](CC2=CC=CC=C2)C(O)=O'."}, {"file": "Exam-05-T10-11-2-prompt.png", "description": "Expected Curtius reaction of the same substrate under SOCl_2, DPPA, BnOH, K_2CO_3 to an unknown product. SMILES/structure strings shown: 'CC(C=C1)=CC=C1CN[C@@H](CC2=CC=CC=C2)C(O)=O', 'O=P(OC1=CC=CC=C1)(N=[N+]=[N-])OC2=CC=CC=C2'."}], "full_mark": 1}, {"sub_id": "Exam-05-T10-12", "tags": ["2.2", "2.4"], "text_open": "10.4.2 One substrate molecule first reacts with one equivalent SOCl_2 to give neutral X containing neither O nor S. Give X.", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-05-T10-13", "tags": ["2.2", "2.3"], "text_open": "10.4.3 Give three classes of sulfur-containing key intermediates from X to product.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-05-T10-14", "tags": ["2.1", "2.2"], "text_open": "10.4.4 Write the overall equation. DBU need not be included.", "text_images": [], "full_mark": 2}]} | |
| {"problem_id": "Exam-06-T1", "header": "Problem 1: The Immortal Red Dragon. The development of alchemy depended on the joint investigations of scholars in China and abroad. In Tiangong Kaiwu, Song Yingxing recorded a method for purifying the “Red Dragon.” Low-purity “Red Dragon” was first “calcined together for five double-hours” in a sealed iron vessel to obtain a silvery-white elemental metal that is liquid at room temperature. The metal was then reacted in the proportion “one jin of metal to two jin of shitingzhi (a nonmetallic element),” yielding “fourteen liang of vermilion and three liang five qian of secondary vermilion.” Both “vermilion” and “secondary vermilion” are high-purity forms of the “Red Dragon.”", "header_images": [], "subquestions": [{"sub_id": "Exam-06-T1-1", "tags": ["2.1"], "text_open": "1-1-2 Write the equation for the reaction that produces the silvery-white elemental metal.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-06-T1-2", "tags": ["2.1", "4.2"], "text_open": "1-2 Throughout the history of alchemy, many scholars explored the chemical properties of the “Red Dragon.” Because it is chemically stable, subduing the immortal “Red Dragon” was a longstanding goal of alchemists. (i) Boiling a mixture of “wood ash” and “lime” gives strongly basic “caustic potash.” Boiling this with excess sulfur produces a dark-red mixed solution called “liver of sulfur.” When the “Red Dragon” is placed in this solution, it “completely merges with the red holy water.” Adding acid generates foul-smelling “dragon's breath” and instantly turns the originally clear liquid turbid (reaction 1). After the reaction, a large amount of soft, pale yellow-brown powder remains at the bottom. (ii) The “Green Lion” is a salt with formula mass 278.02. When strongly heated in a furnace, it first releases clear water and then “roars” out pungent “white smoke,” which condenses into a highly corrosive viscous liquid called “oil of vitriol” (reaction 2). Heating “oil of vitriol” with “saltpeter,” a potassium salt, produces dense “red fumes” that condense into slightly yellow “strong water.” Adding “sal ammoniac,” an ammonium salt, to the “strong water” gives the “King's Sword,” meaning aqua regia. When the King's Sword pierces the “Red Dragon,” it dies amid “red fumes”; its body is completely consumed into clear water, leaving only “yellow grease” floating on the surface (reaction 3). The “yellow grease” is the same substance as shitingzhi. Write the equations for reactions 1-3. Note: reaction 1 comprises two reactions, and reaction 2 is the overall reaction by which the “Green Lion” forms “oil of vitriol.”", "text_images": [], "full_mark": 9}]} | |
| {"problem_id": "Exam-06-T3", "header": "Problem 3: Protection of Isocyanates (12 points, 5%)\nHMDI [OCN(CH_2)_6NCO] is a common monomer for polymerization reactions.", "header_images": [], "subquestions": [{"sub_id": "Exam-06-T3-1", "tags": ["1.7", "1.6"], "text_open": "3.3 Researchers recently developed a supramolecular method for protecting HMDI. The pentagonal-prismatic product P5A, obtained by Lewis-acid-catalyzed reaction of diethoxybenzene with paraformaldehyde, forms a 1:1 supramolecular complex with HMDI as shown below.\n3.3.1 Identify possible interactions between P5A and HMDI in the complex other than van der Waals forces.", "text_images": [{"file": "Exam-06-T3-1-1-prompt.png", "description": "Schematic 1:1 supramolecular complex of P5A and HMDI. The OCN-(CH_2)_6-NCO chain of HMDI threads through the cavity of pillar[5]arene P5A. The SMILES of P5A is 'CCOC1=CC(CC2=CC(OCC)=C(CC3=CC(OCC)=C(CC4=CC(OCC)=C(C5)C=C4OCC)C=C3OCC)C=C2OCC)=C(OCC)C=C1CC6=C(OCC)C=C5C(OCC)=C6'."}], "full_mark": 3}, {"sub_id": "Exam-06-T3-2", "tags": ["1.7", "2.8"], "text_open": "3.3.2 After synthesis, the complex can be washed with cyclohexane. When HMDI is needed for polymerization, the complex can be dissociated in DCM or chloroform. Explain why DCM or chloroform can dissociate the complex whereas cyclohexane cannot.", "text_images": [], "full_mark": 2}]} | |
| {"problem_id": "Exam-06-T5", "header": "Problem 5: Layered Conductive Materials", "header_images": [], "subquestions": [{"sub_id": "Exam-06-T5-1", "tags": ["1.3"], "text_open": "5.1 CuSe crystal X has a layered structure. Its tetragonal unit cell is shown below, with dark spheres representing Cu and light spheres representing Se.\n5.1.1 Name the coordination polyhedron formed by the Se atoms around each Cu in crystal X.", "text_images": [{"file": "Exam-06-T5-1-1-prompt.png", "description": "A three-dimensional perspective view of a rectangular-prismatic unit cell drawn with thin dark-gray lines. Axes at the lower left show c vertically upward, b toward the right, and a toward the lower left. Ten larger blue spheres and four smaller green spheres are connected by cylindrical bonds. Blue spheres occupy all vertices of the prism and the face centers of the top and bottom faces. In the upper half, two green spheres are symmetrically positioned near the upper portions of the left and right faces. The left green sphere lies below several blue spheres at the upper left and is bonded to two blue spheres in the upper-left region and the blue sphere at the top center. The right green sphere similarly lies below the upper-right blue spheres and is bonded to two blue spheres in the upper-right region and the top-center blue sphere. In the lower half, two green spheres are symmetrically positioned near the lower portions of the front and back faces. The front green sphere is bonded to the lower-left and lower-right blue spheres of the front face and to the bottom-center blue sphere; the rear green sphere is bonded analogously to the corresponding rear-face blue spheres and bottom-center sphere. Overall, the green spheres lie between upper and lower sets of blue spheres, forming two layers in which green spheres bridge multiple blue spheres."}], "full_mark": 1}, {"sub_id": "Exam-06-T5-2", "tags": ["1.3"], "text_open": "5.1.2 Within each layer, these Cu-Se polyhedra are connected by ( ).\na. sharing vertices\nb. sharing edges\nc. sharing faces", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-06-T5-3", "tags": ["1.3"], "text_open": "5.1.3 State the lattice type of crystal X.", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-06-T5-4", "tags": ["1.3"], "text_open": "5.2 Translate the Cu-Se polyhedral layers in crystal X horizontally so that neighboring Se atoms in the upper and lower layers align along c, then fill all regular square-prismatic voids formed by Se with Tl to obtain tetragonal crystal Y.\n5.2.1 Write the chemical formula of crystal Y in the simplest integer ratio.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-06-T5-5", "tags": ["1.3"], "text_open": "5.2.4 State the lattice type of crystal Y.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-06-T5-6", "tags": ["1.3"], "text_open": "5.4 Replacing each single Cu-Se polyhedral layer in crystal Y with a double layer, in which the two component layers share one layer of Se atoms, gives crystal Z with a conduction mechanism similar to that of Y.\n5.4.1 Write the chemical formula of crystal Z.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-06-T5-7", "tags": ["1.3"], "text_open": "5.4.2 Approximate the 8 Se atoms around Tl in crystal Y as a perfect cube, and assume that the shapes and sizes of all coordination polyhedra remain unchanged in the transformation from Y to Z. State directly the value of unit-cell parameter c for the conventional unit cell of crystal Z.", "text_images": [], "full_mark": 2}]} | |
| {"problem_id": "Exam-06-T8", "header": "Problem 8: Fundamental Organic Chemistry", "header_images": [], "subquestions": [{"sub_id": "Exam-06-T8-1", "tags": ["2.4", "2.7", "2.3"], "text_open": "8.2 Sulfur ylides are important nucleophilic carbon reagents that react with carbonyl compounds, α, β-unsaturated carbonyl compounds, and related substrates. Note: consider thermodynamic and kinetic products. 8.2.1 Compare the two reactions and write the SMILES representation of the major product of each.", "text_images": [{"file": "Exam-06-T8-1-1-prompt.png", "description": "'O=C1CCC(C(C)(C)C)CC1' reacts with '[CH2-][S+](C)C' in THF at 0 °C."}, {"file": "Exam-06-T8-1-2-prompt.png", "description": "'O=C1CCC(C(C)(C)C)CC1' reacts with 'C[S+](=O)(C)[CH2-]' in THF at 65 °C."}], "full_mark": 4}, {"sub_id": "Exam-06-T8-2", "tags": ["2.4", "2.7", "2.3"], "text_open": "8.2.2 Compare the two reactions and write the SMILES representation of the major product of each.", "text_images": [{"file": "Exam-06-T8-2-1-prompt.png", "description": "'O=C1CCC(C(=C)C)C=C1C' reacts with '[CH2-][S+](C)C' in THF at 0 °C."}, {"file": "Exam-06-T8-2-2-prompt.png", "description": "'CC(=C)C1CCC(=O)C(C)=C1' reacts with 'C[S+](=O)(C)[CH2-]' in THF at 50°C."}], "full_mark": 4}, {"sub_id": "Exam-06-T8-3", "tags": ["1.2", "1.7"], "text_open": "8.3 Molecular conformations. 8.3.1 Determine which side is favored in the following conformational equilibrium. 8.3.2 Determine which side is favored in the following conformational equilibrium. Note: combine the answers to 8.3.1-8.3.2 directly, for example, AA.", "text_images": [{"file": "Exam-06-T8-3-1-prompt.png", "description": "8.3.1 Interconversion equilibrium between molecule A and molecule B, whose SMILES are 'O=C1[C@@]2([H])[C@](N(C)CCC2)([H])CCC1' and 'O=C3[C@]4([H])[C@](N(C)CCC4)([H])CCC3', respectively."}, {"file": "Exam-06-T8-3-2-prompt.png", "description": "8.3.2 Interconversion equilibrium between molecule A and molecule B, whose SMILES are 'O=C1[C@@]2([H])[C@](N(C(C)=O)CCC2)([H])CCC1' and 'O=C3[C@]4([H])[C@](N(C(C)=O)CCC4)([H])CCC3', respectively."}], "full_mark": 2}, {"sub_id": "Exam-06-T8-4", "tags": ["4.3"], "text_open": "8.4.2 Examine the following four IR spectra and select from A-D the organic structure that best matches each spectrum. (A) phenol; (B) aniline; (C) N-methylaniline; (D) benzoic acid. Enter the capital letters in the order corresponding to (a)(b)(c)(d), for example, ABCD.", "text_images": [{"file": "Exam-06-T8-4-1-prompt.png", "description": "Four infrared spectra, each with a horizontal axis for vibrational absorption wavenumber in cm^{-1}; the four spectra are labeled (a), (b), (c), and (d) from left to right."}], "full_mark": 4}]} | |
| {"problem_id": "Exam-06-T9", "header": "Problem 9 Aziridines (30 points, 12%). Aziridine is an important electrophilic scaffold with diverse reactivity and broad applications in biomedicine and other fields.", "header_images": [], "subquestions": [{"sub_id": "Exam-06-T9-1", "tags": ["2.4", "2.2"], "text_open": "9.1 Nitrones react with alkynes to afford aziridine structures in high yields. However, when the nitrogen atom of the nitrone bears a tert-butyl group, aziridine intermediate B is converted into stable product C containing a five-membered ring. 9.1.1 Examine the reaction sequence and write the SMILES structures of A, B, and C. Stereochemistry is not required.", "text_images": [{"file": "Exam-06-T9-1-1-prompt.png", "description": "The molecule 'CC(C)(C)[N+]([O-])=C' reacts with the molecule 'COC(=O)C#CC(=O)OC' at 0 °C to form A. A then reacts at 80 °C to form intermediate B, which is ultimately converted to C."}], "full_mark": 4}, {"sub_id": "Exam-06-T9-2", "tags": ["2.3", "2.2"], "text_open": "9.1.2 Identify the type of rearrangement involved in the conversion of A to B.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-06-T9-3", "tags": ["2.3", "2.2"], "text_open": "9.1.3 The following reaction undergoes a process similar to that described above to form an aziridine intermediate. The aziridine moiety then fragments to form a new 1,3-dipole, followed by loss of an oxalate ester group to give the product. Based on this information, write the SMILES structures of the 4 key intermediates in the following reaction. Stereochemistry is not required.", "text_images": [{"file": "Exam-06-T9-3-1-prompt.png", "description": "The molecule 'CC1(C)N=C(C2=CC=CC=C2)C=[N+]1[O-]' reacts with the molecule 'COC(=O)C#CC(=O)OC' in CH_2Cl_2 at rt for 24 h to form the molecule 'CC(C)(N=C1C2=CC=CC=C2)N3C1=C(C(C(OC)=O)=C3C(OC)=O)C(OC)=O'."}], "full_mark": 6}, {"sub_id": "Exam-06-T9-4", "tags": ["2.4", "2.7", "1.2"], "text_open": "9.2 One method for synthesizing aziridines uses a Reformatsky reagent. The following reaction affords aziridine product D in relatively high conversion, together with a small amount of byproduct E, whose absolute configuration is not labeled. 9.2.1 Write the SMILES structure of aziridine product D, including stereochemistry.", "text_images": [{"file": "Exam-06-T9-4-1-prompt.png", "description": "'[H][C@@]1(N=C1C2=CC=CC=C2)C' reacts with 'BrC(C(OCC)=O)(C)C' in the presence of Zn in Et_2O/PhH (1:1) at 60°C for 2-5 h to form D and E. The SMILES of E is 'O=C(N3)C(C4(N(C4([H])C)C(C5=CC=CC=C5)=C3C)C6=CC=CC=C6)(C)C'. In the structure of E, the methyl-substituted carbon atom of the aziridine ring is labeled a."}], "full_mark": 2}, {"sub_id": "Exam-06-T9-5", "tags": ["2.2", "1.2"], "text_open": "9.2.2 The formation of E is known to proceed through an aziridine structure. Write the SMILES structures of the key intermediates in the conversion of the substrates to E, including stereochemistry.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-06-T9-6", "tags": ["1.2"], "text_open": "9.2.3 Use S/R notation to specify the absolute configuration of carbon atom a in product E.", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-06-T9-7", "tags": ["2.2", "1.5", "2.3"], "text_open": "9.3 Studies show that aziridine scaffolds can also be synthesized without adding an external dipole. Under photochemical conditions, an α-azidocinnamate forms a small-ring intermediate F whose structure resembles that of the nitrogen-containing substrate in the reaction in Problem 9.2. F can subsequently form dimeric and trimeric products. 9.3.1 During the oligomerization of F, photochemical ring opening first generates dipole G. Using the structure of the dimeric product, analyze G and write the SMILES structures of its reasonable resonance forms that satisfy an 8-electron configuration.", "text_images": [{"file": "Exam-06-T9-7-1-prompt.png", "description": "The substance whose structure in the figure has the SMILES '[N-]=[N+]=N/C(C(OC)=O)=C\\C1=CC=CC=C1' reacts under hν in acetone for 1 h at 30°C to form a product with the SMILES 'O=C(C2=N[C@H](N3[C@]2([C@H]3C4=CC=CC=C4)C(OC)=O)C5=CC=CC=C5)OC' and a trimer."}], "full_mark": 3}, {"sub_id": "Exam-06-T9-8", "tags": ["2.4", "2.2"], "text_open": "9.3.2 The trimer is formed when the photoexcited dimer undergoes addition with monomer F. On this basis, write the SMILES structure of the trimeric product. Stereochemistry is not required.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-06-T9-9", "tags": ["2.4", "2.3", "2.7"], "text_open": "9.3.3 The following azide is known to give different major products under thermal and photochemical conditions, although the two products have similar ring systems. Based on the information above, write the SMILES structures of the major products under thermal and photochemical conditions, respectively.", "text_images": [{"file": "Exam-06-T9-9-1-prompt.png", "description": "Structure of the azide substrate. The corresponding SMILES is 'O=C(/C(N=[N+]=[N-])=C/C1=CC=CC=N1)OCC'."}], "full_mark": 2}, {"sub_id": "Exam-06-T9-10", "tags": ["2.3", "2.4"], "text_open": "9.3.4 The following reaction synthesizes an aziridine scaffold through an ionic mechanism. Its design resembles that of the reaction above in that a dipolar intermediate is generated in situ, and the product contains a bridged ring system. On this basis, write the SMILES structures of the dipolar intermediate and the final product. Stereochemistry is not required.", "text_images": [{"file": "Exam-06-T9-10-1-prompt.png", "description": "The substance whose structure in the figure has the SMILES 'S=C(SC(C(O)=O)C1=CC=CC=C1)C2=CC=CC=C2' is treated sequentially under 1) Ac_2O/Et_3N and 2) 'CC3N=C3C4=CC=CC=C4'; the product is indicated by a question mark."}], "full_mark": 4}]} | |
| {"problem_id": "Exam-06-T10", "header": "Problem 10 Radical Cyclization and Photochemical Reactions (29 points, 12%). Compared with conventional ionic cyclization reactions, radical processes generally feature mild reaction conditions, broad functional-group compatibility, and, in some highly strained ring-forming processes, lower transition-state barriers. They have therefore attracted broad interest in natural-product synthesis and the construction of complex molecular scaffolds.", "header_images": [], "subquestions": [{"sub_id": "Exam-06-T10-1", "tags": ["2.2", "2.4", "1.2"], "text_open": "10.1 Radical cascade cyclization and ring expansion constitute an important strategy in modern organic synthesis for constructing medium-sized rings containing 8~11 members. The Dowd-Beckwith reaction, which efficiently converts a cyclic ketone scaffold into a medium-ring enone system, is a representative radical ring-expansion reaction. 10.1.1 Using the information above and the reaction shown below, write the SMILES structures of the key radical intermediates and the final product, paying attention to stereochemistry.", "text_images": [{"file": "Exam-06-T10-1-1-prompt.png", "description": "The molecule 'O=C(CCC[C@H]1[Sn](CCCC)(CCCC)CCCC)[C@]1(CCCCI)C' reacts under Bu_3SnH, AIBN, and PhH, reflux."}], "full_mark": 5}, {"sub_id": "Exam-06-T10-2", "tags": ["2.2", "2.4", "1.2"], "text_open": "10.1.2 Changing the configuration of one stereogenic center in the substrate causes a corresponding change in the product structure. Write the SMILES structure of the final product obtained from the substrate shown below under the same reaction conditions, and write the SMILES structure of the key intermediate conformation that accounts for the product stereochemistry.", "text_images": [{"file": "Exam-06-T10-2-1-prompt.png", "description": "Dowd-Beckwith substrate in which the configuration of a stereogenic center has been changed. The corresponding SMILES is 'O=C(CCC[C@@H]1[Sn](CCCC)(CCCC)CCCC)[C@]1(CCCCI)C'."}], "full_mark": 4}, {"sub_id": "Exam-06-T10-3", "tags": ["2.2", "2.4"], "text_open": "10.1.3 When the substrate is modified further, an unexpanded-ring product is observed in addition to the target medium-ring enone. Using the reaction below, write the SMILES structures of this unexpanded-ring product and the key intermediates through which it is formed.", "text_images": [{"file": "Exam-06-T10-3-1-prompt.png", "description": "The molecule 'O=C(CCC1)C([H])(CCCCI)[C@H]1[Sn](CCCC)(CCCC)CCCC' reacts under Bu_3SnH, AIBN, and PhH, reflux."}], "full_mark": 3}, {"sub_id": "Exam-06-T10-4", "tags": ["2.2", "2.4", "2.7"], "text_open": "10.2 Radicals can also be generated photochemically. For example, under irradiation, FeCl_{3} can undergo ligand-to-metal charge transfer, an LMCT process, to generate chlorine radicals capable of hydrogen-atom abstraction. 10.2.1 Under the standard reaction conditions, two products A and B are obtained in a ratio of 1: 1.4. When the temperature is raised to 60°C, the product ratio changes to 1: 4. Using this information, write the SMILES structures of the key intermediates leading to A and B, respectively. If the two pathways share a common intermediate, write that intermediate only once. Note: A and B are constitutional isomers.", "text_images": [{"file": "Exam-06-T10-4-1-prompt.png", "description": "The molecule 'CC(C(C)(C)C)=O' reacts with the molecule 'C=CC(OCC1=CC=CC=C1)=O' under FeCl_3 (25 mol%), a 390 nm LED, and MeCN, N_2, rt, 0.3 M, 36 h to form A and B."}], "full_mark": 7}, {"sub_id": "Exam-06-T10-5", "tags": ["2.7", "2.3", "1.7"], "text_open": "10.2.2 Keep the standard reaction conditions unchanged, but replace only the ketone substrate with the structure shown below. Determine whether the formation ratio of B relative to A increases or decreases, and explain why.", "text_images": [{"file": "Exam-06-T10-5-1-prompt.png", "description": "Structure of the replacement ketone substrate in 10.2.2. The corresponding SMILES is 'CC(C(C)C)=O'."}], "full_mark": 4}, {"sub_id": "Exam-06-T10-6", "tags": ["2.4", "2.3"], "text_open": "10.2.3 Keep the standard reaction conditions unchanged, but replace one of the alkene substrates with the compound shown below. Experimentally, the reaction does not produce the expected acyclic product but instead gives two cyclized products. Ignoring stereochemistry, write the SMILES structures of these two products.", "text_images": [{"file": "Exam-06-T10-6-1-prompt.png", "description": "Structure of the replacement alkene substrate in 10.2.3. The corresponding SMILES is 'N#C/C(C#N)=C/C1=CC=CC=C1'."}], "full_mark": 4}]} | |
| {"problem_id": "Exam-07-T3", "header": "Synthesis and Applications of Contrast Agents\nThe rapid expansion of medical magnetic resonance imaging has driven the development of contrast agents, most of which are complexes of element 64, Gd. Gd(III) is highly paramagnetic, making it an excellent candidate relaxation agent; among species that exist stably as hydrated ions, it is one of the most strongly magnetic.", "header_images": [], "subquestions": [{"sub_id": "Exam-07-T3-1", "tags": ["2.4", "1.4"], "text_open": "3.3 In one method for preparing a common Gd(III) contrast agent, aminoacetic acid is reacted with 1,2-dibromoethane in a 1:1 ratio in aqueous solution to give a tetrabasic acid X. Addition of X to $[Gd(H_2O)_8]^{3+}$ gives an anion A with a monocapped square-antiprismatic coordination geometry, one ligand of which is deprotonated X. Write the SMILES representations of X and A.", "text_images": [], "full_mark": 5}]} | |
| {"problem_id": "Exam-07-T4", "header": "Determination of Sulfur Dioxide in Wine\nThe sulfur dioxide content of a wine sample is determined spectrophotometrically as follows. I. Construct a calibration curve. Accurately transfer 0.00 mL, 0.20 mL, 0.50 mL, 1.00 mL, 2.00 mL, and 3.00 mL of sulfur dioxide working standard solutions (containing 0.0 μg, 2.0 μg, 5.0 μg, 10.0 μg, 20.0 μg, and 30.0 μg sulfur dioxide, respectively) into 25 mL stoppered test tubes, and add formaldehyde-buffered absorbing solution to 10.00 mL. Then add, in sequence, 0.5 mL of 3 g/L ammonium sulfamate solution, 0.5 mL of 1.5 mol/L sodium hydroxide solution, and 1.0 mL of 0.5 g/L pararosaniline hydrochloride solution (structure shown below). Mix thoroughly and allow to stand for 20 min. Measure the absorbance of each standard solution at 579 nm using a UV–visible spectrophotometer. The following data are obtained:\n\n| $\\text{SO}_2$ content $m / \\mu\\text{g}$ | 0.0 | 2.0 | 5.0 | 10.0 | 20.0 | 30.0 |\n|---|---|---|---|---|---|---|\n| Absorbance $A$ | 0.003 | 0.062 | 0.159 | 0.312 | 0.618 | 0.928 |\n\nII. Accurately weigh 0.5000 g of the wine sample and add formaldehyde-buffered absorbing solution to 100 mL. After shaking for 2 h, take 10 mL of the solution and perform the spectrophotometric determination as in Step I. Repeat the measurement three times; the absorbances A’ are 0.637, 0.638, and 0.635. In China, the required sulfur dioxide content of alcoholic beverages is below $0.2500 \\ \\mathrm{g\\,dm^{-3}}$. Take the density of the wine to be $0.9 \\ \\mathrm{g\\,cm^{-3}}$. The structure of pararosaniline hydrochloride is 'NC1=CC=C(/C(C2=CC=C(N)C=C2)=C3C=CC(C=C\\3)=[NH2+])C=C1.[Cl-]'.", "header_images": [{"file": "Exam-07-T4-header-1.png", "description": "Structural diagram of pararosaniline hydrochloride; the SMILES representation is 'NC1=CC=C(/C(C2=CC=C(N)C=C2)=C3C=CC(C=C\\3)=[NH2+])C=C1.[Cl-]'."}], "subquestions": [{"sub_id": "Exam-07-T4-1", "tags": ["2.4"], "text_open": "4-1 The principle used in this titration is that A, formed by the addition reaction of sulfur dioxide with formaldehyde, forms a blue-violet complex with pararosaniline hydrochloride. Its absorbance is proportional to the sulfur dioxide concentration. Write the SMILES representation of A.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-07-T4-2", "tags": ["4.1", "2.8"], "text_open": "4-2 Why must 0.5 mL of 1.5 mol/L sodium hydroxide solution be added to adjust the pH before measurement? (No more than 50 words.)", "text_images": [], "full_mark": 2}]} | |
| {"problem_id": "Exam-07-T7", "header": "The Chemistry of As, Sb, and Bi\nThe pnictogens are a familiar group of elements. N and P are well known for their wide range of applications in organic chemistry; As, Sb, and Bi can likewise bond to carbon and form many interesting compounds. ", "header_images": [], "subquestions": [{"sub_id": "Exam-07-T7-1", "tags": ["1.4"], "text_open": "7.1 The positive trivalent oxidation state is common among the pnictogens. Their lone pairs and vacant d orbitals give them Lewis basicity and acidity of varying strength. 7.1.1 In sharp contrast to $NR_3$, the inertness of the 6s electrons makes $BiR_3$ a rather weak base with poor coordination ability, although it can sometimes form complexes. The neutral complex A and the monocation B both contain a $BiPh_3$ ligand. In A, the mass fractions of Bi and the central metal Cr are 33.05 % and 8.22 %, respectively; in B, the mass fractions of Bi and the central metal Fe are 33.86 % and 9.04 %, respectively. The central metals in both A and B obey the 18-electron rule, and both complexes contain CO ligands. Write the chemical formulas of A and B.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-07-T7-2", "tags": ["1.4", "1.5"], "text_open": "7.1.2 Considering the structures of A and B, explain why the Bi-M bonds in them are stable. (No more than 50 words.)", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-07-T7-3", "tags": ["1.4", "2.4"], "text_open": "7.1.3 Compared with $BiR_3$, $SbR_3$ is slightly less acidic but markedly more basic, which enables a wider variety of reactions. Mixing equimolar amounts of $Me_2SbBr$, $Me_3Sb$, and $MeSbBr_2$ does not give mixed products from methyl-group exchange; instead, a 2:1 ionic compound is obtained. Both its cation and anion contain two Sb atoms. The cation contains no Br, and the anion has a center of symmetry. Write the SMILES representations of the cation and anion separately, and state the geometry at each metal center.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-07-T7-4", "tags": ["1.1", "1.4"], "text_open": "7.2 Reaction of $SbF_2OTf$ or $SbF(OTf)_2$ with the ligand $^{Dipp}BIAN$ (structure shown below) in $CH_2Cl_2$ gives the five-coordinate Sb complex $SbF_2(^{Dipp}BIAN)OTf$ or $SbF(^{Dipp}BIAN)(OTf)_2$, respectively.\n7.2.1 Use VSEPR theory to predict the molecular geometry of $SbF(^{Dipp}BIAN)(OTf)_2$.", "text_images": [{"file": "Exam-07-T7-4-1-prompt.png", "description": "Structure of $^{Dipp}BIAN$: 'CC(C)C(C=CC=C1C(C)C)=C1/N=C2C3=CC=CC4=C3C(C\\2=N/C5=C(C(C)C)C=CC=C5C(C)C)=CC=C4'."}], "full_mark": 2}, {"sub_id": "Exam-07-T7-5", "tags": ["1.4", "1.5"], "text_open": "7.2.2 The Sb-O bond in $SbF_2({}^{Dipp}BIAN)OTf$ is longer than that in $SbF({}^{Dipp}BIAN)(OTf)_2$. Briefly explain why.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-07-T7-6", "tags": ["2.4"], "text_open": "7.2.3 Replacing $F^-$ with $N_3^-$ greatly decreases the stability of the product. After $Sb(N_3)(^{Dipp}BIAN) (OTf)_2$ is left overnight in a protic solvent, its proton NMR spectrum shows that it has decomposed to form a new cation. This cation contains no Sb and contains a special five-membered ring in addition to the five-membered carbon ring in $^{Dipp}BIAN$. Write the SMILES representation of this cation; a textual description may be used to supplement the structure.", "text_images": [], "full_mark": 2}]} | |
| {"problem_id": "Exam-07-T9", "header": "Bridged Polycyclic Heterocycles Involving Unexpected Rearrangements", "header_images": [], "subquestions": [{"sub_id": "Exam-07-T9-1", "tags": ["2.2", "2.4"], "text_open": "9.1 The reaction below is catalyzed by AgOTf.\n9.1.1 The reaction initially forms a nitrogen-containing heterocyclic intermediate. Write the SMILES representations of the 4 key intermediates and of the product formed after the 1,3-dipolar cycloaddition. Add a textual description if necessary.", "text_images": [{"file": "Exam-07-T9-1-1-prompt.png", "description": "'O/N=C/C1=CC=CC=C1C#CC2=CC=CC=C2' reacts with 'O=S(OC1=CC=CC=C1[Si](C)(C)C)(C(F)(F)F)=O' under AgOTf (10mol%), TBAF (3.0equiv), THF, 50°C conditions."}], "full_mark": 10}, {"sub_id": "Exam-07-T9-2", "tags": ["2.2"], "text_open": "9.1.2 However, the product described in 9.1.1 is not observed; the product formed after its further rearrangement is shown below. Write the SMILES representation of 1 key intermediate in the reaction.", "text_images": [{"file": "Exam-07-T9-2-1-prompt.png", "description": "'C12=C([C@H]3C(C=CC=C4)=C4O[C@@H]2C(C5=CC=CC=C5)=N3)C=CC=C1'"}], "full_mark": 2}, {"sub_id": "Exam-07-T9-3", "tags": ["2.2"], "text_open": "9.2 Some steps of the reaction below are very similar to those in 9.1. PhIO (iodosylbenzene) is a hypervalent iodine oxidant commonly used for the oxidative dearomatization of electron-rich arenes in the presence of a nucleophilic solvent. 9.2.1 Write the SMILES representation of the key OMe-containing electrophilic species formed by reaction of the starting material with PhIO in MeOH.", "text_images": [{"file": "Exam-07-T9-3-1-prompt.png", "description": "'CC1=CC=C(NS(C2=CC=C(C)C=C2)(=O)=O)C(C#CC3=CC=CC=C3)=C1' first reacts with iodosylbenzene, PhI=O, in methanol at room temperature for 5min, and then reacts with 'O/N=C/C1=CC=CC=C1C#CC2=CC=CC=C2' (1.1 equiv) under AgOTf (10 mol%) in dichloroethane at room temperature."}], "full_mark": 2}, {"sub_id": "Exam-07-T9-4", "tags": ["2.2", "2.4", "4.3"], "text_open": "9.2.2 The key electrophilic species then cyclizes to give a key intermediate, which undergoes a [3 + 3] cycloaddition with an intermediate from 9.1.1. Write the SMILES representations of the cyclized key intermediate and the cycloaddition product. Hint: The product contains a pyrrole ring system and has 1H NMR (400 MHz, $CDCl_3$) δ 7.95-7.93 (m, 2 H), 7.51-7.41 (m, 9 H), 7.31-7.27 (m, 2 H), 7.23 (d, J = 8.1 Hz, 2 H), 7.10 (d, J = 8.1 Hz, 2 H), 6.94 (d, J = 10.4 Hz, 1 H), 6.89-6.87 (m, 1 H), 5.94 (s, 1 H), 5.70 (s, 1 H), 5.68 (d, J = 10.3 Hz, 1 H), 4.66 (s, 1 H), 3.20 (s, 3 H), 2.31 (s, 3 H), 1.21 (s, 3 H).", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-07-T9-5", "tags": ["2.2", "2.4"], "text_open": "9.2.3 However, when the product above is stirred in 1,4-dioxane at 90 °C for 2 hours, it undergoes a reaction similar to that in 9.1. Write the SMILES representations of the 2 key intermediates and the final product.", "text_images": [], "full_mark": 6}, {"sub_id": "Exam-07-T9-6", "tags": ["2.2"], "text_open": "9.4 EWG is an acyclic electron-withdrawing group. By inference from 9.1 and 9.2, write the SMILES representation of 1 key tetracyclic intermediate in the reaction below.", "text_images": [{"file": "Exam-07-T9-6-1-prompt.png", "description": "'O/N=C/C1=CC=CC=C1C#CC2=CC=CC=C2' reacts with 'C=C=C-EWG' under 5 mol% AgOTf catalysis in DMSO, first at 60 °C for 20min and then at room temperature for 12h, to give 'O=C1C[C@@H]2C3=CC=CC=C3[C@@H](N=C2C4=CC=CC=C4)[C@@H]1[EWG]'."}], "full_mark": 2}]} | |
| {"problem_id": "Exam-08-T1", "header": "Problem 1 Peierls's Wondrous Gift\nIn the 18 century, Austrian mineralogist Müller von Reichenstein discovered a new substance while studying a gold ore. It had a silvery-white metallic luster but was brittle and a poor conductor; it burned with a blue flame and gave off an unusual odor resembling rotten radishes. German chemist Martin Klaproth later confirmed that it was a new element and named it after the Latin word for Earth. Interestingly, another member of the same group discovered decades later was named after the Moon to complement Earth. This element, denoted E, can be obtained by reduction of either its oxide A or its sodium oxoacid salt B; E has the same oxidation state in A and B. An early preparation heated B with carbon black. The simple hydride C of E is a foul-smelling, highly unstable gas whose aqueous solution is weakly acidic. Passing C into $\\mathrm{FeCl_3}$ solution gives a black precipitate. Another oxide D of E is an orange solid. When D dissolves in concentrated aqueous NaOH, the resulting anhydrous sodium salt contains 46.64% E by mass.", "header_images": [], "subquestions": [{"sub_id": "Exam-08-T1-1", "tags": ["2.1"], "text_open": "1-2-1 Write the equation for heating B with carbon black to obtain elemental E.", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-08-T1-2", "tags": ["2.1"], "text_open": "1-2-3 Write the equation for dissolution of D in concentrated aqueous NaOH.", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-08-T1-3", "tags": ["1.3", "1.6"], "text_open": "1-4-1 Structural studies of $\\alpha$-E show infinite one-dimensional zigzag chains in which each E atom bonds to its two neighboring E atoms, forming continuous $(E)_n$ chains with $n\\rightarrow\\infty$. Elemental halogen X is widely used for surgical disinfection. E and X form F, G, and H with E:X ratios of 1:1, 2:1, and 1:4, respectively. F retains the $(E)_n$ chain. Both E and X have two chemical environments, each in a 1:1 ratio. The coordination numbers of the two E atoms are in a 3:4 ratio, and one X atom forms only one covalent bond to E. G can be viewed as portions of two long F chains joined by additional E-E bonds to form an infinite six-membered-ring structure, and the E coordination numbers in G are the same as in F. H exists as a tetramer with two E environments, and every E atom is six-coordinate. Give SMILES descriptions of the structural motifs of F and G.", "text_images": [], "full_mark": 6}, {"sub_id": "Exam-08-T1-4", "tags": ["1.1", "1.6"], "text_open": "1-4-2 Give the SMILES structure of the H tetramer.", "text_images": [], "full_mark": 3}]} | |
| {"problem_id": "Exam-08-T6", "header": "Lithium or Sodium?\nBalancing fluctuations in renewable-electricity generation through electrochemical energy storage is essential to achieving carbon-peaking and carbon-neutrality goals. Lithium-ion batteries dominate the market, but constraints on lithium resources have stimulated research into sodium-ion batteries. Their operating principles are similar, yet their material structures and performance differ substantially. This problem considers two representative layered-oxide cathodes.", "header_images": [], "subquestions": [{"sub_id": "Exam-08-T6-1", "tags": ["3.2"], "text_open": "6.2 The following electrochemical data are given. NCM811 has an average discharge voltage of 3.8 V versus metallic lithium and an actual reversible specific capacity of $200\\,\\mathrm{mAh\\cdot g^{-1}}$. The P2 sodium material has an average discharge voltage of 3.2 V versus metallic sodium and an actual reversible specific capacity of $130\\,\\mathrm{mAh\\cdot g^{-1}}$. The volumetric energy density is $E_v=\\rho\\times C\\times U$, where $\\rho$ is density, $C$ specific capacity, and $U$ average voltage.\n6.2.1 Calculate the theoretical volumetric energy density of each cathode and their ratio $\\lambda$ (lithium-based/sodium-based).", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-08-T6-2", "tags": ["3.2"], "text_open": "The diffusion coefficient D of ions in a layered material and the temperature T obey the Arrhenius equation. For P2-type $Na_{0.67}Ni_{0.33}Mn_{0.67}O_2$, the experimentally measured pre-exponential factor is $D_0 = 5.0 × 10^{-4} cm^2·s^{-1}$ and the activation energy is E_a = 0.35 eV. The Arrhenius equation is $D = D_0 e^{-\\frac{E_a}{RT}}$.\n6.3.2 The lithium-ion diffusion activation energy in NCM811 is approximately \\(0.48\\,\\mathrm{eV}\\). If the pre-exponential factors have the same order of magnitude, compare sodium- and lithium-ion diffusion rates in such layered structures at the same temperature and briefly explain. (No more than 100 words.)", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-08-T6-3", "tags": ["1.7"], "text_open": "6.4 Sodium-ion batteries are viewed as a potential complement to lithium-ion batteries because sodium is abundant and the batteries offer good low-temperature performance and relatively high safety, but commercialization still faces challenges.\n6.4.1 Briefly explain, in terms of intrinsic material properties, why increasing the volumetric energy density of sodium-ion batteries may be difficult. (No more than 100 words.)", "text_images": [], "full_mark": 2}]} | |
| {"problem_id": "Exam-09-T1", "header": "Material-Balance Analysis of an Electrolytic Process\n\"If this silvery-white lightweight metal could be produced on a large scale at low cost, transportation and building structures would undergo a qualitative transformation.\" In the late 19th century, industry fiercely competed to develop a method for producing this lightweight metal. At almost the same time, chemists Charles M. Hall and Paul Héroult independently proposed, on opposite sides of the Atlantic, a high-temperature molten-salt electrolysis route: a reddish-brown ore A (whose principal component is A') is purified and then electrolytically reduced in a molten-salt bath to obtain the silvery-white metal B, thereby laying the foundation of the modern metallurgical industry for B. Natural ore A is usually reddish brown and often contains silicon dioxide impurities. During electrolysis, the mineral flux C is added to lower the melting point of the system, and a carbonaceous anode is used. The anode is continuously consumed, releasing gas D. Industrial electrolytic cells normally operate at atmospheric pressure and at approximately 955℃. Before Hall-Héroult electrolysis, ore A must be purified by the Bayer process. This process mainly comprises the following steps: first, crushed ore A is stirred with a 30 wt% NaOH solution at 473 K and 3.5 MPa for 2 h, converting it into the soluble complex E; the solution is then cooled to 320 K, seeded with particles approximately 75 μm in size (50 g·L^{-1}), and allowed to stand for 48 h, precipitating F with a relatively narrow particle-size distribution that facilitates subsequent calcination; finally, F is calcined in air at 1370 K for 6 h to regenerate high-purity A'. When mineral flux C is the principal component of the molten-salt bath, the bath usually contains two key substances: solid G, which must be replenished periodically, and compound H. Solid G is added periodically to maintain their molar ratio in the bath. If $Na_2CO_3$ impurity enters the bath, $Na_2CO_3$ reacts with G, releasing a small amount of gas D and producing compound H, which remains soluble in the molten-salt bath, together with another by-product. Define the bath ratio as BR = n(H) / n(G) = 2.50", "header_images": [], "subquestions": [{"sub_id": "Exam-09-T1-1", "tags": ["2.1", "3.6"], "text_open": "1.2 On a certain day, 15.0 kg of $Na_2CO_3$ impurity enters the molten-salt bath. Assume that all H produced dissolves in the bath and all D escapes. 1.2.1 Calculate the mass of A' additionally produced by the side reaction that day.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-09-T1-2", "tags": ["2.1"], "text_open": "1.2.2 Calculate the additional mass of G that should be added that day to restore BR to 2.50.", "text_images": [], "full_mark": 4}]} | |
| {"problem_id": "Exam-09-T4", "header": "Kjeldahl Nitrogen Determination\nDetermination of nitrogen content has long been an important topic in analytical chemistry. In 1883, the Danish chemist Johan G.C.T. Kjeldahl invented the Kjeldahl method. The method mainly comprises digestion, alkalization, and titration: sulfuric-acid digestion converts the nitrogen in a sample into an ammonium salt; alkalization releases ammonia by distillation; and the absorbed ammonia is quantified indirectly by acid-base titration. The organic nitrogen content of a solid sample is to be determined. Answer the following questions based on the procedures described.", "header_images": [], "subquestions": [{"sub_id": "Exam-09-T4-1", "tags": ["4.1", "4.5"], "text_open": "4-1-1 First operation: digestion. Weigh $5.000 \\ \\mathrm{g}$ of the sample into a dry $100 \\ \\mathrm{mL}$ Kjeldahl flask. Add $0.4 \\ \\mathrm{g}$ of copper sulfate, $6 \\ \\mathrm{g}$ of potassium sulfate, and $20 \\ \\mathrm{mL}$ of anhydrous sulfuric acid. Swirl gently, place a small funnel in the flask mouth, and set the flask at an angle of 45° on an asbestos gauze with a small opening. Heat slowly. When the contents turn black and foaming has completely ceased, increase the heat to keep the liquid gently boiling. After the liquid becomes blue-green, clear, and transparent, continue heating for another $1 \\ \\mathrm{h}$. Remove the Kjeldahl flask and cool it to room temperature. Carefully add $20 \\ \\mathrm{mL}$ of water, quantitatively transfer the contents to a $100 \\ \\mathrm{mL}$ volumetric flask, rinse the inner wall of the Kjeldahl flask with a small amount of water, and combine the rinsings in the volumetric flask. Dilute to volume with water, mix, and reserve for use. Conduct a blank experiment in parallel. From the apparatus shown below, select the Kjeldahl flask suitable for digestion. The choices are labeled (a), (b), (c), and (d).", "text_images": [{"file": "Exam-09-T4-1-1-prompt.png", "description": "Image corresponding to option (a): a pear-shaped flask with a short neck and a ground-glass stopper."}, {"file": "Exam-09-T4-1-2-prompt.png", "description": "Image corresponding to option (b): a flask with a long neck, a pear-shaped bottom, and a ground-glass stopper."}, {"file": "Exam-09-T4-1-3-prompt.png", "description": "Image corresponding to option (c): an Erlenmeyer flask with a conical body, a flat bottom, and a ground-glass stopper."}, {"file": "Exam-09-T4-1-4-prompt.png", "description": "Image corresponding to option (d): a volumetric flask with a narrow neck, pear-shaped bottom, an annular calibration mark on the neck, and a ground-glass stopper."}], "full_mark": 1}, {"sub_id": "Exam-09-T4-2", "tags": ["4.1", "4.5"], "text_open": "4-2 Alkalization. Assemble the Kjeldahl distillation apparatus as shown below. Fill vessel 1 with water to 2/3 of its volume. Add several glass beads, several drops of methyl red, and several milliliters of sulfuric acid until the solution is red. Heat the water in the steam generator to boiling and maintain boiling. Add $10.0 \\ \\mathrm{mL}$ of $20 \\ \\mathrm{g/L}$ boric acid solution and 3 drops of mixed indicator A to the receiving flask, and immerse the lower end of apparatus 8 beneath the liquid surface in vessel 9. Using a pipette, accurately transfer $5.0 \\ \\mathrm{mL}$ of sample digest through apparatus 4 into reaction chamber 5. Rinse the small cup with $10 \\ \\mathrm{mL}$ of water and allow the rinsing to flow into reaction chamber 5, then tightly insert the rod-shaped glass stopper in apparatus 4. Slowly add $400 \\ \\mathrm{g/L}$ sodium hydroxide solution through apparatus 4 into reaction chamber 5 and immediately seal the system. After distilling for $15 \\ \\mathrm{min}$, move the distillate receiver so that the liquid surface is below the lower end of the condenser. Continue distilling for $1 \\ \\mathrm{min}$. Verify with pH paper that the distillate is neutral, rinse the outside of the lower end of the condenser with a small amount of water, and remove the receiving flask. Which of the following statements about the experimental operations and their purposes are correct?\n(a). Methyl red is added to apparatus 1 to generate acidic vapor that reacts with ammonia in apparatus 5 to form an ammonium salt.\n(b). After sodium hydroxide is added, the glass stopper must be inserted tightly and a water seal formed immediately to prevent loss of ammonia.\n(c). The principal purpose of apparatus 6 is to separate liquid water from the steam generated in apparatus 1.\n(d). After alkalization and collection for one digest have been completed, the next reaction may be started by opening screw clamp 7, closing screw clamp 3, and replacing the solution in vessel 9 with distilled water, in that order, to discharge the waste liquid and rinse the reaction chamber.", "text_images": [{"file": "Exam-09-T4-2-1-prompt.png", "description": "This black-and-white line drawing shows an assembly of laboratory glassware. Three main sections on the left, center, and right are connected by tubing, and the numbers 1 through 9 identify specific components. Left section: number 1 points to the liquid inside a long-neck round-bottom flask, represented by a horizontal dashed line. The bottom of the flask rests on a square base. Number 2 points to this base, which has a circular knob at the center of its front face and four feet underneath. The mouth of the round-bottom flask is sealed with a stopper. A very long straight vertical glass tube passes through the center of the stopper, extending below the liquid surface inside the flask and far above the flask mouth. A short right-angle glass tube also passes through the stopper and is connected at its end to U-shaped flexible tubing. Center section: the U-shaped flexible tubing from the left connects to a narrow inlet above the central apparatus. The main central apparatus is inclined from lower left to upper right and contains an outer jacket and an inner structure. Number 4 points to an upward-facing vertical inlet at the upper-right end of the inclined apparatus; it has a stopper through which a short straight tube is inserted. Number 5 points to an elongated cavity or inner-tube structure inside the inclined apparatus. Number 6 points to the enlarged spherical lower-left end of the inclined apparatus. Number 7 points to a narrow tube extending from the spherical end below number 6, connected to a short length of flexible tubing fitted with a clamp (pinchcock). Number 3 points to another narrow tube extending from the upper-middle part of the outer jacket, likewise connected to a short length of flexible tubing fitted with a clamp (pinchcock). The upper-right end of the inclined apparatus connects to the right section through an inverted U-shaped bent glass tube. Right section: number 8 points to a vertically oriented straight condenser. The end of the inverted U-shaped bent glass tube enters the top of the condenser. Two horizontal side-arm connections extend to the right from the outer jacket of condenser 8. A right-pointing arrow beside the upper side arm indicates outward flow, and a left-pointing arrow beside the lower side arm indicates inward flow. Number 9 points to an Erlenmeyer flask directly beneath the condenser. The flask contains liquid, represented by a horizontal dashed line. The inner straight tube of condenser 8 extends downward through the mouth of the flask and below the liquid surface inside it."}], "full_mark": 2}]} | |
| {"problem_id": "Exam-09-T6", "header": "Mg-Air Batteries\nWith their high specific energy density, stable operating voltage, simple device construction, and excellent safety, magnesium-air batteries have become candidates for emergency backup power in hospitals, emergency power for disaster relief, portable field energy-storage devices, and automotive power supplies.", "header_images": [], "subquestions": [{"sub_id": "Exam-09-T6-1", "tags": ["2.8", "3.4", "3.6"], "text_open": "6-2-2 An ideal anode film both suppresses hydrogen evolution at the anode and does not impede interfacial electrochemical reactions. A film that is both \"dense and porous\" has therefore become a research objective. The K_{sp} values of hydroxides of selected elements are shown below. From Cu, La, and Sr, select the metallic element that can be added to the Mg-alloy electrode, and explain your choice. Given K_{sp}[Mg(OH)_2] = 5.61 × 10^{-12}, K_{sp}[Cu(OH)_2] = 2.2 × 10^{-20}, K_{sp}[La(OH)_3] = 2.0 × 10^{-19}, and K_{sp}[Sr(OH)_2] = 3.2 × 10^{-4}.", "text_images": [{"file": "Exam-09-T6-1-1-prompt.png", "description": "This is a plot with logarithmic coordinates. Vertical axis: $\\boldsymbol{[M^{n+}]}$ (unit: mol/L), on a logarithmic scale ranging from $10^{-13}$ to $10^{1}$. Horizontal axis: $\\boldsymbol{pH}$, ranging from 0 to 14. Solubility-equilibrium curves for various metal hydroxides and metal oxides are plotted and labeled with different line styles, including solid and dashed lines. The substances include the following. Hydroxides: $Sn(OH)_2$, $Fe(OH)_2$, $Fe(OH)_3$, $Ga(OH)_3$, $Sn(OH)_4$, $Pb(OH)_4$, $Ce(OH)_4$, $Sc(OH)_3$, $In(OH)_3$, $Al(OH)_3$, $Dy(OH)_3$, $Y(OH)_3$, $Sm(OH)_3$, $Eu(OH)_3$, $Ce(OH)_3$, $Nd(OH)_3$, $Gd(OH)_3$, $La(OH)_3$, $Ni(OH)_2$, $Mn(OH)_2$, $Mg(OH)_2$, $Cd(OH)_2$, $Co(OH)_2$, $Zn(OH)_2$, $Pb(OH)_2$, and $Ca(OH)_2$. Oxides: $Fe_2O_3$, $FeO$, $MnO_2$, $Ga_2O_3$, $PbO_2$, $SnO_2$, $ZrO_2$, $In_2O_3$, $Al_2O_3$, $Sc_2O_3$, $Nd_2O_3$, $MgO$, $Dy_2O_3$, $SnO$, $ZnO$, $Sm_2O_3$, $Gd_2O_3$, $Y_2O_3$, $NiO$, $NiO_2$, $PbO$, $CoO_2$, and $MnO$. Overall trend of the curves: as pH increases, the corresponding metal-ion concentration $\\boldsymbol{[M^{n+}]}$ decreases."}], "full_mark": 4}, {"sub_id": "Exam-09-T6-2", "tags": ["2.8", "3.6"], "text_open": "6-2-3 In practical use, additional Ca is incorporated into the Mg alloy. Explain why. (K_{sp}[Ca(OH)_2] = 5.5 × 10^{-6})", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-09-T6-3", "tags": ["2.8", "3.6"], "text_open": "6-3 Investigation of second-phase effects\n\nAlloying is an effective means of modifying materials and can significantly improve the discharge performance of magnesium alloys. The type, morphology, and distribution of the second phase, i.e., another phase distributed discontinuously within the matrix phase, directly affect material performance. When a new metal is added to the base electrode material during alloy-electrode manufacture, a second phase forms. Microgalvanic couples between the second phase and matrix phase undergo galvanic corrosion. Galvanic corrosion occurs when two different metals are in contact while simultaneously exposed to an electrolyte, forming a spontaneous cell in which the more active metal acts as the anode and corrodes. The rate of galvanic corrosion depends on the contact area between the two metals.\n\n6-3-1 The anode efficiency is calculated as $P = \\frac{W_f}{\\Delta W}$, where $\\Delta W$ is the actual mass loss and $W_f$ is the theoretical mass loss, calculated from $W_f = \\frac{I \\times t}{F \\times \\sum\\limits_{i=1}^{n} \\frac{X_i \\times N_i}{M_i}}$. Here, $F$ is the Faraday constant, $X_i$ is the mass fraction of each element in the magnesium anode, $N_i$ is the corresponding number of electrons transferred by each element in the electrochemical process, and $M_i$ is the atomic mass of each element.\n\nAlloy materials A and B have the following compositions:\n\n| Alloy material | La | Ce | Ca | Al | Zn | Mg |\n| --- | --- | --- | --- | --- | --- | --- |\n| A | / | / | / | 6% | 3% | 91% |\n| B | 0.15% | 0.45% | 0.3% | 6% | 3% | 90.1% |\n\nFor $\\Delta W$, $\\Delta W_A = 36.27\\,\\text{mg}$ and $\\Delta W_B = 30.01\\,\\text{mg}$. Calculate the anode efficiencies of A and B at a current density of $0.5\\,\\text{mA/cm}^2$, duration of $6\\,\\text{h}$, and area of $14\\,\\text{cm}^2$.", "text_images": [], "full_mark": 6}, {"sub_id": "Exam-09-T6-4", "tags": ["2.8", "3.6"], "text_open": "6-3-2 Subsequent research proposed that the difference in efficiency between alloy materials A and B originates from a change in the second phase. After addition of the new metals, the second phase changes from the particulate and planar morphologies in A to the rod-like and relatively continuous β-Mg_{17}Al_{12}/τ-Mg_{32}(Al, Zn)_{49} and Al_{2}(La, Ce) structures in B. Explain the difference in efficiency on this basis.", "text_images": [], "full_mark": 2}]} | |
| {"problem_id": "Exam-09-T7", "header": "Semiconductors and Electrochemistry\nSemiconductors are materials whose electrical conductivity at room temperature lies between that of conductors and insulators. Common semiconductor materials include silicon, germanium, and gallium arsenide. Based on band theory and Fermi statistics, doping with group V or group III elements, or with other materials that can donate electrons (donors) or accept electrons (acceptors), can regulate the Fermi level and carrier concentration of a semiconductor and thereby its various properties. Semiconductor carriers generally comprise electrons and holes. In this problem, involving an n-type semiconductor in which electrons are the majority carriers, only electrons need be considered. Nonuniform doping can create a carrier-concentration gradient in a semiconductor. The resulting diffusion causes charge separation and a built-in electric field; the built-in field opposes diffusion until equilibrium is reached. This is fundamentally similar to the electrochemical picture in which a chemical-potential difference drives charge transfer, which in turn produces a potential difference that opposes further charge transfer, ultimately establishing an electrode potential at equilibrium. By comparing a concentration-gradient-doped semiconductor with the electromotive force of an electrochemical concentration cell, this problem explores the universality of diffusion-electric-field equilibrium.\n7.1 Built-in electric field and diffusion equilibrium in a semiconductor\nConsider an n-type semiconductor at temperature T = 300 K with electron concentration varying with position x as $n(x)=N_{d0}\\exp(-x/L)$. Here $N_{d0}=1\\times10^{18}\\ \\mathrm{cm}^{-3}$, $L=0.1\\ \\mu\\mathrm{m}$, $D_n=225\\ \\mathrm{cm}^2\\mathrm{/s}$, $e=1.6\\times10^{-19}\\ \\mathrm{C}$, and $k=1.38\\times10^{-23}\\ \\mathrm{J/K}$. These quantities are, respectively, the donor concentration at x = 0 (which, under the approximation used here, may be regarded as the electron concentration at x = 0), characteristic length, electron diffusion coefficient, elementary charge, and Boltzmann constant.", "header_images": [], "subquestions": [{"sub_id": "Exam-09-T7-1", "tags": ["3.6"], "text_open": "7.2 Electrochemistry. 7.2.1 When an iron sheet is immersed in water, an electrical double layer forms near the iron electrode, as shown below. Briefly explain how the structure forms, and qualitatively sketch the absolute value of the potential as a function of distance from the metal surface, taking the potential infinitely far from the electrode to be zero.", "text_images": [{"file": "Exam-09-T7-1-1-prompt.png", "description": "This two-dimensional diagram, titled \"Figure 9.4 Schematic Structure of the Electrical Double Layer,\" shows the spatial distribution of an electrical double layer from left to right. At the far-left edge is a vertical row of minus signs (short horizontal lines). Immediately to their right is a vertical solid line shaded with diagonal hatching on its left side; the words \"Electrode surface\" are set vertically to the left of this line. A short distance to the right is a parallel vertical dashed line bearing a neat vertical series of small circles, each containing a \"+\" sign. The narrow interval between the solid line and this middle dashed line is marked at the bottom by a horizontal upward brace, beneath which the spacing symbol $d$ and the label \"Compact layer\" appear. Extending rightward from the middle dashed line to another parallel dashed boundary at the far right is a wider rectangular region. Several similar small circles containing \"+\" signs are scattered irregularly within it. This wider interval is likewise marked at the bottom by a horizontal brace, beneath which the spacing symbol $\\delta$ and the label \"Diffuse layer\" appear."}], "full_mark": 5}]} | |
| {"problem_id": "Exam-10-T3", "header": "Question 3 Anticoagulant Polymers (12 points, 5%). Anticoagulation is the prevention or treatment of thrombotic disease by inhibiting blood coagulation. Traditional natural anticoagulants, such as heparin, have limitations including structural complexity and substantial batch-to-batch variation. Developing synthetic anticoagulants with well-defined structures and controllable properties has therefore become an important research direction in medicinal chemistry and materials science.", "header_images": [], "subquestions": [{"sub_id": "Exam-10-T3-1", "tags": ["2.4", "2.5"], "text_open": "3.1.1 ACVA, with the condensed structural formula $\\mathrm{HOOCCH_2CH_2C(CN)(CH_3)N=NC(CN)(CH_3)CH_2CH_2COOH}$, is a water-soluble radical initiator and is used in a much smaller amount than reagent BM1429 in the synthesis. Write the SMILES strings of the major and minor polymer products, explicitly showing the end-group structures. The 2 products differ only in their end groups.", "text_images": [{"file": "Exam-10-T3-1-1-prompt.png", "description": "Polymerization of sodium 4-vinylbenzenesulfonate, whose SMILES may be written as 'C=CC1=CC=C(C=C1)S(=O)([O-])=O.[Na+]', in the presence of BM1429, with SMILES 'O=C(O)CCSC(=S)SC(C)C(=O)O', ACVA, and water at 80 °C for 4 h."}], "full_mark": 4}, {"sub_id": "Exam-10-T3-2", "tags": ["2.2", "2.3", "2.5"], "text_open": "3.1.2 Continue to use the reaction scheme from 3.1.1. Besides becoming an end group in the product, reagent BM1429 reversibly binds to the radical polymer chain, thereby lowering the radical concentration in the system and controlling the reaction. Write the SMILES string of the intermediate formed by this reversible binding. The end group remote from the reaction center need not be shown.", "text_images": [], "full_mark": 2}]} | |
| {"problem_id": "Exam-10-T5", "header": "Structural Chemistry of Two-Dimensional Transition-Metal Dichalcogenides and Their Derivatives (26 points, 12%). Two-dimensional transition-metal dichalcogenides, or TMDs, generally have layered structures held together by van der Waals forces. Their symmetry and physical properties change substantially as the number of layers decreases.", "header_images": [], "subquestions": [{"sub_id": "Exam-10-T5-1", "tags": ["1.3"], "text_open": "Structural background: $WSe_2$ has a molybdenite-type hexagonal layered structure. The figure below is a projection along $a$; large spheres represent W and small spheres Se. One unit cell contains two layers along $c$.\n5-1-2 Explain how W and Se occupy the hexagonal plane lattice in a projection along the c axis, and give the projected relationship within one unit cell.", "text_images": [{"file": "Exam-10-T5-1-1-prompt.png", "description": "A two-dimensional projection of a vertical rectangular unit cell outlined by thin black lines. Axes at the lower left show c vertically upward and b horizontally to the left. The figure contains two larger gray spheres and eight smaller green spheres joined by slender cylindrical bonds. An upper gray sphere lies inside the upper half of the cell near its right boundary and is connected to four green spheres at the upper-left, lower-left, upper-right, and lower-right positions. The two green spheres on the left lie within the cell, whereas the two on the right lie outside the right boundary. A lower gray sphere lies inside the lower half near the left boundary and is likewise connected to four green spheres at the upper-left, lower-left, upper-right, and lower-right positions. The two green spheres on the left lie outside the left boundary, whereas the two on the right lie within the cell. The two gray spheres are not directly connected; together they form two separate four-coordinate units offset vertically and oriented in opposite horizontal directions."}], "full_mark": 2}]} | |
| {"problem_id": "Exam-10-T6", "header": "The 2025 Nobel Prize in Chemistry was awarded to Susumu Kitagawa, Richard Robson, and Omar M. Yaghi for their contributions to research on metal-organic frameworks, or MOFs. Chemists recently prepared a new Cr-MOF material, CPM-243, that resists strong acids and bases and remains stable over approximately 16 pH units under extreme conditions ranging from pH < 0 to pH > 14, redefining the feasibility of Cr-MOFs in the search for the most chemically stable MOFs.", "header_images": [], "subquestions": [{"sub_id": "Exam-10-T6-1", "tags": ["1.4", "2.4"], "text_open": "6-2 Cr-MOFs are currently known for MOFs based on $\\mathrm{Cr_3O}$ trimers, such as MIL-88(Cr), generally with dicarboxylate ligands such as terephthalate. Write the SMILES string of a single $[\\mathrm{Cr_3O(RCOO)_6}]$ metal-cluster node in this MOF. Use 'COC(C)OC' to represent $\\mathrm{RCOO^-}$ and disregard the charge on the node.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-10-T6-2", "tags": ["1.4", "2.1"], "text_open": "6-3-2 Known Cr-trimer MOFs have pendant ligands, such as F^{-}/OH^{-}/$\\mathrm{H_2O}$, and correspondingly low-connectivity frameworks. In CPM-243(Ac), chemists instead used the tripyridyl ligand tpt to provide three additional intercluster cross-links through Cr-N bonds, producing a 9-connected MOF framework. The synthesis was performed by adding chromium(III) acetate hydrate, $\\mathrm{Cr(Ac)_3 \\cdot 2H_2O}$, terephthalic acid, $\\mathrm{H_2bdc}$, 2,4,6-tris(4-pyridyl)-1,3,5-triazine, tpt, and glacial acetic acid, $\\mathrm{Hac}$, to a polytetrafluoroethylene liner. After adding 5.0 g of water and stirring for one hour, the vessel was sealed and heated for 3 days in an oven preheated to $220^\\circ\\mathrm{C}$. Given that CPM-243(Ac) has the formula $[(\\mathrm{Cr_3O})(\\mathrm{bdc})_x(\\mathrm{tpt})_y(\\mathrm{ac})]$, first determine $x$ and $y$ from the structure, and then write the reaction equation.", "text_images": [{"file": "Exam-10-T6-2-1-prompt.png", "description": "Structure of the organic compound tpt, with the SMILES 'C1(C2=CC=NC=C2)=NC(C3=CC=NC=C3)=NC(C4=CC=NC=C4)=N1'."}], "full_mark": 4}]} | |
| {"problem_id": "Exam-10-T9", "header": "", "header_images": [], "subquestions": [{"sub_id": "Exam-10-T9-1", "tags": ["2.3", "2.4", "2.7"], "text_open": "9-1-1 In radical reactions of allylic compounds using NBS or HBr, substitution and addition compete, and different reagents show different selectivities. Write the major products of the following two reactions.", "text_images": [{"file": "Exam-10-T9-1-1-prompt.png", "description": "Two organic reaction schemes. In the first, 'CC1(C)C=CCCC1' reacts with NBS under irradiation. In the second, 'CC(C1CCCCC1)=C' reacts with HBr and $\\mathrm{(BzO)_2}$."}], "full_mark": 4}, {"sub_id": "Exam-10-T9-2", "tags": ["2.2", "2.3"], "text_open": "9-2-1 Radical reactions can also achieve efficient intramolecular N-C bond coupling. Write the key intermediates in the following reaction.", "text_images": [{"file": "Exam-10-T9-2-1-prompt.png", "description": "Organic reaction scheme in which 'C1(CN2CCCCC2)NCCCC1' reacts with NCS under irradiation to give 'N12CCCCC1CN3C2CCCC3'."}], "full_mark": 5}, {"sub_id": "Exam-10-T9-3", "tags": ["2.4"], "text_open": "9-3-2 Write the structure of the major product of the following reaction. Disregard stereochemistry.", "text_images": [{"file": "Exam-10-T9-3-1-prompt.png", "description": "Organic reaction scheme in which 'IC1C(OC/C=C/C)OCCC1' reacts with triethylborane and oxygen in water and hypophosphorous acid."}], "full_mark": 2}, {"sub_id": "Exam-10-T9-4", "tags": ["2.2", "2.3", "2.4"], "text_open": "9-3-4 The following reaction undergoes efficient triethylborane-catalyzed cyclization to give two isomers, A1 and A2, of which A2 is the major product. Write the SMILES string of A2 and those of the key intermediates leading to A2.", "text_images": [{"file": "Exam-10-T9-4-1-prompt.png", "description": "Organic reaction scheme in which 'CC1(C)C(O1)/C=C/CCCC#C' reacts with triethylborane and triphenyltin hydride in benzene to give A1, 'OC(C)(C)/C=C/C(CCC/1)C1=C[Sn](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4', and A2."}], "full_mark": 10}]} | |
| {"problem_id": "Exam-11-T3", "header": "A photoresist is a polymeric material sensitive to light of specific wavelengths. Upon irradiation, it undergoes specific chemical reactions that substantially change its solubility in a developer, thereby forming the desired pattern on the surface of a semiconductor material. As a key patterning material in integrated-circuit fabrication, advances in photoresist chemistry directly determine the limits of process miniaturization and the industry's capacity for technological self-reliance.", "header_images": [], "subquestions": [{"sub_id": "Exam-11-T3-1", "tags": ["2.1", "2.5"], "text_open": "3.1 The novolak-diazonaphthoquinone (Novolak-DNQ) system is the most widely used photoresist for the g-line (436 nm) and i-line (365 nm). One component is a high-molecular-weight linear phenolic resin, and the other is diazonaphthoquinone, whose structure is shown below.\n3.1.1 The phenolic resin prepared under acidic conditions is a low-molecular-weight linear prepolymer. The curing agent hexamethylenetetramine may be added to an aqueous solution of the prepolymer to cure and cross-link it upon heating. Identify which substrate must be used in excess when preparing the acid-catalyzed phenolic resin, write the decomposition equation of the curing agent upon heating, and on this basis state the role of the curing agent.", "text_images": [{"file": "Exam-11-T3-1-1-prompt.png", "description": "'O=C1C2=CC=CC=C2C(S(=O)(O[Ar])=O)=CC1=[N+]=[N-]'"}], "full_mark": 5}, {"sub_id": "Exam-11-T3-2", "tags": ["1.6"], "text_open": "3.1.2 State the principal intermolecular forces between the two components of the Novolak-DNQ system.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-11-T3-3", "tags": ["2.3", "2.4"], "text_open": "3.1.3 In the Novolak-DNQ system, the photosensitive component undergoes rearrangement upon irradiation to form unstable structure A, which then forms B in the alkaline developer N(CH_3)_4^+OH^-, changing the solubility of the photoresist. Write the SMILES representations of A and B.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-11-T3-4", "tags": ["2.4", "2.5"], "text_open": "3.2 PVCi is a high-resolution photoresist. One method for synthesizing PVCi is shown below.\n3.2.1 The first step is a conventional free-radical polymerization of a double bond, and all subsequent steps are functional-group substitutions. Write the SMILES representations of X, Y, Z, and PVCi. The terminal-group structures need not be specified; use [*] to denote the connection sites at the two ends of the polymer backbone.", "text_images": [{"file": "Exam-11-T3-4-1-prompt.png", "description": "X (C_4H_6O_2) reacts under AIBN/CH_3OH conditions to give Y; Y reacts under CH_3OH/NaOH conditions to give Z; Z reacts with 'O=C(Cl)/C=C/C1=CC=CC=C1' to give PVCi."}], "full_mark": 6}, {"sub_id": "Exam-11-T3-5", "tags": ["2.3", "2.5"], "text_open": "3.2.2 Write the SMILES representation of the structure formed by the interchain reaction of PVCi upon irradiation at 360 nm. The terminal-group structures and degrees of polymerization need not be specified; use [*] and [**] to denote connection sites in the polymer chains.", "text_images": [], "full_mark": 2}]} | |
| {"problem_id": "Exam-11-T5", "header": "Problem 5: Familiar Structures from an Unfamiliar Perspective (18 points, 7%)", "header_images": [], "subquestions": [{"sub_id": "Exam-11-T5-1", "tags": ["1.3"], "text_open": "5.2 Crystal Y has the chemical formula QF_y. Its projections along the a and c directions are shown below; black spheres represent Q ions and white spheres represent F^-. 5.2.1 Give the value of y.", "text_images": [{"file": "Exam-11-T5-1-1-prompt.png", "description": "The image contains two two-dimensional projections of a crystal structure. The left panel is a rectangular view of the b-c plane. Axes at the lower left show c vertically upward, b horizontally to the right, and a perpendicular to the page. A large black sphere is located at each of the four corners, with two-dimensional fractional coordinates (b, c) = (0.00, 0.00), (1.00, 0.00), (0.00, 1.00), and (1.00, 1.00). Two large black spheres lie inside the unit cell at approximately (0.33, 0.33) and (0.67, 0.67). The small white spheres in the unit cell include 4 single spheres at approximately (b, c) = (0.00, 0.50), (0.67, 0.17), (1.00, 0.50), and (0.33, 0.83), and 3 pairs of overlapping spheres at approximately (b, c) = (0.17, 0.17), (0.50, 0.50), and (0.83, 0.83). The right panel is a parallelogram view of the a-b plane. Axes at the lower left show a horizontally to the right, b diagonally toward the upper left, and c perpendicular to the page; the angle between a and b is approximately 120°. A large black sphere is located at each of the four corners, with two-dimensional fractional coordinates (a, b) = (0.00, 0.00), (1.00, 0.00), (0.00, 1.00), and (1.00, 1.00). Two large black spheres lie inside the unit cell at approximately (0.33, 0.67) and (0.67, 0.33). All the small white spheres in the unit cell are single spheres, 11 in total, at approximately (0.00, 0.50), (0.50, 0.00), (0.33, 0.17), (0.17, 0.33), (0.83, 0.17), (0.50, 0.50), (0.17, 0.83), (1.00, 0.50), (0.83, 0.67), (0.67, 0.83), and (0.50, 1.00)."}], "full_mark": 2}]} | |
| {"problem_id": "Exam-11-T7", "header": "Problem 7: Investigation and Isolation of Complexes of Group $\\mathrm{IVA}$ Elements. Transition-metal $\\pi$ complexes are key intermediates in many transition-metal-catalyzed reactions and have been widely used as catalysts and synthetic reagents. Recently, the first main-group-metal $\\pi$ complex was successfully isolated.", "header_images": [], "subquestions": [{"sub_id": "Exam-11-T7-1", "tags": ["1.4", "1.5"], "text_open": "7-1 Alkene complexes of main-group elements are difficult to synthesize. Bonding between a metal and an unsaturated hydrocarbon can be described by the Dewar-Chatt-Duncanson model, as shown below. Following the example, draw a diagram illustrating the orbital interactions in an alkene $\\pi$ complex of a heavier main-group element.", "text_images": [{"file": "Exam-11-T7-1-1-prompt.png", "description": "Left section: a red letter M appears at the top. Directly below M is a molecular structure containing a $\\mathrm{C=C}$ double bond, enclosed by a parallelogram to indicate that it lies in a plane. A vertical dashed line connects M to the center of the $\\mathrm{C=C}$ bond. Each carbon atom of the double bond is bonded to two R groups, whose three-dimensional orientations are indicated by solid and hashed wedges. Below is the label “M = Transition-metals.” The middle section shows two sets of orbitals, one above the other. Below is a $\\mathrm{C-C}$ framework with attached R groups. Each carbon bears an “8”-shaped orbital, and the two orbitals have the same phase, with both lower lobes unshaded. Above is a four-lobed metal orbital whose two horizontal lobes are shaded and whose two vertical lobes are unshaded. Its lower vertical lobe faces the carbon orbitals below. Two upward arrows run from the shaded upper lobes on the carbon atoms to the horizontal lobes of the metal orbital. The label below reads “$\\sigma$-donation.” 3. The right section likewise shows two sets of orbitals, one above the other. The orbitals on the two carbon atoms of the lower $\\mathrm{C-C}$ framework have opposite phases: the left orbital is unshaded below and shaded above, while the right orbital is shaded below and unshaded above. The metal orbital above has four lobes with diagonally distributed phases: the upper-right and lower-left lobes are shaded, while the upper-left and lower-right lobes are unshaded. Two downward arrows point from the two lower lobes of the metal orbital to the corresponding upper lobes of the carbon orbitals. The label below reads “$\\pi$-backdonation.”"}], "full_mark": 2}, {"sub_id": "Exam-11-T7-2", "tags": ["1.4", "1.5"], "text_open": "7-2 Zero-valent lead complex 1 undergoes a combination reaction with $\\mathrm{P_4}$ in toluene at room temperature to form lead bis(phosphene) $\\pi$ complex 2, in which $\\mathrm{Pb}$ is coordinated by two phosphenes containing $\\mathrm{P=P}$ bonds. Write the SMILES representation of 2; [R] may be used. The only atoms coordinated to $\\mathrm{Pb}$ are $\\mathrm{P}$ atoms.", "text_images": [{"file": "Exam-11-T7-2-1-prompt.png", "description": "Reactant SMILES: '[R]C1[Ge](C2=C(C=CC=C2C3=C(C=C(C=C3C(C)C)C(C)C)C(C)C)C4=C(C=C(C=C4C(C)C)C(C)C)C(C)C)[Pb][Ge](C5=C(C6=C(C=C(C=C6C(C)C)C(C)C)C(C)C)C=CC=C5C7=C(C=C(C=C7C(C)C)C(C)C)C(C)C)C1[R]'. The number “1” is shown beside it. A right-pointing arrow appears to its right, with 1.0 equiv $\\mathrm{P_4}$ above the arrow and toluene, RT, 1 h below it. The arrow points to the number 2. C[R]-C[R] denotes a C2B12 unit."}], "full_mark": 3}, {"sub_id": "Exam-11-T7-3", "tags": ["1.4"], "text_open": "7-4 A synthetic route to a complex containing a ligand of a Group $\\mathrm{IVA}$ element is as follows. At ambient temperature, $\\mathrm{AuCl}$ reacts with triethylphosphine in a 1:1 ratio to give complex A. At room temperature, ligand X (shown below) and $\\mathrm{KC_8}$ are added to a tetrahydrofuran solution of A, affording the desired highly symmetric precursor B, which contains three $\\mathrm{Au}$ atoms and has a planar framework. Addition of an equimolar amount of A to B gives molecule C, which contains one $\\mathrm{Au-Au}$ bond. Measurements indicate that C is aromatic. Note: this molecule has an extended conjugated system whose electrons originate from the $\\mathrm{d}$ electrons of $\\mathrm{Au}$. The structure of X is 'C[Si](C)(C1([Si](C)(C)C)CCC([Si](C)(C)C)([Si](C)(C)C)[Sn]1)C'. 7-4-1 Write the SMILES representations of B and C.", "text_images": [{"file": "Exam-11-T7-3-1-prompt.png", "description": "SMILES representation of X: 'C[Si](C)(C1([Si](C)(C)C)CCC([Si](C)(C)C)([Si](C)(C)C)[Sn]1)C'"}], "full_mark": 6}, {"sub_id": "Exam-11-T7-4", "tags": ["1.4", "1.5"], "text_open": "7-4-2 In fact, C itself is not aromatic, but polarization can generate an ionic structure whose anion is aromatic. Write the SMILES representations of the anion and cation in this structure, and state the number of $\\pi$ electrons in the extended conjugated system of the anion.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-11-T7-5", "tags": ["1.5", "1.6"], "text_open": "7-4-3 Researchers found that when the coordinating atom in X is $\\mathrm{Sn}$, C shows relatively weak aromaticity; replacing the coordinating atom in X with $\\mathrm{Ge}$ markedly enhances the aromaticity. Explain this observation.", "text_images": [], "full_mark": 2}]} | |
| {"problem_id": "Exam-11-T10", "header": "Chiral Boron Chemistry. Organoboron compounds occupy an important niche in asymmetric synthesis and have become versatile stereospecific building blocks. In general, the boron atom is an electron-deficient center; therefore, the most important intermediates in these reactions are boronate anions (ate complexes).", "header_images": [], "subquestions": [{"sub_id": "Exam-11-T10-1", "tags": ["1.5", "2.2"], "text_open": "10.1 Homologation is a common application of organoboron reagents in organic synthesis. A representative example is the Matteson homologation, whose general reaction pattern is shown below.\n10.1.1 An ate-complex intermediate is formed in the reaction. Write the SMILES representations of the 2 key boron-containing intermediates in the transformation, including stereochemistry.", "text_images": [{"file": "Exam-11-T10-1-1-prompt.png", "description": "The image shows a one-step Matteson homologation. The reactant is 'C[C@H]1[C@H](C)OB(C)O1', and the conditions are 1) CH_2Cl_2, LDA; 2) ZnCl_2; 3) RMgBr. The product is 'C[C@H]1[C@H](C)OB([C@@H](C)[R])O1'."}], "full_mark": 4}, {"sub_id": "Exam-11-T10-2", "tags": ["2.2", "2.7"], "text_open": "10.1.2 Studies show that the addition of ZnCl_2 is the key step controlling the stereospecificity of the product. ZnCl_2 acts as a chelating agent and forms a 5-membered chelate ring, and the principal factor affecting the stability of the migration transition state is 1,3-strain between bulky groups. On this basis, write the SMILES representation of the key intermediate leading to the major product, including stereochemistry.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-11-T10-3", "tags": ["2.7"], "text_open": "10.1.3 If CH_2Cl_2 in the first step is replaced with CH_3CHCl_2, the stereoselectivity of the product will: ( ).\n(a) increase (b) decrease (c) remain essentially unchanged", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-11-T10-4", "tags": ["1.2", "2.4", "2.7"], "text_open": "10.1.4 Based on the information above, write the SMILES representation of the product of the following reaction; the product contains neither silicon nor sulfur. Explain the stereoselectivity of the final step.", "text_images": [{"file": "Exam-11-T10-4-1-prompt.png", "description": "The image shows a three-step reaction. The reactant is 'CC1(C)C(C)(C)OB(CC2=CC=CC=C2)O1'. The first-step conditions are 'ClC([Li])[Si](C)(C)C', THF, -78°C then r.t., ZnCl_2. The second-step conditions are '[Li]C(S[Ar])O[R]', THF, -78°C, ZnCl_2. The third-step condition is H_2SO_4. The final product is unknown and indicated by a question mark."}], "full_mark": 3}, {"sub_id": "Exam-11-T10-5", "tags": ["1.5", "1.6"], "text_open": "10.2 A major advantage of organoboron reagents is the exceptional configurational stability of their chiral carbon atoms, which enables metal-free stereospecific coupling reactions.\n10.2.1 The reactivity of a boron reagent can be tuned by adding different nucleophiles. Among the ate complexes below, the bold C-B bond is most reactive in ( ) and least reactive in ( ).", "text_images": [{"file": "Exam-11-T10-5-1-prompt.png", "description": "The image shows four structures labeled (a), (b), (c), and (d) from left to right. (a) is 'CC1(C)O[B-](C)(OC)OC1(C)C'; (b) is 'CC1(C)O[B-](C)(N(C)C)OC1(C)C'; (c) is 'CC1(C)O[B-](C)(C)OC1(C)C'; and (d) is 'C[B-]1([C@@H]2CCC[C@H]1CCC2)C'. In every structure, a B-CH_3 bond is oriented axially with respect to the five-membered ring and is shown in bold."}], "full_mark": 2}, {"sub_id": "Exam-11-T10-6", "tags": ["1.2", "2.4"], "text_open": "10.2.2 Write the SMILES representations of products A and B of the following reactions, including stereochemistry.", "text_images": [{"file": "Exam-11-T10-6-1-prompt.png", "description": "The image shows two one-step reactions. The reactant in the first is '[H][C@@](C1=CC=CC=C1)(C)B2OC(C)(C)C(C)(C)O2', and the conditions are MeONH_2, nBuLi, then Boc_2O; the unknown product is labeled A. The reactant in the second is 'CC(C(C)(C)O1)(C)OB1[C@](C2=CC=CC=C2)(C)CC', and the conditions are CH_2Cl_2, nBuLi, ZnCl_2, then NaOH/H_2O_2; the unknown product is labeled B."}], "full_mark": 3}, {"sub_id": "Exam-11-T10-7", "tags": ["2.3", "2.7"], "text_open": "10.2.3 An organoboron reagent treated with an aryllithium reagent can trap an electrophile. Experiments show, however, that different electrophiles give different stereochemical outcomes: the ate complex reacts with I_2 with good stereoselectivity, whereas reaction with DIAD results in extensive racemization. The researchers propose that the two reactions proceed by different mechanisms. Explain these observations mechanistically. Note: carbocations are planar, carbon-centered radicals undergo rapid configurational inversion, and carbanions have a high barrier to configurational inversion.", "text_images": [{"file": "Exam-11-T10-7-1-prompt.png", "description": "The image shows a multistep reaction in which the substrate is the same but the conditions differ. The substrate is 'C[C@H](B1OC(C)C(C)(C)O1)CCC2=CC=CC=C2'. The first-step conditions are 4-MeOC_6H_4Li, THF, -78°C, and the first-step product is 'C[C@H]([B-]1(C2=CC=C(OC)C=C2)OC(C)(C)C(C)(C)O1)CCC3=CC=CC=C3.[Li+]'. This product follows two pathways. The upper-right pathway uses I_2 and gives 'C[C@@H](I)CCC1=CC=CC=C1' with 97% enantioselectivity. The lower-right pathway uses DIAD, whose SMILES representation is 'O=C(/N=N/C(OC(C)C)=O)OC(C)C', and gives 'C[C@@H](N(NC(OC(C)C)=O)C(OC(C)C)=O)CCC1=CC=CC=C1' with 13% enantioselectivity."}], "full_mark": 4}, {"sub_id": "Exam-11-T10-8", "tags": ["1.2", "2.2", "2.4"], "text_open": "10.3 Coupling reactions of organoboron reagents can also be conducted in tandem with other reactions, enabling a broader range of synthetic applications.\n10.3.1 The following reaction couples a boron reagent with a thiazole ring. Write the SMILES representations of the product and 3 key boron-containing intermediates, including stereochemistry.", "text_images": [{"file": "Exam-11-T10-8-1-prompt.png", "description": "The image shows a two-step reaction. The substrate is 'O=S1C=C(C)C2=CC=CC=C21'. The reaction conditions are 1) LDA, -78°C; 2) '[Bpin][C@@H](C)C1=CC=CC=C1', followed by Tf_2O, -78°C in the second step. The product is unknown and indicated by a question mark."}], "full_mark": 4}, {"sub_id": "Exam-11-T10-9", "tags": ["2.2", "2.4"], "text_open": "10.3.2 Examine the following reaction and write the SMILES representations of the product and the anionic ate-complex intermediate. Note: the purpose of adding acid is to promote the migration.", "text_images": [{"file": "Exam-11-T10-9-1-prompt.png", "description": "The image shows a multistep reaction. The substrate is 'C=[C]=COC'. The conditions over the first arrow are 1) nBuLi; 2) '[Bpin][C@@H](C)C1=CC=CC=C1'; 3) AcOH. The conditions over the second arrow are NaOH/H_2O_2, THF/H_2O. The product is unknown and indicated by a question mark."}], "full_mark": 4}]} | |
| {"problem_id": "Exam-12-T2", "header": "Problem 2 Schrödinger's Hydrogen-Atom Model\nThe year 2026 marks the 100th anniversary of the publication of Schrödinger's wave mechanics. In Bohr's model of the hydrogen atom, the electron is a particle moving in a fixed circular orbit. Schrödinger's wave mechanics instead treats the electron as having wave-particle duality and therefore as unable to possess simultaneously definite position and momentum. In wave mechanics, the state of motion of a particle is described by a wavefunction. In general, a wavefunction depends on time and position; if it is time-independent, it reduces to a function of position. The wavefunction itself does not directly give definite values of physical quantities such as position or momentum. Rather, it contains information about their probability distributions. Appropriate mathematical operations on the wavefunction yield the probability of finding the particle at a given location, the possible values of its momentum, and the energy of the system.\nHow is the wavefunction obtained? A fundamental postulate of quantum mechanics is that the wavefunction satisfies the Schrödinger equation. This equation is a partial differential equation, and its solutions are wavefunctions. Thus, one first writes the Schrödinger equation for an electron, solves it for a wavefunction, and then uses the resulting wavefunction to determine information about position, momentum, energy, and other quantities.\nIf the potential-energy distribution in space is time-independent, the wavefunction is a function only of position and the equation is called the time-independent Schrödinger equation. For a hydrogen atom, for example, the nucleus may be approximated as fixed; the equation describing the electron is then time-independent, and the electron wavefunction depends on its position coordinates. Position may be expressed in Cartesian coordinates $(x,y,z)$, but spherical polar coordinates $(r,\\theta,\\phi)$ are often used. The physical meanings of $r,\\theta,\\phi$ are shown below, with $r>0,0 \\leq \\theta \\leq \\pi,0 \\leq \\phi \\leq 2\\pi$.\nThe time-independent Schrödinger equation for the hydrogen atom is\n$$-\\frac{\\hbar^2}{2m}\\nabla^2\\psi(r,\\theta,\\phi)-\\frac{Ke^2}{r}\\psi(r,\\theta,\\phi)=E\\psi(r,\\theta,\\phi)$$\nwhere $\\nabla^2$ is the Laplace operator, which in spherical polar coordinates is\n$$\\nabla^2=\\frac{\\partial^2}{\\partial r^2}+\\frac{2}{r}\\frac{\\partial}{\\partial r}+\\frac{1}{r^2}\\left(\\frac{\\partial^2}{\\partial \\theta^2}+\\frac{1}{\\tan\\theta}\\frac{\\partial}{\\partial \\theta}+\\frac{1}{\\sin^2\\theta}\\frac{\\partial^2}{\\partial \\phi^2}\\right)$$\nHere $m$ is the electron mass, $K$ is the electrostatic-force constant, and $\\hbar$ is the reduced Planck constant. $E$ is called the energy eigenvalue, or simply the electron energy. Although the electron's momentum and position are uncertain, its energy is definite.\nNote: $\\partial$ denotes partial differentiation. If $f$ is a function of $x,y$, then $\\frac{\\partial f(x,y)}{\\partial x}$ is defined as the derivative of $f$ with respect to $x$ while the independent variable $y$ is held constant:\n$$\\frac{\\partial f(x,y)}{\\partial x}=\\lim_{\\Delta x\\to 0}\\frac{f(x+\\Delta x,y)-f(x,y)}{\\Delta x}$$\nWhen $f$ is independent of $y$, so that $\\frac{\\partial f}{\\partial y}=0$ identically, $f$ becomes a single-variable function of $x$ and $\\frac{\\partial f}{\\partial x}=\\frac{\\mathrm{d}f}{\\mathrm{d}x}$. A differentiation formula that may be useful is $(x^{\\alpha}e^{kx})'=(kx^{\\alpha}+\\alpha x^{\\alpha-1})e^{kx}$.\nThe time-independent Schrödinger equation for hydrogen has infinitely many solutions. Certain solutions correspond to the atomic orbitals of hydrogen; each orbital is described by a particular wavefunction. The wavefunctions for the orbitals in the first two shells are listed below. 1s: $\\psi_{1s}(r,\\theta,\\phi)=\\frac{1}{\\sqrt{\\pi a_0^3}}e^{-r/a_0}$; 2s: $\\psi_{2s}(r,\\theta,\\phi)=\\frac{1}{4\\sqrt{2\\pi a_0^3}}\\left(2-\\frac{r}{a_0}\\right)e^{-r/(2a_0)}$; $ 2p_x$: $\\psi_{2p_x}(r,\\theta,\\phi)=\\frac{1}{4\\sqrt{2\\pi a_0^3}}\\left(\\frac{r}{a_0}\\right)e^{-r/(2a_0)}\\sin\\theta\\cos\\phi$; $ 2p_y$: $\\psi_{2p_y}(r,\\theta,\\phi)=\\frac{1}{4\\sqrt{2\\pi a_0^3}}\\left(\\frac{r}{a_0}\\right)e^{-r/(2a_0)}\\sin\\theta\\sin\\phi$; $ 2p_z$: $\\psi_{2p_z}(r,\\theta,\\phi)=\\frac{1}{4\\sqrt{2\\pi a_0^3}}\\left(\\frac{r}{a_0}\\right)e^{-r/(2a_0)}\\cos\\theta$. Here $a_0=\\frac{\\hbar^2}{Ke^2m}=52.9\\text{pm}$ is the Bohr radius.", "header_images": [{"file": "Exam-12-T2-header-1.png", "description": "Schematic of spherical polar coordinates: the figure shows a three-dimensional Cartesian coordinate system with origin $O$ and right-handed $x, y, z$ axes. A sphere is centered at O, and point $P$ lies on the sphere in the first octant. The length OP is $r$, and the angle between ray OP and the positive z axis is \\theta. The projection of OP onto the xOy plane makes an angle \\phi with the positive x axis, measured from the positive x axis toward the projection of OP."}], "subquestions": [{"sub_id": "Exam-12-T2-1", "tags": ["3.9"], "text_open": "2.4 Which has the higher energy, the hydrogen 2s or $2p_z$ orbital? Given that the energy of the $2p_z$ orbital is $-3.40\\text{eV}$, calculate from the wavefunction and explain. Note: the energy eigenvalue in the Schrödinger equation represents the orbital energy.", "text_images": [], "full_mark": 7}]} | |
| {"problem_id": "Exam-12-T3", "header": "Problem 3 Pinene in Polymers\nPinene is extracted from pine oil and is an important class of natural products with broad applications in synthetic polymers.", "header_images": [], "subquestions": [{"sub_id": "Exam-12-T3-1", "tags": ["2.4", "2.6"], "text_open": "3.1 Cationic polymerization of \\beta-pinene can produce a high-molecular-weight polymer used in optoelectronic materials. A \\beta-pinene-based polymer synthesis is shown below. 3.1.1 Polymers A and B are known not to contain bridged rings. Write the SMILES structures of polymers A and B, explicitly including the end groups. For each polymer, provide three structures: (a) the repeat unit, using '[X]' and '[Y]' at connections to other repeat units or end groups; (b) the '[X]' end group, using * at the connection to the polymer backbone; and (c) the '[Y]' end group, using * at the connection to the polymer backbone.", "text_images": [{"file": "Exam-12-T3-1-1-prompt.png", "description": "Reaction scheme for 3.1: molecule 'CC1(C)[C@@H]2C(CC[C@H]1C2)=C', labeled \"\\beta-Pinene\" below, reacts with molecule 'CC(Cl)(C)C1=CC=CC=C1' under `EtAlCl_2` to form **A**. **A** then reacts further under `H_2, Pd/Al_2O_3` to give **B**."}], "full_mark": 4}, {"sub_id": "Exam-12-T3-2", "tags": ["2.3", "2.6"], "text_open": "3.1.2 If a polymer chain bearing pinene, or a derivative thereof, at both ends is desired, state how the preceding reaction should be modified.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-12-T3-3", "tags": ["2.3", "2.6", "2.7"], "text_open": "3.2 \\alpha-Pinene (structure shown below) is a biologically renewable feedstock. \\alpha-Pinene itself is difficult to polymerize into a polymer by radical polymerization, but the unsaturated ketone obtained by selective oxidation of \\alpha-pinene (structure shown below) can undergo conventional radical polymerization to give a high-molecular-weight polymer. 3.2.1 From the perspective of monomer structure, explain the difference in radical-polymerization reactivity between \\alpha-pinene and the unsaturated ketone.", "text_images": [{"file": "Exam-12-T3-3-1-prompt.png", "description": "The image shows two chemical structures. On the left is 'CC1(C)[C@@H]2C(C)=CC[C@H]1C2', labeled \"\\alpha-Pinene\" below. On the right is 'CC([C@H](C3)C4)(C)[C@@H]3C(C4=O)=C', labeled \"pinocarvone\" below."}], "full_mark": 2}, {"sub_id": "Exam-12-T3-4", "tags": ["2.4", "2.6"], "text_open": "3.2.2 Structural characterization shows that the backbone of the polymer obtained from the unsaturated ketone contains two different repeat-unit structures. Write the SMILES structures of both repeat units, using an asterisk (*) for the connection sites between repeat units.", "text_images": [], "full_mark": 4}]} | |
| {"problem_id": "Exam-12-T5", "header": "Problem 5 Structure of a Barium Titanate\nA hexagonal-family form of $\\mathrm{BaTiO_3}$ has an interesting structure.\nIn this crystal, all Ba and O atoms together form a special stacking pattern. In a close-packed structure, designate a close-packed layer whose coordination environment above and below is the same as in hexagonal close packing as h, and one whose coordination environment is the same as in cubic close packing as c. For example, double hexagonal close packing (ABACABAC...) can be represented as hchchc.... The joint Ba/O stacking sequence is hcchcch...; considering only Ba gives the same hcchcch... sequence.\nEach Ba is coordinated by 12 O atoms to form a $\\mathrm{[BaO_{12}]}$ tetradecahedral unit. Ti occurs in two chemical environments. In one, two $\\mathrm{TiO_6}$ octahedra share a face to form a $\\mathrm{[Ti_2O_9]}$ unit; in the other, an isolated $\\mathrm{[TiO_6]}$ unit forms, meaning that it does not share a face with another $\\mathrm{TiO_6}$ octahedron, although corner sharing is possible. The ratio of the two types of Ti atoms is 2:1. In the conventional unit cell whose vertices are the Ti atoms of the isolated $\\mathrm{[TiO_6]}$ units, one Ba has coordinates (2/3, 1/3, 0.5967). The unit-cell parameters are a = b = 574.6 pm and c = 1399.9 pm.", "header_images": [], "subquestions": [{"sub_id": "Exam-12-T5-1", "tags": ["1.3"], "text_open": "5.1 Multiple selection: which of the following statements are correct?\n(a) The Ti arrangement resembles hexagonal diamond.\n(b) Each O layer forms a kagome-type \"six-pointed-star array\" like that shown in [Exam-12-T5-1-1-prompt.png].\n(c) If all Ba and O atoms are regarded as jointly forming the hcchcch... close-packed sequence and Ti atoms are treated as interstitial atoms, Ti occupies all octahedral voids formed solely by O atoms.\n(d) This $\\mathrm{BaTiO_3}$ contains 2 distinct Ti-O bonds and 2 distinct Ba-O bonds.", "text_images": [{"file": "Exam-12-T5-1-1-prompt.png", "description": "The image shows a translationally symmetric planar network made of black straight lines and circles and possessing additional symmetries. There are 3 sets of black lines in different directions; within each set, adjacent parallel lines are equally spaced. Together, the three sets partition the plane into closed regular hexagons and equilateral triangles of equal edge length, with a circle at every line intersection. Around each circle, the four polygons appear in the order triangle, hexagon, triangle, hexagon. In the overall connectivity, each of the three edges of every triangle is shared with a hexagonal region, and each of the 6 edges of every hexagon is shared with a hexagonal region."}], "full_mark": 4}, {"sub_id": "Exam-12-T5-2", "tags": ["1.3", "3.9"], "text_open": "5.4 Calculate the second-shortest Ba-Ba distance.", "text_images": [], "full_mark": 5}]} | |
| {"problem_id": "Exam-12-T7", "header": "Problem 7 Pigment Chemistry\nColorful artistic treasures bear witness to human civilization, yet time has stripped some precious artifacts of their original colors. Light exposure is one important cause of these mysterious color changes.", "header_images": [], "subquestions": [{"sub_id": "Exam-12-T7-1", "tags": ["1.3", "1.7"], "text_open": "7.1 Compound A is a red pigment widely used since the Neolithic period, but the red pigment surfaces of many Roman murals are accompanied by gray or black deterioration. Analysis shows that under illumination pigment A absorbs environmental Cl^- and forms white-to-gray chloride B and black elemental substance C; gypsum may also form. Researchers simulated the color change by treating A with NaClO solution under basic conditions. Under light and oxygen, NaClO first forms neutral diatomic species D, a strongly oxidizing key intermediate in ozone-layer depletion. D rapidly oxidizes A to ternary compound E and then further oxidizes E to B. Under illumination, B also gradually forms C together with salt F. A projection of the crystal structure of E is shown below. 7.1.1 Write the chemical formulas of A, B, D, and E.", "text_images": [{"file": "Exam-12-T7-1-1-prompt.png", "description": "The figure contains two crystal-structure visualizations, with the right panel enlarging part of the left. The left panel is a two-dimensional framework. Blue spheres form an approximately close-packed, long-range-ordered pattern. Between some adjacent blue spheres, a red sphere lies near their midpoint and is joined to both by two-tone red-blue segments, red near the red sphere and blue near each blue sphere. Treating blue spheres as polygon vertices and each red sphere plus its two segments as an edge gives two polygon types: nearly equilateral triangles and nonagons with nearly equal edge lengths. Going clockwise, the approximate internal angles of a nonagon are 120°, 240°, 60°, 120°, 240°, 60°, 120°, 240°, and 60°. Every triangle has the same translational environment, as does every nonagon. The triangles and nonagons tile the plane without gaps. A green sphere lies near each of the three blue vertices of every triangle, but not near blue spheres that are not triangle vertices; the green spheres form an approximately kagome pattern. The right panel is a three-dimensional space-filling ball-and-stick model showing the bonding of blue, green, and red spheres in one triangle. Blue spheres occupy the three vertices; green spheres lie below the drawing plane at two blue vertices and above it at the third; red spheres lie approximately at the midpoints of the three sides."}], "full_mark": 7}, {"sub_id": "Exam-12-T7-2", "tags": ["2.1"], "text_open": "7.1.2 Write the ionic equations for oxidation of A to E and of E to B. Note: ignore gypsum formation; the reduction product in both reactions is $Cl^-$. ", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-12-T7-3", "tags": ["2.3"], "text_open": "7.2 Another common red pigment F (note that F in 7.2 is unrelated to the substance denoted F in 7.1) is also a semiconductor material. Its multivalent metal M can undergo interconversion between oxidation states under illumination. Studies show that on the surfaces of Van Gogh's oil paintings, F is photoexcited and its metal M is reduced to a single oxidation state. It then reacts with carbon dioxide released by oily binders such as palmitic acid to form intermediate H and ultimately converts completely into white I. On the surfaces of cave-temple and tomb murals, by contrast, F forms black oxide J under illumination.\n7.2.4 Based on the description in Section 7.2, under otherwise identical illumination, in which solution of equal concentration does F produce the larger photocurrent? ( ). (a) NaHCO_3 solution (b) NaCl solution", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-12-T7-4", "tags": ["2.4"], "text_open": "7.3 The most common red pigment used for dyeing textiles in antiquity was an organic pigment whose color arises from a hydroxyanthraquinone chromophore. The following structure is alizarin, a representative organic red dye: 'O=C(C1=C2C=CC(O)=C1O)C3=CC=CC=C3C2=O'. 7.3.1 Alizarin has two photoconversion pathways: excited-state proton transfer (ESPT) gives structure X_1, while electron transfer (ET) oxidizes it to ionic species X_2. Write the SMILES structures of X_1 and X_2. State in words the position of the radical in X_2.", "text_images": [{"file": "Exam-12-T7-4-1-prompt.png", "description": "The figure shows the structure of alizarin H_2Az: 'O=C(C1=C2C=CC(O)=C1O)C3=CC=CC=C3C2=O'."}], "full_mark": 2}]} | |
| {"problem_id": "Exam-12-T8", "header": "Problem 8 Properties and Applications of Acetal Derivatives\nAn acetal is an organic compound formed when one molecule of an aldehyde or ketone undergoes acid-catalyzed nucleophilic addition with two molecules of alcohol followed by dehydration. Its core structure has two alkoxy groups bonded to the same carbon atom, giving acetals distinctive chemical properties and uses.", "header_images": [], "subquestions": [{"sub_id": "Exam-12-T8-1", "tags": ["1.5", "2.5", "2.7"], "text_open": "8-1 When benzaldehyde is used to protect mannitol (structure shown below; SMILES: 'OC[C@@H](O)[C@@H](O)[C@H](O)[C@H](O)CO'), the reaction ratio is 1:2, whereas with acetone the ratio becomes 1:3. This difference arises because benzaldehyde protects 1,3-diols to form six-membered cyclic acetals, while acetone protects 1,2-diols to form five-membered cyclic ketals. Explain the difference in selectivity.", "text_images": [{"file": "Exam-12-T8-1-1-prompt.png", "description": "The image shows the structure of mannitol, whose SMILES is 'OC[C@@H](O)[C@@H](O)[C@H](O)[C@H](O)CO'."}], "full_mark": 4}, {"sub_id": "Exam-12-T8-2", "tags": ["2.7", "2.8"], "text_open": "8-2-1 Replacing the oxygen atoms in an acetal with sulfur gives a thioacetal, which has different reactivity. Compare the stabilities of the following substances in $1 \\ \\mathrm{mol/L}$ hydrochloric acid and explain your answer.", "text_images": [{"file": "Exam-12-T8-2-1-prompt.png", "description": "Substance A on the left is a ketone with SMILES 'CC1CCCC1=O'; substance B on the right is a thioketone with SMILES 'CC1CCCC1=S'."}], "full_mark": 2}, {"sub_id": "Exam-12-T8-3", "tags": ["2.7", "2.8"], "text_open": "8-2-2 Compare the stabilities of the following substances in $1 \\ \\mathrm{mol/L}$ hydrochloric acid and explain your answer.", "text_images": [{"file": "Exam-12-T8-3-1-prompt.png", "description": "Substance A on the left is a thioketal with SMILES 'C1CCCC12SCCCS2'; substance B on the right is a ketal with SMILES 'C1CCCC12OCCCO2'."}], "full_mark": 2}, {"sub_id": "Exam-12-T8-4", "tags": ["2.4", "2.5"], "text_open": "8-3-1 A proposed synthetic route to M is shown below. Write the SMILES structures of N and M.", "text_images": [{"file": "Exam-12-T8-4-1-prompt.png", "description": "'CC(C)CC=O' reacts with SCCCS to form 'CC(C)CC1SCCCS1'. Subsequent reaction with $\\mathrm{n-BuLi}$ and 'C[Si](C1(SCCCS1)CC2CO2)(C)C' gives compound N. Under $\\mathrm{HgCl_2/MeOH}$ conditions, N is converted to compound M."}], "full_mark": 3}, {"sub_id": "Exam-12-T8-5", "tags": ["2.4", "2.5"], "text_open": "8-3-2 If the reagent in the initial step is replaced with ethylene dithiol, treatment of the thioacetal with butyllithium does not give the expected product. Instead, a side reaction occurs and produces a pungent odor. IR spectroscopy shows that the reaction mixture contains many products, one of which gives an absorption attributable to a carbon-carbon double bond in a hydrocarbon. Based on this information, write 2 products formed in the reaction.", "text_images": [], "full_mark": 3}]} | |
| {"problem_id": "Exam-12-T9", "header": "Reactivity of Nitrogen-Containing Aromatic Rings", "header_images": [], "subquestions": [{"sub_id": "Exam-12-T9-1", "tags": ["1.6", "2.7"], "text_open": "9.1 Pyrrole (1) and pyridine (2) are two common nitrogen-containing aromatic rings with different reactivities.\n9.1.1 Using resonance structures, determine the nucleophilic sites of pyrrole in aromatic electrophilic substitution. Give the SMILES of the major resonance structures of pyrrole.", "text_images": [{"file": "Exam-12-T9-1-1-prompt.png", "description": "1: 'C1=CNC=C1'; 2: 'C1=NC=CC=C1'; 3: '[O-][N+]1=CC=CC=C1'"}], "full_mark": 3}, {"sub_id": "Exam-12-T9-2", "tags": ["1.6", "2.7"], "text_open": "9.1.2 Pyridine can be oxidized by H_2O_2 to **3**. Compare the nucleophilicities of **2** and **3**, and explain your answer.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-12-T9-3", "tags": ["2.2", "2.4"], "text_open": "9.2 A literature report describes a Lewis-acid-controlled electrophilic substitution between indole and triphenyl-1,3,5-triazinane.\n9.2.1 Without a Lewis acid, a short sequence gives final product C; the first step is a thermal pericyclic reaction. Write the SMILES structures of the two key intermediates and product C.", "text_images": [{"file": "Exam-12-T9-3-1-prompt.png", "description": "A ('C12=CC=CC=C1C=CN2') and B ('N1(C2=CC=CC=C2)CN(C3=CC=CC=C3)CN(C4=CC=CC=C4)C1') react in DCE at 80℃ for 24h to give C (C_15H_14N_2)."}], "full_mark": 3}, {"sub_id": "Exam-12-T9-4", "tags": ["2.4"], "text_open": "9.2.2 In the presence of a Lewis acid, C undergoes Lewis-acid-catalyzed fragmentation and rearrangement to give product D. D and C have the same molecular formula. Write the SMILES of D.", "text_images": [{"file": "Exam-12-T9-4-1-prompt.png", "description": "A ('C12=CC=CC=C1C=CN2') and B ('N1(C2=CC=CC=C2)CN(C3=CC=CC=C3)CN(C4=CC=CC=C4)C1') react under Eu(OTf)_3 in DCE at 60℃ for 1h to give D (88%) and E."}], "full_mark": 2}, {"sub_id": "Exam-12-T9-5", "tags": ["2.4"], "text_open": "9.2.3 In the presence of a Lewis acid, the reaction also produces a tetracyclic by-product E containing 2 five-membered rings and 2 six-membered rings. Write the SMILES of E.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-12-T9-6", "tags": ["2.2", "2.4"], "text_open": "9.4 A literature report describes an NHC-catalyzed synthesis of 2-pyrones. Interestingly, replacing the ketone substrate with an ester gives a fused-ring product with a different skeleton.\n9.4.1 The first step in both reactions is activation of substrate F by the NHC. Write the SMILES of the activated intermediate, including the complete N-heterocyclic carbene structure.", "text_images": [{"file": "Exam-12-T9-6-1-prompt.png", "description": "F ('O=C(OC(C)=O)C#CC1=C2C(C=CC=C2)=CC=C1OC') and G ('[R]C(CC1=NC2=CC=CC=C2S1)=O') react under 'CC(C)(C)[C@H]1COCC2=NN(C3=C(C)C=C(C)C=C3C)[C]N21' (10 mol%), base, and solvent. Here R = Ph or R = OEt."}], "full_mark": 1}, {"sub_id": "Exam-12-T9-7", "tags": ["2.2", "2.4"], "text_open": "9.4.3 For R = Ph, write the SMILES structures of the product and the other 2 intermediates besides the answer to 9.4.1. NHC substituents may be represented by R, and stereochemistry is not required. The product is known to contain 5 six-membered rings.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-12-T9-8", "tags": ["2.4"], "text_open": "9.4.4 Write the SMILES structure of the product when R = OEt; stereochemistry is not required. The product is known to contain 4 six-membered rings.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-12-T9-9", "tags": ["1.1"], "text_open": "9.4.5 Determine whether the products of the preceding reactions are chiral.", "text_images": [], "full_mark": 1}]} | |
| {"problem_id": "Exam-13-T1", "header": "Question 1 Vanadium Chemistry. The ionic radius of vanadium is close to those of Fe and Ti, so vanadium commonly occurs in ores associated with Ti and Fe. The principal extraction method first concentrates vanadium in slag, which is roasted to form $FeV_2O_4$. NaCl-$Na_2SO_4$ is then added and melted while oxygen is introduced, converting vanadium into a soluble 1:1 salt in reaction 1. Water leaching gives a vanadium-containing solution, after which an ammonium salt is added to separate the vanadium in reaction 2. Heating the resulting substance at 500 °C gives a simple orange-yellow vanadium oxide product in reaction 3. $VO_2$ has many interesting physical properties, including thermochromism. At atmospheric pressure, $VO_2$ has 2 phases, monoclinic M1 and rutile R, and a phase transition upon heating changes the color of $VO_2$. Thermodynamic data and unit-cell structures for the 2 phases of $VO_2$ near the transition temperature at $p^⊖$ are given below; the large spheres are V and the small spheres are O.\n\nThe first 2 columns of the table are thermodynamic data, and the final six columns are unit-cell parameters.\n\n| Phase | $\\Delta_{f}H_m^{\\ominus}/\\mathrm{kJ\\cdot mol^{-1}}$ | $S_m^{\\ominus}/\\mathrm{J\\cdot mol^{-1}\\cdot K^{-1}}$ | a/pm | b/pm | c/pm | α/° | β/° | γ/° |\n|---|---|---|---|---|---|---|---|---|\n| M1 phase | -714.4 | 61.62 | 575.2 | 452.5 | 538.2 | 90 | 122.6 | 90 |\n| R phase | -710.1 | 74.17 | 455.4 | 455.4 | 285.3 | 90 | 90 | 90 |", "header_images": [], "subquestions": [{"sub_id": "Exam-13-T1-1", "tags": ["4.2"], "text_open": "1.2 Briefly explain why $Na_2SO_4$ is added in the process represented by reaction 1.", "text_images": [], "full_mark": 2}]} | |
| {"problem_id": "Exam-13-T2", "header": "The history of first-generation glucose biosensors dates to 1962, when Clark and Lyons at Cincinnati Children's Hospital immobilized a thin layer of glucose oxidase, GOD, on an oxygen electrode behind a semipermeable membrane permeable only to $O_2$. GOD catalyzes oxidation of glucose by dissolved oxygen in blood, reducing the amount of $O_2$ passing through the membrane. Blood glucose is determined from the resulting change in oxygen-electrode potential.", "header_images": [], "subquestions": [{"sub_id": "Exam-13-T2-1", "tags": ["3.4"], "text_open": "2.2 First-generation glucose biosensors are strongly affected by fluctuations in blood oxygen partial pressure and are readily disturbed by environmental oxygen. Second-generation sensors address this limitation. After GOD oxidizes glucose and becomes reduced, a synthetic mediator M reoxidizes the reduced GOD. The reduced form of M is then oxidized at the measuring electrode to produce an electrical signal from which blood glucose is determined.\nOne glucose biosensor uses ferrocene as M. To test whether electrolyte pH affects the ferrocene redox rate, buffer solutions of different pH values must be prepared. Researchers prepare them by adding $HClO_4$ solution dropwise to $0.10\\ \\mathrm{mol\\cdot L^{-1}}$ K_2HPO_4 solution. The three pKa values of phosphoric acid are 2.15, 7.20, and 12.32, and the HClO_4 concentration is approximately $1.16\\ \\mathrm{mol\\cdot L^{-1}}$. 2.2.1 What volume of $1.16\\ \\mathrm{mol\\cdot L^{-1}}$ HClO_4 solution must be added to 100 mL of $0.10\\ \\mathrm{mol\\cdot L^{-1}}$ K_2HPO_4 solution to prepare a buffer at pH = 6.00?", "text_images": [], "full_mark": 5}]} | |
| {"problem_id": "Exam-13-T5", "header": "Question 5 Crystal Distortions. Under ideal conditions, several classic crystal structures have high symmetry, including cubic perovskite, fluorite, and spinel structures. Under certain conditions, however, some crystal structures distort, breaking the original high symmetry and lowering it to tetragonal, trigonal, orthorhombic, monoclinic, or another crystal system.", "header_images": [{"file": "Exam-13-T5-header-1.png", "description": "A perspective ball-and-stick model of a three-dimensional crystal structure. Axes at the lower left show a in red pointing toward the lower left and into the view, b in green pointing horizontally right, and c in blue pointing vertically upward. The outer framework is a cube with a large light-blue sphere at each of its 8 vertices and thin light-blue rods joining adjacent vertices. A smaller red sphere is embedded near the midpoint of each of the 12 cube edges. The largest green sphere is suspended at the center of the cube and connects radially to the 12 red spheres through 12 thin green rods. The 8 red spheres in the top layer, c = 1, and bottom layer, c = 0, are accompanied by broad white arrows with black borders and bold black coordinates indicating displacement. In the top layer, the left sphere has a rightward arrow and (0.5,0.0098,1); the right sphere has a leftward arrow and (0.5,0.9902,1); the front sphere has a lower-right arrow and (1.0098,0.5,1); and the rear sphere has an upper-right arrow and (-0.0098,0.5,1). In the bottom layer, the left sphere has an outward left arrow and (0.5,-0.0098,0); the right sphere has an outward right arrow and (0.5,1.0098,0); the front sphere has a lower-left arrow and (0.9902,0.5,0); and the rear sphere, seen in perspective below and to the right of the central green sphere, has an upper-left arrow and (0.0098,0.5,0)."}], "subquestions": [{"sub_id": "Exam-13-T5-1", "tags": ["1.3"], "text_open": "5.2 At low temperature, $NaNiO_2$ distorts from a high-symmetry structure into a base-centered monoclinic crystal, as shown below; the large, medium, and small spheres are Na, Ni, and O, respectively. One O atom has coordinates (0.222, 0.5, 0.205), both Ni and Na occupy vertex or face-center positions, and the unit-cell parameters are a = 5.33 Å, b = 2.86 Å, c = 5.59 Å, and β = 110.5°. 5.2.1 Before distortion, the structure follows a close-packing model: O forms the closest-packed array, while alternating Na and Ni layers occupy octahedral holes in the O array. Use ABC for the close-packed O layers, abc for the Na layers, and a’b’c’ for the Ni layers. Represent the close-packing sequence before distortion using letters, giving one repeat unit, and state the Bravais lattice of the undistorted crystal.", "text_images": [{"file": "Exam-13-T5-1-1-prompt.png", "description": "A perspective three-dimensional ball-and-stick model enclosed by an inclined parallelepiped outlined with thin black lines. Axes at the lower left show a in red pointing roughly left, b in green pointing into the page, and c in blue pointing steeply toward the upper right. Three sphere sizes and colors represent the elements: the largest golden spheres are Na, medium silver-gray spheres are Ni, and the smallest red spheres are O. Thin cylindrical bonds are split between the colors of the connected atoms, such as yellow-red Na-O and silver-red Ni-O bonds. The structure is a highly ordered alternating layered arrangement along c. Silver-gray Ni layers lie near the upper and lower faces of the cell. Each Ni connects diagonally to 6 O atoms through 6 silver-red bonds, forming a typical $NiO_6$ octahedron. These edge-sharing octahedra extend laterally in planes perpendicular to c to form robust two-dimensional $NiO_2$ slabs. A layer of large golden Na spheres occupies the central plane in the open space between the 2 $NiO_2$ slabs. Each Na radiates 6 yellow-red bonds in an octahedral geometry, connecting upward to 3 O atoms at the bottom of the upper $NiO_2$ slab and downward to 3 O atoms at the top of the lower $NiO_2$ slab. Thus, rather than forming simple vertical linear units, the interlayer Na atoms act as structural anchors through a three-dimensional network of 6 contacts, 3 above and 3 below, joining the separated silver-red slabs along c into a continuous crystal network."}], "full_mark": 5}, {"sub_id": "Exam-13-T5-2", "tags": ["1.3"], "text_open": "5.2.2 Calculate all distinct Ni-O bond lengths after distortion.", "text_images": [], "full_mark": 4}]} | |
| {"problem_id": "Exam-13-T8", "header": "Question 8 Fundamentals of Organic Chemistry. This problem concerns aldehyde and ketone hydration equilibria, asymmetric addition, neighboring-group participation, and pH-dependent hydrolysis mechanisms.", "header_images": [], "subquestions": [{"sub_id": "Exam-13-T8-1", "tags": ["1.2", "2.7", "2.4"], "text_open": "8.2 Addition reactions of unsymmetrical aldehydes and ketones generally exhibit some stereoselectivity. Select the major product formed in each of the following reactions. 8.2.3", "text_images": [{"file": "Exam-13-T8-1-1-prompt.png", "description": "Question 8.2.3: reduction of tert-butyl-substituted cyclohexanone by (iPrO)_3Al/iPrOH gives two possible cyclohexanol stereoisomers. The 4-tert-butylcyclohexanone reactant has a solid-wedge tert-butyl bond and SMILES 'CC(C)(C)C(CC1)CCC1=O'. A rightward arrow labeled '(iPrO)3Al' above, SMILES 'CC(O[Al](OC(C)C)OC(C)C)C', and iPrOH below, SMILES 'CC(O)C', gives an alcohol with a wavy OH bond and SMILES 'CC(C)(C)[C@H]2CCC(O)CC2'. Below, one diastereomer has a dashed OH bond and SMILES 'CC(C)(C)[C@H]4CC[C@@H](O)CC4'; '(b)' has a solid-wedge OH bond and SMILES 'CC(C)(C)[C@H]3CC[C@H](O)CC3'."}], "full_mark": 1}, {"sub_id": "Exam-13-T8-2", "tags": ["1.2", "2.7", "2.4"], "text_open": "8.2.4", "text_images": [{"file": "Exam-13-T8-2-1-prompt.png", "description": "A bicyclic ketone bearing a bridgehead methyl group has SMILES 'O=C1[C@@]2(C)CC[C@@H](C2)C1'. A rightward arrow labeled LiAlH_4 gives an alcohol with a wavy OH bond, indicating unspecified configuration at the new stereocenter, with SMILES 'OC3[C@@]4(C)CC[C@@H](C4)C3'. Below are two diastereomers: '(a)', with a dashed OH bond, has SMILES 'O[C@@H]5[C@@]6(C)CC[C@@H](C6)C5'; '(b)', with a solid-wedge OH bond, has SMILES 'O[C@H]7[C@@]8(C)CC[C@@H](C8)C7'."}], "full_mark": 1}, {"sub_id": "Exam-13-T8-3", "tags": ["2.3", "4.2"], "text_open": "8.4 Intramolecular catalysis sometimes occurs in aqueous hydrolysis of ketals, esters, and related compounds. This neighboring-group participation allows a group near the reaction center to provide an acidic, basic, or otherwise activating site, producing unusual pH dependence of the rate k. 8.4.1 Kinetic data show that the acetylsalicylate anion hydrolyzes much faster than neutral acetylsalicylic acid. Researchers proposed 2 intramolecular mechanisms, nucleophilic catalysis and general-base catalysis, as shown below. Isotope tracing shows that hydrolysis with H_{2}^{18}O does not produce salicylic acid containing ^{18}O. Which mechanism is correct?", "text_images": [{"file": "Exam-13-T8-3-1-prompt.png", "description": "Two catalytic mechanisms for hydrolysis of acetylsalicylate. In the English-localized upper half, the label is Nucleophilic catalysis:. Acetylsalicylate, SMILES 'CC(OC1=CC=CC=C1C([O-])=O)=O', forms an intermediate containing an ortho oxyanion and mixed-anhydride motif, SMILES 'CC(OC(C1=C([O-])C=CC=C1)=O)=O', which hydrolyzes under an arrow labeled H_2O to salicylic acid, SMILES 'O=C(C1=C(O)C=CC=C1)O'. In the lower half, the label is General-base catalysis:. Acetylsalicylate interacts with H-O-H; 2 curved arrows show the carboxylate oxyanion abstracting a water proton while the water oxygen attacks the acetyl carbonyl carbon. This gives a tetrahedral intermediate bearing methyl, hydroxyl, oxyanion, aryl ether, and ring carboxyl groups, SMILES 'CC(O)(OC1=CC=CC=C1C(O)=O)[O-]', which decomposes to the same salicylic acid, SMILES 'O=C(C1=C(O)C=CC=C1)O'."}], "full_mark": 2}, {"sub_id": "Exam-13-T8-4", "tags": ["1.6", "2.3", "1.7"], "text_open": "8.4.2 Introducing nitro groups at both positions meta to the acetyloxy group of acetylsalicylic acid reverses the hydrolysis mechanism in the resulting dinitro compound, so the other proposed mechanism operates. Give a brief reasonable explanation.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-13-T8-5", "tags": ["2.3", "3.7", "4.2"], "text_open": "8.4.3 The hydrolysis rate k of the cyclic ketal below varies with pH as shown. An anomalous plateau occurs over pH = 2-4. Propose the ketal-hydrolysis mechanism in this region.", "text_images": [{"file": "Exam-13-T8-5-1-prompt.png", "description": "k-pH curve for the cyclic ketal with SMILES 'O=C(C(C=CC=C1)=C1OC2OCCCC2)O'. A rectangular black coordinate frame has inward ticks on all sides and a decorative colored background at the left. The vertical axis is k/×10³ s⁻¹ with labels 0, 10, 20, 30, and 40 from bottom to top. The horizontal axis is pH with consecutive integers 1 through 7. A smooth black curve decreases overall with increasing pH and is relatively flat between pH 2 and 5. Along and near the curve are 12 hollow circular data markers with small center dots."}], "full_mark": 3}, {"sub_id": "Exam-13-T8-6", "tags": ["2.3", "3.7", "4.2"], "text_open": "8.4.4 The rate of ester hydrolysis of the substance below varies with pH as shown. Explain the k-pH curve mechanistically.", "text_images": [{"file": "Exam-13-T8-6-1-prompt.png", "description": "log k-pH curve for ester hydrolysis of the substrate with SMILES 'ClC1=C(OC(C)=O)C(CC2=C(Cl)C(Cl)=CC(Cl)=C2O)=C(Cl)C(Cl)=C1'. A black rectangular frame with inward ticks has vertical axis log k/s⁻¹ labeled -5, -4, -3, -2, and -1 from bottom to top and horizontal axis pH labeled 2, 4, 6, 8, 10, and 12. A continuous black curve with solid and hollow data markers first decreases, reaches a minimum near pH 5-6, then rises, forms a relatively flat plateau around pH 9-11, and bends upward again above pH 12."}], "full_mark": 6}]} | |
| {"problem_id": "Exam-13-T10", "header": "Question 10 Dyotropic Rearrangements. A dyotropic rearrangement is a pericyclic isomerization in which two σ bonds migrate simultaneously within a molecule. Type I and type II rearrangements often show good stereoselectivity and can be used in natural-product synthesis.", "header_images": [], "subquestions": [{"sub_id": "Exam-13-T10-1", "tags": ["2.3", "1.2", "3.9"], "text_open": "10.1 Four-membered-ring lactones are common substrates for dyotropic rearrangements. The following rearrangement constructs a fused lactone containing 3 contiguous stereocenters in one step. 10.1.1 Draw a Newman projection of the reactive conformation of the substrate, viewed along its dashed-wedge bond.", "text_images": [{"file": "Exam-13-T10-1-1-prompt.png", "description": "Dynamic rearrangement equilibria labeled type I and type II. Type I shows an open-chain system in which groups A and B are attached to a carbon chain containing unsaturation; dashed lines indicate delocalization or partial double-bond character, and the equilibrium reversibly shifts the unsaturation or conjugated system along the chain while changing molecular configuration. Type II shows a three-dimensional bridged or cage-like polycycle. In the left isomer, A and B lie on a particular front-right C-C single bond and a C=C bond lies at the rear left. Rearrangement breaks the A/B-bearing bond and forms a new A/B-bearing bond at the front left while shifting the double bond concertedly."}, {"file": "Exam-13-T10-1-2-prompt.png", "description": "Question 10.1: four-membered-ring lactone substrate 'O=C1O[C@@H](C2CCCC2)[C@@H]1OC3=CC=CC=C3' undergoes an MgBr_2-promoted dyotropic rearrangement to give fused lactone '[H]C4(C(C(O5)=O)OC6=CC=CC=C6)C5([H])CCCC4' with 3 contiguous stereocenters."}], "full_mark": 2}, {"sub_id": "Exam-13-T10-2", "tags": ["2.4", "1.2", "2.7"], "text_open": "10.1.2 Write the product SMILES, including stereochemistry.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-13-T10-3", "tags": ["2.3", "2.4"], "text_open": "10.2 The natural product curcumanolide A can be synthesized concisely through the key steps of the following small-organic-molecule-catalyzed asymmetric intramolecular lactonization and dyotropic rearrangement. 10.2.1 Write the SMILES strings of at least 2 key intermediates in the first transformation.", "text_images": [{"file": "Exam-13-T10-3-1-prompt.png", "description": "A multistep synthesis. Reactant 'O=C(CCC1=O)C1(CCCC(O)=O)C' reacts under cat. and a catalyst bearing a dashed-wedge Ph group, SMILES '[C@H]7(CCN8C(SC9=C8C=CC=C9)=N7)C%10=CC=CC=C%10', with TsCl, LiCl, iPr_2NEt below the arrow, to give intermediate 'O=C(O2)[C@H]3CC[C@]4([C@]32CCC4=O)C'. mCPBA then gives A; cat. TMSOTf converts A to B; 2 parallel arrows convert B to curcumanolide A, SMILES 'O=C(/C(C5)=C(C)\\C)O[C@]65[C@H](CC[C@H]6C(OC)=O)C'."}], "full_mark": 4}, {"sub_id": "Exam-13-T10-4", "tags": ["2.8", "1.4", "2.7"], "text_open": "10.2.2 State the role of LiCl in the first transformation.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-13-T10-5", "tags": ["2.4", "1.2", "2.3"], "text_open": "10.2.3 In a later step, Lewis acid TMSOTf catalyzes dyotropic rearrangement of A to B. Write the SMILES strings of A and B, including stereochemistry.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-13-T10-6", "tags": ["1.4", "2.3", "2.4"], "text_open": "10.3 Dyotropic rearrangements can also occur at transition metals. Write the SMILES string of intermediate C in the following transformation.", "text_images": [{"file": "Exam-13-T10-6-1-prompt.png", "description": "A two-step reaction. A reactant bearing central Cp_2Zr and two furan rings, SMILES 'C1([ZrCp2]C2=CC=CO2)=CC=CO1', is heated at 185 °C to intermediate 'C'. C then reacts with MeO_2C-C≡C-CO_2Me, SMILES 'COC(=O)C#CC(=O)OC', at 105 °C to give a benzene product bearing two methoxycarbonyl groups, CO_2Me, and one furan on three consecutive carbons, SMILES 'COC(=O)C1=C(C(=O)OC)C(C2=CC=CO2)=CC=C1'."}], "full_mark": 2}, {"sub_id": "Exam-13-T10-7", "tags": ["2.3", "3.9"], "text_open": "10.4 Type II dyotropic rearrangements generally require aromatization as a driving force. The following transformation may involve a dyotropic mechanism, so researchers tested it using a deuterated substrate. If the reaction follows a dyotropic mechanism, which positions in the product are deuterated? Circle them on the answer sheet.", "text_images": [{"file": "Exam-13-T10-7-1-prompt.png", "description": "A consecutive 2-step reaction. Reactant 'O=C1OC(C2=C1CC3=C(C2)CC3)=O' and deuterated reactant 'C=C4C([2H])C=CC([2H])C4=C', which has 2 wavy bonds labeled D and the English-localized caption 1:1 mixture of cis/trans isomers, are joined by a plus sign and heated at 117 °C to intermediate '[2H]C5C=CC([2H])C6=C5CC(C7)(C(O8)=O)C(C6)(C8=O)CC(C7=C)=C', also bearing 2 wavy D bonds. Heating at 150 °C gives final product 'O=C(O9)C%10%11C(CC(C=CC=C%12)=C%12C%11)(C9=O)CC(C)=C(C)C%10'."}], "full_mark": 2}, {"sub_id": "Exam-13-T10-8", "tags": ["3.9", "1.2", "2.4"], "text_open": "10.5 The following reaction undergoes a dyotropic rearrangement. Write the product SMILES, including stereochemistry.", "text_images": [{"file": "Exam-13-T10-8-1-prompt.png", "description": "The molecule with SMILES 'ClC1=C(Cl)[C@@]2(Cl)[C@]3([H])[C@@]([C@]1(Cl)C2=O)([H])[C@H]4[C@@]5([H])[C@@]6(Cl)C(Cl)=C(Cl)[C@@]([R1]6)(Cl)[C@@]5([H])[C@@H]3[R2]4' is heated; the product is not shown. R1 represents CCl_2 and R2 represents CH_2."}], "full_mark": 3}]} | |
| {"problem_id": "Exam-14-T2", "header": "Electrochemistry of an Oxidoreductase\nLaccase is a copper-containing oxidoreductase that catalyzes the reduction of oxygen to water while oxidizing phenolic compounds and various other substrates. Its catalytic center contains a T1 copper center (electron-acceptor site) and a T2/T3 trinuclear copper cluster (redox site). Because of its high activity and selectivity, laccase is often used as a cathode catalyst in biofuel cells. This problem concerns a laccase from Trametes versicolor, which may be approximated as a spherical protein with diameter d of approximately 5.0 nm. The standard electrode potential of its T1 copper center is $E_{\\mathrm{T1}}^\\circ=+0.78\\ \\mathrm{V}$ (vs. SHE). The laccase is immobilized on different electrodes to investigate its direct electron transfer (DET) and mediated electron transfer (MET) behavior.", "header_images": [{"file": "Exam-14-T2-header-1.png", "description": "A set of colored ribbon and line structures comprising magenta, brown, and green regions.\n\nThe upper-central magenta region includes several ribbon arrows, folded ribbons, and thin lines. The green region on the right includes ribbon arrows, curved lines, and a helical coil near the bottom. The brown region on the left and at the bottom includes several ribbon arrows and thin lines.\n\nVisible labels are His 458, Asp 206, T1 Cu, and T2 & T3 Cu. His 458 is left of center, Asp 206 is upper-right of center, T1 Cu is in the middle-left region, and T2 & T3 Cu is below center.\n\nSeveral small spherical markers in blue, orange, red, purple, and other colors appear near the center. Black dashed circular or elliptical markers surround some spheres and nearby labels. There are no axes, table, legend, or title."}], "subquestions": [{"sub_id": "Exam-14-T2-1", "tags": ["3.7", "4.2"], "text_open": "2.1.2 For random orientations, the electron-transfer rate for the orientation with the T1 copper center nearest the electrode is approximately $1.1\\times10^{13}$ times that for the farthest orientation. Using the figure below, explain why immobilizing the enzyme on the electrode surface is necessary for the redox reaction. The small circle represents the T1 copper center (electron-acceptor site), and the triangle represents the T2/T3 trinuclear copper cluster (redox site). (No more than 50 words.)", "text_images": [{"file": "Exam-14-T2-1-1-prompt.png", "description": "The label Substrate appears at the upper left and Product at the upper right. A long horizontal light-gray rectangle with a dotted texture spans the bottom and is labeled Electrode at the lower right.\n\nA sequence of 5 blue circles is arranged from left to right above the gray rectangle. Each contains a red triangle and a small orange circle. The red triangles have different orientations: No. 1 points approximately upward, No. 2 upper-left, No. 3 left, No. 4 lower-left, and No. 5 downward.\n\nBelow blue circle No. 1 is a white rectangular box labeled e⁻, with a black arrow pointing upward from the box to the blue circle. Two curved arrows above circle No. 1 run from near Substrate toward Product.\n\nBlack dashed horizontal lines extend from several orange dots and connect to vertical double-headed arrows or scale lines on the right. The vertical double-headed arrow near circle No. 1 is labeled d. Similar vertical lines or double-headed arrows extend upward from the gray surface near the remaining circles, indicating progressively larger distances from the gray region to the orange dots."}], "full_mark": 2}]} | |
| {"problem_id": "Exam-14-T5", "header": "Familiar Structures from an Unfamiliar Perspective (II)", "header_images": [], "subquestions": [{"sub_id": "Exam-14-T5-1", "tags": ["1.3"], "text_open": "Structural background: An orthorhombic crystal of $XO_n$ contains long chains of edge-sharing $[XO_6]$ octahedra whose projections coincide along $a$. Two long chains form a small unit, and these units repeat in the $bc$ plane. The figure below is the projection along $a$ (short side $b$, long side $c$).\n5.1.2 Give the value of n.", "text_images": [{"file": "Exam-14-T5-1-1-prompt.png", "description": "White background with a horizontally oriented thin dark-gray rectangular frame near the center. The frame lies behind colored spheres, rod-like connectors, and translucent purple polygonal faces, and some frame edges are obscured by foreground structures.\n\nThere are 24 visible spheres: 16 red and 8 purple or magenta. The red spheres comprise 1 at the far lower left, 1 at the upper left, 2 in the middle-left region, 1 at the lower left, 3 at the top, 2 at the center, 2 in the lower-central region, 2 in the middle-right region, and 2 at the far right. The purple or magenta spheres comprise 1 at the far left, 1 in the middle-left region, 1 upper-left of center, 1 upper-right of center, 1 in the middle-right region, 1 at the far right, 1 lower-left of center, and 1 lower-right of center.\n\nA local structure of 5 spheres appears on the left: 1 magenta sphere at the far left, 1 red sphere lower-left, 1 red sphere upper-left, 1 purple sphere in the middle-left region, and 1 red sphere left of center. Dark-purple or red rod-like connectors form triangular and rhomboidal outlines. A translucent purple face lies in this region, with the spheres and thick connectors drawn above it.\n\nThe upper region contains 6 principal spheres: red spheres at upper-left, upper-center, upper-right, and upper-center, plus purple or magenta spheres upper-left and upper-right of center. Dark-purple or red connectors and a large translucent purple polygonal face join them. The upper-left red sphere lies over the face edge; the upper-center and upper-right red spheres lie on the upper outline; the two purple or magenta spheres lie within or near the polygon boundary; and the red sphere above center obscures parts of the face and connectors.\n\nThe lower-central region contains 6 principal spheres: a middle-left red sphere, a lower-central red sphere, purple spheres lower-left and lower-right of center, another lower-central red sphere, and a lower-left red sphere. Rod-like connectors and a translucent purple polygonal face join them. A foreground lower-central red sphere obscures part of the face; the two purple spheres each connect to several surrounding red spheres; and the lower-left and lower-central red spheres lie along the lower boundary.\n\nA local structure of 6 spheres appears on the right: middle-right red and purple spheres, lower-right and upper-right red spheres, a far-right purple sphere, and a red sphere at the lower right. Red or dark-purple connectors join them. A translucent purple polygon lies between the middle-right red, upper-right red, lower-right red, and middle-right purple spheres. The far-right purple sphere is connected to the red spheres on the right and forms the rightmost point.\n\nThe dark-gray rectangular frame is farthest back; the translucent purple polygonal faces lie in front of it; and the spheres and thicker connectors are mostly in front of the faces. Some connectors are obscured by spheres, and parts of the translucent faces are obscured by red or purple spheres. The structure extends horizontally, with more dispersed ends and denser central, upper, and lower regions."}], "full_mark": 2}, {"sub_id": "Exam-14-T5-2", "tags": ["1.3"], "text_open": "5.1.3 State the lattice type of the crystal.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-14-T5-3", "tags": ["1.3"], "text_open": "5.2.1 Y is in the same group as X, and the structure of $\\mathrm{YO}_n$ is similar to that of $\\mathrm{XO}_n$. $\\mathrm{YO}_n$ is monoclinic, but its unit-cell parameter β = 119.97° is very close to 120°, and a ≈ c. It can therefore be described using a packing scheme analogous to that of a hexagonal crystal: O atoms form an approximate “hexagonal closest packing” along the b axis in an …ABAB… sequence; Y atoms regularly occupy half of the octahedral holes in each layer; and the projections of Y atoms in adjacent layers do not overlap along b. This can be denoted $\\ldots Ac_1Bc_2Ac_1Bc_2\\ldots$. The projection along b is shown below. (The Y atoms on the short diagonal are in the same layer and on the unit-cell surface; those on the long diagonal are in the same layer and inside the unit cell.) Give the number of Y atoms in one unit cell.", "text_images": [{"file": "Exam-14-T5-3-1-prompt.png", "description": "A colored ball-and-stick model on a white background. Behind the main structure is an inclined rectangular or parallelogram frame made of thin dark-gray lines. Portions of the frame are obscured by foreground spheres and connectors.\n\nA three-dimensional direction marker appears at the lower left: a red arrow points upper-left and is labeled a; a blue arrow points right and is labeled c; and a green dot or short arrow near their common origin is labeled b. The angle between the a and c directions is approximately 120°.\n\nThere are 20 visible spheres: 4 larger pale gray-green spheres and 16 red spheres. The 4 pale gray-green spheres lie at the upper-left, upper-right, lower-left, and lower-right positions, forming an approximate quadrilateral. The red spheres are distributed around them and along their connecting paths: 4 at the top, 3 on the left, 4 in the center, 3 at the bottom, and 2 on the right.\n\nThe upper-left pale gray-green sphere is connected to several red spheres: one above by an approximately vertical rod, one upper-left and one lower-left by diagonal rods, one upper-right by a diagonal rod, one middle red sphere lower-right, and one neighboring red sphere lower-left.\n\nThe upper-right pale gray-green sphere is connected to several red spheres: one above by an approximately vertical rod, one upper-center sphere to its upper-left, one central sphere to its lower-left, one below by an approximately vertical red segment, one upper-right, and one lower-right.\n\nThe lower-left pale gray-green sphere is connected to red spheres upper-left, above through an approximately vertical segment, upper-right, lower-right, below, and lower-left.\n\nThe lower-right pale gray-green sphere is connected to red spheres upper-left, upper-right, right, lower-right, below, and lower-left.\n\nThe connectors are mainly gray or pale gray-green, with red portions near some red spheres. Some connectors are obscured by spheres. Overall, the spheres and connectors lie in front of the gray frame, which remains visible through gaps behind the structure. One red sphere at the upper right is paler and appears translucent or highlighted; the remaining red spheres are mostly darker red with light-dark shading."}], "full_mark": 1}]} | |
| {"problem_id": "Exam-14-T9", "header": "Synthesis of Weinreb Amides\nSince their discovery, N-methoxy-N-methylamides have been widely known as Weinreb amides (WAs). WAs exhibit reactivity distinct from that of ordinary amides: they can serve both as excellent acylating reagents and as aldehyde equivalents and are important building blocks in numerous organic syntheses.", "header_images": [], "subquestions": [{"sub_id": "Exam-14-T9-1", "tags": ["2.2", "2.4", "2.5"], "text_open": "9.1 The most common method for preparing Weinreb amides is condensation of a carboxylic acid with N,O-dimethylhydroxylamine. Because amide condensation is thermodynamically unfavorable, an additional coupling reagent is usually required. The carboxylic acid below is to be condensed with N,O-dimethylhydroxylamine to prepare an N-protected amino-acid WA. 9.1.1 DCC is selected as the coupling reagent. Write the SMILES strings of one key intermediate and the coproduct (the product other than the Weinreb amide).", "text_images": [{"file": "Exam-14-T9-1-1-prompt.png", "description": "Carboxylic-acid substrate: 'O=C(N[C@@H]([R])C(O)=O)OCC1=CC=CC=C1'; N,O-dimethylhydroxylamine: 'CNOC'"}, {"file": "Exam-14-T9-1-2-prompt.png", "description": "DCC coupling reagent: 'C1(N=C=NC2CCCCC2)CCCCC1'."}], "full_mark": 3}, {"sub_id": "Exam-14-T9-2", "tags": ["2.4", "2.3", "1.2"], "text_open": "9.1.2 Experimentally, using DCC alone gives a product racemized at the stereogenic carbon because the intermediate in Question 9.1.1 is further converted into a cyclic intermediate that undergoes isomerization and causes racemization. Write the SMILES string of the cyclic intermediate and draw the isomerization mechanism.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-14-T9-3", "tags": ["2.1"], "text_open": "9.2 A trichloromethyl carbinol can undergo a homologation-amination reaction to form a WA containing one additional carbon, as shown below. 9.2.1 Under these conditions, $\\mathrm{(PhSe)_2}$, $\\mathrm{NaBH_4}$, and EtOH generate a complex anion X that contains an Se-B bond but no B-H bond and may be regarded as an equivalent of $\\mathrm{PhSe^-}$. Write the equation for formation of anion X.", "text_images": [{"file": "Exam-14-T9-3-1-prompt.png", "description": "'[R]C(C(Cl)(Cl)Cl)O' reacts with 'CNOC' under $\\mathrm{(PhSe)_2}$, $\\mathrm{NaBH_4}$, NaOH, EtOH, 55°C to form '[R]CC(N(C)OC)=O'"}], "full_mark": 2}, {"sub_id": "Exam-14-T9-4", "tags": ["2.2", "2.4"], "text_open": "9.2.2 Detection of the intermediate below was used to infer the reaction mechanism. The trichloromethyl carbinol substrate first converts under basic conditions into a small-ring intermediate and then into the intermediate shown below. Using the information in Question 9.2.1, write the SMILES strings of 3 key intermediates leading to the final product, excluding the intermediate shown.", "text_images": [{"file": "Exam-14-T9-4-1-prompt.png", "description": "Structure of the intermediate detected in the reaction. SMILES: '[R]C(C(N(C)OC)=O)[Se]C1=CC=CC=C1'"}], "full_mark": 6}, {"sub_id": "Exam-14-T9-5", "tags": ["2.2"], "text_open": "9.2.3 Can (PhSe)_2 be regenerated in this reaction? ( )\n(a) Yes; (b) No", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-14-T9-6", "tags": ["2.4", "2.8", "2.5"], "text_open": "9.2.5 The experiment also produces a byproduct whose molecular mass is approximately 30 lower than that of the major product. Replacing the methyl group bonded to O in N,O-dimethylhydroxylamine with another group can markedly suppress formation of this byproduct, although the reaction time increases accordingly. Using the relevant information, write the SMILES string of one possible modified substrate.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-14-T9-7", "tags": ["2.4"], "text_open": "9.2.6 This experiment must be performed under oxygen-free conditions. If oxygen is present, another byproduct structurally similar to the main product is detected. Given that it contains multiple nitrogen atoms, write its SMILES string.", "text_images": [], "full_mark": 2}]} | |
| {"problem_id": "Exam-14-T10", "header": "Visible-Light Photocatalytic Reactions of Sulfonium Salts\nOn 2023-5-3, researchers from Qingdao University of Science and Technology published a study of visible-light photocatalytic reactions of sulfonium salts. Let the photosensitizer be denoted PC; in the photocatalytic pathway, light excites PC to PC*, which then participates in the reaction.", "header_images": [{"file": "Exam-14-T10-header-1.png", "description": "'CC1=CC=C([S+]2CCCC2)C=C1.F[P-](F)(F)(F)(F)F' reacts with 'C=CC1=CC=CC=C1' under hv, Photocatalyst (mol%), Base (3.0 equiv), DMSO, N2, 24 h, r.t."}], "subquestions": [{"sub_id": "Exam-14-T10-1", "tags": ["2.4", "2.2"], "text_open": "Standard conditions: The sulfonium salt shown reacts with styrene using PC7 (10 mol%) as the photosensitizer, $\\mathrm{Na_2CO_3}$ (3.0 equiv) as the base, and anhydrous DMSO as the solvent under irradiation with a 455 nm blue LED at room temperature for 24 hours.\n10.2.2 Researchers added 1,1-diphenylethylene under the standard conditions to trap a reaction intermediate, giving a product containing 3 rings. Write the SMILES string of this 3-ring product.", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-14-T10-2", "tags": ["2.4", "2.8"], "text_open": "10.2.3 Performing the reaction under the standard conditions but without DMSO gives a different product. Write its SMILES string.", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-14-T10-3", "tags": ["2.2", "2.3", "2.4"], "text_open": "10.3 Write the SMILES strings of the key intermediates and product of this reaction.", "text_images": [], "full_mark": 5}, {"sub_id": "Exam-14-T10-4", "tags": ["2.4", "2.5"], "text_open": "10.4 Treatment of the product from 10.3 with mCPBA (meta-chloroperoxybenzoic acid) gives two additional products. Write their SMILES strings.", "text_images": [], "full_mark": 2}]} | |
| {"problem_id": "Exam-15-T3", "header": "Problem 3 Solid Electrolyte Interphase\nThe solid electrolyte interphase (SEI) is an important research area in modern electrochemistry. During the first charging of a lithium-ion battery, the electrode potential lies below the reduction potential of the electrolyte, so additives and lithium salts in the electrolyte undergo electrochemical reductive decomposition at the solid-liquid interface. Insoluble solid products deposit on the electrode surface and form a passivation film that is electronically insulating but ionically conducting, thereby slowing electrode corrosion. Vinylene carbonate (VC), a classic film-forming additive, reacts preferentially and promotes formation of a high-quality SEI.", "header_images": [{"file": "Exam-15-T3-header-1.png", "description": "The structures in the figure have the following SMILES: VC, 'O=C1OC=CO1'; EC, 'O=C2OCCO2'."}], "subquestions": [{"sub_id": "Exam-15-T3-1", "tags": ["2.2", "2.3", "2.4"], "text_open": "3.3 Studies show that VC preferentially undergoes cleavage of a C-O single bond and is reduced to radical anion A carrying 1 unit of negative charge. It then undergoes radical polymerization to form several polymeric products. 3.3.1 In one pathway, A acts as a chain initiator and reacts with VC to form acyclic polymer B. Write the SMILES of B.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-15-T3-2", "tags": ["2.1", "2.2", "3.6"], "text_open": "3.3.4 In the electrolyte, VC can also be electrochemically reduced to lithium oxalate and an enolate. Write the electrochemical reaction equation; EC participates in the reaction.", "text_images": [], "full_mark": 2}]} | |
| {"problem_id": "Exam-15-T5", "header": "Problem 5 Versatile Rutile. Rutile ($TiO_2$) is a classic structure in the tetragonal crystal system. From a close-packing and interstitial-site perspective, oxygen atoms approximately form hexagonal close packing, and titanium atoms occupy half of the octahedral voids in an ordered manner. The $TiO_6$ octahedra are connected by shared vertices and shared edges. A rutile unit cell and its projection along the c axis are shown below.", "header_images": [{"file": "Exam-15-T5-header-1.png", "description": "The image is a three-dimensional ball-and-stick model of $TiO_2$ on a white background. All spheres have bright highlights and a smooth, glossy three-dimensional appearance. Thin black-gray lines outline a transparent rectangular unit-cell box in perspective. The 9 larger light-blue spheres represent Ti atoms, located at the 8 vertices and the exact center of the box. The 6 smaller bright-red spheres represent O atoms, all inside the cell around the central Ti. Atoms are connected by short two-color cylinders, light blue near Ti and red near O. The central Ti is connected to all 6 red spheres, forming an approximately octahedral $TiO_6$ environment. The left red sphere connects the central Ti and the upper-left-front and lower-left-front vertex Ti atoms; the right red sphere connects the central Ti and the upper-right-back and lower-right-back vertex Ti atoms. The other four red spheres cross around the center: the upper-left-back red sphere connects the upper-left-back vertex Ti, the upper-right-front red sphere connects the upper-right-front vertex Ti, the lower-left-back red sphere connects the lower-left-back vertex Ti, and the lower-right-front red sphere connects the lower-right-front vertex Ti. Natural perspective occlusion creates clear depth. There is no visible text, number, coordinate axis, arrow, legend, unit, or title."}, {"file": "Exam-15-T5-header-2.png", "description": "The image should be reconstructed as a flat two-dimensional vector diagram approximately 365×370 pixels, slightly taller than wide, on a pure-white borderless canvas. Taking the upper-left corner as origin (0,0), x to the right and y downward, first draw an unfilled thin solid gray square approximately 200 pixels per side, aligned horizontally and vertically, with corners near (85,84), (284,84), (284,284), and (85,284). Its line is about 1 pixel wide with slight antialiasing. No spheres or marks sit at the square corners; line midsections remain clear in white gaps, while corners are partly hidden by blue shapes. Place 5 nearly equal slanted rhombic parallelograms, each about 102 pixels per side with acute angles about 65° and obtuse angles about 115°. The 4 outer parallelograms at upper left, upper right, lower left, and lower right share an orientation elongated lower-left to upper-right, with uniform medium-light sky-blue fill, approximately RGB 117,182,224 or #75B6E0. Clockwise, their vertices are [(46,45),(146,23),(124,123),(24,145)], [(245,45),(345,23),(323,123),(223,145)], [(46,245),(146,223),(124,323),(24,345)], and [(245,245),(345,223),(323,323),(223,345)]. They extend well outside the gray square and approach the canvas edges but remain fully visible. The 5th parallelogram lies centrally, rotated approximately 90° relative to the outer 4 and elongated upper-left to lower-right, with clockwise vertices [(124,123),(223,145),(245,245),(146,223)]. Its fill is distinctly lighter, approximately RGB 144,196,230 or #90C4E6. All 5 have flat, texture-free, nearly opaque blue faces outlined by a very thin, approximately 1-pixel light-gray antialiased edge. A faint short gray diagonal divides each face visually into two triangular regions. For the 4 outer shapes the diagonals connect (46,45) to (124,123), (245,45) to (323,123), (46,245) to (124,323), and (245,245) to (323,323); for the center shape the diagonal connects (223,145) to (146,223). The 4 center-shape vertices are shared respectively with the nearest center vertices of the 4 outer shapes: (124,123), (223,145), (146,223), and (245,245). Except for these point contacts, the blue shapes do not overlap substantially. Draw 16 equal red three-dimensional spheres, diameter approximately 24–26 pixels, at all unique parallelogram vertices, centered at (146,23), (345,23), (46,45), (245,45), (124,123), (323,123), (24,145), (223,145), (146,223), (345,223), (46,245), (245,245), (124,323), (323,323), (24,345), and (223,345). The spheres lie above the blue faces and gray lines, hiding line ends. Each has a thin dark-red rim, a bright-red to deep-wine-red surface gradient, and a soft white or pale-pink highlight slightly upper-left of center, giving a smooth glass-or-plastic gloss without a cast shadow. The composition is approximately 180° rotationally symmetric. Outermost spheres leave only about 7–12 pixels of white margin. No text, number, letter, legend, coordinate axis, tick, arrow, symbol, watermark, background texture, or other decoration is present."}], "subquestions": [{"sub_id": "Exam-15-T5-1", "tags": ["1.1", "1.3"], "text_open": "5.2 α-$MnO_2$. α-$MnO_2$ has a rutile-derived structure. Rutile can be viewed as one-dimensional chains formed by edge-sharing $TiO_2$ octahedra, with chains of two different spatial orientations alternating and sharing vertices. α-$MnO_2$ can be viewed as two one-dimensional $MnO_2$ octahedral chains joined by shared edges to form a \"thick one-dimensional chain\"; thick chains of two different spatial orientations alternate and share vertices. The c-axis projection of the polyhedral connectivity is shown. Some thick chains of the same orientation cannot be translated into one another by vectors parallel to the ab plane because they occur at different c-axis heights. The lattice type of α-$MnO_2$ is body-centered tetragonal. 5.2.2 Which symmetry elements does α-$MnO_2$ possess? (a) 4-fold rotation axis; (b) 4-fold rotoinversion axis; (c) $4_1$ screw axis; (d) $4_2$ screw axis.", "text_images": [{"file": "Exam-15-T5-1-1-prompt.png", "description": "The image is a 427×424 pixel two-dimensional diagram on a pure-white background, showing a regular periodic network of purple polygons and red dots, with no axes, text, arrows, or labels. Approximately 72 red nodes are small spheres with a slight three-dimensional red gradient, brighter at the center and darker at the edge. They occupy vertices and connection points of the purple structure. The pattern repeats by approximately 192 pixels both horizontally and vertically, forming a 2×2 arrangement of four identical periodic units. The purple portions comprise approximately 16 large, separated structural units filled with light-to-deeper purple gradients, sometimes darker centrally and lighter at the edges, creating a translucent or three-dimensional effect. Red nodes serve as their vertices or junctions, forming combinations of rhombi, parallelograms, triangles, and irregular polygons. Two principal orientations occur: units extending upper-left to lower-right with roughly horizontal rhombic/parallelogram outlines, and units extending lower-left to upper-right with roughly vertical rhombic/triangular outlines. They alternate and connect through red nodes. In the upper-left region, nodes near (13,84), (51,58), (87,27), (125,51), (87,76), and (52,107) form a purple zigzag polygon extending upper-right. A translated neighboring group connects near (156,85), (182,124), and (205,84). The upper-right is the upper-left translated about 192 pixels right; the lower-left is translated about 192 pixels down; and the lower-right is translated both right and down. Typical red-node spacing is about 30–40 pixels. Connections are implied by polygon edges and vertices without black or other lines. Some nodes lie at junctions of multiple polygons, giving a periodic two-dimensional lattice or repeating coordination-framework appearance. The whole pattern is regular and approximately symmetric horizontally and vertically. Outermost nodes approach but do not touch the edges: top near (148,11) and (341,11), bottom near (85,412) and (278,411), left near (13,84) and (13,276), and right near (397,84) and (413,339). The purple-red structure occupies most of the white canvas and has obvious periodicity."}], "full_mark": 3}, {"sub_id": "Exam-15-T5-2", "tags": ["1.3"], "text_open": "5.2.3 How many neighboring $MnO_6$ octahedra share an edge with each $MnO_6$ octahedron?", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-15-T5-3", "tags": ["1.3"], "text_open": "5.3 Ca-Ti-O crystal. The figure shows the unit cell of a Ca-Ti-O crystal. Small spheres are oxygen, medium spheres titanium, and large spheres calcium. Ti and Ca each have only one chemical environment, while O has three. 5.3.1 Write the chemical formula of the crystal.", "text_images": [{"file": "Exam-15-T5-3-1-prompt.png", "description": "One unit cell contains 28 small red spheres: 8 on edges, 16 on faces, and 6 inside the cell. It contains 12 medium blue spheres: 4 on edges and 8 on faces. It contains 6 large white spheres: 2 on edges and 4 on faces."}], "full_mark": 2}]} | |
| {"problem_id": "Exam-15-T6", "header": "Problem 6 Precipitation-Dissolution Equilibria. Constants at 25 ℃ may be used. Report all calculated results to 3 significant figures and solubilities in mol/L: dissociation constants of $H_2S$, Ka_1 = 1.3×10^{-7}, Ka_2 = 7.1×10^{-15}; Ksp(MnS)=3.2×10^{-11}; Ksp($Mn(OH)_2$)=1.6×10^{-13}.", "header_images": [], "subquestions": [{"sub_id": "Exam-15-T6-1", "tags": ["3.4"], "text_open": "6.3 Accounting for the pH change, calculate the solubility of MnS in pure water.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-15-T6-2", "tags": ["3.4"], "text_open": "6.4 Using the result from 6.3, demonstrate by calculation that $Mn(OH)_2$ precipitate actually forms in the system. Then calculate the solubility of MnS in pure water. Given Ksp($Mn(OH)_2$)=1.6×10^{-13}.", "text_images": [], "full_mark": 6}]} | |
| {"problem_id": "Exam-15-T7", "header": "Problem 7 Low-Valent Compounds of the Aluminum-Group Metals. Group 13 metals, the aluminum group (Al, Ga, In, Tl), can form chain, ring, and cluster compounds characterized by covalent metal-metal bonds and oxidation states below +III. Beginning with $E_2R_4$ compounds, this problem examines their bonding, stability, and reactivity. Chain and ring compounds have an E-E σ single bond (2c-2e) as their framework. After forming three covalent bonds, a metal has 6 valence-shell electrons and is Lewis acidic; the most common form is the $E_2R_4$ unit.", "header_images": [], "subquestions": [{"sub_id": "Exam-15-T7-1", "tags": ["1.4", "1.5"], "text_open": "7.1 Based on the electronic properties of the $E_2R_4$ unit, which of the following can stabilize the $E_2R_4$ unit? A: R is a strongly electron-withdrawing group. B: R is a sterically bulky group. C: ligand L is added to form $R_2(L)E{-}E(L)R_2$.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-15-T7-2", "tags": ["2.1", "2.2", "1.4"], "text_open": "7.2 \"Dihalides\" $E_2X_4$. Among the aluminum-group metals, Ga most readily forms $E_2X_4$, commonly as adduct $X_2(L)E{-}E(L)X_2$. If ligand L is also a halide anion, monoanionic or dianionic complexes form. 7.2.1 In one synthesis, Ga and $I_2$ react in toluene in a 2:1 molar ratio. After cooling to -78°C, 4 eq $NCy_2H$ is added. The mixture reacts for 12 h at room temperature, precipitate is removed by filtration, and the solution is concentrated and cooled to give colorless crystal A in 95% yield. Write the equations for the two steps forming A.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-15-T7-3", "tags": ["1.4", "4.4", "2.4"], "text_open": "7.2.2 Replacing the ligand with the bulky N-heterocyclic carbene IPr and changing the feed ratio to n(Ga) : n($I_2$) : n(IPr) = 2 : 1 : 2 gives ionic compound B. The Ga mass fraction in B is 8.98%, and the principal MS FAB fragment peaks are m/z 389 (100%, $[IPrH]^+$) and 909 (61%, $[Ga_2I_3(IPr)]^+$). Write the SMILES structure of B.", "text_images": [{"file": "Exam-15-T7-3-1-prompt.png", "description": "The IPr ligand in the prompt has SMILES 'CC(C1=C([N+]2=[C-]N(C=C2)C3=C(C=CC=C3C(C)C)C(C)C)C(C(C)C)=CC=C1)C'."}], "full_mark": 2}, {"sub_id": "Exam-15-T7-4", "tags": ["1.4", "2.4"], "text_open": "7.2.3 Passing $NH_3$ into a THF solution of GaBr gives a precipitate. Extraction with DMF and crystallization give crystal C, $C_48H_112Br_18Ga_8N_16O_16$. C is an ionic compound with an anion:cation ratio of 3 : 2, and its anion contains a $Ga_2$ core. Write a chemical formula representing the structure of C.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-15-T7-5", "tags": ["1.4", "2.4"], "text_open": "7.3 $E_2R_4$, where R is alkyl or amino. $E_2R_4$ compounds are usually prepared by reaction of dihalide $E_2X_4L_2$ or subhalide EX with alkyl or amino anions. 7.3.1 Monodentate ligand, silyl: $Ga^{I}[Ga^{III}Cl_4]$ reacts with $Li(THF)_3Si(SiMe_3)_3$ to give D, E, or F depending on feed ratio. n($Ga^{I}[Ga^{III}Cl_4]$) : n($Li(THF)_3Si(SiMe_3)_3$) | product | MS (m/z) | Ga oxidation state | Ga coordination number | Ga chemical environment | Cl element | ideal symmetry. 1 : 1 | tetranuclear D | 1053 (4), 667 (11), 316 (80), 73 (100) | +2 | 4 | only one environment | contains Cl and all Cl environments are identical | $D_{2d}$. 1 : 2 | dinuclear E | 563 (10), 73 (100) | +2 | 3 | | no Cl | $D_{2d}$. 1 : 3 | tetranuclear F | | +1 | 4 | | no Cl | $T_{d}$. Write the SMILES structures of D, E, and F.", "text_images": [], "full_mark": 6}, {"sub_id": "Exam-15-T7-6", "tags": ["1.4", "1.6", "4.3"], "text_open": "7.3.2 Bidentate ligand: the following dpp-bian ligand can coordinate as a monoanion or dianion. Its tripotassium salt $K_3[dpp-bian]$ reacts 1 : 1 with $GaCl_3$ in a reduction-coordination process to give dinuclear H_1. Its zinc salt $[(dpp-bian)ZnI]_2$ reacts with $GaCl_3$ and $K_4[dpp-bian]$ to give dinuclear H_2. Write SMILES structures reflecting their electronic properties, and explain why H_2 absorbs in the near-infrared at 950 nm whereas H_1 does not.", "text_images": [{"file": "Exam-15-T7-6-1-prompt.png", "description": "At far left, the figure shows ligand '[Ar]/N=C(/C2=N/[Ar])C3=CC=CC4=C3C2=CC=C4', labeled \"dpp-bian\", where Ar is 'CC(C)C1=CC=CC(C(C)C)=C1*', and two reaction pathways. In the upper path, `dpp-bian` reacts with `K (ex.)` in `THF` to form $K_4[dpp-bian]$, then with `+0.33 eq dpp-bian` to form $K_3[dpp-bian]$, and then under +$GaCl_3$ and `-3 KCl` gives H1, `[(dpp-bian)Ga-Ga(dpp-bian)]`. In the lower path, `dpp-bian` reacts with Zn (ex.), $I_2$ (0.5 eq), THF to form $[(dpp-bian)ZnI]_2$, then successively with $GaCl_3$ (2 eq) and $K_4[dpp-bian]$ (2 eq) to form H_2, `[(dpp-bian)Zn-Ga(dpp-bian)]`."}], "full_mark": 4}, {"sub_id": "Exam-15-T7-7", "tags": ["1.4", "2.4"], "text_open": "7.3.3 Tridentate ligand: in a suitable tridentate cage-ligand environment, the sixth-period element thallium (Tl) can rarely form a Tl-Tl metal bond. Tridentate ligand $H_3J$ = $(t{-}BuHN)_3SiMe$ can chelate one metal ion between every pair of nitrogen atoms and therefore bridge up to three metal atoms, forming diverse ring or cage structures. Reaction of $H_3J$ with excess n-BuLi in toluene gives dimeric lithium compound $(Li_3J)_2$, in which all Li atoms have identical environments and are three-coordinate to N. Reaction of $(Li_3J)_2$ with excess TlCl in dioxane gives dimeric thallium cluster $(Tl_3J)_2$, in which each Tl is two-coordinate to N and the cluster contains one relatively strong Tl-Tl bond. Write the SMILES structures of $(Li_3J)_2$ and $(Tl_3J)_2$.", "text_images": [], "full_mark": 4}]} | |
| {"problem_id": "Exam-16-T1", "header": "Problem 1: What Is That Smell?\nIn 1803, the British chemist Tennant dissolved crude platinum in dilute aqua regia and found a small amount of black, insoluble powder remaining at the bottom of the vessel. Although early researchers regarded it merely as residual graphite, he suspected that it might contain unknown substances awaiting discovery. Through systematic alternating acid and base treatments, Tennant successfully separated the black residue into two new metallic elements, one of which is metal M. M was named for its distinctive odor and also has an extremely high density and numerous oxidation states.", "header_images": [], "subquestions": [{"sub_id": "Exam-16-T1-1", "tags": ["1.4", "1.7", "2.4"], "text_open": "1.1 M forms several oxides, the best known of which is A. A can be prepared by heating powdered M in a stream of oxygen (Reaction 1). It is a pale-yellow crystal whose melting point is only slightly above body temperature and whose boiling point is only about 400 K. It is appreciably volatile even at room temperature and converts alkenes stereoselectively into cis vicinal diols through a concerted cycloaddition that forms the five-membered cyclic intermediate B. When heated with concentrated hydrochloric acid, A is reduced to complex C of M while gas D is released (Reaction 2). The anion of C has a regular octahedral geometry. M also forms complex ion E with 2,2′-bipyridine (bpy, C_{10}H_{8}N_{2}, molar mass 156.18 g·mol^{-1}). This complex has D_{3} symmetry and exists as Λ/Δ enantiomers; the oxidation state of its central ion is the same as that in C. The mass fraction of M in the anion of C is 47.21%. 1.1.1 Deduce and give the chemical formulas of B-E (give a generic structure for B).", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-16-T1-2", "tags": ["2.3", "2.7", "1.2"], "text_open": "1.3 The dihydroxylation of an alkene by A follows second-order kinetics and proceeds through a concerted [3+2] cycloaddition to form a five-membered cyclic intermediate. 1.3.1 Give the coordination geometry and oxidation state of M in the intermediate. Explain why the product has the cis rather than the trans configuration.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-16-T1-3", "tags": ["4.3", "3.4", "1.4"], "text_open": "1.4 Complex E exhibits a strong MLCT absorption band in the visible region and is deep red when dissolved in acetonitrile. To determine the coordination ratio of M to bpy in E, a student designed a Job plot (method of continuous variations). The total concentration of M and bpy was kept constant at 1.00 × 10^{-4} mol·L^{-1}, their mole ratio was varied, and the absorbance of each solution in acetonitrile was measured at the characteristic wavelength \\lambda_{max} = 480 nm. At this wavelength, neither free M ions nor bpy absorb, although complexes of M with bpy may absorb. A 1.00 cm cuvette was used. $$\\begin{array}{c|ccccccccc}\\text{Sample}&1&2&3&4&5&6&7&8&9\\\\ \\hline \\mathbf{M}\\text{ mole fraction }x(\\mathbf{M})&0.10&0.20&0.30&0.40&0.50&0.60&0.70&0.80&0.90\\\\ \\text{Absorbance }A&0.128&0.248&0.352&0.440&0.488&0.440&0.352&0.248&0.128\\end{array}$$ 1.4.1 Determine the coordination ratio n = [bpy] / [M] in complex ion E under these experimental conditions, show the reasoning, and explain why this coordination ratio is observed in the experiment.", "text_images": [], "full_mark": 3}]} | |
| {"problem_id": "Exam-16-T2", "header": "Problem 2: Thermogravimetric Analysis of a Complex Double Salt\nOn April 28, 2023, researchers from the Institute of General and Inorganic Chemistry published a thermogravimetric study of a double salt. A solution of 2 mL of 85% phosphoric acid diluted with 5 mL of deionized water was added in small portions to a suspension prepared beforehand from 320 mg of calcium carbonate and 5 mL of deionized water. During continuous stirring at room temperature, foaming caused by the release of carbon dioxide was first observed, after which the solution became clear. Another solution containing 560 mg of K_{2}HPO_{4} in 5 mL of deionized water was then added. The resulting solution remained completely clear, and product A was precipitated by adding 15 mL of acetone in five 3 mL portions while stirring continuously. The precipitate was filtered, washed twice with 2.5 mL portions of acetone, and dried under ambient conditions to give A. Thermogravimetric analysis of A gave the result shown below. A is stable at 95°C and completes thermal decomposition at 530 °C. The decomposition occurs in the ranges 95-143°C, 143-240°C, and 240-530°C, each accompanied by a corresponding mass loss. The theoretical mass losses for the three steps are 6.645%, 4.984%, and 6.645%, respectively.", "header_images": [], "subquestions": [{"sub_id": "Exam-16-T2-1", "tags": ["4.6", "4.7", "2.1"], "text_open": "2.1 Using the preparation and thermogravimetric experiments for A, determine the chemical formula of A by appropriate qualitative reasoning and quantitative calculation.", "text_images": [{"file": "Exam-16-T2-1-1-prompt.png", "description": "In this thermogravimetric-differential thermal analysis (TG-DTA) plot, the left vertical axis is mass percentage ($\\text{Mass / \\%}$), ranging from 80 to 100, and corresponds to the dark-red TG curve. As temperature increases, this curve displays several stepwise mass-loss stages and finally levels off at approximately $82\\%$. The right vertical axis is heat flow ($\\text{Heat flow / }\\mu\\text{V}$), ranging from -4 to 2; a downward arrow labeled \"$\\text{endo}$\" indicates that downward deflection is endothermic. The red DTA curve shows four prominent changes with increasing temperature: a sharp, strong downward endothermic peak at 120, a second distinct downward endothermic peak at 186 after recovery, a narrower third downward endothermic peak at 301 after another recovery, and then an upward rise to a broad local exothermic peak at 458, after which the curve trends downward overall."}], "full_mark": 6}, {"sub_id": "Exam-16-T2-2", "tags": ["2.1"], "text_open": "2.2 Write the chemical equations for the three mass-loss steps, expressing A and each intermediate product with formulas containing 2 Ca atoms.", "text_images": [], "full_mark": 3}]} | |
| {"problem_id": "Exam-16-T4", "header": "Problem 4 Determination of Metal Content by Disodium Ethylenediaminetetraacetate Titration. First prepare the standard solutions. Standard zinc titrant (0.0100 mol/L): Weigh 0.6538 g of pure zinc (99.99%) into a 300 mL conical flask fitted with a funnel. Add 15 mL hydrochloric acid. After the vigorous reaction has ceased, heat gently until the zinc has dissolved completely. Cool to room temperature, transfer to a 1000 mL volumetric flask, dilute to the mark with water, and mix. Standard disodium ethylenediaminetetraacetate titrant (0.01 mol/L): Weigh 3.7225 g of disodium ethylenediaminetetraacetate, dissolve it in 400 mL water, transfer to a 1000 mL volumetric flask, dilute to the mark with water, and mix. Standardization: Pipette 25.00 mL of the standard disodium ethylenediaminetetraacetate titrant into a 250 mL conical flask. Add 75 mL water, 10 mL ammoniacal buffer solution (pH = 10), and 3 drops of Eriochrome Black T solution (weigh 0.2 g Eriochrome Black T into a 50 mL beaker, add 15 mL triethanolamine and 5 mL ethanol (95%), and mix). Titrate with the standard zinc titrant to a red endpoint. Three parallel titrations consume 24.85 mL of the standard zinc titrant.", "header_images": [], "subquestions": [{"sub_id": "Exam-16-T4-1", "tags": ["4.1", "4.2"], "text_open": "4.2 Preparation and Pretreatment of the Sample Solution\nPlace the chromium-containing alloy sample (0.9985 g) in a 500 mL beaker, cover with a watch glass, add 10 mL of hydrochloric acid (1+1), and heat to decompose most of the sample. Add 40 mL of concentrated perchloric acid, heat until white fumes appear, and continue heating until the solution volume is 5 mL~10 mL. Remove from heat and cool. Add approximately 100 mL of water and 1 g of ammonium chloride. Neutralize with concentrated ammonia until a precipitate forms, then add hydrochloric acid (1+1) until the precipitate just dissolves, followed by an excess of 5~10 drops. Add hexamethylenetetramine solution (250 g/L) until a precipitate forms and boil. Keep at 80°C~90°C for 15 min~20 min, filter through quantitative filter paper, and wash the precipitate and filter paper with warm hexamethylenetetramine solution (5 g/L). Transfer the precipitate together with the filter paper back to the original beaker. Add 10 mL of concentrated nitric acid to dissolve the precipitate and destroy the filter paper, then add 20 mL of concentrated perchloric acid and heat until white fumes appear. Add concentrated hydrochloric acid in several portions to volatilize the remaining chromium, then continue fuming for approximately 10 min. Remove from heat and cool. 4.2.1 Explain why precipitation is performed with hexamethylenetetramine under heated conditions instead of directly with aqueous ammonia (no more than 100 words).", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-16-T4-2", "tags": ["4.1", "4.2"], "text_open": "4.2.2 Explain why the precipitate is washed with warm hexamethylenetetramine solution (5 g/L) rather than a cold solution (no more than 100 words).", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-16-T4-3", "tags": ["2.1", "4.5"], "text_open": "4.3 Titration of Iron Ions. Add approximately 100 mL of warm water to dissolve the salts, then add 2 mL of sulfosalicylic acid solution (50 g/L). Adjust the solution to pH 1.0–pH 2.2 with hydrochloric acid (1+2) or concentrated ammonia. Titrate with the standard disodium ethylenediaminetetraacetate titrant until the solution changes from red-violet to colorless or pale yellow; 21.64 mL of the standard disodium ethylenediaminetetraacetate titrant is consumed. 4.4 Titration of Aluminum Ions. Continue by adding an appropriate amount of the standard disodium ethylenediaminetetraacetate titrant. Add 20 mL of acetic acid-ammonium acetate buffer solution (pH = 5.5), heat to boiling for 2.5 min~3 min, and add 2~4 drops of xylenol orange solution (1 g/L) as the indicator. Titrate with the standard zinc titrant to a red endpoint (do not calculate the volume). Add 20 mL of ammonium fluoride solution (40 g/L) and boil for 2 min. Remove from heat and cool to room temperature. Add one further drop of xylenol orange solution as the indicator, then titrate with the standard zinc titrant to a red endpoint; 9.24 mL of the standard zinc titrant is consumed.\n4.4.2 Write the reaction equation for the step involving addition of ammonium fluoride (use Y^{4-} for the ethylenediaminetetraacetate tetraanion and HAc for acetic acid).", "text_images": [], "full_mark": 2}]} | |
| {"problem_id": "Exam-16-T5", "header": "Problem 5: Fullerenes. In the 80s of the 20th century, the British astrophysicist Kroto and co-workers discovered a new class of carbon allotropes whose molecular shapes are polyhedra composed of pentagons and hexagons. The most representative example is C_{60}. Its structure resembles the dome of the United States pavilion at the World Exposition in Montreal, Canada, designed by the American architect Buckminster Fuller; these molecules (not only C_{60}) are therefore called fullerenes.", "header_images": [], "subquestions": [{"sub_id": "Exam-16-T5-1", "tags": ["1.3"], "text_open": "5.1 The C_{60} molecule is a 32-faced polyhedron composed of 12 regular pentagonal faces and 20 regular hexagonal faces, and it has high symmetry. Its structure is shown below. 5.1.4 Above 30℃, C_{60} crystals can precipitate from a benzene solution. Their density is 1.65 g/cm^{3}, and the C_{60} molecules adopt cubic close packing in the crystal. Use these data to estimate the van der Waals radius of the C_{60} molecule.", "text_images": [{"file": "Exam-16-T5-1-1-prompt.png", "description": "Structural diagram of C_60: a spherical cage fullerene composed of 12 pentagons and 20 hexagons, with vertices representing carbon atoms and edges representing C-C bonds."}], "full_mark": 4}, {"sub_id": "Exam-16-T5-2", "tags": ["1.1", "1.3", "1.5"], "text_open": "5.3 Smaller fullerenes. In a fullerene molecule, every carbon atom is bonded to three neighboring carbon atoms. From the Lewis-structure perspective, resonance structures can therefore be drawn in which each carbon atom participates in two single bonds and one double bond. In this sense, the smallest possible fullerene could be tetrahedral C_{4} (although it has not yet been synthesized), whereas icosahedral C_{12} is not a fullerene. In this problem (5.3), you need not consider the isolated-pentagon rule. 5.3.1 Which of the following molecule(s) could be fullerenes?\n(a) Octahedral C_6\n(b) Cubic C_8\n(c) C_{12} with the structure of the “truncated cube 3^84^6,” a polyhedron composed of regular triangles and squares whose vertices are the 12 edge midpoints of a cube\n(d) Dodecahedral C_{20}", "text_images": [{"file": "Exam-16-T5-2-1-prompt.png", "description": "The image shows three identical geometric wireframe diagrams side by side, labeled (a), (b), and (c) from left to right. Each diagram consists of a large outer equilateral triangle and four circular nodes: three nodes at the vertices of the large triangle and one node at its center. Three internal lines connect the central node to the three vertices, and three outer edges connect the vertices. In each diagram, two selected connections are drawn as double lines by adding a parallel thin line. In (a), the highlighted connections are the vertical internal line from the center to the top node and the bottom horizontal outer edge. In (b), they are the diagonal line from the center to the lower-right node and the diagonal internal line connecting the top node to the lower-left vertex. In (c), they are the diagonal internal line from the center to the lower-left node and the sloping outer edge from the top node to the lower-right node."}], "full_mark": 2}, {"sub_id": "Exam-16-T5-3", "tags": ["1.3"], "text_open": "5.4 Alkali-Metal Compounds of Fullerenes. Several compounds of C_{60} with alkali metals have been synthesized, some of which exhibit distinctive superconducting properties. We now examine the crystal structure of a compound A of C_{60} and potassium. A belongs to the cubic crystal system. In its cubic unit cell, a C_{60} anion occupies each vertex and the body center, while each pair of opposite faces contains 4 potassium atoms. These potassium atoms form the “cage structure” composed of rhombic and hexagonal faces shown below.\n5.4.2 Two potassium atoms in the crystal have fractional coordinates (1/2, 0.2795, 0) and (0.2205, 1/2, 0). Give the fractional coordinates of all remaining potassium atoms in the unit cell. Do not repeat the two coordinates already given.", "text_images": [{"file": "Exam-16-T5-3-1-prompt.png", "description": "The image shows a unit-cell diagram. Thin black lines outline a cubic unit cell, and purple spheres are located within it, with some pairs connected by purple line segments. At the lower left is a three-dimensional orthogonal coordinate system whose red, green, and blue arrows emerge from a gray central sphere: the red `a` axis points diagonally down-left, the green `b` axis points right, and the blue `c` axis points vertically upward. These axes are parallel to the three sets of unit-cell edges. There are 24 purple spheres distributed over the 6 faces of the cubic unit cell, 4 on each face. The 4 spheres on each face form a rhombus rather than a perfect square; both diagonals of the rhombus are perpendicular to the edges of the square unit-cell face, and the center of the rhombus coincides with the center of that face. Purple line segments occur in two places: along the edges of these rhombi and between the closest pair of purple spheres belonging to rhombi on adjacent unit-cell faces."}], "full_mark": 3}, {"sub_id": "Exam-16-T5-4", "tags": ["1.3"], "text_open": "5.4.3.1 Let the unit-cell edge length be a Å. Calculate the shortest K-K distance.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-16-T5-5", "tags": ["1.3", "1.1"], "text_open": "5.4.3.2 Determine whether the hexagonal faces of the cage-like structure are regular hexagons, and explain.", "text_images": [], "full_mark": 3}]} | |
| {"problem_id": "Exam-16-T6", "header": "Problem 6: Polysulfide Anions\nA sodium-sulfur battery is a high-temperature molten secondary alkali-metal energy-storage battery that uses metallic sodium as the negative electrode, elemental sulfur as the positive electrode, and molten sodium beta-alumina as the solid electrolyte. It offers high energy density, inexpensive raw materials, and suitability for large-scale energy storage, and has become a research focus in electrochemical energy storage in recent years. Charging and discharging proceed through the formation of sodium polysulfides (Na_{2}S_{x}) with different chain lengths.", "header_images": [], "subquestions": [{"sub_id": "Exam-16-T6-1", "tags": ["2.1", "2.4"], "text_open": "6.2 Sodium polysulfides (Na_{2}S_{x}) are key electroactive intermediates in sodium-sulfur batteries, and their anions can be synthesized and observed in the laboratory using common reagents. Under an N_{2} atmosphere, excess sodium tert-butoxide is added to a DMF solution of S_{8}, reducing S_{8} to polysulfide anion A. A disproportionates into another polysulfide anion B containing fewer atoms. Electron paramagnetic resonance (EPR) measurements show that some B anions undergo homolysis in solution to form bent anion C. C is trapped in zeolite cages in lazurite crystals, [(Na,Ca)_{4-8}(AlSiO_{4})_{6}(SO_{4},S,Cl)_{2}], and appears blue in solution. The sodium salt corresponding to polysulfide anion B is a common component of sodium-sulfur batteries. If it is heated to 300°C and melted, anion C dimerizes again to give isomer D of polysulfide anion B; the sulfur atoms in D occupy only two distinct chemical environments. 6.2.1 Give the chemical formulas of A and B.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-16-T6-2", "tags": ["1.6", "2.4"], "text_open": "6.2.2 Give the SMILES structures of C and D, showing the Lewis connectivity, and indicate the formal charge on every sulfur atom.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-16-T6-3", "tags": ["1.6", "2.4"], "text_open": "6.3 Anion C can also be observed when S_{8} is dissolved in liquid ammonia, together with another hydrogen-free anion E. Early studies proposed the following mechanism: first, liquid ammonia undergoes autoionization to form NH_{2}^{-} and NH_{4}^{+}; second, NH_{2}^{-} attacks S_{8}, giving S_{8}NH_{2}^{-} by addition; third, S_{8}NH_{2}^{-} decomposes into cyclic anion F and a stable dibasic acid; fourth, F rearranges to form anion E and S_{8}, while the dibasic acid reacts with S_{8} and NH_{4}^{+} to form (NH_{4})_{2}S_{x}, which is subsequently converted into C. EPR measurements, however, show that anion F is not formed during the reaction. Moreover, the tendency of liquid ammonia to autoionize is extremely small and cannot explain the high solubility of S_{8} in liquid ammonia. Researchers therefore proposed a more plausible mechanism: first, S_{8}NH_{2}^{-} and NH_{3} participate in a proton-transfer equilibrium, forming S_{8}NH_{3} while generating NH_{2}^{-}, which promotes dissolution; second, NH_{3} can also react by nucleophilic substitution to form the small molecule S_{2}NH_{3}; third, S_{8}NH_{2}^{-} decomposes to anion B and intermediate G; fourth, G transfers all of its protons to NH_{3} to form anion H; fifth, H reacts with S_{2}NH_{3} to give anion E and NH_{3}, with all species in this reaction having a stoichiometric coefficient of 1. 6.3.1 Give the SMILES structures of E, F, G, and H, showing the Lewis connectivity. The Lewis structure of E is known to contain two S-S single bonds, and its sulfur atoms occupy two distinct chemical environments.", "text_images": [], "full_mark": 5}, {"sub_id": "Exam-16-T6-4", "tags": ["2.1", "2.2"], "text_open": "6.3.2 In fact, the fifth reaction proceeds in three steps, and both the first and second steps generate SNH_{3}. Write the chemical equations for the three steps in order.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-16-T6-5", "tags": ["3.2", "3.4"], "text_open": "6.4 Anion C is also relatively common in nature and occurs mainly in ores formed under high-temperature, high-pressure conditions. Calculations show that at 350℃ and 0.5 GPa, anion C can exist as a stable form of sulfur. Reaction 1 (unbalanced) is: H_2S(aq.) + SO_4^{2-}(aq.) + H^+(aq.) → C(aq.) + O_2(g.) + H_2O(l.), \\lg K_1 = -9.6 (when the stoichiometric coefficients of C and H^+ are both 1). The standard enthalpies of formation and standard entropies of several species are listed below (ignore the temperature dependence of enthalpy and entropy and ignore microscopic effects): H_2S(aq.): \\Delta_fH_m^⊖ = -39.0 kJ·mol^{-1}, S_m^⊖ = 122.0 J·mol^{-1}·K^{-1}; HS^-(aq.): -17.7, 61.1; HSO_4^-(aq.): -885.8, 126.9; SO_4^{2-}(aq.): -907.5, 17.0; H^+(aq.): 0, 0; O_2(g.): 0, 205.1; H_2O(l.): -285.8, 70.0.\n6.4.2 Derive the equation of the bold sloping line l_1 in the figure below. Hint: f_O2 denotes the fugacity of oxygen, which physically represents the effective pressure of a real gas. The total dissolved sulfur concentration in the figure is 1 wt% (that is, all dissolved sulfur-containing species in 1 kg of solution contain a total of 10 g of S atoms). Assume that the density of water under these conditions is ρ = 1.0 × 10^3 kg/m^3.", "text_images": [{"file": "Exam-16-T6-5-1-prompt.png", "description": "A two-dimensional predominance-diagram-style plot with a white background, black lines, and a rectangular frame. The top and right sides of the frame also have short tick marks. The horizontal axis is labeled “pH,” with major tick labels 1, 2, 3, 4, 5, 6, 7, 8, and 9; the axis extends beyond 9 without a clearly labeled 10. The vertical axis is labeled “$\\log_{10} f_{O_2}$,” with major ticks −22, −24, −26, −28, −30, −32, and −34 from top to bottom. Several solid black lines divide the plot into regions labeled with formulas or letters. The upper-left region is labeled “HSO_4^-,” the upper-right “SO_4^2-,” the lower-left “H_2S,” and the lower-right “HS^-.” Near the center-left is a light-gray irregular polygon labeled with a capital “C.” It has approximately five sides: an upper-left sloping boundary from about pH 1 and $\\log_{10} f_{O_2}$ about −24 to about pH 5.3 and $\\log_{10} f_{O_2}$ about −26; an upper-right sloping boundary from about pH 5.3 and $\\log_{10} f_{O_2}$ about −26 to about pH 7.6 and $\\log_{10} f_{O_2}$ about −28.4; a thinner lower-right boundary from about pH 7.6 and $\\log_{10} f_{O_2}$ about −28.4, extending down-left to about pH 6.0 and $\\log_{10} f_{O_2}$ about −31; a noticeably bold lower boundary from about pH 6.0 and $\\log_{10} f_{O_2}$ about −31, extending up-left to about pH 1 and $\\log_{10} f_{O_2}$ about −26.8; and a left boundary adjoining the vertical axis. A vertical solid line at about pH 5.3 runs from the top border to near the upper vertex of the gray region and separates HSO_4^- from SO_4^2-. Another vertical solid line at about pH 6.0 runs downward from the lower vertex of the gray region to the bottom border and separates H_2S from HS^-. A thinner sloping line also extends down-right from the upper-right vertex of the gray region to near the right border, separating SO_4^2- from HS^-."}], "full_mark": 4}, {"sub_id": "Exam-16-T6-6", "tags": ["2.2", "4.2"], "text_open": "6.5 Anion C also has many applications in organic synthesis. Use the results of the following control experiments to answer Questions 6.5.1-6.5.4. 6.5.1 According to the control experiments, from which species formed by conversion of elemental sulfur might the sulfur in the product originate?", "text_images": [{"file": "Exam-16-T6-6-1-prompt.png", "description": "Control experiments for 6.5.1, showing two parallel reactions. In the upper reaction, 'C1=CC=C(C#CC#CC2=CC=CC=C2)C=C1' (0.2 mmol) reacts with `S` (0.6 mmol) under 1.8 mmol NaOtBu in DMF, N_2, at 25°C to give 'C1=CC=C(C2=CC=C(C3=CC=CC=C3)S2)C=C1' in 65% yield. In the lower reaction, 'C1=CC=C(C#CC#CC2=CC=CC=C2)C=C1' (0.2 mmol) reacts with `S` (0.6 mmol) under 1.8 mmol NaO^tBu and 4.0 equiv TEMPO in ^tBuOH+DMF, N_2, at 25°C to give 'C1=CC=C(C2=CC=C(C3=CC=CC=C3)S2)C=C1' in 5% yield."}], "full_mark": 1}, {"sub_id": "Exam-16-T6-7", "tags": ["2.3", "2.7"], "text_open": "6.5.2 In the control experiments above, no radical intermediate was trapped in the first reaction, indicating that it does not follow a radical mechanism, whereas the third reaction proceeds through a radical mechanism. Explain, from the perspective of molecular structure, why the two reactions follow different mechanisms.", "text_images": [{"file": "Exam-16-T6-7-1-prompt.png", "description": "Control experiments for 6.5.2, showing four parallel reactions. In the first, 'C1=CC=C(C#CC#CC2=CC=CC=C2)C=C1' (0.2 mmol) reacts with `Na_2S · 9H_2O` (0.6 mmol) in DMF, air, at 25°C to give 'C1=CC=C(C2=CC=C(C3=CC=CC=C3)S2)C=C1' in 99% yield. In the second, 'C1=CC=C(C#CC#CC2=CC=CC=C2)C=C1' (0.2 mmol) reacts with `Na_2S · 9H_2O` (0.6 mmol) with 4.0 equiv TEMPO in DMF, air, at 25°C to give 'C1=CC=C(C2=CC=C(C3=CC=CC=C3)S2)C=C1' in 92% yield. In the third, 'C1=CC=C(C#CC2=CC=CC=C2)C=C1' (0.6 mmol) reacts with `K_2S` (0.36 mmol) in DMF, Ar, at 130°C to give 'C1(C2=CC=CC=C2)=C(C3=CC=CC=C3)C(C4=CC=CC=C4)=C(C5=CC=CC=C5)S1' in 65% yield. In the fourth, 'C1=CC=C(C#CC2=CC=CC=C2)C=C1' (0.6 mmol) reacts with `K_2S` (0.36 mmol) with 4.0 equiv TEMPO in DMF, Ar, at 130°C to give only a trace of 'C1(C2=CC=CC=C2)=C(C3=CC=CC=C3)C(C4=CC=CC=C4)=C(C5=CC=CC=C5)S1'."}], "full_mark": 2}, {"sub_id": "Exam-16-T6-8", "tags": ["2.2", "2.3"], "text_open": "6.5.3 Give the SMILES structures of two key intermediates in the third reaction of the control experiments in 6.5.2.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-16-T6-9", "tags": ["2.8", "2.3"], "text_open": "6.5.4 Explain separately why the yields of the following two reactions are substantially lower than that of the third reaction in 6.5.2.", "text_images": [{"file": "Exam-16-T6-9-1-prompt.png", "description": "Control experiments for 6.5.4, showing two parallel reactions. In the upper reaction, 'C1=CC=C(C#CC2=CC=CC=C2)C=C1' (0.6 mmol) reacts with `Na_2S · 9H_2O` (0.36 mmol) in DMF, Ar, at 130°C to give 'C1(C2=CC=CC=C2)=C(C3=CC=CC=C3)C(C4=CC=CC=C4)=C(C5=CC=CC=C5)S1' in 36% yield. In the lower reaction, 'C1=CC=C(C#CC2=CC=CC=C2)C=C1' (0.6 mmol) reacts with `S_8` (0.36 mmol) under NaO^tBu in DMF, Ar, at 130°C to give 'C1(C2=CC=CC=C2)=C(C3=CC=CC=C3)C(C4=CC=CC=C4)=C(C5=CC=CC=C5)S1' in less than 5% yield."}], "full_mark": 2}]} | |
| {"problem_id": "Exam-16-T7", "header": "Problem 7: Baldwin’s Rules\nThis problem concerns the regioselectivity of intramolecular ring-closing reactions, Baldwin’s rules, and their exceptions.", "header_images": [], "subquestions": [{"sub_id": "Exam-16-T7-1", "tags": ["2.3", "2.7", "2.4"], "text_open": "7.1 Give the SMILES structure of the product formed by the intramolecular reaction of the molecule shown below under suitable conditions, and explain the selectivity.", "text_images": [{"file": "Exam-16-T7-1-1-prompt.png", "description": "Substrate in the prompt, with SMILES 'C=C(CC(N)C(OC)=O)C(OC)=O'."}], "full_mark": 2}, {"sub_id": "Exam-16-T7-2", "tags": ["2.3", "2.7", "2.4"], "text_open": "7.2 Baldwin’s rules are important empirical rules for explaining the regioselectivity of ring-closing reactions. Researchers used the reaction below in a classic mechanistic test of these rules. 7.2.1 Give the SMILES structures of the key intermediate in each of the two possible pathways and of the final product, and use Baldwin’s rules to identify the predominant pathway.", "text_images": [{"file": "Exam-16-T7-2-1-prompt.png", "description": "Reaction in the prompt: PhCH=CHCO_2Me reacts with NH_2NH_2 at 65°C."}], "full_mark": 5}, {"sub_id": "Exam-16-T7-3", "tags": ["2.3", "4.2"], "text_open": "7.2.2 Proof by contradiction is often used in scientific research to test reaction mechanisms. Using the substrate shown below as the main starting material, design an experiment to demonstrate that this reaction does not proceed through the pathway disfavored by Baldwin’s rules. State the experimental approach and the basis for the inference.", "text_images": [{"file": "Exam-16-T7-3-1-prompt.png", "description": "Substrate in the prompt, with SMILES 'O=C(/C=C/C1=CC=CC=C1)O[K]'."}], "full_mark": 2}, {"sub_id": "Exam-16-T7-4", "tags": ["2.3", "2.7", "2.4"], "text_open": "7.3 “Exceptions” to Baldwin’s rules. 7.3.1 Give the SMILES structure of the product formed by the intramolecular reaction of the molecule shown below under suitable basic conditions, and explain the selectivity.", "text_images": [{"file": "Exam-16-T7-4-1-prompt.png", "description": "Substrate in the prompt, with SMILES 'BrCCC(C)(C)C(C)=O'."}], "full_mark": 2}, {"sub_id": "Exam-16-T7-5", "tags": ["2.3", "1.6"], "text_open": "7.3.2 The molecule shown below readily undergoes ring closure under suitable basic conditions with the amide nitrogen as the nucleophilic center. Explain why.", "text_images": [{"file": "Exam-16-T7-5-1-prompt.png", "description": "Substrate in the prompt, containing a structure with SMILES 'O=C(N[Ar])CCCBr'."}], "full_mark": 2}, {"sub_id": "Exam-16-T7-6", "tags": ["2.3", "2.7", "2.4"], "text_open": "7.4 Give the SMILES structure of the major product formed by the intramolecular reaction of the molecule shown below under suitable basic conditions, and explain the selectivity.", "text_images": [{"file": "Exam-16-T7-6-1-prompt.png", "description": "Substrate in the prompt, with SMILES 'O=C(C)CC(C(C)=O)CC(C)=O'."}], "full_mark": 2}, {"sub_id": "Exam-16-T7-7", "tags": ["2.3", "2.4", "2.7"], "text_open": "7.5 Drawing on the discussion above, give the SMILES structure of the product formed by the intramolecular reaction of the molecule shown below under suitable conditions.", "text_images": [{"file": "Exam-16-T7-7-1-prompt.png", "description": "Substrate in the prompt, with SMILES 'O=C(CC(OCC)=O)CC(CCl)OC'."}], "full_mark": 2}]} | |
| {"problem_id": "Exam-16-T9", "header": "Problem 9: Effects of Reaction Conditions and Mechanistic Investigation of Electrochemical Decarboxylative Alkylation\nElectrochemical anodic oxidation can use inexpensive and readily available quaternary-ammonium salts of alkyl carboxylic acids as alkyl sources, avoiding the need for additional oxidants and transition-metal catalysts during decarboxylation. The method therefore offers a simple system, mild conditions, and environmental advantages. Researchers used an unsaturated amide as a model substrate to investigate decarboxylative methylation/arylation. Tetrabutylammonium acetate served as both the methyl source and electrolyte, with carbon cloth as the anode, platinum as the cathode, and DMSO as the solvent. Under constant-current electrolysis at 3 mA with 6.0 equivalents of acetic acid as a proton source at 70°C for 9 h (5.0 F/mol), the target oxindole was obtained in 80% yield.", "header_images": [], "subquestions": [{"sub_id": "Exam-16-T9-1", "tags": ["3.5", "2.8", "2.3"], "text_open": "9.1 Further study showed that the acidity of the proton source strongly affects the reaction. When weaker proton sources, such as tert-butanol or hexafluoroisopropanol, or stronger proton sources, such as p-toluenesulfonic acid or trifluoroacetic acid, were used, the yield decreased substantially and the reaction was sometimes almost completely suppressed. Give a reasonable explanation for each observation.", "text_images": [{"file": "Exam-16-T9-1-1-prompt.png", "description": "Standard reaction in the prompt: 'FC(F)(C(C(N(C1=CC=CC=C1)C)=O)=C)F' reacts with `TBAOAc (1.5 equiv.)` using `C (+) / Pt (-)` under `HOAc (6.0 equiv.)`, `DMSO 4 mL, 70°C`, and `I = 3 mA, 5.0 F/mol`, producing 'CCC1(C(N(C2=CC=CC=C21)C)=O)C(F)(F)F'."}], "full_mark": 4}, {"sub_id": "Exam-16-T9-2", "tags": ["2.2", "2.3", "4.2"], "text_open": "9.3 The researchers performed the following experiments to investigate the reaction mechanism. Note: “standard condition” refers to the standard conditions described above, under which the yield is 80%. 9.3.1 To test for a free-radical intermediate, TEMPO was added as a radical trap. The target product did not form at all. Analysis of the reaction mixture together with high-resolution mass spectrometry revealed products formed by trapping with TEMPO. Give the SMILES structures of possible products.", "text_images": [{"file": "Exam-16-T9-2-1-prompt.png", "description": "Mechanistic experiment for 9.3.1: 'CN(C(C(C)=C)=O)C1=CC=CC=C1' reacts with the salt formed from anion 'CC([O-])=O' (shown with a circled minus sign) and cation 'CCCC[N+](CCCC)(CCCC)CCCC' (shown as `N^nBu_4` with a circled plus sign) under `standard condition` with *TEMPO (3.0 equiv.)*."}], "full_mark": 2}, {"sub_id": "Exam-16-T9-3", "tags": ["2.2", "2.3", "2.4"], "text_open": "9.3.3 On the basis of these mechanistic experiments, two pathways remain possible for the ring-closing step of the standard reaction. Give the SMILES structures of the key intermediates that may be involved in the standard reaction and identify the two possible pathways corresponding to the ring-closing step.", "text_images": [], "full_mark": 6}]} | |
| {"problem_id": "Exam-17-T5", "header": "Problem 5 Dicarbon Anions", "header_images": [], "subquestions": [{"sub_id": "Exam-17-T5-1", "tags": ["1.3", "1.4"], "text_open": "5.2 Starting from crystalline $CaC_2$, inserting square Co layers into some of the interlayer spaces and replacing Ca with U gives a tetragonal $UCoC_2$ crystal. Its structure is shown below, where the large spheres represent U, the medium spheres represent Co, and the small spheres represent C. The lattice parameter c = 728 pm.\n5.2.1 Name the coordination polyhedron formed by the C atoms around Co.", "text_images": [{"file": "Exam-17-T5-1-1-prompt.png", "description": "An oblique three-dimensional ball-and-stick rendering of the tetragonal unit cell of $UCoC_2$ on a white background. Thin gray-black lines outline a rectangular-prismatic unit-cell frame elongated vertically along c. There are 10 large spheres, including spheres at face centers on the top and the upper and lower basal faces; 4 medium spheres lie on the side faces, one at an upper-middle position on each left/right face and one at a lower-middle position on each front/back face; and 8 small spheres form upper and lower groups, with two vertical pairs in each group along c joined by short brown sticks. The upper pair is on the front/back faces and the lower pair on the left/right faces. The small spheres are joined to adjacent large spheres by short sticks, forming a layered framework; the medium spheres have no sticks. At lower left, a blue arrow points upward and is labeled c, a green arrow points right and is labeled b, and a red arrow points lower-left and is labeled a. Bonding obscured by occlusion cannot be resolved. Apart from a, b, and c, the image contains no text or legend. According to the caption, the large, medium, and small spheres represent U, Co, and C, respectively. The coordinates of the two carbon atoms in the unit cell are (1/2, 0, 0.155) and (1/2, 0, 0.358)."}], "full_mark": 2}, {"sub_id": "Exam-17-T5-2", "tags": ["1.3"], "text_open": "5.3 A uranium carbide belongs to the cubic crystal system. Its unit-cell projections along the a, b, and c directions are all as shown below. The large spheres represent U and the small spheres represent C. All atoms of the same element are equivalent in the crystal, and the lattice parameter a = 808 pm.\n5.3.1 Write the chemical formula of this crystal.", "text_images": [{"file": "Exam-17-T5-2-1-prompt.png", "description": "Crystal-axis indicator on a white background: a solid red circle is at the lower left with a black letter a superimposed, indicating that the a axis points out of the page toward the viewer. A blue arrow extends upward from the circle and is labeled c above its tip; a green arrow extends rightward and is labeled b to the right of its tip. Apart from a, b, and c, the image contains no text or other graphics."}, {"file": "Exam-17-T5-2-2-prompt.png", "description": "Crystal-axis indicator on a white background: a solid green circle is at the lower left with a black letter b superimposed, indicating that the b axis points out of the page toward the viewer. A red arrow extends upward from the circle and is labeled a above its tip; a blue arrow extends rightward and is labeled c at its right end. Apart from a, b, and c, the image contains no text or other graphics."}, {"file": "Exam-17-T5-2-3-prompt.png", "description": "Crystal-axis indicator on a white background: a solid blue circle is at the lower left with a black letter c superimposed, indicating that the c axis points out of the page toward the viewer. A green arrow extends upward from the circle and is labeled b above its tip; a red arrow extends rightward and is labeled a at its right end. Apart from a, b, and c, the image contains no text or other graphics."}, {"file": "Exam-17-T5-2-4-prompt.png", "description": "Unit-cell projection of the uranium carbide in 5.3 (two-dimensional projection of a ball-and-stick model on a white background): thin black lines form a square unit-cell border, with some spheres lying on the border. The image contains only two sphere types: large light-blue highlighted spheres and small brown spheres. The large spheres occur at the four corners, at the middle of the top and bottom edges, and at several interior positions; some large spheres overlap in pairs. Some small spheres are joined by short brown sticks into paired or chain-like arrangements, while others are isolated. The image contains no text, letters, numbers, arrows, or legend. As specified in the prompt, the large spheres represent U and the small spheres represent C. The prompt states that the projections along the a, b, and c directions are all as shown; this image is that projection panel."}], "full_mark": 3}, {"sub_id": "Exam-17-T5-3", "tags": ["1.3"], "text_open": "5.3.4 One carbon atom in the unit cell has coordinates (1/2, 1/4, 0.536). Calculate the carbon-carbon bond length in this crystal.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-17-T5-4", "tags": ["1.3", "1.7"], "text_open": "5.4 The carbon-carbon bond lengths in ethane, ethene, and ethyne are approximately 154 pm, 134 pm, and 120 pm, respectively. On this basis, predict the negative charge on the dicarbon anion in $CaC_2$, $UCoC_2$, and the uranium carbide in 5.3, respectively. The carbon-carbon bond orders in the three compounds are distinct integers.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-17-T5-5", "tags": ["1.3", "1.7"], "text_open": "5.5 Crystalline $DyCoC_2$ has a structure entirely different from that of $UCoC_2$. Its three-dimensional structure and projection along the c direction are shown below, where the large spheres represent Dy, the medium spheres represent Co, and the small spheres represent C. The only common stable positive oxidation state of Dy is +3. A small sample of $DyCoC_2$ is hydrolyzed in 2 mol/L dilute hydrochloric acid. Gas-chromatographic analysis gives the following mass fractions of the gaseous hydrolysis products: 74% ethane, 13% propane, 12% methane, and 1% ethene; no ethyne is formed.\n5.5.1 Predict the carbon-carbon bond order in $DyCoC_2$.", "text_images": [{"file": "Exam-17-T5-5-1-prompt.png", "description": "Three-dimensional ball-and-stick model of crystalline $DyCoC_2$ (oblique rendering on a white background): thin black lines outline a parallelepiped unit cell. There are 10 large blue-violet highlighted spheres of varying shades: 8 at the 8 corners and 2 at the centers of the top and bottom faces. There are 3 medium dark-blue spheres in a horizontal layer through the middle of the cell, positioned left of center, right of center, and at the center. Approximately 9 small brown spheres surround the medium spheres and are connected to them by dark-blue or brown short sticks; some small spheres bridge two medium spheres, forming an in-layer zigzag network. There are no direct sticks between large spheres or between medium spheres. Some spheres are obscured, so the exact number of brown small spheres cannot be fully determined. A three-axis indicator at the lower left has a blue upward arrow labeled c, a green rightward arrow labeled b, and a red arrow pointing down-left labeled a. Apart from a, b, and c, the image contains no text or legend. As specified in the prompt, the large, medium, and small spheres represent Dy, Co, and C, respectively."}, {"file": "Exam-17-T5-5-2-prompt.png", "description": "Projection of crystalline $DyCoC_2$ along the c direction (two-dimensional projection of a ball-and-stick model on a pale gray-white background): thin black lines form a rectangular unit-cell border. Four large blue spheres lie at the corners; 1 slightly smaller blue sphere lies at the center of the rectangle without any bonds; and 3 medium blue spheres lie at the midpoints of the left and right edges and at the center of the lower edge. There are 10 small brown spheres: 2 terminal spheres outside the rectangle at the upper left and upper right, while the remaining 8 form 4 horizontal dumbbell pairs joined by two parallel short brown sticks. These pairs lie at the upper center, middle left, middle right, and outside the rectangle at the lower center. Each medium blue sphere is connected to 4 brown spheres by two-colored blue-brown short sticks, and adjacent blue spheres and brown pairs form triangular units. A three-axis indicator at the lower left has a red rightward arrow labeled a, a green upward arrow labeled b, and a blue circle bearing the letter c, indicating that the c axis points out of the page toward the viewer. Apart from a, b, and c, the image contains no text or legend. As specified in the prompt, the large, medium, and small spheres represent Dy, Co, and C, respectively."}], "full_mark": 1}, {"sub_id": "Exam-17-T5-6", "tags": ["1.6"], "text_open": "5.5.3 Briefly explain why the experimentally measured carbon-carbon bond length in crystalline $DyCoC_2$ is 138 pm.", "text_images": [], "full_mark": 2}]} | |
| {"problem_id": "Exam-17-T6", "header": "Problem 6 Synthesis of Triphenylmethanol Using a Grignard Reagent\nGrignard reagents are common nucleophiles in organic synthesis. They must be used under anhydrous conditions free of oxygen and $CO_2$, and reactions are usually performed under nitrogen. In a teaching laboratory, however, the synthesis can be carried out with a simpler procedure. In this problem, bromobenzene is used to prepare phenylmagnesium bromide, which is then reacted with methyl benzoate to produce triphenylmethanol.", "header_images": [{"file": "Exam-17-T6-header-1.png", "description": "Experimental apparatus: a three-neck flask is centered, with its left neck closed by a ground-glass stopper. The middle neck holds a funnel-shaped dropping device with a sidearm and stopcock, labeled \"Apparatus 1\". The angled right neck is connected to a spherical condenser sloping upward to the right. Liquid is present in the lower part of the three-neck flask. The labels in the image are Apparatus 1, spherical condenser, three-neck flask, and to drying tube at the top of the condenser."}], "subquestions": [{"sub_id": "Exam-17-T6-1", "tags": ["2.8"], "text_open": "6.2 Synthesis of triphenylmethanol: cool the flask in a cold-water bath while slowly adding a solution of methyl benzoate in anhydrous diethyl ether with stirring. After addition is complete, heat the mixture under reflux for 1 h, cool the flask again in a cold-water bath, and add saturated ammonium chloride solution dropwise to decompose the addition product. Separate the layers and perform steam distillation to obtain the crude product.\n6.2.1 State the purpose of the saturated ammonium chloride workup, and explain why water or hydrochloric acid is not used directly for the workup.", "text_images": [], "full_mark": 3}]} | |
| {"problem_id": "Exam-17-T8", "header": "Problem 8 Fundamentals of Organic Chemistry", "header_images": [], "subquestions": [{"sub_id": "Exam-17-T8-1", "tags": ["1.2"], "text_open": "8.2 Examine the reaction below and answer the question.\n8.2.1 Determine the configuration of the chiral carbon marked with * in the product, using S/R notation.", "text_images": [{"file": "Exam-17-T8-1-1-prompt.png", "description": "'O=C1C[C@H](O[Si](C)(C)C(C)(C)C)C=C1S(=O)(C2=CC=CC=C2)=O'>Me3Al (2.2 equiv.), Toluene, -78 ℃, 30 min>'O=C1C[C@H](O[Si](C)(C)C(C)(C)C)C(C)C1S(=O)(C2=CC=CC=C2)=O'. The chiral carbon marked with * in the product is the ring carbon bonded to the methyl group."}], "full_mark": 2}, {"sub_id": "Exam-17-T8-2", "tags": ["2.8"], "text_open": "8.2.3 When the Lewis acid is changed from $Me_3Al$ to $Et_3Al$ or $^iBu_3Al$, the yield decreases from 86% to 65% and 32%, respectively. Explain the decrease in yield.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-17-T8-3", "tags": ["1.2"], "text_open": "8.3 Consider the two conformations of $FCH_2C(=O)R$ shown below. They interconvert by rotation about the $C_\\alpha$-C(=O) single bond. The relative energy of the conformation on the right is defined as 0.0 kcal/mol.\n8.3.1 Define the energy of the left conformation relative to the right conformation as $\\Delta E = E_{\\text{left conformation}} - E_{\\text{right conformation}}$. Determine the sign of $\\Delta E$.", "text_images": [{"file": "Exam-17-T8-3-1-prompt.png", "description": "The molecular SMILES is '[R]C([CH2]F)=O'. The image shows two conformations connected by a reversible arrow. In the left conformation, F points downward in the plane of the page, COR points upward and to the right in the plane, and the two H atoms project in front of and behind the upper-left region. In the right conformation, F points upward and to the left in the plane, COR points upward and to the right in the plane, and the two H atoms project in front of and behind the lower region. The right conformation is labeled 0.0 kcal/mol; R = NH2, OMe, Me, H."}], "full_mark": 1}, {"sub_id": "Exam-17-T8-4", "tags": ["1.2", "1.6"], "text_open": "8.3.2 For $R = NH_2$, OMe, Me, and H, denote the corresponding values of $\\Delta E$ by A, B, C, and D, respectively. Arrange A, B, C, and D in increasing numerical order.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-17-T8-5", "tags": ["1.2"], "text_open": "8.4 The diagram below shows two conformations, A and B, of a difluoroalkane. Assume that they interconvert only by rotation about a carbon-carbon σ bond, without any change in configuration. Ignore intermolecular interactions and solvent effects, and compare only their intramolecular conformational energies. Determine which of A and B has the lower energy and state the principal reason.", "text_images": [{"file": "Exam-17-T8-5-1-prompt.png", "description": "Both A and B have the SMILES 'C[C@@H](F)C[C@H](C)F'. They are two conformations of the difluoroalkane and are connected by a reversible arrow. In A, the F at C2 points upward on a solid wedge, the F at C4 points downward and to the right on an ordinary single bond, and the methyl group at C4 points upward on a dashed wedge. In B, the F atoms at both C2 and C4 point upward on solid wedges, and the methyl group at C4 points downward and to the right on an ordinary single bond."}], "full_mark": 3}]} | |
| {"problem_id": "Exam-18-T1", "header": "Problem 1 Structural Chemistry of Aluminosilicates\nThe fundamental structural unit of silicates is the $[SiO_4]$ tetrahedron. Sharing corner oxygen atoms between silicate tetrahedra can form finite polynuclear anions, one-dimensional chains, two-dimensional layers, and three-dimensional infinite frameworks. Zeolite molecular sieves are aluminosilicates with regular channels and cage structures; their frameworks can be regarded as three-dimensional networks of corner-sharing $[SiO_4]$ and $[AlO_4]$ tetrahedra.", "header_images": [], "subquestions": [{"sub_id": "Exam-18-T1-1", "tags": ["1.3"], "text_open": "1.1.4 Describe the long-chain silicate structure with general formula $[Si_4O_{11}]_n^{6n-}$, given that it contains 12-membered rings formed by alternating Si and O atoms.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-18-T1-2", "tags": ["1.3"], "text_open": "1.2 Sodalite belongs to the cubic crystal system. In its conventional unit cell, a sodalite cage (β cage) occupies each vertex and the body center, and the cages are connected by sharing hexagonal faces. In addition, each β cage is connected to other β cages by sharing square faces. Each β cage contains four sodium ions arranged as a regular tetrahedron, with a chloride ion at the center of the tetrahedron; each element has only one crystallographically distinct environment.\n1.2.4 In the β cage of sodalite, the number of Al atoms is the maximum allowed by Lowenstein's rule. State the lattice type of sodalite.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-18-T1-3", "tags": ["1.3"], "text_open": "1.3 In Type A zeolite (LTA), a larger α cage forms at the lattice center. An isolated α cage belongs to the Oh point group and has polyhedral notation [4126886]. The LTA framework may be regarded as a three-dimensional infinite structure in which α cages are connected by sharing regular octagonal faces in a “primitive cubic packing” arrangement. 1.3.1 If each LTA unit cell is regarded as containing 1 α cage, state the number of T atoms in one unit cell of Type A zeolite.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-18-T1-4", "tags": ["1.3", "2.1"], "text_open": "1.4 In the laboratory, 4Å molecular sieve A can be synthesized hydrothermally from sodium silicate, sodium aluminate, and sodium hydroxide.\n1.4.3 Write the overall equation for the synthesis of molecular sieve A.", "text_images": [], "full_mark": 2}]} | |
| {"problem_id": "Exam-18-T2", "header": "Problem 2 Fluoride-Ion-Selective Electrode\nThe core sensing element of a fluoride-ion-selective electrode is a LaF$_3$ single-crystal membrane doped with a small amount of Eu$^{2+}$ and sealed into the end of the electrode tube. The tube contains an internal reference solution, generally with fixed concentrations of NaF and NaCl, and an internal Ag/AgCl reference electrode. During doping, Eu$^{2+}$ replaces some La$^{3+}$ lattice sites, creating fluoride-ion vacancies to maintain electroneutrality. When the electrode is immersed in the sample solution, F$^{-}$ undergoes selective exchange at the membrane-solution interface and migrates through the crystal via these vacancies. Because the activities of F$^{-}$ differ across the membrane, the phase-boundary potentials on the two sides also differ; their difference is the membrane potential. Measuring this potential permits rapid and selective determination of fluoride in a sample. The structure is shown below.", "header_images": [{"file": "Exam-18-T2-header-1.png", "description": "Schematic of the apparatus and electrode structures for determining F⁻ in water with a fluoride-ion-selective electrode, comprising panels A, B, and C. Panel A, measurement apparatus: the negative and positive terminals of a high-input-impedance millivoltmeter (mV) are connected by wires to a fluoride-ion-selective electrode and a type 232 saturated calomel electrode with a single salt bridge, respectively. Both electrodes are immersed in liquid consisting of water sample + TISAB at pH≈5.5 in a plastic beaker. A magnetic stir bar with a rotation arrow is at the bottom, and the beaker is placed on a magnetic stirrer. Panel B, enlarged fluoride-ion-selective electrode: a top wire connects through an internal Ag/AgCl reference electrode into the internal reference solution of constant F⁻ activity inside the electrode body. A LaF₃ single-crystal sensing membrane doped with EuF₂ is at the bottom, with F⁻ in the sample solution outside the membrane. Panel C, enlarged type 232 saturated calomel electrode: a wire connects through a platinum wire in contact with Hg to Hg. A filling port, opened during measurement, is beside the platinum wire. Below the Hg are, in sequence, a Hg+Hg₂Cl₂ paste layer, saturated KCl filling solution, and KCl crystals. A porous ceramic liquid junction at the bottom extends into the sample solution. The electrode notation Pt|Hg(l)|Hg₂Cl₂(s)|KCl(saturated)‖sample solution appears at the bottom, with the note “type 232 has a single salt bridge; type 217 has a double salt bridge.”"}], "subquestions": [{"sub_id": "Exam-18-T2-1", "tags": ["3.6"], "text_open": "2.2 For the doped LaF$_3$ single-crystal membrane, the membrane potential is described by the following equation, where $a$ denotes activity and may be approximated by concentration at low F$^{-}$ concentration:\n$$E_{membrane}=\\frac{\\mu_{outside}^{0}-\\mu_{inside}^{0}}{nF}+\\frac{RT}{nF}\\ln\\frac{a_{outside}}{a_{inside}}=k+\\frac{RT}{nF}\\ln\\frac{a_{outside}}{a_{inside}}$$\n2.2.1 Calculate how the millivoltmeter reading changes when the concentration of the ion being measured decreases by a factor of 10.", "text_images": [], "full_mark": 2}]} | |
| {"problem_id": "Exam-18-T6", "header": "Problem 6 Low-Valent Compounds of Group 13 Metals · II\nLow-valent compounds of Group 13 elements such as aluminum, gallium, and indium (in the +1 oxidation state) have attracted widespread interest in synthetic chemistry and materials science because of their distinctive electronic structures and reactivities. These species often exist as oligomers, clusters, or coordination complexes. Their structural diversity arises from weak interactions between metal centers and their sensitivity to ligands. This problem focuses on the synthesis and structures of low-valent gallium and indium compounds and their coordination chemistry with Lewis acids, testing your understanding of cluster electron counting, aromaticity criteria, and coordination geometries.", "header_images": [], "subquestions": [{"sub_id": "Exam-18-T6-1", "tags": ["1.4", "1.6"], "text_open": "Part I: Core Frameworks of Gallium Clusters—Incomplete Cubes and Four-Membered Rings\n6.1 An electron-precise tetragallane anion: Ultrasonically prepared “GaI” reacts with $Li(THF)_3Si(SiMe_3)_3$ in toluene at -78°C. Warming affords compound $A = [Li(THF)_4]^+[R_4Ga_4I_3]^-$ ($R = Si(SiMe_3)_3$). The anion of A has an “incomplete-cube” framework: four Ga atoms and three I atoms occupy four and three vertices of a cube, respectively, forming a $Ga_4I_3$ core. Write the SMILES string for the anion of A. Each framework bond is known to be a 2c–2e bond, and every Ga atom obeys the octet rule.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-18-T6-2", "tags": ["1.4"], "text_open": "6.2 Cyclic tetragallane anions: planar and butterfly conformations\n6.2.1 $ArGaCl_2$ ($Ar$ = 2,6-bis(2,4,6-triisopropylphenyl)phenyl) reacts with excess K in diethyl ether to form $B = K_2[Ga_4Ar_2]$. The $Ga_4$ unit in the anion of B is an almost square four-membered ring. Each $K^+$ coordinates to the $Ga_4$ ring and two benzene rings. Write the SMILES string for B.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-18-T6-3", "tags": ["1.5", "1.7"], "text_open": "6.2.2 $RGaCl_2\\cdot THF$ ($R = Si(t-Bu)_3$) reacts with excess Na (or K) in heptane to form $C = Na_2Ga_4R_4(THF)_2$. The $Ga_4$ core of C adopts a butterfly (nonplanar) conformation, and the Ga–Ga bond lengths are nearly equal (average 2.43 Å) and appreciably shorter than those in the tetrahedrane $Ga_4R_4$ (2.57 Å). Two $Na^+(THF)$ units lie on opposite sides of the ring. Using the Hückel rule (Hückel aromaticity when the number of delocalized electrons $=4n+2$), determine whether the anion of C is aromatic and explain why it is nonplanar, in contrast to planar B.", "text_images": [{"file": "Exam-18-T6-3-1-prompt.png", "description": "The image shows the single-step reaction 'Cl[Ga-](Cl)([O+]1CCCC1)[Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C'>Na (ex.), n-C7H16, 100 °C>'CC([Si](C(C)(C)C)(C(C)(C)C)[Ga]12([Ga@]34[Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)[Ga@@]([Ga]35([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)[Na@@]14C6CCCO6)([Na@@]25C7CCCO7)[Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)(C)C'. In the reactant $RGaCl_2\\cdot THF$ ($R = Si(t-Bu)_3$), THF is connected to Ga by a coordinate bond, represented in the SMILES by charge separation as [O+]/[Ga-]. In the product $C = Na_2Ga_4R_4(THF)_2$, the Ga4 ring adopts a butterfly conformation (the @/@@ stereochemical markers are retained in the SMILES), and the two Na+(THF) units lie on opposite sides of the ring."}], "full_mark": 4}, {"sub_id": "Exam-18-T6-4", "tags": ["1.4", "2.4"], "text_open": "6.4 Propeller-shaped molecule: Lithium amide $LiN(SiMe_3)Dipp$ (2.2 eq) reacts with GaCl (1 eq) in toluene/diethyl ether (3:1) to give the cluster $D = [Li(Et_2O)_4][Ga_8(R)_2(R')_4(R'')_2OLi]$, where $R = N(SiMe_3)Dipp$, $R'$ = NHDipp, and $R''$ = NDipp. In the anion, the two central Ga atoms are bonded only to other Ga atoms. Each of the six peripheral Ga atoms bears one terminal N ligand and is also bonded to a bridging N or O atom. Write the SMILES string for the anion of D; $Li^+$ and solvent molecules need not be shown. The authors propose that this cluster is obtained by substitution of the subhalide shown below, with the principal framework retained during substitution.", "text_images": [{"file": "Exam-18-T6-4-1-prompt.png", "description": "The image shows a single-step reaction with the reactant 'Cl[Ga]12Cl[Ga](Cl)(Cl2)[Ga]3([Ga](Cl)(Cl4)Cl[Ga]54Cl)[Ga]51[Ga]6(Cl)Cl[Ga]3(Cl)Cl6', followed by >LiN(SiMe3)Dipp (2.2 eq), PhMe/Et2O>[D]. The SMILES representation of LiN(SiMe3)Dipp is '[Li]N([Si](C)(C)C)C1=C(C(C)C)C=CC=C1C(C)C'. The reactant on the left is a gallium subhalide cluster containing 8 Ga and 12 Cl atoms: 6 terminal Cl atoms, 4 bridging Cl atoms, and 2 Cl atoms shown as short dashed bridges. The lithium amide $LiN(SiMe_3)Dipp$ and (2.2 eq) are written above the arrow, and PhMe/Et2O below it. The product on the right is identified only by the formula $D = [Li(Et_2O)_4][Ga_8(N(SiMe_3)Dipp)_2(NHDipp)_4(NDipp)_2OLi]$; its structure is not drawn."}], "full_mark": 2}, {"sub_id": "Exam-18-T6-5", "tags": ["1.4", "2.4"], "text_open": "Part III: E→E Coordinate Bonds. Some E–E bonds can be viewed as coordinate bonds formed when a lone pair on a low-valent metal(I) center coordinates to a trivalent metal(III) Lewis acid. Special ligands are required to obtain stable monomeric low-valent metal(I) species. The following questions examine two ligand systems—a tridentate borate ligand and crown ethers—and their stabilization of monovalent gallium and indium.\n6.5 Metastable “GaI” reacts with $[Tp(t-Bu)_2]Na$ to give the monomer $E = [Tp(t-Bu)_2]Ga$. E can react further with $GaI_3$ to form the adduct $F = [Tp(t-Bu)_2]Ga\\rightarrow GaI_3$.\n6.5.1 Write the SMILES strings for E and F.", "text_images": [{"file": "Exam-18-T6-5-1-prompt.png", "description": "The image shows the two-step reaction '[H]B1(N2[N]3=C(C(C)(C)C)C=C2C(C)(C)C)N4[N]([Na]3[N]5=C(C(C)(C)C)C=C(C(C)(C)C)N51)=C(C(C)(C)C)C=C4C(C)(C)C'>GaI>[E]; [E]>GaI3, NEt3>[F]'. On the left is the structure of $[Tp(t-Bu)_2]Na$: B is bonded to one H and one N atom from each of three pyrazole rings. Each five-membered pyrazole ring contains two adjacent N atoms, one C=C bond and one C=N bond within the ring, and a t-Bu group on each of two ring carbon atoms. The other N atom of each ring coordinates to the central Na, with two of the N–Na bonds shown as dashed lines. E and F are indicated only by their letter labels; their structures are not drawn."}], "full_mark": 3}, {"sub_id": "Exam-18-T6-6", "tags": ["1.4", "1.1"], "text_open": "6.6 In addition to $[Tp(t-Bu)_2]$, crown ethers such as 18-crown-6 can also stabilize monovalent aluminum-group metals.\n6.6.1 Soluble $InOTf$ reacts with 18-crown-6 in toluene to give adduct G. Write the SMILES string for G and indicate the coordination geometry of In.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-18-T6-7", "tags": ["2.4", "1.7"], "text_open": "6.6.2 Unlike the parent $InOTf$, G rapidly undergoes oxidative addition with $CHCl_3$ to form the $In(III)$ compound H, whereas the parent $InOTf$ does not react under the same conditions. Write the SMILES string for H and explain why G is more reactive than $InOTf$.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-18-T6-8", "tags": ["2.4", "1.4"], "text_open": "6.6.3 Under different conditions, $[In_2Cl_4]$ can form valence isomers:\nIn toluene, it exists as the ion pair J, in which one of the cation and anion has a regular tetrahedral geometry.\nIn THF, it exists as the symmetric molecule K containing an In–In bond, with each In atom five-coordinate.\nIn the presence of dibenzo-18-crown-6, it exists as the $In(I)\\rightarrow In(III)$ donor–acceptor adduct L.\nDibenzo-18-crown-6:\nWrite the SMILES strings for J, K, and L, and indicate the oxidation state of each In atom.", "text_images": [{"file": "Exam-18-T6-8-1-prompt.png", "description": "The image shows the structure of dibenzo-18-crown-6, whose SMILES representation is 'C12=CC=CC=C1OCCOCCOC(C=CC=C3)=C3OCCOCCO2'. The 6 O atoms and 8 methylene bridges alternate within an 18-membered ring. Each of the 2 benzene rings, drawn in Kekulé form with alternating double bonds, is fused into the crown ring through 2 adjacent carbon atoms. All bonds other than those within the benzene rings are single bonds."}], "full_mark": 9}]} | |
| {"problem_id": "Exam-18-T7", "header": "Problem 7 Rearrangement Reactions", "header_images": [], "subquestions": [{"sub_id": "Exam-18-T7-1", "tags": ["2.2", "2.3"], "text_open": "7.1 The following reaction involves a pericyclic reaction. Write the SMILES strings for 3 key intermediates in the reaction pathway.", "text_images": [{"file": "Exam-18-T7-1-1-prompt.png", "description": "The figure shows a one-step reaction: 'ClC(/C(C)=C/CC)=O'.'CC1=CCCC1'>AlCl3>'O=C1[C@@H](C)[C@@](CC)(C)C2=C1CCC2'. In the product, two carbon atoms bear a methyl and an ethyl group, respectively, on wedged bonds. The @/@@ stereochemical descriptors are retained in the SMILES."}], "full_mark": 6}, {"sub_id": "Exam-18-T7-2", "tags": ["2.2", "2.4"], "text_open": "7.2 In different solvents, a carbene can give different rearrangement products. Under irradiation, the following diazo compound forms a carbene and rearranges to ketene A.\n7.2.1 Write the SMILES string for A.", "text_images": [{"file": "Exam-18-T7-2-1-prompt.png", "description": "The figure shows a one-step reaction: 'O=C1OC2=CC=CC=C2C1=[N+]=[N-]'>hν>[A]. The product is labeled only as [A], and its structure is not shown."}], "full_mark": 2}, {"sub_id": "Exam-18-T7-3", "tags": ["2.4"], "text_open": "7.2.2 In an aqueous $10^{-3}$ mol/L $HClO_4$ solution, A undergoes ring closure to form B. Write the SMILES string for B.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-18-T7-4", "tags": ["2.4", "1.2"], "text_open": "7.2.3 In hexane, A undergoes dimerization to give several products. Write the SMILES strings for C, which contains 4 six-membered rings, and D, which contains 2 five-membered rings and 2 six-membered rings. Both products are centrosymmetric.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-18-T7-5", "tags": ["2.2", "2.4"], "text_open": "7.3 Purpurogallin is a natural pigment formed by oxidation of pyrogallol, followed by dimerization and loss of a small molecule.\n7.3.1 Write the SMILES strings for 2 key intermediates in the second transformation.", "text_images": [{"file": "Exam-18-T7-5-1-prompt.png", "description": "The figure shows a three-step reaction: 'OC1=C(O)C(O)=CC=C1'>[O]>'O=C(C=CC=C1O)C1=O'; 'O=C(C=CC=C1O)C1=O'>H2O>'OC1=CC(CC=CC(O)=C2O)=C2C(O)=C1O'; 'OC1=CC(CC=CC(O)=C2O)=C2C(O)=C1O'>[O]>Purpurogallin. The final product is labeled only by name, and its structure is not shown."}], "full_mark": 2}, {"sub_id": "Exam-18-T7-6", "tags": ["2.4", "1.7"], "text_open": "7.3.2 Purpurogallin is known to be an aromatic compound. Write its SMILES string.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-18-T7-7", "tags": ["2.2", "2.7", "2.3"], "text_open": "7.4 Ozonolysis of allylic alcohols shows unusual selectivity.\n7.4.1 Write the SMILES strings for 2 key intermediates in the first step and 1 key intermediate in the second step.", "text_images": [{"file": "Exam-18-T7-7-1-prompt.png", "description": "The figure shows a two-step reaction: 'OC1(C)C=CC(C)(C)CC1'>O3>'CC1(C(O)OO2)OC2C(C)(C)CC1'; 'CC1(C(O)OO2)OC2C(C)(C)CC1'>NaOH>'O=C1C=CC(C)(C)CC1'."}], "full_mark": 6}, {"sub_id": "Exam-18-T7-8", "tags": ["2.4", "2.5", "2.7"], "text_open": "7.4.2 Write the SMILES string for the product of the following reaction. It contains no cyclic structure.", "text_images": [{"file": "Exam-18-T7-8-1-prompt.png", "description": "The figure shows a one-step reaction: 'C/C(C)=C1[C@@](C)(C[C@H](C)CC/1)O'>1) O3, 2) NaOH>[product not shown]. In the starting material, OH is shown on a hashed wedge, one methyl group on a solid wedge, and the opposite methyl group on a dashed bond. The @/@@ stereochemical descriptors are retained in the SMILES."}], "full_mark": 2}]} | |
| {"problem_id": "Exam-18-T8", "header": "Problem 8 Reactions of Amino-Acid Side Chains\nAmino-acid side chains can undergo chemical reactions when they contain functional groups. These groups include hydroxy, phenolic, thiol, indolyl, imidazolyl, guanidino, methylthio, non-α−amino, and non-α−carboxy groups. Each functional group can react with many reagents, and some of these reactions form the basis of chemical modification of proteins. Stereochemistry is not considered in this problem.", "header_images": [], "subquestions": [{"sub_id": "Exam-18-T8-1", "tags": ["2.4", "2.5"], "text_open": "8.2 Toi reaction: In sodium borate buffer at pH 8~9 and 25~35℃, the guanidino side chain of arginine reacts with 1,2−cyclohexanedione to form condensation product E$^+$. E$^+$ contains a newly formed ring. Its chemical formula is $C_{12}H_{23}N_4O_4^+$, and if only hydrogens attached to carbon are considered, there are 6 distinct chemical environments. Write the SMILES string for E$^+$.", "text_images": [{"file": "Exam-18-T8-1-1-prompt.png", "description": "'O=C(O)C(N)CCCNC(N)=N'.'O=C1CCCCC1=O'>>[E+]. No conditions are written above the arrow, and the product structure is not drawn but is labeled E$^+$. The arginine structure is labeled Arginine."}], "full_mark": 2}, {"sub_id": "Exam-18-T8-2", "tags": ["2.2", "2.5"], "text_open": "8.3 Patchornik reaction: Under mild conditions, the indole ring of a tryptophan residue in a protein can be oxidized by NCS and undergo rearrangement, causing cleavage of the peptide bond. Write the SMILES strings for the two key intermediates between the tryptophan residue and intermediate X.", "text_images": [{"file": "Exam-18-T8-2-1-prompt.png", "description": "The figure shows a two-step reaction: 'O=C(N[R2])C(N[R1])CC1=CNC2=CC=CC=C21'>NCS, H2O>'O=C1C(N[R1])CC2(O1)C3=CC=CC=C3NC2Cl'; 'O=C1C(N[R1])CC2(O1)C3=CC=CC=C3NC2Cl'>NCS, H2O>'O=C1C(N[R1])CC2(O1)C3=CC=CC=C3NC2=O'. The intermediate structure containing Cl is labeled X, while the product on the right has no letter label. R1 and R2 are the original placeholders in the figure."}], "full_mark": 4}, {"sub_id": "Exam-18-T8-3", "tags": ["2.2", "2.5"], "text_open": "8.4.1 Hopkins-Cole reaction: A protein solution is mixed with Hopkins-Cole reagent, whose principal active component is glyoxylic acid. Concentrated sulfuric acid is then carefully added down the wall of the tube so that it forms a separate lower layer. If the protein contains tryptophan, a violet ring appears at the interface between the two liquid layers. However, this reaction is not simple or reliable for quantitative determination of tryptophan because product formation is slow, various color changes occur concurrently, and the products have low stability. Studies have shown that tryptophan can form two differently colored products with glyoxylic acid. When a small amount of aldehyde is used and the reaction temperature is below 15°C, the first colored product is carmine. Write the SMILES strings for three key intermediates between the tryptophan residues and intermediate Y.", "text_images": [{"file": "Exam-18-T8-3-1-prompt.png", "description": "The figure shows a two-step reaction: '[R]C1=CNC2=CC=CC=C21'.'[R]C1=CNC2=CC=CC=C21'.'O=C([H])C(O)=O'>'[R]C1=CNC2=CC=C(C(C(O)=O)C3=CC=C(NC=C4[R])C4=C3)C=C21'; '[R]C1=CNC2=CC=C(C(C(O)=O)C3=CC=C(NC=C4[R])C4=C3)C=C21'>[O], -CO2, -H2O>'[R]C1=CNC2=CC=C(/C=C(C=C3)/C=C4C3=NC=C4[R])C=C21'. The intermediate structure after the first arrow is labeled Y. R is the original placeholder in the figure."}], "full_mark": 6}, {"sub_id": "Exam-18-T8-4", "tags": ["2.2", "2.5"], "text_open": "8.4.2 Raising the temperature to approximately 50°C provides sufficient activation energy to promote a further condensation reaction, forming a second blue-violet product. Write the SMILES strings for two key intermediates between the reactant in the scheme and intermediate Z.", "text_images": [{"file": "Exam-18-T8-4-1-prompt.png", "description": "The figure shows a two-step reaction: 'NC(C(O)=O)CC1=CN=C2C1=C/C(C=C2)=C/C3=CC=C(NC=C4CC(N)C(O)=O)C4=C3'>glyoxylic acid, -H2O>'O=C(C(NC1C(O)=O)CC2=C1N=C3C2=C/C(C=C3)=C/C4=CC=C(NC5=C6CC(NC5C(O)=O)C(O)=O)C6=C4)O'; 'O=C(C(NC1C(O)=O)CC2=C1N=C3C2=C/C(C=C3)=C/C4=CC=C(NC5=C6CC(NC5C(O)=O)C(O)=O)C6=C4)O'>-CO2>'O=C(C(NC1)CC2=C1N=C3C2=C/C(C=C3)=C/C4=CC=C(NC5=C6CC(NC5)C(O)=O)C6=C4)O'. The upper-right structure is labeled Z, and the structure of glyoxylic acid is shown above the first arrow."}], "full_mark": 4}]} | |
| {"problem_id": "Exam-19-T4", "header": "Question 4 Determination of Iron by Titanium(III) Chloride–Potassium Dichromate Titration\nFerromanganese-silicon alloy is widely used as a deoxidizer and alloying additive in steelmaking, and its iron content strongly affects its quality. Potassium dichromate titration is a classical method for determining iron. After acid decomposition of the sample, titanium(III) chloride reduces $Fe^{3+}$ to $Fe^{2+}$; the solution is then titrated with standard potassium dichromate using sodium diphenylamine sulfonate as the indicator. This problem covers sample decomposition, prereduction, removal of interferences, and blank correction.", "header_images": [], "subquestions": [{"sub_id": "Exam-19-T4-1", "tags": ["4.1"], "text_open": "4.1 Sample decomposition\nPlace a 0.5026 g ferromanganese-silicon alloy sample in a platinum dish. Add 10 mL of concentrated nitric acid, then add approximately 2 mL of concentrated hydrofluoric acid dropwise in portions. Heat gently until the sample is completely decomposed. Add 5 mL of concentrated perchloric acid and heat until faint perchloric acid fumes appear, then remove from heat. Transfer the solution to a 250 mL conical flask and continue heating until nearly dry (syrupy), then remove from heat. Add 20 mL of hydrochloric acid (1+1), 1–2 drops of hydrogen peroxide (27%), and approximately 40 mL of water. Dissolve the salts until the test solution is clear, heat to boiling until large bubbles evolve, remove from heat, and cool to room temperature. Hint: high-valent manganese oxides are strong oxidants under acidic conditions and, being insoluble in water, adsorb iron ions and interfere with the determination of iron.\n4.1.2 State the purpose of fuming with perchloric acid and of subsequently continuing to heat the solution until nearly dry.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-19-T4-2", "tags": ["2.1", "4.1"], "text_open": "4.2 Reduction and titration\nDilute the test solution with water to approximately 120 mL and add four drops of sodium tungstate solution (250 g/L). Add titanium(III) chloride solution (1.5%–2%) dropwise until the solution turns blue, then add a further 1–2 drops. With continuous swirling, titrate with potassium dichromate solution (1.250 mmol/L) until the solution becomes colorless, adding slowly near the endpoint; do not record this reading. Immediately add 10 mL of sulfuric acid–phosphoric acid mixture (sulfuric acid:phosphoric acid:water = 2:3:5), then add 3 drops of sodium diphenylamine sulfonate indicator solution (2 g/L). Titrate with standard potassium dichromate solution (2.500 mmol/L) to a violet-blue endpoint and record the consumed volume C.\n4.2.2 Write the reaction equations for the addition of titanium(III) chloride solution and explain why a further 1–2 drops of titanium(III) chloride must be added.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-19-T4-3", "tags": ["4.1", "4.2"], "text_open": "4.2.4 Which statements about the functions of adding the sulfuric acid–phosphoric acid mixture are correct?\nA. It provides the acidity required for titration and prevents hydrolysis of metal ions.\nB. Phosphoric acid forms a colorless complex with iron ions, eliminating interference from their yellow color.\nC. It lowers the potential of the $Fe^{3+}$/$Fe^{2+}$ redox couple, making the titration jump more pronounced.\nD. It decolorizes sodium tungstate, preventing interference with the subsequent titration.", "text_images": [], "full_mark": 3}]} | |
| {"problem_id": "Exam-19-T8", "header": "Question 8 Remarkable Phosphorus Heterocycles\nIntroductory heterocyclic chemistry most often focuses on N, O, and S heteroatoms. Phosphorus, however, has a large valence-shell space and a rich range of coordination numbers and oxidation states. It can form low-coordinate λ species as well as hypercoordinate $\\lambda^5$ species, where λ denotes the number of bonds to P. Phosphorus heterocycles exhibit remarkable structures and properties and are uniquely valuable in studies of low-valent metal ligands and tunable fluorophores. (Stereochemistry need not be considered in this problem.)", "header_images": [], "subquestions": [{"sub_id": "Exam-19-T8-1", "tags": ["1.6", "4.1"], "text_open": "8.1.2 Which of the following measures are commonly used to evaluate the aromaticity of a compound?\n(a) C–X–C bond angle (b) Degree of bond-length equalization (c) Nucleus-independent chemical shift calculation (d) Standard molar enthalpy of formation", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-19-T8-2", "tags": ["2.4"], "text_open": "8.1.4 P–C bonds are highly reactive. Write the SMILES string for product A from the following 1:1 addition reaction using the specified SMILES.", "text_images": [{"file": "Exam-19-T8-2-1-prompt.png", "description": "'C1(C2=CC=CC=C2)=CC(C3=CC=CC=C3)=CC(C4=NC=CC=C4)=P1'.'FC(C#CC(F)(F)F)(F)F'>100°C>[A]. The structure of product A is not drawn; it is labeled only as A."}], "full_mark": 2}, {"sub_id": "Exam-19-T8-3", "tags": ["2.2", "2.4"], "text_open": "8.1.5 The following reaction ultimately gives product B in which the phosphabenzene ring has been disrupted. Write the SMILES strings for Int and B. B has no mirror plane.", "text_images": [{"file": "Exam-19-T8-3-1-prompt.png", "description": "The figure shows a two-step reaction: 'CC(C=C1)=CC=C1C2=CC(C3=CC=CC=C3)=PC(C4=CC=CC=C4)=C2'>PhLi, THF, -78°C to r.t.>[Int]; [Int]>H2O, -LiOH>[B]. The structures of Int and B are not drawn; they are labeled only as Int and B."}], "full_mark": 3}, {"sub_id": "Exam-19-T8-4", "tags": ["2.2", "2.4"], "text_open": "8.2.1 Several routes are available for synthesizing $\\lambda^3$-phosphabenzenes and their derivatives. Early chemists began with phosphole and used relatively harsh conditions. Write the SMILES strings for intermediates C and D in the following synthesis.", "text_images": [{"file": "Exam-19-T8-4-1-prompt.png", "description": "The figure shows a three-step reaction: 'C1(C2=CC=CC=C2)=CC=C(C3=CC=CC=C3)P1C4=CC=CC=C4'>230°C, [1,5]-σ>[C]; [C].'[R]C#C[R]'>>[D]; [D]>230°C, -CPh2>'[R]C1=PC(C2=CC=CC=C2)=CC=C1[R]'. The structures of intermediates C and D are not drawn; they are labeled only as C and D."}], "full_mark": 3}, {"sub_id": "Exam-19-T8-5", "tags": ["2.2", "2.4"], "text_open": "8.2.2 Milder synthetic routes were later developed, beginning with preparation of $\\lambda^5$ phosphorus intermediate a.\n8.2.2.1 Given that F is an ionic compound, write the SMILES strings for intermediates E and F and identify, in order, the types of reactions occurring from F to a.", "text_images": [{"file": "Exam-19-T8-5-1-prompt.png", "description": "The figure shows four steps: 'ClP(/C=C/OCC)N(C)C'>1) 'C=CC[Mg]Cl', THF, -35°C; 2) MeOH, r.t.>[E]; [E]>1) H2O2·CO(NH2)2; 2) MeOTf, DCM, -70°C>[F]; [F]>tBuOK, -70°C>[a]; [a]>[H]>'C1=CP=CC=C1'. The structures of E and F are not drawn and are labeled only as E and F; a has structure 'COP1(OC)=CC=CC=C1'."}], "full_mark": 6}, {"sub_id": "Exam-19-T8-6", "tags": ["2.2", "2.4"], "text_open": "8.3 Phosphenium ions are often regarded as cationic heavy-element analogs of carbenes, but extending classical carbocation rearrangements such as the Wagner–Meerwein rearrangement directly to phosphorus chemistry remains challenging. Researchers proposed that a carbocation generated adjacent to P in a phosphirane might rearrange to a phosphenium ion, opening a new ring-forming pathway. In this problem, $W(CO)_5$ coordinated to P does not participate directly in the transformations.\n8.3.1 The following reaction gives different major products at different temperatures. Write the SMILES strings for G and H, knowing that neither product contains a C=P double bond.", "text_images": [{"file": "Exam-19-T8-6-1-prompt.png", "description": "The figure shows two parallel reactions of the same reactant: 'O=C(C1=CC=CC=C1)[P+]2([W-])CC2.[C-]#[O+].[C-]#[O+].[C-]#[O+].[C-]#[O+].[C-]#[O+]'>AlCl3, toluene, 100°C, 1h>[G]; the same reactant>AlCl3, toluene, 0°C, 1h>[H]. The structures of G and H are not drawn; they are labeled G and H with molecular formulas C14H8ClO5PW and C21H16ClO5PW."}], "full_mark": 4}, {"sub_id": "Exam-19-T8-7", "tags": ["2.4"], "text_open": "8.3.3 Write the SMILES string for the major product of the following reaction, whose mechanism resembles that forming G.", "text_images": [{"file": "Exam-19-T8-7-1-prompt.png", "description": "The figure shows the one-step reaction 'O=C(C1=CC=C(OC)C=C1)[P+]2([W-])C(C)C2.[C-]#[O+].[C-]#[O+].[C-]#[O+].[C-]#[O+].[C-]#[O+]'>AlCl3, toluene, r.t.>product. The product structure is not drawn and no letter label is assigned."}], "full_mark": 2}, {"sub_id": "Exam-19-T8-8", "tags": ["2.4"], "text_open": "8.3.4 To verify the existence of a phosphenium-ion intermediate, the researchers carried out the following reaction. Write the SMILES string for the major product, knowing that it contains four rings.", "text_images": [{"file": "Exam-19-T8-8-1-prompt.png", "description": "The figure shows the one-step reaction 'O=C(C1=C(OC)C=CC=C1)[P+]2([W-])CC2.[C-]#[O+].[C-]#[O+].[C-]#[O+].[C-]#[O+].[C-]#[O+]'>AlCl3, toluene, 60°C, 1h>product. The product structure is not drawn; molecular formula C21H15O6PW is shown."}], "full_mark": 3}]} | |
| {"problem_id": "Exam-19-T9", "header": "Question 9 The Appeal of Self-Immolative Reactions\nFragmentation is an important and elegant class of skeletal-reorganization processes in organic chemistry. Classical fragmentations are usually controlled by specific electronic effects and cleave otherwise inert chemical bonds through concerted electron transfer.", "header_images": [], "subquestions": [{"sub_id": "Exam-19-T9-1", "tags": ["2.2", "2.3"], "text_open": "9.1 Write the SMILES strings for at least 2 key intermediates in the following Eschenmoser–Tanabe fragmentation.", "text_images": [{"file": "Exam-19-T9-1-1-prompt.png", "description": "The figure shows the one-step sequence 'CC1CC2(OC2C(C1)=O)C'>1. TsNHNH2, rt; 2. 55°C, 2h, 50%>'CC(CC(CC#C)C)=O'."}], "full_mark": 4}, {"sub_id": "Exam-19-T9-2", "tags": ["2.4", "2.2"], "text_open": "In recent years, chemists have introduced the concept of fragmentation driven by a cascade of electron transfers into drug delivery and smart sensing, developing self-immolative reactions. Such systems usually consist of a trigger group, a conjugated linker, and a leaving payload. When a specific stimulus activates the trigger, long-range concerted fragmentation begins within the molecule, causing spontaneous disassembly and release of the target molecule.\n9.2 In the following molecule, reduction serves as the specific stimulus. After neutral intermediate A forms, self-immolation occurs immediately. Write the SMILES string for intermediate A.", "text_images": [{"file": "Exam-19-T9-2-1-prompt.png", "description": "The figure shows two steps: 'O=C(NC1=CC=C(CC(NCCN(C)C)=O)C=C1)OCC2=CC=C([N+]([O-])=O)C=C2'>60Co γ-ray, iPrOH/H2O>[A]; [A]>H2O>'O=C(NC1=CC=C(CC(NCCN(C)C)=O)C=C1)OCC2=CC=C(NO)C=C2'.'CO2'.'OCC1=CC=C(NO)C=C1'. Intermediate A is represented only by a bold boxed A and its structure is not shown; CO2 appears as a text label."}], "full_mark": 2}, {"sub_id": "Exam-19-T9-3", "tags": ["2.4", "2.1"], "text_open": "9.3 A slight modification of the substrate in 9.2 allows more complex products to be obtained through a cascade reaction in $H_2O$. The reaction produces 2 inorganic and 2 organic products. Write the SMILES strings for the 2 organic products and the formulas of the 2 inorganic products.", "text_images": [{"file": "Exam-19-T9-3-1-prompt.png", "description": "The figure shows the one-step reaction 'O=C(NC1=CC=C([N+]([O-])=O)C=C1)OCC2=CC=C(B(O)O)C=C2'>H2O2>[products not shown]."}], "full_mark": 6}, {"sub_id": "Exam-19-T9-4", "tags": ["2.4", "1.2"], "text_open": "9.4 Ring formation is fundamental to organic reactions. Certain ring-forming reactions can also serve as specific stimuli that initiate self-immolation.\n9.4.1 In the reaction above, the substrate undergoes Michael addition to give B. Self-immolation of B forms seven-membered-ring compound C. Write the SMILES string for C and the chemical formula of the small inorganic molecule produced.", "text_images": [{"file": "Exam-19-T9-4-1-prompt.png", "description": "The figure shows the one-step reaction 'C=CC(OC1=CC=C(COC(NC2=CC=CC=C2)=S)C=C1)=O'.'SC[C@H](N)C(O)=O'>PBS (PH=7.4, 10mM)>[B]. The cysteine thiol is shown with a dashed wedge to indicate stereochemistry, retained by the @ marker in the SMILES. The product is labeled only B and its structure is not shown."}], "full_mark": 3}, {"sub_id": "Exam-19-T9-5", "tags": ["2.4", "2.3"], "text_open": "9.4.2 Ring formation is not the endpoint; a pericyclic reaction follows. This strategy uses the inverse-electron-demand Diels–Alder reaction (IEDDA) between a trans-cyclooctene derivative (TCO) and a tetrazine derivative to trigger self-immolation. In the reaction below, IEDDA between TCO and tetrazine releases a gas and forms a dihydropyridazine intermediate. After the pericyclic reaction, an N atom of the dihydropyridazine acts as the initiating group for self-immolation, producing 2 aromatic organic compounds. Write the SMILES strings for both products.", "text_images": [{"file": "Exam-19-T9-5-1-prompt.png", "description": "The figure shows the one-step reaction 'CC1=NN=C(C)N=N1'.'O=C(NC1=CC=CC=C1)OC2CCCCC/C=C/2'>>[products not shown]."}], "full_mark": 4}, {"sub_id": "Exam-19-T9-6", "tags": ["2.3", "2.8"], "text_open": "9.4.3 The dihydropyridazine intermediate must undergo a formal σ-migration rearrangement before it can initiate 1,4-elimination, the subsequent self-immolation. Name this rearrangement and explain why it is faster in water than in an organic phase.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-19-T9-7", "tags": ["2.2", "2.4"], "text_open": "9.5 The following substance can undergo a self-immolative reaction through rearrangement.\nWrite the SMILES strings for 1 key rearrangement intermediate containing 2 five-membered rings and the 2 organic molecules produced by self-immolation.", "text_images": [{"file": "Exam-19-T9-7-1-prompt.png", "description": "The figure shows the one-step reaction 'CC(C)(C(C1=CC=CC=C1)O)S(C2=NC3=CC=CC=C3S2)(=O)=O'>10% DMSO/PBS, pH = 7.4, 37°C>[products not shown]."}], "full_mark": 4}, {"sub_id": "Exam-19-T9-8", "tags": ["2.1", "2.4"], "text_open": "9.6 The following polymer readily depolymerizes through self-immolation under acidic conditions. Write the reaction equation.", "text_images": [{"file": "Exam-19-T9-8-1-prompt.png", "description": "The figure shows a cyclic polymer in which three benzene-ring units alternate with O–CH units to form a ring. The visible cyclic skeleton has SMILES 'C1(C2OC3OC(OC4OC(OC5O2)C6=C5C=CC=C6)C7=C4C=CC=C7)=C3C=CC=C1'. A curved bracket marks the repeating portion with subscript n-2, so the full structure is an n-mer of monomer '[*]OC1OC([*])C2=C1C=CC=C2'. H+ is written above the arrow and the product is not shown."}], "full_mark": 3}]} | |
| {"problem_id": "Exam-20-T1", "header": "The Ingenious Green Shoot\nThe heavy element M was first discovered by the British chemist William Crookes (1832-1919) while he was analyzing selenium-containing flue dust sent from a German sulfuric-acid factory. On observing through a spectroscope a bright emerald-green spectral line never seen before, he concluded that he had discovered a new element. Elemental M is silvery white and extremely soft, leaves a dark-gray streak on paper, and rapidly tarnishes when a freshly cut surface is exposed to air. M has two common oxidation states: the lower oxidation state is stable in aqueous solution, whereas compounds in the higher oxidation state readily decompose on heating and are strong oxidants.\nWhen heated in air to approximately 300 °C, M burns to form black hygroscopic solid A. A dissolves in water to give a clear colorless solution containing the strong base B, whose aqueous basicity is close to that of KOH. Passing excess $CO_2$ through a solution of B precipitates colorless crystals C. Mixing C with nitric acid and potassium chromate solution gives yellow precipitate D (reaction 1). Adding hydrogen peroxide to a solution of B produces dark-brown powder E (reaction 2). Heating E with a KI solution acidified with hydrochloric acid gives yellow precipitate F and purple-black solid G (reaction 3). The mass fraction of M in F is 61.69 %.", "header_images": [], "subquestions": [{"sub_id": "Exam-20-T1-1", "tags": ["2.1"], "text_open": "1.1.1 Identify element M and write the chemical formulas of A-G.", "text_images": [], "full_mark": 8}, {"sub_id": "Exam-20-T1-2", "tags": ["2.1", "4.2"], "text_open": "1.2 The chlorides of M have unusual properties. Its lower-valent chloride H is a white solid but gradually turns purple in sunlight. Its higher-valent chloride J can be prepared from H by the combined action of concentrated hydrochloric acid and chlorine. Dissolving J in concentrated hydrochloric acid gives the tetrahedral complex anion K.\n1.2.1 Write the equation for the gradual photochemical conversion of H to the purple material, and explain the composition and origin of the color of the purple product.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-20-T1-3", "tags": ["1.7"], "text_open": "1.2.4 Experimentally, H is much more stable than J. Explain the key atomic-structure effect responsible for this behavior and why it is especially pronounced for M within its group.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-20-T1-4", "tags": ["2.1", "3.4"], "text_open": "1.3 The following experiment is performed to determine the cumulative stability constant $\\beta_4$ of complex anion K formed when J dissolves in hydrochloric acid.\nStep 1: Accurately weigh a quantity of J, dissolve it in 0.500 mol·L$^{-1}$ HCl, and dilute to 50.0 mL. Place a portion of the solution in a 1.00 cm quartz cuvette and measure an absorbance of $A_1 = 0.680$ at 230 nm. At this wavelength, the molar absorptivity of K is $\\varepsilon = 8.50 \\times 10^3\\ \\mathrm{L\\cdot mol^{-1}\\cdot cm^{-1}}$. M ions and $Cl^-$ do not absorb at this wavelength, and at this HCl concentration almost all M is present as K.\nStep 2: Transfer 10.00 mL of the solution above, add 10.00 mL of 0.100 mol·L$^{-1}$ $AgNO_3$ solution, stir thoroughly to complete precipitation, and filter. Measure the absorbance of the filtrate under the same conditions, obtaining $A_2 = 0.306$. After addition of $AgNO_3$, the amount of excess $Ag^+$ is negligible. Ignore dissolution of $AgCl$ and activity-coefficient effects.\n1.3.1 Calculate the equilibrium concentration of free $Cl^-$ in the filtrate after reaction with the $AgNO_3$ solution.", "text_images": [], "full_mark": 5}]} | |
| {"problem_id": "Exam-20-T2", "header": "A Solvent-Responsive Color-Changing Complex\nThe pentacyanoferrate(II) complex ion X is a solvatochromic complex that is highly sensitive to changes in the dielectric constant and hydrogen-bond-donor ability of its solvent environment. Studies show that the color of X changes in opposite directions with solvent composition in the two binary solvent systems methanol/water and formamide (FA)/water.", "header_images": [], "subquestions": [{"sub_id": "Exam-20-T2-1", "tags": ["1.4", "1.6"], "text_open": "2.1 The color of complex ion X arises primarily from metal-to-ligand charge transfer, MLCT.\n2.1.2 In the MLCT mechanism of X, the excited state $X^{*}$ can be represented by a limiting resonance form of the ground-state ion in which one nitrogen atom of the organic ligand is a radical cation. Write the SMILES string of this limiting resonance form and correctly label the oxidation state of Fe.", "text_images": [{"file": "Exam-20-T2-1-1-prompt.png", "description": "X has the SMILES 'COC(C=C1)=CC=C1[N+](C=C2)=CC=C2C3=CC=[N]([Fe+2]([C-]#N)([C-]#N)([C-]#N)([C-]#N)[C-]#N)C=C3'. Complex ion X has an overall charge of two negative units."}], "full_mark": 3}, {"sub_id": "Exam-20-T2-2", "tags": ["1.7"], "text_open": "2.1.3 State which has the larger dipole moment: ground-state X or its excited state $X^{*}$.", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-20-T2-3", "tags": ["3.2", "1.7"], "text_open": "2.2 To quantify the effects of solvent dielectric constant and hydrogen-bond-donor ability on the color of the complex ion, the following linear relationship is used:\n$E_{CT} = E_0 + s\\cdot f(ε) + k\\cdot α$\nwhere ε is the dielectric constant of the solvent, α is the hydrogen-bond-donor acidity of the solvent, the Onsager polarity function is $f(ε)=\\frac{2(ε-1)}{2ε+1}$, and s and k are the corresponding linear coefficients.\n2.2.3 For a binary methanol/water solvent, the fitted linear coefficients are\n$E_0$ = −28.112 kcal·mol^{-1}; s = 123.758 kcal·mol^{-1}; k = −61.985 kcal·mol^{-1}.\nThe dielectric constant ε of the binary methanol/water solvent is approximately linearly related to the mole fraction of methanol $x_{MeOH}$:\n$ε = 79.2 - 43.6\\cdot x_{MeOH}$\nThe hydrogen-bond-donor acidity α of the mixed solvent is approximately independent of solvent composition. Calculate the slope of $E_{CT}$ with respect to $x_{MeOH}$ for the complex in the mixed solvent at $x_{MeOH}$ = 0.50. Hint: for $u = u(v(x))$, $du/dx = (du/dv) \\cdot (dv/dx)$.", "text_images": [], "full_mark": 6}, {"sub_id": "Exam-20-T2-4", "tags": ["1.7", "3.9"], "text_open": "2.3 In a binary formamide (FA)/water solvent, FA is the more polar component. Preferential solvation makes the local solvent mole fraction y around the solute differ from the bulk solvent mole fraction x. The Suppan dielectric-enrichment model is used:\n$\\frac{y_n}{y_p}=\\frac{x_n}{x_p}e^{-Z_{ps}}$\nwhere the subscript n denotes the less polar solvent and p the more polar solvent. In an FA/water mixture, the solvation index is $Z_{ps}=0.412$.\n2.3.2 Define the local enrichment factor of the more polar solvent p as $η_p = y_p / x_p$. Derive an expression for $η_p$ in terms of $x_p$ and $Z_{ps}$, and calculate $η_p$ at $x_p$ = 0.600.", "text_images": [], "full_mark": 5}, {"sub_id": "Exam-20-T2-5", "tags": ["1.6", "1.7"], "text_open": "Experimental observation: in methanol/water, the absorption peak shifts to the blue as the mole fraction of the more polar component, water, increases. In FA/water, the absorption peak shifts to the red as the mole fraction of the more polar component, FA, increases, and preferential solvation occurs.\n2.4 Based on the information and model analysis above, in a methanol/water solvent the ( ) between solute and solvent determines the direction of the solute's absorption-peak shift, and the solvent predominantly stabilizes ( ). In an FA/water solvent, the ( ) between solute and solvent determines the direction of the absorption-peak shift, and the solvent predominantly stabilizes ( ).\n(a) Dipole/electrostatic interactions; (b) hydrogen-bonding interactions; (c) ground-state X; (d) excited-state $X^{*}$", "text_images": [], "full_mark": 2}]} | |
| {"problem_id": "Exam-20-T5", "header": "The Vanishing Sixfold Axis\nCrystals in the hexagonal crystal system are characterized by a sixfold axis along the c direction. In addition to the simplest sixfold rotation axis, for which rotation by 60° about the axis brings the crystal into coincidence with itself, a crystal may contain either of the following two types of sixfold axes:\nFirst, a $6_n$ screw axis, for which rotation by 60° followed by translation along c by n / 6 of the unit-cell length brings the crystal into coincidence with itself.\nSecond, a sixfold rotoinversion axis, for which rotation by 60° followed by inversion through a point on the axis brings the crystal into coincidence with itself.", "header_images": [], "subquestions": [{"sub_id": "Exam-20-T5-1", "tags": ["1.3"], "text_open": "5.3 The symmetry of an individual close-packed layer does not necessarily represent the symmetry of the whole crystal. Consider a crystal described as follows. Its close-packed layers are analogous to those in perovskite: metal M and vacancies occupy the $Ca^{2+}$ and $O^{2-}$ sites, respectively, in a 1:3 ratio. These close-packed layers undergo simple hexagonal stacking along c with a four-layer repeat, and the projections of M along c do not overlap within a repeat. X atoms occupy all trigonal-prismatic voids jointly formed by M atoms and vacancies.\n5.3.1 Write the chemical formula of this crystal.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-20-T5-2", "tags": ["1.3"], "text_open": "5.3.3 State the Bravais lattice type of this crystal.", "text_images": [], "full_mark": 3}]} | |
| {"problem_id": "Exam-20-T6", "header": "Co(Salen) Oxygen Carrier\nMany proteins containing transition-metal ions can bind molecular oxygen to form complexes and reversibly take up and release oxygen for biological processes. Examples include the iron-containing proteins myoglobin and hemoglobin and the copper-containing protein hemocyanin. The Co-Schiff-base complex [Co(II)(Salen)], cobalt N,N'-ethylenebis(salicylideneiminate), is an important representative synthetic model oxygen carrier.", "header_images": [{"file": "Exam-20-T6-header-1.png", "description": "[Co(II)(Salen)] has the SMILES 'C12=C(O[Co]34OC5=CC=CC=C5C=[N]3CC[N]4=C2)C=CC=C1'. The image shows the structure of [Co(II)(Salen)]. The Salen ligand coordinates Co tetradentately through 2 phenoxide O atoms and 2 imine N atoms. The benzene rings are drawn in alternating-single-and-double-bond Kekulé form, both C=N bonds are double bonds, and the central Co atom is represented as four-coordinate by the ring-closure labels [Co]34. No charges or stereochemical markers are shown. The caption [Co(II)(Salen)] appears below the structure."}], "subquestions": [{"sub_id": "Exam-20-T6-1", "tags": ["1.4"], "text_open": "6.1 Depending on the preparation method, [Co(II)(Salen)] can exist in two solid forms: dark-red, gelatinous, active form A and dark-red, crystalline, inactive form B. A rapidly absorbs oxygen at room temperature, whereas B is stable at room temperature and absorbs almost no oxygen. Both are bimolecular dimers.\n6.1.1 A and B dimerize differently: A through Co-Co bonding and B through Co-O bonding. Based on this description, write the actual SMILES strings of A and B, using a textual description if necessary. The Salen ligand may be abbreviated.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-20-T6-2", "tags": ["1.4", "2.4"], "text_open": "6.1.2 Inactive B dissolves in certain solvents, and the resulting solution acquires the ability to absorb oxygen. [Co(II)(Salen)] dissolved in solvent S can react with oxygen by different pathways to form product C with $Co:O_2$ = 1:1 or product D with $Co:O_2$ = 2:1. The Co centers in both C and D are octahedral. Write the SMILES strings of C and D.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-20-T6-3", "tags": ["1.4", "2.8"], "text_open": "6.2 Inactive [Co(II)(Salen)] can be prepared as follows. Add salicylaldehyde and ethylenediamine in stoichiometric amounts successively to a three-neck flask, followed by ethanol. Yellow plate-like crystals form immediately. Assemble the apparatus shown at right, maintain the water bath at approximately 80 °C, and reflux the ethanol until the yellow crystals dissolve completely. After passing nitrogen through the two-way stopcock on the right for some time, use the syringe on the left to inject a solution of $Co(CH_3COO)_2$ slowly through the septum into the three-neck flask. Continue refluxing to obtain a dark-red crystalline precipitate. Stop heating and cool the flask in an ice-water bath while maintaining nitrogen flow and stirring. Collect the crystals on a sintered-glass funnel and dry them by suction. Wash the precipitate successively with deionized water and anhydrous ethanol, then rinse once with acetone and dry by suction to obtain dry red-brown [Co(II)(Salen)].\n6.2.1 What consequences would result from adding excess ethylenediamine? State 2 consequences.", "text_images": [{"file": "Exam-20-T6-3-1-prompt.png", "description": "Apparatus for preparing [Co(II)(Salen)] in 6.2: a three-neck round-bottom flask rests on a base, with a visible liquid level inside. A syringe with plunger is inserted obliquely upward into the left neck. The middle neck is connected to a vertical condenser whose top connects through a bent tube to a U-tube containing liquid at the bottom; the top of the right arm is open. The right neck is connected to an addition component with a stopcock, the two-way stopcock, and a bulb-shaped upper section. The image contains no text labels."}], "full_mark": 2}, {"sub_id": "Exam-20-T6-4", "tags": ["4.1"], "text_open": "6.2.2 What are the purposes of adding deionized water to the U-tube apparatus connected above the condenser? State 2 purposes.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-20-T6-5", "tags": ["4.1"], "text_open": "6.2.3 What problem may occur while cooling the flask in an ice-water bath, and what measure should be taken?", "text_images": [], "full_mark": 3}]} | |
| {"problem_id": "Exam-20-T8", "header": "Fundamentals of Organic Chemistry\nThe rates and selectivities of organic reactions depend not only on electronic structure but also strongly on molecular conformation and spatial arrangement in the transition state.", "header_images": [], "subquestions": [{"sub_id": "Exam-20-T8-1", "tags": ["1.6"], "text_open": "8.1 First consider peroxyacid epoxidation of alkenes from the perspective of frontier molecular orbitals. Examine the general reaction below.\n8.1.2 Under identical reaction conditions, substrate structure affects the rate of epoxidation. Arrange the following compounds in decreasing order of reaction rate.", "text_images": [{"file": "Exam-20-T8-1-1-prompt.png", "description": "General reaction for peroxyacid epoxidation of an alkene: '[R]/C([R])=C([R])/[R]'.'[R]C(OO)=O'>>'[R]C1([R])C([R])([R])O1'. No reaction conditions are labeled on the arrow; R is the placeholder used in the original diagram."}, {"file": "Exam-20-T8-1-2-prompt.png", "description": "The SMILES for A is 'C1CC=CCC1'; the SMILES for B is 'OC1C=CCCC1'; the SMILES for C is 'O=C(OC1C=CCCC1)C'; and the SMILES for D is 'OC1CC=CCC1'."}], "full_mark": 3}, {"sub_id": "Exam-20-T8-2", "tags": ["1.2"], "text_open": "8.2 When the oxygen-transfer reagent is changed, the spatial course of the reaction still strongly affects the epoxidation outcome. Consider the alkene epoxidation involving the dioxirane below.\n8.2.1 For a more structurally complex substrate, its preferred conformation further affects the reaction. Write the SMILES string of the preferred conformation of the actual reactive species derived from the molecule shown under these reaction conditions.", "text_images": [{"file": "Exam-20-T8-2-1-prompt.png", "description": "The molecule is represented by the SMILES 'O=C1[C@H]2[C@H](OC(C)(C)O2)CO[C@@]13OC(C)(C)OC3'. It contains a C=O group and two O—C(CH₃)₂—O ketal fragments; stereochemistry is indicated by solid and dashed wedge bonds."}], "full_mark": 2}, {"sub_id": "Exam-20-T8-3", "tags": ["2.4", "1.2"], "text_open": "8.3 The epoxidation above illustrates how molecular conformation and spatial arrangement in the transition state affect selectivity. Similar stereochemical control can also be used to form carbon-carbon bonds. The enolate alkylation below gives its product with very high stereoselectivity.\n8.3.1 Write the SMILES string of the major alkylation product, explicitly specifying its stereochemistry.", "text_images": [{"file": "Exam-20-T8-3-1-prompt.png", "description": "The substrate is represented by the SMILES 'O=C1OC[C@@H](C(C)C)N1C(CC)=O', with stereochemistry indicated by a solid wedge bond. The arrow is labeled 'LDA, THF, -78°C' above and 'then PhCH2Br' below; no product structure is drawn to the right."}], "full_mark": 2}]} | |
| {"problem_id": "Exam-20-T9", "header": "Problem 9 Prins Cyclization (25 points, 12%)\nThe Prins cyclization is a powerful synthetic transformation that rapidly constructs tetrahydropyran rings. Tetrahydropyran is a common structural unit in many bioactive molecules, including spongistatins and bryostatins. Researchers have used the Prins cyclization in numerous elegant strategies for the total synthesis of natural-product targets.", "header_images": [], "subquestions": [{"sub_id": "Exam-20-T9-1", "tags": ["2.4", "2.2"], "text_open": "9.1 The following is a classic Prins cyclization that gives the final product with high stereoselectivity, where R = alkyl.\n9.1.1 Write the SMILES strings of A-D in the scheme, including stereochemistry. B is a highly strained intermediate.", "text_images": [{"file": "Exam-20-T9-1-1-prompt.png", "description": "Prins cyclization sequence: '[R]C1=CCCCC1'>H2C=O, H2SO4>[A⇌B]; [B]>H2O>[C]; [C]>H2C=O, H2SO4>[D]. In the image, the last arrow points leftward from C to D. A-D are shown only as letter labels and no structures are provided; R is the placeholder used in the original diagram."}], "full_mark": 5}, {"sub_id": "Exam-20-T9-2", "tags": ["2.2", "2.4"], "text_open": "9.1.2 The following cascade comprises a Prins cyclization and a rearrangement and ultimately gives a product containing a five-membered ring. Write the SMILES string of the product and of at least 2 charged key intermediates, including stereochemistry.", "text_images": [{"file": "Exam-20-T9-2-1-prompt.png", "description": "Single-step reaction: 'C=C(C)[C@H]([C@H](C(C)=C)O)O'>Me2C=O, H2SO4>[product not shown]. Stereochemistry of the substrate's two OH groups is shown by wedge bonds and retained by the @ markers in the SMILES."}], "full_mark": 4}, {"sub_id": "Exam-20-T9-3", "tags": ["1.2", "2.7"], "text_open": "9.2 A key advantage of the Prins cyclization is that, starting from a single stereocenter and a defined alkene geometry, it can construct up to four stereocenters on a tetrahydropyran ring with high stereoselectivity. In some cases, however, enantiomeric excess (ee) is lost.\n9.2.1 In the Prins cyclization below, the ee of the major product is lower than that of the starting material, and small amounts of two byproducts are detected.\nPredict the configurations of the chiral carbons in the product by assigning the absolute configurations of chiral carbons a and b in the major product using S/R notation.", "text_images": [{"file": "Exam-20-T9-3-1-prompt.png", "description": "Single-step reaction: 'O[C@H](CC=C)C1=CC=CC=C1' reacts with PhCH2CH2CHO under BF3·OEt2 and AcOH conditions to give the major product, 'O=C(O[C@H]1C[C@@H](CCC2=CC=CC=C2)O[C@@H](C3=CC=CC=C3)C1)C', byproduct 1, 'O=C(OC1CC(C2=CC=CC=C2)OC(C3=CC=CC=C3)C1)C', and byproduct 2, 'O=C(OC1CC(CCC2=CC=CC=C2)OC(CCC3=CC=CC=C3)C1)C'. Starting material M is labeled 87% ee and the major product 67% ee. Chiral carbons a and b are labeled on the ring of the major product; OAc at a is shown on a solid wedge, while CH2CH2Ph at b is shown on a dashed bond."}], "full_mark": 2}, {"sub_id": "Exam-20-T9-4", "tags": ["2.3", "2.2"], "text_open": "9.2.2 The pathway to the major product can undergo racemization through a rearrangement. Experimentally, the enantiomer M’ of substrate M is detected. If M and $PhCH_2CH_2CHO$ are replaced by structure N below, racemization through this rearrangement cannot occur under the same conditions. In light of the byproduct structures and the analysis above, write the SMILES strings of the reactants and products in each step of the mechanism by which M forms M’, and label the reaction conditions. Electron-flow arrows need not be shown.", "text_images": [{"file": "Exam-20-T9-4-1-prompt.png", "description": "Substrate N has the SMILES 'C=CC[C@@H](OC(OC(C)=O)CC1=CC=CC=C1)C2=CC=CC=C2'. N is the structure label in the image."}], "full_mark": 4}, {"sub_id": "Exam-20-T9-5", "tags": ["2.3", "2.7"], "text_open": "9.2.3 Explain why a racemization process analogous to the one above cannot occur when substrate N is used.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-20-T9-6", "tags": ["2.2", "2.4"], "text_open": "9.3 Researchers have developed many broadly useful Prins reactions using unconventional substrates such as alkynols, allenes, and small-ring compounds.\n9.3.1 The reaction below is based on a Prins-type cyclization of a 3,5-diynol with an aldehyde and proceeds through a domino sequence to synthesize a 2,3-dialkylidenetetrahydrofuran framework. The compound shown at the right was detected in the reaction mixture. This compound is not necessarily itself an intermediate and may arise from an intermediate. Based on this information, write the SMILES strings of the key intermediates. Stereochemistry is not required.", "text_images": [{"file": "Exam-20-T9-6-1-prompt.png", "description": "Single-step reaction: 'OCCC#CC#CC1=CC=CC=C1'.'O=CC1=CC=CC=C1'>HBF4·OEt2 (3 equiv), MeOH (3 equiv), DCM, rt, 8 h>'O=C(/C=C1OCCC\\1=C/C2=CC=CC=C2)C3=CC=CC=C3'."}, {"file": "Exam-20-T9-6-2-prompt.png", "description": "The image is labeled 'Detected in the reaction mixture:' and the detected compound has the SMILES 'OCC/C(C(C#CC1=CC=CC=C1)=O)=C\\C2=CC=CC=C2'."}], "full_mark": 4}, {"sub_id": "Exam-20-T9-7", "tags": ["2.4", "2.7"], "text_open": "9.3.2 The following reaction proceeds through a three-membered-ring intermediate and gives the product with high stereoselectivity, governed primarily by stereoelectronic effects. Write the SMILES string of the product, including stereochemistry, and briefly explain the stereoselectivity.", "text_images": [{"file": "Exam-20-T9-7-1-prompt.png", "description": "Single-step reaction: 'C=C([Si](C)(C1=CC=CC=C1)C)C[C@@H](O)C'.'[R]C=O'>TMSOTf>[?]. The product structure is not shown and is represented by '?'. R is the placeholder used in the original diagram and was retained after manual confirmation."}], "full_mark": 4}]} | |
| {"problem_id": "Exam-20-T10", "header": "Photocatalytic Reactions", "header_images": [], "subquestions": [{"sub_id": "Exam-20-T10-1", "tags": ["2.2", "2.4"], "text_open": "10.1 Researchers studied the visible-light photocatalytic difluoroallylation of a silane shown below. The photosensitizer is 4-CzIPN, denoted PC. Under irradiation, PC is excited to PC* and can undergo one-electron reduction to the reduced radical anion PC$^{·-}$. PC$^{·-}$ is unstable and is readily oxidized back to PC. The detailed structure of the photosensitizer need not be considered in any subquestion.\n10.1.1 Based on the reaction mechanism, write the SMILES strings of 1 nitrogen-containing key intermediate and 2 fluorine-containing key intermediates other than PC and its various forms. Quinuclidine is 1-azabicyclo[2.2.2]octane.", "text_images": [{"file": "Exam-20-T10-1-1-prompt.png", "description": "Single-step reaction: 'CC1=CC=C(C(C(F)(F)F)=C)C=C1'.Et3SiH>4-CzIPN, Blue LED; quinuclidine, KHCO3, MeCN>'CC1=CC=C(C(C[Si](CC)(CC)CC)=C(F)F)C=C1'. At the left of the arrow, a p-methylphenyl-C(CF3)=CH2 substrate is joined by a plus sign to Et3SiH; the product is p-methylphenyl-C(=CF2)-CH2SiEt3."}], "full_mark": 6}, {"sub_id": "Exam-20-T10-2", "tags": ["2.4"], "text_open": "10.1.2 When the reaction solvent is changed from the original solvent to THF, a different product is formed. This product still contains 2 fluorine atoms. Write the SMILES string of this byproduct.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-20-T10-3", "tags": ["2.2", "2.4"], "text_open": "10.1.3 Keep the standard reaction conditions unchanged, but replace the para methyl group on the substrate's aryl ring with a para acetyl group (-COCH$_3$). The expected difluoroallylsilane is not obtained; instead, a different major product is formed. Carefully examine the mechanism in 10.1.1, write the SMILES string of this major product, and briefly explain why an analogue of the 10.1.1 product is not formed.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-20-T10-4", "tags": ["2.4"], "text_open": "10.2 The researchers further found that the α-trifluoromethyl alkene derivative above can react with 1,3-dioxolane in a similar visible-light photocatalytic system. Product A is formed predominantly in MeCN, whereas product B is formed predominantly in DMSO. B contains only 1 fluorine atom. Write the SMILES string of B.", "text_images": [{"file": "Exam-20-T10-4-1-prompt.png", "description": "Single-step reaction: 'C=C(C(F)(F)F)C1=CC=C(C2=CC=CC=C2)C=C1'.'O1CCOC1'>4-CzIPN(1 mol%), quinuclidine(20 mol%); NaHCO3(1.5 equiv), Blue LED, Solv., rt, 12 h>. At the left of the arrow, a 4-biphenyl-C(CF3)=CH2 substrate is joined by a plus sign to 1,3-dioxolane; no product structure is drawn to the right."}], "full_mark": 2}]} | |
| {"problem_id": "Exam-21-T1", "header": "Chasing the Rainbow\nA silvery-gray, high-melting metal M has no independent mineral in nature and occurs mainly in molybdenite. Industrially, molybdenite is roasted to concentrate the oxides of M.", "header_images": [], "subquestions": [{"sub_id": "Exam-21-T1-1", "tags": ["3.2"], "text_open": "M forms numerous oxides in a wide range of oxidation states and colors, earning them the name \"chemical rainbow.\" Oxide A contains M in its highest oxidation state. It is a yellow, highly deliquescent crystalline solid and dissolves in water to form a strong acid in which the mass fraction of M is 74.12%. Heating a mixture of A and elemental M in a sealed tube gives red crystals B with a metallic luster. Further heating to about 600 °C converts B into the blue solid M2O5; at still higher temperatures, M2O5 decomposes into the gray solid C. M has a different oxidation state in each of A, B, and C, and the oxidation state of M in C is the one most commonly found in oxides of its group. In the gas phase, the M atoms in A have tetrahedral coordination, whereas in the solid state A consists of corner-sharing octahedra and tetrahedra. Passing $Cl_2$ through a heated mixture of M powder and $KCl$ gives the red trimeric chloride D. 1.4 Thermodynamic data show that the reaction of C with $O_2$ to form B is spontaneous at 400 °C; experimentally, however, B decomposes at 400 °C by the reaction $3B(s)\\rightarrow A(s)+C(s)$. Explain this apparent contradiction.", "text_images": [], "full_mark": 4}]} | |
| {"problem_id": "Exam-21-T2", "header": "Molecular Qubit Frameworks\nMolecular qubit frameworks (MQFs) are among the leading candidate materials for quantum-information chips. Molecular qubits are functional materials that use electron-spin states within molecules to store and process quantum information. Under an applied magnetic field, two controllable quantum states can be selected from the low-energy levels of a magnetic center to serve as the qubit states \"0\" and \"1\". Long-term information storage requires minimizing environmental perturbations to the electron spins and isolating the magnetic centers from one another to obtain a long coherence time. A research team successfully constructed a diluted array of molecular spin qubits by doping trace amounts of $Nd^{3+}$ magnetic spin centers into a rigid yttrium-based MOF.", "header_images": [], "subquestions": [{"sub_id": "Exam-21-T2-1", "tags": ["1.4", "1.1"], "text_open": "2.1 The yttrium-based MOF has the formula $[C_5N_2H_{11}][Y(C_2O_4)_2]$. The cation does not coordinate to $Y^{3+}$. The MOF crystallizes in the monoclinic system with $\\beta$ = 130.6°. Each $C_2O_4^{2-}$ bridges two $Y^{3+}$ ions in a $\\mu_2$-$\\kappa^2$O,O′;$\\kappa^2$O″,O‴ mode, chelating each $Y^{3+}$ ion through two O atoms. Answer the following questions about this compound.\n2.1.1 The cation of this compound contains a six-membered ring, its two nitrogen atoms have identical chemical environments, and it contains no tertiary carbon atom. Write the SMILES string for the cation.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-21-T2-2", "tags": ["1.4"], "text_open": "2.1.2 Give the coordination number of $Y^{3+}$.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-21-T2-3", "tags": ["3.8", "4.7"], "text_open": "2.3 In quantum-information storage, environmental lattice vibrations (phonon scattering) strongly perturb spin qubits and shorten their lifetimes. Theoretical studies show that, in this porous oxalate framework, the spin–lattice relaxation rate $1/T_1$ of $Nd^{3+}$ in the higher-temperature range 6.5K–8.5K is driven primarily by the combined Raman process and Orbach process and obeys the following kinetic equation:\n$\\frac{1}{T_1}=A_{Raman}T^7+A_{Orbach}\\exp\\left(-\\frac{U_{eff}}{T}\\right)$\nHere, $A_{Raman}$ and $A_{Orbach}$ are the prefactors for the respective mechanisms, and $U_{eff}$ is the effective energy barrier of the low-lying excited state arising from strong spin–orbit coupling. The experimentally measured $U_{eff}$ for this material is 32.6 K.\nThe table below gives experimental spin–lattice relaxation times $T_1$ measured for the extremely dilute sample at different temperatures $T$ using a pulsed EPR spectrometer:\n\n| Measurement temperature $T$ / K | Measured spin-relaxation time $T_1$ / μs |\n| --- | --- |\n| 6.8 | 6.359 |\n| 7.2 | 4.623 |\n| 7.6 | 3.252 |\n| 8.0 | 2.347 |\n| 8.5 | 1.612 |\n\n2.3.1 Using the data above, calculate $A_{Raman}$ and $A_{Orbach}$ by linear regression.", "text_images": [], "full_mark": 6}]} | |
| {"problem_id": "Exam-21-T7", "header": "Silver in the Making\nIn the worldview of alchemists, metals were not inert matter but substances that \"grew\" and \"evolved\" within Earth's crust. They believed that all base metals, such as lead and iron, were evolving toward the perfect noble metals, gold and silver. Metal M has the largest atomic number among elements with no definite year or person of discovery. As early as about 1500 CE, the Inca used M, Cu, and Sn to make alloys for the tumi knife.", "header_images": [], "subquestions": [{"sub_id": "Exam-21-T7-1", "tags": ["2.1", "1.1"], "text_open": "7.17.1 When miners and alchemists unearthed M near silver veins, glinting with a faint pink metallic luster, they found it harder than lead, more brittle than tin, and remarkably low-melting. They called M \"silver in the making\" (Tectum Argenti). Adding oxide A of M to excess hydrochloric acid gives a solution of B. Solid B and M are mixed and heated to 325℃, then cooled at 270℃ for 1–2 weeks. Excess B is removed by extraction with benzene, yielding a black solid chloride C containing 16.52% chlorine. C consists of a nine-atom cluster cation D, a mononuclear dianion E, and a dinuclear dianion F; M has the same oxidation state in B, E, and F (reaction 1). Adding excess concentrated sodium hydroxide to a nitrate solution of M may be regarded as forming the simple hydroxide precipitate G. Passing chlorine through this suspension gives the brown, strongly oxidizing powder H (reaction 2).\n7.1.2 Write equations for reactions 1 and 2. Explain why $NaBiO_3$ is a strong oxidizing agent.", "text_images": [], "full_mark": 4}]} | |
| {"problem_id": "Exam-21-T8", "header": "Lithiation Reactions\nFrom the late 30s of the 20th century to the early 50s, Gilmann and Wittig successively discovered lithium–halogen exchange in aryl halides and directed ortho-lithiation of aromatic compounds by substituent heteroatoms. These reactions have become important methods for forming carbon–lithium bonds and constructing complex molecular structures.", "header_images": [], "subquestions": [{"sub_id": "Exam-21-T8-1", "tags": ["2.4"], "text_open": "8.1 Write the SMILES strings for the major products A and B in the reactions below.", "text_images": [{"file": "Exam-21-T8-1-1-prompt.png", "description": "Two parallel reactions share the same substrate: 'ClC1=CC(Br)=CC=C1O[Si](C(C)C)(C(C)C)C(C)C'>1) LDA 1.5eq, THF, -78°C; 2) MeI>[A]; 'ClC1=CC(Br)=CC=C1O[Si](C(C)C)(C(C)C)C(C)C'>1) nBuLi 1.1eq, THF, -78°C; 2) MeI>[B]. The structures of products A and B are not drawn; they are shown only as A and B."}], "full_mark": 2}, {"sub_id": "Exam-21-T8-2", "tags": ["1.4"], "text_open": "8.2 Ortho-lithiation can be viewed as being controlled by two limiting mechanisms: a coordination mechanism and an acid–base mechanism. In most cases, both mechanisms influence the reaction to different extents. A schematic of the coordination mechanism is shown below.\n8.2.1 TMEDA (tetramethylethylenediamine) and butyllithium can form a dinuclear complex with the formula $Li_2Bu_2(TMEDA)_2$, in which both Li atoms are 4-coordinate. Write its SMILES string.", "text_images": [{"file": "Exam-21-T8-2-1-prompt.png", "description": "Schematic of the coordination mechanism: substrate '[H]C1=C(C=CC=C1)CN([R])[R]' forms intermediate '[H]C1=C(C=CC=C1)C[N]([R])([R])[Li]CCCC' through a reversible arrow labeled nBuLi, and the intermediate gives product '[R][N]1([R])CC2=C(C=CC=C2)[Li]1' through a single arrow. A curved arrow runs from the nBu-Li bond to the ortho H atom of the benzene ring. In the product, Li and N are joined by a dashed coordination bond to form a five-membered chelate ring. R2 is the original placeholder in the figure for the two R substituents on N; Li coordination in the intermediate and product is shown by dashed lines."}], "full_mark": 2}, {"sub_id": "Exam-21-T8-3", "tags": ["2.3", "2.7"], "text_open": "The following lithiation gives different products with different additives. No workup has been performed, and each product contains only 1 lithium atom.\n8.2.2 Write the SMILES strings for the major products C and D, and identify the limiting mechanism that may be regarded as controlling each reaction.", "text_images": [{"file": "Exam-21-T8-3-1-prompt.png", "description": "Two parallel reactions share the same substrate: 'COC1=CC=C(CN(C)C)C=C1'>nBuLi>[C]; 'COC1=CC=C(CN(C)C)C=C1'>nBuLi, TMEDA>[D]. The structures of products C and D are not drawn; they are shown only as C and D."}], "full_mark": 4}, {"sub_id": "Exam-21-T8-4", "tags": ["2.4"], "text_open": "8.3 Lithiation reactions are also prone to side reactions, making control of the reaction conditions especially important.\n8.3.1 The following reaction did not give the expected major product. Instead, E, which has the same molecular formula as the expected product, was obtained in 70% yield. Write the SMILES string for E.", "text_images": [{"file": "Exam-21-T8-4-1-prompt.png", "description": "The substrate on the left is 'CC1=C(OC(N(CC)CC)=O)C=CC=C1'. A reaction arrow labeled sec-BuLi, TMEDA, THF above and -78°C-r.t. below points to product E. A box on the right labeled 'Expected product' gives the SMILES representation 'CC1=C(OC(N(CC)CC)=O)C([Li])=CC=C1'. The structure of E is not drawn and is shown only as E."}], "full_mark": 2}, {"sub_id": "Exam-21-T8-5", "tags": ["2.4", "2.2"], "text_open": "8.3.3 The following reaction causes fragmentation of the substrate. Write the SMILES strings for all products.", "text_images": [{"file": "Exam-21-T8-5-1-prompt.png", "description": "The substrate on the left is 'C1(C2=CC=CC=C2)=NOC=C1C3=CC=CC=C3'. A reaction arrow labeled RLi points to '? + ?'; the product structures are not drawn."}], "full_mark": 2}, {"sub_id": "Exam-21-T8-6", "tags": ["2.4"], "text_open": "8.4 Under base-induced conditions, a halogen substituent in a halogenated aromatic compound may migrate to a new substitution site. Such reactions are vividly known as halogen-dance reactions. 8.4.2 Some studies suggest that increasing solvent polarity accelerates the halogen-dance reaction. Write the SMILES string for the major product of the reaction below, knowing that it contains only two halogen atoms.", "text_images": [{"file": "Exam-21-T8-6-1-prompt.png", "description": "The substrate on the left is 'IC1=C(Br)C(F)=CC=C1'. A reaction arrow labeled 1. BuLi, toluene, -100°C above and 2. THF, -75°C; 3. CO2, then HCl below points to the right; the product structure is not drawn."}], "full_mark": 2}]} | |
| {"problem_id": "Exam-21-T9", "header": "Problem 9 Reactions of Sulfur Ylides\nSulfur ylides are important intermediates in organic synthesis with distinctive reactivity. Through the Corey–Chaykovsky reaction, they enable transformations such as cyclopropanation, epoxidation, and aziridination. Recent advances in organocatalytic sulfur-ylide reactions have opened new routes to the enantioselective synthesis of complex molecules.", "header_images": [], "subquestions": [{"sub_id": "Exam-21-T9-1", "tags": ["2.2"], "text_open": "9.1 A sulfur ylide and an α,β-unsaturated ketone undergo the asymmetric cyclopropanation shown below under catalysis by a chiral amine organocatalyst.\n9.1.1 The chiral amine catalyst first undergoes condensation with one substrate. Write the SMILES string for key condensation intermediate A.", "text_images": [{"file": "Exam-21-T9-1-1-prompt.png", "description": "Single-step reaction: 'O=C(/C=S(C)/C)C1=CC=CC=C1'.'O=C(C)/C=C/C1=CC=CC=C1' gives 'O=C([C@@H]1[C@@H](C2=CC=CC=C2)[C@@H]1C(C)=O)C3=CC=CC=C3' under amine-cat (20 mol%), PhCO2H (20 mol%), CHCl3, MS, 30°C, 72h. The two substrates are separated by a plus sign. The sulfur-ylide SMILES is shown in a neutral form without formal charges. The cyclopropane product contains 3 stereocenters, with @/@@ retained in the SMILES; PhC(O)- is shown by a solid wedge and Ph by a dashed wedge. Below the reaction, the chiral amine catalyst amine-cat is shown as 'N[C@@H](C1=CC=CC=C1)[C@H](C2=CC=CC=C2)NC3CCCCC3', with C-NH_2 on a solid wedge and C-NH on a dashed wedge."}], "full_mark": 2}, {"sub_id": "Exam-21-T9-2", "tags": ["2.2", "2.4"], "text_open": "9.2 In addition to conventional cyclopropanation, some sulfur ylides conjugated with double bonds can generate distinctive fused-ring products.\n9.2.1 Write the SMILES strings for the three key intermediates in the following reaction, ignoring stereochemistry.", "text_images": [{"file": "Exam-21-T9-2-1-prompt.png", "description": "Single-step reaction: 'O=C1N(C2=CC=CC=C2)C3=CC=CC=C3C1(OC(OC(C)(C)C)=O)C(C(OCC)=O)=C'.'C[S+](C)C/C=C/C(OCC)=O.[Br-]'>NaOH, CH_3CN>'O=C1N(C2=CC=CC=C2)C3=CC=CC=C3[C@@]14[C@](C5)(C(OCC)=O)[C@@]4([H])C=C5C(OCC)=O'.'O=C1N(C2=CC=CC=C2)C3=CC=CC=C3[C@@]14[C@@](C5)(C(OCC)=O)[C@]4([H])C=C5C(OCC)=O'. The two products are separated by a plus sign and are stereoisomers; @/@@ stereochemical markers are retained in the SMILES."}], "full_mark": 6}, {"sub_id": "Exam-21-T9-3", "tags": ["2.4"], "text_open": "9.2.2 Using the information from the preceding question, write the SMILES string for the product of the following reaction.", "text_images": [{"file": "Exam-21-T9-3-1-prompt.png", "description": "Single-step reaction: 'O=S1(C2=CC=CC=C2C(/C=C/C3=CC=CC=C3)=N1)=O'.'C[S+](C)C/C=C/C(OC)=O.[Br-]'>CsCO3 (3 eq), CH3CN, rt>[product not shown; only ? appears]. Reactant 2 is labeled (2.5 eq)."}], "full_mark": 2}, {"sub_id": "Exam-21-T9-4", "tags": ["2.3", "2.4"], "text_open": "9.2.3 A similar substrate gives a product different from those in the preceding two reactions. Explain why.", "text_images": [{"file": "Exam-21-T9-4-1-prompt.png", "description": "Single-step reaction: 'O=C(C1=CC(Br)=CC=C1S/2)C2=C\\C3=CC=CC=C3'.'C[S+](C)C/C=C/C(OC)=O.[Br-]'>NaH, CHCl_3, 1min>'O=C(OC)C1=CCOC2=C(SC3=CC=C(Br)C=C32)C1C4=CC=CC=C4'."}], "full_mark": 3}, {"sub_id": "Exam-21-T9-5", "tags": ["2.2", "2.4"], "text_open": "9.3.1 Researchers have also developed a DABCO-catalyzed (3+3) cyclization between a 2-(acetoxymethyl)buta-2,3-dienoate and a sulfur ylide, providing a convenient route to 4H-pyrans bearing a vinyl sulfide group. Write the SMILES strings for the four key intermediates in the following reaction, using $NR_3$ in place of DABCO.", "text_images": [{"file": "Exam-21-T9-5-1-prompt.png", "description": "Single-step reaction: 'C=C=C(C(OCC1=CC=CC=C1)=O)COC(C)=O'.'O=C(C[S+](C)C)C1=CC=CC=C1.[Br-]'>20 mol% DABCO, 1.2 eq K2CO3, acetone>'CC1=C(C(OCC2=CC=CC=C2)=O)CC(SC)=C(C3=CC=CC=C3)O1'."}], "full_mark": 8}, {"sub_id": "Exam-21-T9-6", "tags": ["2.3"], "text_open": "9.3.2 Replacing the carbonyl-stabilized sulfur ylide with an ester-stabilized ylide gives a different product. Explain why.", "text_images": [{"file": "Exam-21-T9-6-1-prompt.png", "description": "Single-step reaction: 'C=C=C(C(OCC1=CC=CC=C1)=O)COC(C)=O'.'O=C(C[S+](C)C)OCC.[Br-]' gives 'C#CC1(C(OCC2=CC=CC=C2)=O)CC1C(OCC)=O' under 20 mol% DABCO, 1.2 eq K_2CO_3, acetone."}], "full_mark": 2}, {"sub_id": "Exam-21-T9-7", "tags": ["2.4"], "text_open": "9.3.3 If the allene substrate is replaced by a less reactive alkene, cyclization cannot be completed. Write the SMILES string for the product of the following reaction.", "text_images": [{"file": "Exam-21-T9-7-1-prompt.png", "description": "Single-step reaction: 'C=C(C(OCC)=O)COC(C)=O'.'O=C(C[S+](C)C)C1=CC=CC=C1.[Br-]'>20 mol% DABCO, 1.2 eq K_2CO_3, acetone>[product not shown; only ? appears]."}], "full_mark": 2}]} | |
| {"problem_id": "Exam-22-T2", "header": "Problem 2 Determination of Nickel by EDTA Titration\nNickel is an important strategic metal widely used in stainless steel, high-temperature alloys, and battery materials. Accurate determination of its content is essential for quality control. EDTA complexometric titration is commonly used to determine nickel, but the reaction between $Ni^{2+}$ and EDTA is slow. Back titration (adding excess EDTA first and then back-titrating with a standard $Cu^{2+}$ solution) is therefore often used to improve endpoint sensitivity. This problem concerns the preparation of standard solutions, the principle of back titration, and the elimination of interferences.", "header_images": [], "subquestions": [{"sub_id": "Exam-22-T2-1", "tags": ["2.1"], "text_open": "2.2 Sample Decomposition\nWeigh $m$ = 0.3000 g of the sample into a 250 mL polytetrafluoroethylene beaker. Add 15 mL of concentrated nitric acid and 10 mL of concentrated hydrochloric acid, and heat at low temperature on a hot plate until no obvious reaction remains. If the silicon content of the sample exceeds 1%, add 10–15 drops of concentrated hydrofluoric acid to assist dissolution. Add 10 mL of perchloric acid and continue heating at low temperature until the sample is completely decomposed. Fume with perchloric acid until nearly dry, then remove from the heat. Add 10 mL of hydrochloric acid (1+1), rinse down the beaker wall with water, add water to approximately 50 mL, and gently boil to dissolve the salts. Cool to room temperature under running water, transfer into a 200 mL volumetric flask, dilute to the mark with water, and mix thoroughly.\nIf glassware is used during decomposition, additional Si will be introduced. Write the reaction equation involving $SiO_2$ in the glass.", "text_images": [], "full_mark": 2}]} | |
| {"problem_id": "Exam-22-T3", "header": "Problem 3 Body-Centered Unit Cells and Interstitial Sites", "header_images": [], "subquestions": [{"sub_id": "Exam-22-T3-1", "tags": ["1.3"], "text_open": "3.3 Changing the metal species in the tetragonal crystal above and introducing a slight distortion gives a monoclinic crystal. Its projection along the $b$ direction is shown below, with black spheres representing chromium atoms, gray spheres representing manganese atoms, and white spheres representing phosphorus atoms.\n3.3.1 Give the chemical formula of this crystal.", "text_images": [{"file": "Exam-22-T3-1-1-prompt.png", "description": "Projection of the monoclinic crystal along the $b$ direction. The main drawing is enclosed by a thin solid black, inclined parallelepiped wireframe representing the unit-cell boundary. Its bottom edge is horizontal, and its inclined edges run from upper left to lower right with a slope of approximately -1. The bottom edge is approximately equal in length to its corresponding height. The positive $c$ direction runs from left to right, while the positive $a$ direction runs from lower right to upper left along an inclined edge. The drawing contains 12 solid black spheres representing chromium atoms, 12 gray spheres representing manganese atoms, and 8 white spheres representing phosphorus atoms. Short line segments connect the spheres into a network. The centers of the gray spheres have the following $(a,c)$ coordinates: (0.66936, 0.94746), (0.33064, 0.05254), (0.53460, 0.39020), (0.46540, 0.60980), (0.35230, 0.32464), (0.64770, 0.67536), (0.16936, 0.94746), (0.83064, 0.05254), (0.03460, 0.39020), (0.96540, 0.60980), (0.85230, 0.32464), and (0.14771, 0.67536). The centers of the black spheres have the following $(a,c)$ coordinates: (0.10287, 0.18012), (0.89713, 0.81988), (0.57754, 0.97218), (0.42246, 0.02781), (0.22212, 0.39334), (0.77788, 0.60666), (0.60287, 0.18012), (0.39713, 0.81988), (0.07754, 0.97218), (0.92246, 0.02781), (0.72212, 0.39334), and (0.27788, 0.60666). The centers of the white spheres have the following $(a,c)$ coordinates: (0.53944, 0.33443), (0.46056, 0.66557), (0.29436, 0.24658), (0.70564, 0.75341), (0.03944, 0.33443), (0.96056, 0.66557), (0.79436, 0.24658), and (0.20564, 0.75341). A coordinate legend appears at the lower left: a small solid green circle bears a white lowercase $b$; a red arrow pointing toward the upper left is labeled with a lowercase $a$; and a blue arrow pointing right is labeled with a lowercase $c$. Apart from the legend, the figure contains no text."}], "full_mark": 3}]} | |
| {"problem_id": "Exam-22-T6", "header": "Problem 6 Heteroatoms (23 points, 9%)\nIntroducing heteroatoms often changes chemical reactivity. The following questions examine aromatic substitution, ylide reactions, and amide reduction.", "header_images": [], "subquestions": [{"sub_id": "Exam-22-T6-1", "tags": ["2.7"], "text_open": "6.1 First consider substitution reactions of aromatic systems. Compounds A and B are separately treated with persulfate under basic conditions.\n6.1.1 Is the major substitution product of each of A and B the ortho or the para product?", "text_images": [{"file": "Exam-22-T6-1-1-prompt.png", "description": "The SMILES representation of A is '[O-]C1=CC=CC=C1', and that of B is 'NC1=CC=CC=C1'. A boxed letter A or B appears below the corresponding structure. To the right of each structure is a single-headed arrow pointing right, with $S_2O_8^{2-}$ written above it; the product is not shown. All bonds are solid lines, and no stereochemistry is indicated."}], "full_mark": 2}, {"sub_id": "Exam-22-T6-2", "tags": ["2.2"], "text_open": "6.1.2 Write the SMILES string for 1 key intermediate in the formation of the ortho product.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-22-T6-3", "tags": ["2.4", "2.2"], "text_open": "6.2.2 After comparing the ylide reactions above, consider two related transformations. E reacts separately with a carbonyl compound (an aldehyde or ketone) and with an imine, giving an alkene in each case.\nWrite the SMILES strings for F and at least 3 key intermediates in its formation. Note: some intermediates contain a five-membered ring, and sulfur is present in the products only as elemental sulfur.", "text_images": [{"file": "Exam-22-T6-3-1-prompt.png", "description": "The figure shows two parallel reactions: 'CN(C)/C(SCC(OC)=O)=[N+](C)/C'.'[H]C(C1=CC=CC=C1)=O'>BTMG, THF, -78 ℃, 2h>[F]; and 'CN(C)/C(SCC(OC)=O)=[N+](C)/C'.'[H]/C(C1=CC=CC=C1)=N\\S(C2=CC=C(C)C=C2)(=O)=O'>BTMG, CHCl3, -78 ℃ to rt, 2h>[G]. The letter E is shown below each 'CN(C)/C(SCC(OC)=O)=[N+](C)/C' structure. F and G are represented only by boxed letters, and their structures are not shown. In both reactions, the counterion of E is NTf_2^-, whose structure is not shown."}], "full_mark": 9}]} | |
| {"problem_id": "Exam-22-T7", "header": "Problem 7 Skeletal Rearrangements of Maltol Derivatives\nThe three-dimensional construction of molecular frameworks is crucial in biochemistry and pharmaceutical research. Complex frameworks generally require changing catalysts or reagents, or even carrying out several separate reaction steps. In the reactions below, researchers investigated cascade skeletal rearrangements of maltol derivatives in depth. Simple modification of the substrate affords products with entirely different frameworks, ranging from planar to three-dimensional structures.", "header_images": [], "subquestions": [{"sub_id": "Exam-22-T7-1", "tags": ["2.2", "2.4"], "text_open": "7.1 On heating, maltol ether A undergoes rearrangement to give the key intermediate X. Trapping X with a nucleophile affords product B. When no nucleophile is added, the dimerization product C of X is also detected.\n7.1.1 Formation of the key intermediate X is known to involve only pericyclic reactions and simple tautomerization. Write the SMILES strings for 3 key intermediates in the formation of B. Stereochemistry is not required.", "text_images": [{"file": "Exam-22-T7-1-1-prompt.png", "description": "The figure shows a one-step reaction: 'O=C1C(OCC#C[H])=C(C)OC=C1'>BnSH, DCE, 150℃, 13 h>'OC1=C(O)C(C)=CC(CSCC2=CC=CC=C2)=C1C=O'. The structure on the left is labeled A, and that on the right is labeled B."}, {"file": "Exam-22-T7-1-2-prompt.png", "description": "The figure shows a one-step reaction: 'O=C1C(OCC#C[H])=C(C)OC=C1'>PhCl, 150℃, 13 h>'CC1=C2C(CCC3=C2C(C)=C(O)C(O)=C3C=O)=C(C=O)C(O)=C1O'. The structure on the left is labeled A, and the product on the right is labeled C."}], "full_mark": 4}, {"sub_id": "Exam-22-T7-2", "tags": ["2.2", "2.4"], "text_open": "7.1.2 Write the SMILES string for one key intermediate in the formation of C from X. Stereochemistry is not required.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-22-T7-3", "tags": ["2.4"], "text_open": "7.2 Replacing the nucleophile with styrene gives a reaction similar to that above and affords product D, which contains a five-membered ring. Write the SMILES string for D. Stereochemistry is not required.", "text_images": [{"file": "Exam-22-T7-3-1-prompt.png", "description": "The figure shows a one-step reaction: 'O=C1C(OCC#C[H])=C(C)OC=C1'>'C=CC1=CC=CC=C1' (styrene), 150℃, 13 h>[D]. The structure on the left is labeled A. On the right, only the molecular formula C17H16O3 of product D is given; its structure is not shown."}], "full_mark": 2}, {"sub_id": "Exam-22-T7-4", "tags": ["1.2", "2.3"], "text_open": "7.3 Introducing an allyl group into the maltol substrate above gives substrate E. Under similar conditions, E affords product F and the product G with a new framework. (PMB denotes the p-methoxybenzyl group.)\n7.3.1 The formation of G is known to proceed through an intermediate similar to X. Analyze the three-dimensional conformation of the favored transition state for the step that forms G, and use S/R notation to assign the configuration of the carbon atom in G whose stereochemistry is not labeled.", "text_images": [{"file": "Exam-22-T7-4-1-prompt.png", "description": "The figure shows a one-step reaction: 'O=C1C(OCC#CCC(C)=C)=C(C)OC=C1'>DCE, PMBOH, 150℃, 13 h>'OC1=C(O)C(C)=C(CC(C)=C)C(COCC2=CC=C(OC)C=C2)=C1C=O'+'O=C1C2(O)[C@@](C(CC(C2)=C)=C=C)(C)OC=C1'. The structure on the left is labeled E, and the two products on the right are labeled F and G, respectively."}], "full_mark": 2}, {"sub_id": "Exam-22-T7-5", "tags": ["2.4"], "text_open": "7.3.2 If substrate E in the reaction above is replaced by E’ shown below, determine whether the yield of G’, which has the same framework as G, will increase or decrease. Briefly explain your answer.", "text_images": [{"file": "Exam-22-T7-5-1-prompt.png", "description": "The SMILES representation of E’ is 'O=C1C(OCC#CCC(C)=C)=C(C(C)C)OC=C1'. The structure is labeled E’."}], "full_mark": 3}, {"sub_id": "Exam-22-T7-6", "tags": ["2.3", "2.4"], "text_open": "7.4 Introducing a new group into the substrate allows further cascade cyclization based on the fundamental reaction mode above.\n7.4.1 The following reaction undergoes a Nazarov-type cyclization to form the aromatic product I, which contains multiple fused rings. Write the SMILES string for I and explain why no external Lewis acid catalyst is required.", "text_images": [{"file": "Exam-22-T7-6-1-prompt.png", "description": "The figure shows a one-step reaction: 'O=C1C(OCC#CC2=CSC=C2)=C(C)OC=C1'>DCE, 150℃, 13 h>[I]. The structure on the left is labeled S. On the right, only the molecular formula C13H10O3S of product I is given; its structure is not shown."}], "full_mark": 4}, {"sub_id": "Exam-22-T7-7", "tags": ["2.2", "2.3"], "text_open": "7.4.2 When the substrate shown below is reacted under similar conditions, no product analogous to that in 7.4.1 is observed. Instead, two products, L and O, with complex three-dimensional frameworks are obtained.\nIntermediate J forms by the same process as X. J then undergoes two consecutive concerted reactions, both through six-membered cyclic transition states, to give product L. Alternatively, J can first form M, which undergoes an electrocyclic reaction to give the charge-separated intermediate N. Group migration in N then affords the bridged product O. Based on this information, write the SMILES strings for K, N, and O in the scheme.", "text_images": [{"file": "Exam-22-T7-7-1-prompt.png", "description": "The figure shows a multistep reaction sequence: 'O=C1C(OCC#CC2=C(C=C)C=CC=C2)=C(C)OC=C1'>PhCl, 150℃, 13 h>[J]; [J]>[K]; [K]>'O=C1[C@]([C@@]2(C)OC=C1)(O)CC3=C2C4=CC=CC=C4C=C3' (L); [J]>'C=C1CC(O/C2=C/C=O)C(C=CC=C3)=C3/C1=C(C)/C2=O' (M); [M]>[N]; [N]>[O]. J, K, N, and O are shown only as letter labels, whereas the structures of L and M are drawn. The [C@]/[C@@] descriptors in the SMILES representation of L correspond to the stereochemical orientations shown by wedged bonds in the figure."}], "full_mark": 6}]} | |
| {"problem_id": "Exam-22-T8", "header": "Problem 8 Cascade Cyclizations Involving \\alpha-Oxoketenes (25 points, 11%)\n\\alpha-Oxoketenes are highly reactive intermediates that readily react with various dienes, dipoles, and nucleophiles. Unlike ordinary ketenes, \\alpha-oxoketenes favor [4+2] rather than [2+2] cyclization reactions.", "header_images": [{"file": "Exam-22-T8-header-1.png", "description": "General reaction scheme for an \\alpha-oxoketene with nucleophiles and unsaturated components, using R1, R2, Nu, Y, and Z as placeholders. The \\alpha-oxoketene has SMILES '[R2]C(C([R1])=C=O)=O'. Three arrows originate from this structure. The first points to a molecule with SMILES '[R2]C(C([R1])C([Nu])=O)=O', with NuH written above the arrow. The second points to a molecule with SMILES '[R2]C(C(CC=[Z][Y]1)([R1])C1=O)=O', with a molecule having SMILES 'C=C[Z]=[Y]' shown above the arrow. The third points to a molecule with SMILES '[R2]C(O[Y][Z]1)=C([R1])C1=O', with a molecule having SMILES '[Z]=[Y]' shown above the arrow."}], "subquestions": [{"sub_id": "Exam-22-T8-1", "tags": ["2.2", "2.4"], "text_open": "8.1.1 The Wolff rearrangement is one of the best methods for generating \\alpha-oxoketenes in situ. The resulting intermediate can be trapped immediately by various nucleophiles or participate in cycloaddition reactions, rapidly increasing molecular complexity. For example, dimedone A gives diazoketone B as shown, which rearranges to the corresponding \\alpha-oxoketene C. Write the SMILES string for one key electrically neutral intermediate M in the conversion of A to B, and write the SMILES string for C.", "text_images": [{"file": "Exam-22-T8-1-1-prompt.png", "description": "The figure shows a two-step reaction: 'O=C1CC(C)(C)CC(C1)=O'>TsN3, NEt3, MeCN, rt, 13 h>'O=C1CC(C)(C)CC(C1=[N+]=[N-])=O'. Only “heat” is written above the second arrow, and the product is labeled “\\alpha-oxo ketene C” without a drawn structure."}], "full_mark": 2}, {"sub_id": "Exam-22-T8-2", "tags": ["2.4"], "text_open": "8.1.4 Diazoketone B generally undergoes thermolysis to form a highly reactive \\alpha-oxoketene, which then reacts with nucleophiles. However, B can also add to nucleophiles at room temperature. For example, reaction of A and B in aqueous ammonia gives the tetracyclic compound G, which contains two unusual six-membered rings. The $^{1}$H NMR spectrum of G shows four sets of resonances at $\\delta$ 12.4, 2.6, 2.5, and 1.1. Write the SMILES string for G.", "text_images": [{"file": "Exam-22-T8-2-1-prompt.png", "description": "The figure shows a one-step reaction: 'O=C1CC(C)(C)CC(C1)=O'.'O=C1CC(C)(C)CC(C1=[N+]=[N-])=O'>NH3·H2O>[G]. G is shown only as a label; its structure is not drawn."}], "full_mark": 2}, {"sub_id": "Exam-22-T8-3", "tags": ["2.4"], "text_open": "8.2 Reactions of \\alpha-Oxoketenes (I)\nThe 1,5-benzodiazepin-2-one, or diazepine-type, framework is an important pharmacologically active structure. The following method constructs this framework in a single step.\n8.2.1 An \\alpha-oxoketene contains two electrophilic carbon atoms in a 1,3 relationship and can react with a 1,4-dinucleophile to give the seven-membered diazepine-type compound H1. The reaction is performed under microwave irradiation without additives, proceeds within minutes, and gives moderate to good yields. Write the SMILES string for H1.", "text_images": [{"file": "Exam-22-T8-3-1-prompt.png", "description": "The figure shows a one-step reaction: 'NC1=CC=CC=C1N'.'O=C1CC(C)(C)CC(C1=[N+]=[N-])=O' (B, 1 eq)>Toluene, microwave, 180 °C, 6 min>[H1]. H1 is shown only as a label; its structure is not drawn."}], "full_mark": 2}, {"sub_id": "Exam-22-T8-4", "tags": ["2.4", "2.3"], "text_open": "8.2.2 When the reaction is performed with a seven-membered cyclic ketone diazo compound, the expected product is not obtained; instead, the rearranged imidazolone product H2 forms. Write the SMILES string for H2. Note: lowering the temperature to 130℃ gives reaction product H3, whose structure is similar to H1, and H2 and H3 are constitutional isomers.", "text_images": [{"file": "Exam-22-T8-4-1-prompt.png", "description": "The figure shows a one-step reaction: 'NC1=CC=CC=C1N'.'O=C1CCCCC(C1=[N+]=[N-])=O' (seven-membered cyclic ketone diazo compound, 1 eq)>Toluene, microwave, 180 °C, 3 min>[H2]. H2 is shown only as a label; its structure is not drawn."}], "full_mark": 2}, {"sub_id": "Exam-22-T8-5", "tags": ["2.4", "4.3"], "text_open": "8.3 Reactions of \\alpha-Oxoketenes (II)\n8.3.1 Diels-Alder reaction product: In the following reaction, two reactive species undergo a Diels-Alder reaction to give product J. The IR spectrum of J shows the following peak wavenumbers, IR (neat): 2956, 2866, 1759, 1692, 1409, 1342, 1245, 1144, and 697 cm$^{-1}$. Write the SMILES string for product J.", "text_images": [{"file": "Exam-22-T8-5-1-prompt.png", "description": "The figure shows a one-step reaction: 'O=C1CC(C)(C)CC(C1=[N+]=[N-])=O'.'O=C(N=[N+]=[N-])C1=CC=CC=C1' (5 eq)>Toluene, 130 °C>[J]. J is shown only as a label; its structure is not drawn."}], "full_mark": 2}, {"sub_id": "Exam-22-T8-6", "tags": ["2.4"], "text_open": "8.3.2 Cascade ring closure involving Diels-Alder reactions. The following transformation undergoes one intermolecular Diels-Alder reaction followed by one intramolecular Diels-Alder reaction, constructing the hexacyclic compound K in a single operation. Write the SMILES string for K. Stereochemistry is not required.", "text_images": [{"file": "Exam-22-T8-6-1-prompt.png", "description": "The figure shows a multicomponent reaction: 'CC(CC(C1=[N+]=[N-])=O)(C)CC1=O'.'[H]C(C1=NC(C2=CC=CC=C2)=CO1)=O'.'NCC=C'>Toluene, microwave, 140 °C, 5 min>[K]. K is shown only as a label; its structure is not drawn."}], "full_mark": 2}, {"sub_id": "Exam-22-T8-7", "tags": ["2.2", "2.3", "2.4"], "text_open": "8.3.3 Spirocyclic ring systems can be constructed by reactions of ketenes with imines. The potential-energy diagram for the following reaction is shown below.\nIn the formation of M from C and L, 3 key intermediates, INT1, INT2, and INT3, are involved. They are constitutional isomers, and conversion of INT1 to INT2 involves an s-trans to s-cis conformational change of the diene. Write the SMILES strings for INT1, INT2, and INT3, and identify the type of pericyclic reaction by which INT2 forms product M.", "text_images": [{"file": "Exam-22-T8-7-1-prompt.png", "description": "Reaction potential-energy diagram. The vertical axis is labeled “Relative energy” in kJ/mol, with tick marks at -75, -50, -25, 0, 25, and 50. The horizontal axis is labeled “Reaction coordinate.” In the legend, a solid line denotes pathway I and a dashed line denotes pathway II. Solid pathway I starts at 0 kJ/mol, crosses a 44.8 kJ/mol barrier to intermediate INT1 at 7.6 kJ/mol, crosses a 49.2 kJ/mol barrier to INT2 at 24.2 kJ/mol, and then crosses a 56.9 kJ/mol barrier to the product at -73.4 kJ/mol. Dashed pathway II starts from INT1, crosses a 27.9 kJ/mol barrier to INT3 at -17.3 kJ/mol, and then joins the product at -73.4 kJ/mol."}, {"file": "Exam-22-T8-7-2-prompt.png", "description": "The figure shows a one-step reaction: 'CC(CC(C1=[N+]=[N-])=O)(C)CC1=O'.'C/N=C/C=C/C1=CC=CC=C1'>Toluene, 160 °C, 15 min>'O=C(CC(C)(C)C1)[C@@]21[C@@H](C3=CC=CC=C3)C=CN(C)C2=O'."}], "full_mark": 7}]} | |
| {"problem_id": "Exam-23-T1", "header": "Question 1 Chemistry of Element X (28 points, 12%)\nX was first discovered in 1898 and is widely used in the electronics and electric-light-source industries. It is relatively unreactive and exhibits several oxidation states in its compounds. Common fluorides of X include A, B, and C, all colorless crystals at room temperature, with the mass fraction of fluorine increasing in that order.", "header_images": [], "subquestions": [{"sub_id": "Exam-23-T1-1", "tags": ["2.1"], "text_open": "1.1 A, B, and C can all be prepared by reacting elemental X with fluorine under suitable conditions. A is soluble in water and reacts slowly with it. At 275 ℃, A reacts with CrO_2F_2 to form D, which contains 41.6% Cr. B hydrolyzes rapidly in water and undergoes disproportionation to produce two gases; at 273.15 K and 100 kPa, the gas mixture has a density of 3.91 g·L^-1. Complete hydrolysis of C gives an explosive oxide of X.\n1.1.2 Write the equations for the reaction of A with $CrO_2F_2$ to form D and for the reaction of B with water.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-23-T1-2", "tags": ["2.1"], "text_open": "1.1.3 Dry A and B can both be stored in glass containers, but placing C in a glass or quartz container, whose principal component may be taken as $SiO_2$, is highly dangerous. Use a chemical equation to explain why C cannot be stored in a quartz container.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-23-T1-3", "tags": ["1.4", "1.1"], "text_open": "1.2 In addition to forming covalent compounds with elements such as fluorine and oxygen, X forms a rich variety of ionic compounds. The earliest reported X-containing anion can be prepared as follows: C hydrolyzes and disproportionates under basic conditions to give two gases and octahedral anion E. In addition, C reacts with an equimolar amount of CsF to give F; heating F converts it to G, in which the mass fraction of X is 23.91%.\n1.2.2 In the X-containing anions above, the X–O and X–F distances are close to the sums of the corresponding covalent radii, indicating covalent bonding. In some X-containing anions discovered later, however, the X–halogen distance is substantially longer than the sum of the covalent radii but shorter than the sum of the van der Waals radii. To explain this observation, scientists proposed the concept of an X bond (AeB). When X is bonded to a highly electronegative element or electron-withdrawing group, its electron distribution changes markedly, creating a region of high electrostatic potential along the extension of the $\\sigma$-bond axis, a $\\sigma$ hole. X thereby acts as an electrophilic center that can interact with nucleophiles such as halide ions.\n1.2.2.1 F hydrolyzes slowly in moist air to give H, whose formula may be regarded as an adduct of an oxide of X with $CsF$. In the crystal, the anion of H forms an infinite one-dimensional chain. Write its SMILES string.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-23-T1-4", "tags": ["1.4", "1.1"], "text_open": "1.2.2.2 In the presence of $CsF$, partial hydrolysis of C gives I. I contains 47.93% X by mass, and all X atoms have the same chemical environment. Write the SMILES string for the anion of I.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-23-T1-5", "tags": ["3.2"], "text_open": "1.3 The physical properties of A, B, and C have also been studied extensively. Researchers measured the standard equilibrium constants of the following reactions at several temperatures; all species are gaseous and treated as ideal gases.\n\n| Temperature (℃) / Reaction | 250 | 300 | 350 | 400 |\n| --- | --- | --- | --- | --- |\n| $X + F_2 \\rightarrow A$ | $8.79\\times10^{4}$ | $1.02\\times10^{4}$ | $1.67\\times10^{3}$ | $3.59\\times10^{2}$ |\n| $A + F_2 \\rightarrow B$ | $1.43\\times10^{3}$ | $1.55\\times10^{2}$ | 27.2 | 4.86 |\n| $B + F_2 \\rightarrow C$ | 0.944 | 0.211 | 0.0558 | 0.0182 |\n\n1.3.1 Calculate $\\Delta_r H_m^θ$ and $\\Delta_r S_m^θ$ for each reaction, assuming $\\Delta_r H_m^θ$ and $\\Delta_r S_m^θ$ are temperature-independent.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-23-T1-6", "tags": ["3.4"], "text_open": "1.3.3 At 300 ℃, 1 mol of X and 5 mol of $F_2$ are introduced into a 1 L vessel. Calculate the mole fraction of C in the equilibrium gas mixture.", "text_images": [], "full_mark": 6}]} | |
| {"problem_id": "Exam-23-T3", "header": "Question 3 Reversible-Deactivation Anionic Polymerization (8 points, 4%)\nConventional anionic polymerization of methacrylates, such as ethyl methacrylate shown below, requires highly hazardous organolithium initiators. Because the chain end is extremely reactive, the reaction must be conducted at very low temperature, such as -78 ℃, under rigorously anhydrous and oxygen-free conditions; otherwise severe side reactions terminate the chains.", "header_images": [{"file": "Exam-23-T3-header-1.png", "description": "The SMILES representation of ethyl methacrylate (EMA) is 'O=C(OCC)C(C)=C'."}], "subquestions": [{"sub_id": "Exam-23-T3-1", "tags": ["2.2"], "text_open": "3.1 Chemists recently developed a $CO_2$-mediated reversible-deactivation anionic polymerization (CMAP). The method operates at 90 ℃, uses a stable and safe solid potassium salt as initiator, and produces polymers with a very narrow molecular-weight distribution, or very low dispersity $Đ$, under a $CO_2$ atmosphere. The key conditions are shown below.\n3.1.1 Anionic polymerization initiator A first decarboxylates to form an enolate that initiates polymerization. Write the SMILES string for A.", "text_images": [{"file": "Exam-23-T3-1-1-prompt.png", "description": "'O=C(C(C)=C)OCC'>1.0 equiv. A; 18-crown-6 (1.2 equiv.); TEMPO (0.1 equiv.); 1 atm $CO_2$; DMF, 90°C, 6 h>'O=C(C(CC(C(OCC)=O)(C)C(O[K])=O)(C)C)OCC'. The product is poly(ethyl methacrylate), poly-EMA. Its repeating unit is enclosed in brackets with subscript n at the lower right. The listed product SMILES schematically represents an EtO-C(=O)-C(CH3)2- initiating end connected through -CH2- to a dormant-end unit -C(CH3)(CO2Et)(CO2K)."}], "full_mark": 1}, {"sub_id": "Exam-23-T3-2", "tags": ["2.8"], "text_open": "3.1.2 State the principal chemical role of 18-crown-6 (shown as 18-crown-6 in the reaction scheme) in the system.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-23-T3-3", "tags": ["2.4", "2.3"], "text_open": "3.2 Control of side reactions is also an important topic in anionic polymerization.\n3.2.1 In conventional high-temperature anionic polymerization without $CO_2$ protection, the active chain end of a polymethacrylate readily attacks an upstream carbonyl group in the polymer backbone. This intramolecular backbiting completely terminates the chain. Write the SMILES string for the final cyclic end group formed after backbiting termination of active poly(ethyl methacrylate), poly-EMA; only one chain end need be shown.", "text_images": [], "full_mark": 2}]} | |
| {"problem_id": "Exam-23-T4", "header": "Question 4 Determination of Aluminum by Titration (25 points, 10%)\nAluminum is an important component of light alloys, and its content directly affects strength, corrosion resistance, and processability. EDTA displacement titration is a classical method for determining aluminum. Excess EDTA is first added to complex aluminum ions, and the excess is back-titrated with standard zinc solution. Fluoride is then added to displace EDTA from Al–EDTA, and the released EDTA is titrated with standard zinc solution to determine the aluminum content.", "header_images": [], "subquestions": [{"sub_id": "Exam-23-T4-1", "tags": ["2.1"], "text_open": "4.1 Preparation of Standard Solutions. Disodium ethylenediaminetetraacetate (EDTA) solution (0.02 mol/L): Weigh 7.4448 g EDTA into a 400 mL beaker, dissolve in water, transfer to a 1000 mL volumetric flask, dilute to the mark, and mix. Standard zinc solution (0.02000 mol/L): Weigh 1.6276 g of zinc oxide (primary standard reagent) that has been dried to constant mass at 110 ℃ into a 400 mL beaker. Add 20 mL hydrochloric acid (1 + 1) and heat to dissolve. Add approximately 200 mL water and then add aqueous ammonia (1 + 1) dropwise until Congo red paper changes from blue to violet. Cool, transfer to a 1000 mL volumetric flask, dilute to the mark with water, and mix. 4.2 Dissolution of the Sample. Accurately weigh 1.0008 g of the alloy sample into a 400 mL glass beaker. Add 20 mL~30 mL hydrochloric acid (1 + 1) and heat to dissolve; for samples that are difficult to dissolve, an appropriate amount of concentrated hydrochloric acid or concentrated nitric acid may be added. Add 5 mL concentrated nitric acid dropwise for oxidation, then add 10 mL~30 mL sulfuric acid (1 + 1). Evaporate until sulfuric acid fumes have evolved for 1 min. Remove from heat and cool slightly, add 50 mL water, and heat to dissolve the salts. 4.3 Separation of the Precipitate. Acidity adjustment: Add 100 mL of complexing-agent solution (containing 200 g/L ammonium citrate and 25 g/L ammonium oxalate) to the test solution, and adjust the pH to 4~5 with aqueous ammonia (1 + 1) and sulfuric acid (1 + 8). If a precipitate forms in a tungsten-containing sample, first add concentrated aqueous ammonia until basic and heat to dissolve the tungstic acid precipitate. After cooling, adjust the pH to 4~5 with sulfuric acid (1 + 1) and aqueous ammonia (1 + 1). If the test solution is turbid, heat it to boiling until clear, cool, and readjust the pH to 4~5. Precipitation and dissolution: Transfer the test solution to a 400 mL plastic beaker, add a small amount of paper pulp, and add 100 mL sodium fluoride solution (35 g/L) with stirring. Stir thoroughly for 5 min and let stand for 40 min. Filter through slow filter paper containing a small amount of paper pulp, wash 7~8 times with sodium fluoride solution (5 g/L), and discard the filtrate. Return the precipitate and filter paper to the original glass beaker and immediately add 20 mL hydrochloric acid-boric acid mixed acid (boric acid:hydrochloric acid:water = 15:25:60) to dissolve the precipitate; it must be added immediately, or complete dissolution will be difficult. Add 20 mL~25 mL concentrated nitric acid and boil without covering with a watch glass to destroy the filter paper. Cool slightly, add 7 mL concentrated perchloric acid, and heat to near dryness with fuming. Cool slightly and add 50 mL water to dissolve the salts.\n4.3.3 Write the ionic equation for addition of sodium fluoride in the “Separation of the Precipitate” step. The purpose of adding paper pulp during filtration is (select one).\nA. Increase precipitate purity\nB. Loosen the gelatinous precipitate and prevent passage through the filter\nC. Act as an indicator\nD. Adsorb impurity ions", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-23-T4-2", "tags": ["2.1"], "text_open": "4.3.4 Write the ionic equation for dissolution of the precipitate by the hydrochloric acid–boric acid mixture.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-23-T4-3", "tags": ["4.5"], "text_open": "4.4 Titration\nAccurately add 20.00 mL EDTA solution to the test solution and boil for 2–3 min. Add 2 drops of 2,4-dinitrophenol solution (2 g/L), then add ammonia (1 + 1) dropwise until the solution is pale yellow. Continue boiling for 1–2 min and cool; if the yellow color fades, add ammonia dropwise until pale yellow again. Add 20 mL acetate–ammonium acetate buffer (pH 5.0–5.5) and 4 drops of xylenol orange solution (2 g/L). Titrate with standard zinc solution until violet-red, without recording the reading. Add 1–2 g sodium fluoride, boil for 2–3 min, and cool to room temperature. If the solution is violet-red, add hydrochloric acid (1 + 1) dropwise until the color just disappears. Add another 10 mL acetate–ammonium acetate buffer and 1–2 drops of xylenol orange solution. Titrate with standard zinc solution to a violet-red endpoint; 12.66 mL is consumed.\n4.4.1 Does the EDTA solution require standardization in this experiment? Briefly explain.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-23-T4-4", "tags": ["2.1"], "text_open": "4.4.2 Write the ionic equation for addition of sodium fluoride followed by boiling, using $H_2Y^{2-}$ for the doubly charged ethylenediaminetetraacetate ion.", "text_images": [], "full_mark": 2}]} | |
| {"problem_id": "Exam-23-T5", "header": "Problem 5 Water-Molecule Cages (24 points, 10%)\nThe cubic unit-cell structure of the alloy $Zr_3Au$ is shown below. Au atoms occupy the corners and body center of the cell. Each face contains 2 Zr atoms arranged symmetrically along a direction parallel to a coordinate axis (their three coordinates are 0, 1/2, and either 1/4 or 3/4).", "header_images": [{"file": "Exam-23-T5-header-1.png", "description": "Schematic cubic unit cell of the alloy $Zr_3Au$, drawn as a thin gray wireframe cube. Two kinds of glossy spheres are shown: dark, nearly black spheres near the cube corners and body center, and light gray-white spheres arranged symmetrically near the faces. Thin gray lines connect the spheres. The spheres carry no labels. At lower left is a three-colored coordinate triad: a red arrow points down-left and is labeled with a lowercase a, a green arrow points horizontally right and is labeled b, and a blue arrow points vertically upward and is labeled c. No other text appears in the image."}], "subquestions": [{"sub_id": "Exam-23-T5-1", "tags": ["1.3"], "text_open": "5.1 State the lattice type of crystalline $Zr_3Au$.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-23-T5-2", "tags": ["1.3"], "text_open": "5.2 State the number of nearest, equidistant Zr atoms around each Au atom.", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-23-T5-3", "tags": ["1.3"], "text_open": "5.3 For each Zr atom, state separately the number of nearest, equidistant Au atoms, the number of nearest, equidistant Zr atoms, and the number of next-nearest, equidistant Zr atoms.", "text_images": [], "full_mark": 3}, {"sub_id": "Exam-23-T5-4", "tags": ["1.3"], "text_open": "5.4 Treat the Zr atoms described in 5.2 as the coordinating atoms of Au, and the Zr and Au atoms described in 5.3 as the coordinating atoms of Zr. For each coordination polyhedron in the crystal, connecting its center to all of its vertices produces a number of small tetrahedra. State:\n(1) the number of faces of the coordination polyhedron around an Au atom;\n(2) the number of faces of the coordination polyhedron around a Zr atom;\n(3) the number of coordination polyhedra that contain each small tetrahedron;\n(4) the number of small tetrahedra in one unit cell.", "text_images": [], "full_mark": 7}, {"sub_id": "Exam-23-T5-5", "tags": ["1.3"], "text_open": "5.5 Replace every atomic position in $Zr_3Au$ with a methane molecule and place one water molecule at the center of each small tetrahedron described above. The resulting structure is that of methane hydrate. Write the formula of methane hydrate in the form $xCH_4\\cdot yH_2O$.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-23-T5-6", "tags": ["1.3", "2.1"], "text_open": "5.6 The unit-cell parameter of methane hydrate is $a = 1.20\\ \\mathrm{nm}$. Calculate the maximum theoretical volume of methane gas that can be released from $1\\ \\mathrm{m^3}$ of methane hydrate at 273 K and 1 atm.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-23-T5-7", "tags": ["1.3"], "text_open": "5.7 The cubic crystal of another similar hydrate can be described as follows. Hexadecahedra $[5^{12}6^4]$ whose vertices are water molecules (the notation denotes 12 pentagonal faces and 4 hexagonal faces; the polyhedron has the same symmetry as a regular tetrahedron and is shown below) are joined pairwise by sharing hexagonal faces, forming an infinite diamond-like structure. These hexadecahedra clearly do not fill all space: the four tetrahedral-void positions that are unoccupied in diamond remain. In fact, exactly four pentagonal dodecahedra, also with water molecules at their vertices, fit into each such void. The dodecahedra share pentagonal faces with one another and with the hexadecahedra, and each water molecule in the final structure is shared by four polyhedra. Place one $CCl_4$ molecule at the center of every hexadecahedron and one Xe atom at the center of every dodecahedron.\nStructure of the hexadecahedron $[5^{12}6^4]$\n5.7.1 Write the formula of this crystal in the form $xCCl_4\\cdot yXe\\cdot zH_2O$.", "text_images": [{"file": "Exam-23-T5-7-1-prompt.png", "description": "Wireframe representation of the hexadecahedron $[5^{12}6^4]$. Its vertices are shown as small hollow white circles with thin black outlines, and its edges as thin black lines. Edges facing the viewer are solid, while hidden rear edges are dashed. Alternating hexagonal and pentagonal faces form an approximately spherical cage. The image contains no text, numbers, arrows, or coordinate-axis labels."}], "full_mark": 3}]} | |
| {"problem_id": "Exam-23-T6", "header": "Problem 6 Determination of the Oxygen-Uptake Ratio of an Oxygen Carrier (23 points, 10%)\nA laboratory has prepared inactive Co(Salen) and wishes to determine the oxygen-uptake ratio $n_{O_2}:n_{[Co(II)(Salen)]}$ after the carrier has fully absorbed $O_2$ from its surroundings under specified conditions. Inactive Co(Salen) itself has almost no oxygen-uptake capacity, but it dissolves in coordinating solvents such as DMF or DMSO, and the resulting solution can absorb oxygen. The main procedure is described below:", "header_images": [{"file": "Exam-23-T6-header-1.png", "description": "Schematic apparatus for measuring the oxygen-uptake ratio, with no text, letters, numbers, or arrow labels. At left is a U-shaped gas burette. Its left arm is vertical and bears evenly spaced short, unnumbered graduations. Both arms contain liquid, with the level in the left arm higher than that in the right. In the center is a three-neck round-bottom flask containing liquid in its lower portion and resting above a rectangular base. The left neck connects to a downward-sloping bent addition tube that also contains liquid; its liquid level lies below that in the flask, and its downward-facing open end is not submerged. The middle neck connects to a long vertical tube that bends left at the top and enters the right arm of the gas burette; this connecting tube contains no liquid. The right neck connects to a two-way stopcock assembly with a spherical bulb above it, a short tube at the top, and two symmetric thin rods; this part contains no liquid."}], "subquestions": [{"sub_id": "Exam-23-T6-1", "tags": ["4.1"], "text_open": "6.1 Assembly of the Measurement Apparatus. Accurately weigh 0.1136 g Co(Salen) and transfer it through a dry glass funnel into a clean, dry three-neck flask. Assemble the apparatus as shown below. Apply a layer of petroleum jelly, as thin as possible, to the ground-glass joints of the bent tube, gas inlet tube, and two-way stopcock. Place 5 mL DMF (or DMSO) in the bent tube and connect the charging bent tube to the flask. Use a narrow-tip glass dropper to add 1 mL deionized water to the gas burette, whose graduations increase upward. Inflate the oxygen bag and connect it to the two-way stopcock on the right. Apply pressure to the oxygen bag so that air is expelled through the graduated gas burette. When the oxygen in the bag has been almost completely discharged, close the stopcock. Add approximately 5 mL more deionized water to the graduated gas burette so that the difference between the liquid levels exceeds 2 cm. Observe the liquid levels; the system may be considered to have reached thermal equilibrium when the change in liquid level is less than 0.03 mL over three consecutive minutes.\n6.2 Determination of the Oxygen-Uptake Ratio\nAfter assembling the apparatus, determine the oxygen-uptake ratio of the Co(Salen) sample as follows.\nRecord the laboratory temperature and atmospheric pressure. Remove the oxygen bag, open the stopcock to the atmosphere, and keep the gas burette vertical. When the liquid levels in the two arms are equal (difference less than 0.03 mL), close the stopcock and record the position of the liquid level in the gas burette. Then introduce DMF (or DMSO) into the flask, stir rapidly, and observe the changes in the reactants. Promptly add deionized water to the gas burette with a narrow-tip glass dropper to keep the liquid levels in its two arms equal. Record the liquid-level position every 20~40 seconds until ________. Calculate the actual volume of oxygen absorbed.\n6.2.1 State how the DMF (or DMSO) should be introduced into the flask.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-23-T6-2", "tags": ["4.2"], "text_open": "6.2.2 State the observation that should fill the blank in the procedure above.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-23-T6-3", "tags": ["4.7"], "text_open": "6.2.4 Suppose the preceding group cleaned all of the apparatus with a volatile organic solvent S, which has no coordinating ability, after performing the same experiment. Before starting, a student finds a small amount of liquid solvent S remaining in the three-neck flask. If the apparatus is used directly, will the measured oxygen-uptake ratio be higher or lower than the true value? Briefly explain.", "text_images": [], "full_mark": 3}]} | |
| {"problem_id": "Exam-23-T7", "header": "Problem 7 Reactivity, Preparation, and Applications of Silanes (32 points, 10%)\nSilanes ($Si_nH_{2n+2}$) and their halogenated derivatives are important precursors for high-purity silicon films, semiconductor devices, and solar cells. Unlike hydrocarbons, silanes contain highly reactive Si–Si and Si–H bonds, creating distinctive challenges in their synthesis and purification.", "header_images": [], "subquestions": [{"sub_id": "Exam-23-T7-1", "tags": ["1.2"], "text_open": "7.4.3 Mixing the tridentate ligand pedeta (pentaethyldiethylenetriamine, $Et_2NCH_2CH_2NEtCH_2CH_2NEt_2$) with trichlorosilane in DCM at room temperature gives A = $[(pedeta)H_2SiCl^+]_2[Si_6Cl_{14}^{2-}]$.\n7.4.3.1 How many stereoisomers can the cation of A theoretically have?", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-23-T7-2", "tags": ["1.4"], "text_open": "7.4.3.2 The anion of A has an approximate sixfold rotation axis. Write its SMILES string.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-23-T7-3", "tags": ["2.3"], "text_open": "7.4.3.3 Mechanistic experiments show that adding 1,3-butadiene to this reaction system does not produce detectable 1,1-dichlorosilacyclopent-3-ene. Which of the following is therefore the most likely key one-silicon intermediate?\nA. $:SiCl_2$\nB. $SiCl_3$\nC. $SiCl_3^-$\nD. $SiCl_4^-$", "text_images": [], "full_mark": 1}, {"sub_id": "Exam-23-T7-4", "tags": ["2.3"], "text_open": "7.5 Wurtz coupling. More complex silane molecules can be obtained through Wurtz coupling.\n7.5.1 Mechanistically, which class does the first reaction belong to?\nA. Nucleophilic aromatic substitution\nB. Electrophilic aromatic substitution\nC. Radical substitution\nD. Transition-metal-catalyzed reaction", "text_images": [{"file": "Exam-23-T7-4-1-prompt.png", "description": "The image shows two reaction sequences. The upper row contains three steps: 'C1([Si]2(C3=CC=CC=C3)[Si](C4=CC=CC=C4)(C5=CC=CC=C5)[Si](C6=CC=CC=C6)(C7=CC=CC=C7)[Si](C8=CC=CC=C8)(C9=CC=CC=C9)[Si]2(C%10=CC=CC=C%10)C%11=CC=CC=C%11)=CC=CC=C1'>CF3SO3H>'O=S(O[Si]1(C2=CC=CC=C2)[Si](C3=CC=CC=C3)(C4=CC=CC=C4)[Si](C5=CC=CC=C5)(C6=CC=CC=C6)[Si](C7=CC=CC=C7)(C8=CC=CC=C8)[Si]1(C9=CC=CC=C9)C%10=CC=CC=C%10)(C(F)(F)F)=O'; 'O=S(O[Si]1(C2=CC=CC=C2)[Si](C3=CC=CC=C3)(C4=CC=CC=C4)[Si](C5=CC=CC=C5)(C6=CC=CC=C6)[Si](C7=CC=CC=C7)(C8=CC=CC=C8)[Si]1(C9=CC=CC=C9)C%10=CC=CC=C%10)(C(F)(F)F)=O'>NaBr (3eq), Et2O>[B]; [B]>naphthalene (3eq), Li (6eq), DME, -30 °C>'C1([Si]2(C3=CC=CC=C3)[Si](C4=CC=CC=C4)(C5=CC=CC=C5)[Si]([Si]6(C7=CC=CC=C7)[Si](C8=CC=CC=C8)(C9=CC=CC=C9)[Si](C%10=CC=CC=C%10)(C%11=CC=CC=C%11)[Si](C%12=CC=CC=C%12)(C%13=CC=CC=C%13)[Si]6(C%14=CC=CC=C%14)C%15=CC=CC=C%15)(C%16=CC=CC=C%16)[Si](C%17=CC=CC=C%17)(C%18=CC=CC=C%18)[Si]2(C%19=CC=CC=C%19)C%20=CC=CC=C%20)=CC=CC=C1'. In the lower row, the starting material at left is omitted. An arrow labeled 'HCl (g), AlCl3' above and 'PhH, 50 °C' below leads to an intermediate labeled only C. C is the product obtained by treating the upper-row product (Si10Ph18) with HCl/AlCl3; the connection is omitted because it does not fit on one line. [C]>LiAlH4 (6eq), C6H14/Et2O, -40 °C>'[H][Si]1([H])[Si]([H])([H])[Si]([H])([Si]2([H])[Si]([H])([H])[Si]([H])([H])[Si]([H])([H])[Si]2([H])[H])[Si]([H])([H])[Si]1([H])[H]'."}], "full_mark": 1}, {"sub_id": "Exam-23-T7-5", "tags": ["2.4"], "text_open": "7.5.2 Write the SMILES strings for intermediates B and C.", "text_images": [], "full_mark": 4}, {"sub_id": "Exam-23-T7-6", "tags": ["4.3"], "text_open": "7.5.3 Identify which silicon atom in the product (fused cyclopentasilane) appears farthest upfield in the NMR spectrum, that is, at the smallest chemical shift δ.", "text_images": [], "full_mark": 1}]} | |
| {"problem_id": "Exam-23-T10", "header": "Problem 10 Dearomatization of Benzene Rings (27 points, 13%)\nAromaticity makes the benzene ring structurally stable and provides substantial resonance stabilization. Directly disrupting its aromatic conjugation or cleaving a C–C bond in the ring framework normally requires considerable energy. Dearomatization or ring-opening cleavage of a benzene ring is therefore highly challenging.", "header_images": [], "subquestions": [{"sub_id": "Exam-23-T10-1", "tags": ["2.2"], "text_open": "10.1 In 2024, researchers developed a novel benzene-ring-opening strategy that directly converts a range of aryl azide derivatives into valuable multiply conjugated enynes.\n10.1.1 Write the SMILES string for intermediate A formed before $^tBuONa$ is added.", "text_images": [{"file": "Exam-23-T10-1-1-prompt.png", "description": "Single-step reaction: 'O=C(C1=CC=CC=C1N=[N+]=[N-])C2=CC=CC=C2'>TosNHNH2 (1.2 equiv), TFA (20 mol%), PhCl (0.2 M); then $^tBuONa$ (1.5 equiv), PhCl (0.1 M), 130°C>'N#C/C=C\\C=C/C#CC1=CC=CC=C1'."}], "full_mark": 1}, {"sub_id": "Exam-23-T10-2", "tags": ["2.2", "2.3"], "text_open": "10.1.2 To investigate the mechanism, the researchers performed the following experiments.\nCalculations indicate that after $^tBuONa$ is added, loss of one molecule of gas first produces highly reactive intermediate B. B then reacts with the azide group to form charge-separated tricyclic intermediate C. A concerted reaction of C expels another gas molecule to give the product. Write the SMILES strings for B and C.", "text_images": [{"file": "Exam-23-T10-2-1-prompt.png", "description": "Two parallel mechanistic experiments. Upper reaction: the starting material is labeled only A and its structure is not drawn; conditions are TEMPO (2.0 equiv), $^tBuONa$ (1.5 equiv), PhCl, heat, giving 'N#C/C=C\\C=C/C#CC1=CC=CC=C1' in 70% yield. Lower reaction: 'C1(C(C2=CC=CC=C2)=NN=N3)=C3C=CC=C1'>$^tBuONa$ (1.5 equiv), PhCl, heat>'N#C/C=C\\C=C/C#CC1=CC=CC=C1', 0% yield."}], "full_mark": 3}, {"sub_id": "Exam-23-T10-3", "tags": ["2.4"], "text_open": "10.1.3 Considering the reaction sequence above, write the SMILES strings for the products of the following reactions. (Hint: each reaction proceeds through a high-energy intermediate similar to B.)", "text_images": [{"file": "Exam-23-T10-3-1-prompt.png", "description": "Two parallel reactions, arranged vertically, with neither product drawn. Upper reaction: 'O=C(C1=CC=CC=C1N=[N+]=[N-])C2=CC=C(C)C=C2'>TosNHNH2 (1.2 equiv), TFA (20 mol%), PhCl; then $^tBuONa$ (1.5 equiv), PhCl, heat>[product not shown]. Lower reaction: 'O=C(C)/C(C(C)=O)=C/C#CC1=CC=CC=C1N=[N+]=[N-]'>PhCl (0.05 M), heat>[product not shown]."}], "full_mark": 3}, {"sub_id": "Exam-23-T10-4", "tags": ["2.2"], "text_open": "10.2 In 2025, the group of Professor Daniele Leonori reported oxidative cleavage of electron-rich aromatic rings using photoexcited nitroarenes. (λ = 390 nm denotes irradiation with 390 nm light.)\n10.2.1 Under irradiation, the two substrates react to form a five-membered-ring intermediate Int, whose fragmentation gives the product. Write the SMILES string for Int; the aromatic ring attached to the nitro group may be abbreviated as -Ar.", "text_images": [{"file": "Exam-23-T10-4-1-prompt.png", "description": "Photooxidative-cleavage reaction: 'COC1=CC=CC2=CC=CC=C21'.'O=[N+](C1=CC=CC=C1)[O-].[R]C'>CH2Cl2 (0.1 M), -65°C, 5h, λ = 390 nm>'O=C(OC)C1=CC=CC=C1/C=C\\C=O'. [R] in substrate 2 is the original R-label placeholder."}], "full_mark": 1}, {"sub_id": "Exam-23-T10-5", "tags": ["3.9", "1.6"], "text_open": "10.2.2 The researchers reacted differently substituted nitroarenes with a substrate containing both an aromatic ring and an alkene and found that the substituents affect chemoselectivity. For example, nitroarenes N1 and N2 react with D to give selective oxidation of either the alkene or the aromatic ring.\nStudies of nitroarene excited states show that their lowest triplet state may have either a $^3(n,\\pi^*)$ or $^3(\\pi,\\pi^*)$ electronic configuration. These correspond, respectively, to an n to $\\pi^*$ transition, representable as a dioxygen-diradical species, and a $\\pi, \\pi^*$ transition, involving intramolecular electron transfer from the benzene unit to the nitro group with the unpaired electron delocalized over the entire $\\pi$ system. Based on the electronic properties of the substituents, which state is lower in energy for N1, and which is lower for N2?\n(a) n, $\\pi^*$ state; (b) $\\pi, \\pi^*$ state.", "text_images": [{"file": "Exam-23-T10-5-1-prompt.png", "description": "Three compounds, labeled D, N1, and N2 from left to right. The SMILES for N1 is 'O=[N+](C1=CC=C(OC)C(OC)=C1)[O-]'; for N2, 'O=[N+](C1=CC(C(F)(F)F)=CC(C(F)(F)F)=C1)[O-]'; and for D, 'C/C(C)=C\\CCOC1=CC=CC2=C1C=CC=C2'."}], "full_mark": 2}, {"sub_id": "Exam-23-T10-6", "tags": ["2.7", "3.9"], "text_open": "10.2.3 A triplet excited state with an n, $\\pi^*$ configuration can undergo cycloaddition directly. A triplet excited state with a $\\pi, \\pi^*$ configuration cannot undergo direct concerted cycloaddition; calculations indicate the following pathway. It first forms a complex with the other substrate, in which dispersion is the principal intermolecular attraction. Partial electron transfer between the substrates then gives the nitro group a partial negative charge and the other substrate a partial positive charge, promoting formation of the first C–O bond. Finally, intersystem crossing and ring closure give the five-membered intermediate. Accordingly, a nitroarene excited state with an n, $\\pi^*$ configuration reacts more readily with ( ), whereas one with a $\\pi, \\pi^*$ configuration reacts more readily with ( ). Explain each choice.\n(a) An electron-rich aromatic ring; (b) an isolated double bond.", "text_images": [], "full_mark": 5}, {"sub_id": "Exam-23-T10-7", "tags": ["2.4"], "text_open": "10.2.4 Write the SMILES string for the product formed when D is oxidized by N2 under irradiation.", "text_images": [], "full_mark": 2}, {"sub_id": "Exam-23-T10-8", "tags": ["2.2", "2.3"], "text_open": "10.3 In 2024, Professor Youliang Wang's group achieved thermochemical dearomatization of a benzene ring, constructing a complex ring system in one step.\nExperiments show that when TFA is replaced by deuterated TFA and the reaction is run for 10 h, 99% of the circled hydrogen atom in the product is replaced by deuterium.\n10.3.1 The reaction includes a pericyclic step. Write the SMILES strings for 3–4 key intermediates and name the pericyclic reaction.", "text_images": [{"file": "Exam-23-T10-8-1-prompt.png", "description": "Single-step reaction: 'C=CCN(C1=CC=CC=C1)C(C#CC2=CC=CC=C2)=O'>TFA, 50°C, 12h>'O=C(N1C2)C([H])=C(C31C4C5C=CC3C2C4)C6=C5C=CC=C6'. The hydrogen at C([H]) in the product is circled, corresponding to the deuterium-labeling experiment in the prompt."}], "full_mark": 7}, {"sub_id": "Exam-23-T10-9", "tags": ["2.4"], "text_open": "10.3.2 The reaction also forms two byproducts. E has a ring system similar to that of the major product, whereas F contains only three rings and no bridged ring. Write the SMILES strings for E and F.", "text_images": [], "full_mark": 3}]} | |