dev/dataset.jsonl

{"problem_id": "Exam-01-T5", "header": "Simple Molecules, Complex Reactions", "header_images": [], "subquestions": [{"sub_id": "Exam-01-T5-1", "tags": ["1.4", "1.5"], "text_open": "Label D denotes $N_2O$.\nScientists have developed catalyst X, which converts the toxic gas CO to CO_2 under the action of D. Its formula mass is approximately 471.5, and its structure is shown below, where P represents a P(tBu)_2 group. Write the d-orbital electron configuration of the metal ion in X in an octahedral field, approximating the geometry as a regular octahedron.", "text_images": [{"file": "Exam-01-T5-1-1-prompt.png", "description": "Structural representation of catalyst X: '[H][Ru]12(C#O)(N3C(C4=CC=CC=[N]42)=CC=CC3=C[P]1(C(C)(C)C)C(C)(C)C)C#O'"}], "full_mark": 2, "rubric": ["t_{2g}^{6} e_{g}^{0} (2 points)"], "rubric_images": []}, {"sub_id": "Exam-01-T5-2", "tags": ["2.2", "2.3", "2.4"], "text_open": "Label D denotes $N_2O$.\nBased on experimental observations, the following principal catalytic pathway has been proposed: X reacts with D to form Y, whose formula mass is approximately 487.5. Y rapidly converts into its more stable isomer Z. Z then takes up one molecule of CO while releasing one molecule of CO_2, regenerating X. Write the SMILES structures of Y and Z.", "text_images": [], "full_mark": 4, "rubric": ["Y: 'O[Ru]12(C#O)(N3C(C4=CC=CC=[N]42)=CC=CC3=C[P]1(C(C)(C)C)C(C)(C)C)C#O' (2 points)", "Z: 'CC(C)([P]1(C=C2N(C(C3=CC=CC=[N]34)=CC=C2)[Ru]14(C(O)=O)C#O)C(C)(C)C)C' (2 points). Stereochemistry is not required."], "rubric_images": [{"file": "Exam-01-T5-2-1-rubric.png", "description": "Reference structure of Y: 'O[Ru]12(C#O)(N3C(C4=CC=CC=[N]42)=CC=CC3=C[P]1(C(C)(C)C)C(C)(C)C)C#O'"}, {"file": "Exam-01-T5-2-2-rubric.png", "description": "Reference structure of Z: 'CC(C)([P]1(C=C2N(C(C3=CC=CC=[N]34)=CC=C2)[Ru]14(C(O)=O)C#O)C(C)(C)C)C'"}]}, {"sub_id": "Exam-01-T5-3", "tags": ["2.3", "2.4"], "text_open": "The scientists also found that when THF is used as the solvent, another pathway may operate: X slowly isomerizes to isomer W, which contains two aromatic rings, and W then catalyzes exactly the same process as X. Write the SMILES structure of W.", "text_images": [], "full_mark": 3, "rubric": ["W: 'CC(C)([P]1(CC2=[N](C(C3=NC=CC=C43)=CC=C2)[Ru]14(C#O)(C#O)[H])C(C)(C)C)C' (3 points)."], "rubric_images": [{"file": "Exam-01-T5-3-1-rubric.png", "description": "Reference structure of W: 'CC(C)([P]1(CC2=[N](C(C3=NC=CC=C43)=CC=C2)[Ru]14(C#O)(C#O)[H])C(C)(C)C)C'"}]}]}
{"problem_id": "Exam-01-T7", "header": "Efficiency of Oxidative Phosphorylation", "header_images": [], "subquestions": [{"sub_id": "Exam-01-T7-1", "tags": ["3.2"], "text_open": "A molecular-level hypothesis has been proposed: for every $2e^-$ transferred through the electron-transfer chain, Complex III transports $2HSuc^-$ across the membrane. The opposing concentration gradients that must be overcome in the two transport events are approximately $1:7$ and $1:8$, respectively. Energy losses from other ion-transport processes at Complexes I and IV may be neglected. The inner-side condition is $pH=8$, the outer-side condition is $pH=5$, succinic acid has $pK_{a1,2}=4.2,5.6$, and the temperature is $310\\,\\mathrm{K}$.\nFor this redox reaction, $\\Delta_rG_m^{\\ominus *}\\approx-2.20\\times10^2\\ \\mathrm{kJ\\,mol^{-1}}$; take the magnitude of the available energy as $220\\ \\mathrm{kJ\\,mol^{-1}}$.\nATP synthase can transport only $HSuc^-$ ions on the outer side of the membrane; neither $H_2Suc$ nor $Suc^{2-}$ outside the membrane can pass through ATP synthase. Thus, of the energy stored in $H_2Suc$, $HSuc^-$, and $Suc^{2-}$ during the redox stage, only the fraction associated with external $HSuc^-$ can be utilized. On this basis, calculate $\\eta_{redox}$.", "text_images": [], "full_mark": 3, "rubric": ["$\\eta_{redox} \\approx 0.709 \\times [1 - (RT ln 7 + RT ln 8)/220] \\approx 0.676$ (3 points)"], "rubric_images": []}]}
{"problem_id": "Exam-02-T1", "header": "Problem 1: Carbon Coordination! Since Frankland reported Et_2Zn in 1849, molecules containing carbon-metal bonds have been known and have continually attracted intensive study. Carbon atoms have been found to exhibit a wide variety of electron-donation numbers and coordination numbers when coordinated to metals.", "header_images": [], "subquestions": [{"sub_id": "Exam-02-T1-1", "tags": ["1.4", "1.6"], "text_open": "1-1 For the ligands Ph_3PCPPh_3 (denoted L) and RCR (where R is an alkyl group), state the coordination mode by which each preferentially coordinates to a transition-metal atom.", "text_images": [], "full_mark": 2, "rubric": ["L preferentially coordinates to a π acceptor, whereas the carbene preferentially coordinates to a π donor. (2 points)"], "rubric_images": []}, {"sub_id": "Exam-02-T1-2", "tags": ["1.1", "1.4"], "text_open": "1-2 L reacts with ZnI_2 or CdI_2 in a 1:1 ratio, but the products have different structures. The ZnI_2 complex has 2 mutually perpendicular mirror planes and 1 twofold rotation axis but no inversion center. The CdI_2 complex has only 1 mirror plane and 1 twofold rotation axis but does possess an inversion center. Draw both structures.", "text_images": [], "full_mark": 4, "rubric": ["Structure of the ZnI_2 complex: 'I[Zn]([L])I' (2 points).", "Structure of the CdI_2 complex: '[L][Cd]1(I)I[Cd](I)([L])I1' (2 points)."], "rubric_images": [{"file": "Exam-02-T1-2-1-rubric.png", "description": "Schematic structure of the ZnI_2 complex: 'I[Zn]([L])I'"}, {"file": "Exam-02-T1-2-2-rubric.png", "description": "Schematic structure of the CdI_2 complex: '[L][Cd]1(I)I[Cd](I)([L])I1'"}]}, {"sub_id": "Exam-02-T1-3", "tags": ["2.1", "2.4", "1.4"], "text_open": "1-5 As stated in the introduction, carbon atoms can donate different numbers of electrons in complexes. Reaction of WCl_6 with (CH_3)_3CCH_2Li and Ph_2PCH_2CH_2PPh_2 gives a mononuclear W complex containing 3 carbon-tungsten bonds whose lengths are all distinctly different. Write the equation for its preparation.", "text_images": [], "full_mark": 3, "rubric": ["WCl_6 + 6(CH_3)_3CCH_2Li + Ph_2PCH_2CH_2PPh_2 → 3C(CH_3)_4 + 6LiCl + W(Ph_2PCH_2CH_2PPh_2)[(CH_3)_3CCH_2][(CH_3)_3CCH][(CH_3)_3CC] (3 points)."], "rubric_images": []}]}
{"problem_id": "Exam-02-T3", "header": "Problem 3: Bi-Ion Removal Battery. A bi-ion removal battery (BDI) can intercalate cations and anions from seawater into crystalline phases analogous to those in lithium-ion batteries and store them through conversion reactions in the cathode and anode, enabling efficient seawater desalination with high ion-removal capacity and high charge efficiency. Studies show that a $\\mathrm{NaBi_3O_4Cl_2}$ electrode can simultaneously remove $\\mathrm{Na^+}$ and $\\mathrm{Cl^-}$ from seawater.", "header_images": [{"file": "Exam-02-T3-header-1.png", "description": "On the left is a stacked structure of red blocks. Following the process indicated by the blue arrow “$\\mathrm{-Na^+, -Cl^-}$,” it becomes the central stacked structure of red and orange blocks. The process indicated by the red arrow “$\\mathrm{+Na^+, +Cl^-}$” then converts it to the structure on the right, composed of red and blue blocks. A further process indicated by the blue arrow “$\\mathrm{-Na^+, -Cl^-}$” returns it to the central structure. The legend identifies red blocks as $\\mathrm{NaBi_3O_4Cl_2}$, blue blocks as substance X, and orange blocks as $\\mathrm{Bi}$."}], "subquestions": [{"sub_id": "Exam-02-T3-1", "tags": ["2.1", "3.5"], "text_open": "Substance X is $BiOCl$.\n3-2-3 Write the electrode-reaction equations occurring at the $\\mathrm{NaBi_3O_4Cl_2}$ electrode during charging.", "text_images": [], "full_mark": 4, "rubric": ["$\\mathrm{Na^+ + 3Bi + 2Cl^- + 4H_2O - 9e^- = NaBi_3O_4Cl_2 + 8H^+}$ (2 points; $\\mathrm{Na^+ + 3Bi + 2Cl^- + 8OH^- - 9e^- = NaBi_3O_4Cl_2 + 4H_2O}$ receives 1 point).", "$\\mathrm{Bi + Cl^- + H_2O - 3e^- = BiOCl + 2H^+}$ (2 points)", "$\\mathrm{Bi + Cl^- + 2OH^- - 3e^- = BiOCl + H_2O}$ (1 point).", "Combining the two equations above gives:", "$\\mathrm{3xBi + yNa^+ + (3x-y)Cl^- + (3x+y)H_2O - 9e^- = yNaBi_3O_4Cl_2 + (3x - 3y)BiOCl + (6x + 2y)H^+}$ (4 points)"], "rubric_images": []}, {"sub_id": "Exam-02-T3-2", "tags": ["3.5"], "text_open": "3-3 To construct a complete battery, $\\mathrm{NaTi_2(PO_4)_3@C}$ is used as the electrode on the other side. 3-3-1 State the type of the $\\mathrm{NaTi_2(PO_4)_3@C}$ electrode during seawater desalination.", "text_images": [], "full_mark": 2, "rubric": ["Cathode (2 points)"], "rubric_images": []}, {"sub_id": "Exam-02-T3-3", "tags": ["2.1", "3.5"], "text_open": "3-3-2 Write the electrode-reaction equation occurring at the $\\mathrm{NaTi_2(PO_4)_3@C}$ electrode during seawater desalination.", "text_images": [], "full_mark": 2, "rubric": ["$\\mathrm{NaTi_2(PO_4)_3@C + 2Na^+ + 2e^- = Na_3Ti_2(PO_4)_3@C}$ (2 points)", "Alternatively, $\\mathrm{NaTi_2(PO_4)_3@C + xNa^+ + xe^- = Na(1+x)Ti_2(PO_4)_3@C}$ (2 points)"], "rubric_images": []}]}
{"problem_id": "Exam-02-T5", "header": "Problem 5: Simple Molecules, Complex Reactions. Precise modification of selected sites in a molecule is among the most elegant yet challenging transformations in chemistry, encompassing molecular editing at the single-atom level and biomimetic modification of small molecules.", "header_images": [], "subquestions": [{"sub_id": "Exam-02-T5-1", "tags": ["2.1", "2.2"], "text_open": "5-1 A stable reagent A for introducing a single carbon atom was recently reported. It is prepared as follows: excess PPh_3 reacts with CH_2Br_2 to give compound B (reaction 1); B reacts with OR^- (R is an alkyl group) to give key intermediate C (reaction 2); and C reacts with an equimolar amount of N_2O to ultimately give A (reaction 3). Molecular-orbital analysis confirms that the central carbon atom in A is a **zero-valent carbon atom** stabilized by common ligands. 5-1-1 Write the equations for reactions 1-3.", "text_images": [], "full_mark": 6, "rubric": ["2PPh_3 + CH_2Br_2 = [CH_2(PPh_3)_2]Br_2 (2 points).", "[CH_2(PPh_3)_2]^{2-} + 2OR^- = C(PPh_3) + 2ROH (2 points);", "C(PPh_3)_2 + N_2O = PPh_3CN_2 + OPPh_3 (2 points);"], "rubric_images": []}, {"sub_id": "Exam-02-T5-2", "tags": ["1.6", "1.5"], "text_open": "5-1-2 Write one SMILES representation of A, with the requirement that the carbon atom carry no formal charge in the corresponding Lewis structure. State whether the central P-C-N unit is linear or nonlinear.", "text_images": [], "full_mark": 3, "rubric": ["The SMILES representation of A is '[N-]=[N+]=C=P(C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3' (2 points).", "The central P-C-N unit is nonlinear (1 point)."], "rubric_images": [{"file": "Exam-02-T5-2-1-rubric.png", "description": "Schematic Lewis structure of A, with SMILES '[N-]=[N+]=C=P(C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3'."}]}, {"sub_id": "Exam-02-T5-3", "tags": ["1.5", "1.6"], "text_open": "5-1-3 Write the SMILES representations of all other reasonable resonance structures of A.", "text_images": [], "full_mark": 3, "rubric": ["Other reasonable resonance-structure SMILES for A include '[N-]=[N+]=[C-][P+](C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3', 'N#[N+][C-2][P+](C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3', and 'N#[N+][C-]=P(C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3'. Each reasonable resonance structure is worth 1 point. If extra structures are written, evaluate only the first three."], "rubric_images": [{"file": "Exam-02-T5-3-1-rubric.png", "description": "Other reasonable resonance-structure SMILES for A include '[N-]=[N+]=[C-][P+](C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3', 'N#[N+][C-2][P+](C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3', and 'N#[N+][C-]=P(C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3'. Each reasonable resonance structure is worth 1 point. If extra structures are written, evaluate only the first three."}]}, {"sub_id": "Exam-02-T5-4", "tags": ["2.1", "2.4", "2.2"], "text_open": "5-1-4 Reagent A reacts with benzaldehyde to first form intermediate X after carbon-atom transfer. X can rearrange to stable product X_1 while also giving another product X_2. The molecular mass of X_2 is twice that of X_1. Write the equation for the reaction of A with benzaldehyde to give X_1, and write the SMILES representations of X_1 and X_2.", "text_images": [], "full_mark": 4, "rubric": ["PPh_3CN_2 + PhCHO = PhC≡CH + OPPh_3 + N_2 (2 points).", "The correct SMILES representation of X_1 is 'C#CC1=CC=CC=C1' (1 point).", "The correct SMILES representation of X_2 is 'C1(C=C=C=CC2=CC=CC=C2)=CC=CC=C1' (1 point)."], "rubric_images": [{"file": "Exam-02-T5-4-1-rubric.png", "description": "Structure of X_1, with SMILES 'C#CC1=CC=CC=C1'."}, {"file": "Exam-02-T5-4-2-rubric.png", "description": "Structure of X_2, with SMILES 'C1(C=C=C=CC2=CC=CC=C2)=CC=CC=C1'."}]}, {"sub_id": "Exam-02-T5-5", "tags": ["1.4", "1.5"], "text_open": "5-2 To study activation and conversion of nitrosobenzene (PhNO) during biological metabolism and to achieve reversible redox interconversion among three oxidation states of PhNO in one system, a research group designed a dicobalt system that realizes a biomimetic stepwise four-electron reduction of PhNO. The catalytic cycle is shown below; Cp* denotes the C_5Me_5^- ionic ligand. 5-2-1 Give the oxidation state of Co in N and the electron count around each Co ion. The Co ion in N is low-spin. Draw the d-orbital splitting diagram for Co in a tetrahedral field, approximated as regular tetrahedral although the actual geometry deviates greatly, and give the d-electron configuration.", "text_images": [{"file": "Exam-02-T5-5-1-prompt.png", "description": "Catalytic-cycle diagram. The cycle begins with compound N at the top center. Its cation is '[Cp*][Co]1([N]#CC)[S]([Co]([Cp*])([N]#CC)[S]1C2=CC=CC=C2)C3=CC=CC=C3', with 2 PF_6^- anions. Along the arrow on the right, PhNO is added and 2 MeCN molecules depart to give complex D, represented below as {PhNO}^0. D gains 1 electron to form E, and E can reversibly lose an electron to regenerate D. E gains 1 electron to form F, and F can reversibly lose an electron to regenerate E. Along the left arrow from F, reaction with PhSH and NaBPh_4 eliminates NaSPh and gives complex G; loss of 1 H^+ from G gives F. G reacts with BH_3 to form by-product PhNH_2 and complex H. H reacts with H^+, MeCN, and NH_4PF_6, eliminates 1 H_2O molecule, and returns to the starting structure N, completing the cycle."}], "full_mark": 4, "rubric": ["Co is in the +3 oxidation state (1 point).", "18 electrons (1 point).", "In the schematic d-orbital splitting and electron-configuration diagram for the Co ion, 2 points are awarded only if the d-orbital splitting pattern, electron configuration, and orbital labels are all correct."], "rubric_images": [{"file": "Exam-02-T5-5-1-rubric.png", "description": "Schematic d-orbital splitting and electron configuration of the Co ion. The upper level contains three degenerate t_2 orbitals, two of which each contain one electron of the same spin. The lower level contains two degenerate e orbitals, each filled with two electrons of opposite spin."}]}, {"sub_id": "Exam-02-T5-6", "tags": ["1.4", "2.2"], "text_open": "5-2-2 In intermediates D-H, PhNO and its derived ligands are all bridging ligands and all satisfy the EAN rule. Only in complex F does the ligand adopt an unsymmetrical μ-η^1:η^2 coordination mode. Write the SMILES representations of complexes D, F, G, and H; the anions may be abbreviated.", "text_images": [], "full_mark": 8, "rubric": ["Cation SMILES of D: '[Cp*][Co]1([N](C2=CC=CC=C2)=O3)[S]([Co]3([Cp*])[S]1C4=CC=CC=C4)C5=CC=CC=C5'; the anions are 2PF_6^- (2 points).", "SMILES of F: '[Cp*][Co]([N](C1=CC=CC=C1)2O3)(SC4=CC=CC=C4)[S]([Co]23[Cp*])C5=CC=CC=C5' (2 points).", "Cation SMILES of G: '[Cp*][Co]1([N]([H])(C2=CC=CC=C2)O3)[S]([Co]3([Cp*])[S]1C4=CC=CC=C4)C5=CC=CC=C5'; the anion is BPh_4^- (2 points).", "Cation SMILES of H: '[Cp*][Co]12[S]([Co]([Cp*])([OH]2)[S]1C3=CC=CC=C3)C4=CC=CC=C4'; the anion is 1BPh_4^- (2 points).", "Deduct 1 point if the anion is omitted or incorrect.", "If F is written with both SPh^- ligands bridging, as '[Cp*][Co]([N](C1=CC=CC=C1)2O3)([S]4C5=CC=CC=C5)[S]([Co]234[Cp*])C6=CC=CC=C6', award 1 point."], "rubric_images": [{"file": "Exam-02-T5-6-1-rubric.png", "description": "Structure of D. The cation is '[Cp*][Co]1([N](C2=CC=CC=C2)=O3)[S]([Co]3([Cp*])[S]1C4=CC=CC=C4)C5=CC=CC=C5' and carries 2 units of positive charge; the anions are 2PF_6^-."}, {"file": "Exam-02-T5-6-2-rubric.png", "description": "The structure of F is '[Cp*][Co]([N](C1=CC=CC=C1)2O3)(SC4=CC=CC=C4)[S]([Co]23[Cp*])C5=CC=CC=C5'."}, {"file": "Exam-02-T5-6-3-rubric.png", "description": "Structure of G. The cation is '[Cp*][Co]1([N]([H])(C2=CC=CC=C2)O3)[S]([Co]3([Cp*])[S]1C4=CC=CC=C4)C5=CC=CC=C5' and carries 1 unit of positive charge; the anion is 1BPh_4^-."}, {"file": "Exam-02-T5-6-4-rubric.png", "description": "Structure of H. The cation is '[Cp*][Co]12[S]([Co]([Cp*])([OH]2)[S]1C3=CC=CC=C3)C4=CC=CC=C4' and carries 1 unit of positive charge; the anion is 1BPh_4^-."}, {"file": "Exam-02-T5-6-5-rubric.png", "description": "An alternative structure for F, worth only 1 point, is '[Cp*][Co]([N](C1=CC=CC=C1)2O3)([S]4C5=CC=CC=C5)[S]([Co]234[Cp*])C6=CC=CC=C6'."}]}, {"sub_id": "Exam-02-T5-7", "tags": ["3.5", "3.6", "3.2"], "text_open": "5-2-4 The research group measured reduction potentials, pK_a values, and other data for the relevant species at ambient temperature:\nD + e^- = E, E(1) = -0.075 V\nE + e^- = F, E(2) = -1.08 V\nG = F + H^+, pK_a = 13-16\nTaking the pK_a of G as 13, calculate the Gibbs energy changes for conversion of D to G (reaction a) and homolytic cleavage of the N-H bond in G (reaction b). The ionization energy of a hydrogen atom is 13.6 eV.", "text_images": [], "full_mark": 7, "rubric": ["From ΔG = -zFE: ΔG(1) = -zFE(1) = 7.23 kJ mol^-1 (1 point).", "ΔG(2) = -zFE(2) = 104.2 kJ mol^-1 (1 point).", "From ΔG = -RTlnK: ΔG(3) = -8.314 × 298.15 × ln(10^-13) = 74.2 kJ mol^-1 (1 point).", "Gibbs energy change for reaction a: ΔG(a) = ΔG(1) + ΔG(2) - ΔG(3) = 37.2 kJ mol^-1 (1 point).", "Gibbs energy change for ionization of H: ΔG(4) = I(H) = 13.6 eV × 96485 = 1312 kJ mol^-1 (1 point).", "Gibbs energy change for reaction b: ΔG(b) = ΔG(3) - ΔG(a) - ΔG(4) = -1275 kJ mol^-1 (2 points).", "No credit is awarded if the unit is omitted or incorrect."], "rubric_images": []}]}
{"problem_id": "Exam-02-T6", "header": "Problem 6: Crystal Structures of Metal Superhydrides. On 2023-2-9, researchers from California State University published a study of metal superhydrides.", "header_images": [{"file": "Exam-02-T6-header-1.png", "description": "The image consists of three side-by-side crystal-structure diagrams labeled A, B, and C in black capital letters at their upper left. Each contains a unit-cell frame drawn with thin black lines, a pink ball-and-stick framework, and translucent colored polyhedra or spheres. In panel A, the black lines form an approximately cubic unit cell containing a large translucent blue cage-like polyhedron. Large dark-blue spheres occupy the cube center and all eight vertices. At the center is a polyhedral cage framework made of small pale-pink spheres and pink cylindrical bonds, defining the translucent blue polyhedron. On each face of the cube, pink spheres appear at positions near the middle left, middle right, upper middle, and lower middle. In panel B, thin black lines form an approximately cubic unit cell whose center contains a large translucent reddish-brown cage-like polyhedron. Large green spheres occupy the cube center and the midpoint of each edge. A polyhedral cage framework made of small pale-pink spheres and pink cylindrical bonds defines the translucent reddish-brown polyhedron. The framework contains two characteristic types of structural units. In-plane square units: near the inner side of every unit-cell face is a planar structure parallel to that face; 4 small pale-pink spheres at the upper-left, lower-left, upper-right, and lower-right positions on the side nearer the cell center form a small square. Vertex-directed tetrahedral units: viewed along a unit-cell vertex direction, the 3 pale-pink spheres nearest that vertex among the small squares on three adjacent faces form a triangular base. On the side of this triangle toward the unit-cell vertex is 1 pale-pink sphere outside the center of the triangle and not coplanar with it, with another 1 pale-pink sphere extending outward toward the vertex and bonded to it. These 5 spheres together form a tetrahedron-like unit containing a central atom. In panel C, thin black lines form an approximately hexagonal unit cell containing two overlapping translucent green cage-like polyhedra arranged diagonally from lower left to upper right, one in each of those regions. Each green polyhedron contains one large bright-green sphere near its center. The outer part of panel C is a continuous three-dimensional framework of pale-pink spheres and pink cylindrical bonds. Pink spheres lie around the edges and junction of the two green cages, and the pink bonds form multiple pentagonal, hexagonal, and oblique polygonal rings, resembling two connected or partially overlapping cages. The pink framework of the upper-right cage extends toward the upper-right and right edges of the cell, while that of the lower-left cage extends toward the lower-left and foreground regions. Overall, panel A is dominated by a blue cage polyhedron and blue spheres; panel B by a reddish-brown cage polyhedron and peripheral green spheres; and panel C by two translucent green cage polyhedra and two bright-green interior spheres. All three show the spatial relationship between the pink ball-and-stick framework and the unit-cell frame."}], "subquestions": [{"sub_id": "Exam-02-T6-1", "tags": ["1.3"], "text_open": "6-2 Panel C above shows a Ce superhydride crystal. 6-2-1 Describe how the polyhedra shown are connected, for example, which vertices, edges, or faces they share.", "text_images": [], "full_mark": 3, "rubric": ["They are interconnected by sharing all hexagonal and pentagonal faces (3 points).", "No credit for stating that they share 1 hexagonal face or pentagonal face."], "rubric_images": []}, {"sub_id": "Exam-02-T6-2", "tags": ["1.3"], "text_open": "6-2-3 Give the polyhedral notation for the polyhedron shown in C.", "text_images": [], "full_mark": 2, "rubric": ["Polyhedral notation: 4^{6}5^{6}6^{6} (2 points)"], "rubric_images": []}]}
{"problem_id": "Exam-02-T9", "header": "Problem 9: Selectivity in Chemistry", "header_images": [], "subquestions": [{"sub_id": "Exam-02-T9-1", "tags": ["2.7", "1.2", "2.4"], "text_open": "9-1 Reduction of the following compound with different reducing agents gives different stereochemical outcomes. Write the SMILES representation of major product 9A1 obtained with LiAlH_{4} and that of major product 9A2 obtained with ^{i}Bu_{2}AlH.", "text_images": [{"file": "Exam-02-T9-1-1-prompt.png", "description": "The structure in the image has SMILES 'O=C(C1)C[C@H]2N([C@@H]1CC2)C'."}], "full_mark": 4, "rubric": ["The structures of 9A1 and 9A2 are worth 2 points each; stereochemistry is strictly required.", "The SMILES representation of 9A1 is 'CN1[C@H]2CC[C@@H]1C[C@@H](O)C2'.", "The SMILES representation of 9A2 is 'CN3[C@H]4CC[C@@H]3C[C@H](O)C4'."], "rubric_images": [{"file": "Exam-02-T9-1-1-rubric.png", "description": "The rubric image shows 9A1 and 9A2. The SMILES representation of 9A1 is 'CN1[C@H]2CC[C@@H]1C[C@@H](O)C2', and that of 9A2 is 'CN3[C@H]4CC[C@@H]3C[C@H](O)C4'."}]}, {"sub_id": "Exam-02-T9-2", "tags": ["2.8"], "text_open": "$Zn(OTf)_2$ is used in the selective esterification shown below. It acts as a Lewis acid to activate the ester group and coordinates separately to the two substrates to control the reaction site.\n9-2-2 What amount of Zn(OTf)_{2} should be used in this reaction?", "text_images": [{"file": "Exam-02-T9-2-1-prompt.png", "description": "The image shows a reaction proceeding from left to right. At the far left is 'O=C(O/N=C/c2ncccc2)CCc3ccccc3', labeled **9B** below, joined by a plus sign to 'O[C@@H]1[C@@H](O)[C@H](CO)O[C@@H](OCCCCCCCC)[C@H]1NC(CO)=O' in the center. Above the reaction arrow is “Zn(OTf)_2”; below it are “DMF/THF/DME,” “(1:5:9),” and “30°C, 18h,” in sequence."}], "full_mark": 2, "rubric": ["A catalytic amount (2 points; an equivalent answer is also acceptable)."], "rubric_images": []}, {"sub_id": "Exam-02-T9-3", "tags": ["2.4", "2.7", "1.2"], "text_open": "9-2-3 Write the SMILES representation of the major product, given that only one hydroxy group is esterified.", "text_images": [], "full_mark": 2, "rubric": ["The major product shown in the rubric image (2 points) has SMILES 'CCCCCCCCO[C@H]1[C@H]([C@@H]([C@H]([C@@H](O1)CO)O)O)NC(COC(CCc2ccccc2)=O)=O'."], "rubric_images": [{"file": "Exam-02-T9-3-1-rubric.png", "description": "The rubric image shows the monoesterified major product for 9-2-3, with SMILES 'CCCCCCCCO[C@H]1[C@H]([C@@H]([C@H]([C@@H](O1)CO)O)O)NC(COC(CCc2ccccc2)=O)=O'."}]}, {"sub_id": "Exam-02-T9-4", "tags": ["2.2", "2.3"], "text_open": "9-3-2 Compound 9D is the addition product of acetonitrile and hydrogen peroxide, and 9G contains a peroxide bond. Write the SMILES representations of small molecules 9D and 9F in the catalytic cycle below.", "text_images": [{"file": "Exam-02-T9-4-1-prompt.png", "description": "Structure of compound 9C, with SMILES 'CC1(OC[C@]2(OC[C@H]3OC(C)(O[C@H]3C2=O)C)O1)C'."}, {"file": "Exam-02-T9-4-2-prompt.png", "description": "The image shows a reaction segment in which the far-left compound 'C/C=C/c1ccccc1' reacts to the right under an arrow labeled “9C (15 mol%), H_2O_2 (4 equiv.)” above and “K_2CO_3, CH_3CN” and “0°C, 12h” below."}, {"file": "Exam-02-T9-4-3-prompt.png", "description": "The image is a catalytic cycle. At the top is 'CC1(OC[C@]2(OC[C@H]3OC(C)(O[C@H]3C2=O)C)O1)C', labeled “9C” below. On the right, `CH_3CN + H_2O_2` points left by a straight arrow to “9D.” A curved arrow from “9D” runs down and left into the main cycle, merging with the downward path from “9C,” and both lead to “9E.” Two curved arrows descend from “9E,” one down-right to “9F” and the other down-left to “9G.” A curved arrow from “9G” runs up and left, meeting an upward curved arrow from 'C/C=C/c1ccccc1' at the lower left; together they lead to `[transition state]^‡`. Finally, a curved arrow from `[transition state]^‡` runs up and right to close the cycle at “9C,” with a curved branch along this upward path pointing up-left to the word “product.”"}], "full_mark": 4, "rubric": ["The SMILES representation of 9D is 'CC(OO)=N', and that of 9F is 'CC(N)=O'; 2 points each."], "rubric_images": [{"file": "Exam-02-T9-4-1-rubric.png", "description": "The rubric image gives the SMILES representation of 9D as 'CC(OO)=N' and that of 9F as 'CC(N)=O'."}]}, {"sub_id": "Exam-02-T9-5", "tags": ["2.7", "1.2", "2.4"], "text_open": "9-3-3 Predict the stereochemistry and write the SMILES representation of the product.", "text_images": [], "full_mark": 2, "rubric": ["The product, with SMILES 'C[C@@H]1[C@H](O1)C2=CC=CC=C2', is worth 2 points. No credit if (±) is written or the product is identified as a racemate."], "rubric_images": [{"file": "Exam-02-T9-5-1-rubric.png", "description": "The rubric image shows the asymmetric epoxidation product, with SMILES 'C[C@@H]1[C@H](O1)C2=CC=CC=C2'."}]}]}
{"problem_id": "Exam-03-T2", "header": "Problem 2 Metallabenzenes and Aromaticity. Note: throughout this problem, n-coordinate means that the coordination number of the atom is n.", "header_images": [], "subquestions": [{"sub_id": "Exam-03-T2-1", "tags": ["1.4", "2.2", "2.4"], "text_open": "2-1 The first metallabenzene, A, was prepared from Os(CO)(CS)(PPh_3)_3 and acetylene. Draw the missing structures in the following transformation scheme and give their SMILES. In A-D^+, Os is 6-coordinate, and A contains a three-membered ring.", "text_images": [{"file": "Exam-03-T2-1-1-prompt.png", "description": "Transformation scheme for A-D^+. At the upper left, Os(CO)(CS)(PPh_3)_3 is converted into boldfaced A by a right-pointing arrow labeled \"2C_2H_2\" above and \"- PPh_3\" below. From A, the scheme branches horizontally and vertically. Along the horizontal branch, A is converted into boldfaced B by a right-pointing arrow labeled \"CO\". A downward arrow from B, labeled \"MeI, NaClO_4\" on the right, leads to D^+ClO_4^- at the lower right, with D boldfaced. Along the vertical branch, A and boldfaced C below it are connected by two opposing half-arrows: the downward half-arrow to C is labeled \"HCl\" on the right, and the upward half-arrow to A is labeled \"-HCl\" on the left."}], "full_mark": 8, "rubric": ["2 points for each structure, 8 points total. A: 'O=C[Os]12([P](C3=CC=CC=C3)(C4=CC=CC=C4)C5=CC=CC=C5)([P](C6=CC=CC=C6)(C7=CC=CC=C7)C8=CC=CC=C8)=CC=CC=C1S2'; B: 'S=C1C=CC=C[Os]1([P]2(C3=CC=CC=C34CC=CC=C24)C5=CC=CC=C5)(C=O)(C=O)[P](C6=CC=CC=C6)(C7=CC=CC=C7)C8=CC=CC=C8'; C: 'Cl[Os]1([P](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4)([P](C5=CC=CC=C5)(C6=CC=CC=C6)C7=CC=CC=C7)(C=O)=CC=CC=C1S'; D^+: 'CSC1=CC=CC=[Os]1([P](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4)([P+](C5=CC=CC=C5)(C6=CC=CC=C6)C7=CC=CC=C7)(C=O)C=O'.", "Deduct 0.5 point if the charge on D^+ is omitted."], "rubric_images": [{"file": "Exam-03-T2-1-1-rubric.png", "description": "2-1 Grading structures. Four red chemical structures are arranged horizontally. From left to right: A, O=C[Os]12([P](C3=CC=CC=C3)(C4=CC=CC=C4)C5=CC=CC=C5)([P](C6=CC=CC=C6)(C7=CC=CC=C7)C8=CC=CC=C8)=CC=CC=C1S2; B, S=C1C=CC=C[Os]1([P]2(C3=CC=CC=C34CC=CC=C24)C5=CC=CC=C5)(C=O)(C=O)[P](C6=CC=CC=C6)(C7=CC=CC=C7)C8=CC=CC=C8; C, Cl[Os]1([P](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4)([P](C5=CC=CC=C5)(C6=CC=CC=C6)C7=CC=CC=C7)(C=O)=CC=CC=C1S; and, at the far right, the bracketed structure D^+, CSC1=CC=CC=[Os]1([P](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4)([P+](C5=CC=CC=C5)(C6=CC=CC=C6)C7=CC=CC=C7)(C=O)C=O."}]}, {"sub_id": "Exam-03-T2-2", "tags": ["1.4", "2.2", "2.4"], "text_open": "2-2 Mixing OsCl_2(PPh_3)_3 with a large excess of HC≡CTMS in wet benzene gives a yellow precipitate E and a brown solution. E was identified as a carbyne complex. Three substances were isolated from the solution: F (22% yield), G (4.5% yield), and the osma-benzyne H (30% yield). The authors proposed the key intermediate I, an osmium vinylcarbene complex.\n2-2-1 Draw E and I and give their SMILES. I contains no silicon.", "text_images": [{"file": "Exam-03-T2-2-1-prompt.png", "description": "Reaction and structure scheme for E, F, G, H, and I. On the left, OsCl_2(PPh_3)_3 is converted into boldfaced I by a right-pointing arrow labeled \"H—≡—TMS\" above and \"H_2O\" below. I is then converted into boldfaced E on the right by another right-pointing arrow labeled \"HCl\" above."}], "full_mark": 4, "rubric": ["2 points each for E and I, 4 points total. I: Cl[Os](Cl)([P](C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3)([P](C4=CC=CC=C4)(C5=CC=CC=C5)C6=CC=CC=C6)=C=C; E: CC#[Os](Cl)(Cl)(Cl)([P](C7=CC=CC=C7)(C8=CC=CC=C8)C9=CC=CC=C9)[P](C%10=CC=CC=C%10)(C%11=CC=CC=C%11)C%12=CC=CC=C%12."], "rubric_images": [{"file": "Exam-03-T2-2-1-rubric.png", "description": "2-2-1 Grading structures. Two red chemical structures are arranged horizontally. From left to right: I, Cl[Os](Cl)([P](C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3)([P](C4=CC=CC=C4)(C5=CC=CC=C5)C6=CC=CC=C6)=C=C; and E, CC#[Os](Cl)(Cl)(Cl)([P](C7=CC=CC=C7)(C8=CC=CC=C8)C9=CC=CC=C9)[P](C%10=CC=CC=C%10)(C%11=CC=CC=C%11)C%12=CC=CC=C%12."}]}, {"sub_id": "Exam-03-T2-3", "tags": ["2.2", "2.3", "2.4"], "text_open": "2-2-2 H has molecular formula C_48H_52Si_2P_2Cl_2Os. Draw the reaction intermediates in the conversion of I into H and give their SMILES, together with the SMILES of H.", "text_images": [], "full_mark": 6, "rubric": ["Each red intermediate is worth 2 points. Their SMILES are 'Cl[Os](Cl)(C7=C)(C=C7[Si](C)(C)C)([P](C8=CC=CC=C8)(C9=CC=CC=C9)C%10=CC=CC=C%10)[P](C%11=CC=CC=C%11)(C%12=CC=CC=C%12)C%13=CC=CC=C%13' and 'Cl[Os]1(Cl)([P](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4)([P](C5=CC=CC=C5)(C6=CC=CC=C6)C7=CC=CC=C7)C=C([Si](C)(C)C)C(=C)C([Si](C)(C)C)=C1'. The structure of H is worth 2 points: 'Cl[Os]1(Cl)([P](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4)([P](C5=CC=CC=C5)(C6=CC=CC=C6)C7=CC=CC=C7)#CC([Si](C)(C)C)=C(C([Si](C)(C)C)=C1)C'.", "If the red intermediates are omitted but the blue structures are given, award 1 point. The two blue intermediates have SMILES 'Cl[Os](Cl)([P](C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3)([P](C4=CC=CC=C4)(C5=CC=CC=C5)C6=CC=CC=C6)=CC([Si](C)(C)C)=C=C' and 'Cl[Os]7(Cl)([P](C8=CC=CC=C8)(C9=CC=CC=C9)C%10=CC=CC=C%10)([P](C%11=CC=CC=C%11)(C%12=CC=CC=C%12)C%13=CC=CC=C%13)C=C([Si](C)(C)C)[C+](C)C([Si](C)(C)C)=C7'."], "rubric_images": [{"file": "Exam-03-T2-3-1-rubric.png", "description": "2-2-2 Grading scheme 1. This multistep reaction scheme shows the intermediates and H in red. At the far left is the black starting material I, 'Cl[Os](Cl)([P](C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3)([P](C4=CC=CC=C4)(C5=CC=CC=C5)C6=CC=CC=C6)=C=C'. A right-pointing arrow labeled with '[H]C#C[Si](C)(C)C' leads to the first red five-membered-ring intermediate, 'Cl[Os](Cl)(C7=C)(C=C7[Si](C)(C)C)([P](C8=CC=CC=C8)(C9=CC=CC=C9)C%10=CC=CC=C%10)[P](C%11=CC=CC=C%11)(C%12=CC=CC=C%12)C%13=CC=CC=C%13'. An unlabeled right-pointing arrow then leads to the second red six-membered-ring intermediate, 'Cl[Os]1(Cl)([P](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4)([P](C5=CC=CC=C5)(C6=CC=CC=C6)C7=CC=CC=C7)C=C([Si](C)(C)C)C(=C)C([Si](C)(C)C)=C1'. Finally, a right-pointing arrow labeled H^+ gives the red final product H, 'Cl[Os]1(Cl)([P](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4)([P](C5=CC=CC=C5)(C6=CC=CC=C6)C7=CC=CC=C7)#CC([Si](C)(C)C)=C(C([Si](C)(C)C)=C1)C'."}, {"file": "Exam-03-T2-3-2-rubric.png", "description": "2-2-2 Grading scheme 2. The two blue intermediates have SMILES 'Cl[Os](Cl)([P](C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3)([P](C4=CC=CC=C4)(C5=CC=CC=C5)C6=CC=CC=C6)=CC([Si](C)(C)C)=C=C' and 'Cl[Os]7(Cl)([P](C8=CC=CC=C8)(C9=CC=CC=C9)C%10=CC=CC=C%10)([P](C%11=CC=CC=C%11)(C%12=CC=CC=C%12)C%13=CC=CC=C%13)C=C([Si](C)(C)C)[C+](C)C([Si](C)(C)C)=C7'."}]}, {"sub_id": "Exam-03-T2-4", "tags": ["1.6", "3.9"], "text_open": "2-3 In 2020, researchers reported an example of a nonplanar aromatic transition-metal heterocycle. The HOMO of the butadiene ligand combines with the d orbitals of the two Fe atoms to form a molecular orbital, producing stabilization and aromatic character.\n2-3-1 State the total number of d electrons contributed by the two Fe atoms to the aromatic orbital.", "text_images": [{"file": "Exam-03-T2-4-1-prompt.png", "description": "Structure of a nonplanar aromatic transition-metal heterocycle (SMILES: 'CC1=C([Fe]23[N](C)(C)CC[N]3(C)C)([Si](C)(C)C)[Fe](Br)(Br)C2([Si](C)(C)C)=C1C'). The center is a complex network comprising a planar four-carbon butadiene skeleton and two Fe atoms positioned above and below the plane. Both solid and dashed connections are shown. In the conventional solid-line bonding, the upper Fe is bonded to both N atoms of a TMEDA ligand above it; short strokes on the N atoms denote methyl groups. The lower Fe is bonded to two Br atoms, directed downward and to the right. The dashed lines depict multicenter coordination: two dashed lines extend downward from the upper Fe to the two TMS-substituted carbon atoms, and two dashed lines extend upward from the lower Fe to the same two TMS-substituted carbon atoms."}], "full_mark": 1, "rubric": ["Each Fe contributes 1 electron, for a total of 2 electrons. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-03-T2-5", "tags": ["1.6", "1.7", "3.9"], "text_open": "2-3-3 Explain why the ground-state spin multiplicity of this tetrahedral Fe(II) complex is 0. (No more than 100 words.)", "text_images": [], "full_mark": 2, "rubric": ["The Fe(II) centers interact electronically through the butadiene dianion ligand by antiferromagnetic coupling, so their unpaired-electron spins are opposite. (2 points)", "An answer of \"broken symmetry\" also receives full credit. (2 points)"], "rubric_images": []}, {"sub_id": "Exam-03-T2-6", "tags": ["1.6", "3.9"], "text_open": "2-4 For rare-earth metals, however, synthesizing analogous nonplanar aromatic heterocycles is difficult because of the d^0 electron configuration and the energy mismatch of the d orbitals. In 2024, researchers further reduced the butadiene ligand and prepared the first nonplanar aromatic rare-earth heterocycle. (The left figure shows the 1,3-butadiene dianion; the right figure shows the 2-butene tetraanion obtained after reduction.) Explain how the reduced ligand overcomes each of the two difficulties above. (No more than 60 words for each point.)", "text_images": [{"file": "Exam-03-T2-6-1-prompt.png", "description": "Rare-earth heterocycle structure scheme. A right-pointing arrow labeled \"Two-electron reduction\" connects two molecular structures. The reactant on the left is enclosed in black brackets bearing a ^{2-} charge and is labeled \"1,3-butadiene dianion\" below. Inside is an s-cis 1,3-butadiene-core derivative whose core may be represented as '[*]C=C([*])C([*])=C[*]'. The carbon skeleton is highlighted by thick gray lines. The top and bottom carbon-carbon bonds are double bonds, each drawn as a thick gray line parallel to a thin gray line, whereas the central vertical carbon-carbon bond is a single thick gray line. A dashed line extends outward from each of the four carbon vertices to denote an unspecified substituent. The product on the right is similarly enclosed in brackets bearing a ^{4-} charge and labeled \"2-butene tetraanion\" below. Its spatial arrangement is the same, but the skeleton is highlighted in thick blue. The central vertical carbon-carbon bond is now double, and the top and bottom bonds are single, corresponding to the core SMILES '[*]CC([*])=C([*])C[*]'. Each of its four vertices likewise bears an outward dashed substituent marker."}, {"file": "Exam-03-T2-6-2-prompt.png", "description": "Nonplanar aromatic rare-earth heterocycle. A complex bimetallic structure is fully enclosed by a thick purple circle; directly below it, bold red text reads \"RE = Sc, Sm, Lu, Gd\". Inside the circle on the right are two metal centers, one above the other, each shown in bold red brackets as \"[RE]\". On the left, thick blue lines depict the ligand skeleton corresponding to the 2-butene tetraanion in the preceding figure (SMILES '[*]CC([*])=C([*])C[*]', where the asterisks denote unspecified substituents). The ligand contains a vertical carbon-carbon double bond on the left, shown by a thick blue line and a thin black line. Each carbon of this double bond is joined by a thick blue single bond extending rightward to a terminal carbon node. All four carbon nodes carry outward dashed substituent markers. Each of the upper and lower terminal carbon nodes is connected by two separate thick blue lines to both the upper and lower [RE] centers."}], "full_mark": 4, "rubric": ["The reduced ligand has two additional electrons and supplies 6 electrons to the molecular orbital, compensating for the inability of RE to contribute d electrons. (2 points)", "After reduction, the higher-energy Ψ_3 orbital interacts with the d orbitals and is better energy-matched to the higher-energy d orbitals of RE. (2 points)"], "rubric_images": []}]}
{"problem_id": "Exam-03-T10", "header": "Problem 10 Synthesis of Indigoid Compounds (25 points)", "header_images": [], "subquestions": [{"sub_id": "Exam-03-T10-1", "tags": ["2.4", "2.2"], "text_open": "10-1 In 1909, H. Friedländer noted that an excellent purple pigment mentioned in the Bible—the pigment that the Romans called Tyrian purple after the Phoenician port of Tyre in present-day Lebanon—was in fact 6,6'-dibromoindigo, F. This precious early dye was extracted from small purple sea snails, the murex; as many as 12000 snails were required to prepare 1.5 g of dye. After nearly a century of continuing research into its synthesis, Pinkney and Chammers proposed a new method, which Iming and coworkers improved into an efficient route. 10-1-1 Give the SMILES of intermediate E.", "text_images": [{"file": "Exam-03-T10-1-1-prompt.png", "description": "10-1 Synthetic route. This multistep scheme begins at the upper left with 'CC1=CC=C(N)C=C1', which reacts under HNO_3 and H_2SO_4 to give intermediate B. B reacts under 1) NaNO_2, HBr and 2) CuBr to give C. C then reacts under CrO_3 and (CH_3CO)_2O to give D. In the second row, D reacts under H_2SO_4 and H_2O to give E. Finally, E reacts under NaOH and H_2O, (CH_3)_2CO to give the product F at the lower right, with SMILES 'O=C1NC2=C(C=C(Br)C=C2)C1=C3NC4=C(C=C(Br)C=C4)C3=O'."}], "full_mark": 2, "rubric": ["The SMILES of intermediate E is 'BrC(C=C1[N+]([O-])=O)=CC=C1C=O'. (2 points)"], "rubric_images": [{"file": "Exam-03-T10-1-1-rubric.png", "description": "The grading figure gives intermediate E with SMILES 'BrC(C=C1[N+]([O-])=O)=CC=C1C=O'."}]}, {"sub_id": "Exam-03-T10-2", "tags": ["2.7", "1.6"], "text_open": "10-1-2 Explain the regioselectivity of the conversion of A into B. (No more than 50 words.)", "text_images": [], "full_mark": 2, "rubric": ["The amino group is protonated and becomes an electron-withdrawing meta-directing group. (1 point) The methyl group is an ortho/para-directing group. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-03-T10-3", "tags": ["2.1", "2.2"], "text_open": "10-2 In the route above, conversion of E into F is clearly the key step. In fact, as early as 1882, Baeyer used this reaction to synthesize indigo in one step from o-nitrobenzaldehyde. 10-2-1 Write the chemical equation for this reaction, representing o-nitrobenzaldehyde and indigo by their SMILES.", "text_images": [{"file": "Exam-03-T10-3-1-prompt.png", "description": "One-step synthesis. At the far left is 'O=CC1=CC=CC=C1[N+]([O-])=O', labeled G below. Under NaOH and H_2O, (CH_3)_2CO, it gives the product at the far right, 'O=C1NC(C=CC=C2)=C2/C1=C3NC4=C(C=CC=C4)C/3=O', labeled H below."}], "full_mark": 2, "rubric": ["2NaOH + 2(CH_{3})_{2}CO + 2'O=CC1=CC=CC=C1[N+]([O-])=O' → 2CH_{3}COONa + 4H_{2}O + 'O=C1NC(C=CC=C2)=C2/C1=C3NC4=C(C=CC=C4)C/3=O'. (2 points) An incorrectly balanced equation earns only 1 point. An ionic equation earns only 1 point. No credit is awarded if the acid-base conditions are incorrect."], "rubric_images": [{"file": "Exam-03-T10-3-1-rubric.png", "description": "The o-nitrobenzaldehyde in the equation has SMILES 'O=CC1=CC=CC=C1[N+]([O-])=O'."}, {"file": "Exam-03-T10-3-2-rubric.png", "description": "The indigo in the equation has SMILES 'O=C1NC(C=CC=C2)=C2/C1=C3NC4=C(C=CC=C4)C/3=O'."}]}, {"sub_id": "Exam-03-T10-4", "tags": ["2.2", "2.3"], "text_open": "10-2-2 The final step of the reaction is shown below. Considering the byproducts formed in the equation in 10-2-1, give the SMILES of the four key intermediates in the conversion of G into M. For intermediates related by proton gain, loss, or transfer, only one need be given.", "text_images": [{"file": "Exam-03-T10-4-1-prompt.png", "description": "10-2-2 Question figure. A reaction proceeds from left to right. At the far left is 'O=C1C=NC2=C1C=CC=C2', labeled M below, followed by a plus sign and the middle molecule 'OC1=C(NC2=C1C=CC=C2)O', labeled N below. A one-way right-pointing reaction arrow leads to the product at the far right, 'O=C5NC(C=CC=C6)=C6/C5=C7NC8=C(C=CC=C8)C/7=O', labeled H below, indicating that M and N react to form H."}], "full_mark": 6, "rubric": ["6 points total, 1.5 points for each of the four key intermediates. Their SMILES are 'CC(CC(O)C1=CC=CC=C1[N+]([O-])=O)=O', 'CC(C1C(O)C2=C(C=CC=C2)[N+]1(O)[O-])=O', 'CC(C1C(C2=C(C=CC=C2)N1O)=O)=O', and 'O=C1CN(C2=C1C=CC=C2)O'. The fourth intermediate may instead be drawn in its carbonyl form, with SMILES 'ON1C2=C(C=CC=C2)C(O)C1C(O)([O-])C'. For each intermediate, another form obtainable by proton transfer, gain, or loss also receives credit."], "rubric_images": [{"file": "Exam-03-T10-4-1-rubric.png", "description": "Five chemical structures are shown side by side from left to right. The first four are red and the fifth, at the far right, is blue. A blue \"or\" separates the fourth and fifth structures. The red structures have SMILES 'CC(CC(O)C1=CC=CC=C1[N+]([O-])=O)=O', 'CC(C1C(O)C2=C(C=CC=C2)[N+]1(O)[O-])=O', 'CC(C1C(C2=C(C=CC=C2)N1O)=O)=O', and 'O=C1CN(C2=C1C=CC=C2)O'. The blue structure has SMILES 'ON1C2=C(C=CC=C2)C(O)C1C(O)([O-])C'."}]}, {"sub_id": "Exam-03-T10-5", "tags": ["2.2", "2.5"], "text_open": "10-3 Nitrogen plays a key role in the reaction above, but analogous substrates and methods are unsuitable for obtaining the sulfur-substituted product. After the nineteen-eighties, Scottish chemists synthesized thioindigo, K, using the reaction below. The reaction still proceeds through an intermediate containing a fused six- and five-membered ring system. It is known that contraction of the six-membered ring into the five-membered ring does not occur by ring opening followed by ring closure. Answer the following questions. 10-3-1 In what form does the oxygen atom of the sulfoxide depart?", "text_images": [{"file": "Exam-03-T10-5-1-prompt.png", "description": "Structure/reaction description for synthesis of thioindigo K. Sulfoxide J gives thioindigo K under NaOAc/HOAc. The SMILES of J is 'BrC(C1=O)=C([S+](C2=C1C=CC=C2)[O-])Br'; thioindigo K may be represented as 'O=C1C2=C(S/C1=C3SC4=C(C\\3=O)C=CC=C4)C=CC=C2'."}], "full_mark": 1, "rubric": ["AcO^{-}. (1 point; AcOH is also acceptable.)"], "rubric_images": []}, {"sub_id": "Exam-03-T10-6", "tags": ["2.3", "2.7"], "text_open": "10-3-2 During ring contraction, which atom is lost? (   ) (a) the carbon atom nearer the carbonyl group    (b) the carbon atom nearer the sulfoxide group", "text_images": [], "full_mark": 2, "rubric": ["(b) (2 points)"], "rubric_images": []}, {"sub_id": "Exam-03-T10-7", "tags": ["2.2", "2.3"], "text_open": "10-3-4 Give the SMILES of the six key intermediates in the reaction. For intermediates related by proton gain, loss, or transfer, only one need be given. A substitution proceeding by addition-elimination or by an S_{N1} process counts as only one step.", "text_images": [], "full_mark": 9, "rubric": ["9 points total; each of the six key intermediates in the grading figure is worth 1.5 points. Their SMILES are 'O=C1C(Br)=C(OC(C)=O)[S+]([O-])C2=C1C=CC=C2', 'O=C1C(Br)=C([O-])[S+](OC(C)=O)C2=C1C=CC=C2', 'O=C1C2=C(C=CC=C2)[S+]3C(C13Br)=O', 'O=C1C2=C(C=CC=C2)SC1(Br)C(OC(C)=O)=O', 'OC1=C(SC2=C1C=CC=C2)Br', and 'O=C1C(C2(Br)SC3=C(C=CC=C3)C2=O)SC4=C1C=CC=C4'. For each intermediate, another form obtainable by proton transfer, gain, or loss also receives credit. The fifth intermediate may also be drawn in its carbonyl form."], "rubric_images": [{"file": "Exam-03-T10-7-1-rubric.png", "description": "The six key intermediates in the synthesis of thioindigo K have SMILES 'O=C1C(Br)=C(OC(C)=O)[S+]([O-])C2=C1C=CC=C2', 'O=C1C(Br)=C([O-])[S+](OC(C)=O)C2=C1C=CC=C2', 'O=C1C2=C(C=CC=C2)[S+]3C(C13Br)=O', 'O=C1C2=C(C=CC=C2)SC1(Br)C(OC(C)=O)=O', 'OC1=C(SC2=C1C=CC=C2)Br', and 'O=C1C(C2(Br)SC3=C(C=CC=C3)C2=O)SC4=C1C=CC=C4'."}]}]}
{"problem_id": "Exam-04-T6", "header": "Problem 6 Preparation of TPP. TPP is 5,10,15,20-tetraphenylporphyrin (M = 614.7 g·mol-1) and can be prepared rapidly in quantity by the following method: 'O=CC1=CC=CC=C1' + 'C1=CNC=C1' --('CCC(=O)O', reflux, 30 min)--> 'C1(/C=CC(/C(C2=CC=CC=C2)=C(N/3)/C=CC3=C(C(C=C/4)=NC4=C5\\C6=CC=CC=C6)/C7=CC=CC=C7)=N1)=C(C8=CC=CC=C8)\\C9=CC=C5N9'.", "header_images": [{"file": "Exam-04-T6-header-1.png", "description": "The image shows the synthesis of a porphyrin. PhCHO and pyrrole appear as reactants on the left, separated by a plus sign. Propionic acid is drawn above a right-pointing reaction arrow, and \"reflux, 30 min\" is written below it. The product on the right is 'C1(/C=CC(/C(C2=CC=CC=C2)=C(N/3)/C=CC3=C(C(C=C/4)=NC4=C5\\C6=CC=CC=C6)/C7=CC=CC=C7)=N1)=C(C8=CC=CC=C8)\\C9=CC=C5N9'."}], "subquestions": [{"sub_id": "Exam-04-T6-1", "tags": ["4.1"], "text_open": "6-1 Pretreatment of reagents: the benzaldehyde and pyrrole used in the reaction must be dried and purified. 6-1-1 Select the drying agents suitable for pyrrole. (a) P_4O_10    (b) anhydrous Na_2SO_4    (c) concentrated H_2SO_4    (d) CaH_2", "text_images": [], "full_mark": 2, "rubric": ["(b), (d). (1 point each, 2 points total; any incorrect selection receives no credit.)"], "rubric_images": []}, {"sub_id": "Exam-04-T6-2", "tags": ["4.1"], "text_open": "6-1-2 Select the solid drying agent that must be removed by filtration before distillation. (a) CaO (b) anhydrous Na_2SO_4 (c) Na metal (d) CaH_2", "text_images": [], "full_mark": 2, "rubric": ["(b) (2 points)"], "rubric_images": []}]}
{"problem_id": "Exam-05-T4", "header": "Problem 4 Phosphoric Acid and Phosphate Chemistry. Phosphates are important in industrial production, food additives, and other fields, so measuring their content and understanding their properties are significant.", "header_images": [], "subquestions": [{"sub_id": "Exam-05-T4-1", "tags": ["2.1"], "text_open": "4-1 Aqueous sodium phosphate dodecahydrate, Na_3PO_4·12H_2O, is strongly basic and is used in washing powders and fat-saponification agents. It is prepared by reacting calcium dihydrogen phosphate with sodium carbonate and then adding sodium hydroxide solution. Write the preparation equation.", "text_images": [], "full_mark": 2, "rubric": ["Ca(H_2PO_4)_2+Na_2CO_3 + 4NaOH + 20H_2O = 2[Na_3PO_4·12H_2O] + CaCO_3. (2 points)"], "rubric_images": []}, {"sub_id": "Exam-05-T4-2", "tags": ["4.5"], "text_open": "Titration notation: Add a volume $V_0$ of standard HCl of concentration $c_1$ to the sample, then titrate with standard NaOH of concentration $c_2$. The first jump near pH 4.0 consumes a NaOH volume $V_1$, and continuing to the second jump near pH 8.8 consumes an additional volume $V_2$.\n4-2-2 What relation among V_0, V_1, V_2, c_1, and c_2 indicates that the sample contains Na_2HPO_4? Express it symbolically.", "text_images": [], "full_mark": 2, "rubric": ["$V_0c_1-V_1c_2<2V_2c_2$, or equivalently $V_0c_1<c_2(V_1+2V_2)$. (2 points)"], "rubric_images": []}, {"sub_id": "Exam-05-T4-3", "tags": ["4.5", "4.7"], "text_open": "4-2-3 Calculate the mass fraction ω(%) of Na_3PO_4 in the sample. Significant figures are required.", "text_images": [], "full_mark": 4, "rubric": ["First determine that $30 × 1.057 - 9.60 × 1.032 < 2 × 10.75 × 1.032$, showing that the sample contains Na_2HPO_4. (2 points)", "$\\omega =\\frac{30.00\\times 1.057-9.60\\times 1.032-10.75\\times 1.032\\times {{10}^{-3}}\\times 163.94}{2.013}=87.22\\%$. (2 points; directly obtaining n(Na_3PO_4) = 0.01071 mol without the determination also earns 2 points. Deduct 1 point for incorrect significant figures.)"], "rubric_images": []}, {"sub_id": "Exam-05-T4-4", "tags": ["1.7", "2.1"], "text_open": "4-3-2 Phosphorus atoms in different oxidation states can also be linked in catenated phosphoric acids. H_4P_2O_5 contains linked P(Ⅱ) and P(Ⅳ). It has two hydrolysis pathways under basic conditions. Only the first generates gas, and each gives only one phosphorus-containing product. Write the ionic equations for both pathways.", "text_images": [], "full_mark": 4, "rubric": ["Pathway one: (HP_2O_5)^(3-) + OH^- = (P_2O_6)^(4-) + H_2. (2 points)", "Pathway 2: (HP_2O_5)^(3-) + OH^- = 2(HPO_3)^(2-). (2 points)", "Writing reactants or products in acid form earns 1 point per equation. Answers containing (P_2O_5)^(4-) or (PO_3)^(3-) receive no credit."], "rubric_images": []}, {"sub_id": "Exam-05-T4-5", "tags": ["2.4"], "text_open": "4-3-3 Hydrolysis of cyclic H_6P_6O_12 is complex. First, one P-P bond cleaves to form a linear catenated hexaphosphoric acid by disproportionation; then terminal P-P bonds cleave stepwise. How many ways can the linear hexaphosphoric acid hydrolyze to catenated pentaphosphoric acid? Give the products of each. Use P for a phosphorus-containing group and a superscript to denote oxidation state; for example, catenated diphosphoric acid is P^4-P^4.", "text_images": [], "full_mark": 4, "rubric": ["Three. (1 point)", "P^1 + P^4-P^3-P^3-P^3-P^4. (1 point) P^3 + P^2-P^3-P^3-P^3-P^4. (1 point) P^5 + P^2-P^3-P^3-P^3-P^2. (1 point)", "Deduct 1 point for an incorrect hydrolysis pathway and 1 point if the monophosphorus product is omitted."], "rubric_images": []}]}
{"problem_id": "Exam-05-T7", "header": "Problem 7 Explosive Chemistry. An explosion rapidly converts a system's energy into mechanical work through gas expansion, usually with intense heat, light, and sound. Explosive chemistry appears both as a laboratory hazard and in high-density energy-storage materials.", "header_images": [], "subquestions": [{"sub_id": "Exam-05-T7-1", "tags": ["2.4"], "text_open": "7-1-2 Isopropanol deterioration is modeled by acetone reacting in acid with equal numbers of H_2O_2 molecules: two acetone molecules give DADP and 3 give TATP. Both are highly symmetric peroxides. Give their SMILES.", "text_images": [], "full_mark": 2, "rubric": ["7-1-2 DADP: 'CC1(C)OOC(C)(C)OO1'; TATP: 'CC1(C)OOC(C)(C)OOC(C)(C)OO1'. (1 point each)"], "rubric_images": [{"file": "Exam-05-T7-1-1-rubric.png", "description": "DADP and TATP structures as listed. SMILES/structure strings shown: 'CC1(C)OOC(C)(C)OO1', 'CC1(C)OOC(C)(C)OOC(C)(C)OO1'."}]}, {"sub_id": "Exam-05-T7-2", "tags": ["2.2", "2.3"], "text_open": "7-1-3 One TATP first gives cyclic A plus acetone; A then gives acetone, dipolar B, and gas. Write 2 equations with A and B as SMILES.", "text_images": [], "full_mark": 2, "rubric": ["7-1-3 Reaction 1: 'CC1(C)OOC(C)(C)OOC(C)(C)OO1>>CC2(C)OOOC(C)(C)OO2.CC(C)=O'. (1 point)", "Reaction 2: 'CC1(C)OOOC(C)(C)OO1>>C/C(C)=[O+]/[O-].CC(C)=O.O=O'. (1 point)", "No credit unless A and B are SMILES. 'CC1(OO1)C' and 'CC([O+])([O-])C' are not accepted for B."], "rubric_images": [{"file": "Exam-05-T7-2-1-rubric.png", "description": "Reaction 1: 'CC1(C)OOC(C)(C)OOC(C)(C)OO1>>CC2(C)OOOC(C)(C)OO2.CC(C)=O'."}, {"file": "Exam-05-T7-2-2-rubric.png", "description": "Reaction 2: 'CC1(C)OOOC(C)(C)OO1>>C/C(C)=[O+]/[O-].CC(C)=O.O=O'."}, {"file": "Exam-05-T7-2-3-rubric.png", "description": "Incorrect B: 'CC1(OO1)C'."}, {"file": "Exam-05-T7-2-4-rubric.png", "description": "Incorrect B: 'CC([O+])([O-])C'."}]}, {"sub_id": "Exam-05-T7-3", "tags": ["1.6", "1.7"], "text_open": "7-1-4 DADP is unexpectedly much more stable than M. Lone-pair donation into a particular antibonding orbital controls the difference. Give limiting resonance-form SMILES for each and explain. No more than 100 words.", "text_images": [{"file": "Exam-05-T7-3-1-prompt.png", "description": "'CC1(C)CCC(C)(C)OO1'."}], "full_mark": 4, "rubric": ["DADP: 'CC1(C)OOC(C)(C)OO1' ↔ '[O-]OC(C)(C)O/[O+]=C(C)/C'; M: 'CC1(C)CCC(C)(C)OO1' ↔ '[CH2-]CC(C)(C)O/[O+]=C(C)/C', or the listed methyl-anion alternative. (1 point each; carbocation forms accepted.) (2 points total.) Additional structure strings specified: 'CC1(C)CCC(C)(C)OO1', 'CC2=[O+]OC(C)(C)CC2.[CH3-]'.", "O is more electronegative than C, so the O^- limiting form of DADP is more stable and gives greater stabilization. Equivalently, σ*(C-O) is lower than σ*(C-C) and hyperconjugates better with the O lone pair, reducing lone-pair repulsion. (2 points)"], "rubric_images": [{"file": "Exam-05-T7-3-1-rubric.png", "description": "DADP and M resonance pairs as listed. SMILES/structure strings shown: 'CC1(C)OOC(C)(C)OO1', '[O-]OC(C)(C)O/[O+]=C(C)/C', 'CC1(C)CCC(C)(C)OO1', '[CH2-]CC(C)(C)O/[O+]=C(C)/C'."}, {"file": "Exam-05-T7-3-2-rubric.png", "description": "Alternative M resonance pair with 'CC2=[O+]OC(C)(C)CC2.[CH3-]'. SMILES/structure strings shown: 'CC1(C)CCC(C)(C)OO1'."}]}, {"sub_id": "Exam-05-T7-4", "tags": ["1.7"], "text_open": "7-2-2 If H_2O_2 is enclosed in the CL-20 cage, does sensitivity rise or fall? Explain briefly. (No more than 100 words.)", "text_images": [], "full_mark": 2, "rubric": ["7-2-2 It rises. (1 point) Heated H_2O_2 generates radicals that initiate rapid chain decomposition of CL-20. (1 point; radical generation or initiation of decomposition is sufficient.)"], "rubric_images": []}, {"sub_id": "Exam-05-T7-5", "tags": ["2.2"], "text_open": "7-3 Thermal decomposition of condensed-phase CL-20 may proceed through a series of complex transformations. Initially, cleavage of an N-N bond in CL-20 generates NO_2, which then reacts with CL-20 and its remaining fragments through several pathways: (1) NO_2 oxidizes CL-20 to G. G may be represented by the formula CN-CHO. Homolytic bond cleavage and hydrogen-radical transfer by G produce the common small molecules H and I, of which H contains nitrogen. (2) NO_2 reacts with CL-20 to form J, which decomposes to a hydroxyl radical and K; K contains no carbon. 7-3-1 Give the SMILES representations of G (without formal charges), I, and J.", "text_images": [], "full_mark": 5, "rubric": ["7-3-1 G: '[C]=NC([H])=O'; I: '[C-]#[O+]'; J: 'O=NO[H]'. 5 points total: G and J are worth 2 points each, and I is worth 1 point. Each structure earns credit only if it is entirely correct."], "rubric_images": [{"file": "Exam-05-T7-5-1-rubric.png", "description": "G SMILES. SMILES/structure strings shown: '[C]=NC([H])=O'."}, {"file": "Exam-05-T7-5-2-rubric.png", "description": "I SMILES. SMILES/structure strings shown: '[C-]#[O+]'."}, {"file": "Exam-05-T7-5-3-rubric.png", "description": "J SMILES. SMILES/structure strings shown: 'O=NO[H]'."}]}, {"sub_id": "Exam-05-T7-6", "tags": ["1.5"], "text_open": "7-3-2 Give two other reasonable resonance forms of G.", "text_images": [], "full_mark": 2, "rubric": ["'[C-]#[N+]C([H])=O' and '[C]=[N+]=C([H])[O-]'. (1 point each) (2 points total.)"], "rubric_images": [{"file": "Exam-05-T7-6-1-rubric.png", "description": "The two resonance forms. SMILES/structure strings shown: '[C-]#[N+]C([H])=O', '[C]=[N+]=C([H])[O-]'."}]}, {"sub_id": "Exam-05-T7-7", "tags": ["2.2", "2.3"], "text_open": "7-3-3 Intermediate X rearranges intramolecularly and then cleaves to two common small molecules. Give the other key intermediate and final products.", "text_images": [{"file": "Exam-05-T7-7-1-prompt.png", "description": "Lewis structure of X: 'C=N[N+]([O-])=O'."}], "full_mark": 3, "rubric": ["Intermediate '[O-]CN=[N+]=O'. (1 point)", "Products '[H]C([H])=O' and '[N-]=[N+]=O'. (1 point each)"], "rubric_images": [{"file": "Exam-05-T7-7-1-rubric.png", "description": "Red mechanism: X rearranges to '[O-]CN=[N+]=O', then cleaves to formaldehyde and nitrous oxide, with curved electron arrows shown. SMILES/structure strings shown: 'C=N[N+]([O-])=O', 'C=N[N+]([O-])=O', '[O-]CN=[N+]=O', '[H]C([H])=O', '[H]C([H])=O', '[N-]=[N+]=O', '[N-]=[N+]=O'."}]}, {"sub_id": "Exam-05-T7-8", "tags": ["3.7", "3.8", "4.7"], "text_open": "7-4 N-N bond cleavage is the rate-determining step in the early thermal decomposition of crystalline CL-20. To determine its activation energy $E_a$, researchers simulated CL-20 decomposition in a rigid container and obtained the initial $NO_2$ formation rates below. $T\\ (\\mathrm{K})$: 1488, 1969, 2547, 3124. $k\\ (\\mathrm{s^{-1}})$: $1.28\\times10^9$, $1.54\\times10^9$, $1.74\\times10^9$, $1.89\\times10^9$. 7-4-1 Treat N-N cleavage under the simulated conditions as a first-order reaction obeying the Arrhenius law. Calculate $E_a$ and the pre-exponential factor $A$, both to 2 significant figures.", "text_images": [], "full_mark": 6, "rubric": ["From the Arrhenius equation $k=Ae^{-E_a/(RT)}$, linearization gives $\\ln k=-\\dfrac{E_a}{RT}+\\ln A$. (1 point; either the original or linearized equation earns credit.)\n\n| \\(1/T \\, (\\text{K}^{-1})\\) | \\(6.720 \\times 10^{-4}\\) | \\(5.079 \\times 10^{-4}\\) | \\(3.926 \\times 10^{-4}\\) | \\(3.201 \\times 10^{-4}\\) |\n|---|---|---|---|---|\n| \\(\\ln k\\) | 20.97 | 21.16 | 21.28 | 21.36 |\n\n(1 point; at least one type of data must be processed, otherwise no credit is awarded for this step.) $\\ln k=-1.10\\times10^3/T+21.7$. (2 points) Therefore $E_a=9.2\\ \\mathrm{kJ\\cdot mol^{-1}}$ and $A=2.7\\times10^9\\ \\mathrm{s^{-1}}$. (2 points)"], "rubric_images": []}, {"sub_id": "Exam-05-T7-9", "tags": ["3.7", "4.7"], "text_open": "7-4-3 The simulated pre-exponential factor is four orders of magnitude below experiment for CL-20 under real explosion conditions. Give 1 principal reason based on the setup. No more than 100 words.", "text_images": [], "full_mark": 1, "rubric": ["The simulation uses a rigid container of constant volume, whereas the volume of a real explosion increases rapidly; equivalently, the fixed volume keeps the local concentration of gaseous products higher than in a real explosion. (1 point; the complete explanation is required.)"], "rubric_images": []}]}
{"problem_id": "Exam-05-T8", "header": "Problem 8 Basic Organic Chemistry", "header_images": [], "subquestions": [{"sub_id": "Exam-05-T8-1", "tags": ["1.2", "1.6"], "text_open": "8.2 Apply conformational analysis analogous to n-butane. 8.2.1 Rank the shown conformations of 1-butene from high to low energy.", "text_images": [{"file": "Exam-05-T8-1-1-prompt.png", "description": "3 conformations of C=CCC. A has a left double bond and a zigzag chain ending in an ordinary vertical-up bond. B has a similar skeleton but an upward solid-wedge terminal bond. C has a solid-wedge terminal bond directed lower right."}], "full_mark": 2, "rubric": ["B > A > C. (2 points; if n pairwise swaps give the correct order, award 2 - n.)"], "rubric_images": []}, {"sub_id": "Exam-05-T8-2", "tags": ["1.2", "1.6"], "text_open": "8.2.2 Rank the shown conformations of allyl alcohol from high to low energy.", "text_images": [{"file": "Exam-05-T8-2-1-prompt.png", "description": "4 conformations of C=CCO. A ends in an ordinary upward OH bond; B in an upward solid-wedge OH bond; C in a rightward solid-wedge OH bond of different orientation; D ends directly in OH by an ordinary line without wedge stereochemistry."}], "full_mark": 2, "rubric": ["D > B > C > A. (2 points; if n pairwise swaps give the correct order, award 2 - n.)"], "rubric_images": []}, {"sub_id": "Exam-05-T8-3", "tags": ["1.2", "1.6"], "text_open": "8.2.3 Rank the shown conformations of 2-methyl-1-butene from high to low energy.", "text_images": [{"file": "Exam-05-T8-3-1-prompt.png", "description": "4 conformations of 'C=C(C)CC'. A has the terminal methyl cis to the double bond by an ordinary upward bond; B has an upward solid wedge; C a lower-right solid wedge; D an ordinary trans arrangement."}], "full_mark": 2, "rubric": ["D > B > A > C. (2 points; if n pairwise swaps give the correct order, award 2 - n.)"], "rubric_images": []}, {"sub_id": "Exam-05-T8-4", "tags": ["1.2", "1.6"], "text_open": "8.2.4 Rank the shown conformations of methylallyl alcohol from high to low energy.", "text_images": [{"file": "Exam-05-T8-4-1-prompt.png", "description": "4 conformations of 'C=C(C)CO'. A has OH cis to the double bond by an ordinary line; B an upward solid-wedge OH; C a rightward solid-wedge OH; D an ordinary trans double-bond/OH arrangement."}], "full_mark": 2, "rubric": ["D > B > C > A. (2 points; if n pairwise swaps give the correct order, award 2 - n.)"], "rubric_images": []}, {"sub_id": "Exam-05-T8-5", "tags": ["1.5", "1.6"], "text_open": "8.3.3 Which has the largest X-C-O bond angle, X = CF_3, CH_3, or CBr_3?", "text_images": [], "full_mark": 1, "rubric": ["B. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-05-T8-6", "tags": ["1.6"], "text_open": "8.3.4 Which of X-C-O (X = CF_3, CH_3, CBr_3) has the smallest bond angle?", "text_images": [], "full_mark": 1, "rubric": ["A. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-05-T8-7", "tags": ["1.7"], "text_open": "8.4 Ring strain can strongly affect reaction rates and equilibria. 8.4.1 Rank the substances from greatest to least ring strain.", "text_images": [{"file": "Exam-05-T8-7-1-prompt.png", "description": "3 compounds: A, 'C1CC=CC1'; B, 'C1=CC=CC1'; C, 'C=C1CCCC1'."}], "full_mark": 2, "rubric": ["A > C > B (2 points). If the order is not completely correct and n swaps are required to obtain the correct order, award 2 - n points, with a minimum score of 0."], "rubric_images": []}, {"sub_id": "Exam-05-T8-8", "tags": ["1.7"], "text_open": "8.4.2 Select the substance that reacts faster with NaBH_4.", "text_images": [{"file": "Exam-05-T8-8-1-prompt.png", "description": "2 compounds: A, 'O=C1CCCC1'; B, 'O=C1CCCCC1'."}], "full_mark": 1, "rubric": ["B. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-05-T8-9", "tags": ["1.7"], "text_open": "8.4.3 Select the substance with the greater equilibrium enol fraction.", "text_images": [{"file": "Exam-05-T8-9-1-prompt.png", "description": "2 compounds: A, 'O=C1C(C(OCC)=O)CCC1'; B, 'O=C1C(C(OCC)=O)CCCC1'."}], "full_mark": 1, "rubric": ["B. (1 point)"], "rubric_images": []}]}
{"problem_id": "Exam-06-T4", "header": "Problem 4 Determination of Tin by Potassium Iodate Titration. Tin is commonly determined by potassium iodate titration. The procedure includes sample decomposition, masking and separation, reductive fusion with zinc powder, reduction with aluminum powder, and titration of tin(II) ions with potassium iodate.", "header_images": [], "subquestions": [{"sub_id": "Exam-06-T4-1", "tags": ["2.1"], "text_open": "4.2 Sample pretreatment. Place 0.2026 g of the sample in a 250 mL beaker, add 2 g of potassium chlorate and 20 mL of concentrated hydrochloric acid, and digest at low temperature for 20 min. Add 60 mL of hydrochloric acid (1+1), 1 g of hydrazinium sulfate, and 0.3 g of potassium bromide, then heat and evaporate for 50 min. Remove from the heat and allow to cool slightly. Cover with a watch glass and add hydrogen peroxide dropwise until brownish-red bromine appears. Add 10 mL of EDTA solution (100 g/L) and 5 mL of beryllium sulfate solution (40 g/L), and neutralize with aqueous ammonia to pH = 9-10. Add a small amount of paper pulp and filter. Wash the precipitate with an aqueous ammonia-ammonium nitrate wash solution. Transfer the precipitate together with the filter paper to a crucible, dry, ash, and cool. 4.2.1 Write the ionic equation for the reduction of arsenic acid by hydrazine in hydrochloric acid medium, by which arsenic is removed through volatilization.", "text_images": [], "full_mark": 2, "rubric": ["Ionic equation: 2H_3AsO_4 + N_2H_6^{2+} + 6Cl^{-} + 4H^{+} = 2AsCl_3↑ + N_2↑ + 8H_2O (2 points)", "An equation written using N_2H_5^{+} is also acceptable: 2H_3AsO_4 + N_2H_5^{+} + 6Cl^{-} + 5H^{+} = 2AsCl_3 + N_2 + 8H_2O."], "rubric_images": []}, {"sub_id": "Exam-06-T4-2", "tags": ["4.1", "4.7"], "text_open": "Reductive-fusion conditions: Add a zinc powder-borax-boric acid mixed flux to the crucible and cover it with additional mixed flux and NaCl. Place it in a furnace at 750 ℃, raise the temperature to 820 ℃, fuse for 30 min, then remove and cool.\n4.3.3 Explain how an excessively long fusion time or excessively high temperature affects the determination.", "text_images": [], "full_mark": 2, "rubric": ["① Zinc and tin may be lost by volatilization or oxidized by air.", "② Reduction may be incomplete, giving a result that is too low.", "③ The crucible may be corroded, introducing impurities.", "Full credit for 3 correct points, 1 point for 2 correct points, and no credit for only 1 correct point."], "rubric_images": []}, {"sub_id": "Exam-06-T4-3", "tags": ["2.1", "4.5"], "text_open": "4.4.2 Write the ionic equations for the titration process.", "text_images": [], "full_mark": 2, "rubric": ["IO_3^{-} + 5I^{-} + 6H^{+} = 3I_2 + 3H_2O (1 point)", "or 8I^{-} + IO_3^{-} + 6H^{+} = 3I_3^{-} + 3H_2O", "or HIO_3 + 5I^{-} + 5H^{+} = 3I_2 + 3H_2O", "I_2 + Sn^{2+} = Sn^{4+} + 2I^{-} (1 point)", "or Sn^{2+} +I_3^{-} = Sn^{4+} + 3I^{-}", "or SnCl_4^{2-} + I_2 + 2Cl^{-} = SnCl_6^{2-} + 2I^{-}", "or SnCl_4^{2-} + I_3^{-} + 2Cl^{-} = SnCl_6^{2-} + 3I^{-}", "A combined equation is also acceptable:", "3Sn^{2+} + IO_3^{-} + 6H^{+} = 3Sn^{4+} + I^{-} + 3H_2O (2 points)", "or 3SnCl_4^{2-} + IO_3^{-} + 6H^{+} +6Cl^{-} = 3SnCl_6^{2-} + I^{-} + 3H_2O", "or HIO_3 + 3Sn^{2+} + 5H^{+} = I^{-} + 3Sn^{4+} + 3H_2O"], "rubric_images": []}]}
{"problem_id": "Exam-07-T6", "header": "The Phases and Properties of Water and the Evolution of Temperature-Scale Definitions\nWater is the source of life, and its unique physicochemical properties play a crucial role in nature. The phase diagram below describes the phase equilibria of water at different temperatures and pressures. ", "header_images": [{"file": "Exam-07-T6-header-1.png", "description": "Schematic physicochemical phase diagram drawn with black lines on a white background. The origin of the rectangular coordinate system is labeled “O”; the vertical axis points upward and is labeled “p,” and the horizontal axis points rightward and is labeled “T/K.” The value “273.16” appears toward the left of the horizontal axis, with a vertical dashed line rising from it to the triple point, which is also labeled “O.” Four lines meet at triple point O: a solid solid-gas coexistence curve bending down and left to point B; a slightly curved solid-liquid coexistence line with negative slope extending up and left to point C; dashed line OD extending leftward; and a solid liquid-gas coexistence curve bending up and right to point A, the critical point. Point P lies on solid curve OA. A double-slash break symbol “//” appears midway along this curve, with a corresponding “//” break on the right side of the horizontal axis. A vertical dashed line fq passes through point P on curve OA, with f above and q below. Above and to the right of point A is a right-angled boundary extending rightward and upward; the top of its vertical segment is labeled “E,” and the right end of its horizontal segment is labeled “F.” The regions are labeled as follows: “ice” to the left of line OC, “water” between OC and OA, “water vapor” below OA and OB, and “supercritical water,” written vertically, inside the upper-right right-angled region bounded by E and F."}], "subquestions": [{"sub_id": "Exam-07-T6-1", "tags": ["3.1"], "text_open": "6.1.2 State the physical meaning of curve OD.", "text_images": [], "full_mark": 2, "rubric": ["Line OD is the metastable equilibrium line between supercooled liquid water and water vapor, or equivalently the vapor-pressure curve of supercooled water (2 points)."], "rubric_images": []}]}
{"problem_id": "Exam-07-T10", "header": "The Remarkable Role of LiCl", "header_images": [], "subquestions": [{"sub_id": "Exam-07-T10-1", "tags": ["2.4"], "text_open": "10.1.1 Write the SMILES representation of the reaction product, given that the product contains 1 fewer carbon atom than the reactant.", "text_images": [{"file": "Exam-07-T10-1-1-prompt.png", "description": "'COC(C=C1)=CC=C1C2=CC=C(OC)N=C2' is reacted under LiCl/DMF conditions with microwave irradiation."}], "full_mark": 2, "rubric": ["'O=C1C=CC(C2=CC=C(OC)C=C2)=CN1' (2 points)"], "rubric_images": [{"file": "Exam-07-T10-1-1-rubric.png", "description": "'O=C1C=CC(C2=CC=C(OC)C=C2)=CN1'"}]}, {"sub_id": "Exam-07-T10-2", "tags": ["2.4", "2.2"], "text_open": "10.1.2 Based on 10.1.1, write the SMILES representation of the reaction product and the SMILES representation of 1 key intermediate that illustrates the role of LiCl, given that the product contains 1 fewer carbon atom than the reactant.", "text_images": [{"file": "Exam-07-T10-2-1-prompt.png", "description": "'COC1=C(OC)C=C(C=O)C=C1' is reacted under LiCl/DMF conditions with microwave irradiation."}], "full_mark": 4, "rubric": ["Product: 'OC1=C(OC)C=C(C=O)C=C1' (2 points)", "Intermediate: '[Cl-].[Li+].O=CC1=CC(OC)=C(OC)C=C1', in which the Li ion is coordinated simultaneously by the O atoms of the two methoxy groups and by Cl^-, giving a three-coordinate geometry. (2 points)"], "rubric_images": [{"file": "Exam-07-T10-2-1-rubric.png", "description": "'OC1=C(OC)C=C(C=O)C=C1'"}, {"file": "Exam-07-T10-2-2-rubric.png", "description": "Intermediate: the two ether O atoms coordinate to Li to form the complex '[Cl-].[Li+].O=CC1=CC(OC)=C(OC)C=C1'"}]}, {"sub_id": "Exam-07-T10-3", "tags": ["2.4", "1.2", "2.8"], "text_open": "10.1.3 The starting material is an optically active chiral ester. Write the SMILES representation of the reaction product. State the advantage of this method over methods such as hydrolysis with a strong base.", "text_images": [{"file": "Exam-07-T10-3-1-prompt.png", "description": "'O=C([C@@H](C1=CC2=CC=C(C=C2C=C1)OC)C)OC' is reacted under LiCl/DMF conditions with microwave irradiation."}], "full_mark": 4, "rubric": ["'OC([C@H](C)C1=CC2=CC=C(OC)C=C2C=C1)=O' (2 points)", "The configuration at the α stereocenter is retained without racemization. (2 points)"], "rubric_images": [{"file": "Exam-07-T10-3-1-rubric.png", "description": "'OC([C@H](C)C1=CC2=CC=C(OC)C=C2C=C1)=O'"}]}, {"sub_id": "Exam-07-T10-4", "tags": ["2.2", "2.3"], "text_open": "10.2 The following reaction proceeds through a radical process, in which the fluorinating reagent is a single-electron oxidant.\n 10.2.1 Write the SMILES representations of 1 key intermediate associated with the fluorinating reagent and 2 key radical intermediates that contain no halogen atoms and only one six-membered ring.", "text_images": [{"file": "Exam-07-T10-4-1-prompt.png", "description": "'NC1=CC=CC=N1' and 'ClC[N+]12CC[N+](CC2)(F)CC1' · $2\\text{BF}_4^{-}$) react under LiCl/DMF conditions to give the product with the SMILES representation 'NC1=CC=C(Cl)C=N1'."}], "full_mark": 6, "rubric": ["'ClC[N+]1(CC2)CC[N+]2CC1' (note: a nitrogen-centered radical cation) · $\\text{BF}_4^{-}\\text{F}^{-}$ (2 points).", "'NC1=C[CH+][C]C=N1' (2 points).", "'N=C1C=C[C]C=N1' (2 points)"], "rubric_images": [{"file": "Exam-07-T10-4-1-rubric.png", "description": "'ClC[N+]1(CC2)CC[N+]2CC1' (note: a nitrogen-centered radical cation) · $\\text{BF}_4^{-}\\text{F}^{-}$"}, {"file": "Exam-07-T10-4-2-rubric.png", "description": "'NC1=C[CH+][C]C=N1'."}, {"file": "Exam-07-T10-4-3-rubric.png", "description": "'N=C1C=C[C]C=N1'"}]}, {"sub_id": "Exam-07-T10-5", "tags": ["2.4"], "text_open": "10.2.2 Write all by-products of the reaction; give the SMILES representation for the organic product.", "text_images": [], "full_mark": 6, "rubric": ["'ClC[N+]1(CC2)CCN2CC1' · $\\text{BF}_4^{-}$ (2 points)", "HF (2 points)", "$LiBF_4$ (2 points)"], "rubric_images": [{"file": "Exam-07-T10-5-1-rubric.png", "description": "'ClC[N+]1(CC2)CCN2CC1' · $\\text{BF}_4^{-}$"}]}]}
{"problem_id": "Exam-08-T8", "header": "Conformational Stability and Electronic Effects\nThis problem tests a systematic understanding of basic conformational analysis and the origins of molecular stability. Compare conformations accessible through bond rotation and answer in terms of steric effects, hyperconjugation, dipole-dipole interactions, electronegativity differences, and related factors.", "header_images": [], "subquestions": [{"sub_id": "Exam-08-T8-1", "tags": ["1.2", "1.7"], "text_open": "For 8.1–8.3 and 8.5–8.6, describe in words the most stable conformation of each molecule shown, specifying in particular whether each substituent is equatorial or axial.\n8.1", "text_images": [{"file": "Exam-08-T8-1-1-prompt.png", "description": "'CC(C)(C)N1CN(C(C)(C)C)CN(C(C)(C)C)C1'"}], "full_mark": 2, "rubric": ["Most stable conformation: a trisubstituted nitrogen-containing six-membered ring whose skeleton lies approximately in the plane of the page. It contains three nitrogen atoms, 1 at the left, one at the upper right, and one at the lower right. Each nitrogen bears a tert-butyl (tBu) substituent, and all three tBu groups project outward from the ring. Two tBu groups are equatorial and one is axial. (2 points)"], "rubric_images": [{"file": "Exam-08-T8-1-1-rubric.png", "description": "A trisubstituted nitrogen-containing six-membered ring whose skeleton lies approximately in the plane of the page. It contains three nitrogen atoms, 1 at the left, one at the upper right, and one at the lower right. Each nitrogen bears a tert-butyl (tBu) substituent, and all three tBu groups project outward from the ring. Two tBu groups are equatorial and one is axial."}]}, {"sub_id": "Exam-08-T8-2", "tags": ["1.2", "1.7"], "text_open": "8.2", "text_images": [{"file": "Exam-08-T8-2-1-prompt.png", "description": "'CN1N(C)CN(C)N(C)C1'"}], "full_mark": 2, "rubric": ["Most stable conformation: a six-membered ring containing four nitrogen atoms in an approximately chair conformation. Two adjacent N atoms lie on the left and another adjacent pair on the right; carbon atoms at the top and bottom connect the pairs. Each N bears an outward-pointing substituent. The methyl substituents on N atoms at positions 1 and 4 are equatorial, whereas those on N atoms at positions 2 and 5 are axial. (2 points)"], "rubric_images": [{"file": "Exam-08-T8-2-1-rubric.png", "description": "A six-membered ring containing four nitrogen atoms in an approximately chair conformation. Two adjacent N atoms lie on the left and another adjacent pair on the right; carbon atoms at the top and bottom connect the pairs. Each N bears an outward-pointing substituent. The methyl substituents on N atoms at positions 1 and 4 are equatorial, whereas those on N atoms at positions 2 and 5 are axial."}]}, {"sub_id": "Exam-08-T8-3", "tags": ["1.2", "1.7"], "text_open": "8.3", "text_images": [{"file": "Exam-08-T8-3-1-prompt.png", "description": "'CN1N(CC2=CC=CC=C2)CN(C)N(CC3=CC=CC=C3)C1'"}], "full_mark": 2, "rubric": ["Most stable conformation: a six-membered chair containing four nitrogen atoms, arranged as adjacent pairs on the left and right sides. Both Bn substituents are axial, while the other two methyl substituents are equatorial. (2 points)"], "rubric_images": [{"file": "Exam-08-T8-3-1-rubric.png", "description": "A six-membered chair containing four nitrogen atoms, arranged as adjacent pairs on the left and right sides. Both Bn substituents are axial, while the other two methyl substituents are equatorial."}]}, {"sub_id": "Exam-08-T8-4", "tags": ["1.2", "1.6", "1.7"], "text_open": "8.4 Explain, in terms of steric hindrance, hyperconjugation, dipole-dipole interactions, electronegativity differences, and related effects, why the anti-staggered conformation predominates in 1,2-dihaloethanes when the halogen is Cl or Br, whereas the gauche-staggered conformation predominates when the halogen is F. (No more than 100 words.)", "text_images": [], "full_mark": 5, "rubric": ["Cl and Br have large atomic radii. Although C-Cl and C-Br bonds are long, the atoms themselves are bulky, and bringing the two halogens close together in the gauche conformation causes strong van der Waals repulsion. (1 point)", "Because F is strongly electronegative, the C-F antibonding orbital is low in energy and is a good electron acceptor. (1 point)", "In the gauche conformation, a C-H bond and a C-F bond are antiperiplanar, giving effective orbital overlap that substantially lowers the system's energy. (1 point)", "The two strongly polar bonds subtend an angle of 60°, producing substantial dipole-dipole repulsion. (1 point)", "The stabilization from hyperconjugation exceeds the destabilization from dipole-dipole repulsion. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-08-T8-5", "tags": ["1.2", "1.7"], "text_open": "8.5", "text_images": [{"file": "Exam-08-T8-5-1-prompt.png", "description": "'O=C(C)N1CCCC[C@H]1C'"}], "full_mark": 2, "rubric": ["Most stable conformation: an N-acyl-substituted nitrogen-containing cyclic skeleton. At the left is an acetyl fragment, $\\mathrm{CH_3}$-C(=O)-N-, and to the right is a nitrogen-containing ring attached through N. The ring adopts an overall chair conformation. The acetyl group is approximately equatorial, while the methyl substituent at the $\\alpha$ position to N is axial. (2 points)"], "rubric_images": [{"file": "Exam-08-T8-5-1-rubric.png", "description": "An N-acyl-substituted nitrogen-containing cyclic skeleton. At the left is an acetyl fragment, $\\mathrm{CH_3}$-C(=O)-N-, and to the right is a nitrogen-containing ring attached through N. The ring adopts an overall chair conformation. The acetyl group is approximately equatorial, while the methyl substituent at the $\\alpha$ position to N is axial."}]}, {"sub_id": "Exam-08-T8-6", "tags": ["1.2", "1.7"], "text_open": "8.6", "text_images": [{"file": "Exam-08-T8-6-1-prompt.png", "description": "'O=C(C)N1[C@@H](C)CCC[C@H]1C'"}], "full_mark": 2, "rubric": ["Most stable conformation: an N-acyl-substituted nitrogen-containing cyclic skeleton. At the left is an acetyl fragment, $\\mathrm{CH_3}$-C(=O)-N-, and to the right is a nitrogen-containing ring attached through N. The ring adopts an overall chair conformation. The acetyl group is approximately equatorial, and both methyl groups at positions $\\alpha$ to N are axial. (2 points)"], "rubric_images": [{"file": "Exam-08-T8-6-1-rubric.png", "description": "An N-acyl-substituted nitrogen-containing cyclic skeleton. At the left is an acetyl fragment, $\\mathrm{CH_3}$-C(=O)-N-, and to the right is a nitrogen-containing ring attached through N. The ring adopts an overall chair conformation. The acetyl group is approximately equatorial, and both methyl groups at positions $\\alpha$ to N are axial."}]}]}
{"problem_id": "Exam-08-T10", "header": "Brook Rearrangement\nNote: Unless otherwise specified, cyclic structures involved in the Brook-rearrangement process need not be shown for intermediates in this problem. The Brook rearrangement is a nucleophilic rearrangement centered on silyl migration and is commonly used to construct carbon-oxygen and carbon-carbon bonds.", "header_images": [], "subquestions": [{"sub_id": "Exam-08-T10-1", "tags": ["1.2", "2.2", "2.4"], "text_open": "10.1 A typical Brook rearrangement proceeds as shown below.\n10.1.1 Give the SMILES structures of intermediates A and B.", "text_images": [{"file": "Exam-08-T10-1-1-prompt.png", "description": "Under basic conditions, '[R][C@@](O)([2H])[Si]([R])([R])[R]' reversibly forms [A]; [A] reversibly forms [B]; and protonation of [B] gives '[H][C@@]([R])([2H])O[Si]([R])([R])[R]'."}], "full_mark": 2, "rubric": ["A: '[O-][C@]([R])([2H])[Si]([R])([R])[R]'; B: '[R][C-]([2H])O[Si]([R])([R])[R]'. (1 point each.) No credit is awarded unless retention of configuration at the carbon atom is indicated."], "rubric_images": [{"file": "Exam-08-T10-1-1-rubric.png", "description": "A ('[O-][C@]([R])([2H])[Si]([R])([R])[R]') and B ('[R][C-]([2H])O[Si]([R])([R])[R]')."}]}, {"sub_id": "Exam-08-T10-2", "tags": ["2.4"], "text_open": "10.2 Complete the following reactions using the information above. Ignore stereochemistry.\n10.2.1 Give the SMILES structure of product C.", "text_images": [{"file": "Exam-08-T10-2-1-prompt.png", "description": "'O=C([Si](C)(C)C)C1=CC=CC=C1' reacts with 'FC(F)(F)[Si](C)(C)C' in the presence of $[\\mathrm{nBu}_4\\mathrm{N}][\\mathrm{Ph}_3\\mathrm{SnF}_2]$ to give C."}], "full_mark": 2, "rubric": ["C: 'F/C(F)=C(O[Si](C)(C)C)/C1=CC=CC=C1'. (2 points)"], "rubric_images": [{"file": "Exam-08-T10-2-1-rubric.png", "description": "C ('F/C(F)=C(O[Si](C)(C)C)/C1=CC=CC=C1')."}]}, {"sub_id": "Exam-08-T10-3", "tags": ["2.4"], "text_open": "10.2.2 Give the SMILES structure of product D. Hint: D contains a TMS group but no O-Si bond.", "text_images": [{"file": "Exam-08-T10-3-1-prompt.png", "description": "'BrC1CC1CC2=CC=CC=C2' is treated with tBuLi in $\\mathrm{Et_2O}$ at -78 °C, then with 'O=C([Si](C)(C)C)C1=CC=CC=C1' from -78 °C to room temperature for 30 min, followed by $\\mathrm{NH_4Cl}$ and $\\mathrm{H_2O}$, to give D."}], "full_mark": 2, "rubric": ["D: 'O=C(CC(CC1=CC=CC=C1)C[Si](C)(C)C)C2=CC=CC=C2'. (2 points)"], "rubric_images": [{"file": "Exam-08-T10-3-1-rubric.png", "description": "D ('O=C(CC(CC1=CC=CC=C1)C[Si](C)(C)C)C2=CC=CC=C2')."}]}, {"sub_id": "Exam-08-T10-4", "tags": ["2.4"], "text_open": "10.3 The Brook rearrangement has broad applications and can be used to construct an eight-membered ring stereoselectively. The Brook rearrangement occurs during formation of E, and G contains two fewer seven-membered rings than F.\n10.3.1 Ignoring stereochemistry, give the SMILES structures of intermediates E and F.", "text_images": [{"file": "Exam-08-T10-4-1-prompt.png", "description": "'O=C([Si](C)(C)C(C)(C)C)/C=C/C(C)(C)C' reacts with '[Li]OC1=CCCCC=C1' in THF at -80 °C to give E. E reacts with 'O=S(N1C(C2=CC=CC=C2)O1)(C3=CC=CC=C3)=O' under NaHMDS conditions to give F. F is treated with $\\mathrm{Pb(OAc)_4}$ in MeOH/benzene to give G ('O=C(C1CCCC(C=O)C(C(C)(C)C)/C=C1/O[Si](C)(C)C(C)(C)C)OC'; the stereochemistry of its three stereogenic carbon atoms is not shown)."}], "full_mark": 2, "rubric": ["E: 'O=C(C1)[C@H]2CCC[C@@H]1[C@@H](C(C)(C)C)C=C2O[Si](C)(C)C(C)(C)C'; F: 'O=C(C1O)C2CCCC1[C@H](C(C)(C)C)C=C2O[Si](C)(C)C(C)(C)C'. (1 point each; stereochemistry is not required.)"], "rubric_images": [{"file": "Exam-08-T10-4-1-rubric.png", "description": "E: 'O=C(C1)[C@H]2CCC[C@@H]1[C@@H](C(C)(C)C)C=C2O[Si](C)(C)C(C)(C)C'; F: 'O=C(C1O)C2CCCC1[C@H](C(C)(C)C)C=C2O[Si](C)(C)C(C)(C)C'."}]}, {"sub_id": "Exam-08-T10-5", "tags": ["1.2"], "text_open": "10.3.2 Give a SMILES structure that shows the relative stereochemistry of the stereogenic centers in G.", "text_images": [], "full_mark": 2, "rubric": ["G: 'O=C([C@H]1CCC[C@@H](C=O)[C@@H](C(C)(C)C)/C=C1/O[Si](C)(C)C(C)(C)C)OC'. (2 points; no credit if the stereochemistry is incorrect.)"], "rubric_images": [{"file": "Exam-08-T10-5-1-rubric.png", "description": "G ('O=C([C@H]1CCC[C@@H](C=O)[C@@H](C(C)(C)C)/C=C1/O[Si](C)(C)C(C)(C)C)OC')."}]}, {"sub_id": "Exam-08-T10-6", "tags": ["1.5", "2.3"], "text_open": "10.4 In recent years, research on Brook rearrangements has expanded beyond silicon systems, and phosphorus-mediated rearrangements have received increasing attention. The following is a new two-step, one-pot synthesis of tetrasubstituted furans using a Brook-like rearrangement.\n10.4.1 Briefly explain why P(V), rather than P(III), is used for the Brook-like rearrangement. (No more than 100 words.)", "text_images": [{"file": "Exam-08-T10-6-1-prompt.png", "description": "'[R1]C(O)(P(OC)(OC)=O)C#C[R2]' reacts with '[R3]C([H])=O' under tBuOK (10 mol%), DMF, -60 °C, 3 h to give H. H then reacts with 'O=C(N1I)CCC1=O' (NIS) to give '[R1]C1=C(I)C([R2])=C([R3])O1'."}], "full_mark": 1, "rubric": ["A phosphate ester formed by P(V) can accept electron attack to form a pentacoordinate intermediate. P(III), with its lone pair, is predominantly nucleophilic and is less able to accept electron attack. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-08-T10-7", "tags": ["2.2", "2.3", "2.4"], "text_open": "10.4.2 Give the SMILES structure of H and at least two charged intermediates formed on the pathway to H.", "text_images": [], "full_mark": 4, "rubric": ["H: '[R1]C(OP(OC)(OC)=O)=C=C(C(O)[R3])[R2]'. (2 points; no other structure is accepted.)", "Two charged intermediates in the formation of H are '[R1]C(C#C[R3])(P(OC)(OC)=O)[O-]' and '[R1][C-](OP(OC)(OC)=O)C#C[R3]'. (1 point each; reasonable resonance forms also earn credit.)"], "rubric_images": [{"file": "Exam-08-T10-7-1-rubric.png", "description": "H: '[R1]C(OP(OC)(OC)=O)=C=C(C(O)[R3])[R2]'."}, {"file": "Exam-08-T10-7-2-rubric.png", "description": "First: '[R1]C(C#C[R3])(P(OC)(OC)=O)[O-]'; second: '[R1][C-](OP(OC)(OC)=O)C#C[R3]'."}]}, {"sub_id": "Exam-08-T10-8", "tags": ["2.4"], "text_open": "10.4.3 I is a phenanthrene derivative. Using the information above, give the SMILES structure of product I.", "text_images": [{"file": "Exam-08-T10-8-1-prompt.png", "description": "'[Ar]C#CC1=CC=CC=C1C2=CC=CC=C2C(C(OCC)=O)=O' is treated with HPO(OiPr)_2 and P2-tBu (10 mol%) in refluxing EtOAc for 7 h to give I."}, {"file": "Exam-08-T10-8-2-prompt.png", "description": "P2-tBu: 'CN(P(N=P(N(C)C)(N(C)C)N(C)C)(N(C)C)=NC(C)(C)C)C'."}], "full_mark": 4, "rubric": ["I: '[Ar]C(OP(OC(C)C)(OC(C)C)=O)C1=C(C=CC=C2)C2=C(C=CC=C3)C3=C1C(OCC)=O'. (3 points)"], "rubric_images": [{"file": "Exam-08-T10-8-1-rubric.png", "description": "I: '[Ar]C(OP(OC(C)C)(OC(C)C)=O)C1=C(C=CC=C2)C2=C(C=CC=C3)C3=C1C(OCC)=O'."}]}]}
{"problem_id": "Exam-09-T5", "header": "Reviewing the Old to Learn the New: Revisiting a Hexagonal Laves Phase\nA magnesium-zinc alloy belongs to the hexagonal crystal system, with unit-cell parameters a = 0.5157 nm and c = 0.8373 nm. Projections of the unit cell along different directions are shown below. The large light-colored spheres represent Mg atoms, and the small dark-colored spheres represent Zn atoms. Interestingly, the projections along the a and b directions are identical. One Zn atom has coordinates (0.830, 0.170, 0.250). ", "header_images": [{"file": "Exam-09-T5-header-1.png", "description": "The image presents, side by side, three two-dimensional projections of a crystal structure on white backgrounds, each enclosed by thin solid gray lines and composed of small dark-gray spheres and large light-gray spheres. Each projection includes blue coordinate axes with arrows. The left projection is a vertically elongated rectangular unit-cell projection showing the bc plane. At its lower-left corner, the coordinate axes indicate the b axis horizontally to the right and the c axis vertically upward. A total of 10 small dark-gray spheres are visible in the rectangle: four at the vertices, two at the midpoints of the vertical edges, and 4 within the cell arranged from bottom to top in a staggered left-right-left-right zigzag along the vertical direction. Another 4 large light-gray spheres are distributed inside, arranged from bottom to top in the opposite right-left-right-left sequence. The middle projection immediately beside it is a rectangular projection of the ac plane. Its lower-left axes indicate the c axis vertically upward but the a axis horizontally to the left. This view is completely identical to the left view in cell shape and in the number, positions, and visual arrangement of all dark- and light-gray spheres: 4 dark spheres at the vertices, 2 at the midpoints of the vertical edges, 4 additional dark spheres staggered left and right inside, and 4 large light spheres staggered right and left. The rightmost projection shows the ab plane. Its lower-left coordinate axes indicate the a axis horizontally to the right and the b axis obliquely upward to the left, forming an obtuse angle and defining a parallelogram unit-cell outline with a horizontal base and sides inclined upward to the left. A small dark-gray sphere lies at each of the four vertices. Another 6 small dark-gray spheres are arranged in a ring inside the cell: one each near the left and right interiors of the top horizontal edge, one each near the left and right interiors of the bottom horizontal edge, and one near the middle interior of each inclined side. Among these 6 internal dark spheres, only 2 large light-gray spheres lie in diagonal relation. One occupies an open region in the upper-left part of the cell and is partially surrounded by 3 internal dark spheres; the other occupies the lower-right region and is likewise partially surrounded by 3 internal dark spheres."}], "subquestions": [{"sub_id": "Exam-09-T5-1", "tags": ["1.3"], "text_open": "5.2 Calculate the shortest Zn-Zn distance. ", "text_images": [], "full_mark": 6, "rubric": ["The possible shortest Zn-Zn distances may occur between Zn1-Zn2, Zn2-Zn3, or Zn2-Zn4. First write their coordinates: Zn1(0, 0, 0), Zn2(0.830, 0.170, 0.250), Zn3(0.830, 0.660, 0.250), Zn4(0.830, -0.340, 0.250) (1 point). Calculate the Zn1-Zn2 distance, retaining one extra significant figure during the calculation: $d_a = |0.830 - 1| \\times 0.5157 \\text{ nm} = 0.08767 \\text{ nm}$, $d_b = |0.170 - 0| \\times 0.5157 \\text{ nm} = 0.08767 \\text{ nm}$, $d_c = |0.250 - 0| \\times 0.8373 \\text{ nm} = 0.2093 \\text{ nm}$, and $d(\\text{Zn1-Zn2}) = \\sqrt{d_a^2 + d_b^2 - 2 d_a d_b \\cos \\theta + d_c^2}$ $= \\sqrt{0.08767^2 + 0.08767^2 - 2 \\times 0.08767^2 \\times (-1/2) + 0.2093^2} = 0.259 (\\text{nm})$ (2 points). Calculate the Zn2-Zn3 distance: $d(\\text{Zn2-Zn3}) = (0.660 - 0.170) \\times 0.5157 \\text{ nm} = 0.253 (\\text{nm})$ (1 point). Calculate the Zn2-Zn4 distance: $d(\\text{Zn2-Zn4}) = 2 \\times (0.170 - (-0.340)) \\times 0.5157 \\text{ nm} = 0.263 (\\text{nm})$ (1 point). Therefore, the shortest Zn-Zn distance is Zn2-Zn3 and equals 0.253nm (1 point)."], "rubric_images": []}]}
{"problem_id": "Exam-09-T8", "header": "Problem 8 Fundamentals of Organic Chemistry", "header_images": [], "subquestions": [{"sub_id": "Exam-09-T8-1", "tags": ["1.1", "1.5"], "text_open": "Let the bond angles shown below be α and β, respectively, and define γ = α - β. Rank the following substances in decreasing order of γ.", "text_images": [{"file": "Exam-09-T8-1-1-prompt.png", "description": "A: 'O=C1OCC1'; B: 'O=C1OCCC1'; C: 'O=C1OCCCC1'. The image labels the two bond angles α and β, where α is the carbonyl O-C-C bond angle and β is the ester O-C-O bond angle."}], "full_mark": 3, "rubric": [" ABC (3 points)", "Each blank is worth 1 point; award that 1 point if correct, and that 1 point is not awarded if incorrect."], "rubric_images": []}, {"sub_id": "Exam-09-T8-2", "tags": ["2.3", "3.1", "3.7"], "text_open": "Reaction background: The rate constant of the reaction shown below is significantly greater in acetone than in ethanol.\nCompared with the reaction in ethanol, how do the Gibbs free energies of the reactant ground state and transition state change relative to one another in acetone, thereby decreasing the activation energy? (a) Only the transition-state energy decreases. (b) Only the reactant energy increases. (c) The reactant energy decreases and the transition-state energy also decreases. (d) None of the above (write the correct answer as well).", "text_images": [{"file": "Exam-09-T8-2-1-prompt.png", "description": "1-Bromopropane reacts with sodium iodide in a solvent: 'BrCCC' gives 'ICCC' under NaI and Solv. conditions."}], "full_mark": 4, "rubric": [" (d) (2 points). The energies of both the reactant and transition state are higher than in ethanol, but the reactant energy increases by more than the transition-state energy (2 points)."], "rubric_images": []}]}
{"problem_id": "Exam-09-T10", "header": "Indium-Catalyzed Alkyne-Migration Reaction\n4-Alkynyl-2-quinolone derivatives occur widely in bioactive molecules with antibacterial, antitumor, and other activities. Indium-catalyzed alkyne migration provides a new route for efficiently synthesizing these compounds. Using ortho-alkynyl benzamides as substrates, the method efficiently constructs a 4-alkynyl-2-quinolone framework under $\\mathrm{In(OTf)_3}$ catalysis accompanied by selective alkyne migration. Analyze the reaction process and answer the questions.", "header_images": [], "subquestions": [{"sub_id": "Exam-09-T10-1", "tags": ["2.2", "2.3"], "text_open": "10.1 Alkyne migration involves transposition of an alkynyl group along a carbon chain, allowing interconversion between internal and terminal alkynes. A classic example is the Alkyne Zipper Reaction, which uses KAPA reagent to move an alkyne to the terminus. Write the SMILES structures of the 3 key uncharged intermediates in this reaction.", "text_images": [{"file": "Exam-09-T10-1-1-prompt.png", "description": "Substrate: 'CCC#CC1=CC=CC=C1'. The reagent is KAPA, represented by SMILES 'NCCCN[K]', in which the terminal -NH_2 chelates to K to form a six-membered ring. Product: 'C#CCCC1=CC=CC=C1'. The figure shows migration of the alkyne from an internal position to the terminus."}], "full_mark": 3, "rubric": ["3 points, 1 point each. Reference SMILES: 'CC=C=C(C1=CC=CC=C1)', 'CC#CCC1=CC=CC=C1', and 'C=C=CCC1=CC=CC=C1'."], "rubric_images": [{"file": "Exam-09-T10-1-1-rubric.png", "description": "'CC=C=C(C1=CC=CC=C1)', 'CC#CCC1=CC=CC=C1', and 'C=C=CCC1=CC=CC=C1'"}]}, {"sub_id": "Exam-09-T10-2", "tags": ["2.8"], "text_open": "10.4 The authors used the ortho-alkynyl benzamide substrate above to study the alkyne-migration process. Analyze the following conversion and answer the questions. 10.4.1 The authors proposed that In(III) also catalyzes isomerization of B to A. The process proceeds smoothly in 1,4-dioxane, giving A in 89 % NMR yield. In contrast, under otherwise identical conditions in chlorobenzene or toluene, only trace A is detected. Explain why.", "text_images": [{"file": "Exam-09-T10-2-1-prompt.png", "description": "Substrate: 'O=C(C(C(C)C)=O)N(C)C1=CC=CC=C1C#CCCCC'; conditions: In(OTf)_3 and 1,4-dioxane (1,4-dioxane). Product A: 'CN1C2=CC=CC=C2C(C#CCCC)=C(C(C)C)C1=O'; product B: 'CN(C([C@@]1(O)C(C)C)=O)C2=CC=CC=C2C1=[C@@]=C([H])CC'."}], "full_mark": 2, "rubric": ["1,4-Dioxane can stabilize the carbocation intermediate and promote the dehydration pathway from B to A (2 points)."], "rubric_images": []}, {"sub_id": "Exam-09-T10-3", "tags": ["1.6", "2.7"], "text_open": "10.4.2 If the catalyst is replaced by a stronger Lewis acid, such as $\\mathrm{Fe(OTf)_3}$ or $\\mathrm{Sm(OTf)_3}$, the yield of product A instead decreases substantially. Explain why.", "text_images": [], "full_mark": 2, "rubric": ["As oxophilic Lewis acids/hard acids, Fe(OTf)_3 and Sm(OTf)_3 coordinate strongly with 1,4-dioxane and thereby inhibit dehydrative isomerization (2 points)."], "rubric_images": []}, {"sub_id": "Exam-09-T10-4", "tags": ["2.4"], "text_open": "10.5 The authors also explored downstream transformations of 4-alkyl-2-quinolone products. Write the SMILES structures of the following 2 products. Pay attention to stereochemistry in reaction 1. Reaction 2 is a ring-expansion reaction of another quinolone derivative.", "text_images": [{"file": "Exam-09-T10-4-1-prompt.png", "description": "Reaction 1 reactant: 'O=C1N(C2=CC=CC=C2C(C#CCCC)=C1C)C'; conditions: C=C[R], Thioxanthen-9-one (0.3 eq.), 390 nm LEDs, CH_3CN, 6-9h; product unknown. Reaction 2 reactant: 'O=C1N(C)C2=CC=CC=C2C(C#CCCC)=C1C3=CC=CC=C3'; conditions: HOTf (2.0 eq.), PhMe, 100°C, 24h."}], "full_mark": 4, "rubric": ["4 points, 2 points each. Reference SMILES: product 1 is 'CN1C2=CC=CC=C2[C@@]3(C#CCCC)[C@@](C[C@@H]3[R])(C)C1=O' (the relative stereochemistry shown must be retained); product 2 is 'CN1C2=CC=CC=C2C3=C(C4=CC=CC=C4C(CCC)=C3)C1=O'."], "rubric_images": [{"file": "Exam-09-T10-4-1-rubric.png", "description": "Product 1 is 'CN1C2=CC=CC=C2[C@@]3(C#CCCC)[C@@](C[C@@H]3[R])(C)C1=O'; product 2 is 'CN1C2=CC=CC=C2C3=C(C4=CC=CC=C4C(CCC)=C3)C1=O'."}]}]}
{"problem_id": "Exam-12-T1", "header": "Problem 1 Metallic Vitamins. Rare-earth elements are core strategic resources in modern materials and chemical industries and are known as \"industrial vitamins.\" Their unique electron configurations give them distinctive physicochemical properties, making them essential materials for emerging industries such as modern semiconductors and new energy. A rare-earth element M has density $7.563 \\ \\mathrm{g \\cdot cm^{-3}}$ and forms an alloy with Co in a fixed ratio for manufacturing strong magnets. It has the following chemical properties. (1) When elemental M is left in air, its metallic luster rapidly disappears and a composite thin layer whose principal component is oxide A forms on its surface. M reacts with hydrochloric acid to form blood-red chloride B. After M has dissolved completely, the blood-red color rapidly fades, producing pale-yellow chloride C and colorless gas D. (2) If M is placed in a tetrahydrofuran solution of an organic compound E containing $90.05\\%$ iodine, reaction at room temperature gives unstable green iodide F and releases gas G. Addition of a small amount of water turns the solution yellow. Gas G is a common plant-ripening hormone.", "header_images": [], "subquestions": [{"sub_id": "Exam-12-T1-1", "tags": ["1.4", "2.4"], "text_open": "1-4-1 Complexes of M also have interesting structures. Addition of 0.5 equivalents of $\\mathrm{P_2Ph_4}$ to $\\mathrm{MCp*_2(THF)_2}$ gives a chain polymer X. After polymer X is separated, the pentane mother liquor is left for several weeks, yielding a yellow-brown complex Y. Y contains two M atoms and a 14-membered macrocycle. The mass fraction of M is $22.19\\%$ in both X and Y. Write the SMILES structure of the repeat unit of X, using [R] for the site connected to the next repeat unit (for example, polypropylene may be written as C([R])C(C)[R]), and the SMILES structure of Y ($\\mathrm{Cp*}$ = pentamethylcyclopentadienyl anion).", "text_images": [], "full_mark": 4, "rubric": ["(2 points each for the structures of X and Y)", "Structure or SMILES corresponding to the scoring figure: X: the polymer repeat-unit SMILES is '[R][Sm]([Cp*])([Cp*])OCCCC[P](C1=CC=CC=C1)(C2=CC=CC=C2)[R]', where R denotes the connection site between polymer monomers. The SMILES of Y is '[*Cp][Sm]1([*Cp])OCCCC[P](C2=CC=CC=C2)(C3=CC=CC=C3)[Sm]([Cp*])([Cp*])OCCCC[P]1(C4=CC=CC=C4)C5=CC=CC=C5'."], "rubric_images": [{"file": "Exam-12-T1-1-1-rubric.png", "description": "X: the polymer repeat-unit SMILES is '[R][Sm]([Cp*])([Cp*])OCCCC[P](C1=CC=CC=C1)(C2=CC=CC=C2)[R]', where R denotes the connection site between polymer monomers. The SMILES of Y is '[*Cp][Sm]1([*Cp])OCCCC[P](C2=CC=CC=C2)(C3=CC=CC=C3)[Sm]([Cp*])([Cp*])OCCCC[P]1(C4=CC=CC=C4)C5=CC=CC=C5'."}]}]}
{"problem_id": "Exam-12-T4", "header": "Problem 4 Titrimetric Analysis of Etching Acids\nWet chemical etching is a major process in the semiconductor-chip industry. It uses acidic substances to corrode a substrate surface and thereby achieve etching. A mixture of hydrofluoric acid, fluorosilicic acid, and nitric acid is a common etchant. A rapid and accurate laboratory method for determining the contents of these three acids in an etchant is described below.", "header_images": [], "subquestions": [{"sub_id": "Exam-12-T4-1", "tags": ["4.2"], "text_open": "Procedure background: Accurately weigh 3.028 g of the sample. Place 1.502 g of it in a plastic beaker to prepare test solution A, and use the remaining sample to prepare test solution B by the same procedure. During preparation, add 10 mL of saturated $KNO_3$ solution and anhydrous ethanol, mix, dilute with anhydrous ethanol, and let stand for approximately 20 min until the precipitate has completely settled into a separate layer. Label this as test solution A. Prepare test solution B from the remaining sample by the same procedure.\n4.1.2 State the purpose of adding saturated KNO_3 solution.", "text_images": [], "full_mark": 2, "rubric": ["Adding KNO_3 precipitates SiF_6^{2-}, preventing interference from hydrolysis of SiF_6^{2-} (2 points)."], "rubric_images": []}, {"sub_id": "Exam-12-T4-2", "tags": ["2.1", "4.2"], "text_open": "4.2 Titrimetric analysis. Add 30 mL of water and 3 drops of phenolphthalein indicator to test solution A. Titrate with 0.5018 $\\mathrm{mol \\cdot L^{-1}}$ standard NaOH solution until the solution becomes faint pink; 24.39 mL of the standard NaOH solution is consumed. Add 30 mL of water and 2 drops of methyl-orange indicator to test solution B. Adjust the pH with NaOH solution until the solution just turns bright yellow. Using a fluoride-ion combination electrode as the measuring electrode, titrate with 0.08273 $\\mathrm{mol \\cdot L^{-1}}$ standard La(NO_3)_3 solution; precipitate forms continuously during the titration. At the endpoint, 25.11 mL of La(NO_3)_3 solution is consumed. Weigh 0.5003 g of the original sample into a beaker containing 15 mL of 0.5 $\\mathrm{mol \\cdot L^{-1}}$ NaOH solution, add water to 60 mL, and verify with pH paper that the pH is greater than 13. Heat slowly to boiling, maintain a gentle boil for 5 min, and cool to room temperature. Add 2 drops of methyl-red indicator, adjust with nitric acid until red, and then adjust with NaOH solution until the solution just turns bright yellow. Using a fluoride-ion combination electrode as the indicator electrode, titrate with 0.08273 $\\mathrm{mol \\cdot L^{-1}}$ standard La(NO_3)_3 solution. At the endpoint, 15.28 mL is consumed. 4.2.1 Write all reaction equations occurring in Sections 4.1 and 4.2; write repeated reactions only once.", "text_images": [], "full_mark": 4, "rubric": ["$2KNO_3 + H_2SiF_6 = 2HNO_3 + K_2SiF_6$ (1 point)  $NaOH + HNO_3 = NaNO_3 + H_2O$ (0.5 point)  $NaOH + HF = NaF + H_2O$ (0.5 point)  $La(NO_3)_3 + 3NaF = 3NaNO_3 + LaF_3$ (1 point)  $H_2SiF_6 + 8NaOH = Na_2SiO_3 + 6NaF + 5H_2O$ (1 point)"], "rubric_images": []}, {"sub_id": "Exam-12-T4-3", "tags": ["4.7", "4.5"], "text_open": "4.2.2 Calculate the mass fractions of HF, H_2SiF_6, and HNO_3 in the sample. Note: the fluoride-ion electrode indicates the endpoint by the abrupt change in fluoride-ion concentration.", "text_images": [], "full_mark": 6, "rubric": ["Let the mass fractions of HF, $H_2SiF_6$, and $HNO_3$ be $w_1$, $w_2$, and $w_3$, respectively.", "$0.5018\\ \\mathrm{mol\\cdot L^{-1}}\\times24.39\\ \\mathrm{mL}=1.502\\ \\mathrm{g}\\times\\left(\\frac{w_1}{20.008}+\\frac{2w_2}{144.106}+\\frac{w_3}{63.018}\\right)$", "$3\\times0.08273\\ \\mathrm{mol\\cdot L^{-1}}\\times25.11\\ \\mathrm{mL}=(3.028-1.502)\\ \\mathrm{g}\\times\\frac{w_1}{20.008}$", "$3\\times0.08273\\ \\mathrm{mol\\cdot L^{-1}}\\times15.28\\ \\mathrm{mL}=0.5003\\ \\mathrm{g}\\times\\left(\\frac{w_1}{20.008}+\\frac{6w_2}{144.106}\\right)$", "The system of equations is worth 3 points, 1 point for each equation. Solving the 3 equations simultaneously gives $w_1=8.171\\%, w_2=8.397\\%, w_3=18.27\\%$. The answers are worth 3 points, 1 point each. Significant figures are required. Intermediate results must contain at least 4 significant figures; deduct 1 point for any significant-figure error, only once in total."], "rubric_images": []}]}
{"problem_id": "Exam-12-T10", "header": "Phenonium-Ion Chemistry\nPhenonium ions are highly reactive cationic intermediates whose applications were long limited by difficult preparation and poor reaction control. In recent years, however, physical-organic studies and new strategies have greatly expanded their synthetic potential.", "header_images": [], "subquestions": [{"sub_id": "Exam-12-T10-1", "tags": ["2.2"], "text_open": "10.1 Kinetic measurements provide an indirect method for studying phenonium-ion intermediates. Consider the following solvolysis in glacial acetic acid. 10.1.1 Two mechanisms are possible: one-step path A and two-step path B. Write the SMILES structure of the intermediate in path B.", "text_images": [{"file": "Exam-12-T10-1-1-prompt.png", "description": "The figure shows a reaction in which 'CC1=CC=C(S(OCCC2=CC=CC=C2)(=O)=O)C=C1' reacts in HOAc to give 'O=C(OCCC1=CC=CC=C1)C'."}], "full_mark": 2, "rubric": ["The intermediate in path B is shown in the rubric figure. (2 points; a resonance structure also earns credit)", "Structure or SMILES corresponding to the rubric figure: 'C1=CC=CC2(CC2)[CH+]1'"], "rubric_images": [{"file": "Exam-12-T10-1-1-rubric.png", "description": "'C1=CC=CC2(CC2)[CH+]1'"}]}, {"sub_id": "Exam-12-T10-2", "tags": ["2.3", "3.7"], "text_open": "10.1.3 Solvolysis kinetics were similarly studied for multiple precursors capable of forming phenonium ions, giving two different kinetic patterns, Figure 1 and Figure 2. Briefly explain mechanistically the behavior in Figure 2.", "text_images": [{"file": "Exam-12-T10-2-1-prompt.png", "description": "Figures 1 and 2 show two kinetic plots for phenonium-ion precursors. Each has only a solid left vertical axis and bottom horizontal axis, forming an L-shaped frame with no numerical ticks. Both vertical axes are labeled $\\lg(k_X/k_H)$ and both horizontal axes Hammett $\\sigma$ scale. The left plot contains two intersecting dashed straight lines. One has a steep negative slope, is labeled $k_\\Delta$-line at its upper start, and slope: $\\rho_\\Delta$ near its middle. The other has a very shallow negative slope, crosses the lower half and intersects the first line near the lower left; it is labeled $k_s$-line at its left end and slope: $\\rho_s$ on the right. The right plot contains only the steep negatively sloped dashed line, similarly labeled $k_\\Delta$-line and slope: $\\rho_\\Delta$."}], "full_mark": 2, "rubric": ["For both electron-withdrawing and electron-donating substituents, the mechanism proceeds through a phenonium-ion intermediate. (1 point)", "This indicates that the carbocation is relatively stable. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-12-T10-3", "tags": ["2.3", "3.7"], "text_open": "10.1.4 The following two substrates correspond to the two cases in 10.1.3. Which has the kinetic behavior in Figure 1? (      ). Which has the behavior in Figure 2? (      ).", "text_images": [{"file": "Exam-12-T10-3-1-prompt.png", "description": "Two phenonium-ion precursors, (a) and (b), are shown. With no benzene substituent, (a) is 'CC1=CC=C(S(OC(C)CC2=CC=CC=C2)(=O)=O)C=C1' and (b) is 'CC1=CC=C(S(OC(C)(C)CC2=CC=CC=C2)(=O)=O)C=C1'. Either may be substituted on the benzene ring, so benzene-X denotes a substituent of unspecified position on the ring."}], "full_mark": 2, "rubric": ["(b) (1 point)", "(a) (1 point)"], "rubric_images": []}, {"sub_id": "Exam-12-T10-4", "tags": ["2.2"], "text_open": "10.2 In recent years, chemists have developed various synthetic strategies to overcome the limitations on phenonium-ion formation. The use of hypervalent iodine reagents marked an important turning point for phenonium ions in synthesis; the core strategy is to introduce a “super” leaving group. The following highly regioselective difluorination has been reported. (Amine: an amine compound used to regulate the acid-base conditions of the system.) Hypervalent iodine reagent X is first generated in situ during the reaction.\n10.2.2 Write the SMILES structures of 2 key reaction intermediates, excluding X.", "text_images": [{"file": "Exam-12-T10-4-1-prompt.png", "description": "The figure shows a one-step reaction. Reactant 'C=C(C)C1=CC=C([N+]([O-])=O)C=C1' is treated with 'CC1=CC=C(I)C=C1' (10 mol%), 'ClC[N+]1(CC2)CC[N+]2(F)CC1' (1.5 eq.), Amine:HF ≈ 1:9, in CHCl_3 at r.t., giving 'CC(F)(F)CC1=CC=C([N+]([O-])=O)C=C1'."}], "full_mark": 2, "rubric": ["The two key intermediates are shown in the rubric figure. 2 points total, 1 point each.", "Structures or SMILES corresponding to the rubric figure: the 2 key intermediates excluding X are 'CC(CI(F)[Ar])(F)C1=CC=C([N+]([O-])=O)C=C1' on the left and 'CC1(F)CC21C=CC([N+]([O-])=O)=C[CH+]2' on the right."], "rubric_images": [{"file": "Exam-12-T10-4-1-rubric.png", "description": "The 2 key intermediates excluding X are 'CC(CI(F)[Ar])(F)C1=CC=C([N+]([O-])=O)C=C1' on the left and 'CC1(F)CC21C=CC([N+]([O-])=O)=C[CH+]2' on the right."}]}, {"sub_id": "Exam-12-T10-5", "tags": ["2.7"], "text_open": "10.2.3 Briefly explain the regioselectivity of the final step.", "text_images": [], "full_mark": 1, "rubric": ["$F^-$ attacks the electron-deficient carbon atom and therefore attacks the carbon bearing F. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-12-T10-6", "tags": ["2.4"], "text_open": "10.2.4 Complete the following reaction.", "text_images": [{"file": "Exam-12-T10-6-1-prompt.png", "description": "The figure shows a one-step reaction. Reactant 'O=[N+](C1=CC=C2C(CCC/C2=C\\C(OCC)=O)=C1)[O-]' is treated with 'CC1=CC=C(I)C=C1' (20 mol%), mCPBA (1.5 eq.), Amine:HF = 1:7, in CHCl_3 at r.t. The product is unknown and indicated by ?."}], "full_mark": 2, "rubric": ["The reaction product is shown in the rubric figure. (2 points)", "Structure or SMILES corresponding to the rubric figure: 'CC1=CC=C2C(CCCC(F)(F)C2C(OCC)=O)=C1'"], "rubric_images": [{"file": "Exam-12-T10-6-1-rubric.png", "description": "'CC1=CC=C2C(CCCC(F)(F)C2C(OCC)=O)=C1'"}]}, {"sub_id": "Exam-12-T10-7", "tags": ["2.2"], "text_open": "10.3 Diazo reagents have also been used to generate phenonium-ion intermediates. Examine the following reaction. 10.3.1 Write the SMILES structure of the phenonium-ion intermediate.", "text_images": [{"file": "Exam-12-T10-7-1-prompt.png", "description": "The figure shows a one-step reaction between 'BrCC1=CC=C(OC)C=C1' and 'CCC(C(OCC1=CC=CC=C1)=O)=[N+]=[N-]' under SnBr_4 (50 mol%) in CH_2Cl_2 from -78℃ to r.t. The product is unknown."}], "full_mark": 1, "rubric": ["The phenonium-ion intermediate is shown in the rubric figure. (1 point)", "Structure or SMILES corresponding to the rubric figure: 'COC1=C[CH+]C2(CC2(C(OCC3=CC=CC=C3)=O)CC)C=C1'"], "rubric_images": [{"file": "Exam-12-T10-7-1-rubric.png", "description": "'COC1=C[CH+]C2(CC2(C(OCC3=CC=CC=C3)=O)CC)C=C1'"}]}, {"sub_id": "Exam-12-T10-8", "tags": ["2.2", "2.4"], "text_open": "10.3.2 The favored transition state for opening the phenonium ion is known to be a six-membered cyclic transition state. Write the SMILES of the major product; stereochemistry is not required.", "text_images": [], "full_mark": 1, "rubric": ["The major product is shown in the rubric figure. (1 point)", "Structure or SMILES corresponding to the rubric figure: 'BrCC(CC)(C(OCC1=CC=CC=C1)=O)C2=CC=C(OC)C=C2'"], "rubric_images": [{"file": "Exam-12-T10-8-1-rubric.png", "description": "'BrCC(CC)(C(OCC1=CC=CC=C1)=O)C2=CC=C(OC)C=C2'"}]}, {"sub_id": "Exam-12-T10-9", "tags": ["1.1", "2.2"], "text_open": "10.4 Another strategy for obtaining phenonium ions is to synthesize long-lived carbocations. Examine the following reaction; configurations of stereogenic carbons are not marked in the product. 10.4.1 Write the SMILES structures of the key intermediates, including stereochemistry.", "text_images": [{"file": "Exam-12-T10-9-1-prompt.png", "description": "The figure shows a one-step reaction. Reactant 'O=C(/C(C(OC)=O)=C/C1=CC=CC=C1)/C(C)=C(C2=CC=CC=C2)\\CC' is treated with Cu(SbF_6)_2 (1 eq.) in CH_2Cl_2 from r.t. to 40°C, giving 'O=C1C(C(OC)=O)=CC(C2=CC=CC=C2)(CC)C1(C)C3=CC=CC=C3'."}], "full_mark": 6, "rubric": ["The key intermediates are shown in the rubric figure. 6 points total. Deduct 1 point for each incorrect organic skeleton, up to 4 points. Chelating coordination between the organic skeleton and copper must be shown, and it must be stated in words or otherwise indicated that copper bears additional ligands besides the organic skeleton; otherwise deduct 1 point once. For stereochemical errors, deduct 1 point once.", "Structures or SMILES corresponding to the rubric figure, from left to right: '[H][C@@]1(C2=CC=CC=C2)[C@](C3=CC=CC=C3)(CC)[C+](C)C4=C1C(OC)=O[#29]O4', '[H][C@@]1(C2=CC=CC=C2)[C@]3(CC)[C@](C34[CH+]C=CC=C4)(C)C5=C1C(OC)=O[#29]O5', '[H][C@]1([C+]([C@@](C2=CC=CC=C2)(C3=C1C(OC)=O[Cu]O3)C)CC)C4=CC=CC=C4', and '[H][C@@]12[C@@]([C@@]23[CH+]C=CC=C3)(CC)[C@@](C)(C4=CC=CC=C4)C5=C1C(OC)=O[#29]O5'. The Cu catalytic center is denoted [Cu], indicating that in addition to chelation by the organic skeleton, the Cu ion bears other ligands that are irrelevant to the reaction."], "rubric_images": [{"file": "Exam-12-T10-9-1-rubric.png", "description": "The four intermediates from left to right are '[H][C@@]1(C2=CC=CC=C2)[C@](C3=CC=CC=C3)(CC)[C+](C)C4=C1C(OC)=O[#29]O4', '[H][C@@]1(C2=CC=CC=C2)[C@]3(CC)[C@](C34[CH+]C=CC=C4)(C)C5=C1C(OC)=O[#29]O5', '[H][C@]1([C+]([C@@](C2=CC=CC=C2)(C3=C1C(OC)=O[Cu]O3)C)CC)C4=CC=CC=C4', and '[H][C@@]12[C@@]([C@@]23[CH+]C=CC=C3)(CC)[C@@](C)(C4=CC=CC=C4)C5=C1C(OC)=O[#29]O5'. The Cu catalytic center is abbreviated [Cu], indicating that Cu bears additional ligands irrelevant to the reaction."}]}, {"sub_id": "Exam-12-T10-10", "tags": ["1.1", "2.4", "2.7"], "text_open": "10.4.2 If the phenyl group on the right side of the substrate is replaced with 2,4,6-trimethoxyphenyl, abbreviated Ar, a product with different relative substituent positions is obtained. Write its SMILES, including stereochemistry.", "text_images": [{"file": "Exam-12-T10-10-1-prompt.png", "description": "Schematic of the 2,4,6-trimethoxyphenyl group, represented by SMILES 'COC1=CC(OC)=CC(OC)=C1[R]', where [R] is the connection to another group."}], "full_mark": 2, "rubric": ["The product structure using the Ar abbreviation is shown in the rubric figure. (2 points)", "Structure or SMILES corresponding to the rubric figure: '[Ar]C1=C(C(OC)=O)C([C@](C)(C2=CC=CC=C2)[C@H]1CC)=O'"], "rubric_images": [{"file": "Exam-12-T10-10-1-rubric.png", "description": "'[Ar]C1=C(C(OC)=O)C([C@](C)(C2=CC=CC=C2)[C@H]1CC)=O'"}]}]}
{"problem_id": "Exam-13-T9", "header": "Question 9 Meldrum's Acid. Meldrum's acid is 2,2-dimethyl-1,3-dioxane-4,6-dione and has distinctive reactivity and broad synthetic applications.", "header_images": [], "subquestions": [{"sub_id": "Exam-13-T9-1", "tags": ["2.3", "3.9"], "text_open": "9.2 On heating, the following Meldrum's-acid derivative decomposes through fragmentation and a series of other reactions to give the products shown. Draw the mechanisms leading to the two products and identify the type of each pericyclic reaction.", "text_images": [{"file": "Exam-13-T9-1-1-prompt.png", "description": "Thermal decomposition of the Meldrum's-acid derivative O=C(OC(C)(C)OC/1=O)C1=C(O)/C gives products O=C2C=C(C)OC(C)(C)O2 and O=C3C=C(C)OC(C3C(C)=O)=O."}], "full_mark": 8, "rubric": ["Award 4 points for the mechanisms and 4 points for the reaction types.", "Answer SMILES: 'CC(O)=C1C(=O)OC(C)(C)OC1=O'; 'CC(=O)C1=C(O)OC(C)(C)OC1=O'; 'CC(=O)C'; 'CC(=O)C(=C=O)C(=O)O'; 'CC(=O)C=C=O'; 'CC1=CC(=O)OC(C)(C)O1'; 'CC(=O)C1C(=O)C=C(C)OC1=O'."], "rubric_images": [{"file": "Exam-13-T9-1-1-rubric.png", "description": "Three rows of reaction equations. Row 1: 'CC(O)=C1C(=O)OC(C)(C)OC1=O' undergoes a 1,5-hydrogen shift, also labeled or [1,5]-σ rearrangement, to give 'CC(=O)C1=C(O)OC(C)(C)OC1=O'. This intermediate undergoes a retro-Diels-Alder reaction, also labeled or retro-[4+2] cycloaddition, with loss of acetone, - Me_2CO, SMILES 'CC(=O)C', to give the open-chain ketene carboxylic acid 'CC(=O)C(=C=O)C(=O)O'. Loss of - CO_2 gives O=C=CH-C(=O)CH_3, SMILES 'CC(=O)C=C=O'. Row 2: one molecule of 'CC(=O)C=C=O' and one acetone molecule, 'CC(=O)C', undergo a Diels-Alder reaction, also labeled or [4+2] cycloaddition, to give 'CC1=CC(=O)OC(C)(C)O1'. Row 3: 2 molecules of 'CC(=O)C=C=O' undergo the same cycloaddition to give 'CC(=O)C1C(=O)C=C(C)OC1=O'."}]}, {"sub_id": "Exam-13-T9-2", "tags": ["1.2", "2.7"], "text_open": "9.3 As shown below, Cu/Ag bimetallic catalysis of a Meldrum-acid derivative and a terminal alkyne in the presence of chiral ligand L^{*} gives a stereospecific product that can be transformed into many derivatives. The reagent loadings are CuI 5 mol%, AgOAc 6.5 mol%, and L^{*} 6.5 mol%. 9.3.1 When R_{1} = R_{2} = Ph, give the absolute configuration of the stereogenic carbon in product P.", "text_images": [{"file": "Exam-13-T9-2-1-prompt.png", "description": "Reactants 'O=C(OC(C)(C)OC/1=O)C1=C\\[R1]' and '[R2]C#C', joined by a plus sign, react under CuI, AgOAc and H_2O above the arrow and DCM, L* below to give product P, SMILES '[R1][C@@H]1CC(OC([R2])=C1)=O'. A right brace branches P into 3 pathways. Path A uses excess LiAlH_4 at -78 °C with $R_1 = 3-BrC_6H_4, R_2 = CH_2OBn$. Path B uses mCPBA, p-TsOH with $R_1 = Ph, R_2 = CH_2OBn$. Path C uses $BnNH_2$, p-TsOH with $R_1 = Ph, R_2 = (CH_2)_4OTBS$."}], "full_mark": 1, "rubric": ["R (1 point)."], "rubric_images": []}, {"sub_id": "Exam-13-T9-3", "tags": ["1.4", "2.3", "2.4"], "text_open": "9.3.2 Write the SMILES strings of at least 3 key metal-containing intermediates in formation of P from the 2 substrates. Ligands on the metal may be omitted; each intermediate contains only 1 type of metal.", "text_images": [], "full_mark": 6, "rubric": ["Any 3 key metal-containing intermediate SMILES are sufficient, 2 points each.", "Answer SMILES: '[R2]C#C[Cu].O=C1OC(C)(C)OC(O)=C1C(C#C[R2])[R1]'; 'O=C(OC(C)(C)O2)C(C(C#C[R2])[R1])C2=O'; 'O=C3OC(C)(C)OC4=[O+]C(C)=C([Ag])C([R1])C43'."], "rubric_images": [{"file": "Exam-13-T9-3-1-rubric.png", "description": "Answer metal-containing intermediates, in order: '[R2]C#C[Cu]'; 'O=C1OC(C)(C)OC(O)=C1C(C#C[R2])[R1]'; 'O=C(OC(C)(C)O2)C(C(C#C[R2])[R1])C2=O', whose C-C triple bond coordinates [Ag]; and 'O=C3OC(C)(C)OC4=[O+]C(C)=C([Ag])C([R1])C43'."}]}, {"sub_id": "Exam-13-T9-4", "tags": ["2.4", "2.6"], "text_open": "9.3.3 A, C_{19}H_{21}BrO_{3}, contains a hemiketal; B, C_{19}H_{18}O_{4}, contains a five-membered ring; and C, C_{22}H_{25}NO_{2}, contains a spiro ring. Using the hints, write the SMILES strings of A, B, and C obtained from their corresponding P products.", "text_images": [], "full_mark": 6, "rubric": ["Award 2 points each for the SMILES strings of A, B, and C.", "Answer SMILES: A 'BrC1=CC=CC(C2CCOC(COCC3=CC=CC=C3)(O)C2)=C1'; C 'O=C(CC(CC45CCCCO5)C6=CC=CC=C6)N4CC7=CC=CC=C7'; B 'O=C(C8)OC(C(COCC9=CC=CC=C9)=O)C8C%10=CC=CC=C%10'."], "rubric_images": [{"file": "Exam-13-T9-4-1-rubric.png", "description": "Answer structures A, B, and C. A contains a hemiketal and bromoaryl group, B contains a five-membered lactone/carbonyl motif, and C contains a nitrogen spirocycle. SMILES: A 'BrC1=CC=CC(C2CCOC(COCC3=CC=CC=C3)(O)C2)=C1'; C 'O=C(CC(CC45CCCCO5)C6=CC=CC=C6)N4CC7=CC=CC=C7'; B 'O=C(C8)OC(C(COCC9=CC=CC=C9)=O)C8C%10=CC=CC=C%10'."}]}, {"sub_id": "Exam-13-T9-5", "tags": ["2.3", "2.4", "2.6"], "text_open": "9.4 The following multicomponent cascade reaction of Meldrum's acid gives the product with high diastereoselectivity. Because Meldrum's acid is acidic, its own acidic proton promotes the reaction spontaneously, including a protonation step. 9.4.1 Write the SMILES strings of 4 key intermediates; stereochemistry need not be shown.", "text_images": [{"file": "Exam-13-T9-5-1-prompt.png", "description": "Multicomponent cascade reaction of Meldrum acid, 'O=C(OC(C)(C)O5)CC5=O', an α,β-unsaturated aldehyde, 'O=C/C(C)=C/C=C/C1=CC=CC=C1', and MeOH at 75 °C in a sealed tube to give polycyclic product '[H]C2(C(C=C3)C4=CC=CC=C4)[C@@]3(OC([C@@H]2C(OC)=O)=O)C'. The product has 2 wavy bonds indicating unspecified stereochemistry."}], "full_mark": 8, "rubric": ["The rubric gives SMILES for 4 key intermediates, 2 points each; stereochemistry is not required.", "Answer SMILES: 'O=C1OC(C)(C)OC(/C1=C\\C(C)=C\\C=C\\C2=CC=CC=C2)=O'; '[OH+]=C3OC(C)(C)OC(/C3=C\\C(C)=C\\C=C\\C4=CC=CC=C4)=O'; 'CC(O5)(C)OC(O[H])=C(C6[C+](C)C=CC6C7=CC=CC=C7)C5=O'; 'CC(OC8=C9C%10C(C=CC%10C%11=CC=CC=C%11)(C)O8)(C)OC9=[OH+]'; 'CC(OC%12=C%13C%14C(C=CC%14C%15=CC=CC=C%15)(C)O%12)(C)OC%13=O'; 'O=C(OC%16(C)C%17C(C%18=CC=CC=C%18)C=C%16)C%17=C=O'; 'O=C(OC)C%19=C(OC%20(C)C%19C(C%21=CC=CC=C%21)C=C%20)OC(C)(O)C'."], "rubric_images": [{"file": "Exam-13-T9-5-1-rubric.png", "description": "At upper left, molecules 'O=C1OC(C)(C)OC(/C1=C\\C(C)=C\\C=C\\C2=CC=CC=C2)=O' and '[OH+]=C3OC(C)(C)OC(/C3=C\\C(C)=C\\C=C\\C4=CC=CC=C4)=O' are separated by the English-localized blue word or. At upper right is 'CC(O5)(C)OC(O[H])=C(C6[C+](C)C=CC6C7=CC=CC=C7)C5=O'. At lower left, 'CC(OC8=C9C%10C(C=CC%10C%11=CC=CC=C%11)(C)O8)(C)OC9=[OH+]' and 'CC(OC%12=C%13C%14C(C=CC%14C%15=CC=CC=C%15)(C)O%12)(C)OC%13=O' are separated by blue or. At lower right, 'O=C(OC%16(C)C%17C(C%18=CC=CC=C%18)C=C%16)C%17=C=O' and 'O=C(OC)C%19=C(OC%20(C)C%19C(C%21=CC=CC=C%21)C=C%20)OC(C)(O)C' are separated by blue or."}]}, {"sub_id": "Exam-13-T9-6", "tags": ["1.2", "2.7", "2.4"], "text_open": "9.4.2 Write the SMILES string of the product, including the stereochemistry indicated by the 2 wavy bonds.", "text_images": [], "full_mark": 4, "rubric": ["Write the product SMILES shown in the rubric and include the stereochemistry of the two wavy bonds (4 points).", "Answer SMILES: '[H][C@]1([C@@H](C=C2)C3=CC=CC=C3)[C@@]2(OC([C@@H]1C(OC)=O)=O)C'."], "rubric_images": [{"file": "Exam-13-T9-6-1-rubric.png", "description": "A bicyclic structure drawn mainly in blue and red. A five-membered lactone on the left is fused to a five-membered cyclopentene on the right. The lactone contains one O and one carbonyl; one carbon bears a blue solid-wedge methoxycarbonyl group, MeO-C=O. At the ring-junction bridgeheads, the upper carbon bears a blue solid-wedge methyl group, Me, and the lower carbon bears a red solid-wedge hydrogen, H. The adjacent cyclopentene carbon bears a red solid-wedge phenyl group, Ph. The SMILES with stereochemistry is '[H][C@]1([C@@H](C=C2)C3=CC=CC=C3)[C@@]2(OC([C@@H]1C(OC)=O)=O)C'."}]}]}
{"problem_id": "Exam-14-T3", "header": "Frustrated Noncovalent Organic Glass\nAs amorphous materials crosslinked through noncovalent interactions, supramolecular glasses are constrained by a trade-off between glass-forming ability and mechanical properties. Inspired by frustrated Lewis pairs, researchers designed and prepared a “frustrated noncovalent organic glass” in which the noncovalent interaction sites were altered. This increased the mechanical toughness of the supramolecular glass by a factor of 4.3 while providing high interfacial adhesion, enabling its use as an impact-resistant coating.", "header_images": [], "subquestions": [{"sub_id": "Exam-14-T3-1", "tags": ["2.4", "2.6"], "text_open": "3.1 A preparation route for a common supramolecular glass A is shown below. Write the SMILES strings of intermediates X1 and X2 and the repeat unit of polymer A (terminal groups are not required; use [*] to mark the connection sites between repeat units). EDC is 1-ethyl-3-(3-dimethylaminopropyl)carbodiimide, an organic dehydrating agent.", "text_images": [{"file": "Exam-14-T3-1-1-prompt.png", "description": "Reaction sequence: 'O=C1C(OCC)=C(OCC)C1=O' reacts with 'NCCO' under [$\\mathrm{Et_3N}$, EtOH, 70°C] to form X1; X1 then reacts with 'C=CCCCCCCCCC(O)=O' under [DMAP, EDC, $\\mathrm{CH_2Cl_2}$] to form X2; finally, X2 reacts with 'SCOCCOCS' under [AIBN, DMF, 70°C] to form A."}], "full_mark": 6, "rubric": ["Structure of X1: 'O=C1C(NCCO)=C(C1=O)NCCO' (2 points)", "Structure of X2: 'O=C1C(NCCOC(CCCCCCCCC=C)=O)=C(C1=O)NCCOC(CCCCCCCCC=C)=O' (2 points)", "Structure of polymer A; the repeat unit is 'O=C1C(NCCOC(CCCCCCCCCCSCCOCCOCCS[*])=O)=C(NCCOC(CCCCCCCCCC[*])=O)C1=O', where * denotes a connection site between repeat units. (2 points)"], "rubric_images": [{"file": "Exam-14-T3-1-1-rubric.png", "description": "Structure of intermediate X1: 'O=C1C(NCCO)=C(C1=O)NCCO'"}, {"file": "Exam-14-T3-1-2-rubric.png", "description": "Structure of intermediate X2: 'O=C1C(NCCOC(CCCCCCCCC=C)=O)=C(C1=O)NCCOC(CCCCCCCCC=C)=O'"}, {"file": "Exam-14-T3-1-3-rubric.png", "description": "Structure of polymer A; the repeat unit is 'O=C1C(NCCOC(CCCCCCCCCCSCCOCCOCCS[*])=O)=C(NCCOC(CCCCCCCCCC[*])=O)C1=O', where * denotes a connection site between repeat units."}]}, {"sub_id": "Exam-14-T3-2", "tags": ["1.7", "1.6", "2.4"], "text_open": "3.2.1 Replacing 2-hydroxyethylamine (ethanolamine) in the above route with the structure shown below, while leaving all other steps unchanged, gives “frustrated noncovalent organic glass” B. The polymer chains of B and A differ only by the isopropyl substituent. The repeat units of both polymers A and B form rings through hydrogen bonding, but the numbers of hydrogen bonds and the groups involved differ: A forms a 10-membered ring and B a 7-membered ring. Write the SMILES strings corresponding to schematic hydrogen-bonded structures between repeat units of A and B, respectively; the hydrogen bonds may be represented as ordinary chemical bonds.", "text_images": [{"file": "Exam-14-T3-2-1-prompt.png", "description": "SMILES string of the structure that replaces 2-hydroxyethylamine (ethanolamine): 'N[C@H](C(C)C)CO'"}], "full_mark": 4, "rubric": ["Hydrogen-bonding diagram for A (2 points)", "Hydrogen-bonding diagram for B (2 points)", "SMILES strings shown: 'O=C1C(N2CCOC([*])=O)=C(N(CCOC([*])=O)[H]/O=C3C(NCCOC([*])=O)=C(C/3=O/[H]2)NCCOC([*])=O)C1=O', where * represents part of the polymer chain and may be omitted from the drawing; 'O=C1C(N2[H]/O([H]N3[C@@H](COC([*])=O)C(C)C)=C([*])\\OC[C@@H](C(C)C)N(C(C4=O)=C(N([C@@H](COC([*])=O)C(C)C)[H]/5)C4=O)[H]O5=C([*])/OC[C@H]2C(C)C)=C3C1=O', where * represents part of the polymer chain and may be omitted from the drawing."], "rubric_images": [{"file": "Exam-14-T3-2-1-rubric.png", "description": "Hydrogen-bonding diagram in which repeat units of polymer A form a 10-membered ring: 'O=C1C(N2CCOC([*])=O)=C(N(CCOC([*])=O)[H]/O=C3C(NCCOC([*])=O)=C(C/3=O/[H]2)NCCOC([*])=O)C1=O', where * represents part of the polymer chain and may be omitted from the drawing."}, {"file": "Exam-14-T3-2-2-rubric.png", "description": "Hydrogen-bonding diagram in which repeat units of polymer B form a 7-membered ring: 'O=C1C(N2[H]/O([H]N3[C@@H](COC([*])=O)C(C)C)=C([*])\\OC[C@@H](C(C)C)N(C(C4=O)=C(N([C@@H](COC([*])=O)C(C)C)[H]/5)C4=O)[H]O5=C([*])/OC[C@H]2C(C)C)=C3C1=O', where * represents part of the polymer chain and may be omitted from the drawing."}]}, {"sub_id": "Exam-14-T3-3", "tags": ["1.7"], "text_open": "3.2.2 The chain structures of polymers A and B can be represented by the figures below. Which structure corresponds to frustrated noncovalent organic glass B? (      )\n(a) Left figure;\t(b) Right figure", "text_images": [{"file": "Exam-14-T3-3-1-prompt.png", "description": "Multiple short gray rod-like elements and curved gray lines. A tightly spaced, approximately horizontal group of rods appears in the upper-left region, with parallel offsets between rods. Two inclined groups in the upper-right run overall from upper left to lower right, again with parallel offsets. An approximately horizontal group appears near the lower center. Each rod has conspicuously wider protrusions at both ends. Fine gray curves connect the rod groups, bending, crossing, and reversing direction; a longer loop-like curved line is visible on the right."}, {"file": "Exam-14-T3-3-2-prompt.png", "description": "Four groups of short gray rod-like elements connected by curved gray lines. The upper-left group contains approximately 5 vertically stacked parallel rods; the upper-right group contains approximately 4; the lower-left and lower-right groups each contain approximately 5 stacked parallel rods and are slightly inclined. Multiple gray curves connect the four groups, with crossing, bending, and branching line relationships visible near the center."}], "full_mark": 1, "rubric": ["(a) (1 point)"], "rubric_images": []}, {"sub_id": "Exam-14-T3-4", "tags": ["1.7"], "text_open": "3.2.3 If A and B have the same degree of polymerization, give the ratio of the numbers of hydrogen bonds they form (as the simplest integer ratio).", "text_images": [], "full_mark": 2, "rubric": ["1:2 (2 points)"], "rubric_images": []}]}
{"problem_id": "Exam-14-T6", "header": "Kinetic Proofreading in Enzyme Reactions\nIncorrect enzyme-amino acid binding can occur during protein synthesis, so organisms require proofreading mechanisms to ensure accurate enzyme recognition. Studies show that aminoacyl-tRNA synthetases (aaRSs) use kinetic proofreading to reduce the fraction of incorrect products. In this problem, C denotes the correct amino acid and D the incorrect amino acid.", "header_images": [], "subquestions": [{"sub_id": "Exam-14-T6-1", "tags": ["2.9", "2.4"], "text_open": "6.2.1 In fact, aaRS uses a kinetic proofreading mechanism that can be represented by the elementary-reaction pathway below. ATP can activate amino-acid functional groups. For example, under enzyme catalysis ATP reacts with the carboxyl group of phenylalanine, releasing pyrophosphate and forming an intermediate in which AMP is linked to the carboxyl group. The structures of ATP and phenylalanine are shown below. Write the SMILES string of the ATP-activated phenylalanine product.", "text_images": [{"file": "Exam-14-T6-1-1-prompt.png", "description": "Reaction scheme on a white background with black text and arrows.\n\nE + aa + ATP on the left is connected to E·aa·ATP on the right by horizontal reversible arrows labeled k₁ above and k₋₁ below.\n\nA horizontal rightward arrow labeled k₂ connects E·aa·ATP to E·aa·AMP + PPi.\n\nAn upward vertical arrow above E·aa·AMP is labeled k₃ on the left and points to E + aa·AMP.\n\nA downward vertical arrow below E·aa·AMP is labeled k₄ on the left and tRNA on the right and points to E + aa-tRNA + AMP, with aa-tRNA in bold."}, {"file": "Exam-14-T6-1-2-prompt.png", "description": "Adenosine triphosphate, ATP: 'NC1=NC=NC2=C1N=CN2[C@H]3[C@H](O)[C@H](O)[C@@H](COP(OP(OP([O-])([O-])=O)([O-])=O)([O-])=O)O3'; phenylalanine, Phe: '[O-]C([C@@H]([NH3+])CC1=CC=CC=C1)=O'"}], "full_mark": 2, "rubric": ["'NC1=NC=NC2=C1N=CN2[C@H]3[C@H](O)[C@H](O)[C@@H](COP(OC([C@H](CC4=CC=CC=C4)[NH3+])=O)([O-])=O)O3' (2 points; 1 point if the charges are incorrect)"], "rubric_images": [{"file": "Exam-14-T6-1-1-rubric.png", "description": "SMILES string: 'NC1=NC=NC2=C1N=CN2[C@H]3[C@H](O)[C@H](O)[C@@H](COP(OC([C@H](CC4=CC=CC=C4)[NH3+])=O)([O-])=O)O3'"}]}]}
{"problem_id": "Exam-15-T1", "header": "Problem 1 The Immortal Mirror. In 1863, two chemists at the Freiberg Mining Academy in Germany unexpectedly observed a previously unknown spectral line while studying locally mined sphalerite. They assigned it to an undiscovered element M and successfully isolated elemental M from the mineral. At the subsequent Paris Exposition, elemental M was publicly displayed as \"the rarest metal,\" causing a sensation. At that time, M was used to make highly reflective mirrors; today it is more commonly an important semiconductor raw material used in conductive films.", "header_images": [], "subquestions": [{"sub_id": "Exam-15-T1-1", "tags": ["2.1"], "text_open": "1.1 Burning M in air gives a pale-yellow amorphous solid A. Heating A under hydrogen at 400~500°C reduces it to a black crystalline solid B (Reaction 1); dissolving B in dilute hydrochloric acid releases a colorless gas C. Above approximately 700°C, B decomposes to A and elemental M (Reaction 2). At high temperature, A reacts with ammonia to form a yellow semiconductor material D and releases a gas (Reaction 3).\n1.1.2 Write the chemical equations for Reactions 1~3.", "text_images": [], "full_mark": 3, "rubric": ["Reaction 1: $In_2O_3$ + 2$H_2$ = $In_2O$ + 2$H_2O$ (1 point)", "Reaction 2: 3$In_2O$ = $In_2O_3$ + 4In (1 point)", "Reaction 3: $In_2O_3$ + 2$NH_3$ = 2InN + 3$H_2O$ (1 point)"], "rubric_images": []}, {"sub_id": "Exam-15-T1-2", "tags": ["1.4"], "text_open": "1.2 Chloride E of M is a white flaky solid that is extremely hygroscopic and dissolves in water with substantial heat release. In ether solution, E reacts with LiH to form complex F. F decomposes above 0°C and can be used as an in situ reducing agent in organic synthesis. E forms complex E·3THF with the electron-donor ligand THF. Single-crystal X-ray diffraction shows that the M center in this complex is six-coordinate with octahedral geometry. E has a layered crystal structure consisting of close-packed chloride ions with layers of octahedrally coordinated M centers. By contrast, chloride G of the congener whose period differs by only 1 commonly crystallizes as dimer H. 1.2.1 Write the chemical formulas of E~G, write the SMILES of H, and describe its three-dimensional structure in words.", "text_images": [], "full_mark": 5, "rubric": ["E: $InCl_3$ (1 point)", "F: $LiInH_4$ (1 point)", "G: $GaCl_3$ (1 point)", "H: a chloride-bridged $Ga_2Cl_6$ dimer (2 points)"], "rubric_images": [{"file": "Exam-15-T1-2-1-rubric.png", "description": "Structure of H: a chloride-bridged $Ga_2Cl_6$ dimer. Two Ga centers are bridged by two Cl atoms, and each Ga also bears two terminal Cl atoms directed respectively out of and into the drawing plane. Each Ga is tetrahedral. Corresponding SMILES: 'Cl[Ga]1(Cl)Cl[Ga](Cl)(Cl)Cl1'."}]}, {"sub_id": "Exam-15-T1-3", "tags": ["1.5", "1.6"], "text_open": "1.3 In $E\\cdot3THF$, the $M$–Cl bond length is approximately 242 pm and the $M$–O bond length is approximately 225 pm. The ionic radius of $M$ is 80 pm, the ionic radius of $Cl^-$ is 181 pm, and the van der Waals radius of O in THF is approximately 152 pm.\n1.3.2 Use the bond-length data to compare the bonding character of the M-Cl and M-O bonds, including their relative ionic/covalent character, and explain.", "text_images": [], "full_mark": 3, "rubric": ["The In-Cl bond has relatively greater covalent character than the In-O bond. (1 point)", "The In-Cl bond length, 242 pm, is less than the sum of the ionic radii, 80 + 181 = 261 pm. The 19 pm difference is approximately 7.28%, indicating some covalent contribution to the In-Cl bond. The In-O bond length, 225 pm, is less than the sum of the relevant radii, 80 + 152 = 232 pm. The 7 pm difference is approximately 3.01%, also indicating some covalent contribution. Comparison of these differences shows that In-Cl is shortened to a greater relative extent and therefore has greater covalent character. (2 points)", "If the element is identified incorrectly but the explanation is reasonable, credit is still awarded."], "rubric_images": []}]}
{"problem_id": "Exam-15-T4", "header": "Problem 4 Determination of Chloride by the Volhard Method. ① Pipette 20.00 mL of 0.1 mol/L standard silver nitrate titrant into a 250 mL conical flask. Add 30 mL water, 5 mL nitric acid solution, and 2 mL saturated ferric ammonium sulfate solution. While shaking, titrate with standard potassium thiocyanate titrant until a pale brown-red color appears and persists for 1 min. Record the volume of standard potassium thiocyanate consumed: V_{1} = 19.88 mL. ② Weigh 0.10 g sodium chloride that has been ignited at 500 ℃ ~ 600 ℃ to constant mass, accurate to 0.1 mg. Dissolve it in approximately 40 mL water in a beaker and transfer to a 100 mL volumetric flask. Add 5 mL nitric acid solution (1+3), then add 25.00 mL of 0.1 mol/L standard silver nitrate solution while shaking vigorously. Dilute to the mark with water and mix. Let stand in the dark for 5 min, filter through rapid filter paper, and discard the first 10 mL of filtrate. Accurately pipette 50.00 mL filtrate into a 250 mL conical flask, add 2 mL of 1 mol/L ferric ammonium sulfate solution, and titrate with standard potassium thiocyanate while shaking until a pale brown-red color appears and persists for 1 min. Record V_{2} = 4.01 mL. ③ Measure 50 mL test solution into a 100 mL colorimetric tube and otherwise follow the same procedure as in ②. Record V_{3} = 2.98 mL.", "header_images": [], "subquestions": [{"sub_id": "Exam-15-T4-1", "tags": ["4.5", "2.1"], "text_open": "4.1 Calculate the concentrations of the standard silver nitrate and potassium thiocyanate titrant solutions.", "text_images": [], "full_mark": 5, "rubric": ["Let the exact concentration of $AgNO_3$ solution be c1 and that of KSCN solution be c2.", "In step ①, $Ag^+$ and $SCN^-$ react in a 1 : 1 ratio:", "\\[{{c}_{1}}\\times 20.00\\text{ mL}={{c}_{2}}\\times 19.88\\text{ mL}\\]", "Therefore:", "\\[{{c}_{2}}=\\frac{20.00}{19.88}{{c}_{1}}\\] (1 point)", "From step ②, the mass of the primary standard NaCl is m = 0.1000 g, and M(NaCl) = 58.44 g/mol.", "The amount of NaCl is \\[n(\\text{C}{{\\text{l}}^{\\text{-}}})=\\frac{0.1000\\text{ g}}{58.44\\text{ g/mol}}=1.71116\\text{ mmol}\\] (1 point).", "In the 100 mL volumetric flask, 25.00 mL $AgNO_3$ is added; the excess $Ag^+$ is determined by back-titration of the remaining solution.", "Titration of 50 mL filtrate consumes 4.01 mL KSCN solution, so the excess $Ag^+$ in the 100 mL solution consumes 8.02 mL KSCN solution.", "From the material balance:", "\\[25.00\\cdot {{c}_{1}}=1.71116+8.02\\cdot {{c}_{2}}\\] (1 point).", "Solving simultaneously gives:", "\\[c(\\text{AgN}{{\\text{O}}_{\\text{3}}})={{c}_{1}}\\approx 0.1011\\text{ mol/L}\\] (1 point).", "\\[c(\\text{KSCN})={{c}_{2}}=1.006036\\times 0.10106\\approx 0.1017\\text{ mol/L}\\] (1 point)."], "rubric_images": []}, {"sub_id": "Exam-15-T4-2", "tags": ["4.1", "4.7"], "text_open": "4.4.2 Explain why the first 10 mL of filtrate is discarded.", "text_images": [], "full_mark": 1, "rubric": ["To eliminate adsorption of solutes such as $Ag^+$ ions by the filter paper, ensuring that the subsequently pipetted 50.00 mL filtrate accurately represents the concentration of the diluted solution."], "rubric_images": []}, {"sub_id": "Exam-15-T4-3", "tags": ["4.2", "4.5"], "text_open": "4.5 If addition of standard silver nitrate produces a colloidal solution, what additional operation is required before filtration?", "text_images": [], "full_mark": 2, "rubric": ["Heat the solution in a boiling-water bath until a coagulated silver chloride precipitate appears. (1 point) Remove it and cool rapidly to room temperature in cold water. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-15-T4-4", "tags": ["4.5", "2.1", "4.7"], "text_open": "4.6 Calculate the $Cl^-$ concentration in the sample.", "text_images": [], "full_mark": 3, "rubric": ["In step ③, the total amount of $Ag^+$ added is \\[25.00\\text{ mL}\\times 0.10106\\text{ mol/L}=2.5265\\text{ mmol}\\] (1 point)", "Titration of 50.00 mL filtrate consumes 2.98 mL standard KSCN solution.", "Therefore, the total excess $Ag^+$ is \\[2\\times 2.98\\text{ mL}\\times 0.10167\\text{ mol/L}=0.6060\\text{ mmol}\\] (1 point)", "The total amount of $Cl^-$ in the test solution is \\[n(\\text{C}{{\\text{l}}^{\\text{-}}})=2.5265-0.6060=1.9205\\text{ mmol}\\]", "The original $Cl^-$ concentration in the test solution is:", "\\[c(\\text{C}{{\\text{l}}^{\\text{-}}})=\\frac{1.9205\\text{ mmol}}{50.00\\text{ mL}}\\approx 0.03841\\text{ mol/L}\\] (1 point)"], "rubric_images": []}]}
{"problem_id": "Exam-15-T8", "header": "Problem 8 Analysis of Substituent Effects. This problem compares the magnitudes of -ΔG° for a series of substituents or substrates in the conformational or isomerization reactions shown.", "header_images": [], "subquestions": [{"sub_id": "Exam-15-T8-1", "tags": ["1.2", "3.2"], "text_open": "Reaction (a) below is shared by 8.3–8.6. In 8.3–8.5, compare the shared Reaction (a) with Reaction (b) shown in the corresponding subproblem and select the reaction with the larger -ΔG°.\n8.3", "text_images": [{"file": "Exam-15-T8-1-1-prompt.png", "description": "Prompt figure for part a: the image shows a change from left to right. On the left is a six-membered ring in a chair conformation with an axial Me group on the right-hand carbon; its SMILES is 'CC1CCCCC1'. A one-way arrow points right. On the right is the same chair-shaped ring with an equatorial Me group on the right-hand carbon; its SMILES is likewise 'CC1CCCCC1'."}, {"file": "Exam-15-T8-1-2-prompt.png", "description": "Prompt figure for part b: the image shows a change from left to right. On the left is a six-membered ring in a chair conformation bearing a carbonyl (=O) at the lower-left vertex and an axial Me group on the right-hand carbon; its SMILES is 'CC1CC(=O)CCC1'. A one-way arrow points right. On the right is the same chair-shaped ring with the same lower-left carbonyl and an equatorial Me group on the right-hand carbon; its SMILES is likewise 'CC1CC(=O)CCC1'."}], "full_mark": 2, "rubric": ["a (2 points)"], "rubric_images": []}, {"sub_id": "Exam-15-T8-2", "tags": ["1.2", "3.2"], "text_open": "In 8.4, Reaction (b) is a base-catalyzed isomerization.\n8.4", "text_images": [{"file": "Exam-15-T8-2-1-prompt.png", "description": "The image shows a change from left to right. On the left is a six-membered chair bearing an equatorial tert-butyl group at the lower-left carbon, an axial Me group at the right-hand carbon, and a carbonyl (=O) at the lower-right vertex; its SMILES is 'CC1CCC(CC1=O)C(C)(C)C'. A one-way arrow points right. On the right is the same chair with the equatorial tert-butyl group and lower-right carbonyl unchanged, but with the Me group equatorial; its SMILES is likewise 'CC1CCC(CC1=O)C(C)(C)C'."}], "full_mark": 2, "rubric": ["a (2 points)"], "rubric_images": []}, {"sub_id": "Exam-15-T8-3", "tags": ["1.2", "3.2"], "text_open": "In 8.5, Reaction (b) is a base-catalyzed isomerization.\n8.5", "text_images": [{"file": "Exam-15-T8-3-1-prompt.png", "description": "Prompt figure for option (b): the image shows a change from left to right. On the left is a six-membered chair bearing an equatorial tert-butyl group at the lower-left carbon, an axial $CHMe_2$ group at the right-hand carbon, and a carbonyl (=O) at the lower-right vertex; its SMILES is 'CC(C)C1CCC(CC1=O)C(C)(C)C'. A one-way arrow points right. On the right is the same chair with the equatorial tert-butyl group and lower-right carbonyl unchanged, but with the $CHMe_2$ group equatorial; its SMILES is likewise 'CC(C)C1CCC(CC1=O)C(C)(C)C'."}], "full_mark": 2, "rubric": ["a (2 points)"], "rubric_images": []}, {"sub_id": "Exam-15-T8-4", "tags": ["1.2", "3.2"], "text_open": "8.6 Arrange -ΔG° for the following reactions in decreasing order.", "text_images": [{"file": "Exam-15-T8-4-1-prompt.png", "description": "Prompt figure for b: the image shows a change from left to right. On the left is a six-membered chair heterocycle containing a ring oxygen near the lower-right position and bearing an axial Me group on the right-hand carbon; its SMILES is 'C[C@H]1OCCCC1'. A one-way arrow points right. On the right is the same chair heterocycle with the ring oxygen in the same position and an equatorial Me group; its SMILES is 'C[C@@H]1OCCCC1'."}, {"file": "Exam-15-T8-4-2-prompt.png", "description": "Prompt figure for c: the image shows a change from left to right. On the left is a six-membered chair heterocycle containing a ring oxygen at the lower-left position and bearing an axial Me group on the right-hand carbon; its SMILES is 'C[C@H]1COCCC1'. A one-way arrow points right. On the right is the same chair heterocycle with the ring oxygen in the same position and an equatorial Me group; its SMILES is 'C[C@@H]1COCCC1'."}, {"file": "Exam-15-T8-4-3-prompt.png", "description": "Prompt figure for d: the image shows a change from left to right. On the left is a six-membered chair heterocycle containing a ring oxygen at the far-left position and bearing an axial Me group on the right-hand carbon; its SMILES is 'CC1CCOCC1'. A one-way arrow points right. On the right is the same chair heterocycle with the ring oxygen in the same position and an equatorial Me group; its SMILES is likewise 'CC1CCOCC1'."}, {"file": "Exam-15-T8-4-4-prompt.png", "description": "The image shows a change from left to right. On the left is a six-membered chair heterocycle containing an NH group near the lower-right position and bearing an axial Me group on the right-hand carbon; its SMILES is 'C[C@H]1NCCCC1'. A one-way arrow points right. On the right is the same chair heterocycle with the NH group in the same position and an equatorial Me group; its SMILES is 'C[C@@H]1NCCCC1'."}], "full_mark": 3, "rubric": ["bedac (3 points)", "3 points for a completely correct answer."], "rubric_images": []}]}
{"problem_id": "Exam-15-T9", "header": "Problem 9 Ketene Chemistry. Ketenes are highly reactive, unstable intermediates with diverse synthetic methods and broad industrial applications.", "header_images": [], "subquestions": [{"sub_id": "Exam-15-T9-1", "tags": ["2.3", "2.4"], "text_open": "9.2 Thermal electrocyclic ring opening of cyclobutenones is an important method for generating conjugated polyenyl ketene intermediates. Under thermal conditions, the following substrate undergoes a cascade reaction to give fused-ring product A. 9.2.1 Write the SMILES structure of product A, including stereochemistry.", "text_images": [{"file": "Exam-15-T9-1-1-prompt.png", "description": "The substrate has structure SMILES 'OC(C1=O)(C(OC)=C1OC)C2=C(COC)CCC2'. It is a mixture of enantiomers whose stereocenter is the carbon bearing the hydroxy group, with a wavy bond between that carbon and the hydroxy group. Under 120°C microwave conditions it gives A."}], "full_mark": 2, "rubric": ["'[H][C@]1(CCC2)[C@]2(COC)C(C(OC)=C(C1=O)OC)=O' (2 points)."], "rubric_images": [{"file": "Exam-15-T9-1-1-rubric.png", "description": "Rubric structure for A: corresponding SMILES '[H][C@]1(CCC2)[C@]2(COC)C(C(OC)=C(C1=O)OC)=O'."}]}, {"sub_id": "Exam-15-T9-2", "tags": ["2.3", "2.4"], "text_open": "9.2.2 Write the SMILES structure of B.", "text_images": [{"file": "Exam-15-T9-2-1-prompt.png", "description": "The image shows a reaction sequence and reagent structure. At far left is compound 'OC(C1=O)(C(OC)=C1OC)C2=C(OC)C(C(C)(C)C)CCC2', a stereoisomeric mixture with 2 stereocenters: the carbon bearing hydroxy and the carbon bearing tert-butyl; wavy bonds connect the two stereogenic carbons respectively to hydroxy and tert-butyl. At `120 °C, microwave`, it forms **B**. **B** then reacts under `DMP` to give 'CC(C)(C)C(CCCC34C(C(OC)=C(C4=O)OC)=O)=C3OC', labeled **C**. At far right is the DMP reagent, with SMILES 'O=C(OI(OC(C)=O)(OC(C)=O)5OC(C)=O)C6=C5C=CC=C6' and label \"DMP\"."}], "full_mark": 2, "rubric": ["'CC(C)(C1=C(OC)C2(C(C(OC)=C(OC)C2O)=O)CCC1)C' (2 points)."], "rubric_images": [{"file": "Exam-15-T9-2-1-rubric.png", "description": "Rubric structure for B: corresponding SMILES 'CC(C)(C1=C(OC)C2(C(C(OC)=C(OC)C2O)=O)CCC1)C'."}]}, {"sub_id": "Exam-15-T9-3", "tags": ["2.3", "2.4"], "text_open": "A slight substrate modification leads through a different pathway to a product of another structure. Thermal treatment of the following substrate generates a ketene that reacts differently; product B is ultimately oxidized to spirocyclic C. 9.2.3 Write the SMILES structures of the 2 key intermediates in conversion of this substrate to B.", "text_images": [], "full_mark": 4, "rubric": ["'CC(C)(C)C1CCCC(/C(O)=C(C(OC)=C=O)/OC)=C1OC' and 'CC(C)(C)C(CCCC23C(O)=C(C(OC)=C3O)OC)=C2OC' (2 points each)."], "rubric_images": [{"file": "Exam-15-T9-3-1-rubric.png", "description": "The two key intermediates in conversion to B have SMILES 'CC(C)(C)C1CCCC(/C(O)=C(C(OC)=C=O)/OC)=C1OC' and 'CC(C)(C)C(CCCC23C(O)=C(C(OC)=C3O)OC)=C2OC', respectively."}]}, {"sub_id": "Exam-15-T9-4", "tags": ["2.3", "2.4"], "text_open": "9.3 The following photolysis gives mainly products D and E. Mechanistically, the substrate is photoexcited and then cleaves a C-C bond to form diradical F. Radical coupling then forms ketene intermediate G, which is ultimately converted to D and E. 9.3.1 Write the SMILES structures of F and G.", "text_images": [{"file": "Exam-15-T9-4-1-prompt.png", "description": "The image shows a photochemical reaction. Molecule 'O=C1[C@@](N(N([C@H]2C=C1)C(OCC)=O)C(OCC)=O)(C=C2)OC' reacts under hv in `MeOH` (SMILES 'CO') to give two products. The left product, 'O=C(C[C@@H]3[C@]([C@]3([H])N(N4C(OCC)=O)C(OCC)=O)([H])C=C4OC)OC', is labeled **D**. The right product, 'O=C5C[C@]6([H])[C@]([C@]6([H])N(N7C(OCC)=O)C(OCC)=O)([H])[C@@]5([H])C7(OC)OC', is labeled **E**. A plus sign separates the products."}], "full_mark": 4, "rubric": ["F: 'O=[C]/C=C\\[C@@H]1N(N([C](C=C1)OC)C(OCC)=O)C(OCC)=O' or '[H][C@@](N(N2C(OCC)=O)C(OCC)=O)([CH]C=C=O)[CH]C=C2OC' (2 points); G: '[H][C@@](N(N3C(OCC)=O)C(OCC)=O)(C4C=C=O)C4C=C3OC' (2 points)."], "rubric_images": [{"file": "Exam-15-T9-4-1-rubric.png", "description": "Rubric structures for F and G: F is a diradical intermediate with SMILES 'O=[C]/C=C\\[C@@H]1N(N([C](C=C1)OC)C(OCC)=O)C(OCC)=O' or '[H][C@@](N(N2C(OCC)=O)C(OCC)=O)([CH]C=C=O)[CH]C=C2OC'; G is the ketene intermediate with SMILES '[H][C@@](N(N3C(OCC)=O)C(OCC)=O)(C4C=C=O)C4C=C3OC'."}]}, {"sub_id": "Exam-15-T9-5", "tags": ["2.3", "2.4"], "text_open": "9.3.2 If MeOH is replaced with deuterated $CD_3OD$, write the SMILES structures of deuterated products D-d and E-d.", "text_images": [], "full_mark": 4, "rubric": ["D-d: 'O=C(C([2H])[C@]1([H])[C@]([C@]1([H])N(N2C(OCC)=O)C(OCC)=O)([H])C=C2OC)OC([2H])([2H])[2H]' (2 points); E-d: 'O=C3C([2H])[C@]4([H])[C@]([C@]4([H])N(N5C(OCC)=O)C(OCC)=O)([H])[C@@]3([H])C5(OC([2H])([2H])[2H])OC' (2 points)."], "rubric_images": [{"file": "Exam-15-T9-5-1-rubric.png", "description": "Rubric structures for D-d and E-d: D-d SMILES 'O=C(C([2H])[C@]1([H])[C@]([C@]1([H])N(N2C(OCC)=O)C(OCC)=O)([H])C=C2OC)OC([2H])([2H])[2H]'; E-d SMILES 'O=C3C([2H])[C@]4([H])[C@]([C@]4([H])N(N5C(OCC)=O)C(OCC)=O)([H])[C@@]3([H])C5(OC([2H])([2H])[2H])OC'."}]}, {"sub_id": "Exam-15-T9-6", "tags": ["2.3", "2.4"], "text_open": "9.4 Radical reactions can also construct fused-ring centers bearing a quaternary carbon, as shown below. The reaction is known to proceed through a ketene radical intermediate. Write the SMILES structures of any 2 key intermediates.", "text_images": [{"file": "Exam-15-T9-6-1-prompt.png", "description": "The image shows a reaction equation. 'C#CCCCCCC', drawn as $C_6H_13$ attached to a carbon-carbon triple bond, reacts with 'C45=NCCCN4CCCCC5'. Above the arrow are $^{n}Bu_3SnH$, AIBN; below are CO (80 atm) and benzene, 80°C. The product is 'O=C1N2C(N3CCC2)(CCCCC3)C=C1CCCCCC'."}], "full_mark": 4, "rubric": ["4 rubric intermediates for 9.4 (4 points; any 2 suffice): first, 'CCCCCC/[C]=C/[Sn](CCCC)(CCCC)CCCC'; second, 'O=[C]/C(CCCCCC)=C\\[Sn](CCCC)(CCCC)CCCC' (equivalent structure 'O=C=C([CH][Sn](CCCC)(CCCC)CCCC)CCCCCC'); third, '[O-]/C([N+]1=C(CCCCC2)N2CCC1)=C([CH][Sn](CCCC)(CCCC)CCCC)\\CCCCCC'; fourth, 'O=C(N3CCCN4C35CCCCC4)[C](C5[Sn](CCCC)(CCCC)CCCC)CCCCCC'."], "rubric_images": [{"file": "Exam-15-T9-6-1-rubric.png", "description": "4 rubric intermediates for 9.4: 'CCCCCC/[C]=C/[Sn](CCCC)(CCCC)CCCC'; 'O=[C]/C(CCCCCC)=C\\[Sn](CCCC)(CCCC)CCCC', with equivalent structure 'O=C=C([CH][Sn](CCCC)(CCCC)CCCC)CCCCCC'; '[O-]/C([N+]1=C(CCCCC2)N2CCC1)=C([CH][Sn](CCCC)(CCCC)CCCC)\\CCCCCC'; and 'O=C(N3CCCN4C35CCCCC4)[C](C5[Sn](CCCC)(CCCC)CCCC)CCCCCC'."}]}, {"sub_id": "Exam-15-T9-7", "tags": ["2.3", "2.4"], "text_open": "9.5 Ketenes may also react with other highly reactive intermediates. In the following reaction, active species I forms together with ketene H; H and I react with dimethyl acetylenedicarboxylate to give the product. 9.5.1 Write the SMILES structure of I.", "text_images": [{"file": "Exam-15-T9-7-1-prompt.png", "description": "The image shows a three-component reaction. Molecules 'O=C(C(C6=CC=CC=C6)=C7C8=CC=CC=C8)C7=O', 'CC(O1)(C)N=NC1(OC)OC', and vertically drawn 'O=C(C#CC(OC)=O)OC', shown as a carbon-carbon triple bond bearing $CO_2Me$ at both ends, are separated by plus signs. They react to form complex spirocyclic product 'O=C(C(C2=CC=CC=C2)=C(C3=CC=CC=C3)C45OC(C(C(OC)=O)=C5C(OC)=O)(OC)OC)C4(OC)OC'."}], "full_mark": 2, "rubric": ["'CO[C]OC' (2 points)."], "rubric_images": [{"file": "Exam-15-T9-7-1-rubric.png", "description": "Rubric structure for I: dimethoxycarbene, with SMILES 'CO[C]OC'."}]}, {"sub_id": "Exam-15-T9-8", "tags": ["2.3", "2.4"], "text_open": "9.5.2 Write the SMILES structures of the key intermediates in conversion of H and I to the product.", "text_images": [], "full_mark": 4, "rubric": ["Key intermediates for 9.5.2 (4 points; any 1 of the first 2 may be written): first, 'O=C=C(C1=CC=CC=C1)/C(C2=CC=CC=C2)=C([O-])\\C(OC)=[O+]C'; second, 'O=C(C(C3=CC=CC=C3)=C(C4=CC=CC=C4)C5=O)C5(OC)OC'; third, 'O=C(/C(C(OC)=[O+]C)=[C-]\\C(OC)=O)OC'."], "rubric_images": [{"file": "Exam-15-T9-8-1-rubric.png", "description": "Rubric key intermediates in conversion of H and I to product: first, 'O=C=C(C1=CC=CC=C1)/C(C2=CC=CC=C2)=C([O-])\\C(OC)=[O+]C'; second, 'O=C(C(C3=CC=CC=C3)=C(C4=CC=CC=C4)C5=O)C5(OC)OC'; third, 'O=C(/C(C(OC)=[O+]C)=[C-]\\C(OC)=O)OC'."}]}, {"sub_id": "Exam-15-T9-9", "tags": ["1.4", "2.2"], "text_open": "9.6 Ketenes may also be connected with transition-metal carbonyl complexes. 9.6.1 Write the by-product formed in the first conversion.", "text_images": [{"file": "Exam-15-T9-9-1-prompt.png", "description": "The image shows a multistep reaction and a group-equivalence definition. The initial reactant is formed by coordination of the carbon-carbon double bond in ligand 'O=C(/C=C/C1=NC=CC=C1)[Fc]' to iron carbonyl fragment '[Fe](C#O)(C#O)(C#O)(C#O)', i.e. $Fe(CO)_4$. Under `MeLi, CO`, it forms an intermediate in which two carbon-carbon double bonds of the ligand coordinate to iron carbonyl fragment '[Fe](C#O)(C#O)(C#O)', i.e. $Fe(CO)_3$; the full intermediate SMILES is '[Fc]C(/C=C/C1=NC=CC=C1)=C=O.[O+]#[C-][Fe]([C-]#[O+])[C-]#[O+]'. Heating under `Δ` then gives product **J**, with molecular formula $C_19H_15FeNO$. At far right, `Fc ≡` is defined as the ferrocenyl structure."}], "full_mark": 1, "rubric": ["$MeCO_2Li$ (1 point)"], "rubric_images": []}, {"sub_id": "Exam-15-T9-10", "tags": ["1.4", "2.4"], "text_open": "9.6.3 Using the molecular-formula clue, write the SMILES structure of final product J.", "text_images": [], "full_mark": 2, "rubric": ["'[Fc]C1=CC=C(C=CC=C2)N2C1=O' (2 points)."], "rubric_images": [{"file": "Exam-15-T9-10-1-rubric.png", "description": "Rubric structure for J: corresponding SMILES '[Fc]C1=CC=C(C=CC=C2)N2C1=O'."}]}]}
{"problem_id": "Exam-16-T3", "header": "Problem 3: Photopromoted Polymerization", "header_images": [], "subquestions": [{"sub_id": "Exam-16-T3-1", "tags": ["2.2", "2.3", "2.4"], "text_open": "3.1 On 2023-3-24, researchers from Istanbul Technical University published a study of visible-light-induced cationic step-growth polymerization. 3.1.1 In a direct initiation system for cationic polymerization, an onium salt can absorb light directly to initiate the reaction. An advantage of this initiation mode is that it generates an acidic environment in situ, further promoting the reaction. On this basis, give all products of the initiation reaction shown below.", "text_images": [{"file": "Exam-16-T3-1-1-prompt.png", "description": "The figure shows a reaction in which the species on the left is a salt composed of a circled negatively charged `PF_6^-` anion and a circled positively charged cation; its corresponding SMILES is '[Ar][I+][Ar].F[P-](F)(F)(F)(F)F'. The salt reacts to the right under `hv` and `300 nm` shown above the arrow and `R-H` shown below the arrow."}], "full_mark": 4, "rubric": ["ArI (1 point), HPF_{6} (1 point), R· (1 point), Ar· (1 point)", "If more than 4 products are listed, only the first 4 are considered; incorrect entries do not incur deductions."], "rubric_images": []}, {"sub_id": "Exam-16-T3-2", "tags": ["2.2", "2.4"], "text_open": "3.1.2 Give the SMILES structures of 1 key active intermediate derived from the monomer and of the product in this reaction.", "text_images": [{"file": "Exam-16-T3-2-1-prompt.png", "description": "The figure shows a reaction in which the leftmost molecule is 'C1=CSC=C1', followed by the label \"(2 equiv)\". It reacts with a salt shown at the upper right, composed of a circled negatively charged `PF_6^-` anion and a circled positively charged cation; the corresponding SMILES of the salt is 'C1([I+]C2=CC=CC=C2)=CC=CC=C1.F[P-](F)(F)(F)(F)F'. The reaction proceeds to the right under `hv` shown below the arrow."}], "full_mark": 4, "rubric": ["'C1=CC=C[S+]1' (2 points), 'C1(C2=CC=CS2)=CC=CS1' (2 points)"], "rubric_images": [{"file": "Exam-16-T3-2-1-rubric.png", "description": "Rubric structure 1: a thiophene radical cation/key active intermediate. The structure in the image has the SMILES 'C1=CC=C[S+]1'."}, {"file": "Exam-16-T3-2-2-rubric.png", "description": "Rubric structure 2: the bithiophene product. The structure in the image has the SMILES 'C1(C2=CC=CS2)=CC=CS1'."}]}]}
{"problem_id": "Exam-17-T2", "header": "Problem 2 Determination of Paracetamol by Cyclic Voltammetry and Study of Its Electrochemistry\nCyclic voltammetry (CV) is an important electrochemical technique for investigating electrochemical reaction mechanisms and performing quantitative analysis. It uses a three-electrode system consisting of a working electrode, a reference electrode, and an auxiliary electrode. A triangular-wave potential that varies linearly with time is applied to the working electrode while the response current is recorded. One triangular-wave scan completes both an oxidation scan and a reduction scan, and the resulting current–potential curve is called a cyclic voltammogram. A substance that can undergo oxidation or reduction at an electrode within the scanned potential range is called an electroactive substance. A typical cyclic voltammogram exhibits an oxidation peak (anodic peak) and a reduction peak (cathodic peak). Under suitable conditions, the peak currents ($I_{pa}$ and $I_{pc}$) are proportional to the concentration of the electroactive substance and form the basis of quantitative analysis. The peak potentials ($E_{pa}$ and $E_{pc}$) and the peak-potential separation ($\\Delta E_p = E_{pa} - E_{pc}$) can be used to assess reaction reversibility and the number of electrons transferred. Changing the scan rate ($v$) can distinguish diffusion control from adsorption control and reveal the kinetics of subsequent chemical reactions.\nParacetamol (APAP, also known as acetaminophen) is a commonly used antipyretic analgesic. Studies show that APAP undergoes electrochemical oxidation at an electrode surface, and its oxidation product NAPQI undergoes subsequent chemical reactions in acidic media. A research group used cyclic voltammetry to investigate the electrochemical behavior of APAP systematically.\nAt room temperature, the peak current ($I_p$, in A) for a reversible electrochemical reaction in a quiescent solution at a planar electrode is related to the scan rate ($v$, in V/s), the concentration of the electroactive substance ($C$, in $\\mathrm{mol/cm^3}$), and other parameters by the Randles–Sevcik equation:\n$$I_p = 2.69 \\times 10^5 n^{3/2} A C D^{1/2} v^{1/2}$$\nn: number of electrons transferred in the reaction.\nA: effective area of the working electrode (unit: $\\mathrm{cm^2}$).\nD: diffusion coefficient of the electroactive substance (unit: $\\mathrm{cm^2/s}$), reflecting how rapidly the substance moves through solution.\nThe figure below is a typical cyclic voltammogram. Curve 1 is the background signal, and curve 2 is the normal response signal.", "header_images": [{"file": "Exam-17-T2-header-1.png", "description": "Typical cyclic voltammogram. The horizontal axis is potential, with tick marks of approximately 0.2, 0.1, 0, -0.1, and -0.2 from left to right; the vertical axis is current. The peak-current and peak-potential symbols $I_{pc}$, $I_{pa}$, $E_{pa}$, and $E_{pc}$ are labeled. Two closed curves are labeled 1 and 2. As stated in the question, curve 1 is the background signal and curve 2 is the normal response signal."}], "subquestions": [{"sub_id": "Exam-17-T2-1", "tags": ["4.7"], "text_open": "2.2 At a scan rate of $40\\ \\mathrm{mV\\cdot s^{-1}}$ in a pH = 2.2 buffer, the research group obtained the following cyclic-voltammetry data for standard APAP solutions:\n\n| APAP concentration/(mmol/L) | 0.10 | 0.40 | 1.00 | 3.00 | 5.00 |\n| ----------------- | ----- | ----- | ----- | ----- | ----- |\n| Anodic peak current $I_{pa}$/μA | 2.15 | 5.82 | 13.46 | 38.74 | 63.52 |\n\nUsing concentration as the horizontal coordinate and anodic peak current as the vertical coordinate, obtain the linear regression equation of $I_{pa}$ against APAP concentration and calculate the expected peak current for an APAP solution with a concentration of $3.50\\ \\mathrm{mmol\\cdot L^{-1}}$.", "text_images": [], "full_mark": 5, "rubric": ["With concentration as $x$ (unit: $\\mathrm{mmol\\cdot L^{-1}}$) and peak current as $y$ (unit: μA), least-squares fitting gives the linear regression equation $y = 12.55x + 0.90$, $R^2 = 1.0000$ (3 points). When $x = 3.50\\ \\mathrm{mmol\\cdot L^{-1}}$: $I_{pa} = 12.55 \\times 3.50 + 0.90 = 44.8\\ \\mu\\mathrm{A}$ (2 points; the number of significant figures is not required)."], "rubric_images": []}, {"sub_id": "Exam-17-T2-2", "tags": ["2.3"], "text_open": "2.3 To investigate the oxidation mechanism of APAP, the group performed CV scans at $40\\ \\mathrm{mV\\cdot s^{-1}}$ in media of different acidity (pH = 6.0, pH = 2.2, and $1.8\\ \\mathrm{mol\\cdot L^{-1}}\\ H_2SO_4$) and observed the following:\nAt pH = 6.0, both oxidation and reduction peaks appear, but $I_{pa}$ ≠ $I_{pc}$; at pH = 2.2, only the oxidation peak appears and the reduction peak disappears; under strongly acidic conditions ($1.8\\ \\mathrm{mol\\cdot L^{-1}}\\ H_2SO_4$), the reduction peak reappears.\n2.3.1 Explain why the reduction peak disappears at pH = 2.2.", "text_images": [], "full_mark": 2, "rubric": ["At pH = 2.2, the APAP oxidation product NAPQI rapidly undergoes addition of water to form an electrochemically inactive NAPQI hydrate. This substance cannot be reduced within the scanned potential range, so the reduction peak disappears."], "rubric_images": []}, {"sub_id": "Exam-17-T2-3", "tags": ["2.3"], "text_open": "2.3.2 What chemical reaction is indicated by the reappearance of the reduction peak under strongly acidic conditions?", "text_images": [], "full_mark": 2, "rubric": ["Under strongly acidic conditions, the NAPQI hydrate is further hydrolyzed to benzoquinone, which can be reduced at the electrode; the reduction peak therefore reappears (2 points)."], "rubric_images": []}]}
{"problem_id": "Exam-17-T3", "header": "Problem 3 Aqueous Self-Assembly and Competitive Recognition by Bowl-Shaped Molecules\nMolecular recognition in water is often governed simultaneously by electrostatic interactions, hydrogen bonding, solvation effects, and shape complementarity, so the actual outcome of supramolecular assembly can differ from a simple prediction based on charge complementarity. Tribenzotriquinacene (TBTQ) derivatives are rigid bowl-shaped molecules whose frameworks have distinct concave and convex faces. Introducing ionic groups around the rim of a TBTQ framework can confer water solubility and enable charge-controlled self-assembly.\nAs shown, compound A is a cationic TBTQ bowl bearing 6 ammonium groups, compound B is an anionic TBTQ bowl bearing 6 carboxylate groups, and compound C is mesitylene-α, α′, α″-trisulfonate, a trianionic guest bearing three $CH_2SO_3^-$ groups on an aromatic ring.", "header_images": [{"file": "Exam-17-T3-header-1.png", "description": "Compound A (TBTQ hexaammonium salt, labeled $A\\cdot Cl_6$): a bowl-shaped TBTQ framework in which each ring bears 3-aminopropoxy chains on 2 adjacent aromatic carbon atoms, for a total of 6 $NH_3^+$ groups and 6 $Cl^-$ ions. Its SMILES string is '[NH3+]CCCOC1=CC(C2C3(C4C5=C(C=C(C(OCCC[NH3+])=C5)OCCC[NH3+])C3C6=CC(OCCC[NH3+])=C(OCCC[NH3+])C=C62)C)=C4C=C1OCCC[NH3+].[Cl-].[Cl-].[Cl-].[Cl-].[Cl-].[Cl-]'."}, {"file": "Exam-17-T3-header-2.png", "description": "Compound B (TBTQ hexacarboxylate, labeled $B\\cdot (NH_4)_6$): a bowl-shaped TBTQ framework in which each ring bears carboxymethoxy chains on 2 adjacent aromatic carbon atoms, for a total of 6 $CO_2^-$ groups and 6 $NH_4^+$ ions. Its SMILES string is '[O-]C(COC(C=C1C2C3(C)C(C4=CC(OCC([O-])=O)=C(C=C4C3C5=CC(OCC([O-])=O)=C(C=C52)OCC([O-])=O)OCC([O-])=O)C1=C6)=C6OCC([O-])=O)=O.[NH4+].[NH4+].[NH4+].[NH4+].[NH4+].[NH4+]'."}, {"file": "Exam-17-T3-header-3.png", "description": "Compound C (trisodium benzene-1,3,5-tris(methylsulfonate), labeled $Na_3\\cdot C$): its SMILES string is '[O-]S(=O)(CC1=CC(CS(=O)([O-])=O)=CC(CS(=O)([O-])=O)=C1)=O.[Na+].[Na+].[Na+]'."}], "subquestions": [{"sub_id": "Exam-17-T3-1", "tags": ["1.7"], "text_open": "3.1 Based only on the charge complementarity and bowl-shaped structures of A and B, describe in words the AB capsule assembly that would be most readily expected. No specific structure needs to be drawn.", "text_images": [], "full_mark": 2, "rubric": ["The two bowl openings face each other, and the positive charges of the ammonium groups around the rim of A attract the negative charges of the carboxylate groups around the rim of B (2 points), as illustrated by:"], "rubric_images": [{"file": "Exam-17-T3-1-1-rubric.png", "description": "Schematic answer for the capsule assembly in 3.1. On a pure white background, two similarly sized hemispheres are arranged vertically, centered, and separated by a narrow horizontal white gap without touching. The upper hemisphere is sky blue, with its curved face upward and flat face downward, and bears a black plus sign (+) at the center of its front curved surface. The lower hemisphere is red, with its flat face upward and curved face downward like a bowl, and bears a black minus sign (-) at the center of its front curved surface. Their flat faces are parallel and face each other, together approximating a complete sphere. Both hemispheres have smooth three-dimensional shading, highlights, and gloss without outlines. The figure contains no words, letters, numbers, arrows, or explicit chemical structures."}]}]}
{"problem_id": "Exam-17-T4", "header": "Problem 4 Gravimetric Determination of Tungsten (17 points, 8%)\nTungsten is an important strategic metal that has an irreplaceable strengthening role in cemented carbides, high-speed steel, and specialty steels. Accurate determination of tungsten content is essential for quality control of alloy materials. In the classical gravimetric method, tungsten is converted to a tungstic-acid precipitate, ignited to tungsten trioxide, and weighed. Accurate tungsten content is obtained by correcting for interference from coexisting elements such as molybdenum, chromium, and vanadium. This problem concerns precipitation and separation of tungsten, correction for impurities, and special treatment of niobium-containing samples, and tests understanding of the principles of gravimetric analysis and strategies for eliminating interference.\nThe hydrochloric acid used in this problem is diluted 1:1, the nitric acid is concentrated, and every reported mass has been corrected for the tare mass of the crucible.", "header_images": [], "subquestions": [{"sub_id": "Exam-17-T4-1", "tags": ["2.1"], "text_open": "4.1 Procedure for samples without niobium\nA sample of mass $m$ is weighed into a beaker. Hydrochloric acid and nitric acid are added, and the mixture is heated gently to dissolve the sample and evaporated to a syrup to remove nitrogen oxides. Hydrochloric acid is added to dissolve the salts, the solution is diluted with water and brought to the boil, and cinchonine hydrochloride solution is added slowly to precipitate all tungsten as tungstic acid. The precipitate is aged at 70-80℃ for 4 h and then filtered and washed with a cinchonine hydrochloride wash solution.\nThe precipitate and filter paper are placed in a platinum crucible, carbonized, and ignited at 750-800℃. The residue is moistened with water, hydrofluoric acid and sulfuric acid are added, and the mixture is heated to volatilize silicon. It is then ignited to constant mass, giving a mass $m_1$ of impure $WO_3$.\nSodium carbonate is added to the platinum crucible for fusion. After cooling, the melt is extracted with hot water and filtered, and the residue is ignited and weighed to give $m_2$. The filtrate is analyzed spectrophotometrically for molybdenum, chromium, and vanadium, which are converted to the corresponding oxide masses $m_3$, $m_4$, and $m_5$.\n4.1.1 Write the equation for oxidation of elemental tungsten to tungstic acid, $H_2WO_4$, in hydrochloric acid/nitric acid.", "text_images": [], "full_mark": 2, "rubric": ["$W + 2HNO_3 = H_2WO_4 + 2NO\\uparrow$ (2 points)"], "rubric_images": []}, {"sub_id": "Exam-17-T4-2", "tags": ["4.7", "2.1"], "text_open": "4.1.6 During aqueous extraction after sodium carbonate fusion, $WO_3$ is converted to $Na_2WO_4$ and enters solution, whereas $TiO_2$ and $Fe_2O_3$ remain insoluble. Analysis of a tungsten-steel sample gives the following data: $m$ = 0.5123 g, $m_1$ = 0.0842 g, $m_2$ = 0.0018 g, $m_3$ = 0.0023 g, $m_4$ = 0.0015 g, and $m_5$ = 0.0012 g. Calculate the mass fraction of tungsten in the sample.", "text_images": [], "full_mark": 2, "rubric": ["$w_W = (m_1-m_2-m_3-m_4-m_5)/m \\times 0.7930 \\times 100\\% = 12.0\\%$ (2 points). Deduct 1 point if the answer is not reported to 3 significant figures."], "rubric_images": []}]}
{"problem_id": "Exam-17-T7", "header": "Problem 7 Low-Valent Cluster Compounds of Group 13 Metals\nLow-valent compounds of Group 13 elements such as aluminum and gallium can form metal clusters with diverse structures. Their bonding modes and electronic structures are important for understanding the nature of metal-metal bonds. This problem concerns low-valent gallium and aluminum halides and their assembly reactions with ligands, focusing on structural analysis, oxidation-state calculations, and cluster reactivity.", "header_images": [], "subquestions": [{"sub_id": "Exam-17-T7-1", "tags": ["2.1"], "text_open": "Synthesis and structures of $E_3$ units: Ga-Ga bonds are relatively stable and can support extended compounds.\n7.1 Chain-like $E_3$ structure: reaction of freshly prepared LiM with 1.67 eq GaI affords the trigallane C = $Ga_3M_3I_2$.\n7.1.1 Write the reaction equation for the formation of C from LiM and GaI.", "text_images": [{"file": "Exam-17-T7-1-1-prompt.png", "description": "Two-step reaction: '[dipp]N=C=N[dipp]'>t-BuLi (1eq), n-hexane, 0 °C>intermediate Li+M-, where M denotes the ligand and the SMILES for M- is '[dipp]/N=C([N-][dipp])/C(C)(C)C'. The M- anion has an N,N'-bis(dipp) amidinate framework with t-Bu bonded to the central carbon. The second step is '[dipp]/N=C([N-][dipp])/C(C)(C)C'>GaI (1.67eq), Toluene, -78 °C>C, where C = $Ga_3M_3I_2$ and its SMILES is 'I[Ga]1([Ga]2([Ga]3(I)N([dipp])C(C(C)(C)C)=[N]3[dipp])N([dipp])C(C(C)(C)C)=[N]2[dipp])N([dipp])C(C(C)(C)C)=[N]1[dipp]'. Three Ga atoms form a chain through Ga-Ga single bonds, each terminal Ga bears one I, and each Ga is chelated by one M ligand. No wedge or dashed bonds are shown."}], "full_mark": 2, "rubric": ["$3LiM + 5GaI = Ga_3M_3I_2 + 2Ga + 3LiI$ (2 points)"], "rubric_images": []}, {"sub_id": "Exam-17-T7-2", "tags": ["2.1"], "text_open": "7.1.2 Write the equation for the substitution reaction of trigallane C with $NaBEt_3H$ to form Et(M)Ga-Ga(M)-Ga(M)H.", "text_images": [], "full_mark": 2, "rubric": ["$Ga_3M_3I_2 + NaBEt_3H = Et(M)Ga-Ga(M)-Ga(M)H + NaI + BEt_2I$ (2 points)", "$Ga_3M_3I_2 + NaBEt_3H = Et(M)Ga-Ga(M)-Ga(M)H + NaBEt_2I_2$ is also acceptable."], "rubric_images": []}, {"sub_id": "Exam-17-T7-3", "tags": ["2.4", "1.6"], "text_open": "7.2 Cyclic $E_3$ structure: cyclotrigallanes have intriguing electronic properties. Reaction of $(Mes_2C_6H_3)Li$ with $GaCl_3$ gives $(Mes_2C_6H_3)GaCl_2$, which is reduced with potassium metal in diethyl ether to afford the cyclic gallium compound D = $K_2[(Mes_2C_6H_3)Ga]_3$. This compound has a nearly equilateral triangular $Ga_3$ core, with an average Ga-Ga bond length of 2.426 Å that is markedly shorter than a typical single bond, indicating a higher bond order. The two potassium ions lie on opposite sides of the $Ga_3$ plane. Write the SMILES string for D and determine whether its anion is aromatic.", "text_images": [{"file": "Exam-17-T7-3-1-prompt.png", "description": "Two-step reaction: $GaCl_3$ reacts with LiAr (1.0 eq) in $Et_2O$ at -78 °C; $K^0$ (4.0 eq) is then added and the mixture is reacted at rt. for 5 days to give $K_2(ArGa)_3·(Et_2O)_2$. The structure defining Ar = $(Mes_2C_6H_3)$ appears at the right: the top carbon of the central benzene ring, drawn in an alternating-double-bond Kekulé form, bears a wavy bond marking the attachment point to Ga, and each of the two ortho carbons bears a 2,4,6-trimethylphenyl group (Mes). The encodable structure of Ar is represented by the SMILES '[*]C1=C(C2=C(C)C=C(C)C=C2C)C=CC=C1C3=C(C)C=C(C)C=C3C', where [*] is the attachment point corresponding to the wavy bond."}], "full_mark": 3, "rubric": ["Structure of D: 'CC1=CC(C)=CC(C)=C1C(C=CC=C2C3=C(C)C=C(C)C=C3C)=C2[Ga-]4=[Ga-]([Ga]4C5=C(C6=C(C)C=C(C)C=C6C)C=CC=C5C7=C(C)C=C(C)C=C7C)C8=C(C9=C(C)C=C(C)C=C9C)C=CC=C8C%10=C(C)C=C(C)C=C%10C.[K+].[K+]'. The two K+ ions coordinate on opposite sides of the Ga3 triangular plane (2 points). The anion is aromatic (1 point)."], "rubric_images": [{"file": "Exam-17-T7-3-1-rubric.png", "description": "Reference structure of D: the anion [(Mes2C6H3)Ga]3 is represented by the SMILES 'CC1=CC(C)=CC(C)=C1C(C=CC=C2C3=C(C)C=C(C)C=C3C)=C2[Ga-]4=[Ga-]([Ga]4C5=C(C6=C(C)C=C(C)C=C6C)C=CC=C5C7=C(C)C=C(C)C=C7C)C8=C(C9=C(C)C=C(C)C=C9C)C=CC=C8C%10=C(C)C=C(C)C=C%10C.[K+].[K+]'. Three Ga atoms form a $Ga_3$ triangle, and each Ga is bonded to one Ar group; each central benzene ring bears Mes groups at the 2,6-positions, and all 6 Mes groups are 2,4,6-trimethylphenyl. The cluster has an overall 2- charge, shown at the center of the $Ga_3$ triangle and represented in the SMILES by two [Ga-] atoms. The two K+ ions coordinate on opposite sides of the Ga3 triangular plane, with their interactions with $Ga_3$ shown by dashed lines. In the SMILES, the Ga3 ring contains one Ga=Ga double bond as an aromatic Kekulé representation, whereas the original drawing shows all three Ga-Ga bonds as solid single lines. No wedge bonds are shown."}]}, {"sub_id": "Exam-17-T7-4", "tags": ["2.4", "1.4"], "text_open": "7.4 Eight-membered-ring structure: gaseous GaI is cocondensed with excess $PEt_3$ in toluene at -196 °C and then warmed slowly, giving F = $Ga_8I_8·6PEt_3$. This cluster has $C_{2h}$ symmetry and contains a nearly planar eight-membered $Ga_8$ ring. Two iodine atoms bridge across the ring to form a four-membered $Ga_2I_2$ ring. All eight Ga atoms are four-coordinate, and the average Ga-Ga bond length is 2.466 Å.\n7.4.1 Write the SMILES string for F.", "text_images": [], "full_mark": 4, "rubric": ["Structure of F = $Ga_8I_8·6PEt_3$: 'I[Ga@]1([P](CC)(CC)CC)[Ga@](I)([P](CC)(CC)CC)[Ga@@](I)([P](CC)(CC)CC)[Ga@@]2(I3)[Ga@](I)([P](CC)(CC)CC)[Ga@@](I)([P](CC)(CC)CC)[Ga@](I)([P](CC)(CC)CC)[Ga@]13I2' (4 points)"], "rubric_images": [{"file": "Exam-17-T7-4-1-rubric.png", "description": "Reference structure of F = $Ga_8I_8·6PEt_3$, represented by the SMILES 'I[Ga@]1([P](CC)(CC)CC)[Ga@](I)([P](CC)(CC)CC)[Ga@@](I)([P](CC)(CC)CC)[Ga@@]2(I3)[Ga@](I)([P](CC)(CC)CC)[Ga@@](I)([P](CC)(CC)CC)[Ga@](I)([P](CC)(CC)CC)[Ga@]13I2'. The 8 Ga atoms form an eight-membered ring through single bonds. Two I atoms bridge opposite Ga atoms across the ring; in the drawing, one bridging I is connected to the two Ga atoms by solid wedge bonds and the other by dashed bonds. Each of the remaining 6 Ga atoms bears one terminal I and one PEt3 (Et3P). The stereochemistry is retained by the @/@@ markers."}]}, {"sub_id": "Exam-17-T7-5", "tags": ["1.7", "2.4"], "text_open": "7.4.2 The Ga/I/$PEt_3$ system forms the eight-membered-ring cluster F ($Ga_8I_8$), whereas the Al/Br/$NEt_3$ system forms the four-membered-ring cluster E ($Al_4Br_4$) under similar conditions. Which explanations are the most reasonable? Select all that apply.\n(A) Ga has a smaller atomic radius than Al and a shorter Ga-Ga bond; formation of a four-membered ring would therefore cause greater ring strain and bond-angle distortion, so an eight-membered ring is favored to relieve strain.\n(B) I has a larger atomic radius than Br and is well suited for bridging; transannular bridges stabilize the eight-membered ring.\n(C) $PEt_3$ is a stronger electron donor than $NEt_3$, increasing the electron density at Ga and shortening the Ga-Ga bonds, which favors formation of a larger ring system.\n(D) The Ga-Ga bonds in the eight-membered ring have a higher bond order than the Al-Al bonds in the four-membered ring, giving greater thermodynamic stability.", "text_images": [], "full_mark": 2, "rubric": ["AB (2 points)"], "rubric_images": []}, {"sub_id": "Exam-17-T7-6", "tags": ["2.4", "1.2"], "text_open": "7.5 Fused-ring structure: attempts to synthesize amine-stabilized $Ga_8I_8·6NEt_3$ were unsuccessful, but partial oxidation unexpectedly gave G = $Ga_8I_8O_2·6NEt_3$. Its structure has $C_i$ symmetry and consists of two nearly planar $Ga_4O$ rings fused through a shared edge of a central four-membered $Ga_2O_2$ ring. All Ga atoms are four-coordinate. Write the SMILES string for G.", "text_images": [], "full_mark": 2, "rubric": ["Structure of G = $Ga_8I_8O_2·6NEt_3$: 'I[Ga@]1(O2[Ga@@](I)([N](CC)(CC)CC)[Ga@](I)([N](CC)(CC)CC)[Ga@]1([N](CC)(CC)CC)I)O3[Ga@@]2(I)[Ga@@](I)([N](CC)(CC)CC)[Ga@](I)([N](CC)(CC)CC)[Ga@@]3(I)[N](CC)(CC)CC' (2 points). Stereochemistry must be shown. The scoring guide includes a second acceptable structure: 'I[Ga@@]1(O2[Ga@@](I)([N](CC)(CC)CC)[Ga@](I)([N](CC)(CC)CC)[Ga@]1([N](CC)(CC)CC)I)O3[Ga@]2(I)[Ga@@](I)([N](CC)(CC)CC)[Ga@](I)([N](CC)(CC)CC)[Ga@@]3(I)[N](CC)(CC)CC'."], "rubric_images": [{"file": "Exam-17-T7-6-1-rubric.png", "description": "Reference structure of G = $Ga_8I_8O_2·6NEt_3$ in the first scoring-guide image, represented by the SMILES 'I[Ga@]1(O2[Ga@@](I)([N](CC)(CC)CC)[Ga@](I)([N](CC)(CC)CC)[Ga@]1([N](CC)(CC)CC)I)O3[Ga@@]2(I)[Ga@@](I)([N](CC)(CC)CC)[Ga@](I)([N](CC)(CC)CC)[Ga@@]3(I)[N](CC)(CC)CC'. Two $Ga_4O$ rings are fused through a shared edge of the central four-membered $Ga_2O_2$ ring. Each of the 8 Ga atoms bears one terminal I, and 6 of the Ga atoms additionally bear NEt3 (Et3N); the two Ga atoms in the central ring have no amine ligand. Wedge/dashed-bond stereochemistry is retained by the @/@@ markers."}, {"file": "Exam-17-T7-6-2-rubric.png", "description": "Reference structure of G = $Ga_8I_8O_2·6NEt_3$ in the second scoring-guide image, represented by the SMILES 'I[Ga@@]1(O2[Ga@@](I)([N](CC)(CC)CC)[Ga@](I)([N](CC)(CC)CC)[Ga@]1([N](CC)(CC)CC)I)O3[Ga@]2(I)[Ga@@](I)([N](CC)(CC)CC)[Ga@](I)([N](CC)(CC)CC)[Ga@@]3(I)[N](CC)(CC)CC'. The framework composition and connectivity are the same as in the first image, but the wedge/dashed orientations of I and NEt3 at some Ga atoms differ. One Ga vertex at the upper right is unlabeled in the image; this vertex is [Ga@@] in the SMILES."}]}, {"sub_id": "Exam-17-T7-7", "tags": ["1.1"], "text_open": "7.6 Tetrahedral type: high-temperature gaseous GaBr is cocondensed with a toluene solution of $NHEt_2$. The resulting dark solution is concentrated and allowed to stand at 25 °C, yielding crystals of cluster H = $Ga_5Br_7(NHEt_2)_5$. H contains a $Ga_5$ core in which the central Ga atom is tetrahedrally surrounded by four Ga atoms.\n7.6.2 In the sterically congested H molecule, the 12 heteroatoms—the halogens and the O/N atoms of the ligands—adopt an arrangement that maximizes their mutual separation. What polyhedron is formed by these 12 vertices?", "text_images": [{"file": "Exam-17-T7-7-1-prompt.png", "description": "Structure of H = $Ga_5Br_7(NHEt_2)_5$: the central Ga is connected by single bonds to four peripheral Ga atoms in a tetrahedral arrangement. The encodable structure of H is represented by the SMILES '[L][Ga](Br)(Br)[Ga]([Ga]([L])(Br)Br)([Ga]([L])([L])Br)[Ga]([L])(Br)Br', where [L] denotes the $NHEt_2$ ligand labeled in the image. A total of 5 L ligands and 7 Br atoms are distributed among the peripheral Ga atoms. The lower right of the image defines L = $NHEt_2$, and all bonds are solid single bonds."}], "full_mark": 1, "rubric": ["Distorted icosahedron or icosahedron (1 point). A regular icosahedron receives no credit."], "rubric_images": []}]}
{"problem_id": "Exam-17-T9", "header": "Problem 9 Stereochemistry\nCareful design of reactant structures and reaction conditions can provide highly selective stereochemical control.", "header_images": [], "subquestions": [{"sub_id": "Exam-17-T9-1", "tags": ["2.4", "1.2"], "text_open": "9.1 Garner's aldehyde is frequently used for stereochemical control in synthesis. The route below prepares it from the natural product S-serine.\n9.1.2 Garner's aldehyde can undergo further nucleophilic addition. Write the SMILES strings of the major products under the different reaction conditions below, representing them as anions and specifying stereochemistry.", "text_images": [{"file": "Exam-17-T9-1-1-prompt.png", "description": "Two-step reaction: 'OC[C@@H](C(O)=O)N'>>'OC[C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O'; 'OC[C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O'>DMSO, (COCl)2, DIPEA>'[H]C([C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O)=O'."}, {"file": "Exam-17-T9-1-2-prompt.png", "description": "Two parallel reactions: '[H]C([C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O)=O'>Nu^->[A]; '[H]C([C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O)=O'>Nu^-, [M]>[B]. The products are labeled only [A] and [B]; their structures are not shown."}], "full_mark": 4, "rubric": ["The SMILES for A is '[H][C@@]([O-])([Nu])[C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O', where Nu is the placeholder used in the original diagram.", "The SMILES for B is '[Nu][C@@]([O-])([H])[C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O', where Nu is the placeholder used in the original diagram.", "Each structure is worth 2 points."], "rubric_images": [{"file": "Exam-17-T9-1-1-rubric.png", "description": "The SMILES for A is '[H][C@@]([O-])([Nu])[C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O'; the SMILES for B is '[Nu][C@@]([O-])([H])[C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O'. Nu is the placeholder used in the original diagram."}]}, {"sub_id": "Exam-17-T9-2", "tags": ["2.7", "1.2"], "text_open": "9.1.3 In practice, the two reaction pathways compete. Determine which of the following conditions favor formation of product B, and briefly explain why.\n(1) Solvent: a. THF; b. toluene", "text_images": [{"file": "Exam-17-T9-2-1-prompt.png", "description": "'[H]C([C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O)=O'>nucleophile, solvent, additive>[A]+[B]. The products are labeled only [A] and [B]; their structures are not shown."}], "full_mark": 2, "rubric": ["b. (1 point). THF competitively coordinates the metal cation, reducing the amount of metal cation available for chelation (1 point)."], "rubric_images": []}, {"sub_id": "Exam-17-T9-3", "tags": ["2.7", "1.2"], "text_open": "(3) Additive: a. DMPU; b. $ZnCl_2$", "text_images": [], "full_mark": 2, "rubric": ["b. (1 point). As a Lewis base, DMPU binds the metal cation, reducing the amount of metal cation available for chelation (1 point)."], "rubric_images": []}, {"sub_id": "Exam-17-T9-4", "tags": ["1.2", "2.7"], "text_open": "9.1.4 Garner's aldehyde can also be synthesized from R-serine. In the reaction below, replacing Boc with Ts reverses the selectivity. Explain why.", "text_images": [{"file": "Exam-17-T9-4-1-prompt.png", "description": "'CC1(C)N([R])[C@@H](C=O)CO1'>Et-MgBr, THF>'CC1(C)N([R])[C@@H]([C@@H](O)CC)CO1' + 'CC1(C)N([R])[C@@H]([C@H](O)CC)CO1'. The image also states that the product ratio is 1:9 when R=Boc and 8:1 when R=Ts. R is the placeholder group used in the original diagram."}], "full_mark": 2, "rubric": ["The oxygen atoms of the $SO_2$ sulfonyl group in Ts have low electron density and extremely weak coordinating ability, so a stable chelate cannot form (1 point). The direction of attack therefore more closely follows a nonchelation model, giving the other diastereomer with high selectivity (1 point). Calculations show a larger LUMO energy difference between the two reactive Felkin-Anh conformations of the N-Ts analogue; the orientation of the C-NTs bond thus produces a stronger reactive-conformation preference and enhances anti selectivity."], "rubric_images": []}, {"sub_id": "Exam-17-T9-5", "tags": ["2.4", "1.2"], "text_open": "9.2 Garner's aldehyde can also participate in multicomponent reactions that construct new stereocenters. Write the SMILES string of the product of the reaction below.", "text_images": [{"file": "Exam-17-T9-5-1-prompt.png", "description": "Multicomponent reaction: 'C1(C=CC=C2)=C2NC=C1'.'CC(OC[C@H]1C=O)(C)N1C(OC(C)(C)C)=O'.'O=C1CC(OC(C)(C)O1)=O'>CH3CN, D,L-proline>. No product structure is shown to the right of the arrow."}], "full_mark": 2, "rubric": ["'O=C(OC(C)(C)O1)C(C1=O)[C@@H]([C@@H]2COC(C)(C)N2C(OC(C)(C)C)=O)C3=CNC4=C3C=CC=C4'"], "rubric_images": [{"file": "Exam-17-T9-5-1-rubric.png", "description": "The SMILES string is 'O=C(OC(C)(C)O1)C(C1=O)[C@@H]([C@@H]2COC(C)(C)N2C(OC(C)(C)C)=O)C3=CNC4=C3C=CC=C4'."}]}]}
{"problem_id": "Exam-17-T10", "header": "Problem 10 Cyclobutanation\nCyclobutane frameworks occur widely in bioactive natural products and pharmaceutical molecules. Because a four-membered ring combines appreciable ring strain with a rigid spatial geometry, the development of efficient and selective methods for constructing cyclobutanes has long been an important goal in organic synthesis.", "header_images": [], "subquestions": [{"sub_id": "Exam-17-T10-1", "tags": ["1.2"], "text_open": "10.1.2 A conformation of cyclobutane is shown below. For 1,3-dimethylcyclobutane and 1,2-dimethylcyclobutane, determine whether the cis or trans isomer is more stable, and write the SMILES string of its preferred conformation.", "text_images": [{"file": "Exam-17-T10-1-1-prompt.png", "description": "Cyclobutane is represented by the SMILES '[H][C@]([C@]([H])([H])[C@]([H])1[H])([H])[C@@]1([H])[H]'. The image is a perspective drawing of a puckered cyclobutane conformation, with two bold wedge bonds and 8 labeled H atoms."}], "full_mark": 4, "rubric": ["1,3-Dimethylcyclobutane: the cis isomer is more stable (1 point), with both substituents pseudo-equatorial in the preferred conformation (1 point). 1,2-Dimethylcyclobutane: the trans isomer is more stable (1 point), with both substituents pseudo-equatorial in the preferred conformation (1 point)."], "rubric_images": [{"file": "Exam-17-T10-1-1-rubric.png", "description": "Reference answer: the preferred conformation of cis-1,3-dimethylcyclobutane, with both substituents pseudo-equatorial, has the SMILES 'C[C@H]1C[C@@H](C)C1'. The image is an unlabeled skeletal conformation drawing."}, {"file": "Exam-17-T10-1-2-rubric.png", "description": "Reference answer: the preferred conformation of trans-1,2-dimethylcyclobutane, with both substituents pseudo-equatorial, has the SMILES 'C[C@H]1CC[C@@H]1C'. The image is an unlabeled skeletal conformation drawing."}]}, {"sub_id": "Exam-17-T10-2", "tags": ["2.2"], "text_open": "10.2 Werz and co-workers recently reported a method for constructing cyclobutanone frameworks by exploiting the intrinsic ring strain of cyclopropanol derivatives. Under mild conditions, the reaction couples release of small-ring strain with C-C bond formation, providing a new strategy for cyclobutane synthesis. To investigate the mechanism, the researchers designed deuterium-labeling and substituent-effect experiments. In the deuterium-labeling experiments, sulfonium salts bearing deuterium at different allylic positions were prepared and reacted under identical conditions; the resulting deuterium distributions are shown below. Using Experiment I as an example, write the SMILES strings of all key intermediates consistent with the observations.", "text_images": [{"file": "Exam-17-T10-2-1-prompt.png", "description": "Deuterium-labeling reaction in Experiment I: 'OC1(S(C2=CC=C(C)C=C2)(=O)=O)CC1'.'[2H]/C(C1=CC=CC=C1)=C\\C[S+]2C(C)(CCC2(C)C)C.F[B-](F)(F)F'>LiHMDS, CH2Cl2, -78°C to r.t.>'O=C1CCC1(C=C)C2=CC=CC=C2'. The product region shows the deuterium distribution: 89%, 0% D, 22% D, and 22% D."}, {"file": "Exam-17-T10-2-2-prompt.png", "description": "Deuterium-labeling reaction in Experiment II: 'OC1(S(C2=CC=C(C)C=C2)(=O)=O)CC1'.'CC1(C)CCC(C)(C)[S+]1C/C([2H])=C/C2=CC=CC=C2.F[B-](F)(F)F'>LiHMDS, CH2Cl2, -78°C to r.t.>'O=C1CCC1(C2=CC=CC=C2)C=C'. The product region shows the deuterium distribution: 94%, 94% D, 0% D, and 0% D."}], "full_mark": 7, "rubric": ["Key intermediates: '[2H]/C(C1=CC=CC=C1)=C\\[CH-][S+]2C(C)(C)CCC2(C)C' (1 point); 'O=C1CC1' (1 point); '[2H]C(/C=C/[S+]1C(C)(C)CCC1(C)C)(C2([O-])CC2)C3=CC=CC=C3' (1 point); 'CC1(C)CCC(C)(C)[S+]1[CH-]/C=C(C2(O[2H])CC2)/C3=CC=CC=C3' (2 points); 'CC1(C)CCC(C)(C)[S+]1C([2H])/C=C(C2([O-])CC2)/C3=CC=CC=C3' (2 points)"], "rubric_images": [{"file": "Exam-17-T10-2-1-rubric.png", "description": "Reference key intermediate 1, represented by the SMILES '[2H]/C(C1=CC=CC=C1)=C\\[CH-][S+]2C(C)(C)CCC2(C)C'."}, {"file": "Exam-17-T10-2-2-rubric.png", "description": "Reference key intermediate 2 (cyclopropanone), represented by the SMILES 'O=C1CC1'."}, {"file": "Exam-17-T10-2-3-rubric.png", "description": "Reference key intermediate 3, represented by the SMILES '[2H]C(/C=C/[S+]1C(C)(C)CCC1(C)C)(C2([O-])CC2)C3=CC=CC=C3'."}, {"file": "Exam-17-T10-2-4-rubric.png", "description": "Reference key intermediate 4, represented by the SMILES 'CC1(C)CCC(C)(C)[S+]1[CH-]/C=C(C2(O[2H])CC2)/C3=CC=CC=C3'."}, {"file": "Exam-17-T10-2-5-rubric.png", "description": "Reference key intermediate 5, represented by the SMILES 'CC1(C)CCC(C)(C)[S+]1C([2H])/C=C(C2([O-])CC2)/C3=CC=CC=C3'."}]}, {"sub_id": "Exam-17-T10-3", "tags": ["2.2", "2.4"], "text_open": "10.3 Inspired by this work, other research groups developed sulfur-ylide-mediated ring expansion of donor-acceptor spirocyclopropanes and applied it to the synthesis of spirocyclobutane frameworks.\n10.3.1 Based on the information above, write the SMILES strings of the key intermediate and final product in the reaction below. Stereochemistry is not required.", "text_images": [{"file": "Exam-17-T10-3-1-prompt.png", "description": "Single-step reaction: 'O=C1C2(C(C3=CC=CC=C3)C2)C(C4=CC5=CC=CC=C5C=C41)=O'.'O=C([CH-][S+](C)C)C1=CC=CC=C1'>EtOAc, reflux, 24 h (sulfur ylide 1.5 equiv)>; no product structure is drawn to the right of the arrow."}], "full_mark": 2, "rubric": ["Key intermediate: 'O=C1[C-](CC(C(C(C2=CC=CC=C2)=O)[S+](C)C)C3=CC=CC=C3)C(C4=CC5=CC=CC=C5C=C41)=O' (1 point). Final product: 'O=C1C2(CC(C3=CC=CC=C3)C2C(C4=CC=CC=C4)=O)C(C5=CC6=CC=CC=C6C=C51)=O' (1 point)."], "rubric_images": [{"file": "Exam-17-T10-3-1-rubric.png", "description": "Reference answer: the key intermediate in 10.3.1, represented by the SMILES 'O=C1[C-](CC(C(C(C2=CC=CC=C2)=O)[S+](C)C)C3=CC=CC=C3)C(C4=CC5=CC=CC=C5C=C41)=O'."}, {"file": "Exam-17-T10-3-2-rubric.png", "description": "Reference answer: the final product in 10.3.1, represented by the SMILES 'O=C1C2(CC(C3=CC=CC=C3)C2C(C4=CC=CC=C4)=O)C(C5=CC6=CC=CC=C6C=C51)=O'."}]}, {"sub_id": "Exam-17-T10-4", "tags": ["2.4"], "text_open": "10.3.2 If the spirocyclopropane substrate in 10.3.1 is replaced by the 2’, 3’-unsubstituted spirocyclopropane shown below, the reaction forms a fused-ring byproduct in addition to the desired ring-expanded product. Write the SMILES string of this byproduct.", "text_images": [{"file": "Exam-17-T10-4-1-prompt.png", "description": "The 2’, 3’-unsubstituted spirocyclopropane substrate has the SMILES 'O=C1C2(CC2)C(C3=CC4=CC=CC=C4C=C31)=O'."}], "full_mark": 2, "rubric": ["'O=C1C2=C(OC(C(C3=CC=CC=C3)=O)CC2)C4=CC5=CC=CC=C5C=C41' (2 points)"], "rubric_images": [{"file": "Exam-17-T10-4-1-rubric.png", "description": "Reference answer: the fused-ring byproduct in 10.3.2, represented by the SMILES 'O=C1C2=C(OC(C(C3=CC=CC=C3)=O)CC2)C4=CC5=CC=CC=C5C=C41'."}]}, {"sub_id": "Exam-17-T10-5", "tags": ["1.2", "2.7"], "text_open": "10.3.3 The reaction is in fact stereoselective. For the reaction in 10.3.1, assume that the substrate is racemic. Determine whether the major product is racemic, and determine whether the two substituents on the cyclobutane ring are cis or trans, explaining your answer.", "text_images": [], "full_mark": 3, "rubric": ["The product is racemic (1 point). The substituents are trans (1 point). In the cis pathway, Ph and COPh are crowded on the same face, whereas trans ring closure has less steric hindrance (1 point)."], "rubric_images": []}]}
{"problem_id": "Exam-18-T3", "header": "Problem 3 Latent Monomers and Controlled Metathesis Polymerization\nControlling the temporal and spatial sequence of polymerization is an important topic in conventional polymer synthesis. Chemists have generally developed “latent catalysts” that switch between inactive and active states in response to external stimuli such as light or heat. Recent research, however, has introduced a new “latent monomer” strategy: isomerization of a photoswitchable molecule generates an active monomer in situ and on demand, enabling efficient control of polymerization.\nThis problem concerns a classic class of highly reactive ring-opening metathesis polymerization (ROMP) monomers: norbornadiene (NBD) derivatives. Under ultraviolet irradiation of a specific wavelength, they undergo intramolecular cycloaddition to form highly strained quadricyclane (QC) structures.", "header_images": [], "subquestions": [{"sub_id": "Exam-18-T3-1", "tags": ["2.4"], "text_open": "3.1 Synthesis and Photothermal Isomerization of the Latent Monomer\nMonomer A, an NBD derivative, can be prepared by reacting cyclopentadiene with diethyl acetylenedicarboxylate at room temperature. Under 313 nm ultraviolet irradiation, A is quantitatively converted into latent monomer B, a QC derivative. Adding silica-coated gold nanobipyramids (AuBPs) to B and irradiating at 850 nm with infrared light regenerates A.\n3.1.1 Write the SMILES strings for A and B.", "text_images": [], "full_mark": 3, "rubric": ["'O=C(OCC)C1=C(C(OCC)=O)[C@@H]2C=C[C@H]1C2' and 'O=C(OCC)C12C3(C(OCC)=O)C4C3C1C2C4'. (1 point for the structure of A and 2 points for the structure of B)"], "rubric_images": [{"file": "Exam-18-T3-1-1-rubric.png", "description": "A has SMILES 'O=C(OCC)C1=C(C(OCC)=O)[C@@H]2C=C[C@H]1C2', and B has SMILES 'O=C(OCC)C12C3(C(OCC)=O)C4C3C1C2C4'."}]}, {"sub_id": "Exam-18-T3-2", "tags": ["2.4"], "text_open": "3.3 The complete inertness of latent monomers toward the catalyst at room temperature enables the synthesis of complex polymer topologies that are difficult to obtain by conventional metathesis polymerization. The researchers replaced the ethoxycarbonyl groups, $-COOC_2H_5$, in B with pent-4-enyloxycarbonyl groups, $-COO(CH_2)_3CH=CH_2$, to give monomer F. F was mixed with the Grubbs catalyst and reacted at 40℃, releasing gas X and giving a soluble linear polymer G with double bonds in its main chain. G was coated onto a microscope slide placed over a masked LED and then heated to 80℃, rapidly converting the film into an insoluble crosslinked network H. Write the SMILES strings for polymers G and H.", "text_images": [], "full_mark": 4, "rubric": ["G is a polymer whose repeat unit has the SMILES string 'O=C(OCCCC=C)C12C3(C(OCCCC=C)=O)C4C3C1C2C4'. Two repeat units are connected through their terminal double bonds, with one isolated double bond between each pair of repeat units. In the drawing, enclose the repeat unit in square brackets and place n as a subscript at the lower right. (2 points each for the structures of G and H; deduct 1 point if the degrees of polymerization of the upper and lower chains in H are both labeled m or both labeled n)"], "rubric_images": [{"file": "Exam-18-T3-2-1-rubric.png", "description": "G is a polymer with repeating-unit structure 'O=C(OCCCC=C)C12C3(C(OCCCC=C)=O)C4C3C1C2C4'. Two repeat units are connected through their terminal double bonds, with one isolated double bond between each pair of units. The repeating unit is enclosed in square brackets with a subscript n at the lower right."}, {"file": "Exam-18-T3-2-2-rubric.png", "description": "H is a polymer with repeating-unit structure 'C=CC(CC1C=C)C(C(OCCCC=C)=O)=C1C(OCCCC=C)=O'. Two repeat units are connected through double bonds attached to the five-membered ring, with one isolated double bond between each pair of units, and also through double bonds attached to the ester groups, again with one isolated double bond between each pair of units. The two connection modes have different degrees of polymerization: m for the former and n for the latter. Two pairs of square brackets enclose the two chains and carry subscripts n and m, respectively."}]}]}
{"problem_id": "Exam-18-T4", "header": "Problem 4 Complexometric Titration\nEthylenediaminetetraacetic acid (EDTA) is an effective chelating ligand widely used in chemical analysis. Approximately 20 g of a standard reagent is roughly weighed, 250mL of water is added, and the mixture is heated to dissolve the solid, cooled, and mixed thoroughly. The measured pH of the solution is 4.2. The stepwise ionization constants of EDTA are shown below.\n\n|  | $K_{a1}$ | $K_{a2}$ | $K_{a3}$ | $K_{a4}$ | $K_{a5}$ | $K_{a6}$ |\n| --- | --- | --- | --- | --- | --- | --- |\n|  | $10^{-0.9}$ | $10^{-1.6}$ | $10^{-2.0}$ | $10^{-2.67}$ | $10^{-6.16}$ | $10^{-10.26}$ |", "header_images": [], "subquestions": [{"sub_id": "Exam-18-T4-1", "tags": ["4.5", "3.4"], "text_open": "The standard solution is then standardized. Weigh 0.2500 g of working-reference ZnO that has been ignited to constant mass in a furnace at 800 ℃. Moisten it with a small amount of water, dissolve it in 2 mL of 20% hydrochloric acid solution, and add 100 mL of water. Adjust the solution to pH 8 with hexamethylenetetramine (HMTA) solution, add 10 mL of $(CH_2)_6N_4$-$(CH_2)_6N_4$·HCl buffer to adjust the pH to 10, and then add 5 drops of Eriochrome Black T indicator solution. Titrate with the prepared EDTA sodium-salt solution until the color changes from purple to pure blue. Conduct a blank determination in parallel. After blank correction, the volume of EDTA consumed is $V = 15.35\\ \\mathrm{mL}$.\n4.2.2 The buffering ability of a buffer solution is commonly measured by its buffer capacity, denoted β. It is defined as the amount db in mol of strong base required to raise the pH of 1L of buffer solution by dpH units, with the mathematical expression\n$\\beta = \\frac{db}{dpH} = -2.30[H^+] \\times \\frac{db}{d[H^+]}$\nGiven $pK_b((CH_2)_6N_4) = 4.9$, calculate the buffer capacity of a pH = 10 $(CH_2)_6N_4$-$(CH_2)_6N_4$·HCl buffer in which the analytical concentration of $(CH_2)_6N_4$ is 0.1000 mol/L. Hint: express $b$ as a function of $[H^+]$ and differentiate with respect to $[H^+]$ to obtain $\\frac{db}{d[H^+]}$. B may be used to represent $(CH_2)_6N_4$ in the calculation.", "text_images": [], "full_mark": 4, "rubric": ["In this problem, the HMTA-HMTA·HCl buffer may be treated as the sum of a $c$ mol/L HMTA·HCl solution and a $b$ mol/L monobasic strong-base solution. In the HMTA-HMTA·HCl buffer, addition of strong base causes $HB^+ + OH^- \\rightleftharpoons B + H_2O$. The proton-balance equation is\n$b = [OH^-] + [B] - [H^+]$.\nUsing the distribution fraction gives\n$b = -[H^+] + \\frac{K_w}{[H^+]} + \\frac{cK_a}{[H^+] + K_a}$. (1 point; either of the two equations above is sufficient.)\nDifferentiating with respect to $[H^+]$ gives\n$\\frac{db}{d[H^+]} = -1 - \\frac{K_w}{[H^+]^2} - \\frac{cK_a}{([H^+] + K_a)^2}$. (2 points)\nSubstitution into the original equation gives\n$\\beta = -2.30[H^+] \\times \\frac{db}{d[H^+]} = 2.30[H^+]\\{1 + \\frac{K_w}{[H^+]^2} + \\frac{cK_a}{([H^+] + K_a)^2}\\} = 0.02307$. (1 point)\nFull credit is also given for using $b \\approx [B] - [B]_0 = [HB^+]_0 - [HB^+]$ to obtain $\\frac{db}{d[H^+]} = \\frac{d([B] - [B]_0)}{d[H^+]} = -\\frac{cK_a}{([H^+] + K_a)^2}$.\nA correct result of $\\beta = 0.0228$ obtained using the Henderson-Hasselbalch equation is also awarded full credit."], "rubric_images": []}, {"sub_id": "Exam-18-T4-2", "tags": ["4.5", "3.4"], "text_open": "4.2.3 Neglecting volume changes during standardization, the change in solution pH over the entire process must not exceed 0.2 pH unit to ensure an accurate endpoint. Determine the minimum value of $[(CH_2)_6N_4]$.", "text_images": [], "full_mark": 4, "rubric": ["During standardization, $H_2Y^{2-} + Zn^{2+} = ZnY^{2-} + 2H^+$.\nThus, the amount of $H^+$ added to the solution during standardization is $n(H^+) = 2n(ZnO) = 0.006144\\ \\mathrm{mol}$. (1 point)\nFrom the definition of buffer capacity, the condition in the question becomes $\\beta \\geq -\\frac{da}{dpH} = \\frac{0.006144}{0.2} = 0.03072$. (1 point)\nBecause $[H^+]$ is very small under these conditions, $\\beta = 2.30[H^+]\\{1 + \\frac{K_w}{[H^+]^2} + \\frac{cK_a}{([H^+] + K_a)^2}\\} \\approx \\frac{2.30cK_a[H^+]}{([H^+] + K_a)^2}$. (1 point)\nSubstitution and solution of the equation give $c \\geq 0.09632\\ \\mathrm{mol/L}$. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-18-T4-3", "tags": ["4.1"], "text_open": "4.3 During standardization, why is the $(CH_2)_6N_4$-$(CH_2)_6N_4$·HCl buffer not added directly to adjust the solution to pH=10?", "text_images": [], "full_mark": 1, "rubric": ["The pH must be adjusted gradually to prevent formation of a ZnO precipitate that would interfere with standardization. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-18-T4-4", "tags": ["4.5", "1.4"], "text_open": "4.4 The EDTA standard solution standardized in this problem can be used to determine the concentrations of Cu2+ and Mg2+ in a solution by either of the following methods. Method 1: Masking Cu2+ ions. Adjust the solution to pH 5.5 with an HMTA buffer and add excess thiosulfate to mask Cu2+. In this reaction, thiosulfate acts as both a reducing agent and a ligand. Method 2: Selective determination of $Cu^{2+}$. Adjust the solution to pH 5 with an HMTA buffer. Using 5-Br-PADAT as the indicator, selectively determine $Cu^{2+}$ with an EDTP standard solution whose principal component is disodium ethylenediaminetetrapropionate. 4.4.2 Based on the data below, explain why EDTP shows high selectivity for $Cu^{2+}$.\n\n|  | $Cu^{2+}$ | $Zn^{2+}$ | $Cd^{2+}$ | $Mg^{2+}$ |\n| --- | --- | --- | --- | --- |\n| Ionic radius/pm | 73 | 74 | 95 | 72 |\n| $\\lg K_{M-EDTP}$ | 15.4 | 7.8 | 6.0 | 1.8 |\n| $\\lg K_{M-EDTA}$ | 18.8 | 16.50 | 16.46 | 8.7 |", "text_images": [], "full_mark": 2, "rubric": ["$Cu^{2+}$ exhibits the Jahn-Teller effect in an octahedral field and adopts a distorted octahedral geometry. (1 point) EDTP is a chelating ligand that forms a relatively strained macrocycle on coordination to a metal ion, making the complexes less stable. (1 point) The distorted octahedral geometry relieves some of this strain, making Cu-EDTP more stable."], "rubric_images": []}]}
{"problem_id": "Exam-19-T7", "header": "Question 7 Fundamental Organic Chemistry", "header_images": [], "subquestions": [{"sub_id": "Exam-19-T7-1", "tags": ["1.2", "1.6"], "text_open": "7.2 Consider the two conformations of $FCH_2CH_2X^+$ shown below, which interconvert by rotation about the C–C single bond. The relative energy of the conformation on the right is defined as 0.0 kcal/mol. Define the relative energy of the left conformation with respect to the right as $\\Delta E=E_{left\\ conformation}-E_{right\\ conformation}$.\n7.2.2 Let the values of $\\Delta E$ for $X^+ = NH_3^+$, $OH_2^+$, and 1-pyridyl (1-pyridinium) be A, B, and C, respectively. Arrange A, B, and C in increasing order.", "text_images": [{"file": "Exam-19-T7-1-1-prompt.png", "description": "The figure shows two conformations of $FCH_2CH_2X^+$ connected by a reversible-equilibrium arrow. The left conformation has SMILES 'FCC[X+]', and the right conformation has SMILES 'FCC[X+]'. They are conformations of the same molecule and therefore share the same SMILES. In the left conformation, the C–X+ bond is shown as a thick wedge. The right conformation is an extended zigzag arrangement in which X+ and F are not anti, and 0.0 kcal/mol is written below it. The bottom label reads $X^+ = -NH_3^+, -OH_2^+, 1-pyridyl$."}], "full_mark": 2, "rubric": ["B < A < C (2 points).", "Credit is awarded only for the completely correct order."], "rubric_images": []}, {"sub_id": "Exam-19-T7-2", "tags": ["4.3"], "text_open": "7.3 Comparison of stretching wavenumbers\n7.3.1 Which of the following 2 compounds has the higher carbonyl stretching wavenumber in the infrared spectrum?", "text_images": [{"file": "Exam-19-T7-2-1-prompt.png", "description": "The SMILES of A is '[H]C(C1=CC=CC=C1)=O'; the SMILES of B is '[H]/C(C1=CC=CC=C1)=[O+]/[B-](F)(F)F'. B is the benzaldehyde–BF3 adduct. A ⊕ sign appears next to the carbonyl oxygen and a ⊖ sign next to F3B, corresponding to [O+] and [B-] in the SMILES. O–B is a single bond."}], "full_mark": 1, "rubric": ["A (1 point)."], "rubric_images": []}, {"sub_id": "Exam-19-T7-3", "tags": ["1.2", "1.7"], "text_open": "7.4 Drug molecules usually must adopt a specific conformation when binding to a target protein. If the dominant free-state conformation of a drug is already close to its bound conformation, the free-energy cost of conformational reorganization upon binding is reduced, which can increase binding affinity. Indinavir binds most effectively to HIV protease when its central carbon chain adopts an extended zigzag conformation. The figure below shows two epimers, A and B, of an indinavir analog. Determine which isomer is more likely to bind the target protein more strongly.", "text_images": [{"file": "Exam-19-T7-3-1-prompt.png", "description": "The SMILES of A is 'O=C(N[C@@H]1C(C=CC=C2)=C2C[C@H]1O)[C@H](CC3=CC=CC=C3)[C@@H](F)[C@H](O)CN(CC4)[C@H](C(NC(C)(C)C)=O)CN4CC5=CC=CN=C5'; the SMILES of B is 'O=C(N[C@@H]1C(C=CC=C2)=C2C[C@H]1O)[C@H](CC3=CC=CC=C3)[C@H](F)[C@H](O)CN(CC4)[C@H](C(NC(C)(C)C)=O)CN4CC5=CC=CN=C5'. A and B are epimers of an indinavir analog with the same skeleton, differing only in the orientation of F at the central-chain CH(F) carbon. F is shown with a wedge in A and a dashed bond in B; the SMILES preserve this stereochemical difference as [C@@H](F) and [C@H](F), respectively."}], "full_mark": 2, "rubric": ["A (2 points)."], "rubric_images": []}, {"sub_id": "Exam-19-T7-4", "tags": ["1.2"], "text_open": "7.5 Consider the three representative conformations A, B, and C shown below. Arrange their relative energies in increasing order.", "text_images": [{"file": "Exam-19-T7-4-1-prompt.png", "description": "The figure shows three representative conformations A, B, and C of dimethoxymethane ($CH_3OCH_2OCH_3$), with reversible-equilibrium arrows between A and B and between B and C. All three have the same SMILES: A is 'COCOC', B is 'COCOC', and C is 'COCOC'. Their conformational differences are expressed by bond orientations that SMILES cannot encode. A is a planar zigzag with both O–CH3 bonds drawn as ordinary single bonds. In B, the left O–CH3 bond is a solid wedge pointing downward and the right O–CH3 bond is an ordinary single bond extending toward the upper right. In C, the left O–CH3 bond is a solid wedge pointing downward and the right O–CH3 bond is a dashed bond pointing downward."}], "full_mark": 2, "rubric": ["C < B < A (2 points).", "Credit is awarded only for the completely correct order."], "rubric_images": []}]}
{"problem_id": "Exam-20-T7", "header": "Nitric Oxide Synthase and Its Artificial Models\nNitric oxide, NO, is an important signaling molecule involved in regulating many physiological processes in mammalian cardiovascular, immune, and nervous systems. In the human body, NO is synthesized under the catalysis of the heme enzyme nitric oxide synthase (NOS). Researchers are also working to reproduce this biological synthetic pathway using artificial models outside the body.", "header_images": [], "subquestions": [{"sub_id": "Exam-20-T7-1", "tags": ["2.1", "1.6"], "text_open": "7.1 Arginine is a natural amino acid involved in multiple metabolic pathways in the human body, including synthesis of the signaling molecule nitric oxide and the urea cycle. The structure of arginine in the cellular environment is shown below.\n7.1.1 In the final step of the urea cycle, arginine is a reactant in a reaction that releases urea and forms amino acid A. No redox process occurs. Write the reaction equation, representing organic compounds by SMILES strings.", "text_images": [{"file": "Exam-20-T7-1-1-prompt.png", "description": "Arginine has the SMILES '[NH3+][C@@H](CCCNC(N)=[NH2+])C([O-])=O'. Stereochemistry at the α-carbon is shown by a solid wedge bond, and the overall structure is zwitterionic."}], "full_mark": 2, "rubric": ["'[NH3+][C@@H](CCCNC(N)=[NH2+])C([O-])=O' + H2O > '[NH3+][C@@H](CCC[NH3+])C([O-])=O' + 'NC(N)=O'. 2 points total; if the amino group of ornithine in the product is not protonated, award only 1 point."], "rubric_images": [{"file": "Exam-20-T7-1-1-rubric.png", "description": "Reference equation for 7.1.1: '[NH3+][C@@H](CCCNC(N)=[NH2+])C([O-])=O' + H2O > '[NH3+][C@@H](CCC[NH3+])C([O-])=O' + 'NC(N)=O'. The coefficient of every species is 1, and the α-carbon is shown with a solid wedge bond."}]}, {"sub_id": "Exam-20-T7-2", "tags": ["2.2", "2.1"], "text_open": "For the reaction equations below, electrically neutral citrulline in the cellular environment may be represented by the SMILES string '[NH3+][C@@H](CCCNC(N)=O)C([O-])=O'.\n7.1.3 In the biological pathway for nitric oxide synthesis, arginine is the starting amino acid and reacts with 1 equivalent of NADPH and 1 equivalent of $O_2$ to form intermediate B and water (reaction 1). B then reacts further with $O_2$ and NADPH to form citrulline and nitric oxide, NO (reaction 2). The oxidized form of NADPH in organisms is $NADP^+$. Based on this information, infer and write the SMILES string of B and write the equation for reaction 2. The detailed structures of NADPH/$NADP^+$ need not be represented by SMILES.", "text_images": [], "full_mark": 3, "rubric": ["Intermediate: '[NH3+][C@@H](CCCN/C(N)=N\\O)C([O-])=O' or '[NH3+][C@@H](CCCNC(NO)=[NH2+])C([O-])=O' (1 point).", "2'[NH3+][C@@H](CCCN/C(N)=N\\O)C([O-])=O' + 2O2 + NADPH + H+ > 2'[NH3+][C@@H](CCCNC(N)=O)C([O-])=O' + 2NO + 2H2O + NADP+ or 2'[NH3+][C@@H](CCCNC(NO)=[NH2+])C([O-])=O' + 2O2 + NADPH > 2'[NH3+][C@@H](CCCNC(N)=O)C([O-])=O' + 2NO + 2H2O + NADP+ + H+ (2 points total). Balancing with $OH^-$/$H_2O$ is also acceptable."], "rubric_images": [{"file": "Exam-20-T7-2-1-rubric.png", "description": "Intermediate B in the first representation has the SMILES '[NH3+][C@@H](CCCN/C(N)=N\\O)C([O-])=O'."}, {"file": "Exam-20-T7-2-2-rubric.png", "description": "Intermediate B in the second representation has the SMILES '[NH3+][C@@H](CCCNC(NO)=[NH2+])C([O-])=O'."}, {"file": "Exam-20-T7-2-3-rubric.png", "description": "Reference equation for reaction 2, first version: 2'[NH3+][C@@H](CCCN/C(N)=N\\O)C([O-])=O' + 2O2 + NADPH + H+ > 2'[NH3+][C@@H](CCCNC(N)=O)C([O-])=O' + 2NO + 2H2O + NADP+."}, {"file": "Exam-20-T7-2-4-rubric.png", "description": "Reference equation for reaction 2, second version: 2'[NH3+][C@@H](CCCNC(NO)=[NH2+])C([O-])=O' + 2O2 + NADPH > 2'[NH3+][C@@H](CCCNC(N)=O)C([O-])=O' + 2NO + 2H2O + NADP+ + H+."}]}, {"sub_id": "Exam-20-T7-3", "tags": ["2.4", "1.4"], "text_open": "7.2 The enzymatic mechanism above shows that guanidino groups and their derivatives play important roles in biological NO synthesis. Accordingly, oxidation of guanidine-related ligands by various oxidants to generate NO in vitro has been studied extensively as a strategy for artificial NOS models. Osmium-centered complexes can undergo pH-dependent aerobic oxidation at ambient temperature and pressure to release nitric oxide. The precursor complex is shown below.\n**7.2.1** Interestingly, the precursor complex reacts with a large excess of $CN^-$ (20 eq.) to form a new cyanamide complex **X**. During the process, the tridentate Schiff-base ligand isomerizes into a bidentate ligand containing a five-membered ring, accompanied by hydride transfer; finally, the Os center undergoes a one-electron oxidation. The resulting complex ion **X** has a meridional (mer) structure and an overall charge of two negative units. Draw the structure of complex ion **X** and state the oxidation state of the central Os.", "text_images": [{"file": "Exam-20-T7-3-1-prompt.png", "description": "The precursor complex has the SMILES 'Cl[Os]12([N](C3=CC=CC=C3O1)=CC4=C(O2)C=CC=C4)([OH2])#N'. The central Os is six-coordinate and bears chloride, water, and an N,O-containing ligand."}], "full_mark": 4, "rubric": ["Complex ion X has the SMILES '[H]N(C#N)[Os]1(C#N)(C#N)(C#N)OC2=CC=CC=C2C3=[N]1C4=CC=CC=C4O3' and is an overall 2- anion. The structure of X is worth 3 points: 1 point for the correct bidentate ligand, 1 point for the correct cyanamide ligand, and 1 point for the correct meridional arrangement of cyanide ligands. Deduct 1 point if the charge is not shown, and deduct 1 point if an unnecessary (±) label is added.", "Oxidation state of Os: +3 or III (1 point)"], "rubric_images": [{"file": "Exam-20-T7-3-1-rubric.png", "description": "Complex ion X has the SMILES '[H]N(C#N)[Os]1(C#N)(C#N)(C#N)OC2=CC=CC=C2C3=[N]1C4=CC=CC=C4O3' and is an overall 2- anion, with the total charge shown outside the brackets."}]}, {"sub_id": "Exam-20-T7-4", "tags": ["2.4", "2.2"], "text_open": "7.2.2 Complex ion X rapidly undergoes addition with $NH_2OH\\cdot HCl$ to form monoanionic complex C. Infrared spectroscopy indicates that C contains a C=N double bond outside an aromatic ring. C reacts with $O_2$ by different pathways at different pH values.\nAt pH = 6.8, C reacts with 0.5 equivalent of $O_2$ to form complex ion D, which undergoes linkage isomerization of the ligand to form E. Further oxidation of E by $O_2$ gives small molecule M and diamagnetic complex ion F containing a nitrosyl ligand. Single-crystal diffraction gives an N-O bond length of 1.393 Å in C and 1.158 Å in D.\nAt pH = 12, reaction of C with $O_2$ gives complex ion G. G is structurally similar to D and carries the same charge, but its relative molecular mass is 1 less than that of D. G can rapidly lose a proton to form X and release NO, or react with $OH^-$ to form dianionic complex ion H and likewise release NO.\nIn this problem, abbreviate ligand groups that do not participate in subsequent reactions. For example, the precursor complex in 7.2.1 may be written as [Os]≡N. Based on this information, infer and write the SMILES strings of C-H and M.", "text_images": [], "full_mark": 11, "rubric": ["C: '[#76]/[N]([H])=C(NO)\\N', an overall 1- anion (2 points).", "D: '[#76]/[N]([H])=C(N=O)\\N', an overall 1- anion (2 points).", "E: '[#76][N](/C(N)=N/[H])=O', an overall 1- anion (1 point).", "F: '[#76]N=O', an overall 1- anion (1 point).", "G: '[#76]/N=C(N=O)\\N', an overall 1- anion (2 points).", "H: '[#76]N([H])C(N)=O', an overall 2- anion (2 points).", "M: 'NC#N' (1 point).", "As required in the prompt, ligand groups that do not participate in subsequent reactions are abbreviated in these answers. Expanding [Os] to the full structure is also acceptable."], "rubric_images": [{"file": "Exam-20-T7-4-1-rubric.png", "description": "C has the SMILES '[#76]/[N]([H])=C(NO)\\N' and is an overall 1- anion, with the total charge shown outside the brackets."}, {"file": "Exam-20-T7-4-2-rubric.png", "description": "D has the SMILES '[#76]/[N]([H])=C(N=O)\\N' and is an overall 1- anion, with the total charge shown outside the brackets."}, {"file": "Exam-20-T7-4-3-rubric.png", "description": "E has the SMILES '[#76][N](/C(N)=N/[H])=O' and is an overall 1- anion, with the total charge shown outside the brackets."}, {"file": "Exam-20-T7-4-4-rubric.png", "description": "F has the SMILES '[#76]N=O' and is an overall 1- anion, with the total charge shown outside the brackets."}, {"file": "Exam-20-T7-4-5-rubric.png", "description": "G has the SMILES '[#76]/N=C(N=O)\\N' and is an overall 1- anion, with the total charge shown outside the brackets."}, {"file": "Exam-20-T7-4-6-rubric.png", "description": "H has the SMILES '[#76]N([H])C(N)=O' and is an overall 2- anion, with the total charge shown outside the brackets."}, {"file": "Exam-20-T7-4-7-rubric.png", "description": "M has the SMILES 'NC#N'."}, {"file": "Exam-20-T7-4-8-rubric.png", "description": "Reference structure with the [Os] abbreviation expanded, represented by the SMILES 'C[Os]1(C#N)(C#N)(C#N)OC2=CC=CC=C2C3=[N]1C4=CC=CC=C4O3'. The central Os is bonded to a methyl group, three cyanide ligands (C≡N), and an O,N-bidentate ligand, the deprotonated form of 2-hydroxyphenylbenzoxazole."}]}, {"sub_id": "Exam-20-T7-5", "tags": ["1.4"], "text_open": "7.2.3 Among complexes C-H, identify those in which the oxidation state of the central Os differs from that in X. Give their labels.", "text_images": [], "full_mark": 2, "rubric": ["F, G", "2 points total. Deduct 1 point for each extra or omitted choice, down to a minimum of zero."], "rubric_images": []}, {"sub_id": "Exam-20-T7-6", "tags": ["2.1"], "text_open": "7.2.4 Write the equation for formation of H from C with release of NO at pH = 12. Ligand groups not participating in the reaction may be abbreviated.", "text_images": [], "full_mark": 2, "rubric": ["4'[#76]/[N]([H])=C(NO)\\N' + 3O2 + 4OH- > 4'[#76]N([H])C(N)=O' + 4NO + 6H2O (2 points for the reaction equation)."], "rubric_images": [{"file": "Exam-20-T7-6-1-rubric.png", "description": "Reference equation for 7.2.4: 4'[#76]/[N]([H])=C(NO)\\N' + 3O2 + 4OH- > 4'[#76]N([H])C(N)=O' + 4NO + 6H2O. C is a 1- anion and H is a 2- anion."}]}]}
{"problem_id": "Exam-21-T3", "header": "Preparation and Properties of a Thermochromic Copper(II) Complex\nMaterials whose compounds exhibit thermochromism through structural phase transitions are widely used in anti-counterfeiting applications. To synthesize the thermochromic complex $[Cu(deen)_2](ClO_4)_2$ (deen denotes N,N-diethylethylenediamine), hydrated copper perchlorate must first be prepared, followed by introduction of the deen ligand.", "header_images": [], "subquestions": [{"sub_id": "Exam-21-T3-1", "tags": ["1.4", "4.3"], "text_open": "Hydrated copper perchlorate can be prepared by a precipitation–dissolution method. Dissolve $CuSO_4\\cdot 5H_2O$ in water, add $NaOH$ solution with stirring, and heat slowly on a hot plate until the precipitate turns black. Maintain gentle boiling for 5–10 min, allow the mixture to cool slightly, and collect the precipitate by suction filtration. Rinse it with a small amount of deionized water, then use a glass rod to transfer the precipitate into an evaporating dish containing 10 mL of water. Add perchloric acid, gently guide the precipitate into the solution with the glass rod, and stir until it dissolves completely to form a copper perchlorate solution. Concentrate the solution by gentle heating to about 2 mL. On cooling, crystals of hydrated copper perchlorate are obtained. The preparation of $[Cu(deen)_2](ClO_4)_2$ begins with the hydrated copper perchlorate crystals obtained above. Adding absolute ethanol to these crystals gives a clear blue solution. In a dry beaker, dissolve N,N-diethylethylenediamine in ethanol to obtain a clear pale-yellow solution. Slowly add the ethanolic deen solution to the ethanolic copper perchlorate solution with stirring; the mixture immediately turns deep blue-violet. Stir for 2–3min, then allow it to stand for about 10 min. A large amount of pink precipitate forms at the bottom of the vessel while the supernatant remains deep blue-violet. Collect the product by suction filtration and dry it to obtain $[Cu(deen)_2](ClO_4)_2$. 3.2.2 Dissolving the red $[Cu(deen)_2](ClO_4)_2$ in water again gives a deep-blue solution. Explain why. (No more than 50 words)", "text_images": [], "full_mark": 4, "rubric": ["$H_2O$ coordinates at the axial sites of $[Cu(deen)_2]^{2+}$, changing the coordination geometry of Cu(II) from square planar toward elongated octahedral. (2 points)", "The energy of the associated d–d transition decreases and the absorption undergoes a red shift, causing the complementary color to shift toward blue and giving a blue appearance. (2 points)"], "rubric_images": []}, {"sub_id": "Exam-21-T3-2", "tags": ["1.4"], "text_open": "3.3 In a thermochromism experiment on $[Cu(deen)_2](ClO_4)_2$, a small amount of the $[Cu(deen)_2](ClO_4)_2$ complex is held between pieces of adhesive tape and heated with a hair dryer. The sample changes from pink to blue-violet and returns to pink on cooling.\n3.3.1 Based on crystal field theory, describe in words as necessary the structural change of the $[Cu(deen)_2]^{2+}$ unit upon heating and cooling. The cooled unit may be approximated as centrosymmetric, whereas after heating it is no longer centrosymmetric.", "text_images": [], "full_mark": 3, "rubric": ["Before heating, the coordination geometry is square planar (1 point); in the high-temperature phase, the $CuN_4$ unit undergoes a slight tetrahedral distortion out of the plane. (2 points)", "The answer must show that the two ligands are not coplanar after heating. It need not be drawn as an ideal tetrahedron, although an ideal tetrahedral depiction is not penalized."], "rubric_images": [{"file": "Exam-21-T3-2-1-rubric.png", "description": "Rubric diagram for 3.3.1 showing the structures of the $[Cu(deen)_2]^{2+}$ unit after heating and cooling. Both the left and right structures contain a central Cu coordinated by four N atoms. Two N atoms are $-NH_2$ termini (each labeled $H_2$ next to N, at upper left and lower right), and two are $-NEt_2$ termini (each bearing two ethyl zigzag chains). The structure to the left of the reversible arrow is the preheating structure: all four Cu coordination bonds lie in one plane and Cu has square-planar geometry. In the heated structure on the right, the two bonds to the right of Cu are wedge bonds pointing behind and in front of the plane (a dashed wedge above and a solid bold wedge below), while the other two remain ordinary solid lines, indicating a slight tetrahedral distortion of four-coordinate Cu. A reversible arrow connects the structures, labeled \"Heating\" above and \"Cooling\" below. No charge is shown in the diagram."}]}, {"sub_id": "Exam-21-T3-3", "tags": ["4.6"], "text_open": "3.4 To investigate the thermochromism of $[Cu(deen)_2]X_2$ complexes formed with monovalent counteranion $X^-$, differential scanning calorimetry is performed on $[Cu(deen)_2](ClO_4)_2$ and $[Cu(deen)_2](NO_3)_2$. Multiple interactions occur between the cations and anions in both complexes. The results are shown below. For both complexes, the color change is associated with a phase-transition temperature, and the chemical composition remains unchanged across the color change. The temperature at the peak maximum is defined as the transition temperature.\n3.4.1 From the graph, directly give the thermochromic transition temperatures of $[Cu(deen)_2](ClO_4)_2$ and $[Cu(deen)_2](NO_3)_2$.", "text_images": [{"file": "Exam-21-T3-3-1-prompt.png", "description": "Untitled coordinate plot. The horizontal axis is labeled \"Temperature/°C\" with major ticks at 50, 100, 150, and 200 and minor ticks at 25, 75, 125, and 175. The vertical axis is labeled \"Thermogravimetry\" vertically and has no numerical ticks. Two solid black curves are shown. The legend at lower left uses short line segments for \"$ClO_4$(a)\" and \"$NO_3$(b),\" and the curves are labeled (a) and (b). Curve (a), corresponding to $[Cu(deen)_2](ClO_4)_2$, has a lower baseline and a sharp upward peak at about 45–50°C, remains flat thereafter, and drops steeply after about 200°C. Curve (b), corresponding to $[Cu(deen)_2](NO_3)_2$, has a higher baseline, a broad upward peak at about 110°C, a sharp upward peak at about 145°C, another smaller upward peak at about 180°C, and a steep drop after about 200°C. No arrows are shown."}], "full_mark": 4, "rubric": ["$[Cu(deen)_2](ClO_4)_2$ : 45 ℃ (any value from 40 ~ 50 ℃ is accepted) (2 points). No credit is awarded if any other peak is identified as the transition peak.", "$[Cu(deen)_2](NO_3)_2$: 145 ℃ (any value from 135–150 ℃ is accepted) (2 points)"], "rubric_images": []}, {"sub_id": "Exam-21-T3-4", "tags": ["1.4", "4.6"], "text_open": "3.4.2 Using the information above, explain the origin of the difference between the transition temperatures of $[Cu(deen)_2](ClO_4)_2$ and $[Cu(deen)_2](NO_3)_2$. (No more than 100 words)", "text_images": [], "full_mark": 4, "rubric": ["① Difference in coordination ability: $NO_3^-$ coordinates more strongly than $ClO_4^-$ and may partially coordinate to Cu(II), so structural distortion of $[Cu(deen)_2](NO_3)_2$ requires more energy and occurs at a higher transition temperature. (2 points)", "② The hydrogen-bond network formed by $NO_3^-$ and ligand N-H groups is stronger and more ordered than that in the $ClO_4^-$ system. This makes the low-temperature structure of $[Cu(deen)_2](NO_3)_2$ more stable, or raises the energy required for the phase transition, resulting in a higher transition temperature. (2 points)", "Anion size (steric effects) and lattice energy may also matter, but they were not addressed above. Answers based on these factors—for example, that bulky $ClO_4^-$ gives a loosely packed lattice and sufficient free space for configurational distortion of $[Cu(deen)_2]^{2+}$, whereas dense packing of $NO_3^-$ suppresses such distortion—may receive 1–2 points at the grader's discretion."], "rubric_images": []}]}
{"problem_id": "Exam-21-T4", "header": "Determination of Manganese by Ammonium Nitrate Oxidation Titrimetry\nManganese is an important element in steels and alloys, and its content directly affects their strength, hardness, and wear resistance. Ammonium nitrate oxidation titrimetry is a classical method for determining manganese: ammonium nitrate oxidizes Mn(II) to Mn(III) in a phosphoric acid medium, and the product is then titrated with a standard ammonium iron(II) sulfate solution. This problem addresses standardization of the titrant, control of the oxidation conditions, and calculation of the result.", "header_images": [], "subquestions": [{"sub_id": "Exam-21-T4-1", "tags": ["1.4"], "text_open": "4.2 Sample treatment and titration\nWeigh 0.2026 g of sample into a 300 mL conical flask. Add 15 mL of concentrated phosphoric acid and 5 mL of concentrated nitric acid, and heat until the sample dissolves completely. Continue heating and evaporating until the surface becomes still and faint fumes just begin to appear (about 220–240 °C), then remove the flask from the heat. Immediately add 3.0 g ± 0.2 g of ammonium nitrate. While swirling the flask, use a rubber-bulb pipette to blow tangentially across the mouth of the flask to remove nitrogen oxides until the yellow fumes disappear completely. Allow the mixture to cool slightly, add 60 mL of distilled water, mix thoroughly, and cool to room temperature. Titrate with the standard ammonium iron(II) sulfate solution until the solution is pale red. Add 2 drops of N-phenylanthranilic acid indicator solution (2 g/L) and continue titrating to a bright-yellow endpoint. The standard ammonium iron(II) sulfate solution consumed is 16.48 mL, and the blank consumes 0.84 mL.\n4.2.2 Give the predominant form of manganese after oxidation by ammonium nitrate. (Hint: the ligand is dihydrogen phosphate.)", "text_images": [], "full_mark": 2, "rubric": ["$Mn(H_2PO_4)_3$ (manganese(III) tris(dihydrogen phosphate), tris(dihydrogenphosphato)manganese(III), manganese dihydrogen phosphate, or another equivalent expression) (2 points)"], "rubric_images": []}, {"sub_id": "Exam-21-T4-2", "tags": ["2.1"], "text_open": "4.2.4 Write the net ionic equation for the titration. In the products, the metal ions are present as free ions rather than complexes.", "text_images": [], "full_mark": 2, "rubric": ["$Mn(H_2PO_4)_3 + Fe^{2+} + 3H^+ = Mn^{2+} + Fe^{3+} + 3H_3PO_4$ (2 points)"], "rubric_images": []}]}
{"problem_id": "Exam-21-T6", "header": "Pharmacokinetics and Molecular Design (II)\nPharmacokinetics quantitatively studies the absorption, distribution, metabolism, and excretion of drugs in living organisms and uses mathematical principles and methods to describe how plasma drug concentration changes with time. Chemical-kinetic modeling provides a systematic understanding of drug metabolism in vivo. This problem concerns the oxidative metabolism of diclofenac in the body.", "header_images": [], "subquestions": [{"sub_id": "Exam-21-T6-1", "tags": ["3.7"], "text_open": "The Michaelis equation is $r=r_{max}[S]/(K_m+[S])$.\nExperiments show that $[S]$ and $r$ for the 4′-hydroxylation of diclofenac catalyzed by human liver microsomes obey the following relationship:\n| Group | $[S]_1$ / $\\mu\\mathrm{M}$ | $r_1$ / $(\\mathrm{pmol \\cdot min^{-1} \\cdot mg^{-1}})$ | $[S]_2$ / $\\mu\\mathrm{M}$ | $r_2$ / $(\\mathrm{pmol \\cdot min^{-1} \\cdot mg^{-1}})$ |\n| --- | --- | --- | --- | --- |\n| Low concentration | 0.90 | 39.27 | 1.80 | 72.00 |\n| High concentration | 90.0 | 392.73 | 180.0 | 411.43 |\n\n6.2.4 Under the same conditions, the 5-hydroxylation has $K_m = 43\\ \\mu\\mathrm{M}$ and $r_{max} = 15.4\\ \\mathrm{pmol \\cdot min^{-1} \\cdot mg^{-1}}$. Intrinsic clearance $CL_{int}$ is defined by $CL_{int} = r_{max}/K_m$. Calculate $CL_{int}$ for the 4′- and 5-hydroxylation pathways and briefly explain the kinetic meaning of $CL_{int}$. Hint: the drug concentration in the liver is much lower than $K_m$.", "text_images": [], "full_mark": 5, "rubric": ["From the given Michaelis equation, $r = r_{max}[S]/(K_m + [S])$, which can be rearranged to $[S]/r = (1/r_{max})[S] + K_m/r_{max}$. Let $x = [S]$ and $y = [S]/r$; then $y = ax + b$, where $a = 1/r_{max}$ and $b = K_m/r_{max}$. (1 point)", "Using the experimental data provided:\n| x | 0.9 | 1.8 | 90 | 180 |\n| y | 0.02292 | 0.02500 | 0.2292 | 0.4375 |\nLinear regression gives $y = 0.00231480x + 0.0208339$, with $R^2 \\approx 1.000$. Therefore, $r_{max} = 1/a = 432\\ \\mathrm{pmol \\cdot min^{-1} \\cdot mg^{-1}}$ and $K_m = b/a = 9.00\\ \\mu\\mathrm{M}$. (1 point)", "$CL_{int}(4′) = 432/9.00 = 48.0\\ \\mu\\mathrm{L \\cdot min^{-1} \\cdot mg^{-1}}$ (1 point); $CL_{int}(5) = 15.4/43 = 0.358\\ \\mu\\mathrm{L \\cdot min^{-1} \\cdot mg^{-1}}$ (1 point).", "Kinetic meaning: when the drug concentration $[S]$ is extremely low, $[S]$ in the denominator of the Michaelis equation can be neglected, giving $r \\approx \\left(\\frac{r_{max}}{K_m}\\right) \\cdot [S]$. Under these conditions, $CL_{int}$ represents the first-order rate constant. (1 point; describing it as the clearance capacity per unit amount of enzyme at low concentration is also accepted.)"], "rubric_images": []}, {"sub_id": "Exam-21-T6-2", "tags": ["2.9", "3.7"], "text_open": "6.2.5 The kinetic parameters for the individual steps in the pathways above are shown below:\n| Pathway | Substrate → product | Enzyme | $CL_{int}$ / $(\\mu\\mathrm{L}\\cdot\\mathrm{min}^{-1}\\cdot\\mathrm{mg}^{-1})$ |\n| --- | --- | --- | --- |\n| 4′-hydroxylation pathway | DF → 4′-OH-DF | CYP2C9 | / |\n| 4′-hydroxylation pathway | 4′-OH-DF → A | CYP2C9 | 358 |\n| 5-hydroxylation pathway | DF → 5-OH-DF | CYP3A4 | / |\n| 5-hydroxylation pathway | 5-OH-DF → B | CYP2C9 | 103 |\n\nThe researchers also compared detoxification of the quinone products from the two hydroxylation pathways. The measured curves are shown below, where \"conjugate\" denotes a detoxification product.\nExplain why both steps of the 5-hydroxylation pathway have lower $CL_{int}$ values than those of the 4′-hydroxylation pathway, yet the 5-hydroxylation pathway is more toxic. (No more than 100 words)", "text_images": [{"file": "Exam-21-T6-2-1-prompt.png", "description": "Coordinate plot containing two side-by-side panels labeled A and B at upper left. Both horizontal axes are labeled 'GSH concentration (mM)' with ticks at 0, 2, 4, and 6; both vertical axes are labeled 'Conjugate formation (%)' with ticks at 0, 50, and 100. Each panel contains one solid black curve with filled circular data markers, some with vertical error bars. In panel A, the curve rises rapidly from near the origin as GSH concentration increases, approaches 100% at about 2 mM, and remains on a plateau. In panel B, the curve rises more gradually from near the origin and approaches 100% at about 5–6 mM, without a clear plateau. There is no legend or title."}], "full_mark": 2, "rubric": ["① In the 5-hydroxylation pathway, once 5-OH-DF is formed it is more readily converted to toxic B. In contrast, the first step of the 4′ pathway is relatively faster, so 4′-OH-DF is more likely to accumulate and can enter detoxification pathways such as conjugative metabolism. (1 point)", "② The graph shows that quinone A forms abundant adduct at a lower GSH concentration, indicating that it is captured more readily by GSH. Quinone B requires a higher GSH concentration to reach a similar level and is therefore more likely to escape GSH detoxification and bind covalently to cellular proteins. (1 point) Thus, although the 5′ pathway has lower metabolic flux, it is more toxic."], "rubric_images": []}, {"sub_id": "Exam-21-T6-3", "tags": ["1.7", "2.9"], "text_open": "6.3 Diclofenac has a half-life of about 1.0h in humans and is cleared mainly through CYP2C9-catalyzed 4′-hydroxylation of an aromatic ring. The structurally similar fenclorac is more resistant to aromatic hydroxylation and has a half-life exceeding 20h. Its structure is shown below.\nFrom a structural perspective, explain why fenclorac has a longer duration of action. (No more than 50 words)", "text_images": [{"file": "Exam-21-T6-3-1-prompt.png", "description": "The SMILES representation of fenclorac is 'OC(CC1=C(OC2=CC=C(Cl)C=C2Cl)C=CC=C1)=O'."}], "full_mark": 2, "rubric": ["In fenclorac, the principal 4′-hydroxylation site of diclofenac is occupied by a para Cl substituent, so there is no 4′-H available for hydroxylation. (2 points) Its oxidative metabolic clearance is therefore lower, giving a longer half-life and duration of action.", "An answer also noting that the bridging atom in fenclorac is O, so formation of a quinonoid structure by oxidation cannot eliminate the formal charge, whereas diclofenac has a bridging N atom that can form an imine upon oxidation, receives 1 point."], "rubric_images": []}]}
{"problem_id": "Exam-22-T1", "header": "Problem 1 Chemistry of the Rare-Earth Elements\nRare-earth elements are indispensable strategic resources for modern high technology, new energy, advanced materials, defense, and related fields. Often called the “vitamins of industry” and a “treasure house of new materials,” they comprise scandium, yttrium, and all the lanthanides. For convenience, the symbol Ln is commonly used in coordination and inorganic chemistry to denote the lanthanides or the entire rare-earth series. This problem examines rare-earth chemistry in detail.\nNote: Different letters or alphanumeric combinations in this problem may represent the same element or chemical formula.", "header_images": [], "subquestions": [{"sub_id": "Exam-22-T1-1", "tags": ["1.3"], "text_open": "1.2 Consider a rare-earth element M1 as an example of rare-earth chemistry. M1 has two common oxidation states in its compounds. Elemental M1 is a silvery-white metal that slowly tarnishes in air. M1 dissolves in dilute hydrochloric acid, releasing colorless gas A and giving a pale blue-green solution (Reaction 1). Adding dilute sulfuric acid dropwise to this solution immediately produces white precipitate B, which is insoluble in dilute nitric acid. After B is filtered, washed, and dried, it is ignited to constant mass at 800 °C under Ar; the white precipitate becomes dark-brown powder C (Reaction 2), an oxide of M1. If the mass of B is 2.480 g, 1.760 g of C is obtained after ignition. A separate portion of B is heated with dilute hydrochloric acid and bromine water until it dissolves (Reaction 3), during which the solution changes from pale blue-green to colorless (or very pale pink). The solution is divided into two portions. Excess aqueous ammonia and oxalic acid are added separately to give white gelatinous precipitate D (with aqueous ammonia) and white precipitate E (with oxalic acid), respectively. Filtering and igniting either D or E at 900 °C likewise gives C. C is mixed with Y2O3 in a certain ratio and calcined at high temperature; product F emits bright red fluorescence under ultraviolet irradiation. 1.2.3 Write ionic equations for Reactions 1 and 3 and a chemical equation for Reaction 2.", "text_images": [], "full_mark": 6, "rubric": ["Reaction 1: $Eu + 2H^+ \\rightarrow Eu^{2+} + H_2$ (2 points)\nReaction 2: $2EuSO_4 \\rightarrow Eu_2O_3 + SO_2 + SO_3$ (2 points)\nReaction 3: $2EuSO_4 + Br_2 \\rightarrow 2Eu^{3+} + 2Br^- + 2SO_4^{2-}$ (2 points)"], "rubric_images": []}, {"sub_id": "Exam-22-T1-2", "tags": ["1.4"], "text_open": "1.4 Coordination Compounds of Rare-Earth Elements\n1 equivalent of elemental rare-earth metal M5 and 2.8 equivalents of ethylene glycol monomethyl ether are added together to a toluene solution to form X. Addition of 2,6-dimethylphenol then precipitates orange crystals Y. Y is filtered and washed; upon slight heating it becomes a different crystal Z. Further characterization provides the following information:\n① The effective magnetic moment of X is 7.9 $\\mu_B$.\n② Crystal diffraction shows that Y belongs to the orthorhombic crystal system, with $a$ = 12.748 Å, $b$ = 12.900 Å, and $c$ = 29.220 Å. The unit cell contains 2 formula units, and the theoretical density of Y is 1.428 $\\mathrm{g/cm^3}$.\n③ Elemental analysis of Z gives the following experimental mass fractions: $\\omega(C)$ = 39.60%, $\\omega(O)$ = 18.86%, and $\\omega(H)$ = 5.74%. Crystal diffraction shows that Z belongs to the tetragonal crystal system, with $a$ = 14.837 Å and $c$ = 15.117 Å. The crystal contains 2 formula units, and the density of Z is 1.1854 times the theoretical density of Y.\n④ The anion of Z contains an $[M5_4O_4]$ cluster framework and exactly two types of ligands. One ligand adopts two different coordination modes. All M5 atoms are 7-coordinate, and a particular conformation has 1 twofold rotation axis.\n⑤ The oxidation state of M5 is the same in X, Y, and Z.\n1.4.2 Determine the chemical formulas of X, Y, and Z, showing your calculations. In your answer, the ethylene glycol monomethyl ether anion may be abbreviated as $mee^-$, and the 2,6-dimethylphenoxide anion as $dmp^-$.\n", "text_images": [], "full_mark": 9, "rubric": ["Y is orthorhombic, with unit-cell volume $V_Y = 12.748 \\times 12.900 \\times 29.220 = 4805.2\\ \\mathrm{Å}^3$. From the density equation $\\rho_Y = ZM / (N_AV)$, with $Z = 2$ for both Y and Z:\nFirst calculate $M_Y$: $\\frac{M_Z}{M_Y} = \\frac{\\rho_Z V_Z}{\\rho_Y V_Y} = \\frac{1.1854\\rho_Y V_Z}{\\rho_Y V_Y} = 1.1854 \\times \\frac{3327.80}{4805.2} = 0.8209$. (1 point for determining the molar mass of Y)\n$M_Y = \\frac{1.428 \\times 4805.2 \\times 10^{-24} \\times 6.022 \\times 10^{23}}{2} = 2066\\ \\mathrm{g/mol}$.\nThus, $M_Y - M_Z$ = 369 g/mol, approximately the mass of 4 toluene molecules, so Y is a toluene solvate of Z. (1 point)\nFor Z, $\\omega(C)$ = 39.63% and $\\omega(O)$ = 18.85%. Taking $M_Z$ = 1697 g/mol:\nNumber of C atoms: $n(C) = (1697 \\times 0.3960) / 12.01 \\approx 56$\nNumber of O atoms: $n(O) = (1697 \\times 0.1886) / 16.00 \\approx 20$\nNumber of H atoms: $n(H) = (1697 \\times 0.0574) / 1.008 = 96.6$\n(1 point for determining the numbers of C, O, and H atoms in Z; integers or decimals are accepted.)\nThe hydrogen mass fraction from elemental analysis often has a relatively large error. The system contains only two types of ligands: ① the ethylene glycol monomethyl ether anion, $mee^-$: $C_3H_7O_2^-$, containing 3 C and 2 O atoms; ② the 2,6-dimethylphenoxide anion, $dmp^-$: $C_8H_9O^-$, containing 8 C and 1 O atom.\nLet the numbers of $dmp^-$ and $mee^-$ ligands be x and y, respectively. Then $\\begin{cases} 8x + 3y = 56 \\\\ x + 2y = 20 \\end{cases}$, giving x = 4 and y = 8. Thus, Z contains 4 $dmp^-$ and 8 $mee^-$ ligands. (1 point for determining the numbers of ligands in Z; it is not necessary to distinguish dmp from dmpH or mee from meeH.)\nLet Z contain z metal atoms of atomic weight $M_{M5}$. From mass conservation:\n$z \\cdot M_{M5} = M_Z - 4M_{dmp} - 8M_{mee} - 4M_H = 1697 - 4 \\times 121.17 - 8 \\times 75.09 - 4 \\times 1.008 \\approx 608$ (g/mol).\nTry z = 4: $M_{M5}$ = 608/4 = 152. Atomic weight 152 corresponds to europium (Eu). In fact, the statement that the anion of Z contains an $[M5_4O_4]$ cluster framework establishes that z ≥ 4.\nMagnetic-moment verification: $\\mu_{eff}$ = 7.9 $\\mu_B$, in close agreement with the theoretical value of 7.94 $\\mu_B$ for $Eu^{2+}$ ($4f^7$); experimental values are generally 7.6~8.0 $\\mu_B$. Therefore, M5 = Eu in the +2 oxidation state. (2 points for identifying M5. Any reasonable procedure that identifies M5 earns at least 6 points for the entire subquestion.)\nZ = $H_4[Eu_4(mee)_8(dmp)_4]$ (1 point)\nY = $H_4[Eu_4(mee)_8(dmp)_4]\\cdot 4C_7H_8$ (1 point)\nX = $[Eu(mee)_2]_n$ (1 point; $Eu(mee)_2$ is also accepted)"], "rubric_images": [{"file": "Exam-22-T1-2-1-rubric.png", "description": "Equation: $\\frac{M_Z}{M_Y} = \\frac{\\rho_Z V_Z}{\\rho_Y V_Y} = 1.1854 \\times \\frac{3327.80}{4805.2} = 0.8209$."}, {"file": "Exam-22-T1-2-2-rubric.png", "description": "Equation: $M_Y = \\frac{1.428 \\times 4805.2 \\times 10^{-24} \\times 6.022 \\times 10^{23}}{2} = 2066\\ \\mathrm{g/mol}$."}, {"file": "Exam-22-T1-2-3-rubric.png", "description": "Equation: $\\begin{cases} 8x + 3y = 56 \\\\ x + 2y = 20 \\end{cases}$."}]}, {"sub_id": "Exam-22-T1-3", "tags": ["1.4"], "text_open": "1.4.3 Write the SMILES strings for the cation and anion of Z. You may treat all $\\sigma$ coordinate bonds as ordinary single bonds and represent the structures using SMILES.", "text_images": [], "full_mark": 4, "rubric": ["Structure: the cation consists of 4 hydrogen ions, $(H^+)$; the anion has SMILES 'CO1CCO[Eu]1234([dmp])O5(CCO6C)[Eu]789(OCCO8C)([dmp])O2(CCO9C)[Eu]%10%11%12(O(C)CCO%12)([dmp])O3(CCO4C)[Eu]5%136(O(C)CCO%13)([dmp])O%107CCO%11C', where dmp denotes a dmp ligand, and the anion carries a $4-$ charge. The anion has an approximately cubic [Eu_4O_4] core. All dmp ligands are terminal. The mee ligands fall into two classes: 4 each provide 1 O face-bridging group and 1 linear bridging group, with the terminal O bearing a methyl group, whereas the other 4 coordinate only through 2 terminal O atoms.", "This subquestion uses deductions. Deduct 1 point if, and only if, some mee ligands in the anion have the wrong “direction,” with the methyl-bearing O drawn as the face-bridging group and the non-methyl-bearing O as the linear bridge. Any other error in the anion structure loses all points assigned to the anion. Deduct 1 point if the cations are absent or incorrect, and 1 point if the charge of the anion is not shown. The minimum score is 0."], "rubric_images": [{"file": "Exam-22-T1-3-1-rubric.png", "description": "Scoring diagram. $(H^+)_4$ is shown on the left, and the anion is shown on the right. The anion has SMILES 'CO1CCO[Eu]1234([dmp])O5(CCO6C)[Eu]789(OCCO8C)([dmp])O2(CCO9C)[Eu]%10%11%12(O(C)CCO%12)([dmp])O3(CCO4C)[Eu]5%136(O(C)CCO%13)([dmp])O%107CCO%11C' and an approximately cubic [Eu_4O_4] core. A $4-$ charge is written at the upper right of the anion. The 4 dmp ligands are represented by the label 'dmp', with their structures omitted."}]}, {"sub_id": "Exam-22-T1-4", "tags": ["2.1"], "text_open": "1.5 Electrochemistry of the Rare-Earth Elements\nThe following standard electrode potentials are given for acidic medium at 25 °C:\n$O_2(g) + 4H^+ + 4e^- \\rightarrow 2H_2O$    $E° = +1.229\\ \\mathrm{V}$\n$M1^{n+} + e^- \\rightarrow M1^{(n-1)+}$    $E° = -0.35\\ \\mathrm{V}$\n1.5.1 At pH = 0 and 25 °C, calculate the common logarithm $\\lg K$ of the standard equilibrium constant for oxidation of $M1^{(n-1)+}$ to $M1^{n+}$ by $O_2$.", "text_images": [], "full_mark": 3, "rubric": ["Reaction: $4Eu^{2+} + O_2 + 4H^+ \\rightarrow 4Eu^{3+} + 2H_2O$\n$E° = E_+ - E_- = 1.229\\ \\mathrm{V} - (-0.35\\ \\mathrm{V}) = 1.579\\ \\mathrm{V}$ (1 point)\n$\\Delta_r G° = -nFE° = -4 \\times 96485\\ \\mathrm{C\\cdot mol^{-1}} \\times 1.579\\ \\mathrm{V} = -6.095 \\times 10^5\\ \\mathrm{J\\cdot mol^{-1}}$ (1 point)\n$\\ln K = -\\frac{\\Delta_r G°}{RT} = \\frac{6.095 \\times 10^5}{8.314 \\times 298} = 246$\n$\\lg K = \\lg e \\times \\ln K = 106.8$ (1 point). Writing M1 instead of Eu in the procedure and result incurs no deduction."], "rubric_images": [{"file": "Exam-22-T1-4-1-rubric.png", "description": "Equation: $E° = E_+ - E_- = 1.229\\ \\mathrm{V} - (-0.35\\ \\mathrm{V}) = 1.579\\ \\mathrm{V}$."}, {"file": "Exam-22-T1-4-2-rubric.png", "description": "Equation: $\\Delta_r G° = -nFE° = -4 \\times 96485\\ \\mathrm{C\\cdot mol^{-1}} \\times 1.579\\ \\mathrm{V} = -6.095 \\times 10^5\\ \\mathrm{J\\cdot mol^{-1}}$."}, {"file": "Exam-22-T1-4-3-rubric.png", "description": "Equation: $\\ln K = -\\frac{\\Delta_r G°}{RT} = \\frac{6.095 \\times 10^5}{8.314 \\times 298} = 246$."}, {"file": "Exam-22-T1-4-4-rubric.png", "description": "Equation: $\\lg K = \\lg e \\times \\ln K = 106.8$."}]}, {"sub_id": "Exam-22-T1-5", "tags": ["2.1"], "text_open": "1.5.2 The partial pressure of oxygen in air is $p(O_2)$ = 0.21 bar. Use a calculation to determine whether $M1^{(n-1)+}$ will be oxidized by oxygen in air when pH = 4.00 and $[M1^{(n-1)+}]$ = $[M1^{n+}]$ = 1.00 mol·L-1.", "text_images": [], "full_mark": 2, "rubric": ["$E(O_2) = E° + \\frac{RT}{4F}\\ln\\frac{[H^+]^4 p(O_2)}{p°} = 0.982\\ \\mathrm{V}$ (1 point)\nBecause the concentrations are equal, the $Eu^{3+}/Eu^{2+}$ electrode potential is $E = E° = -0.35\\ \\mathrm{V}$. The cell electromotive force is $\\Delta E = 0.982\\ \\mathrm{V} - (-0.35\\ \\mathrm{V}) = 1.33\\ \\mathrm{V} > 0$, and $\\Delta G < 0$. Therefore, under these conditions, $Eu^{2+}$ is still oxidized spontaneously by oxygen in air. (1 point for calculating the correct electromotive force and reaching the correct conclusion.)\nIf the calculated result is incorrect, no point is awarded for the conclusion even if it is correct. Writing M1 instead of Eu in the procedure and result incurs no deduction."], "rubric_images": [{"file": "Exam-22-T1-5-1-rubric.png", "description": "Equation: $E(O_2) = E° + \\frac{RT}{4F}\\ln\\frac{[H^+]^4 p(O_2)}{p°} = 0.982\\ \\mathrm{V}$."}, {"file": "Exam-22-T1-5-2-rubric.png", "description": "Equations: $\\Delta E = 0.982\\ \\mathrm{V} - (-0.35\\ \\mathrm{V}) = 1.33\\ \\mathrm{V} > 0$ and $\\Delta G < 0$."}]}]}
{"problem_id": "Exam-22-T5", "header": "Unlike most covalent compounds, whose structures can be described using 2c-2e bonds, polyhedral clusters are often electron-deficient systems and require multicenter bonding models.\nWade's rules, more generally known as the polyhedral skeletal electron pair theory (PSEPT), are an important tool for understanding cluster structures. Let $n$ denote the number of atoms or isolobal fragments that form the cluster framework. Subtracting the electrons used in terminal two-center two-electron bonds, nonskeletal lone pairs, and the internal bonding of ligands from the total valence-electron count gives the number of skeletal electrons. Skeletal electron-pair counts of $n+1$, $n+2$, $n+3$, and $n+4$ correspond, respectively, to closo, nido, arachno, and hypho structures. A closo cluster is a complete triangulated polyhedron with $n$ vertices, whereas nido, arachno, and hypho clusters may be regarded as the corresponding closo parent polyhedron with one, two, or three vertices removed, respectively. For example, in $B_5H_9$, 5 BH fragments contribute 10 electrons and 4 additional hydrogen atoms contribute another 4, giving 14 skeletal electrons or 7 skeletal electron pairs. Since the skeletal electron-pair count is $n+2$, the cluster is nido.\nThis problem uses low-valent clusters of Group 13 metals (Al, Ga, In, and Tl) to examine cluster synthesis, structure, bonding, and reactivity.", "header_images": [], "subquestions": [{"sub_id": "Exam-22-T5-1", "tags": ["1.6"], "text_open": "5.1 $B_4Cl_4$ and $B_4(t\\text{-}Bu)_4$ have tetrahedral cores, whereas the acetamido-substituted boron compound $B_4(NHAc)_4$ adopts a butterfly-shaped cyclic structure. Explain why.", "text_images": [], "full_mark": 2, "rubric": ["Because of B-N π bonding, the B atoms are relatively electron-rich. (2 points)"], "rubric_images": []}, {"sub_id": "Exam-22-T5-2", "tags": ["2.1"], "text_open": "5.2 Write equations for the following reactions.\n5.2.1 $AlCl\\cdot x\\,Et_2O$ ($x>1$) reacts with $MgCp^*_2$ at low temperature in toluene/diethyl ether to form $Al_4Cp^*_4$.", "text_images": [], "full_mark": 2, "rubric": ["$4\\,AlCl\\cdot x\\,Et_2O + 2\\,MgCp^*_2 = Al_4Cp^*_4 + 2\\,MgCl_2(OEt_2)_2 + (4x-4)\\,Et_2O$ (2 points)"], "rubric_images": []}, {"sub_id": "Exam-22-T5-3", "tags": ["2.1"], "text_open": "5.2.2 $GaI$ reacts with $Li(THF)_3Ge(SiMe_3)_3$ in toluene to form $[GaGe(SiMe_3)_3]_4$.", "text_images": [], "full_mark": 2, "rubric": ["$4\\,GaI + 4\\,Li(THF)_3Ge(SiMe_3)_3 = [GaGe(SiMe_3)_3]_4 + 4\\,LiI(THF)_3$ (2 points)"], "rubric_images": []}, {"sub_id": "Exam-22-T5-4", "tags": ["2.4"], "text_open": "5.4 Tetrahedral clusters not only have unusual structures but also show diverse oxidation chemistry, giving different products under different conditions. Draw the structures of products A, B, D, and E below. 5.4.1 Oxidation of [GaSi(t-Bu)_3]_4 with oxygen gives cubic cluster A = [OGaSi(t-Bu)_3]_4.", "text_images": [], "full_mark": 3, "rubric": ["A is the cubic cluster $[OGaSi(t\\text{-}Bu)_3]_4$. (3 points)"], "rubric_images": [{"file": "Exam-22-T5-4-1-rubric.png", "description": "The reference structure of A has SMILES '[Si]([Ga]12O3[Ga]4(O1[Ga]5([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)O4[Ga]3([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)O52)[Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)(C(C)(C)C)(C(C)(C)C)C(C)(C)C'. It has a $Ga_4O_4$ heterocubane framework in which 4 Ga and 4 O atoms alternately occupy the vertices of a cube, forming 12 Ga-O edges. Each Ga is bonded to one $Si(t\\text{-}Bu)_3$ group, giving $[OGaSi(t\\text{-}Bu)_3]_4$. All bonds are single bonds, and no charges are shown."}]}, {"sub_id": "Exam-22-T5-5", "tags": ["2.4"], "text_open": "5.4.2 Further hydrolysis of A gives product B, which contains only one ring and has only one chemical environment for its Ga atoms. Write the SMILES string for B. If necessary, describe its detailed structure in no more than 50 words. Multiple reasonable structures may be possible; writing any one reasonable SMILES string is sufficient.", "text_images": [], "full_mark": 2, "rubric": ["Draw either of the following structures. First: 'O[Ga]1([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)[OH][Ga](O)([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)[OH][Ga](O)([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)[OH][Ga](O)([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)[OH]1'. Second: '[OH2+][Ga-]1([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)O[Ga-]([OH2+])([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)O[Ga-]([OH2+])([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)O[Ga-]([OH2+])([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)O1'."], "rubric_images": [{"file": "Exam-22-T5-5-1-rubric.png", "description": "One reasonable structure of B, and the most accurate reference answer, has SMILES 'O[Ga]1([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)[OH][Ga](O)([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)[OH][Ga](O)([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)[OH][Ga](O)([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)[OH]1'. It is an eight-membered $Ga_4O_4$ ring in which 4 Ga atoms and 4 $\\mu_2$-OH groups alternate. Each Ga is bonded to one $Si(t\\text{-}Bu)_3$ group and one terminal OH group. All bonds are single bonds, and no charges are shown."}, {"file": "Exam-22-T5-5-2-rubric.png", "description": "Another acceptable structure of B has SMILES '[OH2+][Ga-]1([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)O[Ga-]([OH2+])([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)O[Ga-]([OH2+])([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)O[Ga-]([OH2+])([Si](C(C)(C)C)(C(C)(C)C)C(C)(C)C)O1'. It is an eight-membered $Ga_4O_4$ ring, with the ring oxygen atoms labeled O. Each Ga is bonded to one $Si(t\\text{-}Bu)_3$ group and one water ligand. The SMILES uses charge-separated $[OH2+]$ and $[Ga-]$ forms; the 4 water ligands are labeled OH2/H2O in the figure."}]}, {"sub_id": "Exam-22-T5-6", "tags": ["2.4"], "text_open": "5.4.3 Oxidation of $C = [InC(SiMe_3)_3]_4$ with o-nitrosotoluene gives a cubic cluster and product D, which contains two benzene rings. Write the SMILES string for aromatic product D and, if necessary, describe its detailed structure in no more than 50 words.", "text_images": [], "full_mark": 2, "rubric": ["'CC1=C(/N=N/C2=C(C)C=CC=C2)C=CC=C1'"], "rubric_images": [{"file": "Exam-22-T5-6-1-rubric.png", "description": "The reference structure of D has SMILES 'CC1=C(/N=N/C2=C(C)C=CC=C2)C=CC=C1'. It has an azobenzene framework with an N=N double bond; /N=N/ in the SMILES denotes the trans configuration. Each benzene ring bears one methyl substituent drawn as a short line without a text label, and each methyl group is ortho to the position bonded to nitrogen. All remaining bonds are single bonds, and no charges are shown."}]}, {"sub_id": "Exam-22-T5-7", "tags": ["2.4"], "text_open": "5.4.4 Oxidation of C with 1,2-thiirane gives a trigonal-bipyramidal cluster E. Write the SMILES string for E and, if necessary, describe its detailed structure in no more than 50 words.", "text_images": [], "full_mark": 2, "rubric": ["'C[Si](C([In]12[In]34(C([Si](C)(C)C)([Si](C)(C)C)[Si](C)(C)C)[In]15(C([Si](C)(C)C)([Si](C)(C)C)[Si](C)(C)C)[In]23(C([Si](C)(C)C)([Si](C)(C)C)[Si](C)(C)C)[S]54)([Si](C)(C)C)[Si](C)(C)C)(C)C'"], "rubric_images": [{"file": "Exam-22-T5-7-1-rubric.png", "description": "The reference structure of E has SMILES 'C[Si](C([In]12[In]34(C([Si](C)(C)C)([Si](C)(C)C)[Si](C)(C)C)[In]15(C([Si](C)(C)C)([Si](C)(C)C)[Si](C)(C)C)[In]23(C([Si](C)(C)C)([Si](C)(C)C)[Si](C)(C)C)[S]54)([Si](C)(C)C)[Si](C)(C)C)(C)C'. It has an $In_4$ tetrahedral framework with 6 In-In edges and one S atom bridging a face in a $\\mu_3$ mode. Each In is bonded to one $C(SiMe_3)_3$ group, labeled C(SiMe3)3 in the figure, giving a trigonal-bipyramidal cluster. All bonds are single bonds, and no charges are shown."}]}, {"sub_id": "Exam-22-T5-8", "tags": ["2.2"], "text_open": "5.6 Reaction of $GaI$ with $Li(THF)_3R$ ($R = SiMe(SiMe_3)_2$) in toluene gives the octahedral cluster $G = [GaR]_6$. In contrast, reaction of $GaI$ with $Na(THF)_2R'$ ($R' = Si(t\\text{-}Bu)_3$) in toluene forms the octahedral cluster $H = [Ga_6R'_4Bn_2]^{2-}$, where Bn denotes benzyl.\n5.6.1 State the source of the benzyl groups in H.", "text_images": [], "full_mark": 1, "rubric": ["The solvent toluene is deprotonated to generate benzyl anions. (1 point)"], "rubric_images": []}, {"sub_id": "Exam-22-T5-9", "tags": ["1.7"], "text_open": "5.6.3 Explain why G is a highly distorted octahedron whereas H is nearly a regular octahedron.", "text_images": [], "full_mark": 3, "rubric": ["$[GaR]_6$ has 6 skeletal electron pairs (sep), fewer than the 7 sep required for a closo octahedron (1 point). The electron deficiency removes the degeneracy and gives rise to a Jahn-Teller effect (2 points), ultimately causing severe distortion of the octahedron."], "rubric_images": []}, {"sub_id": "Exam-22-T5-10", "tags": ["1.7"], "text_open": "5.8 Electron-rich cluster compounds: When the number of ligands in a cluster is much smaller than the number of metal atoms, its structure approaches that of a bare metal-atom cluster; such species are called electron-rich cluster compounds. A representative example is the metalloid cluster $K = Ga_{84}[N(SiMe_3)_2]_{20}^{4-}$. This cluster contains 84 Ga atoms, 64 of which are “bare.” Its structure has three layers: a central $Ga_2$ dumbbell with a bond length of 235 pm, approaching a $Ga\\equiv Ga$ triple bond; a middle $Ga_{32}$ football-shaped tube; and an outer assembly of three $Ga_{10}$ rings surrounded by 20 $N(SiMe_3)_2$ ligands.\nCompare K with the fullerene $Sc_2@C_{84}$, and identify which structural units in K correspond to the encapsulated $Sc_2$ and to the $C_{84}$ carbon cage, respectively.", "text_images": [{"file": "Exam-22-T5-10-1-prompt.png", "description": "Unlabeled three-dimensional ball-and-stick molecular model. The overall structure is a vertically symmetric multilayer assembly. The upper and lower layers each contain a polyhedron made of dark-blue spheres, approximately 12 in total. A larger central polyhedron is formed by approximately 12 purple spheres, and the outermost framework contains approximately 24 orange-yellow spheres. Spheres of the same color are connected by solid lines of the same color, with radial connections between layers. The drawing represents the layered structure of the metalloid cluster $K = Ga_{84}[N(SiMe_3)_2]_{20}^{4-}$ described in the question."}], "full_mark": 2, "rubric": ["The central $Ga_2$ unit can be regarded as a “molecule” encapsulated within the $Ga_{32}$ cage, analogous to $Sc_2$ in $Sc_2@C_{84}$. (1 point)", "The $Ga_{32}$ unit is a hollow structure analogous to a fullerene carbon cage. (1 point)"], "rubric_images": []}]}
{"problem_id": "Exam-23-T8", "header": "Problem 8 Norbornane (24 points, 11%)\nBecause of their rigid frameworks and fixed conformations, norbornane and its derivatives have become classic models in physical organic chemistry for studying stereoelectronic effects, neighboring-group participation, and nonclassical carbocations. The structure and carbon numbering are shown below.", "header_images": [{"file": "Exam-23-T8-header-1.png", "description": "Skeletal structure of norbornane (bicyclo[2.2.1]heptane), with the carbon atoms labeled 1, 2, 3, 4, 5, 6, and 7. Its SMILES is '[C@@H]12CC[C@H](C2)CC1'."}], "subquestions": [{"sub_id": "Exam-23-T8-1", "tags": ["1.6", "2.4"], "text_open": "8.2 A substituent X can be introduced at C2 of norbornane, giving the exo-X and endo-X diastereomers.\n8.2.1 When the substituent is hydroxy, which is more stable, the exo or endo configuration? Explain.", "text_images": [{"file": "Exam-23-T8-1-1-prompt.png", "description": "Two norbornane frameworks with substituent X attached to a ring carbon. The left structure is labeled 'exo' below; X extends up-right by a single bond, and the halogen is equatorial. Its SMILES is '[F,Cl,Br,I][C@H]1[C@@H](C2)CC[C@@H]2C1'. The right structure is labeled 'endo'; X extends vertically downward, and the halogen is axial. Its SMILES is '[F,Cl,Br,I][C@@H]1[C@@H](C2)CC[C@@H]2C1'. Here [F,Cl,Br,I] is the ChemDraw export representation of substituent X as a list of halogen atoms."}], "full_mark": 3, "rubric": ["The exo configuration is more stable (1 point).", "In the endo configuration, the C–O bond and C6-endo-H have a 1,3-diaxial spatial relationship and repel each other (1 point).", "In the exo configuration, the C1–C6 bond can hyperconjugate with the C–O bond (1 point).", "Award 1 point for the steric difference and 1 point for the hyperconjugative effect. If the stability order is incorrect, award no explanation points."], "rubric_images": []}, {"sub_id": "Exam-23-T8-2", "tags": ["1.6", "2.4"], "text_open": "8.2.3 Compare the acidities of exo-2-norbornanecarboxylic acid and endo-2-norbornanecarboxylic acid in aqueous solution, and explain the difference.", "text_images": [], "full_mark": 3, "rubric": ["exo-2-Norbornanecarboxylic acid is more acidic (1 point).", "The endo carboxylate anion lies on the concave face of the molecule, where the surrounding framework hinders the approach of water. Poor solvation destabilizes the anion, so its conjugate acid is weaker (2 points).", "Award 1 point for identifying steric hindrance and 1 point for stating that the exo carboxylate is well solvated by water and its negative charge is effectively stabilized. If the acidity order is incorrect, award no explanation points."], "rubric_images": []}, {"sub_id": "Exam-23-T8-3", "tags": ["2.4", "2.3"], "text_open": "8.2.4 Write the SMILES string for the major product of the following reaction.", "text_images": [{"file": "Exam-23-T8-3-1-prompt.png", "description": "Two-step reaction. The substrate is a norbornane derivative with SMILES 'Br[C@@H]1[C@@H](C2)CC[C@@H]2[C@@H]1[2H]': bromine is attached to one ring carbon and deuterium [2H] to the adjacent ring carbon. Two successive rightward arrows follow the substrate; the first is labeled 'LDA' and the second 'm-CPBA'. No structure is drawn after either arrow."}], "full_mark": 2, "rubric": ["Major product: (2 points)"], "rubric_images": [{"file": "Exam-23-T8-3-1-rubric.png", "description": "Major product: an oxygen atom bridges the two adjacent ring carbons that originally bore Br and D in the norbornane framework, forming a three-membered epoxide ring. The oxygen is labeled 'O'. Its SMILES is 'C1C[C@H]2C[C@@H]1[C@@H]3O[C@H]23'."}]}, {"sub_id": "Exam-23-T8-4", "tags": ["2.3", "2.4"], "text_open": "8.3 The distinctive framework gives norbornane and norbornene derivatives exceptionally rich reactivity.\n8.3.1 The two compounds below give the same product upon solvolysis in acetic acid. Identify which compound solvolyzes faster and write the SMILES string for the product.", "text_images": [{"file": "Exam-23-T8-4-1-prompt.png", "description": "Two norbornane frameworks labeled A (left) and B (right). In A, OTs (p-toluenesulfonate) extends vertically downward by a single bond; its SMILES is 'CC1=CC=C(S(O[C@@H]2[C@@H](C3)CC[C@@H]3C2)(=O)=O)C=C1'. In B, OTs extends diagonally up-right; its SMILES is 'CC1=CC=C(S(O[C@H]2[C@@H](C3)CC[C@@H]3C2)(=O)=O)C=C1'. The OTs group occupies the same position in both structures but has a different stereochemical orientation."}], "full_mark": 3, "rubric": ["B (1 point)", "Product structure: (2 points; award only 1 point if the product is not identified as racemic; an endo product receives no credit)"], "rubric_images": [{"file": "Exam-23-T8-4-1-rubric.png", "description": "Product: a norbornane ring carbon bears OAc (acetoxy), with '(±)' at lower right indicating a racemate. Its SMILES is 'O=C(O[C@H]1[C@@H](C2)CC[C@@H]2C1)C'."}]}, {"sub_id": "Exam-23-T8-5", "tags": ["2.3", "2.4"], "text_open": "8.3.2 Which of the following compounds solvolyzes faster in acetic acid? Explain using an important intermediate.", "text_images": [{"file": "Exam-23-T8-5-1-prompt.png", "description": "Two norbornene frameworks labeled A (left) and B (right). Each ring contains one C=C bond, and p-toluenesulfonate is attached to bridge carbon C7. In A, TsO extends down-left by a single bond; its SMILES is 'CC1=CC=C(S(O[C@H]2[C@H]3CC[C@@H]2C=C3)(=O)=O)C=C1'. In B, OTs extends up-right; its SMILES is 'CC1=CC=C(S(O[C@@H]2[C@H]3CC[C@@H]2C=C3)(=O)=O)C=C1'. The structures otherwise differ only in the orientation of the sulfonate group."}], "full_mark": 2, "rubric": ["A (1 point)", "Key intermediate: (1 point; its enantiomer also receives credit)"], "rubric_images": [{"file": "Exam-23-T8-5-1-rubric.png", "description": "Key intermediate (first image, red): its SMILES can be written as '[C@H]12CC[C@@H]([CH+]2)C=C1'. The double bond conjugatively stabilizes the carbocation at the bridgehead carbon, as indicated by dashed lines."}, {"file": "Exam-23-T8-5-2-rubric.png", "description": "Key intermediate (second image, blue): its SMILES can be written as '[C@H]12CC[C@@H]([C@@H]23)C3[CH+]1'."}]}, {"sub_id": "Exam-23-T8-6", "tags": ["2.3", "2.4"], "text_open": "8.3.3 In the reaction below, an optically active substrate racemizes under the action of $KO^tBu$. Write the SMILES string for a key intermediate that explains the racemization.", "text_images": [{"file": "Exam-23-T8-6-1-prompt.png", "description": "One-step reaction. The substrate is a norbornanone derivative whose carbonyl oxygen is labeled 'O'; the ring carbon adjacent to the carbonyl carbon bears two methyl groups, shown as two short single bonds. Its SMILES is 'CC1(C)[C@H]2CC[C@H](C2)C1=O'. The arrow is labeled 'KOtBu, HOtBu' above (t is superscript) and '250°C' below. The product at right has the same framework as the substrate and is marked '(±)' at lower right to indicate a racemate. Thus, the substrate is chiral and the product is racemic."}], "full_mark": 2, "rubric": ["Key intermediate: (2 points)"], "rubric_images": [{"file": "Exam-23-T8-6-1-rubric.png", "description": "Key intermediate: in the norbornanone framework, the former carbonyl carbon is bonded to a negatively charged oxygen, labeled 'O' with a circled minus sign. That carbon also has an additional single bond to another ring carbon (C2), and C5 retains two methyl groups. Its SMILES is 'CC1(C)C2CC3C(C2)C13[O-]'."}]}, {"sub_id": "Exam-23-T8-7", "tags": ["2.4", "1.6"], "text_open": "8.3.4 Based on the structure in 8.3.3, predict which hydrogen in the substrate is the most acidic. Give a reasonable explanation.\n(a) $H_a$\n(b) $H_b$\n(c) $H_c$\n(d) $H_d$\n(e) $H_e$", "text_images": [{"file": "Exam-23-T8-7-1-prompt.png", "description": "The same substrate as in 8.3.3: the carbonyl oxygen is labeled 'O', and the ring carbon adjacent to the carbonyl carbon bears two methyl groups. Five hydrogen positions are additionally labeled: $H_a$ and $H_b$ are attached to the same ring carbon (C2), $H_c$ to bridgehead carbon C4, and $H_d$ and $H_e$ to bridge carbon C7. The hydrogen-labeled SMILES is 'CC1(C)[C@@]2([Hc])C[C@]([Ha])([Hb])[C@H]([C@@]2([He])[Hd])C1=O', where [Ha] through [He] are placeholders for the image labels $H_a$ through $H_e$."}], "full_mark": 4, "rubric": ["(a) (2 points)", "The bonding orbital of the C2–H bond overlaps effectively with the antibonding orbital of the C=O bond. Donation of the C2–H bonding electrons into the carbonyl π* orbital increases the acidity of this hydrogen (2 points)."], "rubric_images": []}]}