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"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": []
}
]
}
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