ChemArgus public release (dataset-only): dev 25% (with rubrics) + test 55% (rubrics stripped); private 20% never published

.gitattributes

@@ -1,3 +1,4 @@
+* text=auto eol=lf
 *.7z filter=lfs diff=lfs merge=lfs -text
 *.arrow filter=lfs diff=lfs merge=lfs -text
 *.avro filter=lfs diff=lfs merge=lfs -text

.gitignore

@@ -0,0 +1,16 @@
+# dependencies
+node_modules/
+
+# evaluation artifacts
+runs/
+
+# python
+__pycache__/
+*.pyc
+
+# never published: Private Set (held out for anti-cheating final checks)
+private/
+
+# local tooling, not part of the public dataset presentation
+harness/
+metadata/

dev/instance_ids.json

@@ -0,0 +1,50 @@
+[
+  "Exam-01-T5",
+  "Exam-01-T7",
+  "Exam-02-T1",
+  "Exam-02-T3",
+  "Exam-02-T5",
+  "Exam-02-T6",
+  "Exam-02-T9",
+  "Exam-03-T2",
+  "Exam-03-T10",
+  "Exam-04-T6",
+  "Exam-05-T4",
+  "Exam-05-T7",
+  "Exam-05-T8",
+  "Exam-06-T4",
+  "Exam-07-T6",
+  "Exam-07-T10",
+  "Exam-08-T8",
+  "Exam-08-T10",
+  "Exam-09-T5",
+  "Exam-09-T8",
+  "Exam-09-T10",
+  "Exam-12-T1",
+  "Exam-12-T4",
+  "Exam-12-T10",
+  "Exam-13-T9",
+  "Exam-14-T3",
+  "Exam-14-T6",
+  "Exam-15-T1",
+  "Exam-15-T4",
+  "Exam-15-T8",
+  "Exam-15-T9",
+  "Exam-16-T3",
+  "Exam-17-T2",
+  "Exam-17-T3",
+  "Exam-17-T4",
+  "Exam-17-T7",
+  "Exam-17-T9",
+  "Exam-17-T10",
+  "Exam-18-T3",
+  "Exam-18-T4",
+  "Exam-19-T7",
+  "Exam-20-T7",
+  "Exam-21-T3",
+  "Exam-21-T4",
+  "Exam-21-T6",
+  "Exam-22-T1",
+  "Exam-22-T5",
+  "Exam-23-T8"
+]

dev/tasks/Exam-01-T5/problem.json

@@ -0,0 +1,70 @@
+{
+  "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'"
+        }
+      ]
+    }
+  ]
+}

dev/tasks/Exam-01-T7/problem.json

@@ -0,0 +1,20 @@
+{
+  "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": []
+    }
+  ]
+}

dev/tasks/Exam-02-T1/problem.json

@@ -0,0 +1,60 @@
+{
+  "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": []
+    }
+  ]
+}

dev/tasks/Exam-02-T3/problem.json

@@ -0,0 +1,58 @@
+{
+  "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": []
+    }
+  ]
+}

dev/tasks/Exam-02-T5/problem.json

@@ -0,0 +1,175 @@
+{
+  "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": []
+    }
+  ]
+}

dev/tasks/Exam-02-T6/problem.json

@@ -0,0 +1,39 @@
+{
+  "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": []
+    }
+  ]
+}

dev/tasks/Exam-02-T9/problem.json

@@ -0,0 +1,124 @@
+{
+  "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'."
+        }
+      ]
+    }
+  ]
+}

dev/tasks/Exam-03-T10/problem.json

@@ -0,0 +1,149 @@
+{
+  "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'."
+        }
+      ]
+    }
+  ]
+}