[ { "name": "IChO-2025_7", "background": { "context": "The petroleum industry plays an important role in the UAE economy. The oil called 'Dubai Crude' is a benchmark for global oil prices. Crude oil is a complex mixture of organic compounds, mainly hydrocarbons, along with smaller amounts of S, N, and O-containing compounds.\n\nThe hydrocarbon composition of oil is characterised by the PONA number. The abbreviation stands for $P$ - paraffins (alkanes), $O$ - olefins (alkenes), $N$ - naphthenes (cycloalkanes), and $A$ - aromatics (arenes). One method to determine the composition of oil is gas chromatography–mass spectrometry (GC–MS)." }, "points": 22 }, { "name": "IChO-2025_7.1", "modality": "image + text", "type": "Structure Construction", "evaluation": "Structure Match", "points": 8, "field": "", "source": "IChO-2025", "question": { "context": "Below are mass spectra (electron impact ionisation, $I$ – relative intensity) of four hydrocarbons 1–4 of Dubai Crude oil. They include linear alkane P, linear alkene O (no cis–trans isomers), monosubstituted cycloalkane N, and arene A. 1'–4' are selected fragmentation products. Give the SMILES string** for each compound (1–4) and each main fragment (1'–4'). If you cannot determine 1–4, write letter code (P/O/N/A) and molecular formula. Hint: species $3'$ is formed by a McLafferty rearrangement.", "requirement": "Your response should list SMILES for 1, 2, 3, 4, 1', 2', 3', and 4'.", "images": [ "images/7/7.1_1.png", "images/7/7.1_2.png", "images/7/7.1_3.png", "images/7/7.1_4.png" ] }, "answer": [ { "step": 1, "content": "1 = ethylcyclohexane, SMILES: CCC1CCCCC1, class N, formula C8H16", "points": 1 }, { "step": 2, "content": "2 = n-octane, SMILES: CCCCCCCC, class P, formula C8H18", "points": 1 }, { "step": 3, "content": "3 = 1-octene, SMILES: C=CCCCCCC, class O, formula C8H16", "points": 1 }, { "step": 4, "content": "4 = ethylbenzene, SMILES: CCC1=CC=CC=C1, class A, formula C8H10", "points": 1 }, { "step": 5, "content": "1' = cyclohexyl cation fragment, SMILES: C1CCC[CH+]C1, formula C6H11+", "points": 1 }, { "step": 6, "content": "2' = propyl or isopropyl cation, SMILES: CC[CH2+], formula C3H7+", "alternative_smiles": "C[CH+]C", "points": 1 }, { "step": 7, "content": "3' = allylic McLafferty fragment from oct-1-ene, SMILES: C=CCCC+, formula C4H7+", "points": 1 }, { "step": 8, "content": "4' = benzyl/tropylium cation, SMILES: [CH2+]C1=CC=CC=C1, formula C7H7+", "alternative_smiles": "C1=CC=CC=C[CH+]1", "points": 1 } ], "grading": { "criteria": { "for_compounds_1_to_4": { "full": "1 pt per correct SMILES match.", "partial": [ { "condition": "SMILES incorrect but molecular formula correct.", "points": 0.25 }, { "condition": "Compound belongs to the same P/O/N/A class.", "points": 0.25 }, { "condition": "No SMILES provided but molecular formula given.", "points": 0.25 }, { "condition": "Only the correct class letter (P/O/N/A) provided.", "points": 0.25 } ] }, "for_fragments_1prime_to_4prime": { "full": "1 pt per correct SMILES fragment (including ion).", "partial": [ { "condition": "SMILES incorrect but molecular formula correct.", "points": 0.25 }, { "condition": "Charge (“+” or “·+”) is explicitly indicated even if structure incorrect.", "points": 0.25 } ] } } } }, { "name": "IChO-2025_7.2", "modality": "image + text", "type": "Structure Construction", "evaluation": "Structure Match", "points": 2, "field": "", "source": "IChO-2025", "question": { "context": "Two mass spectra of branched alkanes 5 and 6 are shown. Give the SMILES of 5 and 6.", "images": [ "images/7/7.2_5.png", "images/7/7.2_6.png" ] }, "answer": [ { "step": 1, "content": "5 = 2-methylbutane (isopentane). SMILES: CC(CC)C", "formula": "C5H12", "points": 1 }, { "step": 2, "content": "6 = 2,2-dimethylpropane (neopentane). SMILES: CC(C)(C)C", "formula": "C5H12", "points": 1 } ], "grading": { "per_compound": { "5": { "full_credit": "1 pt for correct SMILES (any valid equivalent).", "partial_credit": [ { "condition": "Only the correct molecular formula (C5H12) is given.", "points": 0.5 } ] }, "6": { "full_credit": "1 pt for correct SMILES (any valid equivalent).", "partial_credit": [ { "condition": "Gives 2,2,3,3-tetramethylbutane instead of 2,2-dimethylpropane.", "points": 0.5 }, { "condition": "Only the correct molecular formula (C5H12) is given.", "points": 0.5 } ] } }, "special_case": "If the two structures (5 and 6) are swapped, award a TOTAL of 1 pt." } }, { "name": "IChO-2025_7.3", "modality": "text", "type": "Quantitative Calculation", "evaluation": "Numeric Verification", "points": 3, "field": "", "source": "IChO-2025", "question": { "context": "A $V = 115\\,\\mu\\mathrm{L}$ sample of Dubai Crude (density $\\rho = 871\\,\\mathrm{g\\,dm^{-3}}$) is completely burnt. The gases are passed through $\\mathrm{H_2O_2}$ with excess $\\mathrm{Ba(OH)_2}$ to form a white precipitate of mass $m = 1.395\\,\\mathrm{g}$. Upon acidification with $\\mathrm{HNO_3}$ the precipitate mass decreases by $98.95\\%$. Calculate the sulfur content $w(\\mathrm{S})$ (wt%)." }, "answer": [ { "step": 1, "content": "Only $\\mathrm{BaSO_4}$ remains after acidification. $n(\\mathrm{BaSO_4}) = (1 - 0.9895) \\times \\frac{1.395\\,\\mathrm{g}}{233.39\\,\\mathrm{g\\,mol^{-1}}} = 6.276 \\times 10^{-5}\\,\\mathrm{mol}$, hence $n(\\mathrm{S}) = 6.276 \\times 10^{-5}\\,\\mathrm{mol}$.", "points": 1 }, { "step": 2, "content": "Mass of S in sample: $m(\\mathrm{S}) = n \\cdot M(\\mathrm{S}) = 6.276 \\times 10^{-5}\\,\\mathrm{mol} \\times 32.06\\,\\mathrm{g\\,mol^{-1}} = 2.01 \\times 10^{-3}\\,\\mathrm{g}$.", "points": 1 }, { "step": 3, "content": "Sample mass: $m_{\\text{sample}} = 115 \\times 10^{-6}\\,\\mathrm{dm^3} \\times 871\\,\\mathrm{g\\,dm^{-3}} = 0.100\\,\\mathrm{g}$. Therefore $w(\\mathrm{S}) = 2.01\\%$ (sour oil).", "points": 1 } ] }, { "name": "IChO-2025_7.4", "modality": "multiple choice", "type": "Qualitative Identification", "evaluation": "Selection Check", "points": 1, "field": "", "source": "IChO-2025", "question": { "context": "In addition to sulfur, crude oil contains trace amounts of selenium, which transfers into the water used during refining. In aqueous solutions, selenium exists as $\\mathrm{SeO_3^{2-}}$ and $\\mathrm{SeO_4^{2-}}$. The concentration of $\\mathrm{SeO_3^{2-}}$ can be determined chromatographically by its selective reaction with diamine **X** under acidic conditions, forming piazoselenol **Y**. This reaction is complete only at the right pH range. X and Y elute at different times (t) on the chromatogram shown below. The absorption spectra (1 mAbs corresponds to absorbance $A = 0.001$ ) of X and Y are also provided.", "images": [ "images/7/7.4_1.png", "images/7/7.4_1.png" ], "reaction_scheme": { "reactions": [ { "reactant_name": "X", "reactant_smiles": "NC1=CC=C(N)C(N)=C1", "conditions": [ { "step": 1, "reagents": [ "H2SeO3", "H+" ] } ], "product_name": "Y", "product_smiles": "O=[N+](C1=CC2=N[Se]N=C2C=C1)[O-]" } ] }, "question_text": "Choose the optimal wavelength $(\\lambda)$ for the absorbance measurements in the chromatogram.", "options": { "A": "270 nm", "B": "300 nm", "C": "350 nm", "D": "510 nm" } }, "answer": [ { "correct_option": "C", "content": "The optimal wavelength for absorbance measurements is **350 nm**, as both X and Y absorb the most light at this wavelength.", "points": 1 } ] }, { "name": "IChO-2025_7.5", "modality": "text", "type": "Quantitative Calculation", "evaluation": "Numeric Verification", "points": 8, "field": "", "source": "IChO-2025", "question": { "context": "Acidified water from the refining process, containing $\\mathrm{SeO_3^{2-}}$ and $\\mathrm{SeO_4^{2-}}$ (solution 1), was analysed as follows. A $9.50~\\mathrm{mL}$ sample was mixed with $0.50~\\mathrm{mL}$ of an aqueous solution of X ($300~\\mu\\mathrm{M}$, excess) to form solution 2. The chromatographic peak areas (S) were measured for compounds X and Y, which are proportional to their concentrations. For a blank sample, $S(X) = 0.825$ mAbs·min. For known $\\mathrm{SeO_3^{2-}}$ concentrations, $S(Y) = 1.21 \\times c(\\mathrm{Se})$ (in µM). The ratios $S(X)/S(Y)$ measured at different pH values for four strong acids are given below (assume equilibrium):", "table": { "headers": [ "pH", "HNO3", "HCl", "HBr", "HI" ], "rows": [ [ "3.0", "1.523", "0.634", "0.523", "1.722" ], [ "2.0", "3.334", "0.538", "0.377", "2.223" ], [ "1.5", "5.454", "0.523", "0.368", "5.114" ], [ "1.0", "7.783", "0.523", "0.368", "8.123" ], [ "0", "10.232", "0.581", "0.399", "11.736" ], [ "-0.50", "13.450", "0.782", "0.546", "15.780" ] ] }, "question_text": "Calculate the concentrations of selenium species ($c_{0}(\\mathrm{SeO_3^{2-}})$, $c_{0}(\\mathrm{SeO_4^{2-}})$ in µM) in solution 1." }, "answer": [ { "step": 1, "content": "At high [H+] (HNO3), $\\mathrm{SeO_3^{2-}}$ is oxidised to $\\mathrm{SeO_4^{2-}}$, leading to a high $S(X)/S(Y)$ ratio (minimal formation of Y). Conversely, HI reduces $\\mathrm{SeO_3^{2-}}$ and $\\mathrm{SeO_4^{2-}}$ to Se, again giving high ratios. HCl and HBr give optimal conditions (1 pt).", "points": 1 }, { "step": 2, "content": "From the blank: $S(X) = 0.825$ mAbs·min for $c(X) = 15.0~\\mu\\mathrm{M}$ (since $300~\\mu\\mathrm{M} \\times 0.50~\\mathrm{mL} / (9.50 + 0.50)~\\mathrm{mL} = 15.0~\\mu\\mathrm{M}$). Therefore, the proportionality constant $k = 0.825 / 15.0 = 0.0550~\\mathrm{mAbs·min·\\mu M^{-1}}$ (1 pt).", "points": 1 }, { "step": 3, "content": "For HCl medium: total Se species $= 15.0~\\mu\\mathrm{M}$.\n$$[X]_{\\mathrm{HCl}} + [\\mathrm{SeO_3^{2-}}] = 15.0$$\n$$0.523 = 0.0550 [X]_{\\mathrm{HCl}} / (1.21 [\\mathrm{SeO_3^{2-}}])$$ (1 pt)", "points": 1 }, { "step": 4, "content": "For HBr medium, both $\\mathrm{SeO_3^{2-}}$ and $\\mathrm{SeO_4^{2-}}$ are present:\n$$[X]_{\\mathrm{HBr}} + [\\mathrm{SeO_3^{2-}}] + [\\mathrm{SeO_4^{2-}}] = 15.0$$\n$$0.368 = 0.0550 [X]_{\\mathrm{HBr}} / (1.21([\\mathrm{SeO_3^{2-}}] + [\\mathrm{SeO_4^{2-}}]))$$ (1 pt)", "points": 1 }, { "step": 5, "content": "Solving these simultaneous equations:\n$$[\\mathrm{SeO_3^{2-}}] = 1.20~\\mu\\mathrm{M}, \\quad [\\mathrm{SeO_4^{2-}}] = 0.45~\\mu\\mathrm{M}.$$ (1 pt)", "points": 1 }, { "step": 6, "content": "Before dilution correction:\n$$c(\\mathrm{SeO_3^{2-}})_0 = 1.20~\\mu\\mathrm{M} \\times 10~\\mathrm{mL}/9.5~\\mathrm{mL} = 1.26~\\mu\\mathrm{M},$$\n$$c(\\mathrm{SeO_4^{2-}})_0 = 0.45~\\mu\\mathrm{M} \\times 10~\\mathrm{mL}/9.5~\\mathrm{mL} = 0.47~\\mu\\mathrm{M}.$$ (1 pt)", "points": 1 }, { "step": 7, "content": "Final result: $c_0(\\mathrm{SeO_3^{2-}}) = 1.26~\\mu\\mathrm{M}$, $c_0(\\mathrm{SeO_4^{2-}}) = 0.47~\\mu\\mathrm{M}$.", "points": 2 } ] } ]