IDEA-AI4S/ChemO huggingface dataset:
IChO-2025_7
Points: 22
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.

The 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).
IChO-2025_7.1
Type: Structure Construction
Points: 8
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
IChO-2025_7.2
Type: Structure Construction
Points: 2
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
IChO-2025_7.3
Type: Quantitative Calculation
Points: 3
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
IChO-2025_7.4
Type: Qualitative Identification
Points: 1
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
IChO-2025_7.5
Type: Quantitative Calculation
Points: 8
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 $= 15.0~\mu\mathrm{M}$.
$$[X]_{\mathrm{HCl}} + [\mathrm{SeO_3^{2-}}] = 15.0$$
$$0.523 = 0.0550 [X]_{\mathrm{HCl}} / (1.21 [\mathrm{SeO_3^{2-}}])$$ (1 pt)
points: 1
step: 4
content: For HBr medium, both $\mathrm{SeO_3^{2-}}$ and $\mathrm{SeO_4^{2-}}$ are present:
$$[X]_{\mathrm{HBr}} + [\mathrm{SeO_3^{2-}}] + [\mathrm{SeO_4^{2-}}] = 15.0$$
$$0.368 = 0.0550 [X]_{\mathrm{HBr}} / (1.21([\mathrm{SeO_3^{2-}}] + [\mathrm{SeO_4^{2-}}]))$$ (1 pt)
points: 1
step: 5
content: Solving these simultaneous equations:
$$[\mathrm{SeO_3^{2-}}] = 1.20~\mu\mathrm{M}, \quad [\mathrm{SeO_4^{2-}}] = 0.45~\mu\mathrm{M}.$$ (1 pt)
points: 1
step: 6
content: Before dilution correction:
$$c(\mathrm{SeO_3^{2-}})_0 = 1.20~\mu\mathrm{M} \times 10~\mathrm{mL}/9.5~\mathrm{mL} = 1.26~\mu\mathrm{M},$$
$$c(\mathrm{SeO_4^{2-}})_0 = 0.45~\mu\mathrm{M} \times 10~\mathrm{mL}/9.5~\mathrm{mL} = 0.47~\mu\mathrm{M}.$$ (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