2025 H2 Chem P1 Q18 — Introduction to Organic Solution

The answer is C (\(\text{CH}_2=\text{CH}_2 + \text{Br}_2 \rightarrow \text{CH}_2\text{BrCH}_2\text{Br}\)).
A nucleophile is an electron-pair donor. In electrophilic addition, the electron-rich C=C \(\pi\) bond of ethene induces a dipole in \(\text{Br}_2\), then its \(\pi\) electron pair attacks the \(\delta+\) Br atom. Ethene donates the electron pair, so the organic reactant behaves as a nucleophile.
• A is wrong: this is free-radical substitution; \(\text{CH}_4\) reacts with \(\text{Cl}\cdot\) radicals and does not donate an electron pair.
• B is wrong: in nucleophilic substitution, \(\text{OH}^-\) is the nucleophile; its lone pair attacks the electron-deficient \(\delta+\) carbon of \(\text{CH}_3\text{I}\), so the organic reactant is the electrophile.
• D is wrong: in nucleophilic addition, \(\text{CN}^-\) is the nucleophile; its lone pair on C attacks the \(\delta+\) carbonyl carbon of \(\text{CH}_3\text{COCH}_3\), so the organic reactant is the electrophile.