2022 H2 Chem P1 Q23 — Halogenoalkane Solution

The answer is B.

Warming with NaOH releases a halide ion only from a halogen bonded to an alkyl (side-chain) carbon. A halogen bonded directly to the benzene ring is not substituted, because the C-X bond is strengthened by partial double bond character. The released halide then gives a silver halide precipitate with \(\text{AgNO}_3\text{(aq)}\).

• 4-chlorobenzyl iodide, \(\text{ClC}_6\text{H}_4\text{CH}_2\text{I}\): the \(\text{CH}_2\text{I}\) iodine is substituted to give \(\text{I}^-\); the ring Cl is not. AgI precipitate forms, and AgI does not dissolve in aqueous ammonia, so the precipitate remains.
• 4-bromobenzyl chloride, \(\text{BrC}_6\text{H}_4\text{CH}_2\text{Cl}\): the \(\text{CH}_2\text{Cl}\) chlorine is substituted to give \(\text{Cl}^-\); the ring Br is not. AgCl precipitate forms, but AgCl dissolves in aqueous ammonia to give a colourless solution, so no precipitate remains.
• Iodobenzene, \(\text{C}_6\text{H}_5\text{I}\): the I is bonded directly to the ring, so no \(\text{I}^-\) is released and no precipitate forms.
• (1-iodoethyl)benzene, \(\text{C}_6\text{H}_5\text{CH(I)CH}_3\): the side-chain I is released as \(\text{I}^-\) (by substitution and elimination). AgI precipitate forms and remains in aqueous ammonia.

Precipitates with \(\text{AgNO}_3\text{(aq)}\): 3 (AgI, AgCl, AgI). Precipitates remaining with \(\text{NH}_3\text{(aq)}\): 2 (both AgI).

• A is wrong: both AgI precipitates remain in aqueous ammonia, so 2 remain, not 1.
• C and D are wrong: iodobenzene gives no precipitate, so only 3 mixtures give precipitates, not 4.