The reaction of 4-bromobenzyl chloride with NaCN in ethanol gives
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4-Bromobenzyl cyanide
The benzylic C–Cl bond reacts easily by nucleophilic substitution, because the benzene ring stabilises the benzylic carbocation by resonance. The aryl C–Br bond has partial double bond character and does not react with \( \ce{CN^-} \). So cyanide replaces only the benzylic chlorine:
\( \displaystyle \ce{Br-C6H4-CH2-Cl + NaCN} \xrightarrow{\text{ethanol}} \ce{Br-C6H4-CH2-CN + NaCl} \)
The product is 4-bromobenzyl cyanide.
Why the other options are incorrect
Option (1) 4-Bromo-2-cyanobenzyl chloride — Incorrect. \( \ce{CN^-} \) does not replace a ring hydrogen, and the reactive C–Cl bond would be left untouched.
Option (3) 4-Cyanobenzyl chloride — Incorrect. It needs the aryl bromine to be replaced, which is the reverse of what happens.
Option (4) 4-Cyanobenzyl bromide — Incorrect. Nothing in the reaction puts bromine on the benzylic carbon; NaCN brings in only \( \ce{CN^-} \).
NCERT (2026-27 edition):
6.7.1 Reactions of Haloalkanes – “For the same reasons, allylic and benzylic halides show high reactivity towards the SN1 reaction.”
6.7.2 Reactions of Haloarenes – “1. Nucleophilic substitution
Aryl halides are extremely less reactive towards nucleophilic substitution reactions due to the following reasons: … C—Cl bond acquires a partial double bond character due to resonance.”