What will be product formed when ammonolysis of benzyl chloride is carried out and the amine so formed reacts with 2 moles of chloromethane?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → N,N-Dimethylphenylmethanamine
In ammonolysis, an alkyl or benzyl halide is heated with ammonia. The halogen is replaced by the –\( \ce{NH2} \) group by nucleophilic substitution. The amine formed is itself a nucleophile, so it can react further with more alkyl halide; each molecule of chloromethane replaces one N–H hydrogen by a methyl group.
\( \displaystyle \ce{C6H5CH2Cl + NH3 -> C6H5CH2NH2 + HCl} \)
\( \displaystyle \ce{C6H5CH2NH2 + 2CH3Cl -> C6H5CH2N(CH3)2 + 2HCl} \)
Naming the product \( \ce{C6H5CH2N(CH3)2} \):
The parent amine \( \ce{C6H5CH2NH2} \) (benzylamine) is phenylmethanamine: a one-carbon chain (methanamine) carrying a phenyl group.
The two methyl groups are on nitrogen, so they get the locants N,N.
NCERT (2026-27 edition):
Preparation of Amines – …
