What is the product formed when one mole of aniline reacts with one mole of chloromethane?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → N-Methylaniline
Amines undergo alkylation with alkyl halides: the lone pair on nitrogen attacks the alkyl halide and one H of the –NH2 group is replaced by the alkyl group.
One mole of chloromethane is enough for one such step:
\( \displaystyle \ce{C6H5NH2 + CH3Cl -> C6H5NHCH3 + HCl} \)
Aniline, a primary amine, becomes the secondary amine N-methylaniline.
Because this product can react further with alkyl halide, a real reaction mixture also holds some of the later products, but the product of the first alkylation step is N-methylaniline.
Why the other options are incorrect
Option (2) N,N-Dimethylaniline — Incorrect. This tertiary amine needs a second alkylation step, with a second mole of chloromethane.
Option (3) N,N,N-Trimethylanilinium chloride — Incorrect. The quaternary ammonium salt is the last stage of successive alkylation and needs three moles of chloromethane.
Option (4) No reaction — Incorrect. The nitrogen lone pair makes aniline a nucleophile, so it does react with an alkyl halide.
NCERT (2026-27 edition):
9.6 Chemical Reactions – “Amines undergo alkylation on reaction with alkyl halides …”
9.4 Preparation of Amines – “… The primary amine thus obtained behaves as a nucleophile and can further react with alkyl halide to form secondary and tertiary amines, and finally quaternary ammonium salt.”