Ethanamine is heated with an excess of bromoethane. Which of the following is the final product of this alkylation?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Tetraethylammonium bromide
In alkylation, the amine attacks the alkyl halide with its unshared electron pair, as a nucleophile.
Each product still has an unshared pair on nitrogen, so it reacts again. With excess bromoethane the alkylation goes on until no hydrogen is left on nitrogen and nitrogen holds four ethyl groups:
\( \displaystyle \ce{C2H5NH2 + C2H5Br -> (C2H5)2NH + HBr} \)
\( \displaystyle \ce{(C2H5)2NH + C2H5Br -> (C2H5)3N + HBr} \)
\( \displaystyle \ce{(C2H5)3N + C2H5Br -> (C2H5)4N^+Br^-} \)
The last product, tetraethylammonium bromide, is a quaternary ammonium salt.
Option (1) N-Ethylethanamine — Incorrect. This secondary amine is formed on the way, but it still has an unshared pair and reacts further with the excess bromoethane.
Option (2) Triethylamine — Incorrect. This tertiary amine is also formed on the way; it too reacts with the excess bromoethane, giving the quaternary salt.
Option (4) N-Ethylethanamide — Incorrect. N-Ethylethanamide is an amide. Amides are formed by acylation, for example of ethanamine with ethanoyl chloride, not by alkylation 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.”