The decreasing order of basicity of following amines in aqueous phase;
(A) N,N-Dimethylmethamine
(B) Methanamine
(C) N-Ethylethanamine
(D) Phenylmethanamine
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (C), (B), (A), (D)
In water, the basicity of an amine depends on three factors acting together: the +I effect of alkyl groups (more alkyl groups push more electron density onto nitrogen), the solvation of the ammonium ion by hydrogen bonding (more N–H bonds give better stabilisation), and steric hindrance to that solvation. Because of these opposing effects, secondary amines are usually the strongest; among methyl amines the tertiary amine is weaker than the primary one. An aryl group withdraws electrons, and even when it is separated from N by a \( \ce{CH2} \) group it lowers the basicity a little.
(A) N,N-Dimethylmethanamine, \( \ce{(CH3)3N} \): tertiary, solvation hindered, pKb 4.22.
(B) Methanamine, \( \ce{CH3NH2} \): primary, well solvated, pKb 3.38.
(C) N-Ethylethanamine, \( \ce{(C2H5)2NH} \): secondary with two ethyl groups, the best balance of +I effect and solvation, pKb 3.00.
(D) Phenylmethanamine, \( \ce{C6H5CH2NH2} \): the electron-withdrawing phenyl group lowers the electron density on N, pKb 4.70.
A smaller pKb means a stronger base, so in decreasing order of basicity:
\( \displaystyle \ce{(C2H5)2NH} \gt \ce{CH3NH2} \gt \ce{(CH3)3N} \gt \ce{C6H5CH2NH2} \)
Shortcut: The secondary amine (C) is the strongest and benzylamine (D) the weakest. Only Option (1) starts with (C) and ends with (D).
NCERT (2026-27 edition):
Basic Character of Amines – …
