The amines which are less basic than aniline are:
A. p-Toluidine
B. 4-Methoxybenzamine
C. 4-Aminobenzoic acid
D. 4-Nitrobenzamine
E. 4-Chlorobenzamine
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → C, D, E only
Aniline is a weaker base than ammonia because the lone pair on nitrogen is delocalised into the benzene ring. A substituent on the ring changes how available this lone pair is: electron releasing groups (–\( \ce{OCH3} \), –\( \ce{CH3} \)) push electron density towards nitrogen and increase basic strength; electron withdrawing groups (–\( \ce{NO2} \), –\( \ce{SO3H} \), –COOH, –X) pull it away and decrease it.
(A) p-Toluidine — More basic than aniline. –\( \ce{CH3} \) is electron releasing (+I, hyperconjugation).
(B) 4-Methoxybenzamine — More basic than aniline. –\( \ce{OCH3} \) releases electrons by resonance (+R) at the para position.
(C) 4-Aminobenzoic acid — Less basic than aniline. –COOH withdraws electrons (−I, −R).
(D) 4-Nitrobenzamine — Less basic than aniline. –\( \ce{NO2} \) is a strong −I, −R group, so it is much less basic.
(E) 4-Chlorobenzamine — Less basic than aniline. –Cl withdraws electrons by its −I effect, which outweighs its +R effect.
NCERT (2026-27 edition):
Arylamines versus Ammonia – “… In case of substituted aniline, it is observed that electron releasing groups like –OCH3, –CH3 increase basic strength whereas electron withdrawing groups like –NO2, –SO3H, –COOH, –X decrease it.”