The most basic amine amongst the following is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → p-toluidine
The basic strength of an aromatic amine depends on how available the lone pair on nitrogen is. In aniline the lone pair is partly drawn into the ring by resonance, so aniline is a weak base. A substituent on the ring changes this:
Electron releasing groups (–\( \ce{CH3} \), –\( \ce{OCH3} \)) push electron density towards nitrogen and increase basic strength.
Electron withdrawing groups (–\( \ce{NO2} \), –COOH) pull electron density away and decrease it.
p-Toluidine (4-methylaniline) carries an electron releasing methyl group, so it is the most basic: p-toluidine (pKb 8.92) > aniline (9.38) > p-aminobenzoic acid > 4-nitroaniline (13.0).
Why the other options are incorrect
Option (1) Aniline — Incorrect. It has no substituent to increase the electron density on nitrogen, so it is less basic than p-toluidine.
Option (2) 4-Nitrobenzenamine — Incorrect. The nitro group withdraws electrons by both −I and −R effects, making the lone pair much less available; it is the weakest base here.
Option (4) p-Amino benzoic acid — Incorrect. The –COOH group is electron withdrawing, so this amine is weaker than aniline.
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.”