Which of the following is the least basic amine?

The correct answer is Option (2) → p-Nitroaniline

• Aniline is basic because the lone pair on nitrogen can accept a proton. An electron-withdrawing group on the ring pulls electron density away from nitrogen and makes the lone pair less available, so it lowers the basic strength. All four amines carry such a group at the para position; the strongest withdrawer gives the least basic amine.

• The order in which these groups raise the acidity of carboxylic acids shows their electron-withdrawing effect: the halogens (–Br, –Cl) are the weakest, then –CN, and –NO₂ is the strongest.

• –NO₂ is the strongest, so p-nitroaniline is the least basic.

Why the other options are incorrect

Option (1) p-Aminobenzonitrile — Incorrect. –CN is a strong electron-withdrawing group, but weaker than –NO₂, so this amine is a little more basic than p-nitroaniline.

Option (3) p-Chloroaniline — Incorrect. Chlorine withdraws electrons far less than –NO₂, so the lone pair stays more available.

Option (4) p-Bromoaniline — Incorrect. Bromine, like chlorine, withdraws electrons far less than –NO₂, so p-bromoaniline is more basic than p-nitroaniline.

NCERT (2026-27 edition):
9.6 Chemical Reactions – “… 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.”

8.9.1 Reactions Involving Cleavage of O–H Bond – “… Electron withdrawing groups increase the acidity of carboxylic acids by stabilising the conjugate base through delocalisation of the negative charge by inductive and/or resonance effects. … The effect of the following groups in increasing acidity order is Ph < I < Br < Cl < F < CN < NO₂ < CF₃”