For the same alkyl group, all the three primary, secondary and tertiary amines are more basic than ammonia in aqueous solutions. Their relative basicity, however, depends upon a combination of three factors: +I-effect of alkyl groups, the stability of the ammonium cation (formed after accepting a proton) due to H-bonding with water molecules and the steric effect of the alkyl groups which tend to reduce the extent of H-bonding. All these three factors are favourable for secondary amines thereby making them the strongest bases. If, however, the alkyl group is bigger than the methyl group, +I-effect of the alkyl group outweighs stability of the ammonium cation due to H-bonding thereby making tertiary amines more basic than the primary amines. Due to delocalization of lone pair of electrons of the nitrogen atom on the benzene ring, aniline is a weaker base than ammonia. The basic strength of the substituted anilines, however, depends upon the nature of the substituent. Whereas electron-donating groups tend to increase, electron-withdrawing groups tend to decrease the basic strength.
Among the following, the strongest base is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → C6H5CH2NH2
An amine is basic because the lone pair on nitrogen accepts a proton. The more available this lone pair, the stronger the base and the smaller its pKb.
In \( \ce{C6H5CH2NH2} \) (benzylamine, phenylmethanamine) the –NH2 group is joined to a \( \ce{CH2} \) group, not to the ring. The lone pair is not in conjugation with the benzene ring, so the compound behaves like an alkylamine: its pKb is 4.70, against 9.38 for aniline.
Why the other options are incorrect
Option (1) C6H5NH2 — Incorrect. In aniline the –NH2 group sits directly on the ring; its lone pair is delocalised over the ring and less available for a proton (pKb 9.38).
Option (2) p-NO2-C6H4NH2 — Incorrect. The –NO2 group is electron withdrawing; it pulls electron density away from nitrogen, so p-nitroaniline is an even weaker base than aniline.
Option (3) m-NO2-C6H4NH2 — Incorrect. The electron withdrawing –NO2 group lowers the basic strength here too, so m-nitroaniline is also weaker than aniline.
NCERT (2026-27 edition):
9.6 Chemical Reactions – “… Larger the value of Kb or smaller the value of pKb, stronger is the base. The pKb values of few amines are given in Table 9.3.”
“… 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.”