Which of the following compounds possess the highest pK_b value?

The correct answer is Option (1) → Aniline (C₆H₅NH₂)

The higher the pK_b, the weaker the base. In aniline the lone pair on nitrogen is delocalised into the benzene ring by resonance, so it is much less available for a proton. Aniline (pK_b 9.38) is the weakest base of the four and has the highest pK_b.

Why the other options are incorrect

• Option (2) Benzylamine, C₆H₅CH₂NH₂ — Incorrect. The \( \ce{CH2} \) group keeps the lone pair away from the ring, so there is no resonance with it; it behaves like an aliphatic amine (pK_b 4.70), a much stronger base.

• Option (3) N-Methylaniline, C₆H₅NHCH₃ — Incorrect. The +I effect of the methyl group raises the electron density on nitrogen a little, so it is slightly more basic than aniline (pK_b 9.30).

• Option (4) N,N-Dimethylaniline, C₆H₅N(CH₃)₂ — Incorrect. Two methyl groups push electrons onto nitrogen; it is more basic than aniline (pK_b 8.92).

NCERT (2026-27 edition):
Arylamines versus Ammonia – “pK_b value of aniline is quite high. … It is because in aniline or other arylamines, the –NH₂ group is attached directly to the benzene ring. It results in the unshared electron pair on nitrogen atom to be in conjugation with the benzene ring and thus making it less available for protonation.”