Q.(i) Between chlorobenzene and benzyl chloride which one gets easily hydrolysed by aqueous NaOH? Why? [2]

Benzyl chloride hydrolyses easily (stable benzylic carbocation); chlorobenzene resists hydrolysis (resonance-strengthened C-Cl bond). $(CH_3)_3C$-$Br$ reacts faster by $S_N1$ than a primary halide because its carbocation is far more stable.

(i) Chlorobenzene vs benzyl chloride hydrolysis: In chlorobenzene, the lone pair on chlorine is delocalised into the aromatic ring through resonance, giving the C-Cl bond partial double-bond character. This makes the bond shorter and stronger than a normal C-Cl single bond, and the sp²-hybridised ring carbon also resists nucleophilic attack (both $S_N1$, since no stable phenyl cation forms, and $S_N2$, due to steric/electronic shielding by the ring). As a result, chlorobenzene needs drastic conditions (high temperature/pressure, or a diazonium-type synthesis route) to hydrolyse.

In benzyl chloride ($C_6H_5CH_2Cl$), the C-Cl bond is an ordinary (not ring-attached) sp³ C-Cl bond, and on ionisation it forms a benzylic carbocation ($C_6H_5CH_2^+$) that is strongly stabilised by resonance delocalisation of the positive charge into the aromatic ring. This makes ionisation (the rate-determining step of $S_N1$) very favourable, so benzyl chloride is hydrolysed easily by aqueous NaOH under mild conditions.

(ii) $S_N1$ rate — $CH_3CH_2CH_2Br$ vs $(CH_3)_3C$-$Br$: $S_N1$ reaction rate depends directly on the stability of the carbocation intermediate formed. $(CH_3)_3C$-$Br$ is a tertiary halide; its ionisation gives a tertiary carbocation, stabilised by hyperconjugation and the +I effect of three methyl groups. $CH_3CH_2CH_2Br$ is a primary halide, whose primary carbocation is highly unstable (and in practice primary halides essentially do not undergo $S_N1$ at all). So $(CH_3)_3C$-$Br$ reacts much faster via the $S_N1$ pathway.

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