The most reactive halide towards SN1 reaction is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → tert-Butyl chloride
In an SN1 reaction the slow step is the loss of the halide ion to form a carbocation.
The more stable the carbocation, the faster it forms, so the reactivity follows 3° > 2° > 1°.
tert-Butyl chloride, \( \ce{(CH3)3CCl} \), is the only tertiary halide here. It gives the 3° carbocation \( \ce{(CH3)3C^+} \), the most stable of the four, so it reacts fastest.
The four options are the chlorides of the four butyl groups, and they follow the same SN1 order as the four isomeric bromobutanes.
Why the other options are incorrect
Option (1) sec-Butyl chloride — Incorrect. \( \ce{CH3CH2CH(Cl)CH3} \) is secondary; its 2° carbocation is less stable than a 3° one, so it reacts more slowly.
Option (3) n-Butyl chloride — Incorrect. \( \ce{CH3CH2CH2CH2Cl} \) is primary and is the least reactive of the four towards SN1.
Option (4) Isobutyl chloride — Incorrect. \( \ce{(CH3)2CHCH2Cl} \) is also primary. The electron-donating inductive effect of its \( \ce{(CH3)2CH-} \) group makes it a little more reactive than n-butyl chloride, but less reactive than a secondary or tertiary halide.
NCERT (2026-27 edition):
6.7.1 Reactions of Haloalkanes – “… Further, greater the stability of carbocation, greater will be its ease of formation from alkyl halide and faster will be the rate of reaction. In case of alkyl halides, 3° alkyl halides undergo SN1 reaction very fast because of the high stability of 3° carbocations.”
“(i) CH3CH2CH2CH2Br < (CH3)2CHCH2Br < CH3CH2CH(Br)CH3 < (CH3)3CBr (SN1) … CH3CH2CH(Br)CH3 is a secondary bromide and (CH3)3CBr is a tertiary bromide. Hence the above order is followed in SN1.”