Which of the following haloalkanes reacts with aqueous KOH most easily?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2-Bromo-2-methyl propane
Aqueous KOH is a polar protic medium, so substitution by the SN1 route is favoured, and the rate depends on how easily the carbocation forms. The order of carbocation stability is 3° > 2° > 1°, so the order of SN1 reactivity is tertiary > secondary > primary.
\( \displaystyle \ce{(CH3)3C-Br -> (CH3)3C+ + Br-} \quad (\text{slow}) \)
\( \displaystyle \ce{(CH3)3C+ + OH- -> (CH3)3C-OH} \quad (\text{fast}) \)
2-Bromo-2-methylpropane is a tertiary bromide; its tert-butyl carbocation is stabilised by three methyl groups, so it reacts most easily.
Why the other options are incorrect
Option (1) 1-Bromobutane — Incorrect. A primary halide; a primary carbocation is too unstable, so it reacts slowest by SN1.
Option (2) 2-Bromobutane — Incorrect. A secondary halide; its 2° carbocation is less stable than the 3° one.
Option (4) 2-Chlorobutane — Incorrect. Also secondary, and the C–Cl bond is stronger than C–Br, so it is even less reactive than 2-bromobutane.
NCERT (2026-27 edition):
Substitution Nucleophilic Unimolecular (SN1) – “… Since the rate of reaction depends upon the slowest step, the rate of reaction depends only on the concentration of alkyl halide and not on the concentration of hydroxide ion. 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.”