Which of the following compounds will undergo \( S_N1 \) reaction at a fastest rate?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2-chloro-2-methylpropane (tert-butyl chloride)
An \( S_N1 \) reaction takes place in two steps. In the slow first step the C–Cl bond breaks and a carbocation is formed; the nucleophile then attacks it quickly. The rate depends only on how easily the carbocation forms, so the more stable the carbocation, the faster the reaction. Alkyl groups stabilise a carbocation by their electron-releasing inductive effect and by hyperconjugation:
\( \displaystyle 3^\circ \gt 2^\circ \gt 1^\circ \)
The four halides are:
(1) 2-chloropentane: secondary
(2) 1-chloropentane: primary
(3) 2-chloro-2-methylpropane: tertiary
(4) 3-chloropentane: secondary
Only the tertiary halide forms a tertiary carbocation:
\( \displaystyle \ce{(CH3)3C-Cl} \xrightarrow{\text{slow}} \ce{(CH3)3C+ + Cl-} \)
\( \displaystyle \ce{(CH3)3C+ + OH-} \xrightarrow{\text{fast}} \ce{(CH3)3C-OH} \)
So 2-chloro-2-methylpropane reacts fastest by the \( S_N1 \) route.
Why the other options are incorrect
Option (1) 2-chloropentane — Incorrect. It gives a secondary carbocation, less stable than a tertiary one.
Option (2) 1-chloropentane — Incorrect. A primary carbocation is the least stable; primary halides react mainly by \( S_N2 \).
Option (4) 3-chloropentane — Incorrect. It also gives a secondary carbocation.
NCERT (2026-27 edition):
Substitution Nucleophilic Unimolecular (SN1) – “… 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.”