Which compound will undergo fastest S_N1 reaction?

The correct answer is Option (1) → 2-bromo-2-methylbutane

In an S_N1 reaction the slow step is the loss of the halide ion to form a carbocation. The more stable the carbocation, the faster this step. Alkyl groups stabilise a carbocation by their +I effect and hyperconjugation, so the order is 3° > 2° > 1°, and tertiary halides react fastest.

Option (1), \( \ce{(CH3)2C(Br)CH2CH3} \), is a tertiary bromide. It ionises to a tertiary carbocation:

    \( \displaystyle \ce{(CH3)2C(Br)CH2CH3 -> (CH3)2C^+CH2CH3 + Br^-} \)

So 2-bromo-2-methylbutane undergoes S_N1 reaction fastest.

Why the other options are incorrect

• Option (2) 3-bromopentane — Incorrect. It is a secondary bromide; the secondary carbocation it forms is less stable than a tertiary one.

• Option (3) 2-bromo-3-methylbutane — Incorrect. It is also secondary. Its carbocation can later rearrange to a tertiary one by a hydride shift, but the slow ionisation step still forms the less stable secondary ion.

• Option (4) 1-bromopentane — Incorrect. It is a primary bromide; primary carbocations are the least stable, so primary halides react by S_N2 rather than S_N1.

NCERT (2026-27 edition):
Substitution Nucleophilic Unimolecular (S_N1) – “… 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 S_N1 reaction very fast because of the high stability of 3° carbocations.”