Arrange the following bromides in the increasing order of their reactivity towards the SN1 reaction.
(A) 1-Bromobutane
(B) 2-Bromobutane
(C) 2-Bromo-2-methylpropane
(D) 1-Bromo-2-methylpropane
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) < (D) < (B) < (C)
An SN1 reaction goes through a carbocation, formed in the slow first step. The more stable the carbocation, the more easily it forms and the faster the reaction. Carbocation stability rises from primary to secondary to tertiary.
(A) 1-Bromobutane, \( \ce{CH3CH2CH2CH2Br} \) — first (least reactive). It is a primary bromide, and its carbocation has only the small electron-donating effect of a straight propyl chain behind it.
(D) 1-Bromo-2-methylpropane, \( \ce{(CH3)2CHCH2Br} \) — second. It is also primary, but the branched (CH3)2CH– group has a greater electron-donating inductive effect, so its carbocation is a little more stable than that from (A).
(B) 2-Bromobutane, \( \ce{CH3CH2CH(Br)CH3} \) — third. It is a secondary bromide; a secondary carbocation is more stable than a primary one.
(C) 2-Bromo-2-methylpropane, \( \ce{(CH3)3CBr} \) — fourth (most reactive). It is a tertiary bromide and gives the most stable carbocation, so it reacts fastest.
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.”
“Of the two primary bromides, the carbocation intermediate derived from (CH3)2CHCH2Br is more stable than derived from CH3CH2CH2CH2Br because of greater electron donating inductive effect of (CH3)2CH- group. Therefore, (CH3)2CHCH2Br is more reactive than CH3CH2CH2CH2Br in SN1 reactions. CH3CH2CH(Br)CH3 is a secondary bromide and (CH3)3CBr is a tertiary bromide.”