Consider the following bromides :
The correct order of reactivity towards \(S_N1\) reaction is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → II > III > I
An SN1 reaction goes through a carbocation, formed in the slow step when the halide ion leaves. The more stable the carbocation, the faster the reaction. Carbocation stability: allylic and benzylic (resonance stabilised) > tertiary > secondary > primary.
II. 3-Bromobut-1-ene — first (most reactive): an allylic secondary halide; the carbocation formed is stabilised by resonance with the C=C bond:
\( \ce{CH2=CH-CH^+-CH3} \leftrightarrow \ce{^+CH2-CH=CH-CH3} \)
III. 2-Bromobutane — second: a secondary halide; its carbocation is stabilised only by the +I effect and hyperconjugation of two alkyl groups.
I. 1-Bromobutane — last: a primary halide; the primary carbocation is the least stable.
Shortcut: The allylic bromide II reacts fastest and the primary bromide I slowest; only Option (2) starts with II and ends with I.
NCERT (2026-27 edition):
Reactions of Haloalkanes – “… Step I is the slowest and reversible. It involves the C–Br bond breaking for which the energy is obtained through solvation of halide ion with the proton of protic solvent. 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. … For the same reasons, allylic and benzylic halides show high reactivity towards the SN1 reaction. The carbocation thus formed gets stabilised through resonance (Unit 8, Class XI).”