The correct order of reactivity of the following alkyl halides toward…

The correct answer is Option (2) → III > IV > I > II

The S_N1 reaction goes through a carbocation, so its rate depends on how stable the carbocation is. For the same carbocation it depends on how easily the halide ion leaves: the C–Br bond is weaker than the C–Cl bond, so \( \ce{Br^{-}} \) leaves more easily.

• III. 3-Bromoprop-1-ene (\( \ce{CH2=CH-CH2Br} \)) — first. It gives the allyl carbocation, which is stabilised by resonance:

      \( \ce{CH2=CH-CH2^{+} <-> ^{+}CH2-CH=CH2} \)

• IV. 2-Bromopropane — second. It gives a secondary carbocation, \( \ce{(CH3)2CH^{+}} \), stabilised by two methyl groups.

• I. Propyl bromide — third. It gives a primary carbocation, \( \ce{CH3CH2CH2^{+}} \), which is much less stable.

• II. Propyl chloride — last. It gives the same primary carbocation, but chloride is a poorer leaving group than bromide.

NCERT (2026-27 edition):
Substitution Nucleophilic Unimolecular (S_N1) – “… 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, 3⁰ alkyl halides undergo S_N1 reaction very fast because of the high stability of 3⁰ carbocations.
For the same reasons, allylic and benzylic halides show high reactivity towards the S_N1 reaction.”

“… For a given alkyl group, the reactivity of the halide, R-X, follows the same order in both the mechanisms R–I> R–Br>R–Cl>>R–F.”