Arrange the following alkyl halides in order of increasing reactivity…

The correct answer is Option (1) → (B), (D), (C), (A)

In an \( S_N1 \) reaction the slow, rate-determining step is the ionisation of the C–X bond to give a carbocation:

    \( \displaystyle \ce{(CH3)3C-X -> (CH3)3C+ + X-}\quad (\text{slow}) \)

All four compounds give the same tert-butyl carbocation, so the rate depends only on how easily the C–X bond breaks. The C–X bond becomes weaker from C–F to C–I, and the larger halide ion is the better leaving group. So the reactivity order of the halides is R–I > R–Br > R–Cl >> R–F.

• (A) \( \ce{(CH3)3C-I} \): weakest C–X bond, most reactive.

• (B) \( \ce{(CH3)3C-F} \): strongest C–X bond, least reactive.

• (C) \( \ce{(CH3)3C-Br} \): second most reactive.

• (D) \( \ce{(CH3)3C-Cl} \): less reactive than the bromide.

In increasing order of reactivity:

    \( \displaystyle \ce{(CH3)3C-F} \lt \ce{(CH3)3C-Cl} \lt \ce{(CH3)3C-Br} \lt \ce{(CH3)3C-I} \)

Shortcut: The fluoride (B) must come first and the iodide (A) last; only Option (1) starts with (B) and ends with (A).

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. … 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.”