Which of the following alkyl halide will undergo S_N1 reaction fastes…

The correct answer is Option (4) → \(Me_3C-I\)

An \( S_N1 \) reaction goes in two steps, and the first, the breaking of the C–X bond to form a carbocation, is the slow, rate determining step:

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

    \( \displaystyle \ce{(CH3)3C^+ + OH^- -> (CH3)3C-OH} \quad \text{(fast)} \)

All four compounds give the same tert-butyl carbocation, so the rate depends only on how easily the halide leaves. The C–X bond becomes weaker and the halide ion a better leaving group as the halogen gets larger:

    \( \displaystyle \ce{R-I} \gt \ce{R-Br} \gt \ce{R-Cl} \gt\gt \ce{R-F} \)

So \( \ce{(CH3)3C-I} \) undergoes the \( S_N1 \) reaction fastest.

Why the other options are incorrect

• Option (1) \(Me_3C-F\) — Incorrect. The C–F bond is the strongest carbon-halogen bond, so the fluoride is the slowest of all.

• Option (2) \(Me_3C-Cl\) — Incorrect. The C–Cl bond is stronger than C–Br and C–I, so the chloride reacts more slowly.

• Option (3) \(Me_3C-Br\) — Incorrect. Bromide is a good leaving group, but iodide is better, so the bromide comes second.

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.”

Note: The same/similar question has been asked in different years by NTA as per the details given below: