Arrange the following compounds in the decreasing order of reactivity…

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

In an S_N1 reaction the slow, rate-deciding step is the ionisation of the C–X bond to a carbocation:

    \( \displaystyle \ce{(C6H5)3C-X -> (C6H5)3C^+ + X^-} \)

All four compounds give the same triphenylmethyl carbocation, so the rate depends only on how easily the C–X bond breaks. Down the group the C–X bond becomes longer and weaker (C–F strongest, C–I weakest), and the halide ion becomes a better leaving group (\( \ce{I^{-}} \) best, \( \ce{F^{-}} \) poorest). So the reactivity order is:

    \( \displaystyle \ce{(C6H5)3C-I} \gt \ce{(C6H5)3C-Br} \gt \ce{(C6H5)3C-Cl} \gt \ce{(C6H5)3C-F} \)

Shortcut: The iodide (A) reacts fastest and the fluoride (D) slowest; only Option (1) starts with (A) and ends with (D).

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.”