Substitution nucleophilic unimolecular \((S_N1)\) reaction is a two step process carried out in polar protic solvents. The reaction \((CH_3)_3C-Br + OH^- \rightarrow (CH_3)_3COH + Br^-\) follows:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → First order
An SN1 reaction takes place in two steps:
Step I (slow): \( \ce{(CH3)3C-Br} \) → \( \ce{(CH3)3C^{+}} \) + \( \ce{Br^{-}} \)
Step II (fast): \( \ce{(CH3)3C^{+}} \) + \( \ce{OH^{-}} \) → \( \ce{(CH3)3C-OH} \)
The slow ionisation (step I) is the rate-determining step and involves only the alkyl halide, so
Rate = k[(CH3)3CBr]
The rate does not depend on \( \ce{[OH^{-}]} \), so the reaction follows first order kinetics (hence “unimolecular”).
Why the other options are incorrect
Option (1) Zero order — Incorrect. The rate depends on the concentration of tert-butyl bromide, so it is not zero order.
Option (3) Second order — Incorrect. Second order kinetics (rate = k[RX][OH−]) is shown by SN2 reactions, in which the nucleophile takes part in the slow step.
Option (4) Pseudo first order — Incorrect. Pseudo first order applies when a second reactant is in large excess (e.g. hydrolysis of ethyl acetate in water); here the rate is truly independent of \( \ce{OH^{-}} \).
NCERT (2026-27 edition):
Substitution Nucleophilic Unimolecular (SN1) – “SN1 reactions are generally carried out in polar protic solvents (like water, alcohol, acetic acid, etc.). The reaction between tert-butyl bromide and hydroxide ion yields tert-butyl alcohol and follows the first order kinetics, i.e., the rate of reaction depends upon the concentration of only one reactant, which is tert-butyl bromide.”