The order of kinetics of S_N1 reactions is:

The correct answer is Option (1) → First

An \(S_N1\) reaction (substitution, nucleophilic, unimolecular) takes place in two steps. In the first, slow step the C–X bond breaks and a carbocation is formed; in the second, fast step the nucleophile attacks the carbocation. For tert-butyl bromide and hydroxide ion:

    \( \displaystyle \ce{(CH3)3C-Br} \xrightarrow{\text{slow}} \ce{(CH3)3C+ + Br^-} \)

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

The rate is decided by the slow step, which involves only the alkyl halide:

    \( \displaystyle \text{Rate} = k[\ce{(CH3)3CBr}] \)

This is first order kinetics. In an \( \mathrm{S_N2} \) reaction, by contrast, the nucleophile attacks in the same single step in which the halide leaves, so the rate is second order, Rate = k[RX][Nu⁻].

NCERT (2026-27 edition):
Substitution Nucleophilic Unimolecular (S_N1)