The reaction (CH3)3CBr + OH– → (CH3)3COH + Br– occurs through SN1 mec…

The correct answer is Option (2) → formation of carbocation and is slowest and reversible.

Nucleophilic substitution in haloalkanes follows one of two mechanisms:

• S_N2: one step; the nucleophile attacks while the halide leaves. Rate = k[RX][Nu⁻]. Favoured by primary haloalkanes; order of reactivity 1° > 2° > 3°.

• S_N1: two steps through a carbocation. Rate = k[RX], independent of the nucleophile. Favoured by tertiary haloalkanes, because the tertiary carbocation is the most stable; order of reactivity 3° > 2° > 1°.

For 2-bromo-2-methylpropane the S_N1 steps are:

    Step I (slow): \( \displaystyle \ce{(CH3)3C-Br <=> (CH3)3C^+ + Br^-} \)
    Step II (fast): \( \displaystyle \ce{(CH3)3C^+ + OH^- -> (CH3)3C-OH} \)

In step I the polar C–Br bond breaks heterolytically. This needs energy (supplied by solvation of the ions in a polar protic solvent), so it is the slowest step and decides the rate of the whole reaction. The carbocation and bromide ion can recombine, so the step is reversible. Step II, the attack of \( \ce{OH^{-}} \) on the carbocation, is fast.

So step I is the formation of the carbocation, and it is the slowest and reversible step.

Why the other options are incorrect

• Option (1) formation of carbocation and is moderate and reversible — Incorrect. The ionisation is not of moderate speed; it is the slowest, rate-determining step.

• Option (3) formation of carbocation and is moderate and irreversible — Incorrect. The step is the slowest, and the ions can recombine, so it is reversible.

• Option (4) formation of carbocation and is fastest and irreversible — Incorrect. This describes step II better; the fast step is the attack of the nucleophile on the carbocation.

NCERT (2026-27 edition):
Substitution Nucleophilic Unimolecular (S_N1) – “S_N1 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.”

“It occurs in two steps. In step I, the polarised C—Br bond undergoes slow cleavage to produce a carbocation and a bromide ion. The carbocation thus formed is then attacked by nucleophile in step II to complete the substitution reaction.”