tert-Butyl bromide reacts with hydroxide ion to give tert-butyl alcohol by the SN1 mechanism. If the concentration of hydroxide ion is doubled while the concentration of tert-butyl bromide is kept the same, how does the rate of the reaction change?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → It remains the same
An SN1 reaction takes place in two steps, and the rate is set by the slower one. The hydroxide ion takes part only in the fast second step.
Step I (slow): the C–Br bond breaks to give a carbocation.
\( \displaystyle \ce{(CH3)3C-Br <=> (CH3)3C+ + Br-} \)
Step II (fast): the carbocation is attacked by the nucleophile.
\( \displaystyle \ce{(CH3)3C+ + OH- -> (CH3)3C-OH} \)
Rate law (first order): \( \text{Rate} = k[\ce{(CH3)3CBr}] \)
[OH−] does not appear in the rate law, so doubling it leaves the rate unchanged.
NCERT (2026-27 edition):
6.7.1 Reactions of Haloalkanes – “…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.”
“Step I is the slowest and reversible. … 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.”