Consider the reaction
(CH3)(C2H5)(C6H5)C–Br+ KOH (aq)→ (CH3)(C2H5)(C6H5)C–OH
Q.25
If the concentration of KOH in the reaction mixture is doubled,
the rate of the reaction will be
Options:
(A) decreased to one-half
(B) the same
(C) increased by two-times
(D) increased by four-times
Effect of KOH Concentration on Reaction Rate in SN1 Reaction
In organic chemistry, the rate of a reaction depends on the reaction mechanism.
This problem analyzes the effect of doubling the concentration of aqueous KOH
on a substitution reaction involving a tertiary alkyl halide.
Given Reaction
(CH3)(C2H5)(C6H5)C–Br+ KOH(aq)→(CH3)(C2H5)(C6H5)C–OH
Correct Answer
(B) The same
Reaction Mechanism Explanation
The given compound is a tertiary alkyl bromide. Tertiary alkyl
halides generally undergo SN1 reactions because
steric hindrance prevents SN2 attack.
In an SN1 reaction, the rate-determining step is the formation of a
carbocation:
Rate ∝ [Alkyl Halide]
The nucleophile (OH−) does not participate in the rate-determining step.
Therefore, changing the concentration of KOH does not affect the reaction rate.
Final Conclusion
Since the reaction follows an SN1 mechanism, doubling the concentration
of KOH has no effect on the rate of reaction.
Correct Option: (B) The same
