Consider the reaction
(CH3)(C2H5)(C6H5)C–Br+ KOH (aq) → (CH3)(C2H5)(C6H5)C–OH
Q.24
The above reaction is an example of
Options:
(A) addition reaction
(B) bimolecular elimination reaction (E2)
(C) unimolecular substitution reaction (SN1)
(D) bimolecular substitution reaction (SN2)
SN1 Reaction: Substitution of Tertiary Alkyl Bromide
Organic reaction mechanisms such as SN1, SN2, E1, and E2 are commonly
tested in competitive exams. This problem involves identifying the
correct mechanism based on the nature of the substrate, solvent,
and reaction conditions.
Given Reaction
(CH3)(C2H5)(C6H5)C–Br+ KOH (aq) → (CH3)(C2H5)(C6H5)C–OH
Step-by-Step Analysis
1. Nature of Substrate
The carbon attached to Br is bonded to three alkyl/aryl groups,
making it a tertiary alkyl bromide.
2. Nature of Solvent and Nucleophile
KOH in aqueous medium provides a polar protic solvent, which
stabilizes carbocations and favors SN1 reactions.
3. Carbocation Stability
The tertiary carbocation formed is highly stable due to
+I effect and resonance stabilization from the phenyl group.
4. Reaction Outcome
Br− leaves first (rate-determining step), followed by
attack of OH− on the carbocation to form alcohol.
Reaction Mechanism
Unimolecular Nucleophilic Substitution (SN1)
Correct Answer
Option (C): Unimolecular substitution reaction (SN1)
SN1 vs SN2 Comparison
| Feature | SN1 | SN2 |
|---|---|---|
| Substrate | Tertiary | Primary |
| Intermediate | Carbocation | None |
| Rate Depends On | Alkyl halide | Alkyl halide + Nucleophile |
| Favored Here | Yes | No |
Conclusion
The reaction involves a tertiary alkyl bromide in aqueous KOH,
proceeds via a stable carbocation intermediate, and results in
substitution. Hence, it clearly follows the SN1 mechanism.
Therefore, the correct answer is Option (C).
