Trans - 2 phenyl 1 - bromocyclopenta ne on reaction with alcoholic KOH produces

Options

A. 4 - phenylcyclopentene
B. 2 - phenylcyclopentene
C. 1 - phenylcyclopentene
D. 3 - phenylcyclopentene

Answer

To solve the question regarding the reaction of trans-2-phenyl-1-bromocyclopentane with alcoholic KOH, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure**:
- Trans-2-phenyl-1-bromocyclopentane consists of a cyclopentane ring with a bromine atom at the first carbon and a phenyl group at the second carbon.
- Draw the cyclopentane ring and place the bromine and phenyl groups accordingly.

2. **Understand the Reaction Conditions**:
- The reaction is carried out with alcoholic KOH. Alcoholic KOH is a strong base, which suggests that it will facilitate elimination reactions rather than substitution reactions.

3. **Mechanism of Reaction**:
- The reaction follows an E2 elimination mechanism. In this mechanism, the base (OH⁻ from alcoholic KOH) abstracts a proton (H) from a carbon adjacent to the carbon bearing the bromine atom.
- Since we have a trans configuration, the hydrogen that will be removed is anti to the bromine.

4. **Elimination of Bromine**:
- The removal of the hydrogen leads to the formation of a double bond between the carbon that lost the hydrogen and the carbon that was attached to the bromine.
- The bromine atom will leave as a bromide ion (Br⁻).

5. **Final Structure**:
- After the elimination, the resulting compound will have a double bond between the first and second carbon atoms of the cyclopentane ring.
- The structure of the final product is 3-phenylcyclopentene, where the phenyl group remains attached to the second carbon.

6. **Conclusion**:
- The final product of the reaction of trans-2-phenyl-1-bromocyclopentane with alcoholic KOH is 3-phenylcyclopentene.

### Final Answer:
The product formed is **3-phenylcyclopentene**.