Trans-2-phenyl-1-bromocyclopentane on reaction with alcoholic KOH produces

Options

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

Explanation

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

### Step-by-Step Solution:

1. **Identify the Structure of the Reactant**:
- Trans-2-phenyl-1-bromocyclopentane consists of a cyclopentane ring with a bromine atom and a phenyl group attached to the second carbon in a trans configuration. The structure can be represented as:
```
Br
|
H--C1--C2--C3
| |
Ph H
```

2. **Understand the Reaction Conditions**:
- The reaction is carried out in the presence of alcoholic KOH, which is a strong base. This environment favors elimination reactions, specifically dehydrohalogenation.

3. **Mechanism of the Reaction**:
- In the presence of alcoholic KOH, the base (OH⁻) abstracts a hydrogen atom from a beta carbon (the carbon adjacent to the carbon bearing the bromine). This process is known as E2 elimination.
- The bromine atom (X) is also removed simultaneously, resulting in the formation of a double bond.

4. **Identify the Beta Hydrogens**:
- In trans-2-phenyl-1-bromocyclopentane, the beta carbons (C2 and C3) have hydrogen atoms that can be eliminated. Since the bromine and the hydrogen that is being removed must be anti to each other (opposite sides), we can choose the appropriate hydrogen to eliminate.

5. **Perform the Elimination**:
- The elimination of HBr occurs, leading to the formation of a double bond between C1 and C2. The resulting product will have a double bond between the two carbons where the bromine and hydrogen were removed.

6. **Draw the Product**:
- The product formed is 3-phenylcyclopentene. The double bond is between C1 and C2, and the phenyl group is still attached to C2. The structure can be represented as:
```
Ph
|
H--C1=C2--C3
|
H
```

### Final Answer:
The product formed from the reaction of trans-2-phenyl-1-bromocyclopentane with alcoholic KOH is **3-phenylcyclopentene**.