In the `S_(N)2` reaction of cis -3- methylcyclo pentyl bromide with alkali the product formed is

Options

A. A cis alcohol
B. A trans alcohol
C. An equimolecular mixture of cis and trans alcohol
D. There is no reaction

Answer

To solve the problem regarding the `S_N2` reaction of cis-3-methylcyclopentyl bromide with alkali, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of the Reactant:**
- The reactant is cis-3-methylcyclopentyl bromide.
- This means we have a cyclopentane ring with a bromine atom attached to one of the carbons and a methyl group at the 3rd position in a cis configuration.

2. **Draw the Structure:**
- Draw the cyclopentane ring and label the carbons.
- Place the bromine atom on one carbon and the methyl group on the adjacent carbon in a cis arrangement.

3. **Understand the `S_N2` Mechanism:**
- The `S_N2` reaction involves a nucleophile attacking the electrophilic carbon from the opposite side of the leaving group (bromine in this case).
- This results in the inversion of configuration at the carbon center where the bromine is attached.

4. **Identify the Nucleophile:**
- In this case, the nucleophile is hydroxide ion (OH⁻) from the alkali (e.g., NaOH).

5. **Perform the Reaction:**
- The hydroxide ion will attack the carbon that is bonded to the bromine from the backside.
- As a result, the bromine will leave, and the hydroxyl group (OH) will take its place.

6. **Determine the Configuration of the Product:**
- Since the original configuration is cis, the inversion during the `S_N2` reaction will lead to the formation of a trans alcohol.

7. **Identify the Product:**
- The product formed will be trans-3-methylcyclopentanol.

8. **Select the Correct Option:**
- Based on the options provided, the correct answer is the one that indicates the formation of a trans alcohol.

### Final Answer:
The product formed in the `S_N2` reaction of cis-3-methylcyclopentyl bromide with alkali is **trans-3-methylcyclopentanol**.