Which of the following cyclopentane derivative is optically inactive ?

Options

A.
B.
C.
D.

Explanation

To determine which cyclopentane derivative is optically inactive, we need to identify the presence of a plane of symmetry in the given compounds. A compound is optically inactive if it has a plane of symmetry, meaning that one half of the molecule is a mirror image of the other half.

### Step-by-Step Solution:

1. **Understand Optical Activity**:
- Optical activity refers to the ability of a compound to rotate plane-polarized light. A compound is optically active if it lacks a plane of symmetry and has chiral centers.

2. **Identify the Compounds**:
- We need to analyze the given cyclopentane derivatives. Typically, these compounds will have different substituents attached to the cyclopentane ring.

3. **Analyze Each Compound for Symmetry**:
- For each compound, check if there is a plane of symmetry:
- **Compound A**: If there are substituents that are not symmetrically placed (e.g., OH above and H below), it lacks a plane of symmetry.
- **Compound B**: Similar analysis as Compound A. If substituents are not arranged symmetrically, it is optically active.
- **Compound C**: If both substituents (e.g., CH3) are positioned symmetrically (both above or below), this compound has a plane of symmetry, making it optically inactive.
- **Compound D**: Check for symmetry. If substituents are different or not symmetrically arranged, it is optically active.

4. **Determine the Optically Inactive Compound**:
- After analyzing all compounds, we find that **Compound C** has a plane of symmetry due to its symmetrical arrangement of substituents. Therefore, it is optically inactive.

5. **Conclusion**:
- The correct answer is **Compound C**, as it possesses a plane of symmetry and is thus optically inactive.