Which of the following compounds shows optical isomerism?

Options

A. `CH_(3)CHCl-CH_(2)-CH_(3)`
B. `CH_(3)-CH_(2)-CHCl-CH_(2)-CH_(3)`
C. `ClCH_(2)-CH_(2)-CH_(2)-CH_(3)`
D. `ClCH_(2)-CH_(2)-CH_(3)`

Explanation

To determine which of the given compounds shows optical isomerism, we need to identify if any of the compounds have chiral centers. A chiral center is typically a carbon atom that is bonded to four different groups, leading to non-superimposable mirror images (enantiomers).

### Step-by-Step Solution:

1. **Identify the Compounds**: We have four compounds to analyze. Let's denote them as A, B, C, and D for simplicity.

2. **Analyze Compound A**:
- Structure: CH3-CH-CL-CH2-CH3
- Check for Chiral Centers: The carbon (CH) is attached to four different groups:
- CH3 (methyl group)
- CL (chlorine)
- CH2 (part of the ethyl chain)
- CH3 (another part of the ethyl chain)
- Since all four groups are different, this carbon is a chiral center.
- Conclusion: Compound A is optically active.

3. **Analyze Compound B**:
- Structure: CH3-CH2-CH-CL-CH2-CH3
- Check for Chiral Centers: The carbon (CH) is attached to:
- CH3 (methyl group)
- CH2 (ethyl group)
- CH2 (another ethyl group)
- CL (chlorine)
- Here, two of the groups are the same (the two CH2 groups). Therefore, this carbon is not a chiral center.
- Conclusion: Compound B is not optically active.

4. **Analyze Compound C**:
- Structure: CL-CH2-CH2-CH2-CH3
- Check for Chiral Centers: The carbon atoms in this compound are either bonded to hydrogen or other carbon chains, and there are no carbons with four different substituents.
- Conclusion: Compound C is not optically active.

5. **Analyze Compound D**:
- Structure: CL-CH2-CH2-CH3
- Check for Chiral Centers: Similar to compound C, there are no chiral centers present as all substituents are not different.
- Conclusion: Compound D is not optically active.

### Final Conclusion:
The only compound that shows optical isomerism is **Compound A**.