Which of the following compounds exhibit optical isomerism?

Options

A.
B. `H_(2)C=C=CH_(2)`
C.
D.

Explanation

To determine which of the given compounds exhibit optical isomerism, we need to identify if any of the compounds contain a chiral carbon atom. A chiral carbon is defined as a carbon atom that is bonded to four different substituents, leading to non-superimposable mirror images (enantiomers).

### Step-by-Step Solution:

1. **Identify the Compounds**: Review the provided compounds to see their structures. We need to analyze each compound for the presence of a chiral carbon.

2. **Check for Chiral Centers**: For each compound, locate the carbon atoms and check if any of them are bonded to four different groups or atoms.

- **Compound 1**:
- Consider a specific carbon atom in the compound.
- This carbon is bonded to a carboxylic acid group (COOH), a hydrogen atom (H), a CH2 group, and a CH group.
- Since all four substituents are different, this carbon is a chiral center.

- **Compound 2**:
- Check the carbon atoms in this compound.
- If any carbon has a double bond or is bonded to identical groups, it cannot be chiral.
- This compound has a double bond, which eliminates the possibility of having a chiral center.

- **Compound 3**:
- Similar to Compound 2, check for double bonds or identical groups.
- If a carbon is bonded to two of the same groups or has a double bond, it cannot be chiral.

- **Compound 4**:
- Again, check for the presence of double bonds or identical groups.
- If the structure shows any double bonds or symmetry, it cannot have a chiral center.

3. **Conclusion**: After analyzing all compounds, only **Compound 1** exhibits optical isomerism due to the presence of a chiral carbon atom with four different substituents. The other compounds do not meet the criteria for optical isomerism.

### Final Answer:
**Only Compound 1 exhibits optical isomerism.**