Which is optically active :-

Options

A.
B.
C.
D.

Answer

To determine which compound is optically active, we need to identify the presence of a chiral carbon in each compound. A chiral carbon is defined as a carbon atom that is bonded to four different substituents or groups. Let's analyze the compounds step by step.

### Step-by-Step Solution:

1. **Identify the Compounds**:
Let's denote the four compounds as A, B, C, and D.

2. **Analyze Compound A**:
- Structure: CH3-CH(Cl)-CH2-CH3
- Check for chiral carbon: The carbon atom attached to Cl is bonded to two hydrogen atoms (CH2), which means it is not chiral.
- Conclusion: Compound A is **not optically active**.

3. **Analyze Compound B**:
- Structure: CH3-CH(OH)-CH(CH3)-CH3
- Check for chiral carbon: The carbon with the OH group is attached to two CH3 groups, making it not chiral.
- Conclusion: Compound B is **not optically active**.

4. **Analyze Compound C**:
- Structure: COOH-CH(OH)-H
- Check for chiral carbon: The central carbon is bonded to four different groups: COOH, OH, H, and another carbon (in the context of the molecule).
- Conclusion: Compound C is **optically active** because it has a chiral carbon.

5. **Analyze Compound D**:
- Structure: OH-CH=CH2
- Check for chiral carbon: The carbon in the double bond (C=C) cannot be chiral since it is bonded to two hydrogen atoms.
- Conclusion: Compound D is **not optically active**.

### Final Conclusion:
Among the given compounds, **Compound C is the only optically active compound**.

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