Which of the following compounds exhibits stereoisomerism?

Options

A. `2`-Methylbutane
B. `3`- Methylbutane
C. `3`-Methylbutanoic acid
D. `2`-Methylbutanoic acid

Explanation

To determine which of the given compounds exhibits stereoisomerism, we will analyze each compound step by step.

### Step 1: Understand Stereoisomerism
Stereoisomerism refers to compounds that have the same molecular formula and connectivity of atoms but differ in the spatial arrangement of those atoms. This can include geometrical isomerism (cis/trans) and optical isomerism (enantiomers).

### Step 2: Analyze Each Compound
1. **2-Methylbutane**:
- Structure: CH3-CH(CH3)-CH2-CH3
- Analysis: This compound does not have a double bond or a ring structure, which means it cannot exhibit geometrical isomerism. Additionally, it does not have any chiral carbon (a carbon atom bonded to four different groups), so it cannot exhibit optical isomerism.
- Conclusion: **Does not exhibit stereoisomerism.**

2. **3-Methylbutane**:
- Structure: CH3-CH2-CH(CH3)-CH3
- Analysis: Similar to 2-methylbutane, this compound does not have a double bond or ring structure, and it also lacks a chiral carbon.
- Conclusion: **Does not exhibit stereoisomerism.**

3. **3-Methylbutanoic Acid**:
- Structure: CH3-CH(CH3)-CH2-COOH
- Analysis: This compound has a carboxylic acid group but still does not have a chiral carbon because the carbon at position 3 is attached to two identical methyl groups.
- Conclusion: **Does not exhibit stereoisomerism.**

4. **2-Methylbutanoic Acid**:
- Structure: CH3-CH(CH3)-CH2-COOH
- Analysis: In this compound, the carbon at position 2 is chiral because it is bonded to four different groups: a carboxylic acid (COOH), a methyl group (CH3), a hydrogen atom (H), and an ethyl group (CH2-CH3). Therefore, this compound can exhibit optical isomerism.
- Conclusion: **Exhibits stereoisomerism.**

### Final Conclusion
Among the given compounds, only **2-Methylbutanoic Acid** exhibits stereoisomerism due to the presence of a chiral carbon atom.