Which of the following compounds exhibits stereoisomerism?

Options

A. `2`-Methyl butene
B. `3`-Methyl butyne
C. `3`-Methyl butanoic acid
D. `2`-Methyl butanoic acid

Explanation

To determine which of the given compounds exhibits stereoisomerism, we need to follow these steps:

### Step 1: Understand Stereoisomerism
Stereoisomerism occurs when compounds have the same connectivity of atoms but differ in the spatial arrangement of those atoms. This can lead to different physical and chemical properties.

**Hint:** Remember that stereoisomers include enantiomers and diastereomers, which arise from chiral centers or geometric arrangements.

### Step 2: Identify Chiral Centers
A chiral center (or chiral carbon) is a carbon atom that is attached to four different groups. The presence of at least one chiral center in a compound typically indicates that it can exhibit stereoisomerism.

**Hint:** Look for carbon atoms that have four different substituents attached to them.

### Step 3: Analyze the Given Compounds
Let’s analyze the given compound, 2-methylbutanoic acid:
- The structure of 2-methylbutanoic acid can be drawn to identify any chiral centers.
- Upon drawing, we find that the second carbon in the chain has four different groups attached (a methyl group, a carboxylic acid group, a hydrogen, and the rest of the carbon chain).

**Hint:** When drawing the structure, label each substituent clearly to see if they are all different.

### Step 4: Confirm the Existence of Stereoisomers
Since 2-methylbutanoic acid has a chiral center, it can exist as two enantiomers (S and R configurations). This confirms that it exhibits stereoisomerism.

**Hint:** Remember that each chiral center can lead to two configurations (R and S), which are mirror images of each other.

### Step 5: Conclusion
Based on the analysis, 2-methylbutanoic acid is the compound that exhibits stereoisomerism due to the presence of a chiral carbon.

**Final Answer:** The correct option for the compound that exhibits stereoisomerism is **2-methylbutanoic acid**.