Which of the following will have one of the stereoisomer meso ?

Options

A. 2-chlorobutane
B. 2, 3-dichlorobutane
C. 2,3-dichloropentane
D. 2-hydroxypropanoic acid

Explanation

To determine which of the given compounds has a stereoisomer that is meso, we need to analyze each compound for the presence of chiral centers and symmetry.

### Step 1: Analyze 2-chlorobutane
- **Structure**: 2-chlorobutane has the following structure:
- CH3-CH(Cl)-CH2-CH3
- **Chirality**: The carbon at position 2 (C-2) is a chiral center because it is attached to four different groups (CH3, Cl, H, and CH2-CH3).
- **Symmetry**: There is no plane of symmetry in 2-chlorobutane.
- **Conclusion**: 2-chlorobutane does not have a meso form.

### Step 2: Analyze 2,3-dichlorobutane
- **Structure**: 2,3-dichlorobutane has the following structure:
- CH3-CH(Cl)-CH(Cl)-CH3
- **Chirality**: Both C-2 and C-3 are chiral centers.
- **Symmetry**: The molecule has a plane of symmetry, which makes it a meso compound.
- **Conclusion**: 2,3-dichlorobutane has a meso form.

### Step 3: Analyze 2-hydroxy pentanoic acid
- **Structure**: 2-hydroxy pentanoic acid has the following structure:
- CH3-CH(OH)-CH2-CH2-COOH
- **Chirality**: The carbon at position 2 (C-2) is a chiral center.
- **Symmetry**: However, the molecule does not have a plane of symmetry.
- **Conclusion**: 2-hydroxy pentanoic acid does not have a meso form.

### Step 4: Analyze 2,4-dimethylpentane
- **Structure**: 2,4-dimethylpentane has the following structure:
- CH3-CH(CH3)-CH2-CH(CH3)-CH3
- **Chirality**: The carbon at position 2 and 4 are both chiral centers.
- **Symmetry**: There is no plane of symmetry in this compound.
- **Conclusion**: 2,4-dimethylpentane does not have a meso form.

### Final Conclusion:
Out of all the compounds analyzed, only **2,3-dichlorobutane** has a meso stereoisomer due to its internal plane of symmetry.

### Answer:
**2,3-dichlorobutane** is the compound that has a meso stereoisomer.

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