Which of the following compounds will give racemic mixture on nucleophillc substitution.
(a) `H_3C-underset(C_2H_5)underset(|) (CH)-Br` (b) `CH_3-underset(C_2H_5)underset(|)overset(Br)overset(|)C-CH_3`
(c) `CH_2-underset(C_2H_5)underset(|)(CH)-CH_2Br`

Options

A. a
B. a,b,c
C. b,c
D. a,c

Explanation

To determine which of the given compounds will yield a racemic mixture upon nucleophilic substitution, we need to analyze each compound for the presence of a chiral center and the mechanism of the reaction (SN1 or SN2).

### Step-by-Step Solution:

1. **Identify Chiral Centers**:
- A chiral center is a carbon atom that is bonded to four different groups. This is crucial for the formation of a racemic mixture, as both enantiomers (R and S) will be produced if the reaction proceeds through a carbocation intermediate (SN1 mechanism).

2. **Examine Option (a)**:
- Compound: `H3C-underset(C2H5)underset(|) (CH)-Br`
- The carbon atom bonded to the bromine (Br) is attached to four different groups: a methyl group (CH3), an ethyl group (C2H5), a hydrogen (H), and the Br itself.
- This carbon is a chiral center.
- Since the reaction will proceed via the SN1 mechanism (due to the formation of a stable carbocation), this will lead to the formation of both R and S enantiomers, resulting in a racemic mixture.

3. **Examine Option (b)**:
- Compound: `CH3-underset(C2H5)underset(|)overset(Br)overset(|)C-CH3`
- The carbon attached to Br is also bonded to two identical methyl groups (CH3).
- Since there are not four different groups attached to this carbon, it is not a chiral center.
- Therefore, this compound will not produce a racemic mixture.

4. **Examine Option (c)**:
- Compound: `CH2-underset(C2H5)underset(|)(CH)-CH2Br`
- The carbon attached to Br is bonded to two hydrogen atoms (CH2) and two different groups (C2H5 and CH).
- This carbon does not have four different substituents, hence it is not a chiral center.
- The reaction will proceed via the SN2 mechanism, which results in inversion of configuration rather than a racemic mixture.

5. **Conclusion**:
- Only option (a) will yield a racemic mixture upon nucleophilic substitution due to the presence of a chiral center and the SN1 mechanism.

### Final Answer:
The compound that will give a racemic mixture on nucleophilic substitution is **(a)** `H3C-underset(C2H5)underset(|) (CH)-Br`.