Which of the following compounds will give racemic mixture on nucleophilic substitution by `OH^(-)` ion?
`(I) CH_3-underset(C_2H_5)underset(|)"CH"-Br`
(II) `CH_3-underset(C_2H_5)underset(|)overset(Br)overset(|)C-CH_3`

Options

A. (I)
B. (II)
C. (I),(II)
D. None

Explanation

To determine which of the given compounds will yield a racemic mixture upon nucleophilic substitution by the `OH^(-)` ion, we need to analyze the structure of each compound and the mechanism of the nucleophilic substitution reaction.

### Step-by-Step Solution:

1. **Identify the Compounds**:
- Compound I: `CH3-C(C2H5)-CH(Br)-C2H5`
- Compound II: `CH3-C(C2H5)(Br)-C(CH3)-C2H5`

2. **Determine the Type of Substitution**:
- Nucleophilic substitution reactions can proceed via two main mechanisms:
- **SN1 Mechanism**: Involves the formation of a carbocation intermediate.
- **SN2 Mechanism**: Involves a direct attack by the nucleophile, leading to inversion of configuration.

3. **Analyze Compound I**:
- The bromine atom is attached to a carbon that is tertiary (due to the presence of two ethyl groups and one methyl group).
- Upon the departure of the bromine, a tertiary carbocation is formed.
- This carbocation is capable of rearranging and can lead to the formation of two different enantiomers when the `OH^(-)` ion attacks.
- Therefore, Compound I will give a racemic mixture.

4. **Analyze Compound II**:
- The bromine atom is attached to a carbon that is also tertiary (due to the presence of two ethyl groups and one methyl group).
- Similar to Compound I, the departure of bromine leads to the formation of a tertiary carbocation.
- However, the presence of two identical groups (the two ethyl groups) means that the attack by the `OH^(-)` ion will not lead to different enantiomers.
- Therefore, Compound II will not give a racemic mixture.

5. **Conclusion**:
- Only Compound I will give a racemic mixture upon nucleophilic substitution by the `OH^(-)` ion.

### Final Answer:
- **Compound I** will give a racemic mixture upon nucleophilic substitution by `OH^(-)` ion.