Which of the following can produce a racemic mixture on monobromination?

Options

A.
B.
C.
D. `CH_(3)-CH_(3)`

Explanation

To determine which compound can produce a racemic mixture upon monobromination, we need to analyze each option based on the presence of different types of hydrogen atoms. A racemic mixture is formed when a reaction produces two enantiomers in equal amounts, which typically occurs when there is a chiral center created during the reaction.

### Step-by-Step Solution:

1. **Understanding Monobromination**:
- Monobromination refers to the substitution of one hydrogen atom with a bromine atom in a hydrocarbon. This process can lead to the formation of different products depending on the structure of the hydrocarbon.

2. **Identifying Chiral Centers**:
- A chiral center is a carbon atom that is bonded to four different substituents. The formation of a racemic mixture typically occurs when a chiral center is generated during the reaction.

3. **Analyzing Each Compound**:
- **Option 1**: If the compound has only one type of hydrogen, monobromination will yield a single product, which will not be a racemic mixture.
- **Option 2**: Similar to option 1, if all hydrogens are equivalent, only one product will be formed, leading to no racemic mixture.
- **Option 3**: If the compound has different types of hydrogens (for example, a secondary carbon and a tertiary carbon), monobromination can lead to the formation of two different products, which can be enantiomers. This would result in a racemic mixture.
- **Option 4**: If the compound does not create a chiral center upon bromination, it will not produce a racemic mixture.

4. **Conclusion**:
- After analyzing the options, the compound in **Option 3** is the one that can produce a racemic mixture upon monobromination because it has different types of hydrogen atoms that can lead to the formation of enantiomers.

### Final Answer:
**Option 3** can produce a racemic mixture on monobromination.