An alkane `C_(6)H_(14)` gives two monochloro derivatives on chlorination. Its possible structure is

Options

A. `CH_(3)CH_(2)CH_(2)CH_(2)CH_(2)CH_(3)`
B. `CH_(3)-underset(CH_(3))underset(|)(C)H-CH_(2)CH_(2)CH_(3)`
C. `CH_(3)-underset(CH_(2)CH_(3))underset(|)(C)H-CH_(2)CH_(3)`
D. `CH_(3)-overset(CH_(3))overset(|)(C)H-overset(CH_(3))overset(|)(C)H-CH_(3)`

Explanation

To solve the question regarding the alkane \( C_6H_{14} \) that gives two monochloro derivatives upon chlorination, we will analyze the possible structures of \( C_6H_{14} \) and determine which one fits the criteria of yielding only two distinct monochloro derivatives.

### Step-by-Step Solution:

1. **Understanding Chlorination**:
- Chlorination of alkanes involves the substitution of hydrogen atoms with chlorine atoms. Each unique hydrogen atom replaced by chlorine will yield a different monochloro derivative.

2. **Identifying the Structure of \( C_6H_{14} \)**:
- The molecular formula \( C_6H_{14} \) corresponds to several structural isomers. We need to find an isomer that has only two distinct types of hydrogen atoms.

3. **Analyzing the Options**:
- We will evaluate the provided options to identify which structure can yield only two monochloro derivatives.

4. **Option D**:
- The structure is \( CH_3CH(CH_3)CH(CH_3)CH_3 \) (2,4-dimethylpentane).
- In this structure, there are two types of hydrogen atoms:
- \( CH_3 \) (methyl groups)
- \( CH \) (the two central carbons)
- Chlorination will replace one of the \( CH \) hydrogens or one of the \( CH_3 \) hydrogens, yielding two distinct monochloro derivatives.

5. **Verifying Other Options**:
- **Option C**: This structure has three different types of hydrogen atoms, leading to three possible monochloro derivatives.
- **Option B**: This structure also has three different types of hydrogen atoms, leading to three possible monochloro derivatives.
- **Option A**: This is a straight-chain alkane (hexane) and will yield multiple monochloro derivatives as well.

6. **Conclusion**:
- The only structure that yields exactly two monochloro derivatives is Option D (2,4-dimethylpentane).

### Final Answer:
The possible structure of the alkane \( C_6H_{14} \) that gives two monochloro derivatives on chlorination is **Option D: 2,4-dimethylpentane**.