An isomer of `C_(6)H_(14)` forms three monochloro derivaties. The isomer may be- (Excluding stereo isomer)

Options

A. neo-pentane
B. n-hexane
C. 2, 3-dimethylbutane
D. iso-hexane

Answer

To solve the problem, we need to identify an isomer of C₆H₁₄ that can form exactly three distinct monochloro derivatives.

### Step-by-Step Solution:

1. **Understanding the Molecular Formula**:
- The molecular formula C₆H₁₄ indicates a saturated hydrocarbon (alkane) with no double bonds or rings. This means it can be a straight-chain or branched-chain alkane.

2. **Identifying Possible Isomers**:
- The isomers of C₆H₁₄ include:
- n-Hexane (straight-chain)
- 2-Methylpentane
- 3-Methylpentane
- 2,2-Dimethylbutane
- 2,3-Dimethylbutane
- 3,3-Dimethylbutane
- We need to determine which of these can produce exactly three monochloro derivatives.

3. **Monochlorination Reaction**:
- Monochlorination involves replacing one hydrogen atom with a chlorine atom using Cl₂ in the presence of light (UV light).
- The number of distinct monochloro products depends on the number of unique hydrogen atoms available for substitution.

4. **Analyzing n-Hexane**:
- Structure: CH₃-CH₂-CH₂-CH₂-CH₂-CH₃
- All hydrogen atoms are equivalent, so monochlorination will yield:
- 1-chlorohexane (from the terminal CH₃)
- 2-chlorohexane (from the second carbon)
- 3-chlorohexane (from the third carbon)
- Total distinct products: **3**

5. **Analyzing 2-Methylpentane**:
- Structure: CH₃-CH(CH₃)-CH₂-CH₂-CH₃
- This structure has different types of hydrogen atoms:
- 1-chloro-2-methylpentane
- 1-chloro-3-methylpentane
- 2-chloro-2-methylpentane
- 2-chloro-3-methylpentane
- Total distinct products: **4**

6. **Analyzing 3-Methylpentane**:
- Structure: CH₃-CH₂-CH(CH₃)-CH₂-CH₃
- This structure also has different types of hydrogen atoms:
- 1-chloro-3-methylpentane
- 2-chloro-3-methylpentane
- 3-chloro-3-methylpentane
- 2-chloro-2-methylpentane
- Total distinct products: **4**

7. **Analyzing 2,2-Dimethylbutane**:
- Structure: CH₃-C(CH₃)(CH₂)-CH₃
- This structure has fewer unique hydrogen atoms:
- 1-chloro-2,2-dimethylbutane
- 2-chloro-2,2-dimethylbutane
- Total distinct products: **2**

8. **Analyzing 2,3-Dimethylbutane**:
- Structure: CH₃-CH(CH₃)-C(CH₃)-CH₃
- This structure has different types of hydrogen atoms:
- 1-chloro-2,3-dimethylbutane
- 2-chloro-2,3-dimethylbutane
- 3-chloro-2,3-dimethylbutane
- Total distinct products: **3**

9. **Analyzing 3,3-Dimethylbutane**:
- Structure: CH₃-CH(CH₃)-C(CH₃)-CH₃
- This structure has fewer unique hydrogen atoms:
- 1-chloro-3,3-dimethylbutane
- 2-chloro-3,3-dimethylbutane
- Total distinct products: **2**

### Conclusion:
The isomer of C₆H₁₄ that can form exactly three monochloro derivatives is **n-hexane** and **2,3-dimethylbutane**.