Compound A `C_(6)H_(12)` forms two monochloro isomers on photochemical chlorination. Its structure can be :

Options

A.
B.
C.
D.

Explanation

To solve the problem of identifying the structure of compound A (C₆H₁₂) that forms two monochloro isomers upon photochemical chlorination, we can follow these steps:

### Step 1: Determine the Degree of Unsaturation
The molecular formula C₆H₁₂ indicates that the compound is likely a saturated hydrocarbon (alkane) or has some degree of unsaturation (like a ring or double bond). The degree of unsaturation can be calculated using the formula:
\[ \text{Degree of Unsaturation} = \frac{(2C + 2 - H)}{2} \]
For C₆H₁₂:
\[ \text{Degree of Unsaturation} = \frac{(2 \times 6 + 2 - 12)}{2} = \frac{2}{2} = 1 \]
This indicates that there is one degree of unsaturation, which could be a ring or a double bond.

### Step 2: Identify Possible Structures
Since there is one degree of unsaturation, we can consider structures that are either cyclic or contain a double bond. The simplest structures to consider are cycloalkanes or alkenes.

1. **Cyclohexane (C₆H₁₂)**: A six-membered ring with no double bonds.
2. **Alkenes**: For example, 1-hexene or 2-hexene, but these typically yield more than two monochloro products due to multiple positions for substitution.

### Step 3: Analyze Chlorination of Cyclohexane
Let’s analyze cyclohexane (C₆H₁₂):
- Cyclohexane has 12 hydrogen atoms, and upon chlorination, the chlorine can replace hydrogen atoms.
- The hydrogen atoms can be classified based on their positions:
- All hydrogen atoms on the ring are equivalent due to symmetry, leading to the formation of only one monochloro product.

### Step 4: Analyze Chlorination of Methylcyclopentane
Next, let’s consider methylcyclopentane:
- This compound has a five-membered ring with one methyl group attached.
- The hydrogens can be classified as:
- Primary hydrogens (on the methyl group)
- Secondary hydrogens (on the ring)
- Tertiary hydrogens (adjacent to the methyl group)

When chlorinating methylcyclopentane:
1. Chlorination at the primary position yields one isomer.
2. Chlorination at the tertiary position yields another isomer.
3. Chlorination at the secondary positions does not yield new isomers due to symmetry.

### Step 5: Conclusion
From the analysis, we conclude that methylcyclopentane (C₆H₁₂) is a suitable candidate that yields exactly two monochloro isomers upon chlorination.

### Final Answer
The structure of compound A (C₆H₁₂) that forms two monochloro isomers on photochemical chlorination is **methylcyclopentane**.