A compound `(P)` having molecular formula `C_(6)H_(10)` contains two `DU`. It yields `Q(C_(6)H_(12))` when reacts with excess of `H_(2)` of the presence of `Ni`. On ozonolysis `P` gives cyclopentanone and compound `Y`.
The compound `Q` gives the number of of monochlorination products

Options

A. `3`
B. `4`
C. `5`
D. `6`

Explanation

To solve the problem step by step, we will analyze the information given about compound P, its reactions, and the resulting compound Q.

### Step 1: Analyze the Molecular Formula and Degree of Unsaturation
- The molecular formula of compound P is \( C_6H_{10} \).
- The degree of unsaturation (DU) is calculated using the formula:
\[ \text{DU} = \frac{(2C + 2 + N - H - X)}{2} \]
where \( C \) is the number of carbons, \( H \) is the number of hydrogens, \( N \) is the number of nitrogens, and \( X \) is the number of halogens.
- For \( C_6H_{10} \):
\[ \text{DU} = \frac{(2 \times 6 + 2 - 10)}{2} = \frac{(12 + 2 - 10)}{2} = \frac{4}{2} = 2 \]
- This indicates that compound P has either two double bonds, one ring and one double bond, or two rings.

**Hint for Step 1:** Remember that the degree of unsaturation helps determine the presence of rings or multiple bonds in the structure.

### Step 2: Determine the Structure of Compound P
- Given that P yields cyclopentanone and another compound Y upon ozonolysis, we can deduce that P likely contains a cyclopentene structure.
- If we assume P is a cyclic compound with a double bond, one possible structure is 1-cyclopentene with a substituent that allows for the formation of cyclopentanone upon ozonolysis.

**Hint for Step 2:** Ozonolysis can help identify the fragments produced, which gives clues about the structure of the original compound.

### Step 3: Reaction of Compound P with Hydrogen
- When P reacts with excess hydrogen (H2) in the presence of nickel, it becomes saturated, resulting in compound Q with the formula \( C_6H_{12} \).
- This indicates that all double bonds in P are reduced, leading to a fully saturated compound Q.

**Hint for Step 3:** The presence of a catalyst like nickel allows for the hydrogenation of double bonds, converting alkenes to alkanes.

### Step 4: Determine the Structure of Compound Q
- The structure of Q can be derived from the structure of P. If P is 1-cyclopentene, Q could be cyclohexane or a substituted cyclohexane depending on the substituents on the cyclopentene.

**Hint for Step 4:** Visualizing the structure of Q helps in identifying potential sites for chlorination.

### Step 5: Count Unique Monochlorination Sites in Compound Q
- To find the number of unique monochlorination products, we need to identify the unique hydrogen atoms in Q that can be replaced by chlorine.
- If Q is cyclohexane, it has 6 hydrogen atoms, but due to symmetry, not all chlorination will yield unique products.
- The unique chlorination sites can be counted as follows:
- If Q is cyclohexane, there are 6 positions, but due to symmetry, there are 2 unique sites (1,2- and 1,3- positions).

**Hint for Step 5:** Consider the symmetry of the molecule when counting unique chlorination sites, as identical positions will yield the same product.

### Conclusion
- The number of unique monochlorination products for compound Q is 4.

**Final Answer:** The compound Q gives 4 unique monochlorination products.