A hydrocarbon `A(C_(10)H_(12))` has no chiral carbon. A gives a white precipitate with ammonical solution of silver nitrate. A on treatment with `H_(2)//pt` gives `B(C_(10)H_(20))`. A on ozonolysis gives `C(C_(8)H_(12)O_(4))` as one product which on heating with soda lime gives `D(C_(6)H_(12))`. `D` on monochlorination with `Cl_(2)//hv` gives `C_(6)H_(11)Cl` as sole isomer. Identify `A` to `D`.

Answer

To solve the problem step by step, we will identify the compounds A to D based on the information provided.

### Step 1: Identify Compound A (C₁₀H₁₂)
Given that A is a hydrocarbon with the formula C₁₀H₁₂ and has no chiral carbon, we can infer that it is likely a cyclic compound. The presence of fewer hydrogen atoms than expected for an alkane suggests that A is a cyclic alkyne. The fact that it gives a white precipitate with ammonical silver nitrate indicates that A is a terminal alkyne.

**Hint:** Look for a cyclic structure that can accommodate a terminal triple bond.

### Step 2: Identify Compound B (C₁₀H₂₀)
When A is treated with hydrogen (H₂) in the presence of a platinum catalyst (Pt), it undergoes hydrogenation to form B, which has the formula C₁₀H₂₀. This indicates that B is a saturated hydrocarbon (alkane).

**Hint:** Remember that hydrogenation adds hydrogen atoms and removes any double or triple bonds.

### Step 3: Identify Compound C (C₈H₁₂O₄)
A undergoes ozonolysis to give product C (C₈H₁₂O₄). Ozonolysis of alkynes typically results in the formation of carboxylic acids. Since C has 8 carbons and 4 oxygen atoms, it likely contains two carboxylic acid groups.

**Hint:** Consider how ozonolysis cleaves the triple bond and adds oxygen to form carboxylic acids.

### Step 4: Identify Compound D (C₆H₁₂)
When C is heated with soda lime, it undergoes decarboxylation to yield D (C₆H₁₂). The presence of only one monochlorination product from D indicates that D is likely a symmetrical compound, such as cyclohexane.

**Hint:** Decarboxylation removes CO₂ from carboxylic acids, reducing the carbon count.

### Step 5: Monochlorination of D
D (C₆H₁₂) is monochlorinated with Cl₂ in the presence of light (hv), yielding a single isomer of C₆H₁₁Cl. This suggests that D is a symmetrical compound, confirming that D is cyclohexane.

**Hint:** The formation of only one isomer during chlorination indicates a symmetrical structure.

### Summary of Compounds
- **A (C₁₀H₁₂)**: A terminal cyclic alkyne, likely 1-cyclodecyne.
- **B (C₁₀H₂₀)**: The hydrogenated product of A, which is a saturated hydrocarbon (decane).
- **C (C₈H₁₂O₄)**: A compound with two carboxylic acid groups, likely 1,2-dicarboxylic acid derived from ozonolysis.
- **D (C₆H₁₂)**: Cyclohexane, which is formed from the decarboxylation of C.

### Final Identification
- **A**: 1-Cyclodecyne
- **B**: Decane
- **C**: 1,2-Dicarboxylic acid (e.g., succinic acid)
- **D**: Cyclohexane