A hydrocarbon, `‘X’(C_(3)H_(4))` decolourise bromine in carbon tetrachloride forming ‘Y’ . ‘X’ gave a red precipitate of a compound ‘Z’ with ammonical cuprous chloride and formed a carbonyl compound with dilute sulphuric acid in the presence of `Hg^(+2)` ions. X, Y, and Z are given by the set -

Options

A. `CH_(2)=C=CH_(2), CH_(2)Br–CBr_(2)–CH_(2)Br, CH_(3)–C-=C–Cu`
B. `CH_(3)–C-=CH, CH_(3)–CBr_(2)–CHBr_(2), CH3–C-=C–Cu`
C. `CH_(3)–C-=CH, CH_(3)–CBr=CHBr, CH_(2)Cu–C-=CH`
D. `CH_(2)=C=CH_(2), CH_(2)Br–CH=CHBr, CH_(3)–C-=C–Cu`

Answer

To solve the problem step by step, we will analyze the given information about hydrocarbon `X (C3H4)` and its reactions.

### Step 1: Identify the structure of hydrocarbon `X`
Given that `X` has the molecular formula `C3H4`, we can calculate its degree of unsaturation (double bond equivalent) using the formula:

\[ \text{Degree of Unsaturation} = \frac{(2C + 2 - H)}{2} \]

Substituting the values:

\[ \text{Degree of Unsaturation} = \frac{(2 \times 3 + 2 - 4)}{2} = \frac{(6 + 2 - 4)}{2} = \frac{4}{2} = 2 \]

This indicates that `X` contains two double bonds or one triple bond.

### Step 2: Determine the type of hydrocarbon
Since `X` decolorizes bromine in carbon tetrachloride, it indicates that `X` is an alkyne (specifically a terminal alkyne). The structure can be represented as:

\[ \text{X} = \text{CH}_3\text{C} \equiv \text{C} \text{H} \]

### Step 3: Identify compound `Y`
When `X` reacts with bromine in carbon tetrachloride, it undergoes an addition reaction, resulting in a dibromo compound. The reaction can be represented as:

\[ \text{CH}_3\text{C} \equiv \text{C} \text{H} + \text{Br}_2 \rightarrow \text{CH}_3\text{CBr}_2\text{CBr}_2\text{H} \]

This compound `Y` is a colorless tetrabromoalkene.

### Step 4: Identify compound `Z`
The hydrocarbon `X` reacts with ammoniacal cuprous chloride, which is a test for terminal alkynes. The reaction produces a red precipitate, which is a copper(I) acetylide. The reaction can be represented as:

\[ \text{CH}_3\text{C} \equiv \text{C} \text{H} + \text{CuCl} + \text{NH}_3 \rightarrow \text{CH}_3\text{C} \equiv \text{C-Cu} + \text{NH}_4\text{Cl} \]

Thus, compound `Z` is `CH3C≡C-Cu`.

### Step 5: Identify the carbonyl compound formed
When `X` is treated with dilute sulfuric acid in the presence of `Hg^(2+)` ions, it forms a carbonyl compound through hydration of the alkyne. The reaction can be represented as:

\[ \text{CH}_3\text{C} \equiv \text{C} \text{H} + \text{H}_2\text{SO}_4 + \text{Hg}^{2+} \rightarrow \text{CH}_3\text{C}(=O)\text{H} \]

This results in the formation of acetone (or propan-2-one).

### Final Summary of Compounds
- **X**: `CH3C≡CH` (1-butyne)
- **Y**: `CH3CBr2CBr2H` (tetrabromoalkene)
- **Z**: `CH3C≡C-Cu` (copper(I) acetylide)