Identify (X),(Y) and (Z) in the following synthetic scheme and write their structures.Explain the formation of labelled formaldehyde `(H_2C^(**)O)` as one of the products when compound (Z) is treated with HBr and subsequently ozonolysed .Mark the `C^(**)` carbon in the entire scheme .
`BaCO_3+H_2SO_4 to (X)` gas [`C^(**)` denotes `C^14`]
`CH_2=CH-Br underset(underset((iii)H_3O^+)((ii))X)overset((i)Mg " " "ether")to(Y)overset(LiAIH_4)to(Z)`

Answer

To solve the problem, we will identify the compounds (X), (Y), and (Z) in the synthetic scheme, provide their structures, and explain the formation of labeled formaldehyde when compound (Z) is treated with HBr and subsequently ozonolysed. We will also mark the C^14 carbon in the entire scheme.

### Step-by-Step Solution:

1. **Identify Compound (X)**:
- The reaction involves barium carbonate (BaCO₃) and sulfuric acid (H₂SO₄).
- The reaction produces barium sulfate (BaSO₄), water (H₂O), and carbon dioxide (CO₂).
- The gas produced in this reaction is carbon dioxide (CO₂).
- Therefore, **Compound (X) is CO₂**.

**Structure of (X)**:
\[ \text{CO}_2 \]

2. **Identify Compound (Y)**:
- The next step involves the reaction of CH₂=CH-Br with magnesium (Mg) in ether, which forms a Grignard reagent.
- The Grignard reagent reacts with (X) which is CO₂.
- The reaction of the Grignard reagent with CO₂ leads to the formation of a carboxylic acid.
- The product formed is CH₂=CH-COOH (acrylic acid).
- Therefore, **Compound (Y) is CH₂=CH-COOH**.

**Structure of (Y)**:
\[ \text{CH}_2=\text{CH}-\text{COOH} \]

3. **Identify Compound (Z)**:
- Compound (Y) is then treated with lithium aluminum hydride (LiAlH₄), which reduces the carboxylic acid to an alcohol.
- The reduction of CH₂=CH-COOH leads to the formation of an unsaturated alcohol, specifically CH₂=CH-CH₂OH.
- Therefore, **Compound (Z) is CH₂=CH-CH₂OH**.

**Structure of (Z)**:
\[ \text{CH}_2=\text{CH}-\text{CH}_2\text{OH} \]

4. **Formation of Labeled Formaldehyde**:
- When compound (Z) is treated with HBr, the hydroxyl group (-OH) is replaced by bromine (Br) through a substitution reaction.
- This gives CH₂=CH-CH₂Br.
- Next, this compound undergoes ozonolysis, which cleaves the double bond.
- The ozonolysis of CH₂=CH-CH₂Br results in the formation of formaldehyde (H₂C=O) and another product (CH₂Br-CHO).
- The formaldehyde produced contains the C^14 carbon, which is marked as C^(**).

**Structure of Labeled Formaldehyde**:
\[ \text{H}_2\text{C}=\text{O} \quad (C^{**} \text{ is the carbon here}) \]

5. **Marking the C^(**) Carbon**:
- The C^(**) carbon in the entire scheme is the carbon that is labeled as C^14 in the formaldehyde product.

### Final Summary:
- **(X)**: CO₂
- **(Y)**: CH₂=CH-COOH
- **(Z)**: CH₂=CH-CH₂OH
- The labeled formaldehyde (H₂C=O) is formed from the ozonolysis of (Z) after treatment with HBr, with the C^(**) carbon being C^14.