Aniline `underset( 0^@ C ) overset( NaNO_2 // HCI ) to ` product `underset( 500 K) overset( H_3 PO_2) to X .` here X will be :

Options

A. `C_6H_5 Cl`
B. `C_6 H_6`
C. `C_6 H_5NHOH`
D. `C_6H_5N_2Cl`

Answer

To solve the given problem, we will follow the sequence of reactions step by step.

### Step 1: Formation of Diazonium Salt
Aniline (C6H5NH2) reacts with sodium nitrite (NaNO2) in the presence of hydrochloric acid (HCl) at 0°C. This reaction leads to the formation of a diazonium salt.

**Reaction:**
\[ \text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + \text{HCl} \rightarrow \text{C}_6\text{H}_5\text{N}_2^+ \text{Cl}^- + \text{H}_2\text{O} \]

**Product:**
The product formed is benzene diazonium chloride (C6H5N2Cl).

### Step 2: Reduction of Diazonium Salt
Next, the diazonium salt (C6H5N2Cl) is treated with phosphorous acid (H3PO2) at a high temperature of 500 K. Phosphorous acid acts as a mild reducing agent.

**Reaction:**
\[ \text{C}_6\text{H}_5\text{N}_2^+ \text{Cl}^- + \text{H}_3\text{PO}_2 \rightarrow \text{C}_6\text{H}_6 + \text{N}_2 + \text{HCl} + \text{H}_3\text{PO}_3 \]

**Product:**
The diazonium group (N2+) is replaced by a hydrogen atom, resulting in the formation of benzene (C6H6).

### Final Answer
The final product X is benzene (C6H6).

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