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 problem step by step, let's break down the reactions involved:

### Step 1: Understand the Reactants
Aniline (C6H5NH2) is an aromatic amine where an amino group (NH2) is attached to a benzene ring.

### Step 2: Diazotization Reaction
When aniline reacts with sodium nitrite (NaNO2) in the presence of hydrochloric acid (HCl) at 0°C, a diazotization reaction occurs. This reaction converts aniline into benzene diazonium chloride (C6H5N2Cl).

**Reaction:**
\[ \text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + \text{HCl} \rightarrow \text{C}_6\text{H}_5\text{N}_2\text{Cl} + \text{NaCl} + 2 \text{H}_2\text{O} \]

### Step 3: Product Formation
The product formed from the diazotization reaction, benzene diazonium chloride (C6H5N2Cl), is stable at low temperatures and can undergo further reactions.

### Step 4: Reaction with Phosphorous Acid
When benzene diazonium chloride is treated with phosphorous acid (H3PO2) at 500 K, it undergoes a reaction that leads to the loss of nitrogen gas (N2) and the formation of benzene (C6H6).

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

### Conclusion
The final product X after the reactions is benzene (C6H6).

### Final Answer
Thus, the value of X is benzene (C6H6).

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