When aniline is treated will sodium nitrite and hydrochloric acid at `0^(@)C`, it gives

Options

A. Phenol and `N_(2)`
B. Diazonium salt
C. Hydrazo compound
D. No reaction takes place

Answer

To solve the question regarding the reaction of aniline with sodium nitrite and hydrochloric acid at 0°C, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**: The reactants in this reaction are aniline (C6H5NH2), sodium nitrite (NaNO2), and hydrochloric acid (HCl).

2. **Understand the Reaction Conditions**: The reaction takes place at a low temperature of 0°C, which is important for the stability of the products formed.

3. **Formation of Nitrous Acid**: When sodium nitrite (NaNO2) is mixed with hydrochloric acid (HCl), it generates nitrous acid (HNO2) in situ:
\[ \text{NaNO}_2 + \text{HCl} \rightarrow \text{HNO}_2 + \text{NaCl} \]

4. **Diazotization Reaction**: Aniline reacts with nitrous acid (HNO2) to form benzene diazonium chloride (C6H5N2Cl). The reaction can be represented as:
\[ \text{C}_6\text{H}_5\text{NH}_2 + \text{HNO}_2 \rightarrow \text{C}_6\text{H}_5\text{N}_2\text{Cl} + \text{H}_2\text{O} \]

5. **Product Characteristics**: Benzene diazonium chloride is a yellow-orange solid and is known to be highly unstable, especially at higher temperatures.

6. **Final Product**: The final product of the reaction is benzene diazonium chloride (C6H5N2Cl), which can be considered a nitrogenous salt.

### Conclusion:
When aniline is treated with sodium nitrite and hydrochloric acid at 0°C, the product formed is benzene diazonium chloride, which can be classified as a nitrogenous salt.

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