When aniline reacts with `NaNO_2` and dil. `HCl` at `0^@-%^2C`, the product formed is

Options

A. Nitroaniline
B. Benzene diazonium chloride
C. Benzene
D. Trinitroaniline

Answer

To solve the question, we will analyze the reaction of aniline with sodium nitrite (NaNO2) and dilute hydrochloric acid (HCl) at a temperature of 0 to 5 degrees Celsius. This reaction is known as diazotization.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactant is aniline, which has the structure C6H5NH2 (a benzene ring with an amino group).
- The other reactants are sodium nitrite (NaNO2) and dilute hydrochloric acid (HCl).

2. **Understand the Conditions**:
- The reaction is carried out at a low temperature (0 to 5 degrees Celsius). This is important because diazotization reactions are typically performed at low temperatures to stabilize the diazonium salt formed.

3. **Write the Diazotization Reaction**:
- When aniline reacts with NaNO2 in the presence of dilute HCl, the amino group (-NH2) of aniline is converted into a diazonium group (-N2+Cl-).
- The reaction can be represented as:
\[ C6H5NH2 + NaNO2 + HCl \rightarrow C6H5N2Cl + NaCl + 2H2O \]
- Here, C6H5N2Cl is benzene diazonium chloride.

4. **Identify the Product**:
- The product formed from this reaction is benzene diazonium chloride (C6H5N2Cl).

5. **Conclusion**:
- Therefore, the product formed when aniline reacts with NaNO2 and dilute HCl at 0 to 5 degrees Celsius is **benzene diazonium chloride**.

### Final Answer:
The product formed is **benzene diazonium chloride**.