N-Methylaniline react with `NaNO_2` and dilute HCl at `0-5^@` C to form.

Answer

To solve the question regarding the reaction of N-Methylaniline with NaNO2 and dilute HCl at 0-5°C, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactant is N-Methylaniline, which has the structure:
\[ \text{C}_6\text{H}_5\text{NHCH}_3 \]
- The other reactants are sodium nitrite (NaNO2) and dilute hydrochloric acid (HCl).

2. **Understand the Reaction Conditions**:
- The reaction takes place at low temperatures (0-5°C). This is important because it affects the stability of the intermediates formed during the reaction.

3. **Formation of Nitrous Acid**:
- When NaNO2 reacts with dilute HCl, it generates nitrous acid (HNO2):
\[ \text{NaNO}_2 + \text{HCl} \rightarrow \text{HNO}_2 + \text{NaCl} \]

4. **Reaction of N-Methylaniline with Nitrous Acid**:
- N-Methylaniline is a secondary amine. When it reacts with nitrous acid (HNO2), it forms a nitroso compound. The reaction can be represented as:
\[ \text{C}_6\text{H}_5\text{NHCH}_3 + \text{HNO}_2 \rightarrow \text{N-nitroso-N-methyl-aniline} + \text{H}_2\text{O} \]
- The product formed is N-nitroso-N-methyl-aniline, which has the structure:
\[ \text{C}_6\text{H}_5\text{N}=\text{N}^+\text{O} + \text{CH}_3 \]

5. **Final Product**:
- The final product is N-nitroso-N-methyl-aniline, which is characterized by the presence of a nitroso group (-N=O) attached to the nitrogen of the amine. The structure can be drawn as:
\[ \text{C}_6\text{H}_5\text{N}=\text{N}^+\text{O} - \text{CH}_3 \]
- This compound is typically yellow in color and is insoluble in water.

### Summary of the Reaction:
- The reaction of N-Methylaniline with NaNO2 and dilute HCl at low temperatures leads to the formation of N-nitroso-N-methyl-aniline.