Aniline can be distinguished from methyl amine by its reaction with

Options

A. p - toluene sulphonyl chloride /KOH
B. (i) `NaNO_2//HCl,0-5^@C`
(ii) alkaline `beta` - naphthol
C. Sn/HCl
D. Acetyl choride

Answer

To distinguish between aniline and methyl amine, we can use the reaction with sodium nitrite (NaNO2) and hydrochloric acid (HCl) to form diazonium salts. Here’s a step-by-step explanation of the process:

### Step-by-Step Solution:

1. **Identify the Compounds**:
- Aniline is C6H5NH2 (benzene amine).
- Methyl amine is CH3NH2.

2. **Reactions with Sodium Nitrite and HCl**:
- When aniline is treated with NaNO2 and HCl at low temperatures (around 0°C), it forms a stable diazonium salt, C6H5N2+Cl− (benzenediazonium chloride).
- The reaction can be represented as:
\[ C6H5NH2 + NaNO2 + HCl \rightarrow C6H5N2^+Cl^- + 2H2O \]

3. **Stability of Diazonium Salts**:
- The diazonium salt formed from aniline is stable at low temperatures and can further react to form phenol when heated with water.
- The reaction is:
\[ C6H5N2^+Cl^- + H2O \rightarrow C6H5OH + N2 + HCl \]

4. **Reaction of Methyl Amine**:
- When methyl amine is treated with NaNO2 and HCl, it forms a diazonium salt as well, but this diazonium salt (CH3N2+) is unstable.
- The reaction can be represented as:
\[ CH3NH2 + NaNO2 + HCl \rightarrow CH3N2^+Cl^- + H2O \]
- This diazonium salt decomposes rapidly, releasing nitrogen gas and forming methanol:
\[ CH3N2^+Cl^- \rightarrow CH3OH + N2 + HCl \]

5. **Conclusion**:
- The key difference is that aniline forms a stable diazonium salt, while methyl amine forms an unstable diazonium salt that decomposes quickly. Therefore, this reaction can be used to distinguish between aniline and methyl amine.