the reduction on benzene diazonium chloride with `SnCl_2 ` and `HCI` acid gives :

Options

A. `C_6H_5NHCH_3`
B. `C_6H_5NHCOCH_3`
C. `C_(6)H_5NHCH_2`
D. `C_6 H_5 NH_2HCI`

Answer

To solve the question regarding the reduction of benzene diazonium chloride with SnCl₂ and HCl, we can break down the process step by step:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants are benzene diazonium chloride (C₆H₅N₂Cl) and the reducing agent SnCl₂ in the presence of HCl.

2. **Understanding the Reaction**:
Benzene diazonium salts can be reduced to form primary amines. In this case, the diazonium group (N₂⁺) will be reduced to an amine (NH₂).

3. **Mechanism of Reduction**:
- The SnCl₂ acts as a Lewis acid and facilitates the transfer of electrons.
- The diazonium ion (C₆H₅N₂⁺) reacts with SnCl₂, where Sn²⁺ donates electrons to the nitrogen atom in the diazonium group, leading to the formation of an amine.

4. **Formation of the Amine**:
- The nitrogen in the diazonium ion is reduced to an amine (NH₂) while the chloride ion (Cl⁻) is released.
- The reaction can be summarized as:
\[ C₆H₅N₂Cl + 2 \, SnCl₂ + 2 \, HCl \rightarrow C₆H₅NH₂ + 2 \, SnCl₄ + H₂ \]
- Here, phenylamine (C₆H₅NH₂) is formed as the main product.

5. **Final Product**:
The final product of the reaction is phenylamine (C₆H₅NH₂).

### Final Answer:
The reduction of benzene diazonium chloride with SnCl₂ and HCl gives phenylamine (C₆H₅NH₂).

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