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 follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactant in this case is benzene diazonium chloride, which has the formula C6H5N2Cl. It consists of a benzene ring (C6H5) attached to a diazonium group (N2+) and a chloride ion (Cl-).

2. **Understand the Reduction Process**:
Reduction is a chemical reaction that involves the gain of electrons or hydrogen, or the loss of oxygen. In this case, we are focusing on the gain of hydrogen.

3. **Identify the Reducing Agent**:
The reducing agent provided in the question is SnCl2 (tin(II) chloride) in the presence of HCl (hydrochloric acid). SnCl2 is a mild reducing agent that can donate electrons to facilitate the reduction of benzene diazonium chloride.

4. **Write the Reaction**:
When benzene diazonium chloride reacts with SnCl2 and HCl, the diazonium group (N2+) is reduced. The reaction can be represented as follows:
\[ C6H5N2Cl + 4H \rightarrow C6H5NH2 + N2 + HCl \]
Here, the diazonium ion is reduced to an amine, specifically aniline (C6H5NH2), and nitrogen gas (N2) is released.

5. **Identify the Product**:
The main product formed from the reduction of benzene diazonium chloride is aniline (C6H5NH2).

6. **Final Answer**:
Therefore, the reduction of benzene diazonium chloride with SnCl2 and HCl gives aniline.