Complete reduction of benzene-diazonium chloride with Zn/HCl gives :

Options

A. Aniline
B. Phenylhydrazine
C. Azobenzene
D. Hydrazobenzene.

Answer

To solve the question regarding the complete reduction of benzene-diazonium chloride with Zn/HCl, we can follow these steps:

### Step 1: Understand the structure of benzene-diazonium chloride
Benzene-diazonium chloride has the following structure:
- It consists of a benzene ring (C6H5) attached to a diazonium group (N2^+), which can be represented as C6H5-N2^+-Cl^-.

### Step 2: Identify the reducing agent
Zinc (Zn) in the presence of hydrochloric acid (HCl) acts as a strong reducing agent. This means it can donate electrons to the benzene-diazonium chloride, facilitating its reduction.

### Step 3: Reduction process
When benzene-diazonium chloride is treated with Zn/HCl, the diazonium group (N2^+) is reduced:
- The first step of reduction converts the diazonium salt into phenyl hydrazine (C6H5-NH-NH2).
- However, since Zn/HCl is a strong reducing agent, further reduction occurs.

### Step 4: Final product formation
The phenyl hydrazine formed is then further reduced to aniline (C6H5-NH2):
- Aniline is the final product of the complete reduction of benzene-diazonium chloride.

### Conclusion
Thus, the complete reduction of benzene-diazonium chloride with Zn/HCl gives **aniline**.

### Final Answer
**Aniline (C6H5-NH2)** is the product formed.

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