Nitrobezene of treatment with zinc dust and aqueous ammonium chloride gives

Options

A. `C_(6)H_(5)N=N-C_(6)H_(5)`
B. `C_(6)H_(5)NH_(2)`
C. `C_(6)H_(5)NO`
D. `C_(6)H_(5)NHOH`

Explanation

To solve the question, we need to determine the product formed when nitrobenzene is treated with zinc dust and aqueous ammonium chloride.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants are nitrobenzene (C6H5NO2), zinc dust (Zn), and aqueous ammonium chloride (NH4Cl).

2. **Understand the Reaction Type**:
- This reaction is a reduction process. Nitrobenzene contains a nitro group (-NO2), which can be reduced to an amine or a hydroxylamine derivative.

3. **Mechanism of Reduction**:
- The zinc dust acts as a reducing agent. It donates electrons to the nitro group in nitrobenzene.
- The nitro group (-NO2) is first reduced to a hydroxylamine (-NHOH) through a series of steps involving the transfer of electrons and protons.

4. **Formation of N-Phenyl Hydroxylamine**:
- The reduction process involves the following steps:
- The nitro group receives electrons from zinc, forming an intermediate.
- This intermediate then reacts with protons (H+) from the ammonium chloride solution, leading to the formation of N-phenyl hydroxylamine (C6H5NHOH).

5. **Final Product**:
- The final product of the reaction is N-phenyl hydroxylamine, represented as C6H5NHOH.

6. **Conclusion**:
- Therefore, the product formed when nitrobenzene is treated with zinc dust and aqueous ammonium chloride is N-phenyl hydroxylamine (C6H5NHOH).

### Answer:
The product formed is **N-phenyl hydroxylamine (C6H5NHOH)**.

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