Reduction of nitroalkanes in neutral medium (e.g. Zn/`NH_(4)Cl`) forms mainly

Options

A. `R-NH_(2)`
B. `R-NHOH`
C. `R-N=N-Cl`
D. All of these

Answer

To solve the question regarding the reduction of nitroalkanes in a neutral medium using Zn and NH₄Cl, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**: The reactants in this reaction are nitroalkanes (R-NO₂), zinc (Zn), and ammonium chloride (NH₄Cl).

2. **Understand the Reaction Conditions**: The reaction takes place in a neutral medium. This means that the pH of the solution does not favor acidic or basic conditions, which can influence the outcome of the reaction.

3. **Recognize the Role of Zinc**: Zinc acts as a reducing agent in this reaction. It donates electrons to the nitro group (NO₂) of the nitroalkane, facilitating its reduction.

4. **Identify the Product of Reduction**: The reduction of nitroalkanes generally leads to the formation of hydroxylamine (R-NHOH). This is a common product when nitro compounds are reduced in a neutral medium.

5. **Write the Overall Reaction**: The overall reaction can be summarized as:
\[ R-NO_2 + Zn + NH_4Cl \rightarrow R-NHOH + by-products \]
Here, R-NHOH represents hydroxylamine.

6. **Conclusion**: Therefore, the main product formed from the reduction of nitroalkanes in a neutral medium using Zn and NH₄Cl is hydroxylamine (R-NHOH).

### Final Answer:
The main product formed is hydroxylamine (R-NHOH).

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