A compound 'A' when treated with `HNO_(3)` (in presence of `H_(2)SO_(4)`) gives compound B, which is the reduced with `Sn` and `HCl` to aniline ? The compound 'A' is

Options

A. Toluene
B. Benzene
C. Ethane
D. Acetamide

Answer

To solve the question step by step, we need to analyze the reactions described and identify the compounds involved.

### Step 1: Understanding the Reaction
The question states that a compound 'A' reacts with `HNO3` in the presence of `H2SO4` to form compound 'B', which is then reduced with `Sn` and `HCl` to form aniline.

### Step 2: Identifying Compound B
The reduction of compound 'B' to form aniline (which is C6H5NH2) suggests that 'B' must contain a nitro group (NO2) because nitro compounds can be reduced to amines. Therefore, compound 'B' is likely nitrobenzene (C6H5NO2).

### Step 3: Determining Compound A
To produce nitrobenzene, we need to consider what compound 'A' could be. The nitration reaction involves treating benzene with a mixture of `HNO3` and `H2SO4`, which leads to the formation of nitrobenzene. Thus, compound 'A' must be benzene (C6H6).

### Step 4: Writing the Reactions
1. **Nitration Reaction**:
\[ \text{C6H6 (Benzene)} + \text{HNO3} + \text{H2SO4} \rightarrow \text{C6H5NO2 (Nitrobenzene)} + \text{H2O} \]

2. **Reduction Reaction**:
\[ \text{C6H5NO2 (Nitrobenzene)} + \text{Sn} + \text{HCl} \rightarrow \text{C6H5NH2 (Aniline)} + \text{by-products} \]

### Conclusion
Based on the above analysis, we conclude that the compound 'A' is benzene.

### Final Answer
The compound 'A' is **Benzene (C6H6)**.

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