Aniline in a set of reactions yielded a product B
(a)`C_(6)H_(5)CH_(2)NH_(2)`
(b)`C_(6)H_(5)NHCH_(2)CH_(3)`
(c)`C_(6)H_(5)NHOH`
(d)`C_(6)H_(5)CH_(2)OH`

Options

A. `C_(6)H_(5)CH_(2)NH_(2)`
B. `C_(6)H_(5)NHCH_(2)CH_(3)`
C. `C_(6)H_(5)NHOH`
D. `C_(6)H_(5)CH_(2)OH`

Explanation

To solve the problem, we need to identify the product formed from aniline (C6H5NH2) through a series of reactions. Let's break down the steps involved in the transformation of aniline to the final product.

### Step-by-Step Solution:

1. **Formation of Diazonium Salt**:
- Aniline (C6H5NH2) is treated with sodium nitrite (NaNO2) and hydrochloric acid (HCl).
- This reaction converts aniline into a diazonium salt, specifically C6H5N2^+Cl^- (C6H5N2Cl).
- **Reaction**:
\[ C6H5NH2 + NaNO2 + HCl \rightarrow C6H5N2^+Cl^- + H2O \]

2. **Nucleophilic Substitution with Copper Cyanide**:
- The diazonium salt is then treated with copper cyanide (CuCN).
- The chloride ion (Cl^-) is replaced by a cyanide group (CN), yielding benzonitrile (C6H5CN).
- **Reaction**:
\[ C6H5N2^+Cl^- + CuCN \rightarrow C6H5CN + N2 + CuCl \]

3. **Reduction to Benzylamine**:
- The next step involves the reduction of benzonitrile (C6H5CN) using hydrogen in the presence of a nickel catalyst (Ni).
- This reduction converts the nitrile group (CN) to an amine group (NH2), resulting in benzylamine (C6H5CH2NH2).
- **Reaction**:
\[ C6H5CN + H2 \xrightarrow{Ni} C6H5CH2NH2 \]

4. **Conversion to Alcohol**:
- Finally, benzylamine (C6H5CH2NH2) is treated with nitrous acid (HNO2).
- In this reaction, the amine group (NH2) is replaced by a hydroxyl group (OH), yielding benzyl alcohol (C6H5CH2OH).
- **Reaction**:
\[ C6H5CH2NH2 + HNO2 \rightarrow C6H5CH2OH + N2 + H2O \]

### Final Product:
The final product after all these reactions is benzyl alcohol, represented by the molecular formula C6H5CH2OH.

### Conclusion:
Thus, the correct answer to the question is option (d) C6H5CH2OH.

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