Identify 'Z' in the sequence `C_6H_5NH_2overset(NaNO_2+HCl)rarrXoverset(CuCN)rarrYoverset(H^(+)//H_2O)rarrZ`

Options

A. `C_6H_5CN`
B. `C_6H_5CONH_2`
C. `C_6H_5COOH`
D. `C_6H_5CH_2NH_2`

Answer

To identify 'Z' in the given sequence of reactions, let's break down the steps involved:

1. **Identify the starting compound**: The starting compound is C₆H₅NH₂, which is aniline. Aniline consists of a benzene ring (C₆H₅) with an amino group (NH₂) attached.

2. **First reaction with NaNO₂ and HCl**: When aniline reacts with sodium nitrite (NaNO₂) in the presence of hydrochloric acid (HCl), it undergoes a diazotization reaction. This reaction converts the amino group (NH₂) into a diazonium group (N₂⁺) while releasing nitrogen gas (N₂) and forming a diazonium salt. The product of this reaction is C₆H₅N₂Cl (benzenediazonium chloride).

3. **Second reaction with CuCN**: The benzenediazonium chloride (C₆H₅N₂Cl) then reacts with copper(I) cyanide (CuCN). In this reaction, the diazonium group is replaced by a cyano group (CN) through a Sandmeyer reaction. The product formed is C₆H₅CN (benzonitrile).

4. **Third reaction with H⁺/H₂O**: The benzonitrile (C₆H₅CN) undergoes hydrolysis when treated with an acid (H⁺) and water (H₂O). During hydrolysis, the cyano group (CN) is converted into a carboxylic acid group (COOH). The final product of this hydrolysis is C₆H₅COOH, which is benzoic acid.

5. **Identify 'Z'**: Therefore, the compound 'Z' at the end of the sequence is benzoic acid (C₆H₅COOH).

### Final Answer:
Z = C₆H₅COOH (benzoic acid)