What is Z in the following reaction sequence?
`C_6H_5NH_2 overset((i) NaNO_2+HCl//273K,(ii) H_3PO_2+H_2O ,(iii) CO ,HCl "anhy".AlCl_3//CuCl)toZ`.

Options

A. `C_(6)H_(5)CO_(2)H`
B. `C_(6)H_(5)OH`
C. `C_(6)H_(5)CHO`
D. `C_(6)H_(6)`

Answer

To determine the compound Z in the given reaction sequence, we will follow the steps outlined in the problem statement.

### Step-by-Step Solution:

1. **Identify the starting compound**:
The starting compound is aniline, which has the formula \( C_6H_5NH_2 \). It consists of a benzene ring (C6H5) with an amino group (NH2) attached.

2. **Diazotization Reaction**:
The first step involves the reaction of aniline with sodium nitrite (NaNO2) and hydrochloric acid (HCl) at low temperatures (273 K). This process converts the amino group (NH2) into a diazonium salt. The reaction can be represented as:
\[ C_6H_5NH_2 + NaNO_2 + HCl \rightarrow C_6H_5N_2^+Cl^- \]
Here, the amino group is transformed into a diazonium group (\( N_2^+ \)) attached to the benzene ring.

3. **Reduction of Diazonium Salt**:
The next step involves treating the diazonium salt with phosphorous acid (H3PO2) and water. This reaction reduces the diazonium salt back to benzene by replacing the diazonium group with a hydrogen atom:
\[ C_6H_5N_2^+Cl^- + H_3PO_2 + H_2O \rightarrow C_6H_6 + N_2 + HCl \]
Thus, we obtain benzene (\( C_6H_6 \)).

4. **Gatterman-Koch Reaction**:
The final step involves the reaction of benzene with carbon monoxide (CO) in the presence of hydrochloric acid (HCl) and anhydrous aluminum chloride (AlCl3) along with copper(I) chloride (CuCl). This is known as the Gatterman-Koch reaction, which introduces a formyl group (CHO) to the benzene ring, yielding benzaldehyde:
\[ C_6H_6 + CO + HCl \xrightarrow{AlCl_3, CuCl} C_6H_5CHO \]
Therefore, the final product Z is benzaldehyde (\( C_6H_5CHO \)).

### Conclusion:
The compound Z in the reaction sequence is benzaldehyde, represented as \( C_6H_5CHO \).