`C_(6)H_(5)N_(2)Cl overset(SnCl_(2)//HCl) (A) to overset(C_(6)H_(5)CHO)to(B)`
The product (B) in the above sequence of reaction is

Options

A. `C_(6)H_(5)-N=N-C_(6)H_(5)`
B. `C_(6)H_(5)-N=CH-C_(6)H_(5)`
C. `C_(6)H_(5)-NH-N=CH-C_(6)H_(5)`
D.

Explanation

To solve the problem, we need to analyze the reactions step by step:

### Step 1: Identify the starting compound
The starting compound is benzene diazonium chloride, which has the formula \( C_6H_5N_2Cl \).

### Step 2: Reaction with SnCl2 and HCl
When benzene diazonium chloride reacts with tin(II) chloride (SnCl2) in the presence of hydrochloric acid (HCl), it undergoes reduction. This reaction converts the diazonium group into an amine group. The product formed in this step is phenylhydrazine, which can be represented as:

\[ C_6H_5N_2Cl + SnCl_2 + HCl \rightarrow C_6H_5NHNH_2 + SnCl_4 + HCl \]

### Step 3: Identify product A
The product A formed from this reaction is phenylhydrazine, which can be represented as \( C_6H_5NHNH_2 \).

### Step 4: Reaction of product A with another compound
Next, product A (phenylhydrazine) reacts with another compound, which is likely an aldehyde or ketone. In this case, it is reacting with benzaldehyde (\( C_6H_5CHO \)). The reaction between phenylhydrazine and benzaldehyde leads to the formation of a phenylhydrazone.

\[ C_6H_5NHNH_2 + C_6H_5CHO \rightarrow C_6H_5N=CHC_6H_5 + H_2O \]

### Step 5: Identify product B
The product formed from this reaction is phenylhydrazone, which can be represented as \( C_6H_5N=CHC_6H_5 \).

### Conclusion
Thus, the product (B) in the above sequence of reactions is phenylhydrazone.

### Final Answer
The product (B) is phenylhydrazone.

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