In the given reaction sequence `C_6H_5-COOHoverset(SOCl_2)rarr(A)overset(NH_2OH)rarr(B)overset(NaOH//Delta)rarr(C),`
'C' will be

Options

A. `C_6H_5COCl`
B. `C_6H_5-overset(O)overset(||)C-NHOH`
C. `C_6H_5-NH_2`
D. `C_6H_5-N=C=O`

Answer

To solve the problem, we will analyze the given reaction sequence step by step.

### Step 1: Identify the starting compound
The starting compound is benzoic acid, which has the formula \(C_6H_5COOH\).

### Step 2: Reaction with SOCl2
When benzoic acid reacts with thionyl chloride (\(SOCl_2\)), it undergoes a conversion to form an acyl chloride. The reaction can be represented as:
\[ C_6H_5COOH + SOCl_2 \rightarrow C_6H_5COCl + SO_2 + HCl \]
This product, \(C_6H_5COCl\), is compound A.

### Step 3: Reaction with NH2OH
Next, compound A reacts with hydroxylamine (\(NH_2OH\)). This reaction leads to the formation of an oxime:
\[ C_6H_5COCl + NH_2OH \rightarrow C_6H_5C(=NOH) + HCl \]
This product, \(C_6H_5C(=NOH)\), is compound B.

### Step 4: Reaction with NaOH and heat
The final step involves heating compound B with sodium hydroxide (\(NaOH\)) and bromine (\(Br_2\)). This is a classic example of the Hofmann degradation reaction, which converts an amide to an amine with one less carbon atom. The reaction can be summarized as:
\[ C_6H_5C(=NOH) + NaOH + Br_2 \rightarrow C_6H_5NH_2 + Na_2CO_3 + NaBr + 2H_2O \]
Here, \(C_6H_5NH_2\) is aniline, which is compound C.

### Conclusion
Thus, the final product C in the reaction sequence is **aniline** (\(C_6H_5NH_2\)).

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