The compound A with the following sequence of | vedclass

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(B) The reaction sequence is as follows:$1. 	ext{Aniline } (C_6H_5NH_2)$ $\xrightarrow{NaNO_2/HCl, 0-5^{\circ}C} 	ext{Benzenediazonium chloride } (C

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Aniline

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(B) The reaction sequence is as follows:$1. 	ext{Aniline } (C_6H_5NH_2)$ $\xrightarrow{NaNO_2/HCl, 0-5^{\circ}C} 	ext{Benzenediazonium chloride } (C_6H_5N_2Cl) 	ext{ (Compound } B	ext{)}$$2. 	ext{Benzenediazonium chloride } (C_6H_5N_2Cl)$ $\xrightarrow{KCN/CuCN} 	ext{Benzonitrile } (C_6H_5CN) 	ext{ (Compound } C	ext{)}$$3. 	ext{Benzonitrile } (C_6H_5CN) \xrightarrow{H_3O^{+}} 	ext{Benzoic acid } (C_6H_5COOH)$Therefore,the compound $A$ is Aniline.

The compound $A$ with the following sequence of reactions gives benzoic acid.
$A$ $\xrightarrow{NaNO_2/HCl} B$ $\xrightarrow{KCN} C$ $\xrightarrow{H_3O^{+}} \text{benzoic acid}$
The compound $A$ is:

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The compound $A$ with the following sequence of reactions gives benzoic acid.$A$ $\xrightarrow{NaNO_2/HCl} B$ $\xrightarrow{KCN} C$ $\xrightarrow{H_3O^{+}} \text{benzoic acid}$The compound $A$ is: