An aromatic compound 'A' of molecular formula C_7H_7ON undergoes a se…

The correct answer is option 4. \(C_6H_6\).

Let us break down the process step by step with explanations for each reaction in the series

Step 1: \(C_7H_7ON\) reacts with \(Br_2 / KOH\) (Hofmann Bromamide Reaction)

The first reaction is a Hofmann Bromamide reaction, which involves the conversion of an amide to a primary amine with the loss of one carbon atom. Thus, \(A\) must be an amide and the only amide possible here is benzamide.

In this reaction:

Benzamide (\(C_7H_7CON\)) reacts with bromine (\(Br_2\)) and potassium hydroxide (\(KOH\)). The amide group (\(-CONH_2\)) is converted into an amine group (\(-NH_2\)) with the loss of one carbon atom.The product of this reaction is aniline (\(C_6H_5NH_2\)).

Thus, compound B is aniline (\(C_6H_5NH_2\)).

Step 2:Aniline (\(C_6H_5NH_2\)) reacts with \(NaNO_2 + HCl\) (Diazotization)

The second reaction is a diazotization reaction, where aniline reacts with sodium nitrite (\(NaNO_2\)) and hydrochloric acid (\(HCl\)) to form benzenediazonium chloride

In this reaction:

Aniline (\(C_6H_5NH_2\)) reacts with sodium nitrite and hydrochloric acid at low temperatures (0°C).

A diazonium salt is formed, specifically benzenediazonium chloride

Thus, Compound C is benzenediazonium chloride

Step 3: Benzenediazonium chloride reacts with ethanol

In the third reaction, benzenediazonium chloride undergoes a reduction reaction when treated with ethanol. The diazonium group (\(-N_2^+\)) is replaced by a hydrogen atom.

In this reaction:

The benzenediazonium chloride reacts with ethanol as the reducing agent. The diazonium group (\(-N_2^+\)) is lost, and a hydrogen atom is substituted in its place. The product formed is benzene (\(C_6H_6\)).

Thus, Compound D is benzene (\(C_6H_6\)).

Hence, the correct answer is option 4. \(C_6H_6\).