An aromatic compound 'A' of molecular formula \(C_7H_7ON\) undergoes a series of reaction as: \(\underset{A}{C_7H_7ON} \overset{Br_2 / KOH}{\longrightarrow} B \overset{NANO_2+ HCl}{\longrightarrow} C \overset{CH_3CH_2OH}{\longrightarrow} D\) Identify the compound D: |
\(C_7H_5N\) \(C_6H_6O\) \(C_6H_7N\) \(C_6H_6\) |
\(C_6H_6\) |
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 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. \(C_6H_6\). |