Question

An amide ‘A’ with molecular formula C$_7$H$_7$ON undergoes Hoffmann Bromamide degradation reaction to give amine ‘B’. ‘B’ on treatment with nitrous acid at 273-278 K forms ‘C’ and on treatment with chloroform and ethanolic potassium hydroxide forms ‘D’. ‘C’ on treatment with ethanol gives ‘E’. Identify ‘A’, ‘B’, ‘C’, ‘D’ and ‘E’ and write the sequence of chemical equations.

Hint:

Hoffmann Bromamide degradation removes the carbonyl group from amides, forming amines. Diazotization of amines with nitrous acid yields phenol derivatives. Carbylamine reaction with chloroform forms isocyanides.

Answer 1

Step 1: Identify the structure of ‘A’ (C$_7$H$_7$ON): The molecular formula C$_7$H$_7$ON suggests the compound is a derivative of an amide. A common structure that fits this molecular formula is benzamide (C$_6$H$_5$CONH$_2$), which contains a benzene ring attached to a carboxamide group. Step 2: Hoffmann Bromamide degradation reaction: In this reaction, benzamide (A) undergoes a degradation when treated with bromine in the presence of alkali (usually NaOH) to give a primary amine. The reaction leads to the loss of one carbon atom from the amide. - The product ‘B’ is aniline (C$_6$H$_5$NH$_2$), formed after the elimination of CO$_2$ from the amide group. \[ \text{C}_6\text{H}_5\text{CONH}_2 \xrightarrow{\text{Br}_2/\text{NaOH}} \text{C}_6\text{H}_5\text{NH}_2 \quad (\text{Aniline}) \] Step 3: Reaction of ‘B’ with nitrous acid: When aniline (B) reacts with nitrous acid (HNO$_2$) at 273-278 K, it undergoes diazotization followed by deamination to form a phenol derivative. - The product ‘C’ is phenol (C$_6$H$_5$OH). \[ \text{C}_6\text{H}_5\text{NH}_2 + \text{HNO}_2 \rightarrow \text{C}_6\text{H}_5\text{OH} + \text{N}_2\text{ (gas)} \] Step 4: Reaction of ‘B’ with chloroform and ethanolic potassium hydroxide: When aniline (B) is treated with chloroform (CHCl$_3$) and ethanolic potassium hydroxide (KOH), it undergoes the Carbylamine reaction, which produces an isocyanide (or isothiocyanate). - The product ‘D’ is phenyl isocyanide (C$_6$H$_5$NC). \[ \text{C}_6\text{H}_5\text{NH}_2 + \text{CHCl}_3 + \text{KOH} \rightarrow \text{C}_6\text{H}_5\text{NC} + \text{KCl} + \text{H}_2\text{O} \] Step 5: Reaction of ‘C’ with ethanol: When phenol (C) reacts with ethanol, it undergoes etherification to form an ethoxy group attached to the benzene ring. - The product ‘E’ is ethyl phenyl ether (C$_6$H$_5$OCH$_2$CH$_3$). \[ \text{C}_6\text{H}_5\text{OH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{HCl}} \text{C}_6\text{H}_5\text{OCH}_2\text{CH}_3 \]