Question

\(R-CONH_{2}+Br_{2}+4NaOH \rightarrow R-NH_{2}+Na_{2}CO_{3}+2NaBr+2H_{2}O\); this reaction is:

• A. Hinsberg reaction
• B. Carbylamine reaction
• C. Sandmeyer reaction
• D. Hofmann bromamide reaction

Hint:

Remember the characteristic reaction: \[ \boxed{ RCONH_2 \xrightarrow[\mathrm{NaOH}]{Br_2} RNH_2 } \] The amine formed contains \[ \boxed{\text{one carbon atom less}} \] than the corresponding amide.

Answer 1

The Correct Option is D

Concept: The Hofmann bromamide reaction (also called Hofmann degradation or Hofmann rearrangement) is a reaction in which a primary amide is converted into a primary amine containing one carbon atom less. The reaction takes place in the presence of bromine and aqueous sodium hydroxide. The general reaction is \[ \boxed{ RCONH_2+Br_2+4NaOH \rightarrow RNH_2+Na_2CO_3+2NaBr+2H_2O } \] Thus, the amide loses its carbonyl carbon as carbonate.

Step 1: Identify the reactants.
The given reactants are \[ RCONH_2,\;Br_2,\;\text{and}\;NaOH. \] A primary amide reacts with bromine in an alkaline medium.

Step 2: Identify the product formed.
The product is \[ RNH_2, \] which is a primary amine. The amine contains one carbon atom less than the original amide. This is the characteristic feature of the Hofmann bromamide reaction.

Step 3: Compare with the given options.

• Hinsberg reaction is used to distinguish primary, secondary and tertiary amines.

• Carbylamine reaction is used for the identification of primary amines.

• Sandmeyer reaction converts diazonium salts into haloarenes.

• Hofmann bromamide reaction converts amides into amines with one less carbon atom.
Therefore, \[ \boxed{\textbf{Option (D)}} \] is the correct answer.