Question

Complete the following equation:
$C{H_3}CON{H_3} + B{r_2} + NaOH \to $

Answer 1

Hint:To answer this question, you must recall the reactions occurring between different organic compounds. The given reaction involves the reaction of a primary amide with bromine and sodium hydroxide.

Complete answer:
When an amide is treated with bromine in an aqueous or alcoholic solution of alkalis like sodium hydroxide or potassium hydroxide, it undergoes Hoffmann Bromamide degradation reaction. The amine is degraded, i.e., loses a carbon atom. As a result the primary amide gets converted into a primary amine. The amine formed would contain one carbon less than the original parent amide.
The reaction can be completed as follows:
$C{H_3}CON{H_2} + B{r_2} + 4NaOH \to C{H_3}N{H_2} + 2NaBr + N{a_2}C{O_3} + 2{H_2}O$

Note:
The mechanism by which the reaction occurs is as follows:
First the hydroxide ion from the alkali sodium hydroxide reacts with the amide. It acts as a base and leaves as a water molecule. Hence, the amide is deprotonated and forms an anion.
The anion now being nucleophilic attacks the bromine molecule causing the heterolytic cleavage of the bromine- bromine bond. N- Bromamide and a bromine anion are formed as a result.
The N- bromamide is again attacked by the hydroxide base and is deprotonated. Again a water molecule is formed along with the anion of the N- bromamide. This anion then undergoes rearrangement where the methyl group bonded to the carbonyl carbon bonds to the nitrogen releasing the bromine anion. An isocyanate intermediate is formed.
The water present in the solution adds on to the isocyanate to form carbamic acid which undergoes decarboxylation, i.e., loses carbon dioxide gas thus, forming a primary amine.