Question: What is produced when \[C{H_3}CON{H_2}\] reacts with \[NaOBr\]? Is \[NaOBr\] a product of (or equiva...

Question

What is produced when $C{H_3}CON{H_2}$ reacts with $NaOBr$ ? Is $NaOBr$ a product of (or equivalent to) $B{r_2} + NaOH$ which are used in Hoffmann bromamide degradation reactions?

Answer 1

Solution

The combination of bromine and aqueous sodium hydroxide is not only a reagent to an organic reaction like Hoffmann bromamide degradation reaction but also used in an inorganic reaction and its product in itself is a reagent for the same organic reaction.

Complete answer:
Hoffmann bromamide degradation reaction is an important organic reaction used in the synthesis of primary amines from corresponding amides. The degradation of the amide results in the removal of the carbonyl group $(C = O)$ from the amide molecule giving an amine with one less carbon atom.
The reagent used to carry out Hoffman degradation reaction is a combination of bromine and aqueous sodium hydroxide $(B{r_2} + NaOH)$ . The hydroxide ion, being basic in nature, abstracts a proton attached to the nitrogen atom of the amide group and it gets substituted by a bromine atom.
$C{H_3}CON{H_2} + B{r_2} + 4NaOH \to C{H_3}N{H_2} + N{a_2}C{O_3} + 2NaBr + 2{H_2}O$
In an aqueous medium the bromine-bromine bond of the atomic molecule is also polarized to some extent that it gets cleaved by the action of sodium hydroxide base resulting in the formation of sodium hypobromite $(NaOBr)$ . Therefore $NaOBr$ is a reagent for Hoffmann bromamide degradation that is prepared in situ by the reaction between bromine and aqueous sodium hydroxide.
$B{r_2} + 2NaOH \to NaBr + NaOBr + {H_2}O$
Thus the product of the reaction between $C{H_3}CON{H_2}$ and $NaOBr$ is methenamine $C{H_3}N{H_2}$ . Yes, $NaOBr$ is a product of $B{r_2} + NaOH$ and is a reagent equivalent to the combination of bromine and aqueous sodium hydroxide.

Note:
This method of producing amines is only effective for the synthesis of primary amines and not secondary or tertiary amines. As the sodium hydroxide present in the reaction medium can also act as a reagent to facilitate the elimination reaction of secondary or tertiary groups as secondary or tertiary carbocations are quite stable. Also it is difficult to migrate groups other than primary alkyls.