Write short notes on Hofmam's bromamide reaction.

Explanation

Hofmann method Mixture of amine is treated with diethyl oxalate where primary amine forms a solid dialkyl oxamide, secondary amine forms a liquid oxamic ester while tertiary amines remain unreacted
`{:(CO-OC_(2)H_(5),,CONHR,),(|,underset(1^(@)"amine")(+2H-NHR)rarr,"|",underset("Ethyl alcohol")(+2C_(2)H_(5)OH)),(underset("Diethyl oxalate")(CO-OC_(2)H_(5)),,underset("Oxamide solid")(CONHR),):}`
`{:(CO-OC_(2)H_(5),,,,),(|,+underset(2^(@)"amine")(H-NR_(2)),rarr,CONR_(2),),(CO-OC_(2)H_(5),,,"|",+underset("Ethyl alcohol")(2C_(2)H_(5)OH)),(,,,underset("Oxamic ester(liquid)")(COOC_(2)H_(5)),):}`
The unreacted tertiary amine is distilled off and the remaining mixture is cooled. The solid oxamide is separated from the liquid oxamic ester by filtration. The two derivatives are then separately treated with strong alkali to regenerate the original amines.