Reaction of propanamide with Br_2 / KOH (aq) | vedclass

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(D) The reaction of an amide with $Br_2$ and an aqueous base $(KOH)$ is known as the $Hoffmann$ $Bromamide$ degradation reaction.In this reaction,the

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Ethylamine

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(D) The reaction of an amide with $Br_2$ and an aqueous base $(KOH)$ is known as the $Hoffmann$ $Bromamide$ degradation reaction.In this reaction,the amide is converted into a primary amine containing one carbon atom less than the original amide.Propanamide is $CH_3CH_2CONH_2$ (a $3$-carbon amide).Upon treatment with $Br_2 / KOH$,it undergoes degradation to form ethylamine $(CH_3CH_2NH_2)$,which is a $2$-carbon primary amine.

Reaction of propanamide with $Br_2 / KOH \ (aq)$ produces :

Similar Questions

Consider the molecule $1,3-dimethyl-1,2,3,4-tetrahydroquinoxaline$ with an amino group at the $5$-position. Upon protonation,which nitrogen atom is the most basic and will be protonated first? Identify the correct structure of the product '$A$' formed after protonation.

Identify substrate $A$ in the following conversion.
$A + \underset{\text{(excess)}}{CH_3 I} \xrightarrow{\Delta} C_2 H_5(CH_3)_3 N^+ I^-$

Carbylamine is formed from aliphatic or aromatic primary amine via which of the following intermediates?

The product of the above reaction is:

Which of the following is the strongest base?

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