Question

For the following reaction:
$C{H}_{3}C{H}_{2}CON{H}_2\stackrel{A}{\to }C{H}_{3}C{H}_{2}N{H}_2\stackrel{B}{\to }C{H}_{3}C{H}_{2}OH$
Choose the correct option(s):

• A. $A$ is $KMn{O}_4$
• B. $A$ is $B{r}_2/NaOH$
• C. $B$ is $HN{O}_2$
• D. $B$ is $KOH$

Answer 1

Answer: (B), (C)

$B$ is $HN{O}_2$

The first reaction is known as Hoffmann degradation of amides.
When an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide, degradation of amide takes place leading to the formation of primary amine. This reaction involving degradation of amide and is popularly known as Hoffmann bromamide degradation reaction. The primary amine thus formed contains one carbon less than the number of carbon atoms in that amide.

The general reaction is as follows:
$\underset{\text{Amide}}{RCON{H}_2}\stackrel{B{r}_2+4NaOH}{\to }\underset{1^{\circ }\text{Amine}}{RN{H}_2}+N{a}_{2}C{O}_3+2NaBr+2H_{2}O$

When ethylamine reacts with $HN{O}_2$, it forms alcohol.



Thus, A and B are $B{r}_2/NaOH,HN{O}_2$ respectively.