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  4. An organic compound A on reacting with NH3 gives B. On heating B gives C. C in the presence of KOH reacts with Br2 to give CH3CH2NH2.A is

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Q. An organic compound $A$ on reacting with $NH_3$ gives $B$. On heating $B$ gives $C. C$ in the presence of $KOH$ reacts with $Br_2$ to give $CH_3CH_2NH_2.A$ is

Amines

Solution:

$A \xrightarrow{NH_3} B \xrightarrow{\Delta} C \xrightarrow{KOH/Br} CH_3CH_2NH_2$
The reaction of $C$ with $KOH/Br_2$ to give amine is called Hofmann bromamide reaction. This reaction is given by acid amides only in which
$R - CH_2 - \overset{\underset{||}{O}}{C}- NH_2$ group undergoes rearrangement along with the loss of $CO_2$ molecule. Thus, the compound $C$ must be acid amide with three carbon atoms. Hence, the compound $C$ is
$CH_3 - CH_2 - \overset{\underset{||}{O}}{C}- NH_2$.
Since, all the options shows that A is an acid and it forms acid amide on reaction with $NH_3$. Thus, acid must contain three carbon atoms. Hence, the compound A is $CH_3CH_2COOH$. The complete series of reaction can be represented as
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