Question

Given,$C{H_3}COOH\xrightarrow[\Delta ]{{N{D_3}}}\left( A \right)\xrightarrow[{KOH}]{{B{r_2}}}\left( B \right)$
$A)C{H_3} - N{D_2}$
$B)C{H_3} - N{H_2}$
$C)C{H_3} - CO - C{H_3}$
D) None

Answer 1

Carboxylic acids upon treatment with ammonia and followed by heating undergoes dehydration i.e.The elimination of water molecules takes the form of amides. Amides upon treatment with bromine in presence of hydroxides form amines which is Hoffmann bromamide degradation.

Complete answer:
Given a molecule is an acetic acid, it is a carboxylic acid with two carbons and thus named as ethanoic acid, but the common name is acetic acid.
When acetic acid treated with ammonia and heating, the compound reacts with ammonia first and forms $C{H_3}COON{H_4}$, but in the given reaction ammonia is not given, deuterated ammonia is given. Thus, the compound treated with deuterated ammonia forms $C{H_3}COONH{D_3}$, further heating of the compound, it loses a water molecule, but here in the lost water molecule one is deuterium and the other is proton.
Thus, the compound A will be $C{H_3}CON{D_2}$, it is a deuterated propenamide, an amide functional group.
When compound A treater with bromine in presence of potassium hydroxide, a base forms amine. But, in this case the deuterated amide is reacted with bromine in presence of potassium hydroxide, which leads to the formation of deuterated amine.
Thus, the compound B will have the molecular formula of $C{H_3}N{D_2}$.
In the given options Option B has the molecular formula of the compound B.

So, the correct answer is “Option B”.

Note:
In the above mechanism, the formation of b from the compound A is a familiar reaction called Hofmann degradation or Hoffmann rearrangement, it deals with the conversion of primary amide to a primary amine. The compound B is a primary amine.