Animated Solution for Chemistry - Organic Chemistry: An organic compound $A$ on reacting with ${\text{NH}}_3$ gives $B$. On heating $B$ gives $C$. $C$ in the presence of $\text{KOH}$ reacts with ${\text{Br}}_2$ to give ${\text{CH}}_3{\text{CH}}_2{\text{NH}}_2$. $A$ is

The beauty of organic chemistry lies in its logical sequences. Every reaction is a clue, and every reagent is a fingerprint pointing to a specific transformation. In this problem, we are presented with a classic reaction sequence that tests our ability to work backwards from a known product to an unknown starting material.

We are given a sequence of transformations:

Our goal is to identify the starting organic compound, $A$. The most strategic way to solve such sequence problems is to start from the end and reverse-engineer the pathway. The final product is ethylamine (${\text{CH}}_3{\text{CH}}_2{\text{NH}}_2$), a primary amine.

Hofmann Bromamide Degradation

Look closely at the final step:

The reagents $\text{KOH}$ and ${\text{Br}}_2$ (often written as $\text{KOBr}$) acting on an unknown compound $C$ to produce a primary amine is the unmistakable signature of the Hofmann bromamide degradation reaction.

This powerful name reaction converts a primary amide into a primary amine. Crucially, it does so by excising the carbonyl carbon ($\text{C=O}$) as a carbonate ion. This means the resulting amine has one carbon atom less than the parent amide.

Since our product, ethylamine (${\text{CH}}_3{\text{CH}}_2{\text{NH}}_2$), contains two carbon atoms, the parent amide $C$ must have contained $2+1=3$ carbon atoms. Therefore, compound $C$ is propanamide:

Now that we have identified $C$, let's move one step back:

Compound $C$ is formed by heating compound $B$. In organic synthesis, heating the ammonium salt of a carboxylic acid drives off a molecule of water to yield an amide. Therefore, compound $B$ must be the ammonium salt corresponding to propanamide. This makes $B$ ammonium propanoate:

Finally, let's uncover $A$:

Compound $B$ is formed by the reaction of $A$ with ammonia (${\text{NH}}_3$). Ammonia is a base, and it reacts with carboxylic acids to form ammonium salts. Since the salt is ammonium propanoate, the starting acid $A$ must be propanoic acid:

By recognizing the signature reagents of the Hofmann degradation and logically stepping backward through the reaction sequence, we have successfully deduced the identity of the starting material. The organic compound $A$ is propanoic acid.