Animated Solution for Chemistry - Organic Chemistry: In the above reactions, products A and B respectively are

The world of organic chemistry is filled with fascinating transformations, and the behavior of amides under different reagents is a perfect example. In this problem, we are presented with a single starting material, $p$-bromobenzamide, and asked to predict the outcomes of two distinct reaction pathways.

Let's embark on this chemical journey and uncover the identities of products A and B!

Our starting molecule is $p$-bromobenzamide. It features a benzene ring with two substituents: a bromo group ($-Br$) at the para position and an amide group ($-CON{H}_2$) at the top.

Because both sets of reagents provided in the question specifically target the amide functional group, the bromo group will remain a silent spectator throughout these transformations. Our entire focus will be on the $-CON{H}_2$ group.

Hofmann Bromamide Degradation

In the first reaction, the amide is treated with $KOBr$ (potassium hypobromite). This reagent is the hallmark of the famous Hofmann bromamide degradation.

This reaction is incredibly unique because it acts as a chemical "step-down" process. It effectively snips out the carbonyl carbon ($C=O$) from the amide group, releasing it as a carbonate ion ($C{O}_3^{2-}$). As a result, the remaining $-N{H}_2$ group attaches directly to the adjacent carbon—in this case, the benzene ring itself.

Because we lose exactly one carbon atom, the $p$-bromobenzamide is converted into $p$-bromoaniline. Thus, our product A is $p$-bromoaniline.

Reduction with $LiAl{H}_4$

The second reaction takes a completely different approach. Here, the amide is treated with $LiAl{H}_4$ (lithium aluminium hydride) followed by acidic hydrolysis ($H_3{O}^{+}$).

$LiAl{H}_4$ is a powerful reducing agent. Unlike the Hofmann degradation, it does not break the carbon skeleton. Instead, it attacks the carbonyl group, completely replacing the double-bonded oxygen with two hydrogen atoms. The $-C(=O)N{H}_2$ group is smoothly reduced to a methylene amine group, $-C{H}_{2}N{H}_2$.

Since the carbon chain length is preserved, the resulting molecule is $p$-bromobenzylamine. Therefore, our product B is $p$-bromobenzylamine.

By carefully analyzing the specific actions of our two reagents, we have successfully identified both products: Product A: $p$-bromoaniline (via a step-down degradation) Product B: $p$-bromobenzylamine (via a direct reduction)

Comparing our findings with the given options, we can confidently conclude that Option (d) is the correct answer. This problem beautifully illustrates how choosing the right reagent can lead to vastly different, yet highly predictable, chemical outcomes!