JEE Chemistry 2023 Question with Solution

Given: Four transformations are shown.

Find: How many of them give aromatic amines.

From the extracted solution:

1. Benzamide with $Br_2 + NaOH$ gives aniline, which is an aromatic amine.
2. In Gabriel phthalimide synthesis, chlorobenzene is a poor substrate for $S_N2$, so no reaction occurs.
3. Nitrobenzene on reduction with $H_2/Pd-C$ gives aniline, which is an aromatic amine.
4. Acetanilide on hydrolysis with dilute $H_2SO_4$ and heat gives aniline + acetic acid.

Thus, aromatic amines are formed in 1, 3 and 4.

Therefore, the number of transformations giving aromatic amines is $3$.

Given: The reactions must be checked one by one for formation of an aromatic amine.

Find: Total number of successful transformations.

1. Benzamide undergoes Hofmann bromamide degradation with $Br_2/NaOH$ to give aniline. This is an aromatic amine.

2. Potassium phthalimide reacts through nucleophilic substitution in Gabriel synthesis, but chlorobenzene does not undergo the required $S_N2$ reaction. So aromatic amine is not formed here.

3. Nitrobenzene on catalytic hydrogenation with $H_2/Pd-C$ is reduced to aniline. This is an aromatic amine.

4. Acetanilide on acidic hydrolysis with dilute $H_2SO_4$ and heat regenerates aniline along with acetic acid. This is again an aromatic amine.

Hence, the successful cases are $1, 3, 4$ only, so the required number is $3$.