JEE Chemistry 2023 Question with Solution

Given: A reaction sequence starts from compound A. In the first step, $\text{Br}_2$ and $\text{CS}_2$ at $0-5^\circ \text{C}$ are used. Then diazotization with $\text{NaNO}_2/\text{HCl}$ occurs, followed by reduction with $\text{H}_3\text{PO}_2$ and heat.

Find: The identity of compound A.

From the solution, the first step is bromination of aniline. It states that aniline undergoes electrophilic substitution with bromine to give:

Then this compound forms a diazonium salt with $\text{NaNO}_2/\text{HCl}$, and on treatment with $\text{H}_3\text{PO}_2$ and heat, the diazonium group is replaced by hydrogen to give:

Therefore, the starting compound that must undergo bromination first is Aniline.

The solution explicitly marks the correct option as D, but the listed options show Aniline as option C. This is a source discrepancy. Based on the worked chemistry, the defensible answer among the given options is Aniline, which corresponds to option C.

Given: Final product shown is tribromobenzene after diazotization and reduction.

Find: Which initial compound can lead to that sequence.

The reagent set $\text{NaNO}_2/\text{HCl}$ indicates diazotization, so the intermediate before this step must contain an amino group on the benzene ring. Reduction of the diazonium salt with $\text{H}_3\text{PO}_2$ replaces the diazonium group by hydrogen.

So, just before reduction, the ring must be tribrominated and carry the diazonium functionality. That means before diazotization, the compound must be tribromoaniline. Such a compound is obtained by bromination of aniline.

Hence the original compound A is Aniline.

Therefore, the correct option from the listed choices is C.