JEE Chemistry 2023 Question with Solution

Given: The starting compound is $p$-nitrotoluene and the sequence is bromination, reduction, diazotization, replacement of diazonium group, and side-chain oxidation.

Find: The correct pair of products $A$ and $E$.

From the solution scheme:

1. On treatment with $Br_2$, bromination occurs to give product $A$ as monobromo nitrotoluene.
2. Sn/HCl reduces the $NO_2$ group to $NH_2$ to form $B$.
3. $NaNO_2/HCl$ at $273\text{-}278 \, \text{K}$ converts the amine into the diazonium salt $C$.
4. $H_3PO_2/H_2O$ replaces the diazonium group by hydrogen to give $D$.
5. Oxidation with $KMnO_4/KOH$ followed by $H_3O^+$ converts the methyl group into $COOH$, giving $E$.

Therefore, the correct option is C.

Given: A multistep aromatic transformation is shown.

Find: Which option matches products $A$ and $E$.

The solution explicitly states "The Correct Option is C" and the worked scheme confirms it.

The key transformations are:

• Bromination introduces one $Br$ substituent in the ring to form $A$.
• Reduction changes $NO_2 \rightarrow NH_2$.
• Diazotization forms the diazonium salt.
• $H_3PO_2/H_2O$ causes deamination, so the diazonium group is replaced by $H$.
• Side-chain oxidation converts $Me \rightarrow COOH$.

Hence the final pair is the one shown in option C according to the provided solution image and statement.