JEE Chemistry 2026 Question with Solution

Given: $A$ has molecular formula $C_8H_9ON$ and gives $B$ with aqueous $Br_2/HO^-$. Compound $B$ undergoes diazotization with aqueous $NaNO_2/HCl$ at $0\text{--}5^\circ \text{C}$ to form $C$, which with $CuCN/NaCN$ gives $D$. Hydrolysis of $D$ gives $E$, and $E$ is also formed by hydrolysis of $A$. Oxidation of $E$ with acidified $KMnO_4$ gives $F$ having two different types of hydrogen atoms.

Find: The structure of $A$.

Aqueous $Br_2/HO^-$ indicates the Hofmann bromamide reaction, so $A$ must be an amide and $B$ must be an amine with one fewer carbon atom.

The molecular formula $C_8H_9ON$ is consistent with a methyl-substituted benzamide.

Diazotization of aromatic amine $B$ with aqueous $NaNO_2/HCl$ at $0\text{--}5^\circ \text{C}$ forms diazonium salt $C$. Reaction with $CuCN/NaCN$ replaces the diazonium group by $-CN$ to give nitrile $D$. Hydrolysis of nitrile $D$ gives carboxylic acid $E$.

Since hydrolysis of $A$ also gives the same acid $E$, $A$ must be the corresponding benzamide derivative.

Oxidation of $E$ with acidified $KMnO_4$ converts the methyl group into another carboxylic acid group. The product $F$ has two different types of hydrogen atoms, which matches a meta-substituted benzenedicarboxylic acid.

Therefore, among the given options, only $m$-methyl benzamide satisfies all the conditions.

The correct option is A.

Given: The sequence is amide $\to$ amine $\to$ diazonium salt $\to$ nitrile $\to$ carboxylic acid $\to$ oxidation product.

Find: Which option fits the full sequence.

1. From aqueous $Br_2/HO^-$, identify Hofmann bromamide degradation:

So $A$ must be a benzamide derivative, not an $N$-substituted amide that would not fit the later sequence.

2. The amine $B$ must be an aromatic primary amine because it gives a diazonium salt with aqueous $NaNO_2/HCl$ at $0\text{--}5^\circ \text{C}$.

3. Sandmeyer replacement gives:

Then hydrolysis gives:

Thus $E$ is a methyl-substituted benzoic acid.

4. Hydrolysis of $A$ also gives the same methyl-substituted benzoic acid, confirming that $A$ is the corresponding methyl benzamide.

5. On oxidation of $E$, the methyl group becomes $-COOH$. The resulting dicarboxylic acid $F$ is said to contain two different types of hydrogen atoms. This excludes the symmetric para isomer and matches the meta isomer.

Hence $A$ is $m$-methyl benzamide.

Therefore, the correct option is A.