JEE Chemistry 2026 Question with Solution

Given: The sequence involves reduction of the nitro group, acetylation of the amine formed, and then bromination.

Find: The mass of B obtained from $137 \, \text{mg}$ of 4-nitrotoluene.

From the reaction sequence:

1. $-NO_2$ is reduced by $\text{Sn/HCl}$ to $-NH_2$.
2. $\text{Ac}_2\text{O}$ converts $-NH_2$ to $-NHCOCH_3$ to form A.
3. $\text{Br}_2/\text{AcOH}$ brominates the ring ortho to the activating acetamido group to form B.

The $-NHCOCH_3$ group is ortho/para directing. Since the para position is occupied by the methyl group, bromine enters the ortho position.

Molar mass of 4-nitrotoluene $(C_7H_7NO_2)$:

So, its molar mass is $137 \, \text{g/mol}$.

Molar mass of B $(C_9H_{10}NOBr)$:

So, its molar mass is $228 \, \text{g/mol}$.

Since the stoichiometry is $1:1$ throughout the sequence:

Therefore, it gives $1 \, \text{mmol}$ of B.

Mass of B:

Given: The starting compound is 4-nitrotoluene and the reaction proceeds to completion.

Find: The product identity of B and its mass.

The first step reduces 4-nitrotoluene to 4-methylaniline. The second step acetylates it to give 4-methylacetanilide (A). In the final step, bromination occurs ortho to the acetamido group because it is an activating ortho/para-directing group, while the para position is already occupied by the methyl group. Thus B is 2-bromo-4-methylacetanilide.

Because no step changes the number of aromatic molecules, $1$ mole of starting material gives $1$ mole of final product under complete conversion. Hence the mass changes only according to molar mass.

Therefore, $137 \, \text{mg}$ of a compound with molar mass $137 \, \text{g/mol}$ corresponds to $1 \, \text{mmol}$, and this produces $1 \, \text{mmol}$ of B, whose molar mass is $228 \, \text{g/mol}$. Hence the mass of B is $228 \, \text{mg}$.

Therefore, the correct option is C.