JEE Chemistry 2026 Question with Solution

Given: A reaction sequence starts from nitrobenzene and asks for the number of bromine atoms in the final product P.

Find: The total number of bromine atoms present in the final product.

From the sequence shown, the key transformations are:

1. Bromination of nitrobenzene
2. Reduction of the nitro group to an amine
3. Bromination with bromine water
4. Diazotization
5. Sandmeyer replacement by bromine

Nitrobenzene contains the $-NO_2$ group, which is meta-directing. Therefore, bromination introduces one bromine atom at the meta position, giving m-bromonitrobenzene.

Reduction with $\mathrm{Sn/HCl}$ followed by neutralisation converts $-NO_2$ into $-NH_2$, so the compound becomes m-bromoaniline.

Now $-NH_2$ is a strongly activating group. With bromine water, substitution occurs at all free ortho and para positions relative to $-NH_2$. In m-bromoaniline, these are positions $2$, $4$, and $6$. Hence, three more bromine atoms are introduced, forming 2,3,4,6-tetrabromoaniline.

Next, the amino group is diazotized and then replaced by bromine in the Sandmeyer reaction. This adds one more bromine atom at the position originally occupied by $-NH_2$.

So, total bromine atoms in the final product are:

Therefore, the final product contains $5$ bromine atoms, so the correct option is B.

Given: The reaction sequence includes bromination, reduction, polybromination, diazotization, and Sandmeyer reaction.

Find: Count bromine atoms present in the final product after all steps.

Step 1: Nitrobenzene undergoes bromination. Since $-NO_2$ is meta-directing, the product is m-bromonitrobenzene.

Bromine count after this step:

Step 2: Reduction using $\mathrm{Sn/HCl}$ changes $-NO_2$ to $-NH_2$. No bromine atom is removed here.

Bromine count remains:

Step 3: Treatment with $\mathrm{Br_2/H_2O}$ causes bromination at all available ortho and para positions of the aniline ring. For m-bromoaniline, the free activated positions are $2$, $4$, and $6$.

Additional bromine atoms added:

Total bromine count now:

Step 4: Diazotization converts $-NH_2$ into a diazonium group.

Step 5: Sandmeyer reaction with $\mathrm{CuBr}$ replaces the diazonium group by bromine.

Additional bromine atoms added:

Final bromine count:

Hence, the final product is pentabromobenzene, and the correct option is B.