JEE Main 2026 — Jan 21, Shift 2 — Chemistry solution

JEE Main2026Jan 21, Shift 2Chemistry

Consider the above sequence of reactions. The number of bromine atom(s) in the final product (P) will be:

Solution

Below are the steps:

(1) Nitrobenzene with $B r_{2} / F e B r_{3}$ gives m-bromonitrobenzene (NO $_{2}$ is a meta-director).

(2) $S n / H C l$ reduces —NO $_{2}$ to —NH $_{2}$ , giving 3-bromoaniline.

(3) After neutralisation, the free amine is available.

(4) $B r_{2} / H_{2} O$ on the highly activated aniline brominates at all open ortho/para positions of NH $_{2}$ . Starting from 3-bromoaniline, this puts Br at positions 2, 4 and 6 — so the aromatic ring now carries 4 Br atoms (one original at 3, plus three new) and an NH $_{2}$ .

(5) $N a N O_{2} / H B r$ at 0–5 °C converts —NH $_{2}$ to the diazonium salt $- N_{2}^{+}$ .

(6) $C u B r / N a B r$ (Sandmeyer) replaces the diazonium with Br, adding a 5th Br on the ring.

Final product P carries 5 Br atoms.

← Back to full paper

Practice more Chemistry

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →