JEE Main 2026 · 21 January · evening · Single correct · +4 / −1
Consider the above sequence of reactions. The number of bromine atom(s) in the final product (P) will be :
- (a) 6
- (b) 5
- (c) 1
- (d) 3
Answer and solution
Answer (b) 5
Solution
Chemistry ma’am · 4 steps
Track where each bromine lands: one meta, three ortho-para to NH₂, and one from Sandmeyer.
- First bromine. -NO₂ is a strong meta director, so Br₂/FeBr₃ gives m-bromonitrobenzene. Sn/HCl then reduces -NO₂ to -NH₃⁺, and neutralising frees m-bromoaniline. That makes 1 Br so far.
- Bromine water. -NH₂ activates the ring so strongly that Br₂/H₂O fills every free ortho and para position with no catalyst. With NH₂ at C1 and Br at C3, the free positions C2, C4 and C6 all take Br. This gives 2,3,4,6-tetrabromoaniline, with 4 Br.
- Sandmeyer step. NaNO₂/HBr at 0 to 5 °C turns -NH₂ into -N₂⁺. CuBr then replaces -N₂⁺ with Br. The ring now carries Br at C1, C2, C3, C4 and C6, and only C5 still has H.
- So. P is pentabromobenzene, C₆HBr₅, with 5 bromine atoms. That is option (b).
The idea to keep. In multistep aromatic problems, redraw the ring after every step. NH₂ with bromine water brominates every free ortho and para spot, and Sandmeyer then swaps the NH₂ itself for Br.
Answer from the NTA final answer key. Solution by PrepWiser.