JEE Main Chemistry Question (2026) — Solution

Step-by-step solution

The reaction sequence is as follows: 1. Bromination of nitrobenzene with Br_2/FeBr_3 results in meta-bromonitrobenzene because the -NO_2 group is meta-directing. (1 Br atom) 2. Reduction of the nitro group using Sn/HCl followed by pH neutralization converts -NO_2 to -NH_2. The product is m-bromoaniline. 3. Reaction with Br_2/H_2O (bromine water) on m-bromoaniline: The -NH_2 group is highly activating and ortho/para directing. In m-bromoaniline, the positions ortho and para to -NH_2 are vacant (positions 2, 4, and 6 relative to -NH_2). Bromine water causes polybromination at all available ortho and para positions. This adds 3 more bromine atoms. (Total 1 + 3 = 4 Br atoms) 4. Diazotization with NaNO_2/HBr at 0-5^ C converts the -NH_2 group into a diazonium salt -N_2^+Br^-. 5. Sandmeyer reaction with CuBr/NaBr replaces the diazonium group with a bromine atom. (Total 4 + 1 = 5 Br atoms) The final product (P) is 1,2,3,4,5-pentabromobenzene. Therefore, the number of bromine atoms in the final product is 5.