JEE Main Chemistry Question (2026) — Solution

Step-by-step solution

The starting compound is 1-ethyl-4-nitrobenzene. Reaction with Br _2/ AlBr _3 is an electrophilic aromatic substitution. The - C _2 H _5 group is activating and ortho/para-directing, while the - NO _2 group is deactivating and meta-directing. Both groups direct the incoming electrophile ( Br ^+) to the position ortho to the - C _2 H _5 group. Thus, compound A is 2-bromo-1-ethyl-4-nitrobenzene. Reaction with Sn / HCl reduces the - NO _2 group to an - NH _2 group, forming compound B (4-amino-2-bromo-1-ethylbenzene). Reaction with NaNO _2/ HCl at 273-278 K converts the - NH _2 group into a diazonium salt, forming compound C (3-bromo-4-ethylbenzenediazonium chloride). Reaction with ethanol ( C _2 H _5 OH ) reduces the diazonium group, replacing it with a hydrogen atom. This forms compound D (1-bromo-2-ethylbenzene). Reaction with alkaline KMnO _4 followed by acidic hydrolysis oxidizes the ethyl side chain (- C _2 H _5) to a carboxylic acid group (- COOH ). The bromine atom remains unaffected. This forms compound E (2-bromobenzoic acid). Answer: