JEE Main 2026 · 28 January · morning · Single correct · +4 / −1
Method used for separation of mixture of products ( B and C ) obtained in the following reaction is
- (a) simple distillation
- (b) sublimation
- (c) fractional distillation
- (d) steam distillation
Answer and solution
Answer (c) fractional distillation
Solution
Chemistry ma’am · 4 steps
Bromine directs the nitro group ortho and para, so B and C are isomers with close boiling points.
- Find A. Benzene with Br₂ and the Lewis acid FeBr₃ undergoes electrophilic bromination. So A is bromobenzene, C₆H₅Br.
- Find B and C. Conc. HNO₃ with conc. H₂SO₄ produces NO₂⁺, which nitrates the ring. Bromine is deactivating but an ortho, para director. So B and C are o-bromonitrobenzene and p-bromonitrobenzene.
- Compare boiling points. The two isomers have the same formula, BrC₆H₄NO₂, and neither can form hydrogen bonds. Their boiling points are different, but they lie close together.
- So. Compounds with close boiling points are separated by fractional distillation. The fractionating column gives repeated vaporisation and condensation, which simple distillation cannot. The answer is (c).
The idea to keep. Identify the real products first. A small boiling point gap needs fractional distillation; steam distillation helps only when one isomer is steam volatile, like o-nitrophenol.
Answer from the NTA final answer key. Solution by PrepWiser.