Q.How will you obtain monobromobenzene from aniline?
You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
Start your 14-day free trial to unlock the full solution →Convert the –NH₂ group into a diazonium salt (NaNO₂/dil. HCl, 0–5 °C), then replace the diazonium group directly by bromine with CuBr/HBr — the Sandmeyer reaction. Two steps, no protecting group: aniline → benzenediazonium chloride → bromobenzene.
The target, monobromobenzene, has no nitrogen at all — so the cleanest strategy is not to brominate the ring while the amino group is still on it, but to use the –NH₂ group itself as the handle: convert it into a diazonium salt and then swap that group for bromine.
Why not just brominate aniline directly? The –NH₂ group is a powerfully activating, ortho/para-directing group. Treating aniline with bromine water doesn't stop at one bromine — it gives 2,4,6-tribromoaniline as a white precipitate. That over-bromination problem is exactly what the diazonium route sidesteps: it never brominates the ring at all.
The standard (NCERT) sequence is:
- Diazotisation: Treat aniline with and dilute at 0–5 °C to form benzenediazonium chloride:
The low temperature matters — diazonium salts decompose readily above ~5 °C.
- Sandmeyer reaction: Add the cold diazonium salt solution to cuprous bromide dissolved in HBr. The diazonium group is replaced by bromine, with nitrogen gas escaping:
(The Gattermann variation — copper powder with HBr — achieves the same substitution.)
The loss of N₂, a supremely stable gas, is what makes this replacement so clean: the reaction is driven forward and the product is a single monosubstituted arene, exactly what we want.
Two classic traps here:
- Diazotising and then reducing with H₃PO₂ gives benzene, not bromobenzene — hypophosphorous acid replaces the diazonium group with hydrogen. To end with a C–Br bond, the diazonium group must be replaced by bromine (CuBr/HBr).
- Direct bromination of aniline gives 2,4,6-tribromoaniline, not a monobromo product — the free –NH₂ group is too strongly activating to stop at one bromine. …
Unlock everything free for 14 days
- Full step-by-step solutions
- Concept-first explanations
- Methods, shortcuts & mistakes
- PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.