Bromination of phenol: practice and mechanism
The OH group activates the ring so strongly that Br₂ in water substitutes at all free ortho and para positions without a catalyst: phenol becomes 2,4,6-tribromophenol.
Example
The OH group activates the ring so strongly that bromine reacts without a Lewis acid, at every free ortho and para position. Phenol gives 2,4,6-tribromophenol, which precipitates as a white solid (test for phenols).
Mechanism step by step
- 1
The OH group pushes electron density into the ring (+M effect); in water some of the even richer phenoxide is present.
- 2
Br₂ is attacked without a Lewis acid, successively at both ortho positions and at para (SEAr).
- 3
2,4,6-Tribromophenol precipitates as a white solid.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- With Br₂ in water it does not stop at monobromophenol. Phenol is so reactive that all three activated positions are brominated. Only with little Br₂ in a non-polar solvent and cold do you get mainly p-bromophenol.
Often confused with: Aromatic bromination (SEAr)
Frequently asked questions
How do I get only monobromophenol?
With one equivalent of Br₂ in a non-polar solvent (CS₂, CCl₄) at low temperature. Mainly p-bromophenol then forms.
Is this a test?
Yes, the white precipitate with bromine water is a classic test for phenol. Aniline reacts similarly (2,4,6-tribromoaniline).