Bromination of Phenol

Phenol reacts with bromine water at room temperature, without a halogen carrier. This is an electrophilic substitution.Observations
• The orange bromine water is decolourised.
• A white precipitate of 2,4,6-tribromophenol forms.EquationC₆H₅OH + 3Br₂ → C₆H₂Br₃OH + 3HBrThree hydrogen atoms on the ring are replaced by bromine atoms, at positions 2, 4 and 6 (numbering the carbon with the -OH group as 1). Each substitution uses one Br₂ and forms one HBr, so three moles of bromine react with each mole of phenol.Why no halogen carrier is needed
A lone pair from the oxygen is partially delocalised into the ring, increasing its electron density. The electron-rich ring polarises the Br₂ molecule, inducing a dipole, so the δ+ bromine atom acts as the electrophile. Benzene's ring has too low an electron density to do this, so benzene needs a halogen carrier and gives only monosubstitution.The formation of a white precipitate with bromine water can be used to show that a compound contains a phenol group.Worked example: what mass of 2,4,6-tribromophenol (M_r 330.7) forms from 0.940 g of phenol (M_r 94.0), assuming complete reaction?
n(phenol) = 0.940 ÷ 94.0 = 0.0100 mol, so 0.0100 mol of product forms.mass = 0.0100 × 330.7 = 3.31 gThis needs 3 × 0.0100 = 0.0300 mol of Br₂.