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Benzene will not react with bromine unless you add a Lewis acid such as FeBr3. Phenol reacts with plain bromine water at room temperature, and does not stop at one bromine: it takes three. That contrast is a favourite exam question, in theory, in prediction and in calculation. This post explains why it happens, how the conditions change the product, how to predict products for substituted phenols, and how to do the mole calculation.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus both list, for phenols, "electrophilic substitution reactions, halogenation, nitration and sulphonation". The JEE (Advanced) 2026 brochure lists "Electrophilic substitution reactions of phenol (halogenation, nitration, sulphonation)". The text is section 7.4.4 of the NCERT Class 12 unit Alcohols, Phenols and Ethers, 2026-27 reprint. Check your year's syllabus.
What does NCERT say happens?
NCERT gives two sets of conditions:
In solvents of low polarity such as CHCl3 or CS2, at low temperature, monobromophenols are formed. Since –OH directs ortho and para, these are 2-bromophenol and 4-bromophenol.
With bromine water, 2,4,6-tribromophenol forms as a white precipitate.
It also explains the missing catalyst: with benzene, FeBr3 polarises the bromine molecule, but "in case of phenol, the polarisation of bromine molecule takes place even in the absence of Lewis acid. It is due to the highly activating effect of –OH group attached to the benzene ring."
Why are positions 2, 4 and 6 the ones that react?
The oxygen of –OH has lone pairs that are delocalised into the ring. The resonance structures place extra negative charge at the two ortho carbons (2 and 6) and the para carbon (4), never at the meta carbons. An electrophile such as Br+ attacks where the electron density is highest. Oxygen also pulls electrons through the σ bond, but the resonance effect is larger, so the ring is more reactive than benzene. Our post on electrophilic substitution in benzene covers directing groups in general.
So in bromine water every activated position reacts. NCERT does not go further into why the solvent matters; for the exam, what you need is the pairing of each set of conditions with its product.
Worked example 1: the mole calculation
0.941 g of phenol is treated with excess bromine water. Find the mass of 2,4,6-tribromophenol formed and the mass of bromine used. Use atomic masses C 12.011, H 1.008, O 15.999 and Br 79.904 from the IUPAC table of standard atomic weights.
Equation: C6H5OH + 3Br2 → C6H2Br3OH + 3HBr.
M(phenol) = 94.11 g/mol, so n = 0.941/94.11 = 0.0100 mol.
M(tribromophenol, C6H3Br3O) = 330.80 g/mol. Mass = 0.0100 × 330.80 = 3.31 g.
Br2 used: 3 × 0.0100 = 0.0300 mol, and M(Br2) = 159.81 g/mol, so 4.79 g.
HBr formed: 0.0300 mol.
The common slip is using a 1 : 1 ratio and getting 1.60 g of bromine. Three substitutions need three Br2 molecules, and each releases one HBr.
Worked example 2: run it backwards
A sample gives 4.96 g of dry tribromophenol. How much phenol was there?
n(precipitate) = 4.96/330.80 = 0.0150 mol.
One phenol gives one tribromophenol, so n(phenol) = 0.0150 mol.
Mass = 0.0150 × 94.11 = 1.41 g.
Worked example 3: predict the product for a substituted phenol
Apply the same rule: bromine goes to every free position ortho or para to –OH.
4-Methylphenol (p-cresol). The para position is taken by CH3. Only 2 and 6 are free, so bromine water gives 2,6-dibromo-4-methylphenol, using two Br2.
2-Methylphenol (o-cresol). Position 2 is taken. Positions 4 and 6 are free, giving 4,6-dibromo-2-methylphenol.
3-Methylphenol (m-cresol). The methyl group sits at a meta position, so 2, 4 and 6 are all free. The product is 2,4,6-tribromo-3-methylphenol.
The methyl group also activates the ring, but here it only blocks or does not block a position. The –OH group is the stronger director and decides where bromine goes.
Common mistakes and how to avoid them
Adding FeBr3 for phenol. It is not needed. Writing it suggests you think phenol behaves like benzene.
Putting bromine at position 3 or 5. Those are meta to –OH and are not activated by resonance.
Writing one product for CS2 conditions. NCERT's word is monobromophenols, plural. Give both the 2- and 4-isomers.
Forgetting HBr. The equation must balance: three HBr for three substitutions.
Confusing it with aniline. Aniline also gives a tribromo product with bromine water, for the same reason: –NH2 is strongly activating. Our post on amines and diazonium salts covers it.
A practice routine
Day 1: Draw the resonance structures of phenol and mark the three carbons that carry negative charge.
Day 2: Predict bromine-water products for 4-nitrophenol, 2,4-dimethylphenol and 3-chlorophenol, giving the number of Br2 used for each.
Day 3: Do three mole calculations, one forwards, one backwards and one with a limiting amount of bromine.
Day 4: Write the conditions and products for halogenation, nitration and sulphonation of phenol from memory, then check NCERT.
If you can follow the resonance argument but cannot reproduce it, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, each with its own lesson.
Today, write the balanced equation for 4-methylphenol with bromine water and check that it uses two Br2, not three.
