ChemistryISC Class 12 · Alcohols, Phenols and Ethers

Reactions of Phenols: ISC Class 12 Chemistry

Updated 15 October 2026 · Fact-checked

Phenol is weakly acidic: it reacts with sodium and NaOH but not with NaHCO₃. Its –OH group strongly activates the ring, so bromination, nitration and sulphonation occur easily at ortho and para positions. Name reactions include Kolbe's (salicylic acid) and Reimer–Tiemann (salicylaldehyde). Neutral FeCl₃ gives a violet colour.

Notice that the –OH group makes phenol react easily with reagents, but the dashed red arrow shows it is too weak an acid to react with NaHCO₃.

Illustration drawn for QuestionBuddy.

Understand Reactions of Phenols

Preparation (ISC list). (i) From benzenediazonium chloride: warming its aqueous solution gives phenol, $C_{6} H_{5} N_{2} C l + H_{2} O \to C_{6} H_{5} O H + N_{2} + H C l$ . (ii) Dow's process: chlorobenzene with aqueous NaOH at 623 K and 300 atm gives sodium phenoxide, then acid. (iii) From benzenesulphonic acid: benzene is sulphonated with oleum, $C_{6} H_{6} + H_{2} S O_{4} \to C_{6} H_{5} S O_{3} H + H_{2} O$ ; the sodium salt is fused with NaOH, $C_{6} H_{5} S O_{3} N a + 2 N a O H \to C_{6} H_{5} O N a + N a_{2} S O_{3} + H_{2} O$ , and the phenoxide is acidified. (iv) Cumene process (manufacture): cumene (isopropylbenzene) is oxidised by air to cumene hydroperoxide, which dilute acid splits into phenol and propanone.

Physical properties. Phenol is a colourless crystalline solid (turning pink on standing in air), with a characteristic smell; it is moderately soluble in water (hydrogen bonding), less than lower aliphatic alcohols because of the large hydrophobic benzene ring.

Acidity. In phenol the O–H oxygen is attached to an sp² carbon, and its lone pair is delocalised into the ring; the phenoxide ion is stabilised by resonance (the negative charge is spread to ortho and para carbons), while an alkoxide ion is not. So phenol (pKa about 10) is much more acidic than ethanol. Electron-withdrawing groups (–NO₂) at ortho/para increase acidity; electron-releasing groups (–CH₃) decrease it. Phenol is weaker than carbonic acid, so it does not liberate CO₂ from NaHCO₃.
• With sodium: $2 C_{6} H_{5} O H + 2 N a \to 2 C_{6} H_{5} O N a + H_{2}$ .
• With NaOH: $C_{6} H_{5} O H + N a O H \to C_{6} H_{5} O N a + H_{2} O$ (alcohols do not react).
• With zinc dust on heating: $C_{6} H_{5} O H + Z n \to C_{6} H_{6} + Z n O$ .

Reactions of the –OH group. Acylation (esterification) with acetyl chloride (in pyridine, which removes HCl) or acetic anhydride gives phenyl acetate: $C_{6} H_{5} O H + C H_{3} C O C l \to C_{6} H_{5} O C O C H_{3} + H C l$ ; $C_{6} H_{5} O H + (C H_{3} C O)_{2} O \to C_{6} H_{5} O C O C H_{3} + C H_{3} C O O H$ . With PCl₅, phenol gives only a poor yield of chlorobenzene, $C_{6} H_{5} O H + P C l_{5} \to C_{6} H_{5} C l + P O C l_{3} + H C l$ , because the C–O bond has partial double-bond character.

Electrophilic substitution. –OH is a strongly activating, ortho-para directing group (its +R effect pushes electron density into the ring), so phenol reacts under much milder conditions than benzene.
• Bromination: in a solvent of low polarity (CHCl₃ or CS₂) at low temperature, monobromophenols form (para major): $C_{6} H_{5} O H + B r_{2} \to C_{6} H_{4} B r O H + H B r$ . With bromine water, a white precipitate of 2,4,6-tribromophenol forms at once: $C_{6} H_{5} O H + 3 B r_{2} \to C_{6} H_{2} B r_{3} O H + 3 H B r$ . In water phenol ionises partly to phenoxide, which is even more strongly activated; no Lewis acid is needed.
• Nitration: dilute HNO₃ at 298 K gives a mixture of o- and p-nitrophenol; concentrated HNO₃ gives 2,4,6-trinitrophenol (picric acid).
• Sulphonation: concentrated H₂SO₄ gives o-hydroxybenzenesulphonic acid as the main product at room temperature and the p-isomer at about 373 K: $C_{6} H_{5} O H + H_{2} S O_{4} \to C_{6} H_{4} (O H) S O_{3} H + H_{2} O$ .
• Kolbe's reaction: sodium phenoxide heated with CO₂ (about 400 K, 4–7 atm) and then acidified gives salicylic acid (2-hydroxybenzoic acid): $C_{6} H_{5} O N a + C O_{2} \to C_{6} H_{4} (O H) C O O N a$ , then $C_{6} H_{4} (O H) C O O N a + H C l \to C_{6} H_{4} (O H) C O O H + N a C l$ . Acetylation of salicylic acid gives aspirin.
• Reimer–Tiemann reaction: phenol warmed with chloroform and aqueous NaOH (about 340 K), then acidified, gives salicylaldehyde (2-hydroxybenzaldehyde); dichlorocarbene is the electrophile. Overall: $C_{6} H_{5} O H + C H C l_{3} + 3 N a O H \to C_{6} H_{4} (O H) C H O + 3 N a C l + 2 H_{2} O$ .

Tests for phenol. (i) Neutral FeCl₃ gives a violet colour (alcohols do not). (ii) Azo dye test: phenol in NaOH couples with benzenediazonium chloride at 273–278 K to give an orange dye, p-hydroxyazobenzene: $C_{6} H_{5} N_{2} C l + C_{6} H_{5} O H \to C_{6} H_{5} N = N C_{6} H_{4} O H + H C l$ . (iii) Bromine water: white precipitate.

Uses. Antiseptic and disinfectant (carbolic acid), manufacture of Bakelite and other resins, dyes, drugs (aspirin, salicylic acid) and picric acid.

Reactions of Phenols: formulas and key facts

Bromine water test: $C_{6} H_{5} O H + 3 B r_{2} \to C_{6} H_{2} B r_{3} O H + 3 H B r$; White precipitate of 2,4,6-tribromophenol.
Kolbe's reaction: $C_{6} H_{5} O N a + C O_{2} 400 K, 4 - 7 a t m C_{6} H_{4} (O H) C O O N a H^{+} salicylic acid$; Ortho carboxylation of the phenoxide ion.
Reimer–Tiemann reaction: $C_{6} H_{5} O H + C H C l_{3} + 3 N a O H \to C_{6} H_{4} (O H) C H O + 3 N a C l + 2 H_{2} O$; Warm (about 340 K), then acidify; gives salicylaldehyde.
Acetylation: $C_{6} H_{5} O H + (C H_{3} C O)_{2} O \to C_{6} H_{5} O C O C H_{3} + C H_{3} C O O H$; Phenyl acetate; acetyl chloride also works (pyridine).
Azo dye test: $C_{6} H_{5} N_{2} C l + C_{6} H_{5} O H N a O H, 273 - 278 K C_{6} H_{5} N = N C_{6} H_{4} O H + H C l$; Orange dye, p-hydroxyazobenzene.
Acidity comparison: $C_{2} H_{5} O H < C_{6} H_{5} O H < H_{2} C O_{3}$; Phenol reacts with NaOH but not with NaHCO₃.

How to solve Reactions of Phenols questions

Phenol questions in ISC are conversions, named reactions, tests and reasons about acidity or reactivity. Treat the molecule as two parts: the acidic –OH group and the activated ring.

1Decide which part reacts: the O–H bond (Na, NaOH, acylating agents), the C–O bond (PCl₅, Zn dust) or the ring (Br₂, HNO₃, H₂SO₄, CO₂ with phenoxide, CHCl₃/NaOH, diazonium salts).
2For ring substitution, give the ortho and para products and note the effect of conditions (bromine water versus Br₂ in CS₂; dilute versus concentrated HNO₃).
3Write named reactions in full with conditions and by-products (Reimer–Tiemann: CHCl₃ + NaOH; Kolbe: CO₂, 400 K, 4–7 atm, then H⁺).
4For tests, give the reagent, the observation with phenol, and the negative result with the other compound.
5For acidity reasons, draw or describe the resonance of the phenoxide ion and compare with an alkoxide ion.
6Balance every equation; count Br₂ and HBr in tribromination and NaOH in Reimer–Tiemann.

Reactions of Phenols in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Kolbe's reaction makes the raw material for aspirin

Suppose a small pharmaceutical unit converts 9.4 g of phenol (M = 94 g mol⁻¹) into sodium phenoxide, heats it with CO₂ under pressure, and then acidifies the product. It isolates 11.0 g of salicylic acid (M = 138 g mol⁻¹), which it will acetylate to make aspirin.

Write the equations, calculate the theoretical and percentage yields of salicylic acid, and name the reagent used to convert it into aspirin.

Try it first, then show the solutionSolution

$C_{6} H_{5} O H + N a O H \to C_{6} H_{5} O N a + H_{2} O$ ; $C_{6} H_{5} O N a + C O_{2} \to C_{6} H_{4} (O H) C O O N a$ ; $C_{6} H_{4} (O H) C O O N a + H C l \to C_{6} H_{4} (O H) C O O H + N a C l$ .
Moles of phenol $= \frac{9.4}{94} = 0.10$ mol, so the theoretical mass of salicylic acid $= 0.10 \times 138 = 13.8$ g.
Percentage yield $= \frac{11.0}{13.8} \times 100 = 79.7%$ .
Acetic anhydride acetylates the –OH group: $C_{6} H_{4} (O H) C O O H + (C H_{3} C O)_{2} O \to C_{6} H_{4} (O C O C H_{3}) C O O H + C H_{3} C O O H$ (aspirin).

Answer: Theoretical yield 13.8 g; percentage yield about 80%. Acetic anhydride converts salicylic acid into aspirin.

Memory hook: Phenoxide plus CO₂ gives salicylic acid; add an acetyl group and you have aspirin.

Tricks to remember Reactions of Phenols

Short memory aids. Say each one aloud once, then use it on a question the same day.

Kolbe = CO₂, Reimer–Tiemann = CHO
Kolbe puts COOH on the ring (Carbon dioxide → carboxylic acid); Reimer–Tiemann puts CHO on (CHCl₃ → aldehyde). Both go ortho.
Use it when: Naming and writing the two reactions.
Phenol: Na yes, NaOH yes, NaHCO₃ no
Phenol is acidic enough for sodium and sodium hydroxide but too weak for sodium hydrogencarbonate (carboxylic acids fizz with it).
Use it when: Distinguishing phenol from carboxylic acids.
Water makes it triple
Bromine in water: three Br atoms (2,4,6-tribromophenol). Bromine in CS₂ in the cold: one Br atom.
Use it when: Products of bromination.

Reactions of Phenols: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Phenol with zinc dust · 1 mark · MCQBoard level

Phenol is heated with zinc dust. The organic product is:
(a) cyclohexanol
(b) benzene
(c) zinc phenoxide only
(d) phenyl acetate

Try it first, then show the solutionSolution

Zinc removes the oxygen atom: $C_{6} H_{5} O H + Z n \to C_{6} H_{6} + Z n O$ .

Answer: (b) benzene.

Example 2: The azo dye test · 2 marks · Short answerBoard level

Describe the azo dye test for phenol. Write the equation and the observation.

Try it first, then show the solutionSolution

Dissolve phenol in NaOH, cool to 273–278 K and add benzenediazonium chloride solution (½ mark).
$C_{6} H_{5} N_{2} C l + C_{6} H_{5} O H \to C_{6} H_{5} N = N C_{6} H_{4} O H + H C l$ (1 mark).
An orange dye (p-hydroxyazobenzene) forms (½ mark).

Answer: Coupling with benzenediazonium chloride in alkali gives an orange azo dye.

Example 3: Acylating phenol · 3 marks · Short answerHarder board level

Write equations for the reaction of phenol with (a) acetyl chloride and (b) acetic anhydride, naming the organic product. Why is phenol usually acylated with an acid chloride or anhydride rather than with acetic acid itself?

Try it first, then show the solutionSolution

(a) $C_{6} H_{5} O H + C H_{3} C O C l pyridine C_{6} H_{5} O C O C H_{3} + H C l$ (1 mark).
(b) $C_{6} H_{5} O H + (C H_{3} C O)_{2} O \to C_{6} H_{5} O C O C H_{3} + C H_{3} C O O H$ ; product phenyl acetate (phenyl ethanoate) in both (1 mark).
The lone pair on phenol's oxygen is delocalised into the ring, so phenol is a weak nucleophile; it reacts well only with the more reactive acid chlorides and anhydrides, and pyridine removes the HCl formed (1 mark).

Answer: Phenyl acetate in both; phenol's oxygen is a poor nucleophile, so a more reactive acylating agent is used.

Example 4: Conditions change the product · 3 marks · Short answerHarder board level

Explain how the products differ when phenol is treated with (a) bromine water and with bromine in CS₂ at 273 K, and (b) concentrated H₂SO₄ at room temperature and at 373 K.

Try it first, then show the solutionSolution

(a) In water phenol partly ionises to phenoxide, which is very strongly activated, and the polar solvent helps polarise Br₂: all three o/p positions are brominated, giving a white precipitate of 2,4,6-tribromophenol (1 mark).
In CS₂ (low polarity) at low temperature, ionisation does not occur and the reaction is controlled: mainly 4-bromophenol with some 2-bromophenol (1 mark).
(b) Sulphonation at room temperature gives mainly o-hydroxybenzenesulphonic acid; at 373 K the more stable p-isomer is the main product (1 mark).

Answer: (a) Tribromophenol in water; monobromophenols in CS₂. (b) Ortho acid at room temperature; para acid at 373 K.

Example 5: From benzene to aspirin · 5 marks · Long answerHOTS / competency

Benzene is converted into A with concentrated H₂SO₄. The sodium salt of A is fused with NaOH and the melt acidified to give B (C₆H₆O). B with CHCl₃ and aqueous NaOH, followed by acid, gives C (C₇H₆O₂). The sodium salt of B heated with CO₂ under pressure and then acidified gives D (C₇H₆O₃), which with acetic anhydride gives E.
(a) Identify A to E. [2½]
(b) Write the equations for the formation of B from the sodium salt of A and for the formation of C. [2]
(c) Name the reactions that give C and D. [½]

Try it first, then show the solutionSolution

(a) A benzenesulphonic acid, B phenol, C salicylaldehyde (2-hydroxybenzaldehyde), D salicylic acid (2-hydroxybenzoic acid), E aspirin (acetylsalicylic acid) (½ mark each).
(b) $C_{6} H_{5} S O_{3} N a + 2 N a O H \to C_{6} H_{5} O N a + N a_{2} S O_{3} + H_{2} O$ , then H⁺ gives phenol (1 mark).
$C_{6} H_{5} O H + C H C l_{3} + 3 N a O H \to C_{6} H_{4} (O H) C H O + 3 N a C l + 2 H_{2} O$ (after acidification) (1 mark).
(c) C: Reimer–Tiemann reaction; D: Kolbe's reaction (½ mark).

Answer: A C₆H₅SO₃H, B C₆H₅OH, C salicylaldehyde, D salicylic acid, E aspirin; Reimer–Tiemann and Kolbe reactions.

Where students lose marks in Reactions of Phenols

The slips examiners see most often, and what to write instead.

Mistake: Incomplete Reimer–Tiemann equation
ISC examiners reported incomplete, unbalanced Reimer–Tiemann equations without by-products (2025), and CH₃Cl used instead of CHCl₃ (2024).
Wrong working
$C_{6} H_{5} O H + C H_{3} C l \to$ salicylaldehyde
Right working
$C_{6} H_{5} O H + C H C l_{3} + 3 N a O H \to C_{6} H_{4} (O H) C H O + 3 N a C l + 2 H_{2} O$
Do this: Use chloroform and alkali; include NaCl and water; mention acidification.
Usually costs: 1 mark.
Mistake: Wrong colour with neutral FeCl₃
Examiners recorded 'orange' or 'pink' instead of violet (2024).
Do this: Phenol gives a violet colour with neutral FeCl₃; the orange colour belongs to the azo dye test.
Usually costs: 1 mark.
Mistake: Missing conditions and by-products in nitration
In 2025 examiners noted conc. H₂SO₄ and water omitted when writing the preparation of picric acid, and unbalanced equations.
Do this: Write all reagents over the arrow and balance: three HNO₃ give three H₂O.
Usually costs: 1 mark.
Mistake: Wrong reason for lower solubility of phenol
Examiners noted reasons such as 'resonance in benzene' (2025).
Do this: Phenol forms hydrogen bonds with water, but its large hydrophobic benzene ring limits solubility compared with lower aliphatic alcohols.
Usually costs: 1 mark.

Exam tips

For each named reaction, learn the reagents, conditions, product name and a balanced equation.
For distinguishing tests, write what happens with both compounds (positive for one, negative for the other).
Remember that phenol needs no Lewis acid for bromination; mention this when asked to compare with benzene.
When asked for acidity order, write it in the direction asked (increasing or decreasing) and use '<' or '>' clearly.
Draw the phenoxide resonance structures when explaining acidity for full marks.

Practice questions on Reactions of Phenols

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Phenol does NOT react with:
(a) sodium
(b) sodium hydroxide
(c) sodium hydrogencarbonate
(d) bromine water
Hint
Compare phenol's acidity with that of carbonic acid.
Answer
(c). Phenol is weaker than carbonic acid and cannot release CO₂.
Q2 · 2 marks · Short answerBoard level
Write the balanced equation for the Reimer–Tiemann reaction and name the product.
Hint
Include NaCl and water as by-products.
Answer
$C_{6} H_{5} O H + C H C l_{3} + 3 N a O H \to C_{6} H_{4} (O H) C H O + 3 N a C l + 2 H_{2} O$ (warm, then acidify). Product: salicylaldehyde (2-hydroxybenzaldehyde).
Q3 · 2 marks · Short answerHarder board level
Ethanol gives a good yield of chloroethane with PCl₅, but phenol gives only a poor yield of chlorobenzene. Explain, with the equation for phenol.
Hint
Think about the strength of the C–O bond.
Answer
$C_{6} H_{5} O H + P C l_{5} \to C_{6} H_{5} C l + P O C l_{3} + H C l$ . In phenol the C–O bond has partial double-bond character (lone pair delocalised into the ring) and the carbon is sp², so the C–O bond is hard to break; in ethanol it is a normal single bond.
Q4 · 3 marks · Short answerHarder board level
Write equations for the preparation of phenol from (a) benzenediazonium chloride and (b) benzenesulphonic acid.
Hint
(b) needs fusion with NaOH and then acid.
Answer
(a) Warm with water: $C_{6} H_{5} N_{2} C l + H_{2} O \to C_{6} H_{5} O H + N_{2} + H C l$ . (b) $C_{6} H_{5} S O_{3} H + N a O H \to C_{6} H_{5} S O_{3} N a + H_{2} O$ ; fuse with NaOH: $C_{6} H_{5} S O_{3} N a + 2 N a O H \to C_{6} H_{5} O N a + N a_{2} S O_{3} + H_{2} O$ ; acidify: $C_{6} H_{5} O N a + H C l \to C_{6} H_{5} O H + N a C l$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Phenol is more acidic than ethanol.
Reason (R): The phenoxide ion is stabilised by delocalisation of its negative charge over the benzene ring, whereas the ethoxide ion is not.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Stability of the conjugate base.
Answer
(a).
Q6 · 3 marks · Short answerHarder board level
Explain why the –OH group makes the benzene ring of phenol (a) more reactive than benzene and (b) ortho-para directing. Why is no Lewis acid needed to brominate phenol?
Hint
Resonance structures of phenol.
Answer
(a) A lone pair on oxygen is delocalised into the ring (+R effect), raising its electron density, so electrophiles attack faster. (b) The resonance structures place negative charge at the ortho and para carbons, so attack occurs there. The ring is so electron-rich that it can polarise Br₂ itself; a Lewis acid is not needed.
Q7 · 3 marks · Short answerHOTS / competency
4.7 g of phenol (M = 94) is treated with excess bromine water. (a) How many moles of Br₂ react? (b) What mass of 2,4,6-tribromophenol (M = 331) is formed if the reaction is complete? (c) Name the by-product and say how it could be detected.
Hint
Use $C_{6} H_{5} O H + 3 B r_{2} \to C_{6} H_{2} B r_{3} O H + 3 H B r$ .
Answer
(a) Phenol $= 0.050$ mol, so Br₂ $= 3 \times 0.050 = 0.15$ mol. (b) $0.050 \times 331 = 16.5$ g. (c) HBr; the solution becomes acidic (turns blue litmus red) and gives a pale yellow precipitate of AgBr with AgNO₃.
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a water-quality chemist tests factory effluent for phenol. A sample gives a violet colour with neutral FeCl₃ solution. Another 100 mL sample, treated with excess bromine water, gives 0.331 g of a dry white precipitate.
(i) What do the two observations show? [1]
(ii) Calculate the mass of phenol in the 100 mL sample. (C₆H₂Br₃OH: M = 331; phenol: M = 94) [1]
(iii) Why must the FeCl₃ solution be neutral for this test? [1]
Hint
1 mol phenol gives 1 mol tribromophenol.
Answer
(i) Violet colour with FeCl₃ indicates a phenol; the white precipitate is 2,4,6-tribromophenol. (ii) Moles of precipitate $= \frac{0.331}{331} = 1.0 \times 1 0^{- 3}$ mol, so phenol $= 1.0 \times 1 0^{- 3} \times 94 = 0.094$ g (94 mg). (iii) The coloured iron-phenoxide complex forms only in neutral solution; acid suppresses phenoxide formation and alkali precipitates Fe(OH)₃.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Product of Kolbe's reaction?Show answer
Salicylic acid (2-hydroxybenzoic acid).
2.Product of Reimer–Tiemann reaction?Show answer
Salicylaldehyde (2-hydroxybenzaldehyde).
3.Colour of phenol with neutral FeCl₃?Show answer
Violet.
4.Product of phenol with bromine water?Show answer
2,4,6-Tribromophenol (white precipitate).
5.Product of phenol heated with zinc dust?Show answer
Benzene.
6.Does phenol react with NaHCO₃?Show answer
No; it is weaker than carbonic acid.

Reactions of Phenols: frequently asked questions

Why is phenol acidic?

Loss of H⁺ gives the phenoxide ion, whose negative charge is delocalised over the benzene ring by resonance, making it relatively stable.

What is the Reimer–Tiemann reaction?

Phenol warmed with chloroform and aqueous sodium hydroxide, followed by acidification, gives salicylaldehyde; a –CHO group enters ortho to –OH.

How is phenol distinguished from ethanol?

Phenol gives a violet colour with neutral FeCl₃ and a white precipitate with bromine water; ethanol does not. Ethanol gives the iodoform test; phenol does not.

What is Kolbe's reaction?

Sodium phenoxide heated with carbon dioxide under pressure (about 400 K, 4–7 atm) and then acidified gives salicylic acid.

Why is phenol more reactive than benzene in electrophilic substitution?

The –OH group donates electron density into the ring by resonance, activating it, especially at the ortho and para positions.