ChemistryCBSE Class 12 · Alcohols, Phenols and Ethers

Reactions of Phenols: Acidity and Electrophilic Substitution

Q: Why is phenol acidic in Class 12 Chemistry?

Because losing a proton gives the phenoxide ion, whose negative charge is spread over the benzene ring by resonance, making it relatively stable.

Q: What is the Reimer–Tiemann reaction?

The reaction of phenol with chloroform and aqueous sodium hydroxide, followed by acidification, which introduces a –CHO group ortho to the –OH, giving salicylaldehyde.

Q: What is Kolbe's reaction?

Sodium phenoxide reacts with carbon dioxide under pressure and heat; acidification gives salicylic acid (2-hydroxybenzoic acid).

Q: How does a nitro group affect the acidity of phenol?

At ortho or para it increases acidity by delocalising the phenoxide's negative charge; at meta it increases acidity less, through the inductive effect only.

Q: Why does phenol give a white precipitate with bromine water?

The –OH group activates all ortho and para positions, so 2,4,6-tribromophenol, which is insoluble in water, forms immediately.

Updated 15 October 2026 · Fact-checked

Phenol is more acidic than alcohols because the phenoxide ion is stabilised by resonance; nitro groups at ortho or para positions raise its acidity further. The –OH group strongly activates the ring towards ortho and para attack, so phenol is easily nitrated and brominated. Its named reactions are Kolbe's reaction (salicylic acid) and the Reimer–Tiemann reaction (salicylaldehyde).

Lower pKa means stronger acid: phenol beats ethanol, and nitro groups, most at ortho and para, make phenol more acidic still.

Illustration drawn for QuestionBuddy.

Understand Reactions of Phenols

Acidity of phenols. Phenol reacts with sodium (like alcohols) and also with aqueous NaOH (unlike alcohols) to give sodium phenoxide: $C_{6} H_{5} O H + N a O H \to C_{6} H_{5} O^{-} N a^{+} + H_{2} O$ . Two reasons make it more acidic than ethanol. First, the oxygen is attached to an $s p^{2}$ carbon, which is more electronegative than an $s p^{3}$ carbon, so the O–H bond is more polar. Second, and more important, the phenoxide ion is stabilised by resonance: the negative charge on oxygen is delocalised over the ortho and para carbons of the ring. Phenol itself also shows resonance, but its structures involve charge separation, so the phenoxide ion gains more. NCERT's $p K_{a}$ values: ethanol 15.9, phenol 10.0.

Effect of substituents. Electron-withdrawing groups such as $- N O_{2}$ at the ortho or para position spread the negative charge of the phenoxide even further (onto the nitro oxygens), so acidity increases: $p K_{a}$ of o-nitrophenol 7.2 and p-nitrophenol 7.1. A meta nitro group withdraws electrons only inductively, so m-nitrophenol ( $p K_{a}$ 8.3) is less acidic than the o and p isomers but still more acidic than phenol. Each additional nitro group increases acidity: 2,4,6-trinitrophenol (picric acid) is a strong acid. Electron-releasing groups such as $- C H_{3}$ destabilise the phenoxide, so cresols ( $p K_{a}$ about 10.1–10.2) are slightly weaker acids than phenol.

Esterification. Like alcohols, phenols form esters with acid chlorides and anhydrides; acetylation of salicylic acid with ethanoic anhydride gives aspirin.

Electrophilic aromatic substitution. The –OH group donates electron density to the ring by resonance (+R), making the ring very reactive and directing electrophiles to ortho and para positions.

*Nitration.* With dilute $H N O_{3}$ at 298 K, phenol gives a mixture of o-nitrophenol and p-nitrophenol. They are separated by steam distillation: o-nitrophenol is steam volatile because of intramolecular hydrogen bonding, while p-nitrophenol is less volatile because its molecules are linked by intermolecular hydrogen bonds. With concentrated $H N O_{3}$ , phenol gives 2,4,6-trinitrophenol (picric acid), but the yield is poor because the ring is also oxidised.
*Halogenation.* In a solvent of low polarity ( $C H C l_{3}$ or $C S_{2}$ ) at low temperature, phenol gives monobromophenols (mainly para). With bromine water, it gives a white precipitate of 2,4,6-tribromophenol at once: $C_{6} H_{5} O H + 3 B r_{2} \to C_{6} H_{2} B r_{3} O H + 3 H B r$ . No Lewis acid catalyst is needed because the ring is so activated, and in water phenol is partly ionised to the even more reactive phenoxide ion.

Kolbe's reaction. Sodium phenoxide is even more reactive than phenol. With carbon dioxide (a weak electrophile) under pressure (about 4–7 atm, 400 K), followed by acidification, it gives 2-hydroxybenzoic acid (salicylic acid) as the main product.

Reimer–Tiemann reaction. Phenol heated with chloroform and aqueous NaOH (about 340 K) introduces a –CHO group at the ortho position. A substituted benzal chloride forms first and is hydrolysed by the alkali; acidification gives salicylaldehyde (2-hydroxybenzaldehyde).

Other reactions. Distilling phenol with zinc dust gives benzene: $C_{6} H_{5} O H + Z n \to C_{6} H_{6} + Z n O$ . Oxidation with chromic acid gives benzoquinone, and phenols slowly darken in air for the same reason.

Reactions of Phenols: formulas and key facts

Acidity order: $ethanol (p K_{a} 15.9) < cresols < phenol (10.0) < m -nitrophenol (8.3) < o - (7.2) \approx p -nitrophenol (7.1)$; Lower $p K_{a}$ = stronger acid; o/p nitro groups stabilise phenoxide by resonance.
Bromine water test: $C_{6} H_{5} O H + 3 B r_{2} H_{2} O C_{6} H_{2} B r_{3} O H ↓ + 3 H B r$; White precipitate of 2,4,6-tribromophenol; no catalyst needed.
Nitration: $C_{6} H_{5} O H dil. H N O_{3}, 298 K o - and p -nitrophenol; conc. H N O_{3} 2,4,6-trinitrophenol$; Separate o and p isomers by steam distillation.
Kolbe's reaction: $C_{6} H_{5} O N a (i) C O_{2}, 4 - 7 atm, 400 K (ii) H^{+} o - H O C_{6} H_{4} C O O H$; Salicylic acid (2-hydroxybenzoic acid).
Reimer–Tiemann reaction: $C_{6} H_{5} O H (i) C H C l_{3}, a q . N a O H, 340 K (ii) H^{+} o - H O C_{6} H_{4} C H O$; Salicylaldehyde (2-hydroxybenzaldehyde).
Reduction with zinc dust: $C_{6} H_{5} O H + Z n Δ C_{6} H_{6} + Z n O$; Converts phenol to benzene.

How to solve Reactions of Phenols questions

Questions cover acidity comparisons, products of substitution reactions, named reactions and conversions. Treat the –OH group as a strong ortho/para director and the phenoxide ion as the key to acidity.

1For acidity, compare the stability of the anions: resonance in phenoxide; extra delocalisation by o/p $N O_{2}$ ; inductive effect only for m- $N O_{2}$ ; destabilisation by $C H_{3}$ .
2Write the order and support it with one line on each effect (and $p K_{a}$ values if remembered).
3For substitution, put new groups at ortho and para; decide mono- or tri-substitution from the conditions (dilute vs concentrated $H N O_{3}$ ; $B r_{2}$ in $C S_{2}$ vs bromine water).
4For Kolbe's reaction, start from sodium phenoxide, add $C O_{2}$ under pressure, then acidify: –COOH enters ortho.
5For Reimer–Tiemann, use $C H C l_{3}$ with aqueous NaOH, then acidify: –CHO enters ortho.
6For conversions, link reactions: phenol → salicylic acid (Kolbe) → aspirin (acetylation); phenol → benzene (Zn dust).
7Explain physical differences (steam volatility, boiling points) by intramolecular vs intermolecular hydrogen bonding.

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.

Bromine water test for phenol

Suppose a water-testing lab checks a factory's waste water for phenol by adding excess bromine water to a sample. A white precipitate forms, which is filtered, dried and weighed.

Name the precipitate and write the equation. If 0.94 g of phenol were present, what mass of precipitate would form? (Phenol 94 g mol⁻¹; 2,4,6-tribromophenol 331 g mol⁻¹)

Try it first, then show the solutionSolution

The –OH group activates the ring so strongly that all three o/p positions are brominated: $C_{6} H_{5} O H + 3 B r_{2} \to C_{6} H_{2} B r_{3} O H + 3 H B r$ .
The white precipitate is 2,4,6-tribromophenol.
Moles of phenol $= 0.94/94 = 0.010$ mol, giving 0.010 mol of tribromophenol (1 : 1).
Mass $= 0.010 \times 331 = 3.31$ g.

Answer: 2,4,6-Tribromophenol; about 3.3 g of precipitate from 0.94 g of phenol.

Memory hook: Phenol plus bromine water: three bromines at once, a white cloud, no catalyst needed.

Tricks to remember Reactions of Phenols

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

Kolbe = COOH, Reimer–Tiemann = CHO
Kolbe's 'K' carries carbon dioxide to make a carboxylic acid; Reimer–Tiemann uses chloroform to make an aldehyde. Both enter ortho.
Use it when: Named reaction questions.
Nitro at o/p pulls hardest
o/p $N O_{2}$ pull by resonance and induction; m- $N O_{2}$ only by induction; $C H_{3}$ pushes back.
Use it when: Ordering acidity of substituted phenols.
Ortho holds hands with itself
o-Nitrophenol's OH holds its own $N O_{2}$ (intramolecular H-bond), so it escapes in steam; para holds neighbours (intermolecular) and stays behind.
Use it when: Steam distillation questions.

Reactions of Phenols: solved examples

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

Example 1: Reimer–Tiemann product · 1 mark · MCQBoard level

Phenol on heating with chloroform and aqueous sodium hydroxide, followed by acidification, gives:
(a) benzoic acid (b) salicylaldehyde (c) salicylic acid (d) benzaldehyde

Try it first, then show the solutionSolution

Reimer–Tiemann reaction introduces –CHO at the ortho position of phenol.

Answer: (b) Salicylaldehyde.

Example 2: Two bromination conditions · 2 marks · Very short answerBoard level

Write the products when phenol reacts with (a) bromine in $C S_{2}$ at low temperature and (b) bromine water. Why do the products differ?

Try it first, then show the solutionSolution

(a) o-Bromophenol and p-bromophenol (p major) (½ mark).
(b) 2,4,6-Tribromophenol, a white precipitate (½ mark).
In water, phenol partly ionises to phenoxide, which is even more strongly activated, and water as a polar solvent helps polarise $B r_{2}$ ; so substitution occurs at all three o/p positions. In non-polar $C S_{2}$ there is no ionisation, so the reaction stops at monosubstitution (1 mark).

Answer: (a) o- and p-bromophenol. (b) 2,4,6-tribromophenol; polar water ionises phenol and polarises bromine.

Example 3: Ordering acidity · 3 marks · Short answerHarder board level

Arrange in increasing order of acidic strength and justify: m-nitrophenol, o-cresol, 2,4,6-trinitrophenol, phenol.

Try it first, then show the solutionSolution

Order: o-cresol < phenol < m-nitrophenol < 2,4,6-trinitrophenol (1 mark).
$C H_{3}$ in o-cresol releases electrons and destabilises the phenoxide slightly ( $p K_{a}$ 10.2 against 10.0 for phenol) (½ mark).
m- $N O_{2}$ withdraws electrons only by its −I effect, so it raises acidity moderately ( $p K_{a}$ 8.3) (½ mark).
In picric acid three $N O_{2}$ groups at ortho and para positions delocalise the negative charge by resonance and −I, making it a strong acid ( $p K_{a}$ about 0.5) (1 mark).

Answer: o-Cresol < phenol < m-nitrophenol < 2,4,6-trinitrophenol.

Example 4: Kolbe and Reimer–Tiemann · 3 marks · Short answerHarder board level

Write the equations, with conditions, for (a) Kolbe's reaction and (b) the Reimer–Tiemann reaction. Name the main product in each case.

Try it first, then show the solutionSolution

(a) Phenol with NaOH gives sodium phenoxide; $C_{6} H_{5} O N a + C O_{2} 4 - 7 atm, 400 K o - H O C_{6} H_{4} C O O N a$ , then $H^{+}$ $o - H O C_{6} H_{4} C O O H$ : salicylic acid (2-hydroxybenzoic acid) (1½ marks).
(b) $C_{6} H_{5} O H + C H C l_{3} + a q . N a O H$ at about 340 K gives an intermediate (substituted benzal chloride) that is hydrolysed by alkali to the sodium salt of salicylaldehyde; acidification gives $o - H O C_{6} H_{4} C H O$ : salicylaldehyde (2-hydroxybenzaldehyde) (1½ marks).

Answer: (a) Salicylic acid. (b) Salicylaldehyde.

Example 5: Nitrophenols: separation and acidity · 5 marks · Long answerHOTS / competency

(a) Phenol is nitrated with dilute nitric acid at 298 K. Name the products and explain how they are separated. (b) Why is p-nitrophenol more acidic than m-nitrophenol? (c) Why is the yield of picric acid poor when phenol is treated directly with concentrated $H N O_{3}$ ? (d) How would you convert phenol into benzene?

Try it first, then show the solutionSolution

(a) o-Nitrophenol and p-nitrophenol. Steam distillation: o-nitrophenol has intramolecular hydrogen bonding, so it does not associate and is steam volatile; p-nitrophenol forms intermolecular hydrogen bonds, so it stays behind (2 marks).
(b) In p-nitrophenoxide the negative charge can be delocalised onto the $N O_{2}$ group by resonance as well as being pulled by its −I effect; in the meta isomer only the −I effect operates (1 mark).
(c) Concentrated $H N O_{3}$ is a strong oxidising agent and the highly activated phenol ring is partly oxidised (and charred) instead of being nitrated (1 mark).
(d) Distil phenol with zinc dust: $C_{6} H_{5} O H + Z n \to C_{6} H_{6} + Z n O$ (1 mark).

Answer: (a) o- and p-nitrophenol, separated by steam distillation. (b) Resonance + −I for para; only −I for meta. (c) Oxidation of the ring. (d) Heat with Zn dust.

Where students lose marks in Reactions of Phenols

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

Mistake: Writing salicylic acid for Reimer–Tiemann (or salicylaldehyde for Kolbe)
Do this: Kolbe: $C O_{2}$ → –COOH. Reimer–Tiemann: $C H C l_{3}$ → –CHO.
Usually costs: 1 mark.
Mistake: Monobromination product with bromine water
Wrong working
Phenol + bromine water → p-bromophenol
Right working
Phenol + bromine water → 2,4,6-tribromophenol (white ppt)
Do this: Monobromination needs $B r_{2}$ in $C S_{2}$ or $C H C l_{3}$ at low temperature.
Usually costs: 1 mark.
Mistake: Explaining acidity by the stability of phenol itself
Do this: Acidity depends on the stability of the anion (phenoxide), not the acid.
Usually costs: ½ to 1 mark.
Mistake: Putting m-nitrophenol as the strongest nitrophenol
Do this: o- and p-nitrophenols are stronger (resonance + −I); meta has only −I.
Usually costs: 1 mark.

Exam tips

Write conditions for named reactions: Kolbe ( $C O_{2}$ , 4–7 atm, 400 K, then $H^{+}$ ); Reimer–Tiemann ( $C H C l_{3}$ , aq. NaOH, then $H^{+}$ ).
Draw at least one resonance structure of phenoxide when explaining acidity.
For nitration, mention steam distillation and the type of hydrogen bonding in one line each.
Use NCERT's $p K_{a}$ values when ordering phenols; they support your reasoning.
In conversions, name intermediates (sodium phenoxide, salicylic acid).

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 on distillation with zinc dust gives:
(a) cyclohexanol (b) benzene (c) benzoic acid (d) benzoquinone
Hint
Zinc removes oxygen.
Answer
(b) Benzene.
Q2 · 2 marks · Very short answerBoard level
Why is phenol more acidic than ethanol? Give two reasons.
Hint
Anion stability and hybridisation.
Answer
(1) The phenoxide ion is stabilised by resonance (negative charge delocalised over the ring), whereas ethoxide is not. (2) In phenol, O is attached to a more electronegative $s p^{2}$ carbon, making the O–H bond more polar; the ethyl group's +I effect makes ethanol's O–H less polar.
Q3 · 3 marks · Short answerHarder board level
Complete: (a) Phenol + conc. $H N O_{3}$ → (b) Phenol + chromic acid → (c) Phenol + NaOH(aq) →
Hint
Trinitration, oxidation, salt formation.
Answer
(a) 2,4,6-Trinitrophenol (picric acid) + $H_{2} O$ (poor yield). (b) Benzoquinone (p-benzoquinone). (c) Sodium phenoxide + water: $C_{6} H_{5} O H + N a O H \to C_{6} H_{5} O N a + H_{2} O$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Phenol is a stronger acid than ethanol.
Reason (R): The ethoxide ion is stabilised by resonance.
(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
Which anion actually has resonance structures?
Answer
(c). Phenol is more acidic, but it is the phenoxide ion, not ethoxide, that is stabilised by resonance.
Q5 · 3 marks · Short answerHarder board level
Explain why the –OH group of phenol is ortho, para-directing and activating, and why phenol reacts with bromine water without a Lewis acid catalyst.
Hint
+R effect of oxygen lone pairs.
Answer
Oxygen's lone pair is delocalised into the ring (+R), raising electron density, especially at ortho and para positions (shown by resonance structures with negative charge there). The ring is so electron-rich that even unpolarised $B r_{2}$ is attacked; in water, phenol also forms some phenoxide, which is more reactive still. So no $F e B r_{3}$ is needed.
Q6 · 1 mark · Very short answerHarder board level
Give a simple chemical test to distinguish phenol from ethanol.
Hint
Bromine water.
Answer
Add bromine water: phenol decolourises it and gives a white precipitate of 2,4,6-tribromophenol; ethanol gives no precipitate. (Neutral $F e C l_{3}$ also gives a violet colour with phenol only.)
Q7 · 3 marks · Short answerHOTS / competency
Show how aspirin can be prepared from phenol in two steps, writing the reagents and conditions for each step.
Hint
Kolbe, then acetylation.
Answer
Step 1 (Kolbe's reaction): phenol → sodium phenoxide (NaOH); $C O_{2}$ at 4–7 atm, 400 K; then $H^{+}$ → salicylic acid. Step 2 (acetylation): salicylic acid + ethanoic anhydride (with a little conc. $H_{2} S O_{4}$ ) → acetylsalicylic acid (aspirin) + ethanoic acid.
Q8 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose a student is given this table of $p K_{a}$ values: phenol 10.0, o-nitrophenol 7.2, m-nitrophenol 8.3, p-nitrophenol 7.1, p-cresol 10.2, ethanol 15.9.
(a) Which compound is the strongest acid in the table? (1 mark)
(b) Why is m-nitrophenol a weaker acid than its o- and p-isomers? (1 mark)
(c) Why is p-cresol a slightly weaker acid than phenol? (1 mark)
(d) Phenol dissolves in NaOH but does not liberate $C O_{2}$ from sodium hydrogencarbonate. What does this tell you about its acid strength? (1 mark)
Hint
Lower $p K_{a}$ , stronger acid.
Answer
(a) p-Nitrophenol ( $p K_{a}$ 7.1). (b) At meta, the $N O_{2}$ group cannot take part in resonance with the phenoxide oxygen; it acts only by −I. (c) The methyl group releases electrons (+I, hyperconjugation), destabilising the phenoxide ion. (d) Phenol is a stronger acid than water but weaker than carbonic acid.

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 the Reimer–Tiemann reaction?Show answer
Salicylaldehyde (2-hydroxybenzaldehyde).
3.Why is o-nitrophenol steam volatile?Show answer
Intramolecular hydrogen bonding.
4.Product of phenol with bromine water?Show answer
2,4,6-Tribromophenol (white precipitate).
5. $p K_{a}$ of phenol (NCERT)?Show answer
10.0.
6.Product of oxidation of phenol with chromic acid?Show answer
Benzoquinone.

Reactions of Phenols: frequently asked questions

Why is phenol acidic in Class 12 Chemistry?

Because losing a proton gives the phenoxide ion, whose negative charge is spread over the benzene ring by resonance, making it relatively stable.

What is the Reimer–Tiemann reaction?

The reaction of phenol with chloroform and aqueous sodium hydroxide, followed by acidification, which introduces a –CHO group ortho to the –OH, giving salicylaldehyde.

What is Kolbe's reaction?

Sodium phenoxide reacts with carbon dioxide under pressure and heat; acidification gives salicylic acid (2-hydroxybenzoic acid).

How does a nitro group affect the acidity of phenol?

At ortho or para it increases acidity by delocalising the phenoxide's negative charge; at meta it increases acidity less, through the inductive effect only.

Why does phenol give a white precipitate with bromine water?

The –OH group activates all ortho and para positions, so 2,4,6-tribromophenol, which is insoluble in water, forms immediately.