ChemistryCBSE Class 12 · Chapter 7

Alcohols, Phenols and Ethers Class 12: Reactions, Mechanisms and Board Answers

Q: How many marks does Alcohols, Phenols and Ethers carry in CBSE Class 12 2026-27?

Unit 7 carries 6 of the 70 theory marks. Expect named reactions, a mechanism or acidity question, and conversions, sometimes as part of a 5-mark organic question.

Q: Are reactions of secondary and tertiary alcohols in the 2026-27 syllabus?

Yes. The 2025-26 text limited chemical properties to primary alcohols, but the 2026-27 syllabus follows the NCERT section headings, and the NCERT chapter covers the reactions of all three classes, such as the Lucas test and their oxidation. Learn them as the book presents them.

Q: Which mechanisms should I learn from this chapter?

The three NCERT mechanisms: acid-catalysed hydration of ethene, acid-catalysed dehydration of ethanol (to ethene, and the SN2 path to ethoxyethane), and the reaction of an ether with HI.

Q: How do I distinguish phenol from ethanol?

Phenol gives a violet colour with neutral FeCl₃ solution and a white precipitate with bromine water; ethanol does neither. Ethanol also gives a yellow precipitate of iodoform with I₂ and NaOH.

Q: Why does Williamson synthesis fail with tertiary halides?

Alkoxide ions are strong bases as well as nucleophiles. With a crowded tertiary halide, SN2 attack is blocked, so the alkoxide removes a β-hydrogen and elimination gives an alkene instead of an ether.

Updated 14 October 2026 · Fact-checked

Alcohols, Phenols and Ethers is Chapter 7 of CBSE Class 12 Chemistry and carries 6 of the 70 theory marks. It covers naming and structure, preparation of alcohols and phenols, acidity, reactions of the O–H and C–O bonds, electrophilic substitution of phenol, Kolbe and Reimer–Tiemann reactions, and ethers by Williamson synthesis.

What Alcohols, Phenols and Ethers is about

Alcohols (R–OH) and phenols (Ar–OH) contain a hydroxyl group; ethers (R–O–R′) have an oxygen joining two carbon groups. Ethanol is the alcohol in drinks and fuel blends, phenol is the starting point for aspirin and plastics, and diethyl ether was once used as an anaesthetic.

The chapter has four parts.

Structure and naming: classification of alcohols as 1°, 2°, 3°, allylic and benzylic; IUPAC names; bond angles (C–O–H in methanol 108.9°, C–O–C in methoxymethane 111.7°) and why the C–O bond in phenol is shorter.

Preparation: alcohols from alkenes (acid-catalysed hydration, hydroboration–oxidation), from carbonyl compounds (reduction, or Grignard reagents) and from acids and esters; phenols from haloarenes, benzenesulphonic acid, diazonium salts and cumene; commercial methanol and ethanol.

Properties and reactions: hydrogen bonding and boiling points; acidity (alcohols < water < phenol, and the effect of –NO₂ and –CH₃ groups); esterification; C–O cleavage with HX (the Lucas test), PX₃ and dehydration (with mechanism); oxidation (primary to aldehyde or acid, secondary to ketone) and dehydrogenation with copper; phenol's ring reactions: nitration, bromination, Kolbe's reaction, Reimer–Tiemann reaction, zinc dust and oxidation.

Ethers: preparation by dehydration and Williamson synthesis (with its limits), physical properties, cleavage by HI (with mechanism) and electrophilic substitution of anisole.

Board questions favour named reactions, mechanisms (hydration of ethene, dehydration of ethanol, ether + HI), acidity orders, tests to distinguish compounds, and multi-step conversions.

Weightage: This chapter is Unit VII, Alcohols, Phenols and Ethers, which carries 6 of the 70 theory marks.Source: CBSE Chemistry (043) syllabus 2026-27, course structure, Unit VII Alcohols, Phenols and Ethers

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 6.3 marks per paper. See the questions

Key concepts

Classification, nomenclature and structure

NCERT section 7.1 Classification; 7.2 Nomenclature; 7.3 Structures of Functional Groups

Alcohols are classified by the carbon carrying –OH: primary (1°), secondary (2°) or tertiary (3°). Allylic alcohols have –OH next to a C=C; benzylic alcohols have –OH on a carbon attached to an aromatic ring. Vinylic alcohols (–OH on a C=C carbon) and phenols (–OH on a ring) have the O–H on an sp² carbon. Mono-, di- and trihydric alcohols have one, two or three –OH groups (ethane-1,2-diol, propane-1,2,3-triol).

IUPAC names: replace -e of the alkane by -ol; number the chain to give –OH the lowest locant (CH₃CH(OH)CH₃ is propan-2-ol). Phenols keep the name phenol, with locants for substituents (2-methylphenol = o-cresol; benzene-1,2-diol = catechol). Ethers are named as alkoxyalkanes: CH₃OCH₂CH₃ is methoxyethane, C₆H₅OCH₃ is methoxybenzene (anisole).

Structure: in alcohols the oxygen is sp³; the C–O–H angle in methanol (108.9°) is a little less than tetrahedral because of lone pair repulsion. In phenol the C–O bond (136 pm) is slightly shorter than in methanol, because of partial double bond character from conjugation of oxygen's lone pair with the ring, and because the carbon is sp². In ethers, the C–O–C angle (111.7° in methoxymethane) is a little more than tetrahedral because the two bulky alkyl groups repel each other.

Worked example

(a) Give the IUPAC names of CH₃CH(OH)CH₂OH and C₆H₅OC₂H₅. (b) Why is the C–O bond length in phenol shorter than in methanol? (2 marks)

Try it first, then show the solutionSolution

(a) Propane-1,2-diol; ethoxybenzene. (1)
(b) The lone pair on oxygen is delocalised into the ring, giving the C–O bond partial double bond character; also the carbon is sp² hybridised. (1)

Answer: (a) Propane-1,2-diol; ethoxybenzene. (b) Partial double bond character (resonance) and sp² carbon.

Preparation of alcohols

NCERT section 7.4.1 Preparation of Alcohols; 7.6 Some Commercially Important Alcohols

From alkenes:

Acid-catalysed hydration: water adds in the Markovnikov way. Mechanism: (1) the alkene is protonated by H₃O⁺ to give a carbocation; (2) water attacks the carbocation; (3) loss of H⁺ gives the alcohol.
Hydroboration–oxidation: diborane adds to the alkene to give a trialkylborane, which is oxidised by H₂O₂ in aqueous NaOH. The overall result is anti-Markovnikov addition of water: propene gives propan-1-ol in excellent yield.

From carbonyl compounds:

Reduction: aldehydes give 1° alcohols and ketones give 2° alcohols, with H₂ over Pt, Pd or Ni, or with NaBH₄ or LiAlH₄.
Carboxylic acids are reduced to 1° alcohols by LiAlH₄ (expensive), so industrially acids are first converted to esters, which are reduced by H₂ with a catalyst.
Grignard reagents add to aldehydes and ketones; hydrolysis of the adduct gives the alcohol. Methanal gives a 1° alcohol, other aldehydes give 2° alcohols and ketones give 3° alcohols.

Commercial alcohols:

Methanol ('wood spirit'): now made by catalytic hydrogenation of CO, $C O + 2 H_{2} \to C H_{3} O H$ (ZnO–Cr₂O₃, 200–300 atm, 573–673 K). It is poisonous: small amounts cause blindness, larger amounts death.
Ethanol: made by fermentation of sugars (molasses, sugarcane, fruits). Invertase converts sucrose to glucose and fructose; zymase in yeast ferments them to ethanol and CO₂. Commercial ethanol is made unfit to drink (denatured) by adding copper sulphate (for colour) and pyridine (a foul-smelling liquid).

Worked example

Show how propene can be converted into (a) propan-2-ol and (b) propan-1-ol. Name the reactions. (2 marks)

Try it first, then show the solutionSolution

(a) Acid-catalysed hydration (Markovnikov): $C H_{3} C H = C H_{2} + H_{2} O H^{+} C H_{3} C H (O H) C H_{3}$ . (1)
(b) Hydroboration–oxidation (anti-Markovnikov): propene + B₂H₆ gives (CH₃CH₂CH₂)₃B, which with H₂O₂/OH⁻ gives $C H_{3} C H_{2} C H_{2} O H$ . (1)

Answer: (a) H₂O/H⁺ (hydration) gives propan-2-ol. (b) B₂H₆ then H₂O₂/OH⁻ (hydroboration–oxidation) gives propan-1-ol.

Preparation of phenols

NCERT section 7.4.2 Preparation of Phenols

1. From haloarenes: chlorobenzene fused with NaOH at 623 K and 320 atm gives sodium phenoxide; acidification gives phenol.
2. From benzenesulphonic acid: benzene is sulphonated with oleum; the sulphonic acid is heated with molten NaOH to give sodium phenoxide, which is acidified.
3. From diazonium salts: aniline is treated with nitrous acid (NaNO₂ + HCl) at 273–278 K to give benzene diazonium chloride, which is warmed with water (or treated with dilute acids): $C_{6} H_{5} N_{2} C l + H_{2} O \to C_{6} H_{5} O H + N_{2} + H C l$ .
4. From cumene (the main industrial method): cumene (isopropylbenzene) is oxidised by air to cumene hydroperoxide, which is treated with dilute acid to give phenol and acetone. Acetone, a valuable by-product, is obtained in large amounts too.
$C_{6} H_{5} C H (C H_{3})_{2} + O_{2} \to C_{6} H_{5} C (C H_{3})_{2} O O H$
$C_{6} H_{5} C (C H_{3})_{2} O O H H^{+}, H_{2} O C_{6} H_{5} O H + C H_{3} C O C H_{3}$

Worked example

Write the equations for the preparation of phenol from cumene. Name the useful by-product. (2 marks)

Try it first, then show the solutionSolution

Air oxidation: $C_{6} H_{5} C H (C H_{3})_{2} + O_{2} \to C_{6} H_{5} C (C H_{3})_{2} O O H$ (cumene hydroperoxide). (1)
Acid treatment: $C_{6} H_{5} C (C H_{3})_{2} O O H H^{+}, H_{2} O C_{6} H_{5} O H + C H_{3} C O C H_{3}$ ; the by-product is acetone (propanone). (1)

Answer: Cumene → cumene hydroperoxide (air) → phenol + acetone (dilute acid). By-product: acetone.

Physical properties and acidity

NCERT section 7.5 Physical Properties; 7.4.3 Chemical Reactions (Acidity of alcohols and phenols)

Boiling points: alcohols and phenols have higher boiling points than hydrocarbons, ethers and haloalkanes of similar mass, because their molecules are linked by intermolecular hydrogen bonds. Boiling points rise with chain length and fall with branching.

Solubility: lower alcohols are miscible with water because they form hydrogen bonds with water; solubility falls as the hydrophobic alkyl part grows.

Acidity of alcohols: alcohols react with active metals (Na, K, Al) to give alkoxides and hydrogen: $2 C H_{3} C H_{2} O H + 2 N a \to 2 C H_{3} C H_{2} O N a + H_{2}$ . Alkyl groups push electrons (+I effect) towards oxygen, making the O–H bond less polar, so acidity is 1° > 2° > 3°. Alcohols are weaker acids than water: alkoxide ions are stronger bases than hydroxide, so $R O^{-} + H_{2} O \to R O H + O H^{-}$ .

Acidity of phenols: phenols react with NaOH as well as Na, so they are stronger acids than alcohols and water. In phenol, the oxygen is attached to an sp² carbon, and the phenoxide ion is stabilised by delocalisation of the negative charge over the ring (with no charge separation). Phenol (pKa 10.0) is about a million times more acidic than ethanol (pKa 15.9).

Substituent effects:

Electron-withdrawing groups (–NO₂) increase acidity, especially at ortho and para positions where they delocalise the negative charge: pKa o-nitrophenol 7.2, m-nitrophenol 8.3, p-nitrophenol 7.1. 2,4,6-trinitrophenol (picric acid) is strongly acidic.
Electron-donating groups (–CH₃) decrease acidity: cresols (pKa 10.1–10.2) are weaker than phenol.

The smaller the pKa, the stronger the acid.

Worked example

Arrange in increasing order of acidic strength and justify: ethanol, phenol, p-nitrophenol, p-cresol, water. (3 marks)

Try it first, then show the solutionSolution

Ethanol is weaker than water because the ethyl group (+I) destabilises the ethoxide ion. (1)
p-Cresol is weaker than phenol because –CH₃ (electron-donating) destabilises the phenoxide; phenol is stronger than water because phenoxide is resonance-stabilised. (1)
p-Nitrophenol is the strongest: –NO₂ at para withdraws electrons and delocalises the negative charge further. Order: ethanol < water < p-cresol < phenol < p-nitrophenol. (1)

Answer: Ethanol < water < p-cresol < phenol < p-nitrophenol

Reactions of alcohols: O–H and C–O cleavage, dehydration

NCERT section 7.4.3 Chemical Reactions (Reactions involving cleavage of O–H and C–O bonds)

O–H cleavage:

With Na or K (see acidity).
Esterification: alcohols and phenols react with carboxylic acids (in the presence of a little concentrated H₂SO₄, reversibly; water is removed as formed), with acid chlorides (in pyridine, which neutralises the HCl) or with acid anhydrides. Acetylation of salicylic acid gives aspirin, an analgesic, anti-inflammatory and antipyretic drug.

C–O cleavage (alcohols only; phenols do this only with zinc):

With HX: $R O H + H X \to R X + H_{2} O$ . The Lucas test uses concentrated HCl + ZnCl₂: alcohols dissolve but the alkyl chlorides do not, so turbidity appears. 3° alcohols give turbidity immediately, 2° after a few minutes, and 1° not at room temperature.
With PBr₃ or PCl₃: alkyl bromides or chlorides.
Dehydration: heating with concentrated H₂SO₄ or H₃PO₄ (or ZnCl₂, alumina) gives alkenes. Ethanol needs conc. H₂SO₄ at 443 K; 2° and 3° alcohols dehydrate more easily: 3° > 2° > 1°.

Mechanism of dehydration of ethanol:
1. Protonation of the alcohol: $C H_{3} C H_{2} O H + H^{+} \to C H_{3} C H_{2} O H_{2}^{+}$ (fast).
2. Loss of water to give a carbocation (slow, rate-determining).
3. Loss of H⁺ from the neighbouring carbon to give ethene.
The acid used in step 1 is released in step 3; ethene is removed as formed to shift the equilibrium.

Worked example

Three test tubes contain propan-1-ol, propan-2-ol and 2-methylpropan-2-ol. How will you identify them using the Lucas reagent? (3 marks)

Try it first, then show the solutionSolution

Add Lucas reagent (concentrated HCl + anhydrous ZnCl₂) to each at room temperature; turbidity appears when the insoluble alkyl chloride forms. (1)
2-Methylpropan-2-ol (3°) gives turbidity immediately; propan-2-ol (2°) gives turbidity after about five minutes. (1)
Propan-1-ol (1°) gives no turbidity at room temperature (only on heating). (1)

Answer: Immediate turbidity: 3° (2-methylpropan-2-ol); in a few minutes: 2° (propan-2-ol); none at room temperature: 1° (propan-1-ol).

Oxidation and dehydrogenation of alcohols

NCERT section 7.4.3 Chemical Reactions (Oxidation)

Oxidation of alcohols removes hydrogen from the O–H and from the carbon bearing it.

Primary alcohols give an aldehyde, which is easily oxidised further to a carboxylic acid by strong oxidants such as acidified KMnO₄ or K₂Cr₂O₇. To stop at the aldehyde, use CrO₃ in an anhydrous medium or pyridinium chlorochromate (PCC), which gives a good yield of aldehyde: $C H_{3} C H = C H C H_{2} O H P C C C H_{3} C H = C H C H O$ .
Secondary alcohols give ketones with chromic anhydride (CrO₃).
Tertiary alcohols resist oxidation; strong oxidants at high temperature break C–C bonds and give a mixture of acids with fewer carbon atoms.

Dehydrogenation with heated copper at 573 K:

1° alcohol → aldehyde: $C H_{3} C H_{2} O H C u, 573 K C H_{3} C H O + H_{2}$
2° alcohol → ketone: $C H_{3} C H (O H) C H_{3} C u, 573 K C H_{3} C O C H_{3} + H_{2}$
3° alcohol → alkene (dehydration): $(C H_{3})_{3} C O H C u, 573 K (C H_{3})_{2} C = C H_{2} + H_{2} O$

In the body, methanol is oxidised to methanal and then methanoic acid, which can cause blindness and death; it is treated with intravenous dilute ethanol, which is oxidised preferentially and gives the kidneys time to remove the methanol.

Worked example

Write the products when vapours of (a) ethanol, (b) propan-2-ol and (c) 2-methylpropan-2-ol are passed over heated copper at 573 K. (3 marks)

Try it first, then show the solutionSolution

(a) Primary: dehydrogenation to ethanal, $C H_{3} C H_{2} O H \to C H_{3} C H O + H_{2}$ . (1)
(b) Secondary: dehydrogenation to propanone, $C H_{3} C H (O H) C H_{3} \to C H_{3} C O C H_{3} + H_{2}$ . (1)
(c) Tertiary: dehydration to 2-methylpropene, $(C H_{3})_{3} C O H \to (C H_{3})_{2} C = C H_{2} + H_{2} O$ . (1)

Answer: (a) Ethanal (b) propanone (c) 2-methylpropene

Reactions of phenols: electrophilic substitution and named reactions

NCERT section 7.4.3 Chemical Reactions (Reactions of phenols)

The –OH group is strongly activating and ortho, para directing, because resonance increases electron density at ortho and para positions.

Nitration: with dilute HNO₃ at 298 K, phenol gives a mixture of o- and p-nitrophenols. They are separated by steam distillation: o-nitrophenol is steam volatile because of intramolecular hydrogen bonding, while p-nitrophenol is less volatile because of intermolecular hydrogen bonding. With concentrated HNO₃, phenol gives 2,4,6-trinitrophenol (picric acid), though the yield is poor; nowadays phenol is first sulphonated with concentrated H₂SO₄ and then nitrated.

Halogenation:

With bromine water, phenol gives a white precipitate of 2,4,6-tribromophenol: $C_{6} H_{5} O H + 3 B r_{2} \to C_{6} H_{2} B r_{3} O H + 3 H B r$ .
With Br₂ in a solvent of low polarity (CHCl₃ or CS₂) at low temperature, monobromophenols form.

The ring is so activated that no Lewis acid is needed; in water, phenol ionises to the even more reactive phenoxide.

Kolbe's reaction: sodium phenoxide (from phenol + NaOH) is even more reactive and reacts with CO₂, a weak electrophile, to give ortho-hydroxybenzoic acid (salicylic acid) after acidification.

Reimer–Tiemann reaction: phenol + chloroform + aqueous NaOH gives a –CHO group at the ortho position. The intermediate substituted benzal chloride is hydrolysed, and acidification gives salicylaldehyde (2-hydroxybenzaldehyde).

With zinc dust: phenol is reduced to benzene: $C_{6} H_{5} O H + Z n \to C_{6} H_{6} + Z n O$ .

Oxidation: with chromic acid, phenol gives benzoquinone; in air it slowly darkens through quinone formation.

Worked example

(a) What happens when phenol is treated with bromine water? Write the equation. (b) Explain why o-nitrophenol is steam volatile but p-nitrophenol is not. (3 marks)

Try it first, then show the solutionSolution

(a) A white precipitate of 2,4,6-tribromophenol forms: $C_{6} H_{5} O H + 3 B r_{2} \to C_{6} H_{2} B r_{3} O H + 3 H B r$ . (1½)
(b) In o-nitrophenol the –OH and –NO₂ groups are close enough to form an intramolecular hydrogen bond, so molecules are not linked to each other and the compound is volatile. (1)
In p-nitrophenol the groups are far apart, so molecules are linked by intermolecular hydrogen bonds, raising the boiling point. (½)

Answer: (a) 2,4,6-Tribromophenol (white precipitate). (b) Intramolecular H-bond in the ortho isomer; intermolecular H-bonds in the para isomer.

Ethers: preparation, properties and reactions

NCERT section 7.7 Ethers (7.7.1 Preparation; 7.7.2 Physical Properties; 7.7.3 Chemical Reactions)

Preparation:

Dehydration of alcohols: ethanol with conc. H₂SO₄ at 413 K gives ethoxyethane; at 443 K it gives ethene. The ether forms by SN2 attack of an alcohol molecule on a protonated alcohol. The method suits 1° alcohols; 2° and 3° alcohols give alkenes.
Williamson synthesis: an alkyl halide reacts with a sodium alkoxide by SN2: $R - X + R^{'} O N a \to R - O - R^{'} + N a X$ . It works for symmetrical and unsymmetrical ethers, and phenols can be converted to their ethers this way. Limitation: the alkyl halide must be primary; with a 3° halide the alkoxide acts as a base and elimination gives an alkene. So to make $(C H_{3})_{3} C - O - C H_{3}$ , use $(C H_{3})_{3} C O N a + C H_{3} B r$ , not $C H_{3} O N a + (C H_{3})_{3} C B r$ .

Physical properties: the C–O bonds are polar, so ethers have a small dipole moment. Their boiling points are close to those of alkanes of similar mass and much lower than those of alcohols, because ethers cannot form hydrogen bonds with each other. They are about as soluble in water as alcohols of similar mass, because ether oxygen can hydrogen-bond with water.

Cleavage by HX: ethers are among the least reactive functional groups, but hot concentrated HI (or HBr) breaks the C–O bond. Order of reactivity: HI > HBr > HCl.

With two primary alkyl groups or a methyl group, the reaction is SN2: the halide attacks the smaller alkyl group of the protonated ether. $C H_{3} O C H_{2} C H_{3} + H I \to C H_{3} I + C H_{3} C H_{2} O H$ .
With a tertiary alkyl group, it is SN1: the stable 3° carbocation forms and the halide ends up on the tertiary carbon. $(C H_{3})_{3} C O C H_{3} + H I \to (C H_{3})_{3} C I + C H_{3} O H$ .
Anisole gives phenol and methyl iodide, because the O–C(aryl) bond has partial double bond character and the sp² carbon cannot undergo substitution: $C_{6} H_{5} O C H_{3} + H I \to C_{6} H_{5} O H + C H_{3} I$ .

Electrophilic substitution of anisole: –OCH₃ is activating and ortho, para directing. Anisole gives p-bromoanisole (major) with Br₂ in ethanoic acid even without a catalyst; Friedel–Crafts alkylation and acylation give o- and p-products; nitration gives o- and p-nitroanisole.

Worked example

(a) Which pair of reactants should be used to prepare 2-methoxy-2-methylpropane by Williamson synthesis, and why? (b) Write the products of the reaction of this ether with HI. (3 marks)

Try it first, then show the solutionSolution

(a) Use sodium tert-butoxide and bromomethane (a primary halide): $(C H_{3})_{3} C O N a + C H_{3} B r \to (C H_{3})_{3} C O C H_{3} + N a B r$ . (1)
With sodium methoxide and tert-butyl bromide, the methoxide acts as a base and elimination gives 2-methylpropene instead. (1)
(b) The tertiary group forms a stable carbocation (SN1), so iodide goes to it: $(C H_{3})_{3} C O C H_{3} + H I \to (C H_{3})_{3} C I + C H_{3} O H$ . (1)

Answer: (a) (CH₃)₃CONa + CH₃Br (primary halide avoids elimination). (b) (CH₃)₃CI + CH₃OH.

Chemistry in real life: Alcohols, Phenols and Ethers

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.

Oxidation and dehydrogenation of alcohols

Suppose a newspaper reports that several people fell ill after drinking illicit liquor contaminated with methanol, and doctors treated them with a drip of dilute ethanol.

Explain why methanol is dangerous and why ethanol is given as treatment.

Try it first, then show the solutionSolution

Inside the body, methanol is converted step by step into methanal and then methanoic acid; these products can cause blindness and even death.
Ethanol competes for the same enzyme and is oxidised preferentially.
This slows the oxidation of methanol, giving the kidneys time to excrete it unchanged.

Answer: Methanol is oxidised to toxic methanal and methanoic acid; dilute ethanol is oxidised in its place, so methanol can be excreted before it harms the patient.

Memory hook: Ethanol takes methanol's place in the queue.

Reactions of alcohols: O–H and C–O cleavage, dehydration

Imagine your grandfather takes an aspirin tablet for a fever and asks you how it is made.

Name the starting phenol derivative and the reaction, and write the equation.

Try it first, then show the solutionSolution

Aspirin is acetylsalicylic acid, made from salicylic acid (2-hydroxybenzoic acid).
The phenolic –OH of salicylic acid is acetylated (an esterification) with acetic anhydride.
$C_{7} H_{6} O_{3} + (C H_{3} C O)_{2} O \to C_{9} H_{8} O_{4} + C H_{3} C O O H$ ; salicylic acid itself comes from phenol by Kolbe's reaction.

Answer: Salicylic acid is acetylated with acetic anhydride: C₇H₆O₃ + (CH₃CO)₂O → C₉H₈O₄ + CH₃COOH.

Memory hook: Phenol → Kolbe → salicylic acid → acetylate → aspirin.

Preparation of alcohols

Suppose a sugar mill sells molasses to a distillery that makes ethanol and sends some of it for use in industry, where it must not be drunk.

Describe how ethanol is made from molasses and how industrial ethanol is made unfit for drinking.

Try it first, then show the solutionSolution

Sucrose in molasses is hydrolysed by invertase to glucose and fructose.
Zymase in yeast ferments these sugars to ethanol and carbon dioxide; air is kept out, otherwise ethanol is oxidised to ethanoic acid.
Industrial alcohol is denatured by adding copper sulphate (to colour it) and pyridine (a foul-smelling liquid).

Answer: Invertase, then zymase in yeast, convert sucrose into ethanol and CO₂; adding CuSO₄ and pyridine denatures it.

Memory hook: Invertase splits, zymase ferments.

Ethers: preparation, properties and reactions

Imagine a lab where a student wants to make 'tert-butyl methyl ether' (2-methoxy-2-methylpropane) and mixes sodium methoxide with tert-butyl bromide, but collects a gas instead.

Explain what went wrong and suggest the right reactants.

Try it first, then show the solutionSolution

Methoxide is a strong base; tert-butyl bromide is a 3° halide, which is crowded and cannot undergo SN2.
So elimination happens instead, giving 2-methylpropene, a gas.
The right choice is sodium tert-butoxide with bromomethane (a primary halide), which reacts by SN2 to give the ether.

Answer: Elimination gave 2-methylpropene; use (CH₃)₃CONa + CH₃Br instead.

Memory hook: In Williamson, keep the halide primary.

Topics in this chapter

Alcohols, Phenols and Ethers formula sheet

Acid-catalysed hydration (Markovnikov): $C H_{3} C H = C H_{2} + H_{2} O H^{+} C H_{3} C H (O H) C H_{3}$; Mechanism: protonation, attack by water, deprotonation.
Hydroboration–oxidation (anti-Markovnikov): $6 C H_{3} C H = C H_{2} + B_{2} H_{6} \to 2 (C H_{3} C H_{2} C H_{2})_{3} B$; Then $(C H_{3} C H_{2} C H_{2})_{3} B + 3 H_{2} O_{2} O H^{-} 3 C H_{3} C H_{2} C H_{2} O H + B (O H)_{3}$ .
Grignard route to alcohols: $H C H O + C H_{3} M g B r \to C H_{3} C H_{2} O M g B r$; Hydrolysis: $C H_{3} C H_{2} O M g B r + H_{2} O \to C H_{3} C H_{2} O H + M g (O H) B r$ . Methanal → 1°, other aldehydes → 2°, ketones → 3°.
Phenol from cumene: $C_{6} H_{5} C (C H_{3})_{2} O O H H^{+}, H_{2} O C_{6} H_{5} O H + C H_{3} C O C H_{3}$; Cumene hydroperoxide is made by air oxidation of cumene.
Phenol from diazonium salt: $C_{6} H_{5} N_{2} C l + H_{2} O warm C_{6} H_{5} O H + N_{2} + H C l$; Diazonium salt from aniline + NaNO₂ + HCl at 273–278 K.
Alcohol + sodium: $2 C H_{3} C H_{2} O H + 2 N a \to 2 C H_{3} C H_{2} O N a + H_{2}$; Acidity: 1° > 2° > 3°; all weaker than water.
Esterification: $C H_{3} C O O H + C_{2} H_{5} O H ⇌ C H_{3} C O O C_{2} H_{5} + H_{2} O$; Conc. H₂SO₄ catalyst; with acid chlorides use pyridine.
Aspirin (acetylation of salicylic acid): $C_{7} H_{6} O_{3} + (C H_{3} C O)_{2} O \to C_{9} H_{8} O_{4} + C H_{3} C O O H$; Salicylic acid + acetic anhydride → acetylsalicylic acid.
Dehydration of ethanol: $C H_{3} C H_{2} O H co n c . H_{2} S O_{4}, 443 K C H_{2} = C H_{2} + H_{2} O$; At 413 K: $2 C H_{3} C H_{2} O H \to C_{2} H_{5} O C_{2} H_{5} + H_{2} O$ .
Dehydrogenation over copper: $C H_{3} C H_{2} O H C u, 573 K C H_{3} C H O + H_{2}$; 2° → ketone; 3° → alkene (dehydration).
Bromination of phenol: $C_{6} H_{5} O H + 3 B r_{2} \to C_{6} H_{2} B r_{3} O H + 3 H B r$; Bromine water gives white 2,4,6-tribromophenol.
Kolbe's reaction: $C_{6} H_{5} O N a + C O_{2} \to H O C_{6} H_{4} C O O N a$; Acidify to get salicylic acid (2-hydroxybenzoic acid).
Reimer–Tiemann reaction (overall): $C_{6} H_{5} O H + C H C l_{3} + 3 N a O H \to H O C_{6} H_{4} C H O + 3 N a C l + 2 H_{2} O$; Simplified overall equation; the product is isolated as salicylaldehyde after acidification.
Phenol with zinc dust: $C_{6} H_{5} O H + Z n \to C_{6} H_{6} + Z n O$; The only C–O cleavage reaction of phenol in this chapter.
Williamson synthesis: $C_{6} H_{5} O N a + C_{2} H_{5} I \to C_{6} H_{5} O C_{2} H_{5} + N a I$; SN2; the alkyl halide must be primary.
Cleavage of ethers by HI: $C_{6} H_{5} O C H_{3} + H I \to C_{6} H_{5} O H + C H_{3} I$; Mixed alkyl ethers: I⁻ attacks the smaller group (SN2), or the 3° group (SN1).
Fermentation: $C_{6} H_{12} O_{6} zymase 2 C_{2} H_{5} O H + 2 C O_{2}$; Sucrose is first split by invertase: $C_{12} H_{22} O_{11} + H_{2} O \to C_{6} H_{12} O_{6} + C_{6} H_{12} O_{6}$ .
Methanol from carbon monoxide: $C O + 2 H_{2} Z n O - C r_{2} O_{3} C H_{3} O H$; 200–300 atm, 573–673 K.

Alcohols, Phenols and Ethers: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Mechanism of acid-catalysed hydration of ethene

Write the mechanism of hydration of ethene to ethanol.

Draw

Step 1: H₂C=CH₂ with a curved arrow from the π bond to H of H₃O⁺, giving CH₃–CH₂⁺ and H₂O.
Step 2: a water molecule's lone pair attacking the carbocation, giving CH₃CH₂–OH₂⁺.
Step 3: a water molecule removing H⁺ from the oxonium ion, giving CH₃CH₂OH and H₃O⁺.

Label

H₃O⁺
Carbocation
Nucleophilic attack by water
Protonated alcohol
Deprotonation

Watch out: Show the acid regenerated at the end, to make clear it is a catalyst.

Resonance stabilisation of the phenoxide ion

Explain the acidity of phenol with resonance structures.

Draw

Phenoxide ion with a negative charge on oxygen.
Curved arrows moving the charge into the ring, giving structures with the negative charge at the two ortho positions and the para position.
Double-headed arrows between the structures.

Label

O⁻
Charge at ortho
Charge at para
Resonance arrows

Watch out: Add one line: phenoxide's structures have no charge separation, so they stabilise the ion more than phenol's structures stabilise phenol.

Mechanism of the cleavage of methoxymethane by HI

Write the mechanism of the reaction of HI with methoxymethane.

Draw

Step 1: CH₃–O–CH₃ taking H⁺ from HI to give the protonated ether CH₃–O⁺(H)–CH₃ and I⁻.
Step 2: I⁻ attacking a methyl carbon from the back (SN2), with the C–O bond breaking to give CH₃I and CH₃OH.
Step 3 (excess HI): CH₃OH protonated, then I⁻ attack gives a second CH₃I and H₂O.

Label

Protonated ether
I⁻ (nucleophile)
SN2 attack
CH₃I
CH₃OH

Watch out: Write SN2 beside step 2; if a tertiary group were present you would show an SN1 carbocation instead.

How the board asks Alcohols, Phenols and Ethers

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Named reactions	Kolbe's reaction, Reimer–Tiemann reaction, Williamson synthesis, hydroboration–oxidation; usually 2 or 3 marks for two or three equations.	Write reactants, reagents and conditions over the arrow, and the final product after acidification where needed.
Mechanisms	Hydration of ethene, dehydration of ethanol to ethene or to ether, reaction of methoxymethane with HI.	Number the steps, use curved arrows, show the protonated intermediate and carbocation (or SN2 attack), and mark the slow step.
Acidity and reasons	Order of acidity, why phenol is more acidic than ethanol, effect of –NO₂ and –CH₃, why ortho-nitrophenol is steam volatile, why ethers have low boiling points.	Talk about the stability of the conjugate base (alkoxide or phenoxide) and the electron effect of the substituent.
Distinguishing tests	Distinguish 1°, 2° and 3° alcohols (Lucas test), phenol and ethanol (FeCl₃ or bromine water), ethanol and methanol (iodoform test, from Chapter 8).	Name the reagent, the observation for each compound and, where possible, an equation.
Conversions and products	Two- or three-step conversions (propene to propan-1-ol, phenol to aspirin, anisole to p-bromoanisole); products of ether cleavage.	One reagent per step; mention conditions (temperature, PCC, Cu at 573 K, dilute or concentrated HNO₃).

Tricks to remember Alcohols, Phenols and Ethers

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

Lucas: Three is quick, Two takes time, One is lazy
Lucas reagent: 3° alcohol turbid immediately, 2° in about five minutes, 1° not at room temperature.
Use it when: Distinguishing classes of alcohols.
Grignard alcohol ladder
HCHO → 1°, RCHO → 2°, R₂CO → 3°. The number of carbon groups on the carbonyl carbon plus one gives the class.
Use it when: Choosing the carbonyl compound for a target alcohol.
413 Ether, 443 Ethene
Lower temperature (413 K) gives the ether; higher temperature (443 K) gives ethene. 'One-three for the ether, four-three for the alkene.'
Use it when: Dehydration conditions.
Kolbe: CO₂ makes COOH; Reimer: CHCl₃ makes CHO
Both put the new group at the ortho position of phenol.
Use it when: Named reaction questions.
Ortho-nitrophenol keeps its hydrogen bond to itself
Intramolecular H-bond = selfish molecule = low boiling, steam volatile. Para shares (intermolecular) and stays behind.
Use it when: Steam distillation reasons.

Alcohols, Phenols and Ethers: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Strongest acid · 1 mark · MCQBoard level

Which of the following is the strongest acid?
(a) Ethanol (b) Phenol (c) p-Nitrophenol (d) p-Cresol

Try it first, then show the solutionSolution

Electron-withdrawing –NO₂ at para stabilises the phenoxide most.
pKa: p-nitrophenol 7.1 < phenol 10.0 < p-cresol 10.2 < ethanol 15.9.

Answer: (c) p-Nitrophenol

Example 2: Steam distillation of nitrophenols · 1 mark · Assertion-reasonBoard level

Assertion (A): o-Nitrophenol can be separated from p-nitrophenol by steam distillation.
Reason (R): o-Nitrophenol has intramolecular hydrogen bonding, while p-nitrophenol has intermolecular hydrogen bonding.
(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.

Try it first, then show the solutionSolution

Both statements are true.
Intramolecular bonding makes the ortho isomer volatile, which is why steam distillation works.

Answer: (a)

Example 3: Ethers from ethanol and conditions · 2 marks · Very short answerHarder board level

What happens when ethanol is heated with concentrated H₂SO₄ at (a) 413 K and (b) 443 K? Write equations. (2 marks)

Try it first, then show the solutionSolution

(a) 413 K: ethoxyethane forms by SN2 attack of ethanol on protonated ethanol: $2 C H_{3} C H_{2} O H \to C_{2} H_{5} O C_{2} H_{5} + H_{2} O$ . (1)
(b) 443 K: dehydration to ethene: $C H_{3} C H_{2} O H \to C H_{2} = C H_{2} + H_{2} O$ . (1)

Answer: (a) Ethoxyethane (b) ethene

Example 4: Three conversions · 3 marks · Short answerHarder board level

How will you convert (a) phenol to salicylic acid, (b) propanone to propan-2-ol, (c) anisole to 4-methoxyacetophenone? (3 marks)

Try it first, then show the solutionSolution

(a) Phenol + NaOH → sodium phenoxide; + CO₂ → sodium salicylate; + H⁺ → salicylic acid (Kolbe). (1)
(b) $C H_{3} C O C H_{3} + H_{2} N i C H_{3} C H (O H) C H_{3}$ (or NaBH₄). (1)
(c) Friedel–Crafts acylation: anisole + CH₃COCl with anhydrous AlCl₃ gives 4-methoxyacetophenone (major) and the 2-isomer. (1)

Answer: (a) Kolbe's reaction (b) catalytic hydrogenation or NaBH₄ (c) Friedel–Crafts acylation.

Example 5: Products of ether cleavage · 3 marks · Short answerHarder board level

Predict the products when each ether is heated with HI and give the mechanism type: (a) CH₃OCH₂CH₃ (b) (CH₃)₃COCH₃ (c) C₆H₅OCH₃. (3 marks)

Try it first, then show the solutionSolution

(a) SN2 at the smaller methyl group: CH₃I + CH₃CH₂OH. (1)
(b) SN1 via the tertiary carbocation: (CH₃)₃CI + CH₃OH. (1)
(c) The aryl C–O bond does not break: C₆H₅OH + CH₃I (SN2 at methyl). (1)

Answer: (a) CH₃I + C₂H₅OH (b) (CH₃)₃CI + CH₃OH (c) C₆H₅OH + CH₃I

Example 6: Alcohols from one alkene, two ways, and their oxidation · 5 marks · Long answerHarder board level

(a) Convert propene to (i) propan-2-ol and (ii) propan-1-ol, naming each reaction. (2)
(b) What happens when propan-1-ol is treated with PCC and when propan-2-ol vapour is passed over Cu at 573 K? (2)
(c) How would the Lucas test distinguish these two alcohols? (1) (5 marks)

Try it first, then show the solutionSolution

(a)(i) Acid-catalysed hydration: CH₃CH=CH₂ + H₂O (H⁺) → CH₃CH(OH)CH₃ (Markovnikov). (1)
(a)(ii) Hydroboration–oxidation: B₂H₆ then H₂O₂/OH⁻ gives CH₃CH₂CH₂OH (anti-Markovnikov). (1)
(b) PCC oxidises propan-1-ol to propanal, CH₃CH₂CHO. (1)
Cu at 573 K dehydrogenates propan-2-ol to propanone: CH₃CH(OH)CH₃ → CH₃COCH₃ + H₂. (1)
(c) Propan-2-ol (2°) gives turbidity in about five minutes; propan-1-ol (1°) gives none at room temperature. (1)

Answer: (a) Hydration and hydroboration–oxidation (b) propanal; propanone (c) 2° turns turbid in minutes, 1° does not.

Example 7: Why phenol reacts with bromine water without a catalyst · 3 marks · Short answerHOTS / competency

Benzene needs FeBr₃ to react with bromine, but phenol gives 2,4,6-tribromophenol instantly with bromine water. In a non-polar solvent at low temperature, phenol gives mainly monobromophenols. Explain both observations. (3 marks)

Try it first, then show the solutionSolution

The –OH group donates a lone pair to the ring by resonance, raising electron density at ortho and para positions, so the ring reacts with Br₂ without a Lewis acid. (1)
In water, phenol partly ionises to phenoxide, which is even more strongly activating, and bromine is polarised by water: all three o- and p-positions are substituted. (1)
In CHCl₃ or CS₂ at low temperature, phenol does not ionise and Br₂ is less polarised, so reaction is controlled and mainly monobromophenols form. (1)

Answer: –OH strongly activates the ring; in water the phenoxide ion is even more reactive, giving the tribromo product; in a non-polar solvent the reaction stops mostly at monosubstitution.

Example 8: Grignard choice · 1 mark · MCQHOTS / competency

Which combination gives butan-2-ol after hydrolysis?
(a) HCHO + CH₃CH₂CH₂MgBr (b) CH₃CHO + CH₃CH₂MgBr (c) CH₃COCH₃ + CH₃MgBr (d) HCHO + (CH₃)₂CHMgBr

Try it first, then show the solutionSolution

Butan-2-ol is CH₃CH(OH)CH₂CH₃, a secondary alcohol, so an aldehyde other than methanal is needed.
CH₃CHO + CH₃CH₂MgBr gives CH₃CH(OMgBr)CH₂CH₃, which hydrolyses to butan-2-ol. (a) and (d) give 1° alcohols; (c) gives a 3° alcohol.

Answer: (b)

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Acidity of phenols

Suppose a student looks up the pKa values of some compounds: ethanol 15.9, phenol 10.0, o-cresol 10.2, m-nitrophenol 8.3, o-nitrophenol 7.2 and p-nitrophenol 7.1. She notes that phenol reacts with NaOH but ethanol does not, and that both react with sodium metal.

(i) · 1 mark
Which compound in the list is the strongest acid?
Answer
Answer: p-Nitrophenol (lowest pKa, 7.1).
A lower pKa means a stronger acid.
(ii) · 1 mark
Why is o-cresol a weaker acid than phenol?
Answer
Answer: –CH₃ is electron-donating and destabilises the phenoxide ion.
The +I effect of the methyl group increases the negative charge on oxygen.
(iii) · 2 marks
Explain why m-nitrophenol is less acidic than o- or p-nitrophenol.
Answer
Answer: At ortho and para, –NO₂ delocalises the negative charge of the phenoxide by resonance; at meta it acts only through the weaker −I effect.
Resonance places the negative charge on ring carbons ortho and para to oxygen, where –NO₂ can take it (1); from the meta position it cannot, so stabilisation is less and the acid is weaker (1).

Making phenol in industry

Suppose a chemical company makes phenol from cumene. Air is blown through cumene to form cumene hydroperoxide, which is then treated with dilute sulphuric acid. The plant sells both phenol and a second product, a common solvent used in nail polish remover. Part of the phenol is then converted into salicylic acid for a medicine factory.

(i) · 1 mark
Name the second product.
Answer
Answer: Acetone (propanone).
Cumene hydroperoxide splits into phenol and acetone with dilute acid.
(ii) · 1 mark
Name the reaction used to convert phenol into salicylic acid.
Answer
Answer: Kolbe's reaction.
Sodium phenoxide reacts with CO₂, then acidification gives salicylic acid.
(iii) · 2 marks
Write the two equations of the cumene process.
Answer
Answer: C₆H₅CH(CH₃)₂ + O₂ → C₆H₅C(CH₃)₂OOH; C₆H₅C(CH₃)₂OOH → C₆H₅OH + CH₃COCH₃ (H⁺, H₂O).
One mark for each equation.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Write the equations for (a) Kolbe's reaction and (b) Reimer–Tiemann reaction. (2 marks) (2 marks)

Step	Marks
(a) Phenol + NaOH gives sodium phenoxide; C₆H₅ONa + CO₂ gives sodium salicylate, and H⁺ gives salicylic acid.	1
(b) Phenol + CHCl₃ + aqueous NaOH gives an intermediate (substituted benzal chloride) that is hydrolysed; acidification gives salicylaldehyde.	1

Write the mechanism of acid-catalysed dehydration of ethanol to ethene. (3 marks) (3 marks)

Step	Marks
Step 1: protonation of ethanol: CH₃CH₂–OH + H⁺ → CH₃CH₂–OH₂⁺ (fast).	1
Step 2: loss of water to give the ethyl carbocation, CH₃CH₂⁺ (slow, rate-determining).	1
Step 3: loss of H⁺ from the adjacent carbon: CH₂=CH₂ forms and H⁺ is regenerated.	1

(a) Explain why phenol is more acidic than ethanol. (b) Arrange o-nitrophenol, m-nitrophenol, phenol and p-cresol in increasing order of acidity, with reason. (c) Write the mechanism of the reaction of HI with methoxymethane. (5 marks) (5 marks)

Step	Marks
(a) Phenoxide ion is stabilised by resonance (negative charge delocalised over the ring); ethoxide is destabilised by the +I effect of the ethyl group.	1.5
(b) p-Cresol < phenol < m-nitrophenol < o-nitrophenol: CH₃ donates electrons; NO₂ withdraws, more effectively at ortho (resonance) than meta (only −I).	1.5
(c) Step 1: CH₃–O–CH₃ + HI → CH₃–O⁺(H)–CH₃ + I⁻ (protonation).	1
Step 2: I⁻ attacks a methyl carbon by SN2: CH₃I + CH₃OH; with excess HI, CH₃OH → CH₃I + H₂O.	1

Where students lose marks in Alcohols, Phenols and Ethers

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

Mistake: Writing anisole + HI → iodobenzene + methanol
Wrong working
C₆H₅OCH₃ + HI → C₆H₅I + CH₃OH
Right working
C₆H₅OCH₃ + HI → C₆H₅OH + CH₃I
Do this: The aryl C–O bond has partial double bond character and the sp² carbon resists substitution; I⁻ attacks the methyl group.
Usually costs: The full equation mark.
Mistake: Using a tertiary halide in Williamson synthesis
Wrong working
CH₃ONa + (CH₃)₃CBr → (CH₃)₃COCH₃ + NaBr
Right working
(CH₃)₃CONa + CH₃Br → (CH₃)₃COCH₃ + NaBr
Do this: The halide must be primary; a 3° halide undergoes elimination.
Usually costs: 1 to 2 marks.
Mistake: Claiming alcohols are more acidic than water
Do this: Alkoxide ions are stronger bases than OH⁻ because of the +I effect of alkyl groups, so alcohols are weaker acids than water; phenol is stronger.
Usually costs: 1 mark in an ordering question.
Mistake: Forgetting the conditions for nitration of phenol
Do this: Dilute HNO₃ at 298 K gives o- and p-nitrophenol; concentrated HNO₃ gives picric acid. Always state which acid.
Usually costs: ½ to 1 mark.
Mistake: Skipping the protonation step in mechanisms
Do this: Every acid-catalysed mechanism in this chapter starts with protonation (of the alkene, the alcohol or the ether). Show it with a curved arrow.
Usually costs: 1 mark in a 3-mark mechanism.

NCERT exercise guide: Alcohols, Phenols and Ethers

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Intext questions and exercises on naming, classification and preparation

What it tests: Classifying alcohols, IUPAC names, structures from names, preparation of alcohols by hydration, hydroboration–oxidation, reduction and Grignard reactions.

How to approach it: For Grignard targets, split the alcohol at the carbinol carbon: one group comes from RMgX, the rest from the carbonyl compound.

Choosing a Grignard pair

Which Grignard reagent and carbonyl compound would you use to make 2-methylpropan-2-ol?

Show the solutionSolution

2-Methylpropan-2-ol is (CH₃)₃COH, a tertiary alcohol, so a ketone is needed.
Propanone (CH₃COCH₃) + CH₃MgBr gives (CH₃)₃COMgBr.
Hydrolysis gives (CH₃)₃COH + Mg(OH)Br.

Answer: Propanone and methylmagnesium bromide, followed by hydrolysis.

Exercises on acidity, physical properties and reactions

What it tests: Acidity orders, boiling points and solubility, Lucas test, dehydration and oxidation products, reactions of phenol, steam volatility of nitrophenols.

How to approach it: For acidity, compare the stability of the conjugate bases; for physical properties, look for hydrogen bonding (inter- or intramolecular).

Boiling points of isomers

Ethanol and methoxymethane have the same molar mass (46 g mol⁻¹), yet ethanol (351 K) boils far above methoxymethane, which is a gas at room temperature. Explain.

Show the solutionSolution

Both have the same molar mass.
Ethanol molecules form intermolecular hydrogen bonds through their O–H groups; methoxymethane has no O–H and cannot.
More energy is needed to separate ethanol molecules, so its boiling point is much higher.

Answer: Hydrogen bonding between ethanol molecules raises its boiling point.

Exercises on ethers and conversions

What it tests: Williamson synthesis and its limitations, reactions of ethers with HI, electrophilic substitution of anisole, multi-step conversions.

How to approach it: In Williamson synthesis put the bulky group on the alkoxide and use a primary halide. For HI cleavage, decide SN2 (attack at the smaller group) or SN1 (tertiary group).

Bromination of anisole

Write the major product of bromination of anisole in ethanoic acid and explain why no catalyst is needed.

Show the solutionSolution

–OCH₃ donates electrons to the ring by resonance, activating it strongly.
It directs to ortho and para; para is favoured because it is less crowded.
C₆H₅OCH₃ + Br₂ → p-BrC₆H₄OCH₃ (major) + HBr.

Answer: 1-Bromo-4-methoxybenzene (p-bromoanisole); the ring is activated enough to react without FeBr₃.

Alcohols, Phenols and Ethers: previous year questions

Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 6.3 marks per paper; 30 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Alcohols, Phenols and Ethers

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026, set 56/1/1, Q8 · 1 mark · MCQ

At low temperature, phenol reacts with dil. $HN O_{3}$ to yield

(A)2, 4, 6-Trinitrophenol
(B)o-Nitrophenol only
(C)p-Nitrophenol only
(D)ortho-and para-nitrophenol

Try it first, then show our solutionSolution

The –OH group activates the ring and directs substitution to the ortho and para positions.
With dilute $HN O_{3}$ at low temperature, phenol gives a mixture of o-nitrophenol and p-nitrophenol. Option (D).
1 mark

Answer: Option (D)

Where the marks are

Gives the o- and p-nitrophenol mixture (1 mark)

See it in the full 2026 board paper, set 56/1/1

2026, set 56/1/1, Q27 · 3 marks · Short answer

Write the reaction involved in the following :
(a) Reimer-Tiemann reaction
(b) Kolbe's reaction
(c) Friedel-Crafts acylation of anisole

Try it first, then show our solutionSolution

(a) Reimer–Tiemann reaction: phenol treated with chloroform in aqueous NaOH introduces –CHO at the ortho position. The intermediate (substituted benzal chloride, then sodium salicylaldehyde) is acidified to give salicylaldehyde:
$C_{6} H_{5} OH CHC l_{3} aq. NaOH o - C_{6} H_{4} (O^{-} N a^{+}) CHO H^{+} o - HO C_{6} H_{4} CHO$ (salicylaldehyde)
1 mark
(b) Kolbe's reaction: sodium phenoxide reacts with $C O_{2}$ (about 400 K, 4–7 atm) to give sodium salicylate, which is acidified to give salicylic acid:
$C_{6} H_{5} O^{-} N a^{+} C O_{2} 400 K, 4 - 7 a t m o - HO C_{6} H_{4} CO O^{-} N a^{+} H^{+} o - HO C_{6} H_{4} COOH$ (2-hydroxybenzoic acid)
1 mark
(c) Friedel–Crafts acylation of anisole:
$C_{6} H_{5} OC H_{3} + C H_{3} COCl anhyd. AlC l_{3} p - C H_{3} O C_{6} H_{4} COC H_{3} (major) + o - C H_{3} O C_{6} H_{4} COC H_{3} (minor) + HCl$
The products are 4-methoxyacetophenone (major) and 2-methoxyacetophenone (minor).
1 mark

Answer: (a) Phenol → salicylaldehyde; (b) sodium phenoxide → salicylic acid; (c) anisole → 4-methoxyacetophenone (major) + 2-methoxyacetophenone (minor)

Where the marks are

Reimer–Tiemann reaction with reagents and product (1 mark)
Kolbe's reaction with conditions and product (1 mark)
Friedel–Crafts acylation of anisole with AlCl₃ and o/p products (1 mark)

See it in the full 2026 board paper, set 56/1/1

2026, set 56/5/1, Q29 · 4 marks · Case-basedCompetency-based

Ethers are prepared by the dehydration of alcohols in presence of protic acids at 413 K. Symmetrical and unsymmetrical ethers can also be prepared by Williamson synthesis. This reaction involves $S_{N} 2$ attack of an alkoxide ion on primary alkyl halide. If a tertiary alkyl halide is used, elimination reaction occurs and an alkene is formed, no ether is formed. C – O bond in ethers are cleaved under drastic conditions with excess of HI. When unsymmetrical ethers react with HI, the alkyl halide is formed from smaller alkyl group. If one of the alkyl group is tertiary, the alkyl halide is formed from the tertiary alkyl group because tertiary alkyl carbocation is more stable than the primary carbocation. Cleavage of alkyl aryl ethers takes place at the alkyl-oxygen bond due to more stable aryl-oxygen bond. The order of reactivity of hydrogen halides is HI > HBr > HCl. Aromatic ethers undergo electrophilic substitution reactions. The alkoxy group attached to the aromatic ring activates the ring towards electrophilic substitution and directs the incoming group to ortho- and para-positions.

Answer the following questions :

(a)
Complete the following equations :
(i) Anisole $+ C H_{3} Cl Anhyd. AlCl_{3} C S_{2}$
(ii) Anisole $Conc. (HNO_{3} + H_{2} S O_{4})$
2 marks
(b)(i)
Write the names of alkyl halide and sodium alkoxide used to prepare tert-butyl ethyl ether.
1 mark
OR
(b)(ii)
Anisole on reaction with HI gives phenol and $C H_{3} - I$ and not methanol and iodobenzene. Justify the statement.
1 mark
(c)
Why is C – O – C bond angle in ethers slightly greater than tetrahedral angle ?
1 mark

Figure in the paper: Both equations in (a) show anisole (benzene ring with OCH3 substituent). See it in the official paper.

Try it first, then show our solutionSolution

(a)(i) Friedel–Crafts alkylation at the o- and p-positions:
$C_{6} H_{5} OC H_{3} + C H_{3} Cl anhyd. AlC l_{3}, C S_{2}$ 2-methoxytoluene (minor) + 4-methoxytoluene (major) + HCl
1 mark
(a)(ii) Nitration at the o- and p-positions:
$C_{6} H_{5} OC H_{3} conc. HN O_{3} + H_{2} S O_{4}$ 2-methoxynitrobenzene (2-nitroanisole, minor) + 4-methoxynitrobenzene (4-nitroanisole, major) + H₂O
1 mark
(b)(i) Use ethyl halide (e.g. bromoethane, $C_{2} H_{5} Br$ ) and sodium tert-butoxide, $(C H_{3})_{3} C - ONa$ :
$(C H_{3})_{3} CONa + C_{2} H_{5} Br \to (C H_{3})_{3} C - O - C_{2} H_{5} + NaBr$
A tertiary halide is not used because it would undergo elimination.
1 mark
(b)(ii) (Alternative) In anisole, the $O - C_{6} H_{5}$ bond is stronger than the $O - C H_{3}$ bond. This is because of resonance and the sp² carbon, which give it partial double-bond character. So after protonation, $I^{-}$ attacks the less hindered $C H_{3}$ group by $S_{N} 2$ and breaks the $O - C H_{3}$ bond, giving phenol and $C H_{3} I$ , not iodobenzene.
(c) The C–O–C bond angle is slightly greater than 109°28′ because of repulsive interaction between the two bulky alkyl groups on oxygen.
1 mark

Answer: (a)(i) 2- and 4-methoxytoluene; (ii) 2- and 4-nitroanisole; (b)(i) ethyl halide + sodium tert-butoxide; (b)(ii) O–C₆H₅ bond stronger (partial double bond), so O–CH₃ cleaves; (c) repulsion between bulky alkyl groups

Where the marks are

(a)(i) o- and p-methoxytoluene (½ + ½ marks)
(a)(ii) o- and p-nitroanisole (½ + ½ marks)
(b)(i) ethyl halide and sodium tert-butoxide (½ + ½ marks) OR (b)(ii) O–C₆H₅ bond stronger due to resonance / partial double-bond character (1 mark)
(c) repulsion between alkyl groups (1 mark)

See it in the full 2026 board paper, set 56/5/1

Every Alcohols, Phenols and Ethers question, by year

For 2027

Sample paper 2026-27: Q15 (1 mark, solved), Q29 (4 marks, solved)

2026

Set 56/1/1: Q8 (1 mark, solved), Q14 (1 mark, solved), Q27 (3 marks, solved)

Set 56/5/1: Q9 (1 mark, solved), Q15 (1 mark, solved), Q29 (4 marks, solved)

Sample paper 2025-26: Q2 (1 mark, solved), Q7 (1 mark, solved), Q21 (2 marks, solved), Q27 (3 marks, solved)

2025

Set 56/1/1: Q8 (1 mark, solved), Q15 (1 mark, solved), Q30 (4 marks, solved)

Set 56/7/1: Q8 (1 mark, solved), Q9 (1 mark, solved), Q13 (1 mark, solved), Q26 (3 marks, solved)

Sample paper 2024-25: Q26 (3 marks, solved), Q32(a) (5 marks, solved)

2024

Set 56/1/1: Q2 (1 mark, solved), Q10 (1 mark, solved), Q11 (1 mark, solved), Q26 (3 marks, solved)

Set 56/5/1: Q3 (1 mark, solved), Q10 (1 mark, solved), Q16 (1 mark, solved), Q23 (3 marks, solved)

Sample paper 2023-24: Q9 (1 mark, solved), Q11 (1 mark, solved), Q13 (1 mark, solved), Q24 (3 marks, solved)

2023

Set 56/1/1: Q10 (1 mark, solved), Q23(a) (2 marks, solved), Q23(b) (2 marks, solved), Q28 (3 marks, solved), Q29(b) (3 marks, solved)

Set 56/5/1: Q2 (1 mark, solved), Q23 (2 marks, solved), Q30(a) (3 marks, solved), Q30(b) (3 marks, solved)

Practice questions on Alcohols, Phenols and Ethers

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
The reagent used to convert a primary alcohol into an aldehyde without further oxidation is:
(a) Acidified KMnO₄ (b) PCC (c) Acidified K₂Cr₂O₇ (d) Conc. HNO₃
Hint
A mild, anhydrous oxidant.
Answer
(b) PCC
Q2 · 1 mark · Assertion-reasonBoard level
Assertion (A): Phenol is more acidic than ethanol.
Reason (R): Phenoxide ion is stabilised by resonance, while 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
Compare the conjugate bases.
Answer
(a)
Q3 · 2 marks · Very short answerBoard level
Write the equations for the preparation of phenol from (a) chlorobenzene and (b) benzene diazonium chloride. (2 marks)
Hint
623 K and 320 atm; warm with water.
Answer
(a) C₆H₅Cl + NaOH (623 K, 320 atm) → C₆H₅ONa, then + HCl → C₆H₅OH + NaCl (1). (b) C₆H₅N₂Cl + H₂O (warm) → C₆H₅OH + N₂ + HCl (1).
Q4 · 2 marks · Very short answerBoard level
Give a chemical test to distinguish (a) phenol and ethanol, (b) propan-2-ol and 2-methylpropan-2-ol. (2 marks)
Hint
FeCl₃; Lucas reagent.
Answer
(a) Phenol gives a violet colour with neutral FeCl₃ (or a white precipitate with bromine water); ethanol does not (1). (b) Lucas reagent: 2-methylpropan-2-ol gives turbidity immediately, propan-2-ol after about five minutes (1).
Q5 · 3 marks · Short answerBoard level
Explain (a) why alcohols have higher boiling points than ethers of similar molar mass, (b) why lower alcohols are soluble in water, and (c) why boiling points of alcohols fall with branching. (3 marks)
Hint
Hydrogen bonding and surface area.
Answer
(a) Alcohols form intermolecular hydrogen bonds; ethers cannot (1). (b) The –OH group forms hydrogen bonds with water molecules (1). (c) Branching reduces surface area and van der Waals attractions (1).
Q6 · 3 marks · Short answerHarder board level
Write the mechanism of acid-catalysed hydration of ethene to ethanol. (3 marks)
Hint
Three steps: protonation, attack, deprotonation.
Answer
Step 1: CH₂=CH₂ + H₃O⁺ → CH₃CH₂⁺ + H₂O (1). Step 2: CH₃CH₂⁺ + H₂O → CH₃CH₂OH₂⁺ (1). Step 3: CH₃CH₂OH₂⁺ + H₂O → CH₃CH₂OH + H₃O⁺ (1).
Q7 · 3 marks · Short answerHarder board level
Give the structures and IUPAC names of the products: (a) phenol + concentrated HNO₃ (b) phenol + Zn dust (c) phenol + chromic acid. (3 marks)
Hint
Picric acid, benzene, a quinone.
Answer
(a) 2,4,6-Trinitrophenol (picric acid) (1). (b) Benzene, C₆H₆ (1). (c) Benzoquinone (cyclohexa-2,5-diene-1,4-dione) (1).
Q8 · 5 marks · Long answerHarder board level
(a) Describe the Williamson synthesis with an example and state its limitation. (2) (b) Why is 2-methoxy-2-methylpropane not made from sodium methoxide and 2-bromo-2-methylpropane? (1) (c) Write the products and mechanism type when 2-methoxy-2-methylpropane is heated with HI. (2) (5 marks)
Hint
SN2 needs a primary halide; a tertiary ether cleaves by SN1.
Answer
(a) Sodium alkoxide + primary alkyl halide by SN2, e.g. C₂H₅ONa + CH₃I → C₂H₅OCH₃ + NaI (1); the halide must be primary, since 2° and 3° halides undergo elimination (1). (b) Methoxide would act as a base on the tertiary halide and give 2-methylpropene by elimination (1). (c) (CH₃)₃COCH₃ + HI → (CH₃)₃CI + CH₃OH (1); SN1, because the tertiary carbocation is stable (1).
Q9 · 3 marks · Short answerHOTS / competency
Suppose you have three unlabelled bottles: butan-1-ol, butan-2-ol and 2-methylpropan-2-ol. Plan tests using (a) Lucas reagent and (b) heated copper at 573 K, giving the expected result for each alcohol. (3 marks)
Hint
Cu gives aldehyde, ketone or alkene depending on the class.
Answer
(a) Lucas: 2-methylpropan-2-ol turbid at once; butan-2-ol in about five minutes; butan-1-ol not at room temperature (1½). (b) Cu at 573 K: butan-1-ol gives butanal (an aldehyde), butan-2-ol gives butanone (a ketone) and 2-methylpropan-2-ol gives 2-methylpropene (an alkene) (1½).
Q10 · 3 marks · Short answerHOTS / competency
Arrange in increasing order of acid strength and justify briefly: water, ethanol, 3-methylphenol, 4-nitrophenol, phenol, 2,4-dinitrophenol. (3 marks)
Hint
Place water between the alcohol and the phenols; then count the –NO₂ groups.
Answer
Ethanol < water < 3-methylphenol < phenol < 4-nitrophenol < 2,4-dinitrophenol (2). Ethoxide is destabilised by the +I effect, so ethanol is weaker than water; phenols are stronger because phenoxide is resonance-stabilised; –CH₃ lowers acidity, and each o/p –NO₂ raises it by delocalising the negative charge (1).
Q11 · 2 marks · Very short answerHarder board level
Why is the reaction of an alcohol with a carboxylic acid carried out in the presence of concentrated H₂SO₄, and why is water removed as it forms? (2 marks)
Hint
The reaction is reversible.
Answer
Concentrated H₂SO₄ acts as a catalyst (and helps remove water) (1). Esterification is reversible, so removing water shifts the equilibrium towards the ester (1).
Q12 · 3 marks · Short answerHOTS / competency
An organic compound A (C₂H₆O) reacts with sodium to give hydrogen, and on heating with concentrated H₂SO₄ at 443 K gives a gas B that decolourises bromine water. A with Cu at 573 K gives C, which reduces Tollens' reagent. Identify A, B and C. (3 marks)
Hint
C₂H₆O with an O–H.
Answer
A: ethanol, CH₃CH₂OH (reacts with Na, so it has O–H) (1). B: ethene, CH₂=CH₂ (dehydration at 443 K; unsaturated) (1). C: ethanal, CH₃CHO (dehydrogenation; an aldehyde reduces Tollens' reagent) (1).

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.What is the C–O–H bond angle in methanol and why is it less than 109.5°?Show answer
108.9°, because of repulsion by the two lone pairs on oxygen.
2.Which reaction converts propene to propan-1-ol?Show answer
Hydroboration–oxidation (B₂H₆, then H₂O₂/OH⁻).
3.What is the by-product in the cumene process for phenol?Show answer
Acetone (propanone).
4.Which is more acidic, phenol or ethanol, and by roughly how much?Show answer
Phenol (pKa 10.0 vs 15.9), about a million times more acidic.
5.What does the Lucas reagent contain?Show answer
Concentrated HCl and anhydrous ZnCl₂.
6.What reagent oxidises a primary alcohol only as far as the aldehyde?Show answer
Pyridinium chlorochromate (PCC) or CrO₃ in an anhydrous medium.
7.What is formed when phenol reacts with bromine water?Show answer
A white precipitate of 2,4,6-tribromophenol.
8.Name the product of Kolbe's reaction.Show answer
Salicylic acid (2-hydroxybenzoic acid).
9.Why do ethers have lower boiling points than isomeric alcohols?Show answer
Ether molecules cannot hydrogen-bond with each other.
10.What are the products of CH₃OCH₂CH₃ with HI?Show answer
CH₃I and CH₃CH₂OH (iodide attacks the smaller methyl group by SN2).

Best order to study it

Learn classification, IUPAC names and the bond angles and lengths of alcohols, phenols and ethers.
Learn every preparation with one equation: hydration, hydroboration–oxidation, reduction, Grignard, and the four routes to phenol.
Study physical properties and acidity carefully, including substituent effects with pKa values.
Learn reactions of alcohols (Na, esterification, HX and Lucas, PX₃, dehydration with mechanism, oxidation, Cu at 573 K) and of phenols (nitration, bromination, Kolbe, Reimer–Tiemann, Zn dust, oxidation).
Finish with ethers: dehydration, Williamson synthesis and its limitation, HI cleavage with mechanism, and anisole's ring reactions.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you write the three-step mechanism of acid-catalysed hydration of ethene?
Can you prepare propan-1-ol and propan-2-ol from propene?
Can you write the cumene route to phenol with both equations?
Can you order ethanol, water, phenol, o-, m- and p-nitrophenol and cresol by acidity?
Can you describe the Lucas test with observations for all three classes?
Can you write the mechanism of dehydration of ethanol to ethene?
Can you write Kolbe's and Reimer–Tiemann reactions and the bromination of phenol?
Can you write the Williamson synthesis of an unsymmetrical ether and predict products of ethers with HI?

Alcohols, Phenols and Ethers: frequently asked questions

How many marks does Alcohols, Phenols and Ethers carry in CBSE Class 12 2026-27?

Unit 7 carries 6 of the 70 theory marks. Expect named reactions, a mechanism or acidity question, and conversions, sometimes as part of a 5-mark organic question.

Are reactions of secondary and tertiary alcohols in the 2026-27 syllabus?

Yes. The 2025-26 text limited chemical properties to primary alcohols, but the 2026-27 syllabus follows the NCERT section headings, and the NCERT chapter covers the reactions of all three classes, such as the Lucas test and their oxidation. Learn them as the book presents them.

Which mechanisms should I learn from this chapter?

The three NCERT mechanisms: acid-catalysed hydration of ethene, acid-catalysed dehydration of ethanol (to ethene, and the SN2 path to ethoxyethane), and the reaction of an ether with HI.

How do I distinguish phenol from ethanol?

Phenol gives a violet colour with neutral FeCl₃ solution and a white precipitate with bromine water; ethanol does neither. Ethanol also gives a yellow precipitate of iodoform with I₂ and NaOH.

Why does Williamson synthesis fail with tertiary halides?

Alkoxide ions are strong bases as well as nucleophiles. With a crowded tertiary halide, SN2 attack is blocked, so the alkoxide removes a β-hydrogen and elimination gives an alkene instead of an ether.

Based on: CBSE Chemistry (043) syllabus 2026-27; NCERT Chemistry Part II, Textbook for Class XII (Reprint 2026-27)