Class 12 Chemistry · NCERT Chapter 7 (Part 2)
~13 min readAlcohols, Phenols and Ethers Class 12 Chemistry Notes
Alcohols, phenols and ethers carry the oxygen functional group at the heart of organic chemistry. This chapter covers nomenclature, the Williamson and hydration preparations, the Lucas test, oxidation of alcohols, the acidic nature of phenols, Kolbe and Reimer–Tiemann reactions and ether cleavage.
Why is phenol a stronger acid than ethanol but weaker than acetic acid?
The phenoxide ion is stabilised by resonance — the negative charge is delocalised over the ring — while the ethoxide ion has only inductive stabilisation. More stabilised conjugate base → stronger acid. But in acetic acid the carboxylate ion is stabilised by two resonance forms, so acetic acid is a stronger acid still. Substituents modify phenol acidity: electron-withdrawing groups (NO₂) raise it, electron-donating groups (CH₃) lower it.
Classification and Nomenclature
An alcohol has –OH on a saturated carbon, a phenol has –OH directly attached to an aromatic ring, and an ether has two organic groups joined by an oxygen atom R–O–R'. Both alcohols and phenols are classified by the number of –OH groups (mono-, di-, tri-).
- Alcohols: primary (1°), secondary (2°), tertiary (3°) — by the carbon bearing the –OH.
- Phenols: monohydric (C₆H₅OH) to polyhydric (catechol, resorcinol, hydroquinone).
- Ethers: symmetrical (CH₃OCH₃) and unsymmetrical (CH₃OC₂H₅).
- IUPAC: alkanols (2-propanol), phenols (2-methylphenol), alkoxy alkanes (methoxymethane).
Name check
MCQ Questions
Preparation of Alcohols and Phenols
- From alkenes (hydration): CH₂=CH₂ + H₂O, catalysed by H₃PO₄/H⁺ — Markovnikov addition gives more substituted alcohol.
- From carbonyls: reduction of aldehydes → 1° alcohols, ketones → 2° alcohols (LiAlH₄/NaBH₄); Grignard + aldehyde/ketone builds higher alcohols added on.
- From haloalkanes: hydrolysis with aqueous NaOH/KOH.
- Fermentation: starch/glucose → ethanol + CO₂.
- Phenols via: aniline diazonium salt + hot water; chlorobenzene + NaOH 623 K (Dow); cumene hydroperoxide (cumene–phenol process); haloarene + NaOH.
Cumene process
MCQ Questions
Physical Properties and Acidity
- Boiling points: alcohols/phenols/ethers rank above haloalkanes of similar mass because of intermolecular hydrogen bonding; ethers have none and boil lower.
- Solubility: lower alcohols and phenols dissolve in water via ROH···H₂O bonds; solubility falls as the alkyl tail grows.
- Density: most monoalcohols lighter than water; glycerol heavier due to more OH groups.
- Boiling order example: CH₃CH₂OH > CH₃OCH₃ despite equal mass — no O–H in the ether.
Acidity of phenols
Phenol (pKa ≈ 10) is acidic enough to react with NaOH but not NaHCO₃; ethanol (pKa ≈ 16) does not react with NaOH. Cause: the phenoxide ion is resonance-stabilised. Electron-withdrawing nitro groups raise acidity strongly (p-nitrophenol pKa ≈ 7.2); electron-donating methyl groups lower it.
MCQ comparison
MCQ Questions
Reactions of Alcohols
- Reaction with metals: 2ROH + 2Na → 2RONa + H₂ (basic alkoxide).
- Reaction with HX (ZnCl₂): forms alkyl halides; Lucas test ranks 3° > 2° > 1°.
- Esterification: ROH + R'COOH ⇌ R'COOR + H₂O (conc. H₂SO₄, reversible).
- Dehydration: 170°C/445K conc. H₂SO₄ → alkene; 140°C/413K → ether.
- Oxidation: 1° → aldehyde → acid (K₂Cr₂O₇/H⁺ or PCC stops at aldehyde); 2° → ketone; 3° no oxidation under mild conditions (needs C–C cleavage).
Oxidation logic
MCQ Questions
Reactions of Phenols and Ethers
- Kolbe's reaction: sodium phenoxide + CO₂ (125°C, 7 atm) → 2-hydroxybenzoic acid (salicylic acid), then + H⁺.
- Reimer–Tiemann: phenol + CHCl₃ + NaOH → salicylaldehyde (aqueous, mild).
- Halogenation: bromine water → 2,4,6-tribromophenol (white ppt); nitration → picric acid with excess HNO₃.
- Friedel–Crafts on phenol takes o-/p- substitution; −OH is strongly activating but ortho/para directing (due to lone pair resonance).
Ether cleavage
R–O–R' + HX (conc. HI/HBr, heat) splits into the smaller alkyl halide + the larger alcohol. Protonation of oxygen then SN1/SN2 attack on the alkyl group gives the products. Also useful for converting an ether back into two hydroxyl compounds.
Exam pairing
MCQ Questions
Importance in Daily Life and Industry
These three classes are everywhere: ethanol fuel and solvents, methanol as feedstock and antifreeze, ethylene glycol in coolants (polyhydric), glycerol in cosmetics and nitroglycerin explosives.
- Phenol: antiseptic/disinfectant (carbolic acid history), monomer for bakelite (with formaldehyde), aspirin precursor.
- Ethers: diethyl ether the classic anaesthetic; MTBE historically a fuel additive; diglyme as solvent in organometallic chemistry.
- Phenolic resins and polycarbonate (from bisphenol-A... made from phenol + acetone, whichever process the cumene route feeds) mould industry.
Remember phenols' dual life
MCQ Questions
NCERT Exercise Questions
From the textbook
Exercises (NCERT Q1–Q3)
- 1Arrange the following in increasing order of boiling point: Ethanol, Propan‑1‑ol, Propan‑2‑ol.
- 2Explain why phenol is more acidic than ethanol.
- 3Write the reaction of ethyl magnesium bromide with formaldehyde and name the product.
Answers at a glance
- 1Ethanol < Propan‑2‑ol < Propan‑1‑ol.
- 2Phenoxide ion is resonance‑stabilised, making phenol more acidic than ethanol.
- 3Propan‑1‑ol is formed.
CBSE Pattern Questions
How CBSE asks this chapter
MCQ Questions
1 Mark Questions
- 1What is denatured alcohol?
- 2Write the reaction of sodium metal with ethanol.
- 3What is the IUPAC name of glycerol?
2 Mark Questions
- 1Explain why the boiling point of ethanol is higher than that of methoxy methane.
- 2How will you convert propene to propan‑2‑ol?
3 Mark Questions
- 1Describe the preparation of phenol from cumene.
- 2Explain why ortho‑nitrophenol is more acidic than ortho‑methoxyphenol.
5 Mark Questions
- 1An organic compound (A) with molecular formula C4H10O on oxidation with acidified potassium dichromate gives a carboxylic acid (B) with molecular formula C4H8O2. Identify A and B.
- 2Explain the following: (i) Phenol gives bromine water test giving white precipitate of 2,4,6‑tribromophenol. (ii) Why does phenol not give the Lucas test?
PDF: Assignments & Tests
Printable study material
Free PDF Downloads & Assignments
Class 12 Chemistry — downloadable assignments and CBSE-pattern tests (PDF).
Topic-wise practice covering the first half of the chapter, with answer key.
Topic-wise practice covering the second half of the chapter, with answer key.
Full CBSE-pattern test (90 minutes, 40 marks) with answer key.
Second full CBSE-pattern test (90 minutes, 40 marks) with answer key.
Quick Revision
Key formulas at a glance
Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.
Hydration of alkene
Markovnikov hydration gives 2° alcohols.
Williamson synthesis
Use 1° RX to avoid E2.
Ether formation vs alkene
413 K → ether; 443 K → alkene.
Lucas test
3° instant, 2° ~5 min, 1° no effect.
Oxidation
Tertiary alcohols resist oxidation.
Kolbe reaction
Electrophile is CO₂ (carboxylation).
Exam Strategy
How this chapter is asked
High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.
- Classification and IUPAC names of alcohols, phenols and ethers — monotrihydric, 1°/2°/3°, alkoxy alkanes.
- Lucas test (ZnCl₂ + conc. HCl): cloudiness speed separates 1°, 2°, 3° alcohols.
- Oxidation: 1°→aldehyde→acid, 2°→ketone, 3° no reaction — and which oxidant (K₂Cr₂O₇/H⁺, Tollen's) is used.
- Esterification in acid and the mechanism; diethyl ether syntheses at 413 K vs 443 K.
- Williamson synthesis conditions (SN2, use 1° halide, alkoxide + no elimination).
- Relative acidity: phenol > ethanol; effect of EWG/EDG on acidity — nitrophenols.
- Kolbe and Reimer–Tiemann reactions of phenoxide with CO₂ and CHCl₃.
- Ether cleavage with HX — which alkyl halide forms first (smaller alkyl).
- Preparation of phenols: from cumene hydroperoxide, aniline diazonium, chlorobenzene.
Practice Numericals
Step-by-step worked examples
Standard exam-level problems solved with clear conceptual breakdowns.
Lucas test prediction
Step-by-Step Solution
- 1Lucas reagent = anhydrous ZnCl₂ in conc. HCl; reacts 3° > 2° > 1°.
- 2Immediate turbidity means the carbocation formed at once → tertiary alcohol.
Final Answer
A tertiary alcohol (e.g. tert-butyl alcohol).
Williamson synthesis route
Step-by-Step Solution
- 1Best: sodium methoxide + ethyl bromide (1° halide → SN2).
- 2CH₃O⁻Na⁺ + CH₃CH₂Br → CH₃O–CH₂CH₃ + NaBr.
- 3Avoid the reverse (ethoxide + methyl iodide works too but methyl halide is best with secondary alkoxides).
Final Answer
CH₃ONa + CH₃CH₂Br → CH₃OCH₂CH₃ (Williamson).
FAQ
Frequently asked questions
How does the Lucas test distinguish primary, secondary and tertiary alcohols?
Add Lucas reagent (conc. HCl + anhydrous ZnCl₂); an alcohol converts to the chloride which is insoluble and clouds the solution. Tertiary alcohols react instantly, secondary alcohols in about 5 minutes, and primary alcohols only on warming. So the speed of turbidity ranks the classes. It fails for very weak alcohols like methanol and those that cannot form carbocations.
Why are phenols more acidic than alcohols?
Compare the conjugate bases. The phenoxide ion spreads its negative charge over the ring by resonance (three equivalent structures), so it is strongly stabilised. The ethoxide ion cannot delocalise and carries the whole charge on oxygen. A more stabilised conjugate base means the equilibrium R–OH + H₂O ⇌ R–O⁻ + H₃O⁺ lies further right for phenols. Electron-withdrawing groups on the ring further stabilise phenoxide and raise acidity; electron-donating groups lower it.
Which conditions favour Williamson synthesis without side products?
Use the alkoxide from the less-hindered alcohol and a primary alkyl halide: SN2 needs an unhindered carbon, and bulky or secondary halides lean to E2 elimination of alkene. Polar aprotic solvent and mild base help. This asymmetry (alkoxide nucleophile + 1° halide) is the key to a clean ether in one step.
What happens when an ether reacts with HI/HBr?
Ethers cleave in strong HX by protonation of oxygen followed by SN1/SN2 attack of the halide on the alkyl group. The smaller alkyl group leaves first to form the alkyl halide, while the larger side becomes the alcohol. Mixed ethers therefore give a specific single alcohol + a specific single haloalkane, tested by the 'major product' question.
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