Alcohols, Phenols and Ethers
Welcome to HSLC Guru! This chapter covers Class 12 Chemistry Chapter 11 — Alcohols, Phenols and Ethers under the ASSEB (Assam State School Education Board) syllabus. Alcohols, phenols, and ethers are three vital classes of oxygen-containing organic compounds. They appear everywhere — from the ethanol in beverages and the methanol used as fuel, to the phenol used as antiseptic, and ethers used as anaesthetics and solvents. In this chapter we discuss their nomenclature, methods of preparation, physical and chemical properties, important reactions like Kolbe and Reimer-Tiemann, ether cleavage, and the industrial production of methanol and ethanol. The complete English-medium notes, question answers, MCQs, fill-in-the-blanks, true/false, and a glossary follow below.
Chapter Summary
Alcohols, Phenols and Classification: Alcohols contain the hydroxyl group (–OH) attached to an sp³ carbon of an alkyl/substituted alkyl group, while phenols have –OH directly attached to an sp² carbon of an aromatic (benzene) ring. Alcohols are classified as primary (1°), secondary (2°) and tertiary (3°) depending on whether the carbon bearing the –OH group is attached to one, two, or three other carbon atoms respectively. They are also classed as monohydric, dihydric, and trihydric based on the number of –OH groups. In IUPAC nomenclature, alcohols end in “-ol” (e.g. ethanol, propan-2-ol) and phenols are named as hydroxy derivatives of benzene (e.g. phenol, 2-methylphenol).
Preparation of Alcohols: Alcohols are prepared from alkenes by acid-catalysed hydration (Markovnikov addition of water) and by hydroboration-oxidation, which gives anti-Markovnikov product without rearrangement. From carbonyl compounds, aldehydes give 1° alcohols and ketones give 2° alcohols on reduction with H₂/Ni or NaBH₄/LiAlH₄. Carboxylic acids and esters give 1° alcohols on reduction with LiAlH₄. Grignard reagents (RMgX) react with aldehydes/ketones to give alcohols of varying class — HCHO gives 1°, other aldehydes give 2°, and ketones give 3° alcohols. Phenol is industrially prepared from chlorobenzene (Dow process), benzenesulphonic acid (alkali fusion), diazonium salts, and the cumene process (most important).
Physical and Chemical Properties: Alcohols and phenols possess intermolecular hydrogen bonding which makes their boiling points unusually high compared to corresponding hydrocarbons and haloalkanes. Lower alcohols are completely miscible with water due to H-bonding. Chemically, alcohols and phenols react with active metals (Na, K) to release H₂ gas. They undergo esterification with carboxylic acids in presence of conc. H₂SO₄. Alcohols undergo oxidation: 1° to aldehydes and then carboxylic acids, 2° to ketones, while 3° alcohols resist oxidation. They also undergo dehydration with conc. H₂SO₄ to alkenes (E1/E2 mechanism). Phenols are more acidic than alcohols because the phenoxide ion is resonance-stabilised; electron-withdrawing groups (–NO₂) further increase acidity, while electron-donating groups (–CH₃, –OCH₃) decrease it.
Special Reactions and Ethers: Phenols undergo Kolbe’s reaction with NaOH and CO₂ to give salicylic acid, and Reimer-Tiemann reaction with CHCl₃/NaOH to give salicylaldehyde. Phenols also undergo electrophilic substitution (bromination, nitration, sulphonation) much more easily than benzene because –OH is ortho/para-directing. Ethers (R–O–R′) are prepared by Williamson synthesis (R–X + R′–O⁻Na⁺) — best for primary alkyl halides since 2°/3° halides give elimination. Ethers undergo cleavage with HI/HBr to give alkyl halide and alcohol/phenol. Methanol (wood spirit) is industrially prepared from synthesis gas (CO + 2H₂ at 200–300 atm/Cu-ZnO-Cr₂O₃), and ethanol is obtained by fermentation of sugars/starch. Methanol is toxic (causes blindness), while ethanol is used in beverages, solvents, and as a fuel additive.
Short Answer Questions (1 Mark)
Q1. What is a primary alcohol?
Answer: An alcohol in which the –OH group is attached to a carbon atom that is itself bonded to only one other carbon atom (or to no carbon, as in methanol) is called a primary (1°) alcohol. Example: ethanol (CH₃CH₂OH).
Q2. Write the IUPAC name of (CH₃)₃C–OH.
Answer: 2-methylpropan-2-ol (a tertiary alcohol).
Q3. Why is phenol more acidic than ethanol?
Answer: The phenoxide ion (C₆H₅O⁻) is resonance-stabilised — the negative charge is delocalised into the benzene ring, making it a weaker base. Hence phenol releases the H⁺ more readily, making it more acidic than ethanol whose ethoxide ion is not resonance-stabilised.
Q4. What is Williamson’s synthesis?
Answer: Williamson’s synthesis is the preparation of ethers by reaction of an alkyl halide (preferably primary) with sodium alkoxide (or aryloxide): R–X + R′–ONa → R–O–R′ + NaX. It is a Sₙ2 reaction.
Q5. Name the catalyst used in the cumene process.
Answer: Cumene (isopropylbenzene) is oxidised in air, then treated with dilute acid to produce phenol and acetone. The conversion uses air (O₂) for oxidation and dilute H₂SO₄ for the hydroperoxide rearrangement.
Q6. Why does methanol cause blindness?
Answer: In the body, methanol is oxidised by alcohol dehydrogenase to formaldehyde and then formic acid, which damages the optic nerve and retina, leading to permanent blindness or even death.
Q7. Give an example of a Grignard reagent.
Answer: Methylmagnesium bromide, CH₃MgBr (or ethylmagnesium iodide, C₂H₅MgI). It is prepared by reaction of an alkyl halide with magnesium turnings in dry ether.
Q8. What is denatured alcohol?
Answer: Denatured alcohol is ethanol made unfit for drinking by adding poisonous substances such as methanol, pyridine, or copper sulphate. It is sold for industrial use without excise duty.
Q9. Name two reagents used to oxidise primary alcohols to carboxylic acids.
Answer: Acidified KMnO₄ and acidified K₂Cr₂O₇ (potassium dichromate) are commonly used. Jones’ reagent (CrO₃ in dil. H₂SO₄/acetone) is also effective.
Q10. What is fermentation?
Answer: Fermentation is the slow chemical decomposition of sugars (e.g. glucose) by enzymes in yeast (zymase, invertase) at 25–30 °C to give ethanol and CO₂: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂.
Short Answer Questions (2-3 Marks)
Q1. Distinguish between hydration and hydroboration-oxidation of alkenes.
Answer: In acid-catalysed hydration, an alkene reacts with water in presence of dilute H₂SO₄ to give an alcohol following Markovnikov’s rule (H adds to C with more H, OH to C with fewer H). It proceeds via carbocation intermediate, so rearrangement may occur. In hydroboration-oxidation, alkene reacts with diborane (B₂H₆) and the resulting trialkylborane is oxidised with H₂O₂/OH⁻ to give an alcohol following anti-Markovnikov addition. It is concerted, so no rearrangement; gives 1° alcohol from terminal alkene.
Q2. Why are alcohols more soluble in water than corresponding alkanes and haloalkanes?
Answer: Alcohols can form intermolecular hydrogen bonds with water molecules through their –OH group, which acts as both hydrogen-bond donor and acceptor. Alkanes and haloalkanes lack a polar O–H bond, so they cannot form hydrogen bonds with water. The energy released by H-bonding compensates for breaking water’s H-bond network, making lower alcohols completely miscible with water.
Q3. Explain the Kolbe’s reaction with example.
Answer: When sodium phenoxide is heated with CO₂ at 125 °C and 4–7 atm pressure, the resulting product is acidified to give salicylic acid (2-hydroxybenzoic acid). The phenoxide ion attacks CO₂ at its ortho position. C₆H₅ONa + CO₂ → C₆H₄(OH)(COONa) → (H⁺) → C₆H₄(OH)(COOH). This is industrially important as salicylic acid is used to make aspirin (acetylsalicylic acid).
Q4. What is Reimer-Tiemann reaction?
Answer: When phenol is treated with chloroform (CHCl₃) and aqueous NaOH at 340 K, followed by acidification, salicylaldehyde (2-hydroxybenzaldehyde) is obtained. The reactive species is dichlorocarbene (:CCl₂), which inserts at the ortho position of phenoxide ion. C₆H₅OH + CHCl₃ + 3NaOH → o-HOC₆H₄CHO + 3NaCl + 2H₂O.
Q5. Why do tertiary alcohols not undergo oxidation easily?
Answer: Oxidation of an alcohol involves removal of H from the carbon bearing the –OH (and an H from O–H). In tertiary alcohols, the carbon bearing –OH has no hydrogen attached — instead it is bonded to three carbon groups. Hence ordinary oxidising agents (KMnO₄, K₂Cr₂O₇) cannot abstract hydrogen and oxidation does not occur. Strong oxidising agents under harsh conditions can break C–C bonds, giving a mixture of ketones and acids.
Q6. Predict the product when ethanol reacts with conc. H₂SO₄ at (a) 413 K and (b) 443 K.
Answer: (a) At 413 K (140 °C), two molecules of ethanol undergo intermolecular dehydration to form diethyl ether: 2C₂H₅OH → C₂H₅–O–C₂H₅ + H₂O. (b) At 443 K (170 °C), intramolecular dehydration occurs giving ethene: C₂H₅OH → CH₂=CH₂ + H₂O. Higher temperature favours alkene formation as it is the kinetic product of E1 elimination.
Long Answer Questions (5-7 Marks)
Q1. Describe the laboratory and industrial preparation of phenol.
Answer: Phenol can be prepared by several routes. (i) From haloarenes (Dow process): chlorobenzene is heated with NaOH at 623 K and 320 atm. The resultant sodium phenoxide is acidified to give phenol: C₆H₅Cl + 2NaOH → C₆H₅ONa + NaCl + H₂O; C₆H₅ONa + HCl → C₆H₅OH + NaCl. (ii) From benzenesulphonic acid: benzene is sulphonated with conc. H₂SO₄, the acid is neutralised and fused with NaOH at 573 K to give sodium phenoxide, which on acidification yields phenol. (iii) From diazonium salts: aniline is diazotised with NaNO₂/HCl at 273 K and the diazonium salt is warmed with water to give phenol and N₂. (iv) Cumene process (industrial): benzene is alkylated with propene to give cumene, which is oxidised by air to cumene hydroperoxide; this is treated with dil. H₂SO₄ to yield phenol and acetone as by-product. The cumene process is most important because both products (phenol and acetone) are industrially valuable.
Q2. Discuss the chemical reactions of alcohols under five major heads.
Answer: (a) Reactions involving cleavage of O–H bond: Alcohols react with active metals (Na, K, Al) to liberate H₂ and form alkoxides — 2R–OH + 2Na → 2R–ONa + H₂. They react with acid chlorides and anhydrides to form esters. (b) Reactions involving cleavage of C–O bond: Alcohols react with HX (e.g. HCl/ZnCl₂ — Lucas reagent) to form alkyl halides; reactivity 3° > 2° > 1°. They react with PCl₃, PCl₅, SOCl₂ to give haloalkanes. (c) Esterification: With carboxylic acids in presence of conc. H₂SO₄, alcohols give esters: RCOOH + R′OH ⇌ RCOOR′ + H₂O. (d) Oxidation: 1° alcohols give aldehydes and then carboxylic acids; 2° alcohols give ketones; 3° alcohols are resistant. PCC (pyridinium chlorochromate) oxidises 1° to aldehyde without further oxidation. (e) Dehydration: Conc. H₂SO₄ at 443 K dehydrates 1° alcohols to alkenes; ease of dehydration: 3° > 2° > 1°. These reactions illustrate the dual reactivity of the C–O–H system.
Q3. Explain why ortho- and para-nitrophenols are more acidic than phenol, while ortho- and para-methoxyphenols are less acidic.
Answer: The acidity of phenols depends on the stability of the phenoxide ion formed after loss of H⁺. Electron-withdrawing groups (like –NO₂) at ortho/para positions stabilise the phenoxide ion by dispersing the negative charge through resonance and inductive (–I, –M) effects. So 4-nitrophenol (pKa ≈ 7.2) is far more acidic than phenol (pKa ≈ 10). 2,4,6-trinitrophenol (picric acid, pKa ≈ 0.4) is even stronger — comparable to mineral acids. Conversely, electron-donating groups (–OCH₃, –CH₃, –NH₂) at ortho/para positions intensify the negative charge on the phenoxide ion through +M/+I effects, destabilising it and making the corresponding phenol less acidic. Therefore, 4-methoxyphenol (pKa ≈ 10.2) is slightly less acidic than phenol. Meta substituents have less effect because they cannot conjugate with the –O⁻.
Q4. Describe Williamson’s synthesis of ethers, mentioning its mechanism, scope, and limitations.
Answer: Williamson synthesis is the most general method to prepare both symmetrical and unsymmetrical ethers. It involves the reaction of an alkyl halide with sodium (or potassium) alkoxide/aryloxide: R–X + R′–O⁻Na⁺ → R–O–R′ + NaX. Mechanism: It is a typical Sₙ2 reaction in which the alkoxide attacks the carbon of the alkyl halide from the side opposite to the leaving group, with simultaneous bond making and bond breaking through a transition state. Scope: It works well for primary alkyl halides combined with any alkoxide (1°, 2°, 3°) — for example, the preparation of methyl tert-butyl ether (MTBE) is best done by reacting CH₃I with (CH₃)₃C–ONa, not the other way round. Aryl ethers (anisoles) are prepared by reacting C₆H₅O⁻Na⁺ with R–X. Limitations: Secondary and tertiary alkyl halides give predominantly elimination (alkene) due to steric hindrance and the strongly basic nature of alkoxides. Aryl halides cannot be used because they do not undergo Sₙ2 reactions easily. Hence the rule: combine the bulkier oxygen with the smaller alkyl halide.
Q5. Discuss the cleavage of ethers with HI/HBr and explain the products formed.
Answer: Ethers are generally inert, but they undergo cleavage of the C–O bond when heated with concentrated HI or HBr (HCl is too weak). Symmetrical ethers give two molecules of alkyl halide and water on cleavage with excess HX: R–O–R + 2HX → 2R–X + H₂O. For unsymmetrical ethers R–O–R′, the products depend on the nature of the alkyl groups. (i) When both groups are primary or secondary, the smaller alkyl group forms the alkyl halide (Sₙ2 mechanism — attack at less hindered C). E.g. CH₃–O–C₂H₅ + HI → CH₃I + C₂H₅OH. (ii) When one group is tertiary, that group forms the alkyl halide (Sₙ1 mechanism — stable carbocation forms at tertiary carbon). E.g. (CH₃)₃C–O–CH₃ + HI → (CH₃)₃C–I + CH₃OH. (iii) For aryl alkyl ethers (anisole), the aryl–O bond is too strong to break, so the alkyl group leaves as alkyl halide and phenol is the other product: C₆H₅–O–CH₃ + HI → C₆H₅OH + CH₃I. The reactivity of HX is HI > HBr > HCl due to nucleophilicity of the halide.
Multiple Choice Questions (MCQs)
Q1. Which of the following is a tertiary alcohol?
(a) Propan-1-ol (b) Propan-2-ol (c) 2-methylpropan-2-ol (d) Butan-1-ol
Answer: (c) 2-methylpropan-2-ol.
Q2. Hydroboration-oxidation of propene gives:
(a) Propan-2-ol (b) Propan-1-ol (c) Acetone (d) Propanal
Answer: (b) Propan-1-ol (anti-Markovnikov product).
Q3. The most acidic compound among the following is:
(a) Phenol (b) o-Nitrophenol (c) p-Nitrophenol (d) 2,4,6-Trinitrophenol
Answer: (d) 2,4,6-Trinitrophenol (picric acid).
Q4. Williamson’s synthesis is best for preparing:
(a) Symmetrical ethers only (b) Ethers from 3° alkyl halides (c) Ethers from 1° alkyl halides (d) Phenols
Answer: (c) Ethers from 1° alkyl halides.
Q5. Reaction of phenol with CHCl₃ and NaOH gives:
(a) Salicylic acid (b) Salicylaldehyde (c) Benzoic acid (d) Anisole
Answer: (b) Salicylaldehyde (Reimer-Tiemann reaction).
Q6. Lucas reagent is:
(a) Anhydrous ZnCl₂ + conc. HCl (b) NaOH + CHCl₃ (c) HNO₃ + H₂SO₄ (d) KMnO₄ in acid
Answer: (a) Anhydrous ZnCl₂ + conc. HCl.
Q7. Methanol is industrially prepared from:
(a) Methane (b) Synthesis gas (CO + 2H₂) (c) Wood (d) Ethene
Answer: (b) Synthesis gas (CO + 2H₂) at 200–300 atm with Cu/ZnO/Cr₂O₃ catalyst.
Q8. Anisole on cleavage with HI gives:
(a) Phenol + methyl iodide (b) Iodobenzene + methanol (c) Phenol + methanol (d) Iodobenzene + methyl iodide
Answer: (a) Phenol + methyl iodide.
Q9. The catalyst used in the cumene process for oxidation is:
(a) V₂O₅ (b) Pt (c) Air/O₂ (d) Ni
Answer: (c) Air/O₂.
Q10. Grignard reagent on reaction with formaldehyde gives:
(a) 1° alcohol (b) 2° alcohol (c) 3° alcohol (d) Carboxylic acid
Answer: (a) 1° alcohol.
Fill in the Blanks
Q1. The IUPAC name of glycerol is __________.
Answer: Propane-1,2,3-triol.
Q2. Phenol on bromination with bromine water gives __________.
Answer: 2,4,6-tribromophenol.
Q3. The reagent used to selectively oxidise 1° alcohol to aldehyde is __________.
Answer: Pyridinium chlorochromate (PCC).
Q4. Ethanol on heating with conc. H₂SO₄ at 443 K gives __________.
Answer: Ethene (CH₂=CH₂).
Q5. The enzyme that converts glucose to ethanol is __________.
Answer: Zymase.
True or False
Q1. Phenol is more acidic than ethanol.
Answer: True.
Q2. Tertiary alcohols are easily oxidised by acidified KMnO₄.
Answer: False — tertiary alcohols resist mild oxidation because the C–OH carbon has no hydrogen.
Q3. Methanol is also called wood spirit.
Answer: True.
Q4. Williamson’s synthesis works best with tertiary alkyl halides.
Answer: False — tertiary alkyl halides give elimination products; primary halides are best.
Q5. Reimer-Tiemann reaction gives salicylic acid as the main product.
Answer: False — Reimer-Tiemann gives salicylaldehyde; Kolbe’s reaction gives salicylic acid.
Glossary
| Term | Meaning |
|---|---|
| Alcohol | Organic compound with –OH group on sp³ carbon |
| Phenol | Compound with –OH directly on aromatic ring |
| Ether | Compound of type R–O–R′ |
| Primary alcohol | –OH on a C bonded to one other C |
| Tertiary alcohol | –OH on a C bonded to three other C atoms |
| Hydroboration | Addition of B₂H₆ across C=C followed by H₂O₂/OH⁻ oxidation |
| Williamson synthesis | R–X + R′ONa → R–O–R′; main ether-prep method |
| Kolbe’s reaction | Phenoxide + CO₂ → salicylic acid |
| Reimer-Tiemann reaction | Phenol + CHCl₃/NaOH → salicylaldehyde |
| Cumene process | Industrial preparation of phenol and acetone |
| Lucas reagent | Anhydrous ZnCl₂ + conc. HCl; tests alcohol class |
| Esterification | Alcohol + carboxylic acid → ester (acid-catalysed) |
| Dehydration | Removal of water from alcohol giving alkene/ether |
| Fermentation | Enzymatic conversion of sugars to ethanol + CO₂ |
| Denatured alcohol | Ethanol made undrinkable by adding methanol etc. |
| Grignard reagent | RMgX prepared in dry ether; adds to C=O |
| Anisole | Methyl phenyl ether (C₆H₅OCH₃) |
| Picric acid | 2,4,6-trinitrophenol — strongly acidic |
| Hydrogen bonding | Attractive force between H of –OH and electronegative atom |
| Ether cleavage | Breakdown of ether by HI/HBr to alkyl halide + alcohol |
Important Reaction Equations
1. Acid-catalysed hydration of ethene: CH₂=CH₂ + H₂O ⟶ (H⁺) ⟶ CH₃CH₂OH (Markovnikov product for unsymmetrical alkenes).
2. Hydroboration-oxidation of propene: 3CH₃CH=CH₂ + (BH₃)₂ ⟶ 2(CH₃CH₂CH₂)₃B; then (CH₃CH₂CH₂)₃B + 3H₂O₂/OH⁻ ⟶ 3CH₃CH₂CH₂OH (propan-1-ol).
3. Grignard reagent with carbonyl: R–MgX + R′CHO ⟶ R′CH(OMgX)R ⟶ (H₂O) ⟶ R′CH(OH)R (secondary alcohol).
4. Reduction of carboxylic acid: RCOOH + 4[H] ⟶ (LiAlH₄) ⟶ RCH₂OH + H₂O. LiAlH₄ is a powerful reductant; NaBH₄ does not reduce –COOH.
5. Esterification: CH₃COOH + C₂H₅OH ⇌ (conc. H₂SO₄) ⇌ CH₃COOC₂H₅ + H₂O. The reaction is reversible and slow without catalyst.
6. Kolbe’s reaction: C₆H₅O⁻Na⁺ + CO₂ ⟶ (125 °C, 5 atm) ⟶ o-HOC₆H₄COONa ⟶ (H⁺) ⟶ Salicylic acid.
7. Reimer-Tiemann reaction: C₆H₅OH + CHCl₃ + 3NaOH ⟶ (340 K) ⟶ o-HOC₆H₄CHO + 3NaCl + 2H₂O.
8. Williamson ether synthesis: CH₃CH₂Br + CH₃ONa ⟶ CH₃OCH₂CH₃ (ethyl methyl ether) + NaBr.
9. Ether cleavage by HI: C₂H₅OC₂H₅ + 2HI ⟶ 2C₂H₅I + H₂O (excess HI gives two molecules of alkyl iodide).
10. Industrial preparation of methanol: CO + 2H₂ ⟶ (Cu/ZnO/Cr₂O₃, 200–300 atm, 573 K) ⟶ CH₃OH.
This concludes the Class 12 Chemistry Chapter 11 — Alcohols, Phenols and Ethers notes for ASSEB students. Practice the question types above to score full marks. For more chapters and notes in English and Assamese medium, keep visiting HSLC Guru. All the best for your Class 12 board examinations!