Classification, nomenclature and structure
Link to Std 11 : In Std 11 you named alkanes and alcohol-type compounds and learnt the shapes of water-like molecules (bent, sp³ oxygen).
Rubbing alcohol, the phenol in antiseptic liquids and the ether once used in anaesthesia are all members of this one family.
Alcohols have –OH on an sp³ carbon, phenols have –OH on a benzene ring, and ethers have an oxygen between two carbon groups (R–O–R′). Alcohols are classed primary, secondary or tertiary by the carbon bearing the –OH. IUPAC: alcohol ends in -ol, ethers are named as alkoxyalkanes.
In the figure : The oxygen is sp³ hybridised in alcohols and ethers; the bond angle is a little more than the water angle (104.5°) in ethers because of the repulsion between bulky groups.
Rule 1: Classification — By the number of –OH groups: mono-, di-, tri-, polyhydric. By carbon: 1°, 2°, 3° alcohols. Allylic alcohol: –OH on an sp³ carbon next to C=C; benzylic alcohol: –OH on an sp³ carbon next to a benzene ring. Ethers: simple (symmetrical, CH₃OCH₃) or mixed (CH₃OC₂H₅).
- Easy : CH₃CH₂CH₂OH is a primary alcohol; (CH₃)₂CHOH is secondary; (CH₃)₃COH is tertiary.
- Medium : Glycol HOCH₂CH₂OH is dihydric; glycerol is trihydric.
- Exam-type : C₆H₅CH₂OH is benzyl alcohol (benzylic) and CH₂=CHCH₂OH is allyl alcohol (allylic).
Rule 2: Nomenclature — Alcohols: select the longest chain containing the –OH, number from the end nearest to –OH, and use -ol (propan-2-ol). Phenols: the parent is phenol, substituents are numbered from the –OH carbon (2-methylphenol is o-cresol). Ethers: alkoxy + alkane, the larger group as the parent (1-methoxypropane; ethoxybenzene).
- Easy : CH₃CH(OH)CH₃ is propan-2-ol (isopropyl alcohol).
- Medium : HOCH₂CH₂OH is ethane-1,2-diol; CH₃OCH₂CH₃ is methoxyethane.
- Exam-type : C₆H₅OCH₃ is methoxybenzene (anisole); 1-methoxy-2-methylpropane is CH₃OCH₂CH(CH₃)₂.
Rule 3: Structure and bond lengths — In alcohols the C–O–H angle is about 109° (sp³ oxygen with two lone pairs). In phenols, the C–O bond is shorter than in alcohols (136 pm versus 143 pm) because the carbon is sp² and the oxygen lone pair is delocalised into the ring. In ethers the C–O–C angle is 111.7°.
- Easy : C–O bond in phenol (136 pm) is shorter than in methanol (142 pm).
- Medium : The oxygen in alcohols is sp³ hybridised.
- Exam-type : Why is the C–O bond in phenol shorter? The carbon is sp² and the C–O bond has partial double bond character through resonance.
Alcohol: OH on sp³ carbon. Phenol: OH on a benzene ring. Ether: C–O–C.
Calling C₆H₅CH₂OH a phenol; the –OH is on a side-chain carbon, so it is an alcohol (benzyl alcohol).
Memory clue: "Ring OH is phenol, chain OH is alcohol".
Spot the difference
| Point | Alcohol | Phenol |
|---|---|---|
| OH attached to | sp³ carbon | sp² ring carbon |
| Acidity | Very weak | Stronger |
| C–O bond | Longer | Shorter |
Practice questions
- 1Classify (CH₃)₃COH as primary, secondary or tertiary.
- 2Write the IUPAC name of CH₃OCH₂CH₃.
- 3What is the name of HOCH₂CH₂OH?
- 4Distinguish an alcohol from a phenol in terms of the carbon bonded to –OH.
- 5Write the structure of 2-methylbutan-2-ol and classify it.
- 6Why is the C–O bond in phenol shorter than in methanol?
- 7Name C₆H₅OCH₂CH₃ and C₆H₅CH₂OH and state which is an ether.
- 8Explain why the C–O–C angle in ethers is greater than the tetrahedral angle.
Show answers (try them yourself first)
- 1Tertiary.
- 2Methoxyethane.
- 3Ethane-1,2-diol (ethylene glycol).
- 4In alcohols the –OH is on an sp³ carbon; in phenols on an sp² carbon of a benzene ring.
- 5(CH₃)₂C(OH)CH₂CH₃; a tertiary alcohol.
- 6In phenol the carbon is sp² and the oxygen lone pair is delocalised in the ring, giving partial double-bond character.
- 7C₆H₅OCH₂CH₃ is ethoxybenzene (an ether); C₆H₅CH₂OH is benzyl alcohol (an alcohol).
- 8The two bulky alkyl groups repel each other, widening the angle from 109.5° to about 111.7°.
Preparation of alcohols and phenols
Link to Std 11 : In Std 11 you learnt addition of water to alkenes (Markovnikov) and the carbonyl group; those reactions are the main routes to alcohols.
Sanitiser alcohol is made on a very large scale by adding water to ethene from petroleum, and phenol and acetone are made together from cumene.
Alcohols are made from alkenes (hydration), from carbonyl compounds (reduction), and from Grignard reagents. Phenol is made from cumene, benzenesulphonic acid, diazonium salts or chlorobenzene.
In the figure : Grignard reagent plus methanal gives a primary alcohol, plus any other aldehyde gives a secondary alcohol, and plus a ketone gives a tertiary alcohol. All end with acid work-up.
Rule 1: Alcohols from alkenes and carbonyl compounds — Hydration: alkene + H₂O (H₂SO₄ catalyst) → alcohol, Markovnikov (propene → propan-2-ol). Hydroboration–oxidation: alkene + B₂H₆, then H₂O₂/OH⁻ → alcohol, anti-Markovnikov (propene → propan-1-ol). Reduction: aldehydes → 1° alcohols, ketones → 2° alcohols with NaBH₄, LiAlH₄ or H₂/Pd; carboxylic acids → 1° alcohols with LiAlH₄.
- Easy : CH₃CH=CH₂ + H₂O/H⁺ → CH₃CH(OH)CH₃.
- Medium : CH₃CHO + NaBH₄ → CH₃CH₂OH; CH₃COCH₃ + NaBH₄ → (CH₃)₂CHOH.
- Exam-type : CH₃COOH + LiAlH₄ (then H₂O) → CH₃CH₂OH.
Rule 2: Alcohols from Grignard reagents — RMgX + HCHO → RCH₂OH (1°); RMgX + R′CHO → RR′CHOH (2°); RMgX + R′COR″ → RR′R″COH (3°); each followed by H₂O/H⁺. Ethylene oxide gives a 1° alcohol with two more carbon atoms.
- Easy : CH₃MgBr + HCHO → CH₃CH₂OH (ethanol, primary).
- Medium : CH₃MgBr + CH₃CHO → (CH₃)₂CHOH (propan-2-ol, secondary).
- Exam-type : CH₃MgBr + CH₃COCH₃ → (CH₃)₃COH (2-methylpropan-2-ol, tertiary).
Rule 3: Phenols — Cumene process: benzene + propene → cumene; oxidation with O₂ → cumene hydroperoxide; acid → phenol + acetone (an important industrial route). From haloarene: chlorobenzene + NaOH at 623 K, 300 atm, then H⁺. From benzenesulphonic acid: fuse with NaOH, then acidify. From diazonium salt: warm with water (ArN₂⁺ → ArOH + N₂ + H⁺).
- Easy : C₆H₅N₂⁺Cl⁻ + H₂O (warm) → C₆H₅OH + N₂ + HCl.
- Medium : Cumene → phenol + propanone: both products are valuable.
- Exam-type : 10 g of benzene (78.11 g mol⁻¹) could give at most 10 ÷ 78.11 × 94.11 = 12.0 g of phenol (94.11 g mol⁻¹).
Grignard + HCHO → 1°; + other aldehyde → 2°; + ketone → 3°.
Forgetting the water/acid work-up: the Grignard reaction first forms an alkoxide, which must be protonated.
Memory clue: "H, aldehyde, ketone → 1, 2, 3" — HCHO, other aldehydes, ketones give primary, secondary, tertiary alcohols.
Spot the difference
| Point | Hydration of alkene | Hydroboration–oxidation |
|---|---|---|
| Orientation | Markovnikov | Anti-Markovnikov |
| Propene gives | Propan-2-ol | Propan-1-ol |
| Reagents | H₂O/H⁺ | B₂H₆, then H₂O₂/OH⁻ |
Practice questions
- 1Which reagent reduces a ketone to a secondary alcohol?
- 2What are the two products of the cumene process?
- 3Name the alcohol formed from CH₃MgBr and HCHO.
- 4How can propan-1-ol be made from propene?
- 5Write the reaction of CH₃MgBr with ethanal and name the product.
- 6How is phenol made from benzene diazonium chloride?
- 7Give a Grignard route to 2-methylpropan-2-ol.
- 8How much phenol can be formed from 10 g of benzene by the cumene process, assuming complete conversion? (78.11 and 94.11 g mol⁻¹)
Show answers (try them yourself first)
- 1NaBH₄ or LiAlH₄ (or H₂/Pd).
- 2Phenol and acetone (propanone).
- 3Ethanol (a primary alcohol).
- 4By hydroboration–oxidation (B₂H₆ followed by H₂O₂/OH⁻), which is anti-Markovnikov.
- 5CH₃MgBr + CH₃CHO → (CH₃)₂CHOMgBr → (CH₃)₂CHOH (propan-2-ol, secondary alcohol).
- 6Warm the diazonium salt solution with water: C₆H₅N₂⁺ + H₂O → C₆H₅OH + N₂ + H⁺.
- 7CH₃MgBr + CH₃COCH₃, then H₂O/H⁺ → (CH₃)₃COH.
- 810 ÷ 78.11 × 94.11 = 12.0 g.
Physical properties and acidity of alcohols and phenols
Link to Std 11 : In Std 11 you learnt hydrogen bonding and how resonance stabilises ions; both explain the properties of these compounds.
Ethanol mixes completely with water, but pentanol does not: the small alcohols are like water, the long-chain ones act like oils.
Alcohols and phenols form hydrogen bonds, so they have much higher boiling points than hydrocarbons, haloalkanes and ethers of similar mass; the lower ones are soluble in water. Acidity: alcohols are very weak acids (weaker than water); phenols are much stronger (but weaker than carboxylic acids) because the phenoxide ion is stabilised by resonance.
In the figure : The bars show pKa values. Phenol is far more acidic than ethanol, and electron-withdrawing groups such as –NO₂ in the o- or p-positions make it stronger still; 2,4,6-trinitrophenol (picric acid) is as strong as many mineral acids.
Rule 1: Boiling point and solubility — Hydrogen bonding between –OH groups raises the boiling point: ethanol (351 K) versus methoxymethane (249 K) and propane (231 K). Within alcohols, bp rises with mass and falls with branching (1° > 2° > 3°). Solubility in water falls as the alkyl group grows because the hydrocarbon part is hydrophobic.
- Easy : C₂H₅OH (351 K) has a higher boiling point than CH₃OCH₃ (249 K): hydrogen bonding.
- Medium : Butan-1-ol boils higher than 2-methylpropan-2-ol (the tertiary alcohol is more branched).
- Exam-type : Why is ethanol miscible with water but hexan-1-ol is not? In ethanol the –OH dominates; in hexanol the large non-polar chain outweighs the –OH.
Rule 2: Acidity of alcohols and phenols — ROH + Na → RONa + ½H₂ (alcohols behave as acids with sodium). Alcohol acidity: 1° > 2° > 3° (alkyl groups push electrons and destabilise the alkoxide). Phenol, C₆H₅OH ⇌ C₆H₅O⁻ + H⁺, is acidic enough to dissolve in NaOH but not in NaHCO₃. Resonance delocalises the negative charge over the ring in the phenoxide ion, but ethoxide ion has no such stabilisation.
- Easy : Phenol dissolves in aqueous NaOH to form sodium phenoxide; ethanol does not react with NaOH.
- Medium : Ka of phenol = 1.3 × 10⁻¹⁰, pKa = −log(1.3 × 10⁻¹⁰) = 9.89.
- Exam-type : For a 0.010 M phenol solution [H⁺] = √(Ka × c) = √(1.3 × 10⁻¹⁰ × 0.010) = 1.14e-06 M, so pH = 5.94.
Rule 3: Effect of substituents on phenol acidity — Electron-withdrawing groups (–NO₂) stabilise the phenoxide ion and increase acidity, most strongly at o- and p-positions; electron-donating groups (–CH₃, –OCH₃) decrease acidity. Order: picric acid > p-nitrophenol > m-nitrophenol > phenol > p-cresol.
- Easy : p-Nitrophenol is more acidic than phenol because NO₂ delocalises the negative charge of the phenoxide.
- Medium : p-Cresol is less acidic than phenol because –CH₃ pushes electrons towards the ring and destabilises the anion.
- Exam-type : Arrange phenol, o-nitrophenol and p-methylphenol in increasing acidity: p-methylphenol < phenol < o-nitrophenol.
Anything that stabilises the phenoxide ion (resonance, electron-withdrawing groups) makes the phenol more acidic.
Thinking alcohols are more acidic than water; they are weaker acids than water, and tertiary alcohols are the weakest.
Memory clue: "Pull makes it sour" — electron-withdrawing groups make phenols more acidic.
Spot the difference
| Point | Ethanol | Phenol |
|---|---|---|
| Ka (approx.) | 10⁻¹⁶ | 1.3 × 10⁻¹⁰ |
| Reacts with NaOH? | No | Yes |
| Anion stabilised by | Nothing (only inductive effect) | Resonance in the ring |
Practice questions
- 1Why does ethanol have a higher boiling point than methoxymethane?
- 2Does phenol dissolve in aqueous NaOH?
- 3Which is more acidic: phenol or ethanol?
- 4Find pKa for phenol (Ka = 1.3 × 10⁻¹⁰).
- 5Arrange phenol, p-nitrophenol and p-cresol in increasing acidity.
- 6Why are 1° alcohols more acidic than 3° alcohols?
- 7Calculate the pH of 0.010 M phenol (Ka = 1.3 × 10⁻¹⁰).
- 8Explain, with resonance, why phenol is more acidic than cyclohexanol.
Show answers (try them yourself first)
- 1Ethanol forms hydrogen bonds between molecules; the ether cannot.
- 2Yes, forming sodium phenoxide.
- 3Phenol.
- 4pKa = 9.89.
- 5p-Cresol < phenol < p-nitrophenol.
- 6Alkyl groups push electrons towards oxygen, making the alkoxide less stable; three groups in 3° alcohols push the most.
- 7[H⁺] = √(1.3 × 10⁻¹² ) = 1.14e-06 M; pH = 5.94.
- 8In phenoxide the negative charge on oxygen is delocalised into the ring (several resonance structures), but in cyclohexoxide it is localised on oxygen, so phenol loses H⁺ more easily.
Chemical reactions of alcohols
Link to Std 11 : In Std 11 you learnt oxidation states in organic compounds and carbocation stability; both decide the products here.
A bottle of wine left open goes sour because the ethanol is slowly oxidised to ethanoic acid (vinegar).
Alcohols react at the O–H bond (acidity, esters) or the C–O bond (substitution, dehydration), and they are oxidised depending on class. 1° → aldehyde → acid; 2° → ketone; 3° resists oxidation.
In the figure : PCC stops at the aldehyde; a strong oxidant carries it to the acid. Secondary alcohols give ketones. Tertiary alcohols have no H on the carbon bearing OH and are not oxidised easily.
Rule 1: Reactions at the O–H bond — With sodium: ROH + Na → RONa + ½H₂. Esterification: ROH + R′COOH ⇌ R′COOR + H₂O (H⁺ catalyst); with acid chlorides or anhydrides the reaction is irreversible (acylation): ethanol + ethanoic anhydride → ethyl ethanoate + ethanoic acid.
- Easy : 2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂.
- Medium : C₂H₅OH + CH₃COOH ⇌ CH₃COOC₂H₅ + H₂O (acid catalyst).
- Exam-type : Aspirin is made by acetylation: salicylic acid (an –OH compound) + acetic anhydride → aspirin.
Rule 2: Reactions at the C–O bond — With HX: ROH + HX → RX + H₂O. Reactivity 3° > 2° > 1° and HI > HBr > HCl. Lucas test (conc. HCl + ZnCl₂): 3° gives turbidity at once, 2° in about 5 minutes, 1° does not turn turbid at room temperature. Dehydration: with conc. H₂SO₄ ethanol gives ethene at 443 K and diethyl ether at 413 K (excess alcohol); the ease of dehydration is 3° > 2° > 1° because carbocation stability follows the same order.
- Easy : (CH₃)₃COH + conc. HCl/ZnCl₂ gives turbidity at once (3°).
- Medium : CH₃CH₂OH → CH₂=CH₂ + H₂O (conc. H₂SO₄, 443 K).
- Exam-type : Which is dehydrated fastest: butan-1-ol, butan-2-ol or 2-methylpropan-2-ol? The tertiary alcohol, via the most stable carbocation.
Rule 3: Oxidation and fermentation — PCC in CH₂Cl₂ converts 1° alcohols to aldehydes; acidified KMnO₄ or K₂Cr₂O₇ converts them to carboxylic acids; 2° alcohols give ketones. Ethanol is manufactured by fermentation: sucrose → glucose + fructose (invertase), then C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ (zymase).
- Easy : CH₃CH₂OH → CH₃CHO (PCC) → CH₃COOH (K₂Cr₂O₇).
- Medium : Propan-2-ol → propanone with CrO₃; butan-2-ol → butanone.
- Exam-type : 100 g of glucose (180.16 g mol⁻¹) gives at most 100 ÷ 180.16 × 2 × 46.07 = 51.1 g of ethanol.
1°: aldehyde then acid. 2°: ketone. 3°: no oxidation. Dehydration and Lucas reaction: 3° > 2° > 1°.
Saying that PCC oxidises a primary alcohol to the acid; PCC stops at the aldehyde.
Memory clue: "1 goes to 2 steps, 2 goes to 1 step, 3 stays" — primary can oxidise twice, secondary once, tertiary not at all.
Spot the difference
| Point | 1° alcohol | 3° alcohol |
|---|---|---|
| Oxidation | Aldehyde then acid | Resists |
| Lucas test | No reaction at room temp. | Immediate turbidity |
| Dehydration | Hardest | Easiest |
Practice questions
- 1Name the product of oxidising propan-2-ol.
- 2What does PCC do to a primary alcohol?
- 3Which alcohol class gives instant turbidity with Lucas reagent?
- 4Write the dehydration of ethanol with conc. H₂SO₄ at 443 K and at 413 K.
- 5Write the reaction of ethanol with ethanoic acid.
- 6How would you distinguish propan-1-ol from 2-methylpropan-2-ol?
- 7Find the maximum ethanol obtainable from 100 g glucose (180.16 g mol⁻¹, ethanol 46.07 g mol⁻¹).
- 8Explain why a 3° alcohol dehydrates more easily than a 1° alcohol.
Show answers (try them yourself first)
- 1Propanone (acetone).
- 2Oxidises it to an aldehyde only.
- 3Tertiary alcohols.
- 4443 K: C₂H₅OH → C₂H₄ + H₂O; 413 K (excess alcohol): 2C₂H₅OH → C₂H₅OC₂H₅ + H₂O.
- 5CH₃CH₂OH + CH₃COOH ⇌ CH₃COOCH₂CH₃ + H₂O (conc. H₂SO₄ catalyst).
- 6Lucas test: 2-methylpropan-2-ol turns turbid at once; propan-1-ol does not at room temperature.
- 7100 ÷ 180.16 × 2 × 46.07 = 51.1 g.
- 8Dehydration proceeds via a carbocation; the 3° carbocation is the most stable, so it forms fastest.
Chemical reactions of phenols
Link to Std 11 : In Std 11 you learnt electrophilic substitution of benzene and the idea of directing groups; –OH is a strong activating group.
The tablet you take for fever (aspirin) is made from phenol through salicylic acid, and picric acid was once a dye and an explosive.
Phenol's ring is made electron-rich by the –OH group, so it undergoes electrophilic substitution at the ortho and para positions much faster than benzene. Phenol also gives Kolbe (salicylic acid) and Reimer–Tiemann (salicylaldehyde) reactions and a violet colour with neutral FeCl₃.
In the figure : Each arrow shows a reagent and a product. Bromine water and concentrated nitric acid replace three hydrogens because the ring is so reactive.
Rule 1: Electrophilic substitution — Nitration: dilute HNO₃ at low T gives o- and p-nitrophenol; conc. HNO₃ gives 2,4,6-trinitrophenol (picric acid). Halogenation: Br₂ in CS₂ at low T gives o- and p-bromophenol (para major); Br₂ water gives a white precipitate of 2,4,6-tribromophenol (a test for phenol).
- Easy : Phenol + 3Br₂ (water) → 2,4,6-tribromophenol (white ppt) + 3HBr.
- Medium : 0.10 mol phenol gives at most 0.10 × 330.8 = 33.1 g of 2,4,6-tribromophenol (330.8 g mol⁻¹).
- Exam-type : 0.10 mol phenol gives at most 22.9 g of picric acid (229.1 g mol⁻¹).
Rule 2: Kolbe and Reimer–Tiemann reactions — Kolbe: sodium phenoxide + CO₂ (400 K, 4–7 atm) → sodium salicylate; acidify → salicylic acid. Reimer–Tiemann: phenol + CHCl₃ + aq NaOH (340 K) → salicylaldehyde through dichlorocarbene :CCl₂, then H⁺. Both go mainly to the ortho position.
- Easy : Salicylic acid + (CH₃CO)₂O → aspirin + CH₃COOH.
- Medium : 13.8 g salicylic acid (138.12 g mol⁻¹) gives at most 13.8 ÷ 138.12 × 180.16 = 18.0 g aspirin (180.16 g mol⁻¹).
- Exam-type : Phenol → salicylaldehyde needs CHCl₃ and NaOH: the electrophile is dichlorocarbene.
Rule 3: Other reactions of phenol — Neutral FeCl₃ gives a violet colour (test for phenols). Zinc dust distillation gives benzene. Oxidation with chromic acid gives benzoquinone. Reaction with NaOH gives phenoxide, which with R–X gives ethers (Williamson).
- Easy : Phenol + Zn (heat) → benzene + ZnO.
- Medium : Phenol with acidified Na₂Cr₂O₇ gives the yellow benzoquinone.
- Exam-type : Phenol + FeCl₃ → violet complex (a quick test; alcohols do not give it).
–OH is ortho/para directing and activating; phenol is brominated to 2,4,6-tribromophenol with bromine water.
Writing the product of phenol with bromine water as monobromophenol; with bromine water it goes to the tribromo compound.
Memory clue: "Bromine water, white ppt; FeCl₃, violet" — two quick tests for phenol.
Spot the difference
| Point | Benzene | Phenol |
|---|---|---|
| Ring reactivity | Moderate | Much higher (–OH activates) |
| Br₂ in water | No reaction | White ppt, tribromo |
| FeCl₃ | No colour | Violet colour |
Practice questions
- 1Name the test that gives a violet colour with phenol.
- 2What is formed when phenol reacts with bromine water?
- 3Name the product of the Kolbe reaction.
- 4Write the Reimer–Tiemann reaction of phenol.
- 5How is aspirin made from phenol?
- 6Why does phenol react faster than benzene in electrophilic substitution?
- 7Find the mass of picric acid formed from 0.10 mol phenol (229.1 g mol⁻¹), assuming complete conversion.
- 8Find the maximum aspirin from 13.8 g of salicylic acid (138.12 and 180.16 g mol⁻¹).
Show answers (try them yourself first)
- 1The ferric chloride (neutral FeCl₃) test.
- 22,4,6-Tribromophenol, a white precipitate.
- 3Salicylic acid (2-hydroxybenzoic acid).
- 4C₆H₅OH + CHCl₃ + 3NaOH → C₆H₄(OH)CHO (salicylaldehyde) + 3NaCl + 2H₂O, via :CCl₂.
- 5Phenol → (NaOH, CO₂; Kolbe) salicylic acid → (acetic anhydride) aspirin.
- 6The oxygen lone pair delocalises into the ring, increasing the electron density at o and p positions.
- 70.10 × 229.1 = 22.9 g.
- 813.8 ÷ 138.12 × 180.16 = 18.0 g.
Ethers and industrial alcohols
Link to Std 11 : In Std 11 you saw that ether-like compounds are fairly inert; they make good solvents for organic reactions, such as Grignard reagents.
Diethyl ether was one of the first anaesthetics, and it is still a standard solvent in laboratories; denatured ethanol is sold as 'spirit'.
Ethers (R–O–R′) are made by dehydration of alcohols (413 K, excess alcohol) or by the Williamson synthesis (alkoxide + primary alkyl halide, SN2). Ethers are cleaved by HI. Methanol is made from synthesis gas; ethanol is made by fermentation.
In the figure : Williamson synthesis uses a primary halide; a tertiary halide gives alkene because the alkoxide acts as a base. HI cleaves the C–O bond of ethers, and in an aryl alkyl ether only the alkyl–O bond breaks.
Rule 1: Preparing ethers — (i) 2ROH → ROR + H₂O (conc. H₂SO₄, 413 K) works only for primary alcohols and gives symmetrical ethers. (ii) Williamson: R–ONa + R′–X → R–O–R′ + NaX; use the primary (or methyl) halide so that SN2 beats elimination. To make a mixed ether with a bulky group, put the bulky group on the alkoxide.
- Easy : CH₃CH₂ONa + CH₃CH₂Br → C₂H₅OC₂H₅ + NaBr.
- Medium : C₆H₅ONa + CH₃Br → C₆H₅OCH₃ (anisole) + NaBr.
- Exam-type : To make methyl tert-butyl ether, use (CH₃)₃CONa + CH₃Br (not CH₃ONa + (CH₃)₃CBr, which gives 2-methylpropene).
Rule 2: Reactions of ethers — Cleavage by HI: R–O–R′ + HI → R–I + R′OH; with excess HI both become iodides. For a 1° alkyl group the halide ion attacks the smaller group (SN2); for a tertiary group it leaves as a carbocation (SN1). Aryl alkyl ether: anisole + HI → phenol + CH₃I. Electrophilic substitution: the –OR group directs ortho/para (anisole gives mainly o- and p-products in nitration, halogenation and Friedel–Crafts).
- Easy : CH₃OCH₂CH₃ + HI → CH₃I + CH₃CH₂OH (iodide attacks the smaller group).
- Medium : (CH₃)₃COCH₃ + HI → (CH₃)₃CI + CH₃OH (tertiary: SN1).
- Exam-type : Anisole + Br₂ (ethanoic acid) → p-bromoanisole (major) + o-bromoanisole.
Rule 3: Methanol and ethanol — Methanol: CO + 2H₂ → CH₃OH (ZnO–Cr₂O₃, 573–673 K, 200–300 atm); it is poisonous (blindness). Ethanol: fermentation of sugars (invertase then zymase, with yeast) and from ethene by hydration. Denatured spirit is ethanol made unfit to drink by adding copper sulphate and pyridine so that it cannot be misused.
- Easy : Industrial methanol is from synthesis gas (CO and H₂).
- Medium : Ethanol from molasses: sucrose → glucose + fructose → ethanol + CO₂.
- Exam-type : Why is denaturing needed? Industrial ethanol is cheap because it is untaxed; denaturing prevents it from being drunk.
Williamson: use a primary alkyl halide. HI cleaves ethers at the alkyl–O bond, with the iodide on the smaller (or 3° via SN1) alkyl group.
Using a tertiary halide in the Williamson synthesis; elimination to an alkene occurs.
Memory clue: "Bulky on the alkoxide, small on the halide" for Williamson synthesis.
Spot the difference
| Point | Dehydration of alcohols | Williamson synthesis |
|---|---|---|
| Type of ether | Symmetrical (mostly) | Symmetrical or mixed |
| Substrate | 1° alcohol | Alkoxide + 1° halide |
| Mechanism | Protonation, SN2 | SN2 |
Practice questions
- 1What is the Williamson synthesis?
- 2Write the equation for the cleavage of anisole with HI.
- 3How is methanol made industrially?
- 4Why is 2-methylpropene formed when (CH₃)₃CBr is treated with CH₃ONa?
- 5How do you prepare ethoxybenzene?
- 6What is denatured alcohol?
- 7Which reagents would you use to prepare methyl tert-butyl ether, and why?
- 8Predict the products of CH₃CH₂OCH(CH₃)₂ with one equivalent of HI and give reasons.
Show answers (try them yourself first)
- 1Reaction of a sodium alkoxide with a primary alkyl halide to give an ether (SN2).
- 2C₆H₅OCH₃ + HI → C₆H₅OH + CH₃I.
- 3From CO and H₂ (synthesis gas) over ZnO–Cr₂O₃ at 573–673 K and 200–300 atm.
- 4A tertiary halide undergoes elimination because the alkoxide acts as a base.
- 5C₆H₅ONa + CH₃CH₂Br → C₆H₅OCH₂CH₃ + NaBr.
- 6Ethanol made unfit for drinking by adding substances such as copper sulphate and pyridine.
- 7Use (CH₃)₃CONa + CH₃Br (primary halide). The reverse pair would give alkene.
- 8Iodide attacks the less hindered ethyl carbon in an SN2 step to give CH₃CH₂I + (CH₃)₂CHOH (the isopropyl carbon is more hindered).