Alcohols, Phenols and Ethers | ISC Class 12 Chemistry Notes

This note covers the classification, structures, naming, preparation, physical properties and reactions of alcohols, phenols and ethers, together with reaction mechanisms, chemical tests, conversions, and the manufacture and uses of methanol and ethanol.

How are alcohols, phenols and ethers classified and named?

A functional group is the atom or group responsible for characteristic chemical reactions. The hydroxyl group, -OH, contains oxygen bonded to hydrogen. In alcohols it attaches to saturated carbon, which forms single bonds. In phenols it attaches directly to an aromatic ring, such as benzene, with delocalised ring electrons.

In general formulae, R and R′ represent alkyl groups, derived from alkanes by removing a hydrogen; Ar represents an aryl group, derived similarly from an aromatic hydrocarbon. An ether has the linkage R-O-R′, with oxygen joining two carbon-containing groups.

What do the classifications mean?

Monohydric, dihydric and trihydric compounds contain one, two and three hydroxyl groups respectively; polyhydric compounds contain several. Saturated, open-chain monohydric alcohols have formula CₙH₂ₙ₊₁OH, where n is the number of carbon atoms. Corresponding saturated open-chain ethers have molecular formula CₙH₂ₙ₊₂O.

Benzylic alcohols have the hydroxyl-bearing carbon next to an aromatic ring; benzyl alcohol, C₆H₅CH₂OH, is therefore an alcohol. Allylic alcohols have that carbon next to a carbon-carbon double bond. The number of hydroxyl groups and the class of their carbon atoms are separate descriptions.

How are systematic names constructed?

IUPAC, the International Union of Pure and Applied Chemistry, provides systematic naming rules. Choose the longest alcohol chain containing the hydroxyl-bearing carbon, number from the end nearer -OH, and replace the alkane ending “e” with “ol”: propan-1-ol. Multiple groups give ethane-1,2-diol or propane-1,2,3-triol.

Phenol is an accepted name. Ortho, meta and para, abbreviated o-, m- and p-, denote 1,2-, 1,3- and 1,4-substitution respectively. The cresols are methylphenols. Benzene-1,2-diol is catechol, benzene-1,3-diol is resorcinol, and benzene-1,4-diol is hydroquinone; each is a dihydric phenol. Name ethers as alkoxy derivatives of the larger parent group: CH₃OCH₂CH₂CH₃ is 1-methoxypropane; C₆H₅OCH₃ is methoxybenzene, commonly anisole.

How do structure and hydrogen bonding affect physical properties?

Hybridisation describes the mixing of atomic orbitals to form bonding orbitals. The labels sp³ and sp² denote mixing one s orbital with three or two p orbitals respectively. Alcohol hydroxyl groups bond to sp³ carbon; phenolic hydroxyl groups bond to sp² ring carbon.

A lone pair is an electron pair not shared in a bond. Oxygen has two lone pairs in these neutral molecules. Alcohol bond angles at oxygen are slightly smaller than the tetrahedral angle because lone pairs repel bonding pairs. Ether bond angles are slightly larger because the two alkyl groups repel one another.

The structures show oxygen lone pairs, bond angles and lengths. The carbon-oxygen distances are 142 pm in methanol, 136 pm in phenol and 141 pm in methoxymethane; pm means picometre.

See Fig. 7.1 in your NCERT textbook

Why do boiling point and solubility differ?

Hydrogen bonding is attraction between hydrogen bonded to a strongly electronegative atom and a suitable electron-rich atom. Intermolecular bonding occurs between molecules. Alcohols and phenols associate through such bonds, raising their boiling points above those of hydrocarbons and ethers of comparable molecular masses.

Alcohol boiling points increase with carbon-chain length. Increased branching lowers boiling point because it reduces surface area and weakens van der Waals attractions, the intermolecular attractions associated with molecular electron distributions. Solubility in water decreases as the water-repelling, or hydrophobic, alkyl or aryl portion becomes larger.

Several lower molecular mass alcohols mix with water in all proportions. Ethers can accept hydrogen bonds from water despite lacking their own O-H bonds. Their weak polarity does not appreciably affect boiling points, which are comparable to those of alkanes of comparable molecular masses.

K is kelvin, the absolute-temperature unit.

Ethoxyethane and butan-1-ol dissolve to almost the same extent: 7.5 and 9 g per 100 mL water respectively; g means gram and mL means millilitre. Pentane is essentially immiscible, meaning it scarcely mixes with water.

Draw R-O-R with two lone pairs on oxygen beside H-O-H. Add a dotted connection from ether oxygen to a hydrogen of water, showing the intermolecular attraction rather than another covalent bond.

How can alcohols be prepared from other organic compounds?

Hydration adds water across an alkene double bond. A catalyst speeds a reaction without being consumed overall. Direct acid-catalysed hydration follows Markovnikov’s rule: hydrogen adds to the double-bond carbon already carrying more hydrogens, while hydroxyl attaches to the other carbon. Thus propene gives propan-2-ol with water and dilute sulphuric acid.

What happens during hydration?

1. The alkene accepts a proton, the hydrogen ion H⁺, from the acidic medium to form a carbocation, an intermediate with positively charged carbon.
2. Water attacks that carbon through an oxygen lone pair. It acts as a nucleophile, an electron-pair donor.
3. The resulting oxygen-containing ion loses a proton, forming the alcohol and regenerating the acid catalyst.

Indirect hydration first adds concentrated sulphuric acid to an alkene, forming an alkyl hydrogensulphate. Hydrolysis, reaction with water that breaks a bond, then forms the alcohol. For ethene: CH₂=CH₂ + H₂SO₄ → C₂H₅HSO₄; C₂H₅HSO₄ + H₂O → C₂H₅OH + H₂SO₄.

Hydroboration-oxidation uses borane, supplied by diborane, followed by hydrogen peroxide in aqueous sodium hydroxide. Boron attaches to the double-bond carbon carrying more hydrogens and is subsequently replaced by hydroxyl. Propene therefore gives propan-1-ol, the orientation opposite to acid-catalysed hydration, in excellent yield.

What other starting materials are useful?

A carbonyl group is C=O. Reduction adds hydrogen or removes oxygen; oxidation reverses these changes in the reactions considered here.

Reduction here adds hydrogen to aldehydes or ketones. A Grignard reagent, RMgX, contains carbon bonded to magnesium; X represents a halogen. Its addition to a carbonyl compound in dry ether, followed by hydrolysis, forms an alcohol with a new carbon-carbon bond.

Methanal gives primary alcohols with Grignard reagents, other aldehydes give secondary alcohols, and ketones give tertiary alcohols. For example, propanone plus methylmagnesium bromide, followed by hydrolysis, gives 2-methylpropan-2-ol. Keep the addition and water-treatment stages separate because water destroys the Grignard reagent.

How do alcohols show acidity and form esters?

A Brønsted acid donates a proton; a Brønsted base accepts one. The polar O-H bond allows alcohols to donate its hydrogen to suitable stronger bases. Oxygen’s lone pairs also allow alcohols to accept a proton, so their role depends on the reacting substance.

Reaction with sodium demonstrates acidity: 2ROH + 2Na → 2RONa + H₂. RONa is a sodium alkoxide, the salt corresponding to removal of the hydroxyl hydrogen. The hydrogen gas comes from O-H cleavage, without removing oxygen from the organic group.

Electron-releasing alkyl groups tend to reduce O-H bond polarity and decrease acidic strength. Alkoxides accept protons from water: RO⁻ + H₂O → ROH + OH⁻. Here RO⁻ is an alkoxide ion and OH⁻ is hydroxide. This explains why aqueous sodium hydroxide is not used to isolate sodium ethoxide from ethanol.

How does acid-catalysed esterification proceed?

Esterification forms an ester, a compound containing the -COO- linkage between organic groups. Ethanol and ethanoic acid react reversibly with a small amount of concentrated sulphuric acid: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O. The double arrow means that forward and reverse reactions can occur.

1. Protonation of the acid’s carbonyl oxygen makes the carbonyl carbon more susceptible to nucleophilic attack.
2. The alcohol oxygen attacks that carbon, producing a tetrahedral intermediate, in which carbon has four single bonds.
3. Proton transfer changes a hydroxyl group into a water group that can leave.
4. Water is eliminated, restoring the carbonyl group in a protonated ester.
5. Loss of a proton gives the ester and regenerates the acid catalyst.

Removing water as it forms favours ester formation. Alcohols also react with acid chlorides, containing -COCl, and acid anhydrides, containing -CO-O-CO-. These reactions introduce an acyl group, a carbonyl group attached to an organic group.

Ethanol with acetyl chloride gives ethyl ethanoate and hydrogen chloride: C₂H₅OH + CH₃COCl → CH₃COOC₂H₅ + HCl. Pyridine, an organic base, neutralises the hydrogen chloride. With acetic anhydride: C₂H₅OH + (CH₃CO)₂O → CH₃COOC₂H₅ + CH₃COOH, using an acid catalyst.

How are alcohols converted into halides and distinguished by Lucas’ test?

Substitution replaces one atom or group with another. Alcohols react with hydrogen halides by replacing -OH with halogen: ROH + HX → RX + H₂O. HX denotes a hydrogen halide, such as hydrogen chloride, HCl, or hydrogen bromide, HBr.

For primary and secondary alcohols, concentrated hydrochloric acid generally requires zinc chloride as catalyst. Tertiary alcohols react readily with concentrated hydrochloric acid at room temperature. Their relative reactivity is tertiary greater than secondary greater than primary. Phenol does not undergo this replacement readily because its ring carbon-oxygen bond has partial double-bond character.

Which chlorinating reagents are used?

Thionyl chloride is preferred because both accompanying products are gases that escape, making separation of the alkyl chloride easier.

What observations identify the alcohol class?

Lucas reagent is concentrated hydrochloric acid with zinc chloride. Alcohols dissolve in it, whereas their alkyl chlorides are immiscible and produce turbidity, a cloudy appearance. For ordinary simple alcohols at room temperature, the speed of cloudiness distinguishes the classes.

How do oxidation and dehydration change alcohols?

Oxidation of primary and secondary alcohols forms a carbon-oxygen double bond by removing hydrogen from oxygen and the hydroxyl-bearing carbon. A primary alcohol gives an aldehyde and then a carboxylic acid, depending on the oxidising reagent and conditions.

Write ethanol oxidation as CH₃CH₂OH + [O] → CH₃CHO + H₂O, then CH₃CHO + [O] → CH₃COOH. The symbol [O] represents an oxygen equivalent supplied by an oxidising agent. Ethanal is the aldehyde and ethanoic acid is the final acid.

Pyridinium chlorochromate, abbreviated PCC, gives aldehydes from primary alcohols in good yield. Chromium trioxide, CrO₃, in an anhydrous medium also allows aldehyde isolation; anhydrous means water-free. Strong oxidants such as acidified potassium permanganate, KMnO₄, produce carboxylic acids.

Secondary alcohols yield ketones. Tertiary alcohols resist ordinary oxidation, but strong oxidants at elevated temperatures can break carbon-carbon bonds and form mixtures of smaller carboxylic acids. They must not be described as incapable of reacting under every condition.

How does dehydration form an alkene?

Dehydration removes water. Heating ethanol with concentrated sulphuric acid at 443 K produces ethene: CH₃CH₂OH → CH₂=CH₂ + H₂O. Secondary and tertiary alcohols dehydrate under milder conditions; ease of dehydration increases from primary to secondary to tertiary.

1. Ethanol accepts a proton on oxygen, forming protonated ethanol, CH₃CH₂OH₂⁺.
2. Water leaves to form a carbocation. In this stepwise treatment, this is the slowest, rate-determining step.
3. The intermediate loses a proton and forms the carbon-carbon double bond of ethene.
4. The acid is regenerated. Removing ethene as it forms drives the equilibrium towards products.

Note: Conditions determine the product. Ethanol with sulphuric acid at 443 K gives ethene; at 413 K, ethoxyethane is the main product. Record the temperature alongside the reagent.

Dehydrogenation removes hydrogen. Passing primary or secondary alcohol vapour over copper at 573 K gives an aldehyde or ketone respectively, with hydrogen. Under these conditions tertiary alcohols undergo dehydration instead. For ethanol: CH₃CH₂OH → CH₃CHO + H₂.

How are methanol and ethanol manufactured and used?

Methanol, CH₃OH, is also called wood spirit because it was formerly obtained by destructive distillation of wood. Today, most methanol is produced by catalytic hydrogenation of carbon monoxide. The Bosch process combines carbon monoxide and hydrogen under high pressure over a mixed oxide catalyst.

The equation is CO + 2H₂ → CH₃OH. Zinc oxide and chromium(III) oxide, ZnO and Cr₂O₃, form the catalyst; conditions are 573 to 673 K and 200 to 300 atm. The unit atm denotes atmosphere, a pressure unit.

Methanol is a colourless liquid boiling at 337 K. It is highly poisonous: small quantities can cause blindness and larger quantities can cause death. It serves as a solvent for paints and varnishes and is chiefly used for making formaldehyde, also called methanal.

How does fermentation produce ethanol?

Fermentation converts sugars into ethanol using enzymes, biological catalysts. The enzyme invertase converts sucrose into glucose and fructose. Zymase, an enzyme system in yeast, then converts these sugars into ethanol and carbon dioxide under anaerobic conditions, meaning absence of air.

C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ + C₆H₁₂O₆, with invertase. The two products are glucose and fructose. Each undergoes the reaction C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ with zymase.

Zymase action is inhibited once the alcohol percentage exceeds 14 percent. If air enters, ethanol can be oxidised to ethanoic acid. Ethanol is a colourless liquid boiling at 351 K and is used as a solvent in paints and for preparing other carbon compounds.

Denaturation makes commercial alcohol unfit for drinking. Copper sulphate supplies colour and pyridine supplies an unpleasant smell. This term does not describe a change from ethanol to methanol. Large quantities of ethanol are also produced by hydration of ethene.

How can one alcohol be converted into another?

Plan conversions through a suitable intermediate. Ethanol can be dehydrated to ethene and hydrated back. To increase the chain by one carbon, convert ethanol to bromoethane, form ethylmagnesium bromide with magnesium in dry ether, add methanal, and hydrolyse the adduct to propan-1-ol.

How is phenol prepared in the laboratory and industry?

Phenol, C₆H₅OH, also called carbolic acid, can be prepared by replacing suitable groups on a benzene ring with hydroxyl. Some preparations initially give sodium phenoxide, C₆H₅ONa, which must be acidified to obtain phenol rather than its salt.

Which routes use benzene derivatives?

1. Dow’s process: heat chlorobenzene with aqueous sodium hydroxide under severe conditions. At 623 K and high pressure, sodium phenoxide forms: C₆H₅Cl + 2NaOH → C₆H₅ONa + NaCl + H₂O. The stated preparation pressure is 320 atm.
2. Acidification: treat the phenoxide with hydrochloric acid: C₆H₅ONa + HCl → C₆H₅OH + NaCl. This protonates the phenoxide ion.
3. Sulphonic acid route: sulphonate benzene with oleum, sulphur trioxide dissolved in sulphuric acid. Convert the product to sodium benzenesulphonate and fuse it with sodium hydroxide: C₆H₅SO₃Na + 2NaOH → C₆H₅ONa + Na₂SO₃ + H₂O. Acidify afterwards.
4. Diazonium salt route: warm benzenediazonium chloride with water: C₆H₅N₂Cl + H₂O → C₆H₅OH + N₂ + HCl. The diazonium group contains two linked nitrogen atoms and is replaced by hydroxyl.

Benzenediazonium chloride is prepared from aniline, C₆H₅NH₂, using sodium nitrite and hydrochloric acid at 273 to 278 K. These reagents generate nitrous acid in the reaction mixture. Warming then brings about hydrolysis and release of nitrogen.

Why is cumene an important starting material?

Cumene is isopropylbenzene, C₆H₅CH(CH₃)₂. Air oxidises it to cumene hydroperoxide, C₆H₅C(CH₃)₂OOH, whose hydroperoxide group contains an O-O-H linkage. Treatment with dilute acid produces phenol and acetone, also called propanone.

The conversion can be recorded as C₆H₅CH(CH₃)₂ + O₂ → C₆H₅C(CH₃)₂OOH, followed by C₆H₅C(CH₃)₂OOH → C₆H₅OH + CH₃COCH₃. Most worldwide phenol production uses cumene. Acetone is a useful by-product obtained in large quantities alongside phenol.

Why is phenol more acidic than an alcohol?

Phenol is a crystalline solid with limited solubility in water. Its hydroxyl group forms hydrogen bonds, but the phenyl portion opposes mixing with water. Phenols slowly oxidise in air to dark-coloured mixtures containing quinones, compounds with carbonyl groups in a conjugated ring system.

Phenol reacts with sodium: 2C₆H₅OH + 2Na → 2C₆H₅ONa + H₂. It also reacts with aqueous sodium hydroxide: C₆H₅OH + NaOH → C₆H₅ONa + H₂O. The second reaction distinguishes its acidity from that of ordinary alcohols.

How does resonance stabilise phenoxide?

Resonance represents electron delocalisation using multiple contributing structures with the same atomic arrangement. Delocalisation spreads charge over more than one atom. In an alkoxide, negative charge remains localised on oxygen; in phenoxide, it is shared through oxygen and the aromatic ring.

The sp² carbon bonded to oxygen is more electronegative than the corresponding sp³ carbon in an alcohol. Electron withdrawal increases O-H bond polarity. Stabilisation of the phenoxide ion further favours proton loss, making phenol more acidic than ethanol.

Electron-withdrawing groups, such as nitro, -NO₂, increase phenol acidity. The effect is more pronounced at ortho and para positions because the resulting negative charge can be effectively delocalised. Electron-releasing alkyl groups, in general, reduce acidity; methylphenols are less acidic than phenol.

An increasing-acidity sequence is propan-1-ol, 4-methylphenol, phenol, 3-nitrophenol, 3,5-dinitrophenol and 2,4,6-trinitrophenol. This combines the effect of the aromatic ring with the number and nature of substituents.

How do pKa values compare the acidic strengths?

A larger $pK_a$ value indicates a weaker acid. The values show that nitrophenols are stronger acids than phenol, whereas cresols and ethanol are weaker acids.

How does phenol undergo substitution and give characteristic tests?

An electrophile accepts an electron pair. In electrophilic aromatic substitution, an electrophile replaces a ring hydrogen. The hydroxyl group activates phenol’s ring by supplying electron density through resonance, especially at the ortho and para positions, so incoming groups favour these positions.

How do bromination and nitration depend on conditions?

With bromine in a low-polarity solvent such as carbon disulphide, CS₂, at low temperature, phenol forms ortho- and para-monobromophenol, with the para product major. A Lewis acid, an electron-pair acceptor, is unnecessary as catalyst because hydroxyl strongly activates the ring.

With bromine water, multiple substitution gives a white precipitate, an insoluble solid: C₆H₅OH + 3Br₂ → C₆H₂Br₃OH + 3HBr. The solid is 2,4,6-tribromophenol. Different conditions therefore give different products from the same phenol starting material.

Dilute nitric acid at 298 K gives ortho- and para-nitrophenol. Concentrated nitric acid gives 2,4,6-trinitrophenol, or picric acid, but the direct reaction has poor yield. A route through phenol-2,4-disulphonic acid, followed by concentrated nitric acid, is used for its preparation.

In ortho-nitrophenol, draw a dotted link between hydroxyl hydrogen and an oxygen of the adjacent nitro group within one molecule. In para-nitrophenol, show hydrogen bonding between separate molecules instead.

The ortho isomer has intramolecular hydrogen bonding, within one molecule, and is steam volatile. The para isomer associates through intermolecular bonding and is less volatile. Steam distillation, separation using water vapour, can therefore separate the two isomers.

What additional reactions and tests are useful?

Sulphonation introduces -SO₃H. Concentrated sulphuric acid at about 293 K favours ortho-phenolsulphonic acid; at about 373 K it favours para-phenolsulphonic acid. For either positional product: C₆H₅OH + H₂SO₄ → HOC₆H₄SO₃H + H₂O.

The ferric chloride test uses neutral iron(III) chloride solution, FeCl₃; phenol gives violet colour through complex formation. In the azo dye test, phenol couples with benzenediazonium chloride in alkaline solution to form coloured para-hydroxyazobenzene. Its azo linkage, -N=N-, joins the two aromatic rings.

What do Kolbe’s reaction and Reimer-Tiemann reaction produce?

Kolbe’s reaction introduces a carboxyl group, -COOH, into the phenolic ring. First form sodium phenoxide using sodium hydroxide. Phenoxide is even more reactive than phenol towards electrophilic substitution and can react with carbon dioxide, a weak electrophile.

Carbon dioxide treatment followed by acidification produces salicylic acid, or 2-hydroxybenzoic acid, as the main product. The sequence is C₆H₅ONa + CO₂ → o-HOC₆H₄COONa, then o-HOC₆H₄COONa + HCl → o-HOC₆H₄COOH + NaCl.

How is an aldehyde group introduced?

In the Reimer-Tiemann reaction, phenol reacts with chloroform, CHCl₃, and aqueous sodium hydroxide. An intermediate bearing a -CHCl₂ group is hydrolysed, and acidification gives salicylaldehyde, or 2-hydroxybenzaldehyde. Its -CHO group is ortho to hydroxyl.

The overall equation, after acid work-up, is C₆H₅OH + CHCl₃ + 3NaOH → o-HOC₆H₄CHO + 3NaCl + 2H₂O. Distinguish this aldehyde product from the carboxylic acid obtained in Kolbe’s reaction; the reagents identify which carbon-containing group enters the ring.

How does phenol react at its hydroxyl group?

Heating phenol with zinc dust removes its oxygen: C₆H₅OH + Zn → C₆H₆ + ZnO. The organic product is benzene. Acetylation, introduction of CH₃CO-, instead retains the phenolic oxygen and converts the hydroxyl group into an ester.

With acetyl chloride in pyridine: C₆H₅OH + CH₃COCl → CH₃COOC₆H₅ + HCl. With acetic anhydride and an acid catalyst: C₆H₅OH + (CH₃CO)₂O → CH₃COOC₆H₅ + CH₃COOH. Both form phenyl ethanoate, also called phenyl acetate.

With phosphorus pentachloride, the formal replacement equation is C₆H₅OH + PCl₅ → C₆H₅Cl + POCl₃ + HCl. Chlorobenzene yield is poor and phosphate ester formation competes, so this is not a satisfactory preparation of chlorobenzene.

Phenol is used in making phenolic resins, dyes and medicinal compounds. Acetylating salicylic acid gives acetylsalicylic acid, or aspirin.

How are ethers prepared and why does reactant choice matter?

The ethereal linkage is C-O-C. Ethers can be prepared from alcohols by dehydration or from an alkoxide and an alkyl halide by Williamson synthesis. Choosing both the substrate and the conditions determines whether substitution or competing elimination predominates.

When does alcohol dehydration give an ether?

At 413 K, ethanol with sulphuric acid gives ethoxyethane as the main product: 2C₂H₅OH → C₂H₅OC₂H₅ + H₂O. One ethanol molecule attacks a protonated ethanol molecule; water leaves and loss of a proton gives the ether.

This is a bimolecular nucleophilic substitution, abbreviated Sₙ2, in which nucleophilic attack and departure of the leaving group occur together. The method is suitable for ethers with unhindered primary alkyl groups. Steric hindrance means crowding that obstructs access to the reacting carbon.

Secondary and tertiary alcohols tend to eliminate water to form alkenes under these acidic conditions. Mixing two different alcohols is unsuitable for selectively preparing an unsymmetrical ether because different alcohol molecules can combine, giving a mixture.

How does Williamson synthesis improve control?

Williamson synthesis uses an alkoxide nucleophile and an alkyl halide: RONa + R′X → ROR′ + NaX. Sodium ethoxide and methyl bromide give methoxyethane: C₂H₅ONa + CH₃Br → C₂H₅OCH₃ + NaBr.

Better results are obtained with a primary alkyl halide. Secondary and tertiary halides favour competing elimination because alkoxides are also strong bases. A tertiary halide with an alkoxide produces an alkene instead of the desired ether.

For tert-butyl ethyl ether, use sodium tert-butoxide with a primary ethyl halide. Reversing the pair to sodium ethoxide and a tertiary butyl halide favours 2-methylpropene. The bulky group can belong to the alkoxide, while the halide supplies an accessible carbon.

Phenoxide also acts as the nucleophile: C₆H₅ONa + CH₃Br → C₆H₅OCH₃ + NaBr. This prepares anisole. It uses substitution at methyl carbon; it does not require displacement at an aromatic carbon.

How do aliphatic ethers react with acids, chlorine and oxygen?

Ethers are relatively unreactive, but their carbon-oxygen bond can be cleaved by concentrated hydrogen iodide, HI, or hydrogen bromide at high temperature. Cleavage means breaking a bond. First oxygen accepts a proton, making departure of an alcohol group easier.

How does hydrogen iodide split an ether?

1. Ether oxygen accepts H⁺, forming an oxonium ion, an ion with positively charged oxygen.
2. Iodide, I⁻, attacks the less hindered alkyl carbon when ordinary primary groups are present. An alcohol and an alkyl iodide form.
3. For methoxyethane: CH₃OC₂H₅ + HI → CH₃I + C₂H₅OH.
4. With excess HI and high temperature, the alcohol also reacts: C₂H₅OH + HI → C₂H₅I + H₂O.

If one group is tertiary, the tertiary iodide forms through a relatively stable carbocation. This follows unimolecular nucleophilic substitution, Sₙ1, where bond breaking precedes nucleophilic attack. Thus methyl tert-butyl ether gives tert-butyl iodide and methanol at the initial cleavage stage.

The general complete cleavage of a dialkyl ether is ROR′ + 2HI → RI + R′I + H₂O. The reactivity order of hydrogen halides is HI greater than HBr greater than HCl. State whether the question concerns initial cleavage or prolonged treatment with excess reagent.

What other reactions are required?

Chlorination of ethoxyethane in the dark substitutes hydrogen at carbons adjacent to oxygen. A representative first-stage equation is CH₃CH₂OCH₂CH₃ + Cl₂ → CH₃CHClOCH₂CH₃ + HCl; further substitution can occur with more chlorine.

On standing in contact with air, ethers can form peroxides, compounds containing an O-O bond. For ethoxyethane, oxidation can form CH₃CH(OOH)OCH₂CH₃, a hydroperoxide. Such products are hazardous because they can decompose violently, especially when concentrated.

Phosphorus pentachloride cleaves the ethereal linkage: C₂H₅OC₂H₅ + PCl₅ → 2C₂H₅Cl + POCl₃. Ethoxyethane is a volatile liquid used as an organic solvent. It formerly had wide use as an inhalation anaesthetic but has been replaced for that purpose by other compounds.

How does anisole combine ether chemistry with aromatic substitution?

Anisole, C₆H₅OCH₃, is a liquid aromatic ether with low water solubility. Its methoxy group, -OCH₃, influences the benzene ring through oxygen’s lone pair. Like hydroxyl in phenol, methoxy activates the ring and directs electrophilic substitution towards ortho and para positions.

Which reactions take place on the ring?

Friedel-Crafts reactions introduce alkyl or acyl groups into an aromatic ring. Aluminium chloride, AlCl₃, acts as a Lewis acid. Anisole bromination does not require iron(III) bromide because the methoxy group already activates the ring strongly.

Which bond breaks with hydrogen iodide?

Heating anisole with hydrogen iodide cleaves the methyl-oxygen bond: C₆H₅OCH₃ + HI → C₆H₅OH + CH₃I. The products are phenol and methyl iodide. The aromatic carbon-oxygen bond has partial double-bond character and is stronger than the methyl-oxygen bond.

Phenol does not continue to an aryl iodide under this treatment. Its sp² ring carbon cannot undergo the substitution required for that conversion.

Aromatic ethers have uses in perfumery and as solvents or intermediates in organic synthesis.

Glossary

• Alcohol — An organic compound with a hydroxyl group bonded to a saturated carbon atom.
• Phenol — A compound whose hydroxyl group is attached directly to an aromatic ring carbon.
• Ether — A compound with oxygen linking two alkyl or aryl groups through single bonds.
• Alkoxide — The negatively charged species formed when an alcohol loses its hydroxyl proton.
• Phenoxide — The resonance-stabilised ion obtained when phenol loses the proton from its hydroxyl group.
• Nucleophile — An electron-pair donor that attacks an electron-deficient centre in a chemical reaction.
• Electrophile — An electron-pair acceptor that reacts at a region of relatively high electron density.
• Carbocation — An organic ion with a positively charged carbon centre that can accept an electron pair.
• Esterification — Formation of an ester, including the acid-catalysed reaction between an alcohol and a carboxylic acid.
• Dehydration — Removal of water from a compound, producing an alkene or ether from alcohols under suitable conditions.
• Resonance — Representation of delocalised electrons by contributing structures that preserve the arrangement of atomic nuclei.
• Fermentation — Enzyme-mediated conversion of sugars into ethanol and carbon dioxide under conditions without air.

Common errors and misconceptions

• Misconception: Every compound with benzene and hydroxyl is a phenol. Correct: Hydroxyl must attach directly to ring carbon; benzyl alcohol has an intervening saturated carbon.
• Misconception: A tertiary alcohol has three hydroxyl groups. Correct: Its hydroxyl-bearing carbon has three carbon neighbours; hydroxyl-group count is a separate classification.
• Misconception: Ether cannot hydrogen-bond with water. Correct: Its oxygen accepts hydrogen bonds from water, although pure ether lacks alcohol-like intermolecular O-H bonding.
• Misconception: Sulphuric acid gives the same ethanol product at every temperature. Correct: Ethoxyethane is the main product at 413 K, while ethene forms at 443 K.
• Misconception: Tertiary alcohols cannot be oxidised under any conditions. Correct: Severe oxidation can cleave carbon-carbon bonds and form smaller carboxylic acids.
• Misconception: Either reactant arrangement works equally well in Williamson synthesis. Correct: A primary alkyl halide favours substitution; a tertiary halide favours elimination.
• Misconception: Anisole with HI produces iodobenzene and methanol. Correct: Methyl-oxygen cleavage gives methyl iodide and phenol.
• Misconception: Kolbe’s and Reimer-Tiemann reactions give the same product. Correct: They introduce carboxyl and aldehyde groups respectively, giving salicylic acid and salicylaldehyde.

Exam-style questions with model answers

Q1. Classify CH₃CH₂OH and C₆H₅OH as an alcohol or phenol, explaining the structural basis for each. [2 marks]

1. CH₃CH₂OH is an alcohol because its hydroxyl group bonds to a saturated carbon in an alkyl group.
2. C₆H₅OH is a phenol because its hydroxyl group bonds directly to a carbon of the benzene ring.

Q2. Propene, CH₃CH=CH₂, is treated separately with water and dilute sulphuric acid, and with borane followed by hydrogen peroxide in aqueous sodium hydroxide. Identify both alcohol products and explain the difference in orientation. [3 marks]

1. Water with dilute sulphuric acid produces propan-2-ol, CH₃CH(OH)CH₃, through acid-catalysed hydration. The hydroxyl group becomes attached to the middle carbon.
2. Borane followed by hydrogen peroxide in aqueous sodium hydroxide produces propan-1-ol, CH₃CH₂CH₂OH, through hydroboration-oxidation, with hydroxyl on the terminal carbon.
3. The first route follows Markovnikov’s rule. In hydroboration, boron attaches to the carbon carrying more hydrogens; replacement of boron by hydroxyl gives the opposite orientation.

Q3. Explain the acid-catalysed esterification of ethanol with ethanoic acid. Give the overall equation and four successive stages of the mechanism. [5 marks]

1. The reversible overall reaction is CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O, catalysed by a small amount of concentrated sulphuric acid. The ester is ethyl ethanoate.
2. The carbonyl oxygen of ethanoic acid accepts a proton. This activates the carbonyl carbon towards attack by the alcohol oxygen.
3. Ethanol attacks through an oxygen lone pair. A tetrahedral intermediate forms, with the original carbonyl carbon now attached through four single bonds.
4. Proton transfer converts a hydroxyl group into a better leaving group. Loss of water restores the carbonyl group, producing a protonated ester.
5. The protonated ester loses a proton to give ethyl ethanoate and regenerate the catalyst. Removing water favours the forward reaction and ester formation.

Q4. Three ordinary simple alcohols are tested with concentrated HCl and ZnCl₂ at room temperature. Sample A clouds immediately, B clouds after a few minutes, and C remains clear. Identify the classes and explain the cloudiness. [4 marks]

1. Sample A is a tertiary alcohol. It forms an alkyl chloride rapidly, so turbidity appears immediately after contact with Lucas reagent.
2. Sample B is a secondary alcohol. Alkyl chloride formation is slower, producing the observed cloudiness after a few minutes.
3. Sample C is a primary alcohol. It does not produce turbidity at room temperature under the stated test conditions.
4. The alcohols dissolve in the reagent, whereas the resulting alkyl chlorides are immiscible. Their separation produces the cloudy appearance used to compare reaction speeds.

Q5. Explain why phenol is more acidic than ethanol, write its reaction with aqueous NaOH, and describe how a para-nitro group changes its acidity. [3 marks]

1. Phenol loses a proton to form phenoxide, whose negative charge is delocalised through oxygen and the aromatic ring. Ethoxide lacks this resonance stabilisation, so ethanol is less acidic.
2. Phenol reacts with aqueous sodium hydroxide according to C₆H₅OH + NaOH → C₆H₅ONa + H₂O, producing sodium phenoxide and water.
3. A para-nitro group withdraws electrons and further stabilises the phenoxide ion through effective delocalisation. It therefore increases acidic strength relative to unsubstituted phenol.

Q6. Phenol is treated separately with bromine water and with bromine in CS₂ at low temperature. State the products and the visible observation expected with bromine water. [2 marks]

1. Bromine water gives a white precipitate of 2,4,6-tribromophenol: C₆H₅OH + 3Br₂ → C₆H₂Br₃OH + 3HBr.
2. Bromine in carbon disulphide at low temperature gives ortho- and para-monobromophenol, with the para isomer as the major product.

Q7. To prepare tert-butyl ethyl ether, compare sodium tert-butoxide plus bromoethane with sodium ethoxide plus tert-butyl bromide. Identify the suitable pair, give its equation and explain the alternative pair’s major product. [3 marks]

1. Sodium tert-butoxide and bromoethane are the suitable pair because the primary carbon of bromoethane is accessible to nucleophilic attack by the alkoxide.
2. The Williamson synthesis equation is (CH₃)₃CONa + C₂H₅Br → (CH₃)₃COC₂H₅ + NaBr. The product is tert-butyl ethyl ether.
3. Sodium ethoxide with tert-butyl bromide gives 2-methylpropene as the major product. The alkoxide acts as a strong base, favouring elimination at the tertiary halide rather than ether-forming substitution.

Q8. Explain the initial cleavage of methoxyethane, CH₃OC₂H₅, by HI, its further reaction with excess HI at high temperature, and the different final products obtained from anisole, C₆H₅OCH₃. Give six separate points including equations and reasons. [6 marks]

1. HI first protonates methoxyethane at oxygen. This forms an oxonium ion and allows an alcohol molecule to act as the departing group.
2. Iodide attacks the less hindered methyl carbon by bimolecular nucleophilic substitution. Initial cleavage gives CH₃OC₂H₅ + HI → CH₃I + C₂H₅OH.
3. With excess HI at high temperature, ethanol undergoes further substitution: C₂H₅OH + HI → C₂H₅I + H₂O.
4. The complete methoxyethane reaction therefore gives methyl iodide, ethyl iodide and water. Overall, CH₃OC₂H₅ + 2HI → CH₃I + C₂H₅I + H₂O.
5. Anisole gives phenol and methyl iodide: C₆H₅OCH₃ + HI → C₆H₅OH + CH₃I. Cleavage occurs at the methyl-oxygen bond after oxygen protonation.
6. The aryl-oxygen bond has partial double-bond character. Phenol’s sp² carbon cannot undergo the substitution needed to form the aryl halide, so phenol does not become iodobenzene.

Key takeaways

• Classify a hydroxyl compound by its attachment site and hydroxyl-group count before predicting its name, acidity or chemical reactions.
• Alcohols form intermolecular hydrogen bonds, while ether oxygen can accept hydrogen bonds from water and thereby assist dissolution.
• Acid-catalysed hydration and hydroboration-oxidation place hydroxyl at different positions in an unsymmetrical alkene such as propene.
• Primary alcohol oxidation can stop at an aldehyde or continue to a carboxylic acid, depending on reagents and conditions.
• Phenoxide resonance stabilisation explains phenol’s stronger acidity, while electron-withdrawing substituents increase acidity further, especially at ortho and para positions.
• Phenol bromination and alcohol dehydration require careful attention to conditions because changing solvent or temperature changes the products.
• Williamson synthesis works best with a primary alkyl halide; tertiary halides favour elimination because alkoxides also act as bases.
• Anisole undergoes ring substitution at ortho and para positions, but HI cleaves its methyl-oxygen bond to give phenol.

Test yourself

What is the systematic name of HOCH₂CH₂OH?

It is ethane-1,2-diol, a dihydric alcohol with one hydroxyl group on each carbon atom.

Why does branching lower the boiling point of isomeric alcohols?

Branching reduces molecular surface area and weakens van der Waals attractions, lowering boiling point within the comparable alcohol series.

What alcohol class does a ketone give with a Grignard reagent followed by hydrolysis?

It gives a tertiary alcohol because the Grignard reagent adds another carbon group to the ketone’s carbonyl carbon.

How do Kolbe’s and Reimer-Tiemann products differ?

Kolbe’s reaction gives salicylic acid, containing -COOH; Reimer-Tiemann reaction gives salicylaldehyde, containing -CHO. Both place the new group ortho to hydroxyl.

Why is ortho-nitrophenol more steam volatile than para-nitrophenol?

Ortho-nitrophenol forms intramolecular hydrogen bonds. Para-nitrophenol forms intermolecular bonds, associating its molecules and reducing volatility.

What are the two products from acid treatment of cumene hydroperoxide?

Phenol and acetone form; acetone is obtained as a useful by-product in large quantities.

What observations would distinguish phenol from ethanol using neutral FeCl₃?

Phenol gives a violet colour with neutral iron(III) chloride solution; ethanol does not give this characteristic phenolic colour.

Why does methoxyethane need excess hot HI for complete cleavage into two alkyl iodides?

Initial cleavage gives methyl iodide and ethanol. The ethanol requires a further reaction with HI to form ethyl iodide and water.