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Alcohols are arguably the most versatile of the A-Level functional groups. The hydroxyl (–OH) substituent sits at the centre of a web of organic transformations: it can be oxidised to aldehydes, ketones or carboxylic acids; it can be dehydrated to alkenes; it can be condensed with a carboxylic acid to form an ester; and it can be combusted as a fuel. This lesson takes a structured tour of that chemistry. We classify alcohols as primary, secondary and tertiary, examine the two industrial routes to ethanol (hydration of ethene and fermentation), develop the contrast between distillation and reflux as the key experimental decision in primary-alcohol oxidation, work through the elimination chemistry that delivers alkenes (with Zaitsev regiochemistry), and finish with esterification and its named products. Practical-skills cues are signposted throughout.
Spec mapping (AQA 7405): This lesson anchors §3.3.5 (alcohols — classification, production, oxidation and elimination). It cross-references L4 of this course (alkenes — hydration of an alkene gives an alcohol), L6 (halogenoalkanes — nucleophilic substitution with aqueous OH⁻ gives an alcohol), and L8 (organic mechanisms master class, where the E1 dehydration mechanism is developed in full). It feeds forward to §3.3.9 (carboxylic acids and derivatives — the oxidation product of a primary alcohol under reflux) and to §3.3.6 (organic analysis — Fehling's solution and Tollens' reagent distinguish aldehydes from ketones, which is how you confirm primary vs secondary in the laboratory). Refer to the official AQA specification document for the exact wording.
Assessment objectives: Classification of an alcohol as primary, secondary or tertiary and recall of the reagents and conditions for each transformation are AO1. Writing balanced equations for oxidation, dehydration and esterification, predicting the products of named reactions, and choosing the correct experimental technique are AO2. Justifying the choice of distillation vs reflux, rationalising why tertiary alcohols resist oxidation by acidified dichromate, and predicting the major product of an elimination with Zaitsev reasoning are AO3.
Alcohols are classified by counting the number of carbon atoms directly bonded to the carbon that bears the –OH group. That carbon — not the molecule as a whole — fixes the class.
| Classification | Carbons bonded to the C–OH | Example | Oxidation behaviour |
|---|---|---|---|
| Primary (1°) | 0 or 1 other carbon | CH₃CH₂OH (ethanol) | Aldehyde → carboxylic acid |
| Secondary (2°) | 2 other carbons | CH₃CH(OH)CH₃ (propan-2-ol) | Ketone only |
| Tertiary (3°) | 3 other carbons | (CH₃)₃COH (2-methylpropan-2-ol) | No reaction with K₂Cr₂O₇ |
Methanol (CH₃OH) is conventionally classified as primary because the OH-bearing carbon is bonded to zero other carbons (only hydrogens). A useful diagnostic question: "How many C–H bonds are there on the carbon that carries the –OH group?" Primary has two (or three for methanol); secondary has one; tertiary has zero. The number of C–H bonds on this carbon is exactly the resource that the oxidation reaction consumes, which is why tertiary alcohols cannot be oxidised by acidified dichromate.
Key Definition: Classification is fixed by the connectivity of the carbon bearing –OH, not by the size or branching pattern of the wider molecule. A long, highly branched chain can still terminate in a primary –CH₂OH group.
The hydroxyl group dominates the physical behaviour of short-chain alcohols:
Ethanol is the most economically important alcohol. Two production routes are studied at A-Level, and you should be able to compare them on conditions, atom economy, sustainability and product purity.
CH₂=CH₂ + H₂O → CH₃CH₂OH
Conditions: steam, concentrated phosphoric acid (H₃PO₄) catalyst on a silica support, 300 °C, 60 atm pressure.
This is the dominant industrial route in petrochemical regions where ethene is abundant (cracked from naphtha or ethane). The reaction is fast, the product is pure (no separation from water-glucose mixtures needed) and the atom economy is 100% — every atom of ethene and water ends up in the product. The drawback is that ethene is itself a fossil-fuel-derived feedstock, so the route is non-renewable. Mechanistically, hydration of ethene proceeds via electrophilic addition (developed in L4): H⁺ from the catalyst adds first, generating a carbocation that is then attacked by water.
C₆H₁₂O₆ → 2 C₂H₅OH + 2 CO₂
Conditions: glucose (typically from cane sugar, sugar beet or starch hydrolysis), yeast (Saccharomyces cerevisiae) providing the enzymes (zymase complex), ~30 °C, anaerobic.
Yeast metabolism converts glucose to ethanol and carbon dioxide. The route uses a renewable feedstock and operates at mild conditions — but it is slow, produces a dilute aqueous solution (~12% ethanol before the yeast is killed by its own product), and so requires distillation to isolate pure ethanol. It is the basis of the bioethanol biofuel industry in countries with large sugar-crop economies (notably Brazil, where ethanol from sugar cane has been a fuel additive for decades).
| Feature | Hydration of ethene | Fermentation |
|---|---|---|
| Feedstock | Ethene (from crude oil) | Glucose (from crops) |
| Renewability | Non-renewable | Renewable |
| Atom economy | 100% | 51.1% (CO₂ is waste) |
| Rate | Fast (continuous process) | Slow (batch process, days) |
| Energy demand | High (300 °C, 60 atm) | Low (30 °C, ambient) |
| Product purity | Pure ethanol | Dilute aqueous; requires distillation |
| Capital cost | High (high-pressure plant) | Low (fermentation vats) |
Exam Tip: A common AO3 question asks you to recommend a route given a context. "Bioethanol for transport fuel in a country with abundant sugar cane" points to fermentation; "high-volume industrial solvent in a petrochemical complex" points to hydration of ethene. Always link the conditions and feedstock to the context.
The A-Level oxidising agent is acidified potassium dichromate(VI) — K₂Cr₂O₇ with dilute sulfuric acid (H₂SO₄). In simplified equations the oxidising agent is written as [O], which represents one oxygen-equivalent of oxidising power. Under the hood, the dichromate ion is reduced:
Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻ → 2 Cr³⁺ + 7 H₂O
The orange Cr₂O₇²⁻ is reduced to green Cr³⁺, and the orange → green colour change is the visual cue that the alcohol has been oxidised. If the colour does not change, no oxidation has occurred — which is itself a diagnostic test for a tertiary alcohol.
A primary alcohol can be oxidised in two distinct ways, and the experimental technique determines which product is isolated.
Partial oxidation to an aldehyde — distillation:
CH₃CH₂OH + [O] → CH₃CHO + H₂O
(ethanol → ethanal)
Heat the primary alcohol with acidified K₂Cr₂O₇ in apparatus configured for distillation, so that the aldehyde — which has a markedly lower boiling point than the parent alcohol, because it cannot hydrogen-bond as a donor — distils out of the flask as soon as it forms. Removing the aldehyde from the oxidising mixture prevents its further oxidation to the carboxylic acid. The dichromate is used in moderate (not excess) amount.
Full oxidation to a carboxylic acid — reflux:
CH₃CH₂OH + 2 [O] → CH₃COOH + H₂O
(ethanol → ethanoic acid)
Heat the primary alcohol with excess acidified K₂Cr₂O₇ under reflux. Reflux is the configuration in which a vertical condenser sits above the reaction flask: vapours rise into the condenser, are returned as liquid to the flask, and the mixture is held at its boiling point for an extended period. The aldehyde intermediate cannot escape, the dichromate is in excess, and a second oxidation step (adding another oxygen) converts the aldehyde to the carboxylic acid. After cooling and acid work-up, the carboxylic acid is isolated.
A secondary alcohol is oxidised to a ketone — and the ketone is the end of the road. Ketones resist further oxidation by acidified dichromate because the carbonyl carbon carries no C–H bond that could be broken (the C–C bonds either side would have to break instead, which dichromate cannot do).
CH₃CH(OH)CH₃ + [O] → CH₃COCH₃ + H₂O
(propan-2-ol → propanone)
Either distillation or reflux works; reflux is conventional because the product is stable to further oxidation. The dichromate colour change (orange → green) confirms the reaction has proceeded.
Tertiary alcohols are not oxidised by acidified dichromate. Oxidation of an alcohol requires the loss of the H atom on the OH-bearing carbon (along with the H of the –OH itself); a tertiary alcohol has no such H. The dichromate solution remains orange — no colour change. This is precisely the experimental test for a tertiary alcohol.
| Alcohol | Distillation product | Reflux product | Dichromate colour |
|---|---|---|---|
| Primary | Aldehyde | Carboxylic acid (excess [O]) | Orange → green |
| Secondary | Ketone | Ketone | Orange → green |
| Tertiary | No reaction | No reaction | Stays orange |
Distillation route: CH₃CH₂OH + [O] → CH₃CHO + H₂O. The ethanal (b.p. 21 °C) distils out as it forms, before the dichromate can attack it a second time.
Reflux route: CH₃CH₂OH + 2 [O] → CH₃COOH + H₂O. Two oxygen-equivalents are required because there are two oxidation steps (alcohol → aldehyde, then aldehyde → carboxylic acid). Use excess dichromate to ensure both steps complete.
CH₃CH(OH)CH₃ + [O] → CH₃COCH₃ + H₂O. Propanone is the product whether the reaction is run under distillation or reflux; there is no second oxidation step.
In distillation, the still-head and condenser are arranged so that vapour rising from the flask leaves the apparatus via a downward-sloping condenser and is collected in a separate receiver. The condenser is positioned above and to the side of the flask. Anything sufficiently volatile to enter the still-head escapes the oxidising environment. This is what stops the aldehyde from being over-oxidised.
In reflux, a vertical condenser sits directly above the flask. Vapour rises into the condenser, cools, condenses, and returns to the flask as liquid. Nothing leaves. The reaction mixture is held at its boiling point indefinitely, and the dichromate has access to every molecule of substrate (and intermediate) until oxidation is complete.
Exam Tip: Distillation vs reflux is one of the most popular six-mark questions in AQA Paper 2. The mark scheme always wants (a) the technique named, (b) the reason — for distillation, "the aldehyde distils off as it forms, preventing further oxidation"; for reflux, "the aldehyde stays in the flask and is oxidised further to the carboxylic acid".
Given an unlabelled alcohol, run two tests:
A primary alcohol therefore gives orange → green AND a positive Fehling's/Tollens' test on the distillate. A secondary alcohol gives orange → green BUT a negative Fehling's/Tollens' on the distillate. A tertiary alcohol gives no dichromate colour change.
Alcohols can be dehydrated — that is, water can be eliminated — to give an alkene. This is the reverse of the acid-catalysed hydration of an alkene from L4 of this course, and the two reactions sit in equilibrium under acid catalysis (the position of equilibrium is shifted in the alkene direction by removing water).
CH₃CH₂OH → CH₂=CH₂ + H₂O (ethanol → ethene)
CH₃CH(OH)CH₃ → CH₃CH=CH₂ + H₂O (propan-2-ol → propene)
The full mechanism is developed in L8 (mechanism master class). In summary:
This is the E1 mechanism — unimolecular elimination, with a carbocation intermediate. The intermediate is stabilised by the same inductive and hyperconjugation effects that you met in L5: tertiary carbocations form fastest, secondary next, primary slowest. Tertiary alcohols therefore dehydrate readily; primary alcohols require harsher conditions.
When more than one alkene can form (because there is more than one β-carbon bearing an H), the more substituted alkene predominates. This is Zaitsev's rule (sometimes anglicised as Saytzeff). The more substituted alkene is more thermodynamically stable, owing to hyperconjugation and inductive stabilisation of the C=C π system by neighbouring alkyl groups.
Worked example — dehydration of butan-2-ol (CH₃CH(OH)CH₂CH₃):
Two β-positions are available — the CH₃ group on one side and the CH₂CH₃ group on the other:
By Zaitsev, but-2-ene is the major product. Typical product distributions from acid-catalysed dehydration of butan-2-ol give roughly 70–80% but-2-ene and 20–30% but-1-ene, with the precise ratio depending on temperature and acid. But-2-ene is itself a mixture of E- and Z-isomers (a separate stereochemical question — see L2 isomerism), with E predominating because it is the lower-energy geometric isomer.
Common Misconception: Alcohol dehydration with H₂SO₄ and halogenoalkane elimination with ethanolic NaOH both produce alkenes, but they are mechanistically different (E1 vs E2) and use opposite conditions (acid vs base). Don't confuse the reagent sets.
When a carboxylic acid and an alcohol are heated together in the presence of a concentrated sulfuric acid catalyst, they undergo a condensation reaction — a small molecule (water) is eliminated — to form an ester.
Carboxylic acid + Alcohol ⇌ Ester + Water
The double-headed arrow is essential: the reaction is reversible and the position of equilibrium is typically only modestly in favour of the ester. To push the yield up, chemists use one or more of: an excess of the cheaper reagent, removal of water as it forms (sometimes by a Dean-Stark apparatus), or concentrated H₂SO₄ (which acts as both catalyst and dehydrating agent, soaking up the water product and shifting equilibrium to the right).
CH₃COOH + CH₃CH₂OH ⇌ CH₃COOCH₂CH₃ + H₂O
(ethanoic acid + ethanol ⇌ ethyl ethanoate + water)
The oxygen of the ester linkage (–COO–) comes from the alcohol. The –OH of the carboxylic acid is lost as part of the water molecule (this has been confirmed by ¹⁸O isotopic-labelling experiments — labelling the alcohol oxygen with ¹⁸O puts the label in the ester product, not in the water).
Ester names take the form alkyl alkanoate:
| Alcohol | Carboxylic acid | Ester | Smell/use |
|---|---|---|---|
| Methanol | Ethanoic acid | Methyl ethanoate | Glue / nail-polish remover |
| Ethanol | Ethanoic acid | Ethyl ethanoate | Solvent, decaf coffee processing |
| Ethanol | Propanoic acid | Ethyl propanoate | Pineapple |
| Propan-1-ol | Methanoic acid | Propyl methanoate | Pear / rum |
| 3-Methylbutan-1-ol | Ethanoic acid | 3-methylbutyl ethanoate | Banana |
Esters are responsible for the characteristic fruity and floral aromas of many fruits and flowers, and are produced industrially as flavourings, fragrances, solvents (ethyl ethanoate, butyl ethanoate) and plasticisers (the phthalate esters that soften PVC).
Exam Tip: When asked to identify the ester formed from a given alcohol and acid, name the alcohol's alkyl group first and the acid's alkanoate stem second. Then draw the ester linkage as R'–COO–R, where R' is the acyl carbon (from the acid) and R is the alkyl carbon (from the alcohol).
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