Whiteboard Chemistry with Joe White

Alcohols

How ethanol is made by fermentation and by hydration, why alcohols are classified 1°/2°/3°, how they oxidise to aldehydes, ketones and carboxylic acids, and how they dehydrate to alkenes.

AQA 7404/7405 Paper 2
C OH H₃C H H H C O O C C C
Building on GCSE

At GCSE you met ethanol as a fuel and drink, made by fermentation. A-Level keeps the –OH functional group but asks far more of it: two industrial routes and their trade-offs, a 1°/2°/3° classification that decides what each alcohol does, controlled oxidation to three different products, and dehydration back to an alkene. The theme for the whole page: one –OH, but the reaction you run — and the alcohol’s class — changes everything.

Classifying alcohols & their properties

Before any reaction, sort the alcohol. Look at the carbon carrying the –OH group and count how many other carbon atoms are attached to it:

Key definitions

A primary (1°) alcohol has the –OH carbon bonded to one other carbon (or none, in methanol). A secondary (2°) alcohol has it bonded to two. A tertiary (3°) alcohol has it bonded to three.

This one label decides how — and whether — the alcohol oxidises later, so get quick at spotting it.

propan-1-ol · primary (1°)

H₃C C H H C H H OH

1 other carbon on the –OH carbon

propan-2-ol · secondary (2°)

H₃C C OH H CH₃

2 other carbons on the –OH carbon

2-methylpropan-2-ol · tertiary (3°)

H₃C C OH CH₃ CH₃

3 other carbons on the –OH carbon

Classify by the carbon bearing the –OH. That count is what the oxidation section hangs on.

Alcohols also behave differently from hydrocarbons physically, because the polar O–H group lets them form hydrogen bonds. That raises their boiling points well above alkanes and alkenes of similar size, and makes short-chain alcohols soluble in water.

Precision points
  • Classify by the –OH carbon, not the longest chain or the whole molecule.
  • Higher boiling point than a comparable alkene is due to hydrogen bonding being stronger than van der Waals forces — never say covalent bonds break when it boils.
  • Tertiary alcohols are the odd one out for oxidation — but examiner reports warn against making “tertiary” the answer to every “no reaction” question. Check which reaction is being blocked.
🧪 Exam-style questions
Q1 [1 mark]

Which statement is not correct for both primary and secondary alcohols? Tick (✓) one box.

Q2 [3 marks]

The boiling point of pentan-2-ol is 119 °C; the boiling point of pent-1-ene is 30 °C. Explain why pentan-2-ol has the higher boiling point.

Show answer

Pentan-2-ol has stronger intermolecular forces (than pent-1-ene). 1 mark

Pent-1-ene has van der Waals (London / induced-dipole) forces only. 1 mark

Pentan-2-ol also has hydrogen bonds between molecules, which are stronger and need more energy to overcome. 1 mark

Do not accept any reference to breaking covalent bonds on boiling.

Source: AQA A-Level Chemistry past papers.

Making ethanol: two routes

Ethanol is made industrially two completely different ways, and AQA loves to make you compare them. One grows it; one builds it from crude-oil feedstock.

Hydration of ethene

Ethene from cracking is reacted with steam over a phosphoric acid (H3PO4) catalyst at about 300 °C and high pressure. The mechanism is electrophilic addition of water across the C=C — the same three arrows as the concentrated-sulfuric-acid route on the alkenes page, except that water itself is the nucleophile attacking the carbocation, so a final H+ is lost from the –OH2+ intermediate. That is exactly the dehydration mechanism further down this page, run in reverse.

It is continuous, fast, gives pure ethanol and has 100% atom economy — but the feedstock is non-renewable (crude oil) and it needs a lot of energy.

Fermentation of glucose

Sugars from plants are fermented by the enzymes in yeast, at about 35 °C, in the absence of air (anaerobic).

The conditions are worth justifying: ~35 °C because the enzymes work too slowly when cold and denature when hot; no air because oxygen lets bacteria oxidise the ethanol to ethanoic acid (vinegar). It uses a renewable feedstock and little energy, but it is slow, runs in batches, and gives a dilute, impure mixture that must be concentrated by fractional distillation.

Hydration of etheneFermentation
Feedstockethene (crude oil) — finitesugars / starch — renewable
Conditions~300 °C, high pressure, H3PO4~35 °C, yeast, anaerobic
Rate & processfast, continuousslow, batch
Productpure ethanoldilute — needs fractional distillation
Atom economy100%lower (CO2 by-product)

Bioethanol & carbon neutrality

A biofuel is a fuel made from recently living material (biomass). Ethanol from fermentation is a biofuel, and is often called carbon-neutral — meaning no net carbon dioxide is added to the atmosphere, because the CO2 released when it burns was taken in by the plant during photosynthesis.

Follow the carbon through three equations — in one, out in two:

Six CO2 in; two out in fermentation plus four out in combustion — six out. On paper, neutral. In practice it is not truly carbon-neutral: the machinery, fertilisers, transport and the distillation itself are usually powered by fossil fuels, and land that could grow food is cleared to grow the fuel crop instead.

hydration of ethene — continuous

CH₂=CH₂ ethene · from cracking H₂O(g) steam H₃PO₄ catalyst 300 °C · ~65 atm continuous flow CH₃CH₂OH ethanol · pure product unreacted gases recycled crude oil · finite 100% atom economy fast · continuous · pure

Fast, pure product, 100% atom economy — but the ethene comes from crude oil

fermentation of glucose — batch

C₆H₁₂O₆(aq) glucose · from plants yeast · ~35 °C anaerobic — no air CO₂ dilute fractional distillation CH₃CH₂OH ethanol sugars · renewable slow · batch dilute → needs distilling

Renewable feedstock — but slow, batch, and the ethanol needs distilling

Two routes to the same molecule — the exam wants the trade-offs: feedstock, rate, purity and atom economy.
🧪 Exam-style questions
Q3 [1 mark]

Which statement is correct about the production and use of ethanol as a biofuel? Tick (✓) one box.

Q4 [4 marks]

Ethanol produced by fermentation of glucose may be regarded as a carbon-neutral fuel. Justify this statement. Include the relevant chemical equations in your answer.

Show answer

Photosynthesis: 6CO2 + 6H2O → C6H12O6 + 6O2 1 mark

Fermentation: C6H12O6 → 2C2H5OH + 2CO2 1 mark

Combustion: 2C2H5OH + 6O2 → 4CO2 + 6H2O 1 mark

The equations show 6 CO2 taken in by photosynthesis and 6 CO2 given out (fermentation + combustion), so there is no net release of CO2. 1 mark

Multiples of the equations are allowed; the final mark depends on the equations showing 6 CO2 in and out.

Source: AQA A-Level Chemistry past papers.

Oxidation of alcohols & Required Practical 5

This is the section the 1°/2°/3° label was for. The oxidising agent is acidified potassium dichromate(VI) (K2Cr2O7 with dilute H2SO4). When it does its job it changes orange solution to green solution (the dichromate(VI) ion Cr2O72− is reduced to green Cr3+). In equations you may use [O] to represent oxygen from the oxidising agent.

Primary alcohols — and the distil-vs-reflux choice

A primary alcohol oxidises in two stages: first to an aldehyde, then to a carboxylic acid. You choose which you get by how you run it:

STAGE 1 ethanol → ethanal · distil now to keep the aldehyde H₃C C H H O H + [O] H₃C C O H + H₂O the ringed O–H and C–H hydrogens leave with the [O] as water — C=O forms STAGE 2 ethanal → ethanoic acid · reflux with excess [O] H₃C C O H + [O] H₃C C O O H no water this time — the [O] inserts into the ringed C–H to make the acid’s O–H
Written out: CH3CH2OH + [O] → CH3CHO + H2O, then CH3CHO + [O] → CH3COOH.
  • Want the aldehyde? Distil it off as it forms, using a limited amount of oxidant — the aldehyde has the lowest boiling point (no hydrogen bonding), so it leaves before it can be oxidised further.
  • Want the carboxylic acid? Reflux with excess oxidant — heating without losing anything keeps the product in the flask to be oxidised the whole way.
DISTILLATION — collect the aldehyde as it forms REFLUX — nothing escapes water out water in condenses to liquid thermometer — bulb level with the side arm still head pear-shaped flask heat ethanol + limited acidified K₂Cr₂O₇ anti-bumping granules ethanal (colourless) collects — open flask → ethanal, out before round two water out water in open at the top — never seal it condenses & drips back heat ethanol + excess acidified K₂Cr₂O₇ anti-bumping granules → carries on to ethanoic acid
The drawing marks live in the details: thermometer bulb level with the side arm, condenser water in at the bottom, and a system that is never sealed — distillation lets the product escape; reflux sends it back.

Secondary & tertiary alcohols

A secondary alcohol oxidises to a ketone, and stops there (there is no easy further oxidation).

H₃C C O H H CH₃ + [O] H₃C C O CH₃ + H₂O the same two ringed hydrogens leave — but now the C=O sits between two carbons
Written out: CH3CH(OH)CH3 + [O] → CH3COCH3 + H2O.

A tertiary alcohol is not oxidised by acidified dichromate — there is no hydrogen on the carbon bearing the –OH to remove — so the mixture stays an orange solution.

Cr₂O₇²⁻ Cr³⁺ PRIMARY (1°) CH₃CH₂OH ethanol [O] CH₃CHO ethanal · aldehyde reflux excess [O] CH₃COOH ethanoic acid distil off as it forms → the oxidation stops here SECONDARY (2°) CH₃CH(OH)CH₃ propan-2-ol [O] CH₃COCH₃ propanone · ketone ✗ no further oxidation TERTIARY (3°) (CH₃)₃COH 2-methylpropan-2-ol no reaction no H on the C–OH carbon to remove stays orange soln
The class of alcohol sets the destination; for a primary alcohol, distil versus reflux sets how far you travel.

Telling an aldehyde from a ketone

Both come from oxidising alcohols, and both contain C=O — but only the aldehyde is easily oxidised further, and two gentle reagents exploit that:

ReagentAldehydeKetone
Tollens’ reagentsilver mirror formsno change
Fehling’s solutionblue solution → brick-red precipitatestays blue solution

tollens’ reagent · warm gently

+ aldehyde silver mirror + ketone no change

The aldehyde reduces Ag+ to silver metal on the glass; a ketone cannot

fehling’s solution · warm gently

+ aldehyde brick-red precipitate + ketone stays blue soln

The aldehyde reduces blue Cu2+ to brick-red Cu2O; a ketone leaves it blue

Both reagents are gentle oxidising agents — only the aldehyde can still be oxidised, so only the aldehyde changes them.

The aldehyde reduces the reagent as it is itself oxidised to a carboxylic acid. A ketone cannot be oxidised this way, so nothing happens.

Required practical 5 — distillation of a product from a reaction

Oxidising a primary alcohol and distilling off the aldehyde as it forms is the classic RP5 set-up: heat the flask, and the low-boiling aldehyde vaporises and condenses into the collection flask before it can be oxidised to the acid. Know the apparatus (pear-shaped flask, still head, thermometer at the side-arm, condenser with water in at the bottom) and why anti-bumping granules are added — both set-ups are drawn, labelled and animated in the figure above. Practical skills are examined across all three papers — see the required practicals page.

🧪 Exam-style questions
Q5 [1 mark]

Which compound reacts to form a ketone when warmed with acidified potassium dichromate(VI)? Tick (✓) one box.

Q6 [1 mark]

Which compound is produced when 1-phenylethanol, C6H5CH(OH)CH3, reacts with acidified potassium dichromate(VI)? Tick (✓) one box.

Q7 [1 mark]

The alcohol CH3CH2CH2OH can be oxidised. Which compound cannot be produced by oxidation of this alcohol? Tick (✓) one box.

Q8 [3 marks]

Propan-1-ol is oxidised to propanoic acid. State a simple chemical test that distinguishes the propanoic acid from the propan-1-ol, and give one observation for the test with each substance.

Show answer

Add sodium carbonate (or sodium hydrogencarbonate) solution. 1 mark

Propanoic acid: effervescence / bubbles (of CO2). 1 mark

Propan-1-ol: no visible change / no reaction. 1 mark

Do not accept a pH-meter reading. An acidified-dichromate test (orange solution → green solution with the alcohol, no change with the acid) also scores.

Source: AQA A-Level Chemistry past papers.

Elimination: dehydrating alcohols to alkenes

Run the hydration reaction backwards and you dehydrate the alcohol: remove a molecule of water and form a C=C double bond. The reagent is a hot concentrated acid catalyst — concentrated H2SO4 or H3PO4, around 170 °C.

It is an acid-catalysed elimination, and the mechanism is three moves:

The dehydration mechanism, arrow by arrow
  • Arrow 1 — protonate: a lone pair on the alcohol’s oxygen attacks an H+, turning –OH into –OH2+ (a good leaving group).
  • Arrow 2 — lose water: the C–O bond breaks, water leaves, and a carbocation forms.
  • Arrow 3 — lose H+: a C–H bond on the next carbon donates its pair to form the C=C, releasing H+ — so the acid is a catalyst.
STEP 1 the O lone pair attacks H⁺ — protonation C H H H C H H O H H + H⁺ from the hot conc acid STEP 2 the C–O bond breaks — water leaves C H H H C H H O H H + –OH₂⁺ is a good leaving group STEP 3 an adjacent C–H pair forms the C=C; H⁺ is released C H H H C H H + H₂O the water has left PRODUCTS ethene + H₂O + H⁺ H⁺ returned — the acid is a catalyst
The reverse of hydration: protonate, lose water, lose a proton. The acid is handed back at the end.

When the alcohol is unsymmetrical, the H+ can be lost from carbons on either side of the carbocation, so you get a mixture of alkenes (including E/Z isomers). Pentan-2-ol, for example, gives pent-1-ene and pent-2-ene.

Why bother? These alkenes can be turned into addition polymers without using monomers made from crude oil — a renewable route into plastics.

🧪 Exam-style questions
Q9 [1 mark]

Which compound can be dehydrated to form an alkene? Tick (✓) one box.

Q10 [1 mark]

Which alcohol, when dehydrated, forms a mixture of alkenes? Tick (✓) one box.

Q11 [5 marks]

Pent-1-ene is formed by the elimination of water from pentan-2-ol. State the reagent and condition for this reaction, and outline the mechanism.

Show answer

Reagent: concentrated sulfuric acid (or concentrated phosphoric acid). 1 mark

Condition: hot / temperature in the range 150–200 °C. 1 mark

Curly arrow from a lone pair on the alcohol O to H+ (protonation). 1 mark

Arrow from the C–O bond to the O on the protonated intermediate (water leaves → carbocation). 1 mark

Arrow from a C–H bond on C1 to the C–C bond, forming the C=C of pent-1-ene. 1 mark

All arrows double-headed. Penalise any extra, contradictory arrows at each stage.

Source: AQA A-Level Chemistry past papers.

Same alcohol, different reactions

The exam trap on this topic is mixing up the reactions. Oxidation and dehydration start from the same alcohol but need different reagents and give different products — and “tertiary” is not the answer to every “why no reaction?”. Read the verb: oxidise, dehydrate, reflux, distil.

Do this to the alcohol…Reagent / conditions…and you get
Oxidise (1°, distil)acidified K2Cr2O7, distil offaldehyde
Oxidise (1°, reflux)acidified K2Cr2O7, excess, refluxcarboxylic acid
Oxidise (2°)acidified K2Cr2O7, warmketone
Dehydratehot conc H2SO4 / H3PO4alkene (+ water)
Precision points
  • Examiner reports flag students confusing dehydration with oxidation — different reagent, different product.
  • They also flag “tertiary alcohol” given as the default answer to any “no reaction” question. A tertiary alcohol resists oxidation, but it dehydrates fine — check which reaction is blocked.
  • For the aldehyde/carboxylic-acid choice, name the technique: distil for the aldehyde, reflux for the acid.
🧪 Capstone questions
Q12 [1 mark]

Which compound produces (CH3)2CHCOCH3 when oxidised? Tick (✓) one box.

Q13 [4 marks]

A student warms 2.0 cm3 of propan-2-ol (density 0.786 g cm−3) with excess acidified potassium dichromate(VI) and collects 0.954 g of propanone (CH3COCH3). Calculate the percentage yield. Give your answer to the appropriate number of significant figures.

Show answer

Mass of propan-2-ol = 2.0 × 0.786 = 1.572 g. 1 mark

Amount of propan-2-ol = 1.572 ÷ 60.0 = 0.0262 mol. 1 mark

Maximum mass of propanone = 0.0262 × 58.0 = 1.52 g. 1 mark

% yield = (0.954 ÷ 1.52) × 100 = 63% (2 sig figs). 1 mark

2 sig figs only, set by the 2.0 cm3 measurement. Allow error-carried-forward at each step.

Source: AQA A-Level Chemistry past papers.

3.3.5 Alcohols — Quick-reference summary
  • Classify by the –OH carbon: 1° (one C attached), 2° (two), 3° (three). Alcohols hydrogen-bond, so higher bp than comparable alkenes.
  • Production — hydration: ethene + steam, H3PO4, ~300 °C; fast, continuous, 100% atom economy, but non-renewable. Fermentation: C6H12O6 → 2C2H5OH + 2CO2; yeast, ~35 °C, anaerobic; renewable but slow and impure (needs fractional distillation).
  • Biofuel: a fuel from living matter; bioethanol is near carbon-neutral (CO2 out balances CO2 taken in by photosynthesis) but not truly, because farming, transport and distillation use energy.
  • Oxidation (acidified K2Cr2O7, orange solution → green solution): 1° → aldehyde (distil off) → carboxylic acid (reflux); 2° → ketone; 3° → no reaction.
  • Aldehyde vs ketone: Tollens’ gives a silver mirror with an aldehyde only; Fehling’s gives a brick-red precipitate with an aldehyde only.
  • Dehydration (hot conc H2SO4/H3PO4): alcohol → alkene + H2O, by acid-catalysed elimination (protonate –OH, lose water → carbocation, lose H+ → C=C). Unsymmetrical alcohols give a mixture of alkenes.

Found an error or have a suggestion?

Help improve these notes by sending feedback.

Want to go deeper?

1-to-1 tuition led by a current AQA examiner.

Alcohol marks turn on precision: the exact colour change for dichromate, distil-versus-reflux for the right oxidation product, the silver mirror that separates aldehyde from ketone, and a dehydration mechanism with every curly arrow in place. Sessions drill these on real AQA past questions.

Enquire now
Ready to get started? Enquire now →