← Unit 4: Structure, synthesis and design

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Topic 1: Properties and structure of organic materials

Predict and explain the products of the oxidation of primary, secondary and tertiary alcohols, the oxidation of aldehydes, and the acid-catalysed esterification of carboxylic acids with alcohols (including hydrolysis as the reverse reaction)

A focused answer to the QCE Chemistry Unit 4 dot point on alcohol oxidation and esterification. Distinguishes primary, secondary and tertiary alcohols by oxidation behaviour, gives the acidified-dichromate / permanganate observation colours, and works through the Fischer esterification of ethanoic acid with ethanol. Includes acid hydrolysis as the reverse reaction.

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What this dot point is asking

QCAA wants you to predict the oxidation products of primary, secondary and tertiary alcohols (and of aldehydes), and to write balanced equations for the Fischer esterification of a carboxylic acid with an alcohol, plus its acid-catalysed reverse (hydrolysis). The dot point is high-yield in IA3 (organic synthesis pathways) and in the EA Paper 2 short response.

The answer

Alcohol oxidation and esterification are the two functional-group transformations that anchor every Unit 4 organic synthesis pathway. Together they convert the alcohol family into aldehydes, ketones, carboxylic acids and esters.

Classifying alcohols

The oxidation behaviour of an alcohol depends on how many carbons are attached to the carbon bearing the -OH group (the alpha carbon).

Class Attached carbons Example
Primary (1 deg) 1 (or 0) C on the alpha carbon propan-1-ol, methanol
Secondary (2 deg) 2 C on the alpha carbon propan-2-ol
Tertiary (3 deg) 3 C on the alpha carbon 2-methylpropan-2-ol

Methanol behaves as a primary alcohol (no C on the alpha carbon counts as zero, but there is at least one H on the C, which is required for oxidation).

Oxidation of primary alcohols

Primary alcohols oxidise in two steps under acidified oxidising conditions (typically acidified potassium dichromate, K2Cr2O7 in dilute H2SO4, or acidified KMnO4).

Step 1: alcohol to aldehyde.

R−CH2−OH+[O]→R−CHO+H2OR-CH_2-OH + [O] \rightarrow R-CHO + H_2O

If the aldehyde is distilled off as it forms (open vessel, gentle heat), the reaction stops here.

Step 2: aldehyde to carboxylic acid.

R−CHO+[O]→R−COOHR-CHO + [O] \rightarrow R-COOH

Under reflux with excess oxidant, the aldehyde is oxidised further to the carboxylic acid.

Observation. Dichromate solution turns orange (Cr2O7^2-) to green (Cr3+). KMnO4 turns purple to colourless (Mn2+).

Example. Ethanol oxidation:

  • distillation product: ethanal (CH3-CHO)
  • reflux product: ethanoic acid (CH3-COOH)

The choice between aldehyde and carboxylic acid depends entirely on the apparatus and amount of oxidant.

Oxidation of secondary alcohols

Secondary alcohols oxidise to ketones in a single step. No further oxidation occurs because the next step would require breaking a C-C bond.

R−CH(OH)−R′+[O]→R−CO−R′+H2OR-CH(OH)-R' + [O] \rightarrow R-CO-R' + H_2O

Observation. Same dichromate colour change (orange to green). The absence of distillable aldehyde plus loss of oxidant colour identifies a secondary alcohol.

Example. Propan-2-ol oxidation: CH3-CH(OH)-CH3 + [O] -> CH3-CO-CH3 (propan-2-one, acetone).

Oxidation of tertiary alcohols

Tertiary alcohols are not oxidised by acidified dichromate or permanganate at the conditions used in school laboratories. Oxidation would require breaking a C-C bond, which is unfavourable.

Observation. No colour change. Dichromate stays orange; permanganate stays purple. This is the diagnostic distinction QCAA tests.

The classification test in practice

A common IA2 design or EA short response gives three alcohols and asks to identify them by oxidation. The decision tree:

  1. Does the oxidant change colour? No -> tertiary. Yes -> proceed.
  2. Is a distillable aldehyde obtained? Yes -> primary (gentle, open). The aldehyde can be tested further (Tollens reagent or further oxidation to acid).
  3. Otherwise (colour change but no aldehyde) -> secondary (product is a ketone).

Esterification: acid + alcohol -> ester + water

A carboxylic acid reacts with an alcohol, with concentrated sulfuric acid as catalyst, to produce an ester and water. The reaction is named after Fischer.

R−COOH+R′−OH⇌H2SO4R−COO−R′+H2OR-COOH + R'-OH \xrightleftharpoons[]{H_2SO_4} R-COO-R' + H_2O

Worked example. Ethanoic acid + ethanol:

CH3−COOH+CH3−CH2−OH⇌H2SO4CH3−COO−CH2−CH3+H2OCH_3-COOH + CH_3-CH_2-OH \xrightleftharpoons[]{H_2SO_4} CH_3-COO-CH_2-CH_3 + H_2O

Ester: ethyl ethanoate. Common smell: pear-drop / nail-polish-remover.

Key features QCAA tests:

  • Equilibrium reaction, not complete. Use a reversible arrow. Typical equilibrium gives 60 to 70 percent ester after several hours of reflux.
  • Concentrated H2SO4 as catalyst. A few drops only. Acts as a Bronsted-Lowry acid catalyst and as a dehydrating agent (removing water shifts the equilibrium right).
  • Reflux conditions. Slow reaction; reflux is required to keep volatile reactants in contact for several hours.
  • Naming convention. Alkyl-from-alcohol + -oate-from-acid. Methanol + ethanoic acid -> methyl ethanoate. Ethanol + methanoic acid -> ethyl methanoate. Reverse the alcohol/acid and the name swaps.

To increase ester yield:

  • Use an excess of the cheaper reactant (Le Chatelier).
  • Remove water as it forms (Dean-Stark trap, or concentrated H2SO4 absorbing water).
  • Remove the ester by distillation as it forms (only if its boiling point is convenient).

Hydrolysis of esters (the reverse reaction)

Heating an ester with dilute aqueous H2SO4 (or dilute aqueous NaOH) cleaves it back to the carboxylic acid and alcohol.

Acid hydrolysis (reverse of Fischer esterification):

R−COO−R′+H2O⇌H+R−COOH+R′−OHR-COO-R' + H_2O \xrightleftharpoons[]{H^+} R-COOH + R'-OH

Equilibrium reaction, driven by excess water in the reverse direction.

Base hydrolysis (saponification) is the same overall transformation but goes to completion because the carboxylic acid is deprotonated to its salt (R-COO- Na+), which cannot recombine with the alcohol:

R−COO−R′+NaOH→R−COO−Na++R′−OHR-COO-R' + NaOH \rightarrow R-COO^- Na^+ + R'-OH

Soap manufacture is the industrial saponification of triglyceride esters with NaOH. Unit 4 mentions this in the biomolecules / polymer section.

Esters in flavour, fragrance and biology

Most short-chain esters are volatile and smell of fruit (ethyl butanoate, pineapple; pentyl ethanoate, banana; octyl ethanoate, orange). The QCAA IA3 research investigation on flavour chemistry is a common context. Larger esters are waxes (beeswax: mainly C30 esters) and fats (triglyceride esters of glycerol with fatty acids).

Common traps

Forgetting "[O]" notation. QCAA marking guides accept [O] as shorthand for the oxidant (Cr2O7^2- + H+ / e- or MnO4- + H+ / e-). Show "[O]" on the reactant side of the arrow.

Writing tertiary alcohol oxidation equations. There is no equation; write "no reaction" explicitly.

Naming the ester reversed. Methyl propanoate is propanoic acid + methanol; propyl methanoate is methanoic acid + propan-1-ol. Different compounds, different smells, different boiling points.

Using a single arrow for esterification. Use a reversible arrow. The equilibrium nature of the reaction is examined in its own right.

Confusing dehydration with oxidation. Concentrated H2SO4 at high temperature dehydrates an alcohol to an alkene (E1 / E2 elimination). Dilute oxidising agents (acidified dichromate) oxidise to aldehyde / ketone. Conditions matter.

In one sentence

Primary alcohols oxidise in two steps (alcohol -> aldehyde -> carboxylic acid) under acidified dichromate or permanganate, secondary alcohols oxidise once (alcohol -> ketone), tertiary alcohols do not oxidise; carboxylic acids react reversibly with alcohols under H2SO4 catalysis to form esters and water (Fischer esterification), and the reverse acid-catalysed hydrolysis returns the original carboxylic acid and alcohol.

Past exam questions, worked

Real questions from past QCAA papers on this dot point, with our answer explainer.

2023 QCAA-style5 marksThree colourless liquids labelled X, Y and Z are propan-1-ol, propan-2-ol and 2-methylpropan-2-ol in some order. Each is heated separately with acidified potassium dichromate. X turns the solution from orange to green and a sweet-smelling product can be distilled off. Y turns the solution from orange to green but no distillable aldehyde forms. Z produces no colour change. (a) Identify X, Y and Z. (b) Write balanced equations for any reactions that occur, using [O] for the oxidising agent.
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A 5-mark answer needs the three identifications with reasoning and the two oxidation equations.

(a) Identification.

Z produces no colour change, so Z is the tertiary alcohol (cannot be oxidised by dichromate). Z = 2-methylpropan-2-ol.

Y turns orange to green but produces no aldehyde. Y is the secondary alcohol (oxidises to a ketone, not an aldehyde, and cannot be over-oxidised). Y = propan-2-ol; product is propan-2-one (acetone).

X turns orange to green and a sweet-smelling distillable product forms. X is the primary alcohol (oxidises to aldehyde, which is volatile and distillable; if heated under reflux it would go further to a carboxylic acid). X = propan-1-ol.

(b) Equations (using [O] for the dichromate oxidant).

X: CH3-CH2-CH2-OH + [O] -> CH3-CH2-CHO + H2O (propan-1-ol -> propanal).

Further oxidation (excess [O], reflux): CH3-CH2-CHO + [O] -> CH3-CH2-COOH (propanal -> propanoic acid).

Y: CH3-CH(OH)-CH3 + [O] -> CH3-CO-CH3 + H2O (propan-2-ol -> propan-2-one).

Z: no reaction.

Markers reward correct primary/secondary/tertiary classification, the dichromate colour cue (orange Cr2O7^2- to green Cr3+), and the two valid equations. The "no reaction" for Z must be stated explicitly.

2022 QCAA-style4 marks(a) Write the balanced equation for the esterification of ethanoic acid with propan-1-ol, naming the ester and stating the catalyst required. (b) Explain why esterification is performed under reflux. (c) Predict what happens to the ester if it is heated with dilute aqueous sulfuric acid.
Show worked answer →

A 4-mark answer needs the equation with named ester, the reflux justification, and the hydrolysis prediction.

(a) Esterification.

CH3−COOH+HO−CH2−CH2−CH3⇌H2SO4CH3−COO−CH2−CH2−CH3+H2OCH_3-COOH + HO-CH_2-CH_2-CH_3 \xrightleftharpoons[]{H_2SO_4} CH_3-COO-CH_2-CH_2-CH_3 + H_2O

Ester: propyl ethanoate. Catalyst: concentrated sulfuric acid (a few drops).

(b) Reflux. Esterification is slow and reaches equilibrium, not completion. Heating speeds the reaction; reflux (boiling with a condenser returning vapour to the flask) lets the mixture stay hot for an extended period without losing the volatile reactants (ethanoic acid bp 118 degrees C; propan-1-ol bp 97 degrees C) or the ester (bp 102 degrees C). Without reflux, the volatile components would evaporate before significant ester forms.

(c) Hydrolysis prediction. Heating the ester with dilute H2SO4 reverses the reaction (acid-catalysed hydrolysis). The ester decomposes back to ethanoic acid and propan-1-ol, the same equilibrium driven in the reverse direction by excess water.

Markers reward the balanced equation with reversible arrow (esterification is an equilibrium), the catalyst, the reflux explanation (volatility plus equilibrium), and the explicit hydrolysis equation or reasoning.

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