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)

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).

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.

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

Step 2: aldehyde to carboxylic acid.

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.

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.

Worked example. Ethanoic acid + ethanol:

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):

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:

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.

Examples in context

Example 1. Cane-juice ethanol oxidation at Bundaberg Distilling Company. Bundaberg uses dilute acidic $\text{K}_2 \text{Cr}_2 \text{O}_7$ in quality-control labs to identify primary alcohol contamination in rum spirit. The dichromate (orange) reduces to $\text{Cr}^{3+}$ (green) as ethanol oxidises to ethanal then to ethanoic acid: $3 \text{CH}_3 \text{CH}_2 \text{OH} + \text{Cr}_2 \text{O}_7^{2-} + 8 \text{H}^+ \rightarrow 3 \text{CH}_3 \text{CHO} + 2 \text{Cr}^{3+} + 7 \text{H}_2 \text{O}$. Continued reflux drives further oxidation to ethanoic acid. By contrast, tertiary alcohols (e.g. 2-methylpropan-2-ol) cannot be oxidised this way - useful for distinguishing alcohol classes by colour change.

Example 2. Ester aroma identification in QCAA IA2 organic brief. A typical task asks students to synthesise three esters by refluxing alcohol + carboxylic acid + concentrated $\text{H}_2 \text{SO}_4$ catalyst. Ethanol + ethanoic acid gives ethyl ethanoate (nail-polish remover smell); methanol + butanoic acid gives methyl butanoate (apple); octanol + ethanoic acid gives octyl ethanoate (orange). The reactions are reversible: $\text{R-COOH} + \text{R'-OH} \rightleftharpoons \text{R-COO-R'} + \text{H}_2 \text{O}$. Concentrated $\text{H}_2 \text{SO}_4$ both catalyses and absorbs water (Le Chatelier favours product). Distinguishing products by odour is a powerful, low-tech ID.

Try this

Q1. Write the balanced equation for the complete combustion of ethanol. [2 marks]

• Cue. $\text{C}_2 \text{H}_5 \text{OH} + 3 \text{O}_2 \rightarrow 2 \text{CO}_2 + 3 \text{H}_2 \text{O}$.

Q2. Describe the oxidation of butan-1-ol with acidified dichromate. State the organic products under (a) distillation and (b) reflux conditions. [4 marks]

• Cue. (a) Distillation: butanal removed before further oxidation. (b) Reflux: butanoic acid.

Q3. $0.50 \, \text{mol}$ ethanol is refluxed with $0.50 \, \text{mol}$ ethanoic acid; equilibrium yield is $66\%$ ethyl ethanoate. (a) Write the equation. (b) Calculate $K_c$ assuming $1 \, \text{L}$. (c) Suggest two ways to push yield higher. [2+3+2 marks]

• Cue. (a) $\text{CH}_3 \text{CH}_2 \text{OH} + \text{CH}_3 \text{COOH} \rightleftharpoons \text{CH}_3 \text{COOC}_2 \text{H}_5 + \text{H}_2 \text{O}$. (b) $K_c = (0.33)^2/(0.17)^2 = 3.77$. (c) Excess one reactant; remove water.