Describe the oxidation of primary, secondary and tertiary alcohols and the products formed.

Classifying alcohols as primary, secondary or tertiary, predicting their oxidation products (aldehyde, carboxylic acid, ketone, or no reaction), the oxidising agents and colour changes, with worked SACE-style examples.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

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What this dot point is asking
Lead worked calculation
Classifying alcohols
Oxidation outcomes
Controlling the primary-alcohol product
Why ketones resist further oxidation
Why it matters

What this dot point is asking

SACE expects you to classify an alcohol, predict its oxidation product(s), name them, and describe the colour change, including how distillation versus reflux controls the primary-alcohol product.

Lead worked calculation

Predicting products for three alcohols

Predict the oxidation products of propan-1-ol, propan-2-ol and 2-methylpropan-2-ol with hot acidified $\text{K}_2\text{Cr}_2\text{O}_7$ .

Classify each

Propan-1-ol: the $\text{-OH}$ carbon is bonded to one other carbon, so primary. Propan-2-ol: bonded to two carbons, so secondary. 2-methylpropan-2-ol: bonded to three carbons, so tertiary.

Apply the oxidation rules

Primary propan-1-ol gives propanal (aldehyde), then propanoic acid (carboxylic acid) on full oxidation. Secondary propan-2-ol gives propanone (a ketone). Tertiary 2-methylpropan-2-ol gives no reaction.

State the observation

The two that react turn the dichromate from orange to green; the tertiary alcohol leaves it orange, a simple test that distinguishes tertiary from primary and secondary alcohols.

Classifying alcohols

Oxidation outcomes

The common oxidising agent is acidified potassium dichromate, $\text{K}_2\text{Cr}_2\text{O}_7$ in dilute sulfuric acid; the dichromate ion is reduced from $\text{Cr}_2\text{O}_7^{2-}$ (orange) to $\text{Cr}^{3+}$ (green).

Alcohol class	First product	Further oxidation	Colour change
Primary	aldehyde	carboxylic acid	orange to green
Secondary	ketone	none (resists)	orange to green
Tertiary	no reaction	no reaction	stays orange

A tertiary alcohol cannot be oxidised because its $\text{-OH}$ carbon has no hydrogen to remove; oxidation would require breaking a $\text{C-C}$ bond, which does not occur under these conditions.

Controlling the primary-alcohol product

Why ketones resist further oxidation

A secondary alcohol oxidises to a ketone and stops there because the carbonyl carbon of a ketone is bonded to two carbon atoms and no hydrogen; further oxidation would again require breaking a $\text{C-C}$ bond. An aldehyde, by contrast, still has a hydrogen on the carbonyl carbon, so it is readily oxidised to a carboxylic acid. This hydrogen-on-the-carbonyl-carbon difference explains why aldehydes are oxidised easily while ketones are not, the basis of tests that distinguish them.

Why it matters

The oxidation of alcohols links alcohols to aldehydes, ketones and carboxylic acids, the building blocks for esters and many syntheses. The class-dependent behaviour and the distillation/reflux control are favourite SACE exam points and a practical way to identify an unknown alcohol.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20225 marksThree alcohols, butan-1-ol, butan-2-ol and 2-methylpropan-2-ol, are each warmed with acidified potassium dichromate. (a) Classify each alcohol. (b) For each, give the name of the organic product(s) or state 'no reaction'. (c) Describe the colour change observed in those that react and explain it.

Show worked answer →

(a) butan-1-ol is primary; butan-2-ol is secondary; 2-methylpropan-2-ol is tertiary. (1 mark)

(b) Butan-1-ol (primary) oxidises to butanal (aldehyde), then to butanoic acid (carboxylic acid) on further oxidation. Butan-2-ol (secondary) oxidises to butanone (ketone). 2-methylpropan-2-ol (tertiary) gives no reaction. (3 marks)

(c) Where reaction occurs, the dichromate is reduced and the solution changes from orange ( $\text{Cr}_2\text{O}_7^{2-}$ ) to green ( $\text{Cr}^{3+}$ ). (1 mark)

SACE 20204 marksEthanol can be oxidised by acidified potassium dichromate. (a) Name the product formed under distillation conditions and the product formed under reflux with excess oxidant. (b) Explain why the apparatus (distillation versus reflux) determines which product is obtained.

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(a) Under distillation the aldehyde ethanal is removed as it forms; under reflux with excess oxidant the further oxidation product ethanoic acid (a carboxylic acid) is obtained. (2 marks)

(b) Ethanal has a lower boiling point than ethanol, so distilling lets it evaporate and be collected before it is oxidised further. Reflux returns the vapour to the flask, keeping the aldehyde in contact with the oxidant so it is oxidised all the way to the carboxylic acid. The apparatus therefore controls the product. (2 marks)