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Oxidation of alcohols (SACE Stage 2 Chemistry)

Syllabus dot point

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

8 min read

Quick answer

Alcohols are classified by the number of carbon atoms bonded to the carbon carrying the -OH\text{-OH} group: primary (one), secondary (two) or tertiary (three). On oxidation with acidified potassium dichromate, a primary alcohol gives an aldehyde then a carboxylic acid, a secondary alcohol gives a ketone, and a tertiary alcohol does not react. The orange dichromate turns green when oxidation occurs.

Jump to a section
  1. What this dot point is asking
  2. Lead worked calculation
  3. Classifying alcohols
  4. Oxidation outcomes
  5. Controlling the primary-alcohol product
  6. Why ketones resist further oxidation
  7. Why it matters
  8. Exam-style questions

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 K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7.

Classify each

Propan-1-ol: the -OH\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

Primary, secondary and tertiary alcohols

The class is set by how many carbon atoms are attached to the carbon bearing the -OH\text{-OH} group. Primary (1): that carbon is bonded to one other carbon (or none). Secondary (2): bonded to two other carbons. Tertiary (3): bonded to three other carbons. The class determines whether and how the alcohol can be oxidised, because oxidation requires a hydrogen on the -OH\text{-OH} carbon.

Oxidation outcomes

The common oxidising agent is acidified potassium dichromate, K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7 in dilute sulfuric acid; the dichromate ion is reduced from Cr2O72−\text{Cr}_2\text{O}_7^{2-} (orange) to Cr3+\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 -OH\text{-OH} carbon has no hydrogen to remove; oxidation would require breaking a C-C\text{C-C} bond, which does not occur under these conditions.

Controlling the primary-alcohol product

Distillation versus reflux

A primary alcohol oxidises in two stages: to an aldehyde, then to a carboxylic acid. The apparatus decides where you stop. Distillation removes the volatile aldehyde as it forms (the aldehyde boils below the alcohol), so it is collected before further oxidation. Reflux with excess oxidant condenses vapours back into the flask, keeping the aldehyde in contact with the oxidant until it is fully oxidised to the carboxylic acid. Same alcohol, same oxidant, different product, set purely by technique.

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 C-C\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.

Misclassifying the alcohol or over-oxidising a ketone

The commonest SACE error is miscounting the carbons on the -OH\text{-OH} carbon and so assigning the wrong class, which leads to the wrong product. Count carefully. A second error is writing that a secondary alcohol (or its ketone product) oxidises to a carboxylic acid; ketones resist further oxidation. A third is forgetting that a primary alcohol's final product under reflux is the acid, not the aldehyde. Always state the colour change (orange to green) for the reacting alcohols.

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 questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2022-style5 marks
Three 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 (Cr2O72−\text{Cr}_2\text{O}_7^{2-}) to green (Cr3+\text{Cr}^{3+}). (1 mark)

SACE 2020-style4 marks
Ethanol 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.
Show worked answer →

(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)

Practise this

Sources & how we know this

Based on the official SACE Board subject outline. Not yet individually reviewed by a person; always check SACE Board and your school's materials for current rules.

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