This chemistry grades 11-12 lesson on oxidation of alcohols notes gives the key ideas first, then worked examples and diagrams. After that, practise with the worksheet on oxidation of alcohols and check yourself with the quiz.
Quick answer: Primary alcohols are oxidised to aldehydes and then carboxylic acids, secondary alcohols to ketones, and tertiary alcohols resist oxidation; acidified dichromate turns from orange to green. Aldehydes give silver mirrors with Tollens’ reagent, ketones do not. Carbonyl compounds undergo nucleophilic addition, for example with cyanide ions.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: A Level 3.3.5
Oxidation of alcohols notes: the key ideas
How are alcohols classified, and how are they oxidised?
An alcohol is primary, secondary or tertiary according to how many carbon atoms are bonded to the carbon carrying the OH group: one (or none), two or three. This matters because oxidation removes a hydrogen from that carbon, and tertiary alcohols have none to lose.
The usual oxidising agent is acidified potassium dichromate(VI). The orange dichromate ion, Cr₂O₇²⁻, is reduced to the green chromium(III) ion, Cr³⁺, so a colour change tells you that oxidation has happened. In equations the oxidant is written as [O].
| Alcohol | Conditions | Product | Colour change |
|---|---|---|---|
| Primary | warm gently, distil the product out at once | aldehyde | orange to green |
| Primary | heat under reflux with excess oxidant | carboxylic acid | orange to green |
| Secondary | heat under reflux | ketone | orange to green |
| Tertiary | warm | no reaction | stays orange |
For example, propan-1-ol gives propanal, then propanoic acid: CH₃CH₂CH₂OH + [O] → CH₃CH₂CHO + H₂O, and CH₃CH₂CHO + [O] → CH₃CH₂COOH. An aldehyde boils at a lower temperature than its alcohol (no hydrogen bonding between its molecules), so distilling it away as it forms stops it being oxidised further.
Ethanol (4.60 g) is oxidised by distillation to ethanal. The yield is 65 %. Find the mass of ethanal.
n(ethanol) = 4.60 / 46.0 = 0.100 mol. Ratio 1 : 1, so theoretical mass = 0.100 × 44.0 = 4.40 g. Actual mass = 0.65 × 4.40 = 2.86 g.
How can you tell an aldehyde from a ketone?
Both contain the carbonyl group C=O, which is polar because oxygen is more electronegative than carbon. An aldehyde has the group –CHO at the end of the chain; a ketone has C=O inside the chain. Only aldehydes are easily oxidised, because the C=O carbon of an aldehyde carries a hydrogen.
A carbonyl group is a carbon atom double-bonded to oxygen, C=O. In an aldehyde it ends the chain (–CHO); in a ketone it sits inside the chain.
| Test | Aldehyde | Ketone |
|---|---|---|
| 2,4-dinitrophenylhydrazine (2,4-DNPH) | orange precipitate | orange precipitate |
| Tollens’ reagent, warm | silver mirror | no change |
| Fehling’s solution, warm | blue to brick-red precipitate | stays blue |
2,4-DNPH detects the carbonyl group but cannot separate aldehydes from ketones. However, the precipitate can be purified and its melting point compared with a data table, which identifies the carbonyl compound.
Do not claim that a ketone gives a silver mirror or that 2,4-DNPH distinguishes aldehydes from ketones. Also remember that a tertiary alcohol does not turn dichromate green.
What happens when carbonyl compounds react with nucleophiles?
The carbon of C=O has a partial positive charge, so it is attacked by nucleophiles. Sodium borohydride, NaBH₄, acts as a source of H⁻ ions and reduces aldehydes to primary alcohols and ketones to secondary alcohols, for example CH₃COCH₃ + 2[H] → CH₃CH(OH)CH₃.
Hydrogen cyanide (made from KCN and acid) adds across C=O by nucleophilic addition. The cyanide ion attacks the carbon, the π electrons move to oxygen, and the oxygen then takes a proton to make a hydroxynitrile. The extra carbon from CN lengthens the chain, and the nitrile can later be hydrolysed to a hydroxy acid.
Nucleophilic addition to C=O: a nucleophile attacks the δ+ carbon, the π bond breaks, and the oxygen becomes O⁻ and is protonated. For an unsymmetrical carbonyl the product has a chiral carbon but forms as a racemic mixture with no optical activity.
Give the product of propanal and HCN and explain why it is optically inactive.
CH₃CH₂CHO + HCN → CH₃CH₂CH(OH)CN, 2-hydroxybutanenitrile. The central carbon has four different groups (H, OH, CN, C₂H₅), so it is chiral. The planar C=O is attacked equally from both faces, giving equal amounts of two enantiomers: a racemic mixture that does not rotate plane-polarised light.
Count the hydrogens on the carbon bearing OH: two means primary (aldehyde, then acid), one means secondary (ketone), zero means tertiary (nothing happens).
- Primary alcohol: aldehyde (distil) then acid (reflux); secondary: ketone; tertiary: no reaction
- Cr₂O₇²⁻ (orange) is reduced to Cr³⁺ (green)
- Tollens’ reagent: silver mirror with aldehydes only; Fehling’s: brick-red precipitate with aldehydes only
- 2,4-DNPH: orange precipitate with any aldehyde or ketone; melting point identifies the compound
- C=O carbon is δ+: nucleophilic addition of CN⁻ gives a hydroxynitrile as a racemic mixture
Oxidising alcohols and testing carbonyl compounds: frequently asked questions
Why do tertiary alcohols not oxidise?
The carbon bearing the OH has no hydrogen to remove, and oxidation would need a C–C bond to break. Under mild conditions the tertiary alcohol therefore stays unchanged and dichromate remains orange.
How do you stop a primary alcohol oxidising all the way to the acid?
Use a slight excess of alcohol and distil the aldehyde out as soon as it forms. The aldehyde has a lower boiling point because it cannot hydrogen-bond to itself.
What does Tollens’ reagent contain?
It is a solution of the diamminesilver(I) ion, [Ag(NH₃)₂]⁺, in alkali. An aldehyde reduces Ag⁺ to silver metal, which coats the tube as a mirror, while the aldehyde is oxidised to a carboxylate.
Why does a ketone resist oxidation by Tollens’ reagent?
A ketone has no hydrogen on the carbonyl carbon, so oxidation would have to break a C–C bond. Mild oxidants cannot do this.
Why is the product of HCN addition usually optically inactive?
The planar carbonyl group is attacked equally from above and below, giving equal amounts of two mirror-image products. Their rotations cancel in the racemic mixture.
