Topic 10 · page 21 of 27

Oxidation of alcohols

Acidified potassium dichromate(VI) oxidises a primary alcohol to an aldehyde and then to a carboxylic acid, a secondary alcohol to a ketone, and a tertiary alcohol not at all. The class of the alcohol decides the outcome.

13 min read10.1810.18(i)10.18(ii)10.18(iii)

Read first: Naming and classifying alcohols

The oxidising agent 10.18

The oxidising agent is potassium dichromate(VI), K₂Cr₂O₇, in dilute sulfuric acid. The orange dichromate ion, Cr₂O₇²⁻, is reduced to the green chromium(III) ion, Cr³⁺. In organic equations the oxidising agent is written [O].

The oxidation of an alcohol removes hydrogen from the carbon bonded to –OH (and from the O–H). What happens next depends on how many hydrogens that carbon had.

Primary alcohols: aldehyde, then carboxylic acid 10.1810.18(i)10.18(ii)

A primary alcohol has two hydrogens on the C–OH carbon, so it can be oxidised twice.

Three rows of skeletal formulae. Primary: propan-1-ol, then [O], then propanal with C=O at the end of the chain, then [O], then propanoic acid with C=O and OH at the end of the chain. Secondary: propan-2-ol, then [O], then propanone with C=O in the middle, and a note that it stops here. Tertiary: 2-methylpropan-2-ol, then [O], then the words no oxidation. — What dichromate(VI) does to each class of alcohol. The C–OH carbon and the new C=O are drawn in orange.

Stage 1: aldehyde. The alcohol loses two hydrogens (one from O–H and one from the carbon) and an aldehyde, –CHO, forms:

\ce{CH3CH2CH2OH + [O] -> CH3CH2CHO + H2O}

Stage 2: carboxylic acid. The aldehyde gains an oxygen atom and becomes a carboxylic acid, $\text{–COOH}$ :

\ce{CH3CH2CHO + [O] -> CH3CH2COOH}

The overall change from alcohol to acid is:

\ce{CH3CH2CH2OH + 2[O] -> CH3CH2COOH + H2O}

Compound	Functional group	Name ending	Example
primary alcohol	–CH₂OH	-ol	propan-1-ol
aldehyde	–CHO	-al	propanal
carboxylic acid	$\text{–COOH}$	-oic acid	propanoic acid

Secondary alcohols: ketones 10.1810.18(iii)

A secondary alcohol has one hydrogen on the C–OH carbon. After it is oxidised there is no hydrogen left on that carbon, so the ketone cannot be oxidised further by dichromate.

\ce{CH3CH(OH)CH3 + [O] -> CH3COCH3 + H2O}

A ketone has a C=O inside the chain, joined to two carbon atoms. Its name ends in -one: propanone, butanone. An aldehyde has the C=O at the end of the chain, so it has a hydrogen on that carbon.

Tertiary alcohols: no oxidation 10.18

A tertiary alcohol has no hydrogen on the C–OH carbon. It cannot lose one, so there is nothing to oxidise by this route, and the solution stays orange. 2-methylpropan-2-ol gives no reaction with acidified dichromate(VI).

Worked exampleName the product

Name the organic product, if any, when each alcohol is warmed with acidified potassium dichromate(VI) and the product is distilled out as it forms, for (a) butan-1-ol, (b) butan-2-ol and (c) 2-methylpropan-2-ol.

(a) Butan-1-ol is primary and the product is distilled out, so it stops at the aldehyde: butanal, CH₃CH₂CH₂CHO.
(b) Butan-2-ol is secondary: butanone, CH₃COCH₂CH₃.
(c) 2-methylpropan-2-ol is tertiary: no reaction.

Check yourself

Name the oxidising agent and describe the colour change.
Write the equation for the oxidation of ethanol to ethanal, using [O].
Write the equation for the oxidation of propan-2-ol, and name the product.
Explain why 2-methylpropan-2-ol is not oxidised.
Write the overall equation for propan-1-ol being oxidised all the way to the carboxylic acid.
Give the class of the alcohol and the product: butan-2-ol with acidified dichromate(VI).

Answers cover these until you have tried the questions

Potassium dichromate(VI) in dilute sulfuric acid. The colour changes from orange to green.
$\ce{CH3CH2OH + [O] -> CH3CHO + H2O}$
$\ce{CH3CH(OH)CH3 + [O] -> CH3COCH3 + H2O}$ . The product is propanone, a ketone.
It is a tertiary alcohol. The C–OH carbon has no hydrogen atom, so it cannot lose one, and it is not oxidised.
$\ce{CH3CH2CH2OH + 2[O] -> CH3CH2COOH + H2O}$
Secondary alcohol. The product is butanone, CH₃COCH₂CH₃.