Aldehydes and ketones both contain the C=O group. Oxidise a primary alcohol and you get an aldehyde, oxidise a secondary alcohol and you get a ketone, and a reducing agent such as NaBH4 takes you straight back down again.
Aldehydes and ketones are carbonyl compounds: both contain C=O. In an aldehyde (RCHO, names ending in -al) the C=O is at the end of the chain with an H attached; in a ketone (RCOR′, names ending in -one) it is bonded to two carbons. A primary alcohol is oxidised by acidified K2Cr2O7 to an aldehyde, which must be distilled off as it forms or it is oxidised further to a carboxylic acid (as happens under reflux). A secondary alcohol gives a ketone, which resists further oxidation. NaBH4 (or LiAlH4) reduces aldehydes back to primary alcohols and ketones back to secondary alcohols.
Build aldehydes and ketones in the simulator, check the C=O position and name each one. Compare their boiling points with the alcohols and alkanes of similar Mr.
| Compound | Formula | Type | Mr | b.p. / °C |
|---|---|---|---|---|
| butane | CH3CH2CH2CH3 | alkane | 58 | −1 |
| propanal | CH3CH2CHO | aldehyde | 58 | 49 |
| propanone | CH3COCH3 | ketone | 58 | 56 |
| propan-1-ol | CH3CH2CH2OH | alcohol | 60 | 97 |
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Which compound is a ketone?
How is ethanal obtained, rather than ethanoic acid, when ethanol is oxidised?
What is formed when propanone is reduced by NaBH4?
Why does propanal (Mr 58) boil lower than propan-1-ol (Mr 60)?
Climb the oxidation ladder in the simulator with a primary and a secondary alcohol. Switch the apparatus between distillation and reflux, then run the reduction with NaBH4.
| Change | Reagent and conditions | Equation |
|---|---|---|
| ethanol to ethanal | acidified K2Cr2O7, warm, distil off at once | CH3CH2OH + [O] → CH3CHO + H2O |
| propan-2-ol to propanone | acidified K2Cr2O7, heat under reflux | CH3CH(OH)CH3 + [O] → CH3COCH3 + H2O |
| ethanal to ethanol | NaBH4 in water or methanol (or LiAlH4 in dry ether) | CH3CHO + 2[H] → CH3CH2OH |
| propanone to propan-2-ol | NaBH4 | CH3COCH3 + 2[H] → CH3CH(OH)CH3 |
An aldehyde has the C=O at the end of the chain, a ketone in the middle. Distillation gives the aldehyde, reflux the acid. Ketones are not oxidised by acidified dichromate, so a ketone never goes on to an acid. In equations, oxidation uses [O] and reduction uses 2[H]; NaBH4 reduces C=O but does not reduce C=C.
Unlocks once the four checks above are done. Worth more XP, written to AS Paper 1 and Paper 2 standard.
Which reagent converts butanone into butan-2-ol?
Which alcohol is oxidised to a ketone by acidified dichromate(VI)?
What mass of propan-2-ol (Mr 60.0) forms when 1.16 g of propanone (Mr 58.0) is completely reduced?
Which equation shows the reduction of ethanal?
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