A primary alcohol is oxidised by acidified potassium dichromate(VI) in two stages: first to an aldehyde, then to a carboxylic acid. The product you get depends on the conditions.Making the aldehyde: distil it off as it forms
Use an excess of the alcohol and gently warm it with the oxidising agent. Aldehydes have lower boiling points than their alcohols, because they cannot form hydrogen bonds with each other. The aldehyde vaporises and is distilled off immediately as it forms, before it can be oxidised further.
CH3CH2OH + [O] → CH3CHO + H2O ethanol → ethanal Making the carboxylic acid: heat under reflux
Use an excess of the oxidising agent and heat under reflux. The aldehyde that forms cannot escape, so it is oxidised further to the carboxylic acid:CH3CHO + [O] → CH3COOH
Overall, using 2[O]:
CH3CH2OH + 2[O] → CH3COOH + H2O ethanol → ethanoic acid Worked example: give the products of oxidising propan-1-ol under each set of conditions.
Distilling off the product as it forms gives propanal, CH3CH2CHO.
Heating under reflux with excess oxidising agent gives propanoic acid, CH3CH2COOH:
CH3CH2CH2OH + 2[O] → CH3CH2COOH + H2O In both cases the orange dichromate(VI) turns green.
Use an excess of the alcohol and gently warm it with the oxidising agent. Aldehydes have lower boiling points than their alcohols, because they cannot form hydrogen bonds with each other. The aldehyde vaporises and is distilled off immediately as it forms, before it can be oxidised further.
Use an excess of the oxidising agent and heat under reflux. The aldehyde that forms cannot escape, so it is oxidised further to the carboxylic acid:
Overall, using 2[O]:
Distilling off the product as it forms gives propanal, CH3CH2CHO.
Heating under reflux with excess oxidising agent gives propanoic acid, CH3CH2COOH: