# oxidation of ethanol

The full equations for the formation of ethanoic acid (Experiment 1) and ethanal (Experiment 2) are
shown below.
Experiment 1: 3CH 3 CH 2 OH + 16H + + 2Cr 2 O 7 2– 3CH 3 COOH + 11H 2 O + 4Cr 3+
Experiment 2: 3CH 3 CH 2 OH + 8H + + Cr 2 O 7 2– 3CH 3 CHO + 7H 2 O + 2Cr 3+
Explain how the different quantities of Na 2 Cr 2 O 7 .2H 2 O and different reaction conditions allow different
organic products to be formed. Density of ethanol = 0.78 g cm –3
In experiment 1 ethanoic acid is being formed because ethanol is being distilled with acidified potassium dichromate, so ethanol will be oxidised to ethanoic acid it won't be further oxidised to ethanal since the vapour won't condense into the flask to be oxidised.

In experiment 2 ethanal is being formed because ethanol is being refluxed with acidified potassium dichromate, so ethanol will be oxidised to ethanal because the condensation can fall back into the flask for further oxidation.

I'm not really sure how different quantities of Na2Cr2O7 would affect it maybe having more Na2Cr2O7 would allow for further oxidation?

I hope this helps
Original post by SpanishGonkey
In experiment 1 ethanoic acid is being formed because ethanol is being distilled with acidified potassium dichromate, so ethanol will be oxidised to ethanoic acid it won't be further oxidised to ethanal since the vapour won't condense into the flask to be oxidised.

In experiment 2 ethanal is being formed because ethanol is being refluxed with acidified potassium dichromate, so ethanol will be oxidised to ethanal because the condensation can fall back into the flask for further oxidation.

I'm not really sure how different quantities of Na2Cr2O7 would affect it maybe having more Na2Cr2O7 would allow for further oxidation?

I hope this helps

Wrong way round.

Distillation gives the aldehyde, further oxidation gives the carboxylic acid.
Original post by charco
Wrong way round.

Distillation gives the aldehyde, further oxidation gives the carboxylic acid.

Thanks for correcting me