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HELP oxidation of alcohols question

In my chemistry hw there is a question which I'm confused about.

There is a substance gemeprost, i.e.


The question asks:

Draw the product when gemeprost is reacted with:
a) acidified K2Cr2O7 under mild conditions (product distilled immediately).
b) excess acidified K2Cr2O7 and heated under reflux.

However, both of the hydroxyl groups on the compound are secondary, so they can only be oxidised to ketones. so why does it make a difference whether the dichromate solution is in excess or not or whether the product is distilled/heated under reflux? no further oxidation can happen anyway.

shouldn't the products in both situations be the same? or am i missing something??
Original post by ♥Samantha♥
In my chemistry hw there is a question which I'm confused about.

There is a substance gemeprost, i.e.


The question asks:

Draw the product when gemeprost is reacted with:
a) acidified K2Cr2O7 under mild conditions (product distilled immediately).
b) excess acidified K2Cr2O7 and heated under reflux.

However, both of the hydroxyl groups on the compound are secondary, so they can only be oxidised to ketones. so why does it make a difference whether the dichromate solution is in excess or not or whether the product is distilled/heated under reflux? no further oxidation can happen anyway.

shouldn't the products in both situations be the same? or am i missing something??


maybe there is oxidative cleavage of the beta-diketone? :dontknow:
Original post by charco
maybe there is oxidative cleavage of the beta-diketone? :dontknow:


idk what that even means. is that AS?
Original post by ♥Samantha♥
idk what that even means. is that AS?


Uh uh, I thought this was University level.

In that case you are correct - they are both secondary and should not oxidise further. :colone:
Original post by charco
Uh uh, I thought this was University level.

In that case you are correct - they are both secondary and should not oxidise further. :colone:


lol no im 16:colondollar: ok i'll put the same answer for both and try clarify it with my teacher before it's due in just in case.
Original post by ♥Samantha♥
In my chemistry hw there is a question which I'm confused about.

There is a substance gemeprost, i.e.


The question asks:

Draw the product when gemeprost is reacted with:
a) acidified K2Cr2O7 under mild conditions (product distilled immediately).
b) excess acidified K2Cr2O7 and heated under reflux.

However, both of the hydroxyl groups on the compound are secondary, so they can only be oxidised to ketones. so why does it make a difference whether the dichromate solution is in excess or not or whether the product is distilled/heated under reflux? no further oxidation can happen anyway.

shouldn't the products in both situations be the same? or am i missing something??


To oxidise a secondary alcohol into a ketone you need to heat under reflux, you cant get the ketone by distillation. So it might be that (a) could be the same, and (b) is the ketone. It seems like the question is meant to catch you out so you get mixed up between the oxidation of primary and secondary alcohols, I might be wrong though.
Original post by Louiseelg0rt
To oxidise a secondary alcohol into a ketone you need to heat under reflux, you cant get the ketone by distillation. So it might be that (a) could be the same, and (b) is the ketone. It seems like the question is meant to catch you out so you get mixed up between the oxidation of primary and secondary alcohols, I might be wrong though.


so do u think in the first one nothing should happen?
Original post by ♥Samantha♥
so do u think in the first one nothing should happen?


Yes because you can only get a ketone when
1) You have the reagent of potassium dichromate(VI) solution and dilute sulphuric acid (i.e. acidified potassium dichromate)
2) heat under reflux

Distillation won't cause it to oxidise. :smile:
Original post by Louiseelg0rt
Yes because you can only get a ketone when
1) You have the reagent of potassium dichromate(VI) solution anddilute sulphuric acid (i.e. acidified potassium dichromate)
2) heat under reflux

Distillation won't cause it to oxidise. :smile:


ok so my class notes agree that you have to heat under reflux. however i don't know why.
Original post by ♥Samantha♥
ok so my class notes agree that you have to heat under reflux. however i don't know why.


Distillation is simply to separate the product of the oxidation and prevent further oxidation.

It works with both aldehydes and ketones, although as you have said there is little chance of the ketone being further oxidised.
I dont have a clue whats going on :smile:
Original post by charco
Distillation is simply to separate the product of the oxidation and prevent further oxidation.

It works with both aldehydes and ketones, although as you have said there is little chance of the ketone being further oxidised.


so won't i still get the ketone if i distill?
Original post by Fez_Shabbir
I dont have a clue whats going on :smile:


thanks for ur helpful comment
Original post by ♥Samantha♥
so won't i still get the ketone if i distill?


yes, ketones distill at a lower boiling temperature than the corresponding alcohol...
Original post by charco
yes, ketones distill at a lower boiling temperature than the corresponding alcohol...


So will the product of part a) have the alcohols oxidised, the same as part b)?
Original post by ♥Samantha♥
So will the product of part a) have the alcohols oxidised, the same as part b)?


I would say so.
Um this is AS....
Original post by ♥Samantha♥
thanks for ur helpful comment

Welcome:cheers:

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