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# Ester exchange

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1. Poly(ethanol) is produced in an ester exchange reaction between poly(ethenyl ethanoate) and methanol

How can this work, though? I thought that in an ester exchange reaction an alcohol group swaps with another alcohol group. Here the alcohol and carboxyllic acid group are swapping??
2. The alcohol groups are swapping with each other. The original alcohol group is the polymer chain, and that is getting replaced with methanol, to leave the alcoholic polymer.
3. (Original post by illusionz)
The alcohol groups are swapping with each other. The original alcohol group is the polymer chain, and that is getting replaced with methanol, to leave the alcoholic polymer.
I don't think this is right, look:
Attached Images

4. (Original post by ilovemath)
I don't think this is right, look:
It is right. Do you know the mechanism for this reaction?
5. (Original post by illusionz)
It is right. Do you know the mechanism for this reaction?
no it is not in my book
i looked for another explanation online but could not find it
6. (Original post by ilovemath)
no it is not in my book
i looked for another explanation online but could not find it
An ester is analagous to an acyl chloride with an OR group instead of a Cl, and they are less reactive. What would the reaction between methanol and an acyl chloride be?
7. (Original post by illusionz)
An ester is analagous to an acyl chloride with an OR group instead of a Cl, and they are less reactive. What would the reaction between methanol and an acyl chloride be?
well I would eliminate the Cl from the acyl and the H from methanol so the link would be
...C-O-C..
8. (Original post by ilovemath)
well I would eliminate the Cl from the acyl and the H from methanol so the link would be
...C-O-C..
So, draw out your polymer. The O-Polymer bit is your OR of the ester, attack the C=O with your methanol, push the arrows in exactly the same way and you should get the product.
9. (Original post by illusionz)
So, draw out your polymer. The O-Polymer bit is your OR of the ester, attack the C=O with your methanol, push the arrows in exactly the same way and you should get the product.
so ester exchange and transesterfication are different then, right?
10. (Original post by ilovemath)
so ester exchange and transesterfication are different then, right?
I've never heard of the term ester exchange before, but they seem to be the same thing. This reaction is transesterification.
11. (Original post by illusionz)
So, draw out your polymer. The O-Polymer bit is your OR of the ester, attack the C=O with your methanol, push the arrows in exactly the same way and you should get the product.
I have now included arrows and it seems wrong, sorry
Attached Images

12. (Original post by ilovemath)
I have now included arrows and it seems wrong, sorry
If the R is your polymer chain, can you not see that they are the same?

NB some of your arrows are a little bit dodgy!
Attached Images

13. (Original post by illusionz)
If the R is your polymer chain, can you not see that they are the same?

NB some of your arrows are a little bit dodgy!
I am a little confused by the lower part of the diagram
first where did the O+ come from? second what is the wierd shape to the left of the O+??
14. (Original post by ilovemath)
I am a little confused by the lower part of the diagram
first where did the O+ come from? second what is the wierd shape to the left of the O+??
Redrawn more clearly.

The O becomes positive as it now has three bonds, and the 'weird shape' is a H3C-CO drawn in shorthand
Attached Images

15. (Original post by illusionz)
Redrawn more clearly.

The O becomes positive as it now has three bonds, and the 'weird shape' is a H3C-CO drawn in shorthand
this is hand drawn by me
is this correct (It looks like it is but just wanna double check
Attached Images

16. (Original post by illusionz)
Redrawn more clearly.

The O becomes positive as it now has three bonds, and the 'weird shape' is a H3C-CO drawn in shorthand
also what would happen in a reaction like

CH3COOC2H5 + HCOOH

how could I do this as I have TWO carbonyl groups?
17. (Original post by ilovemath)
this is hand drawn by me
is this correct (It looks like it is but just wanna double check
Yes although strictly in the step where you lose the H+ it would be removed by a solvent (probably water molecule), rather than just falling off.

The mechanism is probably a bit beyond your course, and in reality alcohols are not very good nucleophiles so you would use either an acid catalyst which protonates the C=O of the ester and makes it more reactive (like an acid chloride), or a base which deprotonates the alcohol and makes that more reactive (better nucleophile).

Still, the best way to understand the reaction is to understand the mechanism
18. (Original post by ilovemath)
also what would happen in a reaction like

CH3COOC2H5 + HCOOH

how could I do this as I have TWO carbonyl groups?
I'm not quite sure what reaction you're trying to do! An ester will not react with a simple carboxylic acid.
19. (Original post by illusionz)
I'm not quite sure what reaction you're trying to do! An ester will not react with a simple carboxylic acid.
oh ok, my revision guide says it will but no mechanism made sense....
anhydries, I presume they just split at the O=C-C=O

e.g: (CH3CO)2O

Having never studied them before in depth I don't know if the two groups have to be the same
if not then how do I know which "part" takes the H bit of the alcohol and which bit takes the O-R bit from the alcohol
20. (Original post by ilovemath)
oh ok, my revision guide says it will but no mechanism made sense....
anhydries, I presume they just split at the O=C-C=O

e.g: (CH3CO)2O

Having never studied them before in depth I don't know if the two groups have to be the same
if not then how do I know which "part" takes the H bit of the alcohol and which bit takes the O-R bit from the alcohol
What sort of product did the revision guide arrive at?

For an anhydride:

Full size: http://i39.tinypic.com/2i97221.png

The bits in circles show you where the atoms in the products come from.

The attacking species ends up in the ester, part of the anhydride is lost to form the acid.

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