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Topic: Oxygen replacement (esters)  (Read 2986 times)

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Offline Rutherford

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Oxygen replacement (esters)
« on: November 12, 2012, 12:50:25 PM »
This is about oxygen replacement. For example:
CH3COOH+CH3OH ::equil:: CH3COOCH3+H2O
a) Why is the oxygen atom from the alcohol on the ester? Why didn't the previous one remain in that position?

If the ester gets hydrolised by NaOH:
CH3COOCH3+NaOH ::equil:: CH3COONa+CH3OH
b)Why does the oxygen atom remain now in the acid (salt)?

Offline Wastrel

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Re: Oxygen replacement (esters)
« Reply #1 on: November 12, 2012, 01:03:30 PM »
The first follows logically from the mechanism.  The second... looks broken.

Offline Rutherford

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Re: Oxygen replacement (esters)
« Reply #2 on: November 13, 2012, 07:24:17 AM »
Why is it so in the mechanism?
How do you mean broken? I found this in a problem where it was asked will the isotope of oxygen from Na18OH be in the alcohol or in the acid salt after the hydrolysis.

Offline easyorganic

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Re: Oxygen replacement (esters)
« Reply #3 on: November 13, 2012, 10:49:30 AM »
I usually teach acyl substitution by way of a "cheater mechanism".  While this is not the actual mechanism, it does provide an easier way for students to predict products.  You should consult your textbook for the complete actual mechanisms.


Offline Rutherford

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Re: Oxygen replacement (esters)
« Reply #4 on: November 13, 2012, 11:10:38 AM »
Okay, I can accept this. What about question b)?

Offline easyorganic

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Re: Oxygen replacement (esters)
« Reply #5 on: November 14, 2012, 09:34:36 AM »
It goes by similar mechanism.  Of course you need to consider the acid/base properties of the acidic product in a basic solution.

Offline fledarmus

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Re: Oxygen replacement (esters)
« Reply #6 on: November 14, 2012, 09:47:27 AM »
It goes by similar mechanism.  Of course you need to consider the acid/base properties of the acidic product in a basic solution.


Which is what Wastrel meant by "broken". This mechanism would suggest that the oxygen should remain with the methyl group in both reactions, which I believe is in fact the case, but is not what is shown in part b above.

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