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Topic: esters from amides?  (Read 3726 times)

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Offline polymerator

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esters from amides?
« on: December 27, 2009, 09:30:50 PM »
Apparently it is not possible to make esters from amides (and an alcohol) according to my book.
But why can't you do it under acidic conditions just like the hydrolysation of an amide?

Protonate carboxyl group, nuc attact from alcohol w. proton loss =>
tetrahedral intermediate
Then
protonate N, loss of amine, protonation of lost amine (drives reaction forward), then regeneration of the carboxyl group.

Offline orgopete

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Re: esters from amides?
« Reply #1 on: December 28, 2009, 12:29:25 AM »
That will work provided it is anhydrous HCl and alcohol.
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Offline polymerator

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Re: esters from amides?
« Reply #2 on: December 28, 2009, 01:22:14 PM »
Alright, thank you very much for your swift reply  :)
Though, I still wonder why this isn't considered an option in my text books. NH3 should be a better leaving group than RO- considering the pKaHs (9 vs. 16). I realize that ROH is a rather bad nucleophile, but is it really that much worse than water? Is there a solvent problem?

Offline orgopete

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Re: esters from amides?
« Reply #3 on: December 29, 2009, 07:54:54 AM »
Not all reactions are in textbooks and not all reactions in textbooks are general reactions, so no importance should be given for an absence per se. An absence of a reaction can mean the reaction fails, for example para nitration of methyl benzoate as the major product. Basically, one shouldn't put too much importance if a reaction is not in a textbook. 

Reactions in which water or an alcohols are nucleophiles are controlled by le Chatlier's principle and not by a kinetic superiority of either. In an amide to ester con :larrow:version, a similar argument can be made. While an amine is a stronger base, hence poorer leaving group, it is also the stronger base in the tetrahedral intermediate. Therefore, the nitrogen will be protonated to a greater extent than the oxygen. A neutral alcohol is the leaving group on acid and not an alkoxide. An ammonium ion giving the free amine is the better leaving group because there will be less oxonium ion than ammonium ion in the intermediate.

In acid, the amine leaving group will be protonated to from an ammonium salt, thus tying up the amine from competitive nucleophilic attack. 
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Offline stewie griffin

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Re: esters from amides?
« Reply #4 on: December 29, 2009, 02:29:00 PM »
The reaction probably isn't listed b/c reactions with anhydrous HCl and alcohols aren't too general. Yeah it would work, but what else is present in your staring materials that can stand up to HCl? Usually there's some other functional groups present that would preclude you from stirring with strong acids. That's my guess as to why it's not listed. But as orgopete said, don't put too much stock into whether or not your book lists it.

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