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This is the mechanism given in my lecture handout:

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I have a problem with the third species along. Surely the negatively charged oxygen would protonate before the other oxygen would. Also, why does this reaction work at all? The amine group is more basic than the "alcohol group" that must leave to form the amide so why wouldn't that protonate first and then just leave (i.e no reaction)?

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  • $\begingroup$ In mechanism we write equilibria which lead to observed product. $\endgroup$ – Mithoron Apr 16 '15 at 19:29
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    $\begingroup$ Protonation / deprotonation by solvent is fast as H+ is just a subatomic particle and is therefore far nimbler than any other ion. Bonds between larger atoms are made and broken slower, because they have more mass and the baggage of inner electron shells, which makes the transition more unwieldy. Any and all reasonably conceivable protonations will exist in fast equilibrium. The rate determining step will either be 1-->2 or 3-->4 (probably the latter, due to the low concentration of species 3 for the reasons you indicate.) $\endgroup$ – Level River St Apr 16 '15 at 22:34
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Surely the negatively charged oxygen would protonate before the other oxygen would.

Most of the time, but not all of the time.

The amine group is more basic than the "alcohol group" that must leave to form the amide so why wouldn't that protonate first and then just leave (i.e no reaction)?

Most of the time, but not all of the time.

Your reasoning is correct, but remember, all of these steps are equilibria. Although what you say will happen most of the time, some of the time the reaction will take the forward route. Presumably the reaction is run with a large excess of ammonia to "force" the reaction in the forward direction.

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    $\begingroup$ Wouldn't protonation of the solvent (which is probably also $\ce{NH3}$) be quicker and hence a more justified mechanism than the proposed one? $\endgroup$ – Jori Apr 16 '15 at 17:15
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    $\begingroup$ @Jori Do you mean proton transfer to the solvent by the first tetrahedral intermediate instead of proton transfer to the alkoxy leaving group? $\endgroup$ – ron Apr 16 '15 at 17:20
  • $\begingroup$ Yes that is what I meant. $\endgroup$ – Jori Apr 16 '15 at 22:04
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    $\begingroup$ @Jori They are all in equilibrium, so they're probably all happening. Protonation of the alkoxy group has the advantage of 1) being intramolecular and 2) making the alkoxy a better leaving group. But who knows what the exact pathway for the proton transfer is? $\endgroup$ – ron Apr 16 '15 at 22:09
  • $\begingroup$ Another way to explain the forward reaction is to remember that attacking an ester is easier than attacking an amide; or that amides are thermodynamically favoured compared to esters. $\endgroup$ – Jan Oct 24 '15 at 23:21

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