[xshadow]

Hi

I 've tried to write some mechanism for a  primary-amide hydrolysis and I've had some doubts while I was writing that mechanism


Here it is:



The point is that when  an -OH group of the  tetrahedral intermediate is protonated it will be a  good leaving  group:

-The other -OH can make a double bond and help the exit of -H2O⁺(the alcoholic group protonate)in order to get the amide (the reagent,and then losing its proton)

- but also the -NH2 could make a double bond with the central carbon also,forming an imine..in a similar way

This is not possible if I have an -NR2


Is this possibile?
Thanks


Ps: I forgot an Hydrogen bonded to the iminic azote

[spirochete]

Your mechanism ends at a tautomer of an amide called an imidic acid. It is basically a reasonable mechanism, but there is very little of the tautomer formed because the amide is much more stable in most contexts. The water itself has little to do with forming the imidic acid (or at least it is not needed to write a mechanism. Your mechanism is still reasonable). You can form the imidic acid just with acid/base catalyst, similar to keto/enol tautomerization. Gain proton, then lose proton, if the conditions are acidic.

In H3O+, the equilbrium gets driven to the carboxylic acid by protonation of the amine leaving group. Products are carboxylic acid + NH4+.

[xshadow]

So theoretically  it can exist but is very instable compared to amide and is usually "neglected"

Thanks

[spirochete]

Yes that's right. I think in the greater scope of chemistry, it's important to think about things like this. In heterocycles such as nitrogenous bases of DNA/RNA it also favors C=O form, but the results were not so obvious. Watson and Crick got it wrong initially, and another chemist corrected them. But on a typical introductory exam, you probably don't get points for creativity. They just want the right answer.

[xshadow]

Yes that's right. I think in the greater scope of chemistry, it's important to think about things like this. In heterocycles such as nitrogenous bases of DNA/RNA it also favors C=O form, but the results were not so obvious. Watson and Crick got it wrong initially, and another chemist corrected them. But on a typical introductory exam, you probably don't get points for creativity. They just want the right answer.

Thanks!!

Is was strange that a lots of sites or textbooks don't say anything about this possibility....or at least consider this possibility and saying that it could happen but is an unstable  product.