Can ketones next to a nitrogen hetero atom tautomerise?

I worry that the electronegative N will stabilise keto form so much that equilibrium will be so far towards keto it would be unreasonable to call the enolisiation process a factor


[![enter image description here](https://i.stack.imgur.com/yYskO.png)](https://i.stack.imgur.com/yYskO.png)




If I understand your question correctly, I think you're asking if an amide can enolize like a ketone. The answer to that question is **yes**. 


![enter image description here](https://i.stack.imgur.com/nvUAl.jpg)


The enol content of most organic compounds is very low and special methods must be used to measure the enol content. Perhaps the question becomes, "**is there more or less enol content** in an amide than in an ester, for example"? 


We can draw resonance structures for an amide (or ester) showing that the nitrogen stabilizes the amide. But above we also drew resonance structures showing how the nitrogen can also stabilize the enol. We can draw similar structures for both an ester and its enol as well. Although in the ester case we wind up putting a positive charge on oxygen in some of the resonance structures, so they might not count for as much as they do in the amide case - but both the starting ester and its enol would both be affected similarly. So as a first approximation I might guess that the enol content in an ester and an amide would be similar.


![enter image description here](https://i.stack.imgur.com/Nlf9Y.jpg)


Indeed the enol content in malonamide is similar to the enol content in diethyl malonate (see page 34 [here](http://etheses.dur.ac.uk/5282/1/5282_2719.PDF?UkUDh:CyT))


![enter image description here](https://i.stack.imgur.com/wiRsF.jpg)


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With a nitrogen adjacent to a carbonyl moiety, the functional group is an amide, and the tautomer involving a $\ce{C=N}$ bond is called an imidic acid. The amide is usually vastly more stable and typically predominates by many orders of magnitude. This can be rationalized by several considerations. First, the two tautomers differ by the exchange of a $\ce{C=O}$ bond with a $\ce{C=N}$ bond and an $\ce{N-H}$ with an $\ce{O-H}$, but the $\ce{C=O}$ bond is particularly strong, making the amide enthalpically more stable. Second, if you consider the resonance structures (i.e., mesomers) available to the two respective tautomers, the amide has a contributing structure in which a negative formal charge is placed on the more electronegative oxygen atom, whereas the corresponding structure in the imidic acid has the charge placed on a less electronegative nitrogen atom. Granted, resonance is not a real physical process, but examination of these structures gives some idea of the distribution of electron density in the frontier orbitals, which too points to favorable increase in stability for the amide. You can apply similar logic to the case of tautomers involving a $\ce{C=C}$ bond, where the ultimate conclusion should be the same (and the difference in stability between tautomers even greater).


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