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In the section on acid-catalysed hydrolysis of nitriles, this webpage has the following line of reasoning;

...the more stable amide tautomer predominates the (following) equilibrium.

enter image description here

(credit)

How are the relative stabilities of these tautomers compared? I made up a reason of the resonance structure for the amide one having negative and positive charges on oxygen and nitrogen respectively, and vice versa for the imidic acid one, but I'd like clarification on that.

There's also another hand-wavy thing I thought about of the double bond being stable on the more electronegative atom, oxygen. Is that correct, too?

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    $\begingroup$ For future reference: If there is an illustration you refer to, take a moment to include it here on ChemSE altogether with a note of credit. The reason: Perhaps the address of the webpage you refer to changes (or is closed for whatever reason). This would yield your question less easy to access, than necessary. Regardless if «amdie» looks like a typo, or not. $\endgroup$ – Buttonwood Apr 28 at 11:50
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The relative stability of the two tautomers (two isomers differing how protons are bound) may be assessed from the equilibrium constant you may formulate for the two. Deriving from above illustration, it would be

enter image description here

to indicate that this is dynamic, with a back and forward reaction at microscopic level. Once formulated as such, you probably recall $\Delta{}_RG = -RT \ln{}K$ to relate between the sample temperature $T$, the universal gas constant $R$, the free reaction enthalpy $\Delta{}_RG$ and the equilibrium constant $K$.

Depending on the substrate, you may use spectroscopic evidence (e.g., $\ce{^1H}$-NMR spectroscopy) to assess the concentration of the two forms. Equally, based on prior work by others, there are databases to either list these values, or provide a prediction (i.e., an educated guess) about $K$. Knowing these equilibria is important to anticipate the likelihood of chemical reactions. It equally offers insight regarding the pharmaceuticals and their uptake, anabolism, interaction with the body (metabolism), and catabolism and eventual elimination, too.

One of the databases available to the public is Tautobase (2020JChemInfModel1085). It is implemented in the freely available DataWarrior program allowing you a search-by-(sub)structure. With some luck, one of the 1680 pairs matches your substrate closely enough, too. In the simplest case one may imagine, however, the amide is strongly favoured:

enter image description here

The database backs this particular entry with the publications 1998JAmChemSoc10359 and 2001JAmChemSoc2681.

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  • $\begingroup$ +1, Thanks for the amazingly well-researched answer. But I was asking about the reasons you make up to explain the observation that we already know, not the values that tell you which the preferred tautomer is themselves. Could you clarify whether my guesses at these reasons are valid? $\endgroup$ – harry Apr 28 at 12:29
  • $\begingroup$ I understood your question «How are the relative stabilities of these tautomers compared?» as a search on how experimental evidence is collected. $\endgroup$ – Buttonwood Apr 28 at 12:32
  • $\begingroup$ @HarryHolmes Neglecting influence by solvent / reaction conditions, for the isolated molecule the total energy including a carbonyl $\ce{C=O}$ may be a different one compared to $\ce{C=C}$ (as in keto-enol equilibria), or $\ce{C=N}$ (as for the case of imidic acid). It would be nice if there is a second perspective including this aspect (and maybe tabulated data like these), too. $\endgroup$ – Buttonwood Apr 28 at 12:56
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    $\begingroup$ Last picture shows different type or tautomerism. $\endgroup$ – Mithoron Apr 28 at 13:10
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    $\begingroup$ @HarryHolmes True, the screen photo is about the putative equilibrium about formamide and amino-ethenol. For the current version of DataWarrior (April 2021) and tautobase (2020), this is the most similar pair with a formide-like pattern. The pair with an imidic acid equally has an amino-ethenol on the other side of the equation. For the energies, I hoped one of the quantum-chem savy readers either a) could point to a compilation, or b) offer an insight / result of computation. $\endgroup$ – Buttonwood Apr 28 at 22:27

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