Which compound will have the most stable enol form?

Question

Which compound will have the most stable enol form, Phloroglucinol or 4-Pyridone?

My Thoughts:

Both in their enol form are aromatic. I am not sure how do I compare their stabilities. The answer given is 4-Pyridone though.

Any help would be appreciated.

Answer 1

The most questions about keto-enol tautomerism can be answered by their NMR spectral data (e.g., evidence of existing enol-form of acetylacetone at room temperature). According to its $\ce{^1H}$-NMR spectral data, it was declared that phloroglucinol, in its neutral form, exists nearly exclusively in its aromatic form (Ref.1). However, their dianion form (e.g., disodium phloroglucinolate) favors keto structures. These results were further analysed and confirmed by kinetic studies (Ref.2).

Wikipedia states that:

4-Pyridone is an organic compound with the formula $\ce{C5H4NH(O)}$. It is a colorless solid. The compound exists in equilibrium with a minor tautomer, pyridin-4-ol.

Martin - マーチン has given here a nice answer using computational evidence for this preference of 4-Pyridone (data are from Ref.3).

Based on these, it is clear that in their neutral forms, phloroglucinol has the most stable enol form, existing in predominantly in phenol form (1,3,5-trihydroxybenzene).

References:

1. R. J. Highet, T. J. Batterham, “The Structure of the Phloroglucinol Dianion,” J. Org. Chem. 1964, 29(2), 475-476 (https://doi.org/10.1021/jo01025a501).
2. Martin Lohrie, Wilhelm Knoche, “Dissociation and keto-enol tautomerism of phloroglucinol and its anions in aqueous solution," J. Am. Chem. Soc. 1993, 115(3), 919-924 (https://doi.org/10.1021/ja00056a016).
3. Peter I. Nagy, Giuliano Alagona, Caterina Ghio, “Theoretical Investigation of Tautomeric Equilibria for Isonicotinic Acid, 4-Pyridone, and Acetylacetone in Vacuo and in Solution,” J. Chem. Theory Comput. 2007, 3(4), 1249–1266 (https://doi.org/10.1021/ct6002252).

Answer 2

In case of pholoroglucinol, its ketoform has 3 $\ce{C=O}$ bond which make its keto form more stable as compared to the keto form of 4-pyridone which has to offer only one $\ce{C=O}$.

Answer 3

On the subject of 4-pyridone, Wikipedia says this:

4-Pyridone is an organic compound with the formula C 5H 4NH(O). It is a colorless solid. The compound exists in equilibrium with a minor tautomer, pyridin-4-ol.

So as far as I know, the aromatic enol form of this guy is lesser in equilibrium content than that of phloroglucinol.

The reason for this is somewhat related to intermolecular H-bonding. An abstract from this paper says:

The N−H···O H-bonding enthalpy between 4-pyridones connected in a chain of H-bonds can achieve 23 kcal/mol for the most central H-bonds, while that between two 4-pyridones is 9.90 kcal/mol based upon DFT calculations on the counterpoise-corrected potential energy surfaces. That the range of enthalpies for N−H···O H-bonds can vary from as little as 2 to as much 23 kcal/mol depends primarily upon the polarizability of whatever internally connects the N−H and CO within the H-bonding molecule, which are two parallel −CC− entities in 4-pyridone. The contribution of covalent or charge-transfer interactions between the π-systems of adjacent 4-pyridones is small.

The following image shows how such a chain may exist

(Red- oxygen, blue- nitrogen, grey- carbon, white- hydrogen)

Hence,the enol content of phloroglucinol would be more than that of 4-pyridone

Also, an interesting point to note about phloroglucinol as well(as per this paper) is that for the neutral compound, the keto tautomer is undetectable spectroscopically. Upon deprotonation, the keto form predominates, possibly due to the conjugation present in the sigma complex formed(but the reasoning is just an educated guess)