2-Pyridone tautomer ratio

Question

The following excerpts from different textbooks about keto–enol tautomerism in pyridones seem to be contradictory. Could someone explain which one is right?

MARCH’S ADVANCED ORGANIC CHEMISTRY pg-103 states that:

In many heterocyclic compounds in the liquid phase or in solution, the keto form is more stable, although in the vapor phase the positions of many of these equilibria are reversed. For example, in the equilibrium between 4-pyridone (149) and 4-hydroxypyridine (150), 149 is the only form detectable in ethanolic solution, while 150 predominates in the vapor phase. In other heterocycles, the hydroxy-form predominates. 2-Hydroxypyridone (151) and pyridone-2-thiol (153) are in equilibrium with their tautomers, 2-pyridone 152 and pyridine 2-thione 154, respectively. In both cases, the most stable form is the hydroxy tautomer, 151 and 153.

Now for the final few lines I think the result is same for both solvents that's why he didn't mention anything about it.

Organic chemistry 2e by Clayden, pg-729 states that:

In fact, 2-hydroxypyridine prefers to exist as the ‘amide’ because that has the advantage of a strong C=O bond and is still aromatic. There are two electrons in each of the C=C double bonds and two also in the lone pair of electrons on the trigonal nitrogen atom of the amide. Delocalization of the lone pair in typical amide style makes the point clearer

but doesn't mention any solvents.

Answer 1

Why not start with the corresponding entry in Wikipedia as suggested in the comment by @Maurice? There, it discerns the solid state

«The predominant solid state form is 2-pyridone. This has been confirmed by X-ray crystallography which shows that the hydrogen in solid state is closer to the nitrogen than to the oxygen (because of the low electron density at the hydrogen the exact positioning is difficult), and IR-spectroscopy, which shows that the C=O longitudinal frequency is present whilst the O-H frequencies are absent.»

from the one of the liquid/in solution:

«The tautomerization has been exhaustively studied. The energy difference appears to be very small. Non-polar solvents favour 2-hydroxypyridine whereas polar solvents such as alcohols and water favour the 2-pyridone.

The energy difference for the two tautomers in the gas phase was measured by IR-spectroscopy to be $2.43$ to $\pu{3.3 kJ/mol}$ for the solid state and $\pu{8.95 kJ/mol}$ and $\pu{8.83 kJ/mol}$ for the liquid state.»

Don't forget $\Delta{}G = -RT \ln{}K$ when accessing the primary references both sections point to, too.

There equally are databases about keto-enol equilibria. The authors of Tautobase for example deposit the extracted data into a public repository on GitHub. 1680 pairs may comfortably read and searched e.g., by (sub)structure with DataWarrior. In the present case, the equilibrium tabulated lists for 2-pyridone in water an estimated $\log{}K = 3.5$ (while the gaseous state is listed with $\log{}K = -0.4$):

Note related earlier posts about this topic here on chemistry.se, like this, or this, too.

References:

Wahl, O.; Sander, T. Tautobase: An Open Tautomer Database. J. Chem. Inf. Model. 2020, 60, 1085-1089; doi 10.1021/acs.jcim.0c00035.

Sander, T.; Freyss, J.; von Korff, M.; Rufener, C. DataWarrior: An Open-Source Program For Chemistry Aware Data Visualization And Analysis. J. Chem. Inf. Model. 2015, 55, 460-473; doi 10.1021/ci500588j.

Answer 2

One factor to consider is the role of hydrogen bonding.

In the gas phase at low pressure there is no hydrogen bonding because the molecules are too isolated; then the hydroxyl form with a pure cyclic conjugation = maximum aromaticity is more stable. In the neat liquid or in a polar solvent, the keto form will have stronger hydrogen bonding due to the carbonyl group being more electron-rich, thus a better proton-acceptor, and the N-H hydrogen with a positive charge on nitrogen acting as a better proton donor for the bond.