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.