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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.

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    $\begingroup$ Wikipedia states that : X-ray crystallography 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 $\endgroup$
    – Maurice
    Dec 31, 2021 at 16:46
  • $\begingroup$ That's an improvement, although you could have cleaned up the awkward new lines when copying from a PDF, as well as the reference numbers in March which are not really helpful when the associated references aren't provided. I've done that for now. Can you add in book titles, editions, and page numbers please. A proper citation should, at the very least, contain sufficient information to find the quoted text. Finally, on-topic, I suspect that the Clayden extract may be a bit simplified. These things can be quite solvent-dependent, AFAIK, and March says as much. $\endgroup$ Dec 31, 2021 at 18:05
  • $\begingroup$ March didn't mention anything about which solvent is used in the tautomerism I asked, same with clayden. But clayden's reasoning seems correct to me though I don't know if March has errors $\endgroup$
    – Math-Wiz
    Dec 31, 2021 at 20:09
  • $\begingroup$ "2-Hydroxypyridone (151) and pyridone-2-thiol (153)" are wrong in the book. They should be 2-Hydroxypyridine (151) and pyridine-2-thiol (153). $\endgroup$
    – Josiah_H
    Jan 1, 2022 at 18:24
  • $\begingroup$ Ig that is a small error of i and o and could be noticed from diagram, but yeah that doesn't answer my doubt $\endgroup$
    – Math-Wiz
    Jan 1, 2022 at 19:05

2 Answers 2

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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 $\log{}K = -0.4$):

enter image description here

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.

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  • $\begingroup$ March states that "In other heterocycles, the hydroxy-form predominates. 2-Hydroxypyridone and pyridone-2-thiol are in equilibrium with their tautomers, 2-pyridone and pyridine 2-thione , respectively. In both cases, the most stable form is the hydroxy tautomer." By which I think he means that solvent has no effect on this as in the line before he stated how solvent affected the other tautomerism and in this he said that it dominates $\endgroup$
    – Math-Wiz
    Jan 1, 2022 at 6:47
  • $\begingroup$ The same database (again, estimated for an aqueous solution) lists for pyridine-2-thiol vs. the thione a $\log{}K = 4.6$ (i.e., thione is the favoured form, too), while pyridine-2-amine is favoured as amine (listed as $\log{}K = -6.5$). Maybe, on occasion, the database's listing of $\log{}K$ - while text books tend to use $pK_a = - \log{}K$ - might pass unnoticed. In line with the database and against «2-Hydroxypyridone and pyridone-2-thiol [...] the most stable form is the hydroxy tautomer» is e.g., Baran's yearly hetchem lecture (video, 25:47). $\endgroup$
    – Buttonwood
    Jan 1, 2022 at 7:34
  • $\begingroup$ Doesn't that video say that keto form is more stable? why you're saying hydroxy or did I get you wrong $\endgroup$
    – Math-Wiz
    Jan 1, 2022 at 11:21
  • $\begingroup$ Likely you got me wrong when I used a quote of yours. $\endgroup$
    – Buttonwood
    Jan 1, 2022 at 13:07
  • $\begingroup$ Didn't get you sir, the video you linked says keto form better due to c=o etc..as mentioned in clayden so which one is right, it didn't mention about solvents ig $\endgroup$
    – Math-Wiz
    Jan 1, 2022 at 17:23
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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.

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  • $\begingroup$ If its given in a question where solvent is not mentioned (I have never seen solvent being given when asked to compare keto-enol forms) then which one to mark, I mean which is the general state where we consider tautomerism? $\endgroup$
    – Math-Wiz
    Jan 1, 2022 at 18:03
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    $\begingroup$ «Neat liquid» is synonym to «the pure compound itself» (i.e., no solvent added); and then (for 2-hydroxypyridine vs 2-pyridone), the form of the pyridone is dominant over the other. $\endgroup$
    – Buttonwood
    Jan 2, 2022 at 18:23

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