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I encountered the following equilibrium constant values for the keto-enol interconversions of phenol in Organic Chemistry (Solomons, Fryhle, Snyder, 3rd Edition): Keto-Enol Phenol Tautomers

I wonder what could explain the differences (though I'm not sure whether they're significant) between the equilibrium constants of the two keto-enol interconversions wherein in one case the double bond of carbon with the oxygen and the hydrogens are oriented in ortho position and in para position in the other.

Edit: Can this difference concern the medium the reaction is catalyzed in or directive influence of groups? Do explanations for equilibrium constant differences exist for other tautomer interconversions or are such queries quite speculative?

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  • $\begingroup$ Consider what is happening during the tautomerism. More so, consider the movement of the proton during the tautomerism and its position relative to the oxygen. $\endgroup$ – LigninPauling Apr 25 at 10:41
  • $\begingroup$ @LigninPauling By your hint, the high opposite charge separation in the 10^11 Keq tautomer and the relative positions of O and Hs on the unsaturated carbons are factors for equilibrium towards the left-hand tautomer. $\endgroup$ – Kartik Apr 25 at 11:47
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I believe that's because in the first step, phenol regains its aromaticity which makes the $K_{eq}$ substantially ($1000$ times) larger than that in the second step where it loses its aromaticity.

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  • $\begingroup$ Isn't phenol already aromatic? And phenol is non-aromatic in both its keto forms; I didn't get where phenol 'regains' its aromaticity. Do you want to say that in the interconversion with the higher equilibrium constant (cyclohexa-2,4-dienone) the reaction proceeds such that phenol regains its aromaticity? $\endgroup$ – Kartik Apr 10 at 5:37
  • $\begingroup$ @Kartik What I'm saying is that phenol is aromatic in its enol form but when you perform the keto conversions we have to break a double bond which makes it lose aromaticity but when you perform the reverse, we re-establish its aromatic properties making it more stable and thus the $K_{eq}$ value is larger. This is also why we hardly ever write phenol in its keto form. $\endgroup$ – Sameer Thakur Apr 10 at 5:41
  • $\begingroup$ Ohhkay. But how does that explain the 10^3 equilibrium constant difference between cyclohexa-1,3-dienone and cyclohexa-2,4-dienone interconversions with phenol's enol? (The two keto isomers are formed in different interconversions.) The double bond is broken in both cases and the phenol is re-formed in similar mechanisms (I guess it's something to do with directive influence or the like). $\endgroup$ – Kartik Apr 10 at 5:47

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