Why is enol content higher in non-polar solvents?

Apparently, for keto-enol tautomerism, the enol content of a given carbonyl compound increases when in a non-polar solvent.

What is the reason for this?

• From my brief search in some textbooks, the matter seems to be much more complicated than your question suggests. Solvent definitely has an important effect on tautomerism. However, this is dependent not only on the polarity, but also on the structure of the carbonyl compound, hydrogen bonding, etc. A rather involved account is given in J. Org. Chem. 1985, 50, 1216; slightly more accessible is Reichardt's Solvent Effects in Organic Chemistry, 3rd ed. (p 106ff.). Oct 8 '17 at 0:03
• Seconding orthocresol. Also, I note that the close votes are given for ‘unclear’ which doesn’t make sense at all as this question is sufficiently clear.
– Jan
Oct 8 '17 at 16:25
• Naive guess: higher dipole moment in carbonyl group compared to enol. Probably more complicated though as suggested in comments above. Enol=hbond donor while keto=hbond acceptor so it would be reasonable to assume that e.g. keto favored in hdonor solvent and enol favored in hacceptor solvent. Just some ideas. Oct 15 '17 at 14:04
• I doubt I am not good as others ,but comparing this thing with cyclic structure is completely unfair because that includes the strain as well as ortho para effect @orthocresol what I feel is the main and strong reason is the hydrogen bonding Aug 12 '20 at 8:43
• @logicalx2 Keto group is a poorer H bond donor ($\beta = 5.3$) than alcohol which is also a donor ($\beta = 5.8$) Apr 26 at 20:28

The type of solvent used is generally not major factor in determining the enol content formed in an keto-enol tautomerism. Generally it is overshadowed by other factors like aromaticity and hydrogen bonding.

But this being said doesn't mean that solvent has no role to play in the enol content. As you mentioned a polar medium prefers the keto form whereas a non-polar medium prefers the enol form.

This is because in a polar solvent, the lone pairs will be already involved in hydrogen bonding with the solvent, making them less available to hydrogen bond with the enol form and thus taking away from the stability of the enol form.

Whereas in a non-polar medium there is no hydrogen bonding with the solvent and thus it is preferable for the molecule to attain an enol form to gain stability via hydrogen bonding.

Sometimes this can cause drastic effects. For example, in benzene, the enol form of 2,4-pentanedione predominates in a 94:6 ratio over the keto form, whereas the numbers almost reverse completely in water.

To quote from Solvent effects on the tautomeric equilibrium of 2,4-pentanedione[1]:

For solvents such as $$\ce{CCl4}$$ and cyclohexane the intramolecular bond of the enol form persists and bulk solvent effects account for the equilibrium enol–keto content. In solvents such as DMSO, disruption of the intramolecular bond occurs and the percentage of enol falls due to unfavorable entropy changes. The enol intramolecular bond is disrupted by the solvents water and methanol. Enol hydrogen bond formation through self-association and with the solvent accounts for the entropy changes upon enolization in these solvents.

Reference:

(1) Spencer, J. N.; Holmboe, E. S.; Kirshenbaum, M. R.; Firth, D. W.; Pinto, P. B. Solvent Effects on the Tautomeric Equilibrium of 2,4-Pentanedione. Can. J. Chem. 1982, 60 (10), 1178–1182.