Which will be the major form among the two tautomeric forms?

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According to me, it should be enol form due to resonance stabilization as well has five membered ring formes througj hydrogen bonding. On the other hand, there is C=O, with greater bond energy in the keto form. Just want a confirmation.

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    $\begingroup$ yeah you are correct!! enol form is more stable. It has resonating structures, a fiver membered ring, hydrogen bonding. Thus making it stable $\endgroup$
    – Prakhar
    Commented Jan 2, 2017 at 18:48
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    $\begingroup$ The enol form has potential to form a giant delocalized pi system granted that everything is sp2 hybridyzed. $\endgroup$
    – turnt
    Commented Jan 3, 2017 at 3:33

1 Answer 1


Whether a diketo compound is present in solution in its diketo or its enol configuration is not something that can be said a priori — it strongly depends on the solvent and whether it is able to hydrogen bond or not.

If you have a solvent that is able to donate hydrogen bonds, the compound has a much greater tendency to adopt the diketo form since all its hydrogen bond accepting lone pairs are saturated with solvent hydrogen bonds. These can adopt a practically linear $\ce{O\bond{...}H-Solv}$ configuration which typically signifies a strong (or low energy) hydrogen bond.

The intramolecular hydrogen bond is rather strained, rather than having a bond angle $\angle (\ce{O\bond{...}H-X}) \approx 180^\circ$ the angle is approximately $108^\circ$. Therefore, it is a lot weaker. This type of hydrogen bond is more favourable if the solvent is not able to donate hydrogen bonds at all. In this case, the enol structure is also slightly more favourable due to the larger conjugated system.

The question can only definitely be answered by NMR studies; depending on what the structures look like a crystal structure may also help.


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