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By applying the logic that phenol exists in enol form rather than keto form as it attains aromatic character, why is it not the case in coumarin?

structure of coumarin

keto and enol form of coumarin

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    $\begingroup$ Which enol form would you suggest? Try drawing it out! That may give you your answer. $\endgroup$ – Jan Sep 10 at 15:12
  • $\begingroup$ If it was amide, the question would make sense... $\endgroup$ – Mithoron Sep 10 at 15:39
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    $\begingroup$ Oxygen is pretty electronegative, and so it doesn't really "like" to be positively charged. Also in general it takes energy to separate charges, so zwitterionic forms aren't favored in general, especially without any proton transfers. $\endgroup$ – Curt F. Sep 10 at 16:56
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    $\begingroup$ There simply is no tautomer possible. There is no keto group in that compound to begin with, and there are no enolisable protons either. What you have drawn are two resonance structures. Both of them are valid configurations, but neither actually exist; see What is resonance, and are resonance structures real? On that note, I don't think the answer by @CurtF. has anything to do with it. $\endgroup$ – Martin - マーチン Sep 11 at 9:06
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    $\begingroup$ The key point (I think) is that if atom nuclei (including hydrogen/protons) don't move between two structures, they aren't tautomers, but are different resonance structures. Then the question is which resonance structure is "closer" to the real structure. Perhaps my comment explains why the left structure is closer to reality, but the key point is that the analogy to phenol tautomers is invalid since here we are dealing with resonance structures instead of tautomers. $\endgroup$ – Curt F. Sep 11 at 16:45
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The answer is as simple as it will be unhelpful to your original question: There is no tautomer possible.

Coumarin does not have a keto group, and there are no enolisable protons either. Coumarin consists of a benzene ring, connected to a heterocycle with the functional group lactone. The molecule is flat and the conjugation extends throughout it.
What you have drawn are two resonance structures. Both of them are valid configurations (there are more), but neither actually exist; see What is resonance, and are resonance structures real? In both of your drawn structures the nuclei positions (in the clamped nuclei approximation) do not move.

The concept which applies here is completely different from keto-enol-tautomrie, where there are two different, observable molecules.

The question which you are probably asking is: Which of the drawn structures is the major contributor to the (so-called) resonance hybrid.

I have run a quick calculation of the DF-B97D3(BJ)/def2-SVP//GFN2-xTB level of theory, which suggests that the conjugation is skewed towards the lactone group. More investigations would be necessary, but the π orbitals should give you a general idea.

pi MO of coumarin (DF-B97D3(BJ)/def2-SVP//GFN2-xTB)

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  • $\begingroup$ You have correctly pointed out what my question should be ! So, can the explanation for skewness towards the lactone form would be Curt's explanation in the earlier comment ? $\endgroup$ – Abhigyan Sep 12 at 13:39
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    $\begingroup$ @Abhigyan Essentially yes. Have a look at the Becke dipol-corrected partial charges of coumarin (DF-B97D3(BJ)/def2-SVP//GFN2-xTB); they clearly indicate that both oxygen atoms will accumulate electron density. Conjugation in this regard rather withdraws electron density from the adjacent ring. Note that this are quite crude simplifications, further study should be undertaken. $\endgroup$ – Martin - マーチン Sep 12 at 13:59

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