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I am having a problem generating the correct tautomer for 1-(propan-2-yl)-1,2,3,5-tetrahydro-6H-pyrazolo[3,4-b]pyridin-6-one. I am told by my collaborator that the one I am generating is wrong.

I am not able to figure out why the structure below is the wrong tautomer.

Keto form of 1-(propan-2-yl)-1,2,3,5-tetrahydro-6H-pyrazolo[3,4-b]pyridin-6-one

Loong
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srinivasan ranga
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3 Answers3

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In keto-enol tautomerism, the equilibrium is mostly towards the keto form, as it should be, since $\ce{C=O}$ is a very strong and stable bond. In some cases however we find that the equilibrium sometimes shifts to enol form. This happens in cases where some additional benefit is done by the enol structure. here are two examples:

A) Removing anti-aromaticity or adding aromaticity to a ring.

B) Extended conjugation formation, providing additional stability.

C) (1,3) dicarbonyls (due to combined effect of H-Bonding and cyclization)

enter image description here

In the structure you have shown, the enol form is more stable because it adds aromatic character to the ring. This additional stability drives the equilibrium towards the enol form.

napstablook
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Maybe the original structure contained the motif of 2-hydroxypyridine. If this guess is correct then a 2-pyridone is one reasonable tautomer, as shown in the first line of the illustration below. And it seems plausible to draw such an isomer for a molecule like 1H-pyrazolo[3,4-b]pyridin-6-ol, too.

enter image description here

Buttonwood
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  • upon generating tautomers using online chemaxon tool, i got this : – srinivasan ranga May 24 '21 at 11:50
  • @srinivasanranga I wasn't yet aware about this generator you possibly refer to. Thank you for this hint. If so, however, pasting the first structure of 1H-pyrazolo[3,4-b]pyridin-6-ol, the program displays a preference for the pyridone motiv ($\ce{C=O}$ and $\ce{NH}$). It is not so clear for me how the program quantifies the results down to a hundredths %, but it is really nice that it lists more than only one alternative. – Buttonwood May 24 '21 at 12:07
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I would say the enol form is favored because it establishes aromaticity of the benzene ring but the fused five-membered ring does not tautomerize as all its other atoms are sp3-hybridized (and therefore unable to do so).

anomeric
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    This assigns much energetic stabilization to aromaticity. This indeed is, comparing cyclohexene, cyclohex-1,3-diene, and (a hypothetic) cyclo-1,3,5-triene against benzene a valuable tool. Note, comparing the equilibria for i) 2-hydroxypyridine vs. 2-pyridone in water, and ii) 2-aminopyridine vs. its imine in water suggest to equally consider hydrogen bonding. According to the Tautobase paper and its public SI (included e.g., in DataWarrior the $\log K$ is +3.5 for the former, and -6.5 for the later. – Buttonwood May 25 '21 at 20:40