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Following up the question Formation of a stable hydrate. I understand that for cyclopropanone the change in hybridization is the reason for stabilization of hydrate since it relieves the angle strain.

So, can this be further be given as a reason to state the hydrate of 2‐indanone (structure below) will be stable? I suppose upon hydration the angle will change from 120° to 109.8°, bringing the angle closer to that of a pentagon that is 108°.

1,3-dihydro-2H-inden-2-one

I am confused if this is appropriate or if it's true. If it's wrong, please explain.

P.S. After going through the comments I understood by considering keto-enol tautomerisation that hydration might not at all be favourable.

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    $\begingroup$ The cyclopentane ring is not perfectly flat. $\endgroup$ – Karl Sep 15 '20 at 6:17
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    $\begingroup$ Need to remember that this compound will be more extensively enolised $\endgroup$ – Waylander Sep 15 '20 at 7:51

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