Why a doubly-fused cyclohexanone is formed instead of bridged cyclohexanone like product which would be more flexible (entropically-favoured)?

Also, the OH group not being at the bridgehead carbon unlike former compound, shouldn't it have a better chance of forming stable alkenone after dehydration as the preferred product formed in Robinson annulation? The following image elaborates my doubt.

Doubt regarding the possible alternative solution to Robinson annulation

Baldwin rule for enolate (source : https://en.wikipedia.org/wiki/Baldwin%27s_rules#Rules_for_enolates)

As per the Baldwin rule for enolate, the enol-endo reaction and enol-exo reaction is favoured for both exo-tet and exo-trig for six/seven-membered chain (from the excerpt is provided as above image).

Both the alternatives follow six-membered exo-trig which is favoured according to above Baldwin rule. (Refer to the image where the carbon chain of both the alternatives is numbered.) The bridged-bicycle compound has 3 faces, where each face has a 6-membered ring when it was visualized through ChemDraw 3D. So, it seems both the alternatives as the product of Robinson annulation is favoured according to modified Baldwin rules.

  • $\begingroup$ What is "better dehydration?" $\endgroup$ – Zhe May 9 '20 at 22:02
  • $\begingroup$ more preferable dehydration mode', hope that helps $\endgroup$ – Swastik May 9 '20 at 22:07
  • $\begingroup$ First of all, the reaction is usually conducted in protic solvent with base. The alkoxides all get protonated. The fused beta-hydroxy ketone gives facile beta-elimination. The protonated, bridged beta-hydroxy ketone cannot eliminate in base (Bredt's Rule). Acid is needed to form the bridged unsaturated ketone. Baldwin's Rules apply to competitive reactions, i.e., which pathway is kinetically faster. $\endgroup$ – user55119 May 9 '20 at 23:08
  • $\begingroup$ One issue is that the drawing on the left does not depict the correct location of the oxygen. It should be in an axial position. Therefore, successive protonation would lead to very facile elimination. $\endgroup$ – Zhe May 10 '20 at 2:20

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