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I was looking into the exam paper of last year when I found this:

Propose arrow-pushing mechanism for the reaction below.

Reaction to be explained

I suppose it's a Wagner–Meerwein rearrangement. (Sorry for the dirty mechanism graph.)

Proposed Wagner-Meerwein rearrangement

Is the answer correct? How would I explain the stereochemical configuration of the reaction? And why doesn't the elimination (the leaving of proton) happen at carbon A, as the cation intermediate is just as stable as the others?

I tried drawing chair conformation graph of the reaction. I marked the possibly important groups in red for convenience.

chair conformation of reactant and product

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  • $\begingroup$ Have you tried drawing the 5 rings in their chair conformations? $\endgroup$ – NotEvans. Aug 6 '16 at 6:51
  • $\begingroup$ @NotWoodward Thank you for the reply! chair conformation graph added. The structure of reactant did inspire me the difference of conformation in the fifth ring, but further explanation didn't come to me yet. $\endgroup$ – Amo Aug 6 '16 at 8:33
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You are correct about the beautiful chain of eight Wagner-Meerwein rearrangements. The stereochemistry of each of these is given by the fact that this arrangement is suprafacial and proceeds under retention, thus a group on the top of the molecule will stay above the ring plane and a group below will stay below. During the rearrangement process, since each intermediate cation is planar, the rings A to D (from where the hydroxy group was to where the double bond will be) will undergo a ring flip each, preserving the general trans-decalin structure for each pair of rings.

Once the final rearrangement has occurred, we arrive at that position in the molecule which features a cis-decalin system: note how in your drawing of the starting material the final ring E is not aligned with all the others but pointing downward. Assuming that final proton, which is eliminated, would actually migrate, we would generate the final carbocation at the cis-decalin junction. The methyl group between the eliminating proton and carbon A is in the wrong orientation to migrate; the carbon belonging to ring E has a much greater tendency because it is better aligned (anti to the proton). That migration would lead to a 6,5-spiro compound and a trisubstituted double bond. That isn’t bad in itself, but the pathway asked for in the question leads to a trans-decalin system by elimination, which features much less strain than a spiro compound. Additionally, a tetrasubstituted double bond is also more stable.

To arrive at the all-trans-polcyclic system, the proton on the junction of rings D and E must eliminate which it is happy to do. Ring D flips, which now turns the cis-decalin system of DE into a trans-decalin system, ring strain is released and the reaction terminates.

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