Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Organic Chemistry Forum for Graduate Students and Professionals => Topic started by: azmanam on July 27, 2009, 07:49:02 AM
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1000th Post!
QUESTION: The authors began with chiral epoxide A and obtained aldehyde B following the conditions below. Provide a mechanism that accounts for this stereo-destructive process. You do not have to explain the double bond geometry in the product. (Bonus: name the named reactions in the mechanism)
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Maybe my proposition is stupid, but... let's try.
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Definitely not stupid. I'm just tickled that people are reading and attempting :)
Your first and last steps are correct. The ones in the middle are not. The silyl group next to an oxygen atom is kinda set up for a particular named reaction. That'll be my hint for now.
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Does it go via a sigmatropic shift?
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The authors do not propose the 6-pi sigmatropic shift, but I like that guess. I think theres a kinetically faster pathway from the alkoxide. Sigmatropics usually require heat (if i'm not mistaken) but BuLi reactions are typically zero or below.
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I thought this could be plausible. :)
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It seems like Brook Rearrangement.
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Yup, Brook, retro-Brook it is. Well done.
http://dx.doi.org/10.1021/jo961265s
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How did the TMS manage to survive - I was under the impression that TMS was fairly acid labile (or am I getting a bit confused with the ether?)
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TMS is rather acid sensitive. They note the loss of silyl group (to give the CH3 capped rearranged product) if workup is not carefully controlled.
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I had few concerns about the answer provided ???
1. The authors of the paper 'propose' the mechanism; but I do not see much of the evidance presented in support of the mechanism. My objection is regarding the step of retro-Brook rearrangement. Under the reaction conditions used I can not understand how that step is thermodynamically feasible [as the driving force for brook rearrangement is higher Si-O bond energy over Si-C].
To me, intermediate 9 (according to the article) looks themodynamically more stable than 10 and 9 should be stable enough to get protonated.
2. Retro-Brook rearrangements I found in literature were all using some other catalyst which was coupling a thermodynamically favorable reaction to make the net :delta:G negative.
3. In the question posted, formation of the other isomer (11, in the article) was not mentioned which I thought could be important in guessing the mechanism.
Thanks.
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I think another factor here is the resonance stability of the alkoxide after the retro Brook. That C-O bond now has significant double bond character through resonance, so you'll have to take that stability into account when considering thermodynamics. And while many retro Brook rearrangements might need catalysis to proceed, perhaps the resonance stability conferred on the molecule after rearrangement is significant in determining the overall stability of the intermediate?
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I think another factor here is the resonance stability of the alkoxide after the retro Brook. That C-O bond now has significant double bond character through resonance, so you'll have to take that stability into account when considering thermodynamics. And while many retro Brook rearrangements might need catalysis to proceed, perhaps the resonance stability conferred on the molecule after rearrangement is significant in determining the overall stability of the intermediate?
I have old eyes. Could you draw that resonance structure for me.
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:)
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eh, too many negative charges.. but you get the idea.