Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Organic Chemistry Forum for Graduate Students and Professionals => Topic started by: azmanam on November 02, 2009, 11:23:01 AM
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G'morning all, Like last week, I encourage you to discuss amongst yourself before I let you know if it's right or not. Not only is part of the fun learning new mechanisms and being exposed to new reactions, but also learning to evaluate strengths and weaknesses of postulated mechanisms. Tactfully, of course :)
QUESTION: Provide a complete arrow pushing mechanism for the following reaction. Predict product and stereochemistry.
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Something like this?
I still haven't predicted the configuration at one of the new stereogenic centres though...
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Hmmm...
How's this for an idea?
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I generally agree with Dan, but:
1. Should the elimination of the silyl group be so complicated? I think the resulted product is definitely an E-alkene due to its stability over the Z-alkene.
2. Two chair-like TSs exsit so the products are a pair of enatiomers. It's the first time that I see the word "dr" and I'm not very sure the meaning of it. Does it mean that the pair of enationmers：their diastereomers = 10：1? I don't think so and just becuse of this the chair-like TSs might not right.
Wating for azmanam to say somthing.
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Two chair-like TSs exsit so the products are a pair of enatiomers.
I think the two products arising from the two chair TSs only vary at one stereogenic centre, so they are diastereoisomers, specifically epimers, not enantiomers - dr = diastereomeric ratio. I think the other possible chair TS will be less favoured, as it places the chain bearing the large silyl group in a (pseudo)axial position (and the methyl equatorial) - and this TS generates the minor diastreoisomer.
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I don't want to get too involved, but I really don't think a cyclic transition state is involved here!
Think about other opportunities for organizing the transition state in the addition step.
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I don't want to get too involved
C'mon! Get involved! That's the name of the game!
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Thanks for your explanation, Dan. The chair-like TSs were also my initial idea, but now I doubt whether it's right, for neither of them shows definite higher stability over ther other. Just see the picture.
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I like your (collective) thinking, and that double bond will be an important component in the reaction... but not at first. I think the hint I'll give is one commonly thrown around at our group meetings when discussing mechanisms involving TMSOTf. TMS is a (relatively) small Lewis acid. We often say that it's acceptable to think of the TMS group as a big proton.
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Sounds interesting! But this afternoon I'll have a test and after that I'll be back.
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:)
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Exhausted. The problem seemed so simple, but actually not only the reacton process but also the stereochemistry are soo complicated!
The ester group's carbonyl oxygen is the most electron-rich atom which should react with TMSOTf fisrt. Then the aldehyde acts as a nucleophilic reagent, which seems so strange, but not impossible. In order not to make the aldehyde uselss, I supposed the aldehyde be a nucleophilic reagent and the subsequent process seemed rather rational so my guess might be right. The stereochemistry of the 5-membered cycle formation is just like that of C-C bond formation in last week' problem.
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neither azmanam nor dan want to say something? i must go to bed soon
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C'mon! Get involved! That's the name of the game!
Hehe - I am pretty certain that I know the answer so I don't want to steal all the fun from the others. :D
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You managed to get the right product... but through the most roundabout way. (and actually, the methyl ester stays intact in the product, so that part of the mechanism is wrong). Whether or not the ester oxygen is a better Lewis base than the aldehyde oxygen, the carbonyl carbon of the aldehyde is more electrophilic and more reactive than the carbonyl carbon of the ester. Thus, the alcohol OTMS (treated as OH) will form the oxocarbenium ion with the aldehyde as the first part of the mechanism.
Interestingly, the diastereoselectivity changes as a) the geometry of the double bond changes from Z to E, and b) the stereochemical relationship between the OTMS and the SiR3 changes from anti to syn. For example, in the E case with a syn relationship, the BOAT transition state is actually favored, due to favorable overlap of the C-Si σ bond with the C-C π* orbital as Dan suggested. We'll call this one complete. Movies, care to add anything? Did you work on this project or in the lab?
The reaction is from the total synthesis of bistramide A:
http://dx.doi.org/10.1021/ol901801h
Background papers, including stereochemical outcome of all 4 alkene/diastereochemical geometries:
http://dx.doi.org/10.1021/ja002087u
http://dx.doi.org/10.1021/ol050982i
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:)
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And I don't understand the " for E/syn case". Should the C-Si σ bond overlap with the C-C π* orbital when the the 6-memebered cycle is being formed? And is the boat-like TS necessary? The cycle can form through chair-like TS and a carbon cation is generated on the alpha-position of TMPS group, then TMPS group is eliminated. (I means that the cycle formation and the elimination of the TMPS group doesn't happen simutaneously)
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I disagree with your statement that an aldehyde oxygen atom is a better nucleophile than ROTMS. Redraw with H+ instead of TMS+. In that scenario, oxocarbenium formation might be easier to see.
And yes, the orbitals need to overlap in the transition state leading to the tetrahydropyran ring. Any time bonds break, it is because electrons have been pumped into a corresponding antibonding orbtial. For the reaction we described, the orbital alignment requirement is met in the chair-like transition state. In the E/syn case, the chair-like transition state has the C-Si σ bond perpendicular to the C-C π* orbital. As a result, there is no favorable overlap, and the THP forming step does not occur. Only through the boat-like transition state is the orbital overlap requirement met and the reaction can occur.
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:)