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Author Topic: Problem of the Week - 11/2/09  (Read 5354 times)

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azmanam

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Problem of the Week - 11/2/09
« on: November 02, 2009, 05:23:01 AM »

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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Dan

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Re: Problem of the Week - 11/2/09
« Reply #1 on: November 02, 2009, 10:09:30 AM »

Something like this?

I still haven't predicted the configuration at one of the new stereogenic centres though...
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Dan

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Re: Problem of the Week - 11/2/09
« Reply #2 on: November 02, 2009, 10:30:34 AM »

Hmmm...

How's this for an idea?
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Heory

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Re: Problem of the Week - 11/2/09
« Reply #3 on: November 03, 2009, 01:59:48 AM »

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.
« Last Edit: November 03, 2009, 02:10:23 AM by Heory »
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Dan

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Re: Problem of the Week - 11/2/09
« Reply #4 on: November 03, 2009, 05:11:58 AM »

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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movies

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Re: Problem of the Week - 11/2/09
« Reply #5 on: November 03, 2009, 11:12:54 AM »

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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Dan

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Re: Problem of the Week - 11/2/09
« Reply #6 on: November 03, 2009, 12:00:32 PM »

I don't want to get too involved

C'mon! Get involved! That's the name of the game!
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Heory

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Re: Problem of the Week - 11/2/09
« Reply #7 on: November 03, 2009, 02:20:41 PM »

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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azmanam

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Re: Problem of the Week - 11/2/09
« Reply #8 on: November 03, 2009, 02:31:01 PM »

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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Heory

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Re: Problem of the Week - 11/2/09
« Reply #9 on: November 03, 2009, 02:38:42 PM »

Sounds interesting! But this afternoon I'll have a test and after that I'll be back.
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Heory

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Re: Problem of the Week - 11/2/09
« Reply #10 on: November 04, 2009, 01:53:33 AM »

 :)
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Heory

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Re: Problem of the Week - 11/2/09
« Reply #11 on: November 04, 2009, 02:21:41 AM »

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.
« Last Edit: November 04, 2009, 02:52:45 AM by Heory »
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Heory

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Re: Problem of the Week - 11/2/09
« Reply #12 on: November 04, 2009, 05:22:48 AM »

neither azmanam nor dan want to say something? i must go to bed soon
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Re: Problem of the Week - 11/2/09
« Reply #13 on: November 04, 2009, 08:36:37 AM »

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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azmanam

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Re: Problem of the Week - 11/2/09
« Reply #14 on: November 05, 2009, 02:37:59 PM »

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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