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Topic: Chrysanthemic acid scheme  (Read 10272 times)

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Offline Rutherford

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Re: Chrysanthemic acid scheme
« Reply #15 on: February 16, 2013, 01:38:34 PM »
That's what bothers me. I can't think of the reaction  :-\.
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Offline discodermolide

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Re: Chrysanthemic acid scheme
« Reply #16 on: February 16, 2013, 01:48:56 PM »
Yes, I didn't think you would get it. OK, the sulfone makes the CH2 group next to it quite acidic, so you can generate a carbanion which attacks what? In a named reaction. I must say that I think the yield of this may be quite low. Also note you have the ethyl ester and are using methoxide as a base so it will transesterify to give product F as the methyl ester. If you can't figure it out I'll post my scheme of what I think goes on.
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Offline Rutherford

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Re: Chrysanthemic acid scheme
« Reply #17 on: February 16, 2013, 02:17:54 PM »
I don't see a way to combine those three compounds  ???.
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Offline discodermolide

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Re: Chrysanthemic acid scheme
« Reply #18 on: February 16, 2013, 02:22:22 PM »
Two compounds. As I said the base abstracts a proton from the CH2 of the sulfone, this adds in  michael reaction to the ester K, the intermediate enolate attacks the carbon carrying the sulfone forming the three membered ring and eliminating the sulfone. Have a look at my scheme.
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Offline Rutherford

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Re: Chrysanthemic acid scheme
« Reply #19 on: February 16, 2013, 02:24:27 PM »
Shouldn't an ethyl ester be obtained?
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Offline discodermolide

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Re: Chrysanthemic acid scheme
« Reply #20 on: February 16, 2013, 02:27:24 PM »
Using NaOMe, I would expect transesterification, if they use methanol as a solvent, if not then the ethyl ester remains.
It doesn't matter as it is hydrolysed anyway.

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Offline Rutherford

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Re: Chrysanthemic acid scheme
« Reply #21 on: February 16, 2013, 02:33:19 PM »
Yes, but structure of F is important, too. As it appears in the previous scheme, where I got that it is an ethyl ester, it has to be the same here. Didn't think of this earlier  ::).
Thanks for the help.
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Offline discodermolide

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Re: Chrysanthemic acid scheme
« Reply #22 on: February 16, 2013, 02:36:33 PM »
No problem.
Then they must have used NaOMe in a solvent like THF or toluene. Then the ethyl ester remains an ethyl ester.
p.s. I have actually made some of these compounds, although not by this methodology. But I can't disclose their structures.
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Offline Rutherford

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Re: Chrysanthemic acid scheme
« Reply #23 on: February 17, 2013, 06:24:30 AM »
Got one issue now. In the attached scheme, G is a natural alcohol. H is the same compound that was present in the previous scheme. What is G then? I have a formula, but I don't recognize this alcohol.
H is:
CC(C)=CCBr
I got that G is:
CC(C)=CCO
Is this a natural alcohol?
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Offline discodermolide

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Re: Chrysanthemic acid scheme
« Reply #24 on: February 17, 2013, 07:59:18 AM »
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Offline Rutherford

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Re: Chrysanthemic acid scheme
« Reply #25 on: February 17, 2013, 08:11:13 AM »
Huh, thanks. It would be a mess if it wasn't.
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Offline Rutherford

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Re: Chrysanthemic acid scheme
« Reply #26 on: February 17, 2013, 10:32:34 AM »
How to draw the (1R,3R)-configuration of chrysanthemic acid? I don't know how to apply the Cahn-Ingold system here. How to do it? I think that the stereocenters are in the ring.
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Offline sjb

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Re: Chrysanthemic acid scheme
« Reply #27 on: February 17, 2013, 10:34:56 AM »
Quote from: Raderford on February 17, 2013, 10:32:34 AM
How to draw the (1R,3R)-configuration of chrysanthemic acid? I don't know how to apply the Cahn-Ingold system here. How to do it? I think that the stereocenters are in the ring.

There's no difference to if it were not in a ring. Which two atoms are the stereocentres?
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Offline Rutherford

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Re: Chrysanthemic acid scheme
« Reply #28 on: February 17, 2013, 10:39:48 AM »
I think that the marked atoms are the stereocenters. Which one is 1R and which one is 3R? How to apply the Cahn-Ingold system?
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Offline discodermolide

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Re: Chrysanthemic acid scheme
« Reply #29 on: February 17, 2013, 10:53:56 AM »
like this?
C/C(C)=C/[C@@]1([H])C(C)(C)[C@]1([H])C(O)=O

(1R,3R)-2,2-dimethyl-3-(2-methylprop-1-en-1-yl)cyclopropanecarboxylic acid

This should be easy for you, we've done enough of them in the past.
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