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Topic: An incredible reaction  (Read 7208 times)

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Offline Heory

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An incredible reaction
« on: November 23, 2009, 08:25:29 AM »
The following decarboxylation reaction looks so incredible, doesn't it? Could anyone explain it?
« Last Edit: November 23, 2009, 08:39:48 AM by Heory »

Offline tmartin

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Re: An incredible reaction
« Reply #1 on: November 23, 2009, 08:45:15 AM »
Does the steroechemistry at the beta position of the carbonyl that leaves stay intact or is it destroyed (as you're indicating?)?


Offline Heory

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Re: An incredible reaction
« Reply #2 on: November 23, 2009, 08:52:52 AM »
I coppied it from my textbook (Hydrolization of penicillin). I don't know whether the steroechemistry stays intact or destroyed, for it's not mentioned in the book. But I coppied the picture, so I thought it's destroyed. However, I'm concerned about the decarboxylation, not the sterochemistry.

Offline stewie griffin

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Re: An incredible reaction
« Reply #3 on: November 23, 2009, 09:10:41 AM »
I bet that the stereocenter is destroyed. I proposed a mechanism below. Basically I think you've got to have something to similar to a beta-keto acid for decarboxylations to occur... thus the proposed ring opening to form a thionium ion (if you can call it that). That destroys the stereochemistry.

Offline tmartin

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Re: An incredible reaction
« Reply #4 on: November 23, 2009, 09:11:34 AM »
Well, stewie beat me to this, but since I drew it all out, I'll post it too.

Offline Heory

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Re: An incredible reaction
« Reply #5 on: November 23, 2009, 09:18:43 AM »
Solved so fast! :)
stewie griffin and tmartin, how smart you are! I must learn hard.

Offline Heory

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Re: An incredible reaction
« Reply #6 on: November 23, 2009, 09:57:25 AM »
The last step breaks Baldwin's rule (5-trig-endo is forbidden). I think B's rule doesn't fit for this situation. Or maybe the mechanism is wrong?

Offline stewie griffin

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Re: An incredible reaction
« Reply #7 on: November 23, 2009, 10:07:49 AM »
I would ignore Baldwin's rules here for the last step. The last step is completely analagous to acid induced acetal formation with a 1,2-diol (ie formation of an acetonide from a diol). 

Offline azmanam

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Re: An incredible reaction
« Reply #8 on: November 23, 2009, 10:12:54 AM »
agreed.  Because of the resonance in the, er, thionium(?) ion, there is significant positive charge character on the carbon atom.  Perhaps it looks more like a carbocation with an empty p-orbital than a 'regular' double bond with full sp2 character?  If it were a typical double bond, the orbital alignment of the nitrogen lone pair into the C(sp2) pi* orbital might not be favorable.  I suspect the resonance structure giving positive charge character to the carbon atom is in play here.
Knowing why you got a question wrong is better than knowing that you got a question right.

Offline movies

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Re: An incredible reaction
« Reply #9 on: November 24, 2009, 02:46:46 PM »
thionium(?) ion

Thiocarbenium ion.  Thionium would be S sigma-bonded to three different things, like in trimethylsulfonium iodide.

Offline jinclean

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Re: An incredible reaction
« Reply #10 on: November 25, 2009, 05:36:12 AM »
i'm confusing about the Thiocarbenium because of its huge polarity and the bonding orbital,the S should ues the d orbit to form the S-C  bond,but the d orbital have the higher energy than the carbon's p orbital.what is the possible configuration in the carbon anion and the thiocarbenium?what is the C(-)-S(+) bond in actually?

Offline stewie griffin

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Re: An incredible reaction
« Reply #11 on: November 25, 2009, 08:47:16 AM »
re: movies
Thanks for the word. For some reason I couldn't remember thiocarbenium when I was typing it all up  ;)
re: jinclean
It's interesting to see how you are thinking about this C=S double bond. I don't see a need to bring in d orbitals on sulfur to explain this. Really what we have here is just a fancy carbocation. So the carbon was initially sp3, then some reaction occurs, and it's now an sp2 hybrid with an empty p orbital (in other words it's a carbocation). The sulfur atom is sp3 hybridized with two sets of lone pairs (just like an oxygen atom in a water molecule or in an ether). To stabilize the carbocation's empty p orbital, we can imagine that sulfur uses one set of it's lone pairs to donate into the empty p orbital. We can represent that stabilization as a second bond between the carbon atom and sulfur atom.
Note that you don't have to draw it as a C=S double bond. That's just one of two resonance structures. If you understand it better to just leave it as a full blown carbocation, you may do so. But you should know that in terms of the resonance rules, it's best to give everyone a full octet. Thus the C=S structure is the better resonance structure of the two. Hopefully the figure will help out a bit.

Offline jinclean

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Re: An incredible reaction
« Reply #12 on: November 26, 2009, 05:26:28 AM »
re:stewie griffin
 thanks for your explaination,but what i am confusing like SMe3(+) I(-) is the  quadrivalence sulfur with the three sigma bonds.that is very strange to me.and in your opinion.the S in the C=S bond uses the SP3 hybird orbit not the P obital? can't the sulfur changes its configurtion to get the better bonding with the carbon?
« Last Edit: November 26, 2009, 05:38:07 AM by jinclean »

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