Organic Chem question: stereochemisty in this Taxol tail synthesis step.
April 13, 2010 9:31 PM   Subscribe

Please explain the stereochemistry of this reaction.

The epoxide opens and the azide alcohol forms. For some reason I am not seeing the stereoselectivity.
posted by jbreyfogle to Science & Nature (9 answers total)
Think about the epoxide as taking up room on one side of the carbon that will be attacked by the alcohol group. The alcohol group has to attack on the opposite side. Does that help?
posted by LittleMissCranky at 9:47 PM on April 13, 2010

@ littlemisscranky

No alcohol is attacking the carbon adjacent to the Ph, its the azide that is the nucleophile. I still don't understand.. :/
posted by jbreyfogle at 9:51 PM on April 13, 2010

@ littlemisscranky

also, in most epoxidations you see, the new substituent is formed trans of the alcohol but on the same side of the alkane bond (similar to Z configuration drawn out, only not with an olefin). It is hard to interpret, the only way I understand the products stereoselectivity is if the reactant expoxide had the carbonyl group and the Ph group eclipsed in a newman projection.
posted by jbreyfogle at 9:57 PM on April 13, 2010

Crud, you're so's been too long since organic chemistry. Don't know what I was thinking.
posted by LittleMissCranky at 10:29 PM on April 13, 2010

@ littlemisscranky


I think I might have figured it out. Hopefully someone can confirm.

The acetic acid displaces the epoxide first, attacking from backside. Since the reaction with AcOH is reversible and fast, the azide is added cis of the epoxide. The addition of AcOH and the formation back to the epoxide is so fast and rapid that it prevents the molecule from rotating and thus the azide adds keeping the same stereochemistry.
posted by jbreyfogle at 10:40 PM on April 13, 2010

Reading on the fly, no time to have a look at the specifics of the rxn, but this former organic chem prof says: get out your model kit, make a model of the reaction center.
posted by Sublimity at 5:30 AM on April 14, 2010 [1 favorite]

Here's a few thoughts on how the azide can end up on the side it does on the beta-C. I'm not a definitive source by far, but I do remember that brainstorming mechanisms with others was always helpful to me:

Two inversions of configuration lead to retention of configuration. The initial treatment of the alpha-epoxyester with the AcOH leads to opening of the epoxide ring. Attack will occur at the beta-C (that C-O epoxy bond breaks), thus the alpha-C will carry the hydroxide (w/oxygen of original epoxide) and will have the configuration seen in the final product. Because the attack occurs anti to the ring, there will be an inversion of configuration at the beta-C.
[note: I'm not sure what the product is at this point. I believe it should be an anti vicinal diol. But something adds anti on the beta-C.]

Azide is a strong nucleophile, and via an Sn2 reaction attacks the beta-C, displacing the new addition. Because of the Sn2 mechanism, there will be inversion of configuration. That is, the azide comes in opposite the previous addition--which is anti the original oxygen. Thus the azide ends up attacking from the same side the alpha-C hydroxide is on, giving the configuration seen in the final product.

Just an idea. I normally wouldn't throw out a half-formed idea like this on askme, but sometimes hearing other students' takes on mechanisms can help nudge your own brain in the right direction. (If you're in a class, I highly recommend working in pairs/trios on this stuff.) Good luck!
posted by neda at 8:02 AM on April 14, 2010

Acetate isn't a very good nucleophile, due to delocalization of the negative charge between the two oxygens. Acetic acid is even worse.

A model would help a fair bit, but it looks like both chiral sites are inverted. (the epoxide is dashed and northward, which would translate into bold and southward, not dashed and southward).

The acetic acid protonates the epoxide, exposing it to nucleophilic ring opening. The carbon adjacent to the ketone is electronegative, while the beta carbon is electron-poor (charges alternate from the ketone; e- poor, rich, poor, etc.). This is why the azide attacks where it does.

For my money, either both up-and both down enantiomers are equivalent (build a model!), or there's a typo.
posted by Orange Pamplemousse at 8:26 AM on April 15, 2010

Mmm, a plastic model shows the expected inversion of stereochem. The ketone should be facing down in the first structure, which is why it looks really weird as written.

There's nothing weird going on, just an acid labile epoxide and a good nucleophile.
posted by Orange Pamplemousse at 8:53 AM on April 17, 2010

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