Sharpless Epoxidation

Epoxidation of an asymmetrical alkene is usually a non-stereospecfic process, yielding a racemic mixture of enantiomers. How do you select for one enantiomer over another?

K. Barry Sharpless figured this puzzle out in 1980, receiving a Nobel Prize in 2001. Go science!

Concept: Important Reagents of Sharpless Epoxidation.

Video Transcript

So now we're going to talk about a form of epoxidation that has some pretty interesting advantages. The name of this reaction is the Sharpless Asymmetric Epoxidation. So the whole point of this reaction is that it's a form of epoxidation that is enantioselective. What that means is that it's going to generate only one of the two possible enantiomers in excess. In fact, it's almost going to perfectly select one enantiomer over another.
Now in order to do this, we're going to use some pretty weird molecules, some pretty weird reagents that overall, I would just ask you to recognize more than memorize because most professors aren't going to get into the nitty-gritty of memorizing every single letter of these reagents. They just want you to know what this reaction is about.
The way this reaction works is that it's going to convert allyl alcohols that means it's an alcohol that has a CH2 and then a double bond. Allyl is a position that says that you're next to a double bond, not directly attached to one. Then it's going to generate a certain epoxide based on the type of tartrate that is used.
These tartrates are basically functional groups that have different chiral centers. What we're going to find is that there's three different possibilities of types of tartrates that I could use in this reaction. I could use the S, S, I could use the R, R. We're talking about these chiral centers right here. So if both of them are S, that's considered a positive tartrate.
Positive if you remember, if you see a little positive sign inside of brackets, what that's talking about is the optical activity. So what that's saying is that the chiral centers are S and S. When you run it through a polarimeter, it's going to rotate light clockwise. That's what the positive means.
Well, the enantiomer of that would mean that both chiral centers are opposite. If you have an R, R tartrate, that's going to be a negative rotation. The reason is because remember that the enantiomer of any chiral center or of any chiral molecule will always have the opposite configuration, the opposite rotation, but of the other configuration. For example, if it was positive 20 degrees, then it would be a negative 20 degrees rotation with the negative DET.
Then finally, we have an R and an S, or an S and an R. This is actually a meso DET, so this one would be actually, since it's meso, this one would have no optical activity. Oops, I'm just going to write no optical activity. This is going back to our chirality chapter. We talked about meso compounds and how they don't rotate plane polarized light. It's impossible to assign a plus or a minus to a meso because it's not going to rotate at all.
This is interesting. We're talking about chiral centers. You're like this sounds a lot like chirality. But what does this have to do with the epoxide? Well, it turns out that you can predict the direction that the epoxide is going to form from what type of enantiomer you're using. It turns out that the positive DET, the positive tartrate, the one that has the positive rotation of light, is going to attack from above. It's going to enantioselectively pick the top part of a double bond to add an epoxide.
Then we've got the negative one. The negative one is going to be the opposite, so it's going to pick – it's going to attack from below. And we would expect the one from below to now form an epoxide below the double bond. So we've got positive is up, negative is down.
Then we've got meso. What do you think about meso? Well, meso would just be both. The reason that meso is both is because this one would be non-enantioselective. Why? Because it doesn't have a preference of top or bottom, so it's just going to be a 50% chance.
Really, we don't really care about the meso one so much and we're not going to use that one synthetically. What we're going to use is positive DET and negative DET as our catalysts to form the upwards epoxide and the downwards epoxide.

Diethyl tartartes (DET) of different optical activities are used to convert allyl alcohols into stereospecific epoxides.  

Concept: General reaction of Sharpless Epoxidation.

Video Transcript

So now you're probably wondering, “Okay, Johnny, how does this actually look in a reaction?” Well, here's the general reaction. So here, as you'll notice, I have an allyl alcohol. This is my allyl alcohol right here. The reason we call it allyl is because it is next to a double bond. It has a CH2 and then it has an OH, so this is allylic. That's a position word.
We have an allylic OH. And when we react it with a peroxide, this is the oxidizing agent. This is what's going to make the O. And then a titanium catalyst. Don't worry too much about the titanium catalyst, you'll just see that there's Ti there, that just stands for titanium. There's a titanium catalyst. And then we use one of the tartrates, either negative or positive.
When you put all those things together what you're going to wind up getting is the epoxide in the place that you want it. If I want my epoxide to face down, then I would use – if I want it to be below the plane, then I would use a negative tartrate. A negative tartrate is going to attack from the bottom of the double bond and it's going to give me my epoxide at the bottom. Now, obviously, that means if the epoxide is facing towards the bottom, then my other substituents must be forced up. Cool.
So far I know these reagents are super confusing, but really I'm not asking much from you. I'm not asking you to memorize them. I'm just saying can you remember that the positive tartrate adds from the top and the negative tartrate adds from the bottom. Is that cool so far?

Concept: How to draw and predict a Sharpless Epoxidation.


Always draw alcohol on the bottom right corner of the double bond. Then determine which epoxide you get according to the DET used.