|Ch. 1 - A Review of General Chemistry||4hrs & 47mins||0% complete||WorksheetStart|
|Ch. 2 - Molecular Representations||1hr & 12mins||0% complete||WorksheetStart|
|Ch. 3 - Acids and Bases||2hrs & 45mins||0% complete||WorksheetStart|
|Ch. 4 - Alkanes and Cycloalkanes||4hrs & 18mins||0% complete||WorksheetStart|
|Ch. 5 - Chirality||3hrs & 33mins||0% complete||WorksheetStart|
|Ch. 6 - Thermodynamics and Kinetics||1hr & 19mins||0% complete||WorksheetStart|
|Ch. 7 - Substitution Reactions||1hr & 46mins||0% complete||WorksheetStart|
|Ch. 8 - Elimination Reactions||2hrs & 24mins||0% complete||WorksheetStart|
|Ch. 9 - Alkenes and Alkynes||2hrs & 10mins||0% complete||WorksheetStart|
|Ch. 10 - Addition Reactions||3hrs & 33mins||0% complete||WorksheetStart|
|Ch. 11 - Radical Reactions||1hr & 57mins||0% complete||WorksheetStart|
|Ch. 12 - Alcohols, Ethers, Epoxides and Thiols||2hrs & 34mins||0% complete||WorksheetStart|
|Ch. 13 - Alcohols and Carbonyl Compounds||2hrs & 14mins||0% complete||WorksheetStart|
|Ch. 14 - Synthetic Techniques||1hr & 28mins||0% complete||WorksheetStart|
|Ch. 15 - Analytical Techniques: IR, NMR, Mass Spect||7hrs & 18mins||0% complete||WorksheetStart|
|Ch. 16 - Conjugated Systems||5hrs & 49mins||0% complete||WorksheetStart|
|Ch. 17 - Aromaticity||2hrs & 24mins||0% complete||WorksheetStart|
|Ch. 18 - Reactions of Aromatics: EAS and Beyond||4hrs & 31mins||0% complete||WorksheetStart|
|Ch. 19 - Aldehydes and Ketones: Nucleophilic Addition||4hrs & 54mins||0% complete||WorksheetStart|
|Ch. 20 - Carboxylic Acid Derivatives: NAS||2hrs & 3mins||0% complete||WorksheetStart|
|Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon||1hr & 56mins||0% complete||WorksheetStart|
|Ch. 22 - Condensation Chemistry||2hrs & 13mins||0% complete||WorksheetStart|
|Ch. 23 - Amines||1hr & 43mins||0% complete||WorksheetStart|
|Ch. 24 - Carbohydrates||5hrs & 56mins||0% complete||WorksheetStart|
|Ch. 25 - Phenols||15mins||0% complete||WorksheetStart|
|Ch. 26 - Amino Acids, Peptides, and Proteins||2hrs & 54mins||0% complete||WorksheetStart|
|Alcohol Nomenclature||5 mins||0 completed|
|Naming Ethers||7 mins||0 completed|
|Naming Epoxides||18 mins||0 completed|
|Naming Thiols||11 mins||0 completed|
|Alcohol Synthesis||8 mins||0 completed|
|Leaving Group Conversions - Using HX||12 mins||0 completed|
|Leaving Group Conversions - SOCl2 and PBr3||13 mins||0 completed|
|Leaving Group Conversions - Sulfonyl Chlorides||8 mins||0 completed|
|Leaving Group Conversions Summary||5 mins||0 completed|
|Williamson Ether Synthesis||4 mins||0 completed|
|Making Ethers - Alkoxymercuration||4 mins||0 completed|
|Making Ethers - Alcohol Condensation||5 mins||0 completed|
|Making Ethers - Acid-Catalyzed Alkoxylation||4 mins||0 completed|
|Making Ethers - Cumulative Practice||10 mins||0 completed|
|Ether Cleavage||8 mins||0 completed|
|Alcohol Protecting Groups||3 mins||0 completed|
|t-Butyl Ether Protecting Groups||6 mins||0 completed|
|Silyl Ether Protecting Groups||4 mins||0 completed|
|Sharpless Epoxidation||10 mins||0 completed|
|Thiol Reactions||6 mins||0 completed|
|Sulfide Oxidation||5 mins||0 completed|
|Physical Properties of Alcohols|
|Acidity/Basicity of Alcohols|
|Active Metals as bases on Alcohols|
|Crown Ether Nomenclature|
|Cyclic Ether Nomenclature|
|Leaving Group Conversions Retrosynthesis|
|Physical Properties of Ethers|
|Williamson Ether Retrosynthesis|
|Synthesis of Phenol Ethers|
|Cleavage of Phenyl Ethers|
|Acidity of Thiols|
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?
Side note: K. Barry Sharpless figured this puzzle out in 1980, receiving a Nobel Prize in 2001. Go science!
Concept #1: Important Reagents of Sharpless Epoxidation.
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 #2: General reaction of Sharpless Epoxidation.
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 #3: How to draw and predict a Sharpless Epoxidation.
Alright so now I just want to add one more little twist to it to make sure that you always get these questions right. If you want to make sure that your alcohol is oriented correctly so you always predict the right product you always want to draw the alcohol on the bottom-Right Corner of the double bond, OK? Now you're going to notice is that in some textbooks they don't use the bottom right corner some textbooks they use the top right corner, some textbooks they tell you other directions but go ahead and use mine just because of the fact that it doesn't matter which corner you use as long as you're consistent and that's going to be the one that translates to the pattern I told you about working, OK? So if you want it another way that's fine but honestly most people really don't understand circles of oxidation so I think this way works really well so we always put the one saying if you have your alcohol oriented wherever you always make sure it has to be in the bottom right corner, Why? Because once it's at the bottom right corner now your double bond is oriented correctly so now when you look at your DET you say OK is this a positive tartrate or a negative tartrate, positive so that means that my epoxide should form above this double bond so that's exactly what we did so now you form your epoxide and notice that whatever is in the front whatever's facing towards the front here let's say this is your line whatever facing towards the front goes on a wedge, OK? And whatever was facing towards the back goes on a dash so that means the bromine and the methyl group going on dash, the alcohol and the H go on the wedge ideally your alcohol should always be going on the wedge, right? Because you always want to face your alcohol towards the front or down and everything else where it's supposed to go, OK? So in this case I would get an epoxide above the plane so an epoxide exactly facing up, OK? Because it's a positive DET so I'm just going to give you guys as an example how would if we had used just so you guys can see what it looks like negative tartrate? Then what would be answer look like, OK? Well I'm just going to tell you right now all of these wedges and dashes don't change because that has to do with how the double bond is oriented the double bond is going to stay the same so I would want to do the same thing of how they are oriented on wedges and dashes the only thing is that if we use negative DET actually I'm going to put it over here so that you guys can see if we use negative DET tartrate instead then what I would get is an Epoxide facing down, OK? And then all these groups where they are at? So that means that I would have my OH facing towards the front I would have my H facing towards the front, OK? Then I would get my methyl in the back and I would get my bromine in the back, OK? And the reason that makes sense is once again think that this double bond is exactly the way it looks and you just popping an epoxide right at the bottom so that means that everything stays in the same place that it was originally same with the positive tartrate, OK? So guys I hope that made sense really a lot of professors don't ask a lot from Sharpless epoxidation, it is a newer reaction discovered in the past twenty or thirty years, OK? So there isn't a whole lot to know I would just suggest knowing this trick and it could maybe get you some points in the test so anyway that's it for now let's go ahead move on to the next topic.
Always draw alcohol on the bottom right corner of the double bond. Then determine which epoxide you get according to the DET used.
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