Preparation of epoxides: Stereochemistry

In the last video, we saw two different ways to form an epoxide from an alkene. The first way was to add a peroxy acid. The second way was to first form a halohydrin using bromine in water and then using sodium hydroxide to start an intramolecular Williamson ether synthesis to form our epoxide. In this video, we'll look at the stereochemistry of epoxide formation for either of these two reactions. And if we start with a cis-alkene, so our hydrogens are on the same side of the double bond, or you can think about the R groups as being on the same side. For the product, the R groups are still going to be on the same side of where the double bond used to be. If you're looking at a trans-alkene-- so I know this is trans because my hydrogens are on opposite sides-- you think about the R groups being on opposite sides of the double bond. And in the product, the R groups are still going to be on opposite sides of where the double bond used to be. So let's look at a reaction involving stereochemistry. So if I start out with an alkene-- and I'm going to put a methyl group here. And I'm going to put an ethyl group over here. And then I'm going to have my hydrogens like that. And if I react this alkene in either of those two reagents-- so I will just go ahead and put ditto marks again. So either of those two are going to give me the same products. If I think about the formation of an epoxide-- so I know I'm going to make an epoxide. And I know I need to have those R groups on opposite sides of where the double bond used to be. So one possible product would be to have the ethyl group coming out at me in space. And then the methyl group would therefore need to be going away from me in space like that. So if that's my product-- at this carbon over on the left-- I must have my hydrogen going away from me in space. And at this carbon over here on the right, I must have my hydrogen coming out at me in space like that. So that's one possible product. I could actually draw another product where my R groups are on opposite sides of where the double bond used to be. So I could have my ethyl group going away from me in space at the carbon on the left, and I could therefore, have my methyl group coming out at me on the carbon on the right like that. And so this would have a hydrogen coming out at me, and then this carbon would have a hydrogen going away from me like that. So I have two possible products. And these are actually enantiomers of each other. So you're actually going to create a racemic mixture, 50% of one enantiomer and 50% of the other one as well because this reaction creates two new chirality centers. So if I take a look at this molecule, these two carbons are chirality centers. So there's stereochemistry at both of those carbons. So when you're drawing the enantiomers, if I start with that molecule on the left, I have my ethyl group coming out at me. All I have to do is reverse that absolute configuration to make the ethyl group going away from me for the enantiomer on the right. If I'm looking at this chirality center over here, I have at my methyl groups going away from me. All I have to do is reverse that absolute configuration, have the methyl group coming out at me. So these two are enantiomers of each other. Let's look at the details of the formation of these two enantiomers, and then we'll also name them using stereochemistry. So if we start with the alkene over here on the left-- so we're starting with that alkene-- and I think about these two carbons-- these two carbons are sp2 hybridized, meaning the molecule will be planar at those two carbons. So I could think about that alkene as being planar at those two carbons. So if I'm going to go ahead and draw that molecule here-- so I'm going to put my methyl group over here on the left and the hydrogen over here. And then I'm going to have a hydrogen over here. And then an ethyl group like that. So it's the same molecule as this one over here on the left. It's Just rotated a little bit. And I know that those two carbons on either side of my double bonds are sp2 hybridized, meaning that that portion of the molecule is planar. So I can think about that portion of the molecule as planar, and I can think, therefore, about the oxygen adding from either side of that plane. The oxygen could add from the top of the plane or the oxygen could add below the plane. So let's go ahead and draw the results of those two different ways of adding the oxygen. So over here on the left, I'm going to think about what happens to the molecule when the oxygen adds to the top of that plane. So if the oxygen adds to the top of the plane, I'm going to go ahead and keep my methyl group and my ethyl group on opposite sides of where the double bond used to be. And, therefore, the two hydrogens are also going to be on opposite sides of where the double bond used to be. And I'm going to show the oxygen adding to the top of the plate. So, therefore, my epoxide would look like that. So that's one possible way to form your epoxide. So let's go ahead think about, what would the epoxide look like if the oxygen added from below that plane? So if the oxygen added from below the plane, once again we need to think about keeping our methyl group and our ethyl group on opposite sides of where the double bond used to be like that. And the hydrogens are also going to be on opposite sides of where the double bond used to be like that. And then we're going to show our oxygen adding to form our epoxide from below the plane. So our epoxide might look like that. So these are my two possible products. And these are enantiomers of each other, but it's kind of hard to see that as we've drawn them right here. So let's see if we can get a different vantage point on our epoxide products. So I'm going to put my eye right here, and I'm going to stare at this epoxide like that. OK. So if I'm staring at that expoxide, I could redraw my epoxide here. So I'm going to put that portion of the molecule like that. Now, if I'm staring at it that way, then I'm going to first focus in on the carbon on my left. So that would be this carbon down here. And I can see that there's an ethyl group coming out at me. So I'm going to go ahead and draw an ethyl group coming out at me in space. And I can see that there's a hydrogen going away from me in space. Hopefully, it's obvious this hydrogen is going away from me in space. So I can represent that. So I'll put my in here to show my hydrogen going away. And now we're going to take a look at the carbon on the right side. So this carbon on the other side right here. This time, I can see that the hydrogen is coming out at me in space, so I'm going to go ahead and draw the hydrogen coming out at me in space like that. And, therefore, this methyl group back here is going away from me in space. So I can go ahead and show that methyl group going away like that. So that's one enantiomer. Let's go ahead and try to redraw the one on the right here. So, again, if I'm staring at this epoxide like that, what do I see? Well, I will see the epoxide as being upside down, so the oxygen will be down here. And I can see that, if I'm looking at the carbon on the left side of my vantage point-- so that would be this carbon down here-- I can see there's an ethyl group coming out at me in space. So here's my ethyl group coming out at me in space. And this hydrogen would be going away from me. So I'll represent that hydrogen going away from me like that. And then, if we move to the carbon on the right side here, I can see that this hydrogen is coming out at me in space. So I can go ahead and draw the hydrogen coming out at me in space. And the methyl group back here would be going away from me in space, so I can have methyl group as a dash like that. So this is one of my products. And, again, it's still not obvious that these are enantiomers. So this is where the model set comes in handy. So if you make this molecule on the right and then you hold this oxygen and you rotate this oxygen up-- much easier to see with the model set right in front of you-- you will find that this molecule is the exact same molecule as the one that we're going to draw right here. So if we rotate it so the oxygen is now pointing upwards, that's actually going to take this hydrogen back here and move it to the front. So when you're drawing that molecule rotated, that hydrogen is going to move to the front, which pushes the ethyl group to the back. So the ethyl group is actually going to the back here. Same thing with this methyl group. This methyl group was in the back. When you rotate it that way, the methyl group is going to end up coming out at you like that. And then that means the hydrogen is going to go away from you in space. So now we can see, hopefully, that these two are enantiomers to each other. And let's go ahead and think about naming this product. So if I were going to name these two molecules, I have to think about how to do it. So I need to find my longest carbon chain, and I want to give my epoxy substituent the lowest number possible. So if I'm going to number my carbon chain, I would make this carbon number one, this carbon number two, three, four, and five like that. So I'm going to name it as a pentane base. So this would be pentane like that. And I can see my epoxy occurs between carbons two and three. So it would be 2,3-epoxypentane like that. So this [INAUDIBLE] would be 2,3-epoxypentane. This one would be too. So I can go ahead and write 2,3-epoxypentane like that. But now I have to think about the stereochemistry at carbons two and three. So that makes things a little bit trickier. So I'm going to look at the enantiomer on the left and I'm going to redraw the enantiomer on the left. So I'm going to redraw this one right down here, and let's see if we can start to assign some stereochemistry. So I have my hydrogen coming out at me. I have my methyl group going away from me like that. This ethyl group is coming out at me. This hydrogen is going away from me. Let's start with this carbon. If I want to figure out the absolute configuration of that carbon, I have to think about what atoms are directly connected to that carbon. So let me go ahead and draw a carbon at here. So that is one of the atoms directly attached to that carbon in blue. So the carbon in blue is directly attached to this carbon, this oxygen, this hydrogen, and this carbon right down here. So those would be the four atoms. So if I'm trying to determine priority, I think about the atomic number. I know that oxygen has the highest atomic number out of those four atoms. So the oxygen's going to get a number one. Hydrogen has the lowest priority so hydrogen gets a number four. So now I have to think about these two carbons, because if there's a tie, I have to think about what those two carbons are attached to. Well, the carbon on the right-- so this carbon right here-- this carbon right here is attached to an oxygen. So let me go ahead and write this here. The carbon on the right is directly attached to an oxygen. It's directly attached-- over here, this oxygen. It's directly attached to a carbon and it's directly attached to a hydrogen. So oxygen, carbon, hydrogen. Let's go ahead and look at the carbon on the left, so this carbon right here. What is that carbon attached to? It's attached to another carbon over here and two hydrogens. So CHH. So the oxygen is going to beat the carbon in terms of atomic number, so this carbon on the right is going to get highest priority. So this carbon is going to get a number two here, and that means this carbon is going to get a number three. So for absolute configuration, my lowest priority group is going away from me and I'm traveling around this way. So I'm going clockwise, so it's an R absolute configuration at that carbon. Let's go ahead and, since that drawing's really busy, let's go ahead and draw it one more time. And we'll figure out the absolute configuration of the other carbon. So let's see. We still have my ethyl group coming out at me. Still have the hydrogen going away from me. Still have this hydrogen coming out at me. Still have this methyl group going away from me like that. So if I'm trying to find the absolute configuration for this carbon, we're going to approach it the same way. Look at the atom directly attached to that carbon. So it would be oxygen, carbon, carbon, hydrogen. So priorities. The oxygen gets highest priority. The hydrogen gets lowest priority. And then this carbon over here is directly attached to an oxygen, so this is going to get second highest priority. And that makes this methyl group over here the third highest priority. So the one, two, three is going around this way, which is counterclockwise. Which makes you think it might be S, but remember that trick that I told you about in an earlier video. It looks S, but this hydrogen is coming out. So all I have to do is switch that, and that takes care of the fact that my lowest priority group is not pointing away from me. So it looks S, but since the hydrogen is coming out at me, I can say with confidence that it's R for an absolute configuration. So we can go ahead and finalize the name. So at carbon two and at carbon three, we have an R absolute configuration. So the name of this enantiomer would be 2R,3R-2,3-epoxypentane. And for this enantiomer over here on the right, I know that it's the enantiomer so I just have to switch the absolute configuration. So this one is going to be 2S,3S-2,3-epoxypentane. And we don't time in this video to go ahead and assign absolute configuration to the enantiomer on the right. But you can go ahead and do so for practice. And you should get 2S,3S for its absolute configuration. Now, this is a problem because, if you're forming a racemic mixture of your epoxide, that's not always what you want to do in organic chemistry. So there are ways to use chiral catalysts that allow you to select out for one of these enantiomers. And we won't cover it in any of these videos since these are more intro organic chemistry videos, but just be aware that it is possible to be selective in which enantiomer you produce.