E2 elimination: Stereospecificity

- [Narrator] The E2 reaction is a stereospecific reaction, which means that the stereochemistry of the substrate determines the stereochemistry of the product because of the mechanism. And for time reasons, I have to assume that you're already familiar with the E2 mechanism and that you understand Newman projections and Sawhorse projections. So the carbon that's bonded to the bromine would be our alpha carbon. The carbon bonded to the alpha carbon would be our beta carbon. And we know in an E2 mechanism, we're going to take this beta proton, our strong base, is gonna take that beta proton. So let's look down that beta-alpha bond here. So we're gonna stare down this bond so we can see a Newman projection. So if you put your eye right here and you look down this bond, you're gonna see a methyl group that's going up and to the right. So down here is our Newman projection, here's our methyl group. The hydrogen would be going up and to the left. So here's our hydrogen in white. Our phenyl group, which remember, just a benzine ring, so this is going to be going down on our Newman projection. And we think about the back carbon, the alpha carbon, we have a bromine which would be going down and to the right. We know there's also a hydrogen bonded to this alpha carbon that'd be going down and to the left. And then we have another phenyl group which would be going straight up in this Newman projection. So here is the phenyl group. We know that the E2 mechanism needs to have an antiperiplanar hydrogen and bromine, so if we look at this Newman projection, we already have that. This hydrogen in white is already antiperiplanar with this bromine, so we already have what we need for the E2 mechanism to occur. On the right here, this is really the same Newman projection. I just turned the whole thing a little bit to the right, so that line that I just drew in magenta would be this line. And so that would be the hydrogen in white. And that would mean the methyl group would be this one right here in yellow. And the phenyl group would be going down and to the left. So talking about the front carbon. For the back carbon, the bromine would be going down now. The phenyl group would be going up and to the right and the hydrogen would be the one in yellow. So again, this is the same Newman projection, just turned a little bit so it's a little bit easier to see our antiperiplanar proton and our bromine. You could draw the product at this point from this Newman projection. You think about a strong base coming along and taking your beta proton. These electrons would move in to form your double bond. At the same time, these electrons come off onto the bromine to form the bromide anion. So let's go ahead and draw the product for this reaction. So from the Newman projection, you need to think about this phenyl group and this methyl group being on the same side of the double bond that formed. So let me draw in our double bond here. So we would have our phenyl group and our methyl group on the same side of the double bond. On the other side of the double bond, we would form, we would have I should say, this hydrogen and this phenyl group. So let's draw those in. So on this back carbon here, that would be the hydrogen, and on the front carbon, I would have my phenyl group. So that is the product of this reaction. Might be a little bit hard to see from this Newman projection, so here' a Sawhorse projection, another viewpoint which might be easier to see. So let's highlight everything again. So this would be our alpha carbon and this would be our beta carbon. So our strong base is gonna take this beta proton and these electrons would move into here to form our double bond, and these electrons come off onto the bromine. So again, the phenyl group and the methyl group would end up on the same side, so here they are. And on the other side, you would have this phenyl group and this hydrogen. So this phenyl group and this hydrogen. So the two phenyl groups are on opposite sides of the double bond, and this would be the E alkine. Let's look at this on a video, and I think it might be really helpful to look at the model forming the alkene. Let's look at our alkyl halide. We can see that the purple groups are the phenyl groups, the red is the bromine and at this carbon, we have a methyl group coming out at us in space and a hydrogen going away from us in space. We're gonna stare down this beta-alpha bond and notice that I'm using the stretchy bonds again so we can better visualize the E2 mechanism. So let's stare down that bond and look at the Newman projection. So here in this Newman projection, we already have our beta hydrogen antiperiplanar with our bromine, so if I rotate it a little bit, there's our Newman projection. I'm also gonna turn to the side a little bit so we can see it from a Sawhorse projection. And our strong base, we think about our strong base coming along and taking this proton. So pretend like the proton goes away, pretend like the bromine goes away, and you can see the double bond that formed. And notice how the two phenyl groups are on opposite sides of the double bond, so that's our product. Let's do another E2 reaction, and notice how similar this is to the last one. The only difference is the stereochemistry at this carbon. Now the hydrogen is coming out at us in space and the methyl group is going away from us in space. So we'll see how changing the stereochemistry affects the stereochemistry of the product. So we need to stare down our bond again, so this would be our alpha carbon, this would be our beta carbon. We're gonna stare down this way just like we did in the previous example. So let's think about the Newman projection that we would have. Now, our hydrogen, I'll highlight in white here, would be going up and to the right on our Newman projection. Our methyl group would be going up and to the left, and our phenyl group would be going down. So here's the phenyl group. So that's our front carbon. Our back carbon would have our bromine going down and to the right. The hydrogen on the alpha carbon would be going down and to the left. And then we have our phenyl group going straight up, so there's our Newman projection. Notice that this time, our beta hydrogen, so right here, this is not antiperiplanar with our bromine. So we're going to need to rotate about our single bond and we'll do that in a minute. First let's just get to this Newman projection on the right. So I'm just gonna draw a line going through this methyl group and this bromine just to help us visualize the fact that here is that same methyl group and then here is that bromine. So it's again, this is the same Newman projection, it's just rotated a little bit to the right. So let me highlight everything. So this would be the hydrogen in white, and we have our phenyl group over here, and then on our back carbon, we would have our hydrogen in yellow, and then we have our phenyl group, so our phenyl groups are still anti to each other. So let's go to the video now so we can rotate about our single bond so we can get into a confirmation that has our beta proton antiperiplanar to our bromine. Let's think about the stereochemistry at this carbon. Now the hydrogen's coming out at us and the methyl group is going away from us. So if we stare down this carbon-carbon bond, we will see our Newman projection. And notice how our beta hydrogen is not antiperiplanar with our bromine. So I'm gonna rotate the whole thing a little bit and then I'm gonna rotate about the single bond so we can get in the proper confirmation. So now our beta proton is antiperiplanar with our bromine. And notice how the two phenyl groups are now gauche to each other. So if I turn a little bit to see things from a Sawhorse projection, we think about taking away this beta proton, so the beta proton goes away, and pretend like the bromine goes away as well. And we can see that for our product, our two phenyl groups end up on the same side of the double bond. Hopefully the video made it clear how to figure out the products. But if you don't have a model set on a test, you have to use your own drawings. So this line in magenta that I drew right here, I forgot to draw it in on this Newman projection, so that's this line. And then when you realize that your proton is not antiperiplanar, you'll have to rotate. You'll have to rotate yourself. And I like to think about rotating the front carbon here. I can take this hydrogen and I'm gonna move it up to here, which would move this methyl group over to this position and it moves the phenyl group over to this position. So the hydrogen in white ends up being vertical. The methyl group ends up going down and to the left and the phenyl group ends up going to the right. So you don't necessarily need a model set to do this. And now we have our hydrogen antiperiplanar with our bromine, so from this Newman projection, you could draw the product. A base takes our beta proton, these electrons move in here and these electrons come off onto our bromine. And so our two phenyl groups would be on the same side of the double bond. Notice how they are gauche to each other in this Newman projection. So if I draw in my double bond, on the back carbon I have a phenyl group. On the front carbon here, I have a phenyl group. And then the groups in the other side of the double bond, I have a hydrogen on the back carbon and a methyl group on the front. So let me draw those in up here. So I have a methyl group on the front carbon and a hydrogen on the back. If you prefer the Sawhorse projection point of view, the base would take our beta proton. These electrons move in to form our double bond, these electrons come off onto the bromine, and we have our two phenyl groups on the same side. And so here they are, they're on the same side. And then we would have our methyl group and our hydrogen on the same side of our double bond. Notice this is the Z alkene, so different stereochemistry for our product as compared to the first reaction. And actually, this reaction would be much slower than the first reaction. And that's because of these two phenyl groups here. So this confirmation, where we have our proton and our bromine antiperiplanar, this is definitely not the most stable confirmation. All right, we have these really bulky phenyl groups gauche to each other, and that means that this would be slower than the first reaction, because this is the confirmation we must have for this mechanism to occur, but it's not a very stable confirmation. If we go back up here to the first reaction that we did, when we had our proton antiperiplanar to our bromine, our two phenyl groups were anti to each other, so they're relatively far away from each other. So this would be a stable confirmation. And so this reaction happens a lot faster than our second reaction. Finally, let's summarize the two reactions that we've seen. So on the left, the stereochemistry at this carbon with our methyl group coming out at us and our hydrogen going away from us, that gave us the E alkene as our product with our two phenyl groups on opposite sides of the double bond. We changed the stereochemistry, so now the hydrogen's coming out at us and the methyl group is going away from us. We got a different product. We got the Z alkene with our two phenyl groups on the same side of our double bond. So that shows you how this is a stereospecific reaction. The E2 mechanism means that the stereochemistry of the substrate determines the stereochemistry of the product.