E2 elimination: Stereoselectivity

- [Instructor] We've already seen that the E2 reaction has a concerted mechanism, the carbon bonded to our halogen is our alpha carbon, the carbon bonded to that carbon would be a beta carbon and we need a beta proton for our mechanism. The hydrogen and the halogen have to be on opposite sides of our bond, here, first I draw in this line, they're on opposite sides. That's said to be anti. And these four atoms need to be nearly in the same plane, the hydrogen, the carbon, the carbon, and the halogen need to be nearly in the same plane, so that's planar. So, we say this has to be antiperiplanar, so let me write this in here, so antiperiplanar. The anti meaning hydrogen and halogen on opposite sides and the planar part meaning those four atoms are in the same plane. Our strong base comes along in our E2 mechanism so a negative charge and takes our beta proton, at the same time these electrons move in here to form our double bond and these electrons come off onto our leaving group. So, let me show these electrons in here in blue, those electrons in blue move in to form our double bond and so we form an alkene. If we're thinking about the stereoselectivity of an E2 mechanism, it can be helpful to look at this mechanism from a different point of view, for example from a Newman projection. So, here we have a Newman projection and we could think about a base coming along in our Newman projection and taking our beta protons, so we're taking this proton here. And then these electrons would move into here, at the same time these electrons come off onto our halogen so it's a little bit easier to see that all four of those atoms are in the same plane if I draw a straight line through here. Or we could look at it from more of a Sawhorse projection, so here is our beta proton and here is one of our carbons, here's another carbon, and here's our halogen, our strong base is going to take our beta proton, and these electrons would move into here to form our double bond and these electrons would come off onto our halogen so that would give us our alkene. So let's take a look at an example where we're thinking about the E2 mechanism and stereoselectivity. Let's say we have this alkyl halide and we react it with sodium ethoxide which is a strong base. The carbon bonded to the halogen is our alpha carbon and the carbons bonded to the alpha carbon are the beta carbons so I'll call this beta one and then we have a beta carbon on the right which I will call beta two. We're going to ignore beta two for right now and just focus on beta one. Beta one carbon has two hydrogens bonded to it so I'll put one on a wedge and one on a dash, and let's highlight those, the one on the wedge I'm going to make green, and the one on the dash I'm going to make white. We're going to stare down that bond, we're gonna stare down this beta one alpha carbon bond. So if you put your eye right here and you stare this way and you stare down that bond, you're going to get a Newman projection. The hydrogen in green would be going up and to the right, so here's the hydrogen in green. The hydrogen in white would be going up and to the left. And then there'd be a methyl group going straight down, so we won't spend too much time on Newman projections since those are covered in earlier videos. On the back carbon he bromine here would be down and to the right, and then there would also be, on this drawing, a hydrogen going away from us in space. On the Newman projection that would be this hydrogen. Then finally there's a methyl group going up in space, so here is our Newman projection staring down that bond. If you look at the hydrogen and the bromine, this hydrogen in white and the bromine, they're already antiperiplanar to each other so if I draw in a line here, you can see that those four atoms are in the same plane, which is what we want. However, to look at it a little bit easier, I've just turned this whole thing a little bit to the right, so this is really the same conformation, I haven't changed the conformation at all, I've just made that magenta line vertical. So, now we have our hydrogen in white is here, our bromine in red is here, and the hydrogen in green would be over here down and to the right. Notice that both of our methyl groups are anti to each other in both of these, so it's really the same Newman projection just turned a little bit so we can see it better. We know in our E2 mechanism our strong base takes our beta proton, so our strong base would be the ethoxide anion, and it's gonna take this beta proton, these electrons move in here, these electrons come off onto our leaving group, and when the double bond forms the two methyl groups end up on opposite sides of the double bond. This actually gives us our trans product, so we would get our product as being trans right here. Maybe it's a little bit easier to see this mechanism from the Sawhorse projection. On the right, so really, again this is not a different conformation, this is just another way of looking at this molecule. Our alpha carbon would be here 'cause here's the bromine, so this would be our beta one carbon, and our strong base would come along and it's going to take this proton, and these electrons would move in here to form our double bond at the same time that the bromine leaves. So, this is our concerted mechanism. When the double bond forms, these two methyl groups end up on opposite sides of the double bond. Let's go to a video, now, where I show you all of this using a model set and it think it's a little bit easier to see that you make the trans product. Also, we'll look at a different conformation and see the other product, so taking the green proton. I forgot to highlight this. This would be the white proton that we are taking, and the green proton would be over here. Here's our alkyl halide, and you can see one hydrogen is green and one hydrogen is white, and we have the red for the bromine. We're going to stare down this carbon-carbon bond and notice I'm using these stretchy bonds to help us out with the E2 mechanism. If we stare down that carbon-carbon bond, we'll see things from a Newman projection. Notice, in this Newman projection, I'm just gonna turn it a little bit so that we get our beta proton and our red bromine antiperiplanar and vertical. Next, think about these two methyl groups on the Newman projection as being anti to each other. I'm gonna turn it a little bit to see things from a Sawhorse perspective. Think about taking this beta proton in the mechanism and pretend like that red bromine goes away, and you can see that now our two methyl groups are on opposite sides of the double bond that formed. This is the trans product. If we go back to our Newman projection, we know there's free rotation around our single bond so I can rotate to get a different conformation to put the green hydrogen antiperiplanar to the red bromine. From this Newman projection, you can see we have these two methyl groups that are now gauche to each other. If I turn it to the side and I take this beta proton here. Again pretend like the bromine goes away. Here you can see that the two methyl groups are on the same side of the double bond that formed. This is the cis product. Here's what we saw in the video. If you're talking about this conformation, our two methyl groups are gauche to each other here. Let me highlight our proton in green, so here's the proton in green, and the bromine is red. If we're taking the proton in green, our strong base which is our ethoxide anion takes this proton. These electrons move into here, these electrons come off onto the bromine, and your two methyl groups end up on the same side, so this give you the cis product. From the Sawhorse projection, if we take this proton, this proton is our green one so let me highlight that, so this is the green proton and the bromine over here would be in red. If we take the green proton, then these electrons move into here, the electrons come off onto our bromine, and the two methyl groups would be on the same side when the double bond forms, so this gives us our cis product. Let me go down here and draw it. This would give us our cis product with our two methyl groups on the same side. If we think about those two different conformations that have protons antiperiplanar, so this conformation had the white proton antiperiplanar, this conformation had the green proton antiperiplanar, the white one gave us the trans product, the green one gave us the cis product. We know that the first conformation is more stable, this conformation is more stable because our bulky methyl groups are anti to each other. Here those two bulky methyl groups are gauche to each other and that's increased steric hindrance. This is the less stable conformation, and therefore this product, our cis product, is not as stable as our trans product. To summarize what we saw for this alkyl halides, we know that our beta one carbon had two protons and we got two different products depending on which proton the base took. We got the trans product and we got the cis product. The trans product turns out to be the major product for this reaction, 'cause this came from the more stable conformation. The cis product is the minor product for this reaction, so this one came from the less stable conformation. This reaction is said to be stereoselective. We have these two stereoisomers and one stereoisomer is favored over the other. The trans product is favored over the cis product, and that's stereoselectivity. We've ignored this beta two carbon until now. This beta two carbon has several protons and it's easy to get one in an antiperiplanar relationship with the bromine, I'll just draw one in really quickly here. We think about a base taking our protons, so our base, our ethoxide anion, takes this proton, these electrons move in here, and these electrons come off onto the bromine. So, we're also going to make another product. I'll put this in parentheses here. We're gonna make this product. Our electrons, in magenta, moved into here to form this double bond. This is what we would get from our beta two carbon. Really in this video we're focused on our beta one carbon and thinking about stereoselectivity, but you can't forget that there can be more than one beta carbon in a reaction.