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E2 mechanism: kinetics and substrate

Mechanism of an E2 elimination reaction. Created by Jay.

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Video transcript

- [Instructor] In an E2 mechanism, we need alkyl halide and we need a strong base. On our alkyl halide, the carbon that's directly bonded to the halogen is the alpha carbon, and the carbon next to the alpha carbon is the beta carbon. And we need a beta hydrogen for this reaction to occur. And all four of those atoms must be in the same plane. So this hydrogen, this carbon, this carbon, and this halogen are all in the same plane, which is why I have the bonds drawn in straight lines here. The hydrogen and the halogen must be on opposite sides, if I draw a little line here through the carbon-carbon bond, and that's said to be anti. So for this mechanism, we need antiperiplanar hydrogen and halogen. The hydrogen and halogen are anti to each other, and they are in the same plane. The mechanism for this reaction is a concerted mechanism. So the strong base comes along and takes this proton, which leaves these electrons, these electrons move into here to form our double bond at the same time that these electrons are coming off onto our halogen, so it's concerted. And let's follow those electrons in red here, the electrons in the carbon-hydrogen bond, moving here to form our double bond. So our product is an alkene. When you look at the kinetics for an E2 mechanism, the overall rates of the reaction is equal to the rate constant times the concentration of the substrate to the first power, and the substrate is your alkyl halides, times the concentration of the base to the first power, so our strong base. And so, this is an elimination reaction that depends on the concentration of both the substrate and the base, and that's why we call this an E2 reaction, an E2 mechanism. The E stands for elimination, and let me go and write that in here. So the E is for elimination. This is an elimination reaction. And the two is because this mechanism is bimolecular, meaning it's dependent on the concentration of two things, the concentration of the substrate and the base. So let's say you increase the concentration of your substrate, which is your alkyl halides, by a factor of two, so increase the concentration of that by a factor of two. Let's say you also increase the concentration of you base by a factor of two. So increase that by a factor of two. What happens to the overall rate of the reaction? So, the rate would be equal to, now you're thinking about two to the first times two to the first, because both of these are to the first power, and we're doubling the concentration of both. So two times two is of course four, so we're gonna increase the rate by a factor of four if you double the concentration of both. So this reaction is similar to an SN2 reaction in terms of the kinetics. Remember, in SN2, the two is there because it's also a bimolecular mechanism. So here's an E2 reaction, and left is our alkyl halides. The carbon that's bonded to our halogen is our alpha carbon. And then all of these carbons around here would be beta carbons. But all of our beta carbons are equivalent. In this case, our strong base is the hydroxide anion, so we need a source of the hydroxide anion here. And the mechanism would be E2. So let me go ahead and redraw out alkyl halide, so I'll put in lone pairs of electrons on the bromine. We know the mechanism is a concerted mechanism. It happens in one step. So let me draw in our hydroxide over here. And we know that hydroxide is going to take a proton from a beta carbon. So I'll draw in a hydrogen here on that beta carbon, and as our base takes this beta proton, the electrons in this bond moving here to form a double bond, at the same time, these electrons come off onto the bromine. So for our product, we would have an alkene. So let me draw in the alkene that would form, and let me highlight our electrons. So these electrons in magenta moved in here to form our double bond, and this is our alkene. So we'd also have the bromide anion. So let me just go ahead and draw this in here, so we'd have our bromide anion with a negative one formal charge. So we could say that these electrons in here in blue came off onto bromine to form bromide anion. And if you take hydroxide, and you add a proton onto hydroxide, you would also have water. So let me draw that in here as well. So your bromide anion, and you would have water. But we're really only thinking about this alkene that forms here. So when we talk about kinetics, we said that our E2 mechanism was similar to an SN2 mechanism. They're both bimolecular. But when you're thinking about the structure of the substrate, E2 is very different from SN2. SN2 reactions do not occur when you had a tertiary substrate, and that was because of steric hindrance. There is too much steric hindrance for the nucleophile to attack. So if you think about hydroxide trying to attack this alkyl halide, there's too much steric hindrance because of these methyl groups here. But when you're thinking about an E2 mechanism, it's different. The hydroxide anion doesn't have to get close to this carbon. It's only taking a proton away, and there's enough room for it to do that. And I think it'll be a little bit more clear if I show you a video, so we can see the difference between those two reactions. Here we have our tertiary alkyl halide with yellow being the halogen, and that's directly bonded to our alpha carbon, and we have three methyl groups around that. So one, two, and three. If we think about this tertiary alkyl halide trying to participate in SN2 mechanism, if the hydroxide anion tried to function as a nucleophile and attack this carbon, we have these bulky methyl groups in the way, which block the hydroxide anion. So there's too much steric hindrance from these relatively bulky methyl groups to attack this carbon. If we think about an E2 mechanism, remember, in an E2 mechanism, we're trying to take a proton from the beta carbons. Here's a beta carbon right here. Let's say we're trying to take this proton. We need to get these four atoms in the same plane. And it's easy just to see that in Newman projection. So if I rotate a little bit, and I turn this, we got a different confirmation. Now we can see that these four atoms are all in the same plane, and that's what we need for an E2 mechanism. And it's pretty easy for hydroxide to come along and take this beta proton right here. There's not much steric hindrance at all, so an E2 mechanism is possible. We've just seen that it's possible for a tertiary alkyl halide to undergo an E2 reaction. And actually, as you go from primary to secondary, to tertiary, you increase in the rate of reactivity in an E2 reaction. To explain why, let's look at mechanism for a secondary and a tertiary substrate, and let's analyze the products and see if we can explain why this is the case. So I'll draw in a beta proton here on my tertiary substrate, and we'll think about our strong base taking this proton. These electrons move in here. These electrons come off onto our halogen. So our product would be this alkene. So our electrons in magenta forms our double bond here. So the same thing for our secondary substrate. I'll draw in a beta proton and we have our strong base that's going to take that beta proton. These electrons move in here, and these electrons come off onto our leaving group. And so, our product for a secondary substrate would look like this, the electrons in magenta formed our double bonds. Now let's analyze our two products. So on the left, this alkene is a monosubstituted alkene. We have one alkyl group bonded to this carbon. On the right, we have a disubstituted alkene, so we have two alkyl groups bonded to this carbon. And as the double bond forms in this concerted mechanism, it is stabilized by the presence of these alkyl groups. So remember from the alkene stability video, the increase substitution of the double bond means increased stability, so this is more stable here. So increased stability. And this alkene has two alkyl groups to stabilize the forming double bond. The alkene on the left has only one alkyl group to stabilize the forming double bonds. That's one way to think about why a tertiary alkyl halide would actually react the fastest. Now, it is possible for a primary alkyl halide to undergo an E2 mechanism, so don't think that a primary won't. As a matter of fact, it will, and we'll talk about examples of that in future videos.