E2 Mechanism - Online Tutor, Practice Problems & Exam Prep

The E2 elimination mechanism occurs when a strong nucleophile interacts with an inaccessible leaving group, favoring beta elimination over backside attack. This concerted reaction requires the beta hydrogen and leaving group to be in an anti-coplanar arrangement, ensuring optimal orbital overlap. The rate of E2 reactions is bimolecular, dependent on both the nucleophile and the alkyl halide. The product formation involves the conversion of two sigma bonds into one pi bond, exemplifying the elimination process. Understanding these principles is crucial for predicting reaction outcomes and mechanisms in organic chemistry.

Drawing the E2 Mechanism.

Okay. So now I want to talk about a mechanism that competes directly with the SN2 substitution, and that's the E2 elimination. So if I were to sum the entire reaction up in one sentence, what I would say is this. I would say an E2 reaction happens when a strong nucleophile reacts with an inaccessible leaving group. Okay? I'm just going to stop right there for a second. Okay? So if you remember back from what we learned about SN2, there's a little bit of a similarity there. Can you guys tell me what condition is similar to the SN2? The nucleophile. Remember that in an SN2 mechanism, you needed a strong nucleophile to start that back side attack. Okay? Same thing with E2. We also want a strong nucleophile. Where the difference comes in is the leaving group. Remember that if you had a very accessible leaving group, what would happen? Backside attack. So remember that backside attack was favored when you have this very accessible leaving group. Okay? Or a very accessible backside. Well, for E2, we prefer an inaccessible leaving group. What that means is that these molecules are going to be bad, generally bad at doing a backside attack, so they're going to prefer to do something else instead. Alright? So what is that other thing? What they're going to do is beta elimination. Okay? Elimination of a beta proton, we're going to talk about that in a second, all in one step. Okay? Let's go ahead and get started. Let's just start the mechanism off and you guys tell me where you think the first arrow is going to come from. Maybe you don't know where it goes, but at least you should be able to tell me where it starts. And that's right. It's going to start at the negatively charged nucleophile just like it did for SN2 because this is a strong nucleophile, so it's going to initiate the contact first. Alright?

So I'm coming over here and you'll notice that I have this nucleophile that wants to do it sees this alkyl halide. There is a very strong dipole there. There's a partial positive right here, and this nucleophile wants nothing more than to give its electrons directly to that positive charge. Okay? So actually don't draw what I just drew yet. I'm just trying to guide you guys through the process. Alright? The nucleophile wants to donate its electrons to that positive, but there's a problem. The problem is that if you'll notice, count that carbon up, what you're going to notice is that this is actually a tertiary alkyl halide. Okay? Do you guys remember what I said about tertiary alkyl halides? Do they have a really good backside? No. They have a terrible backside. Okay? In fact, it's impossible to get through. Okay? Just cannot get anywhere close. So now this nucleophile is frustrated. It's like, well, I'm a strong nucleophile. I want to do backside attack, but I can't. So what am I gonna do? Well, instead it says, okay, instead of acting like a nucleophile and donating my electrons, maybe I can act more like a base. And the way that the bases act is that they are proton acceptors. So it's saying, you know what? It's too difficult to do this backside attack, so instead let me just pull off a proton. Okay? And by pulling off the proton, maybe I can donate my electrons that way. So we're going to go ahead and erase this arrow and that's not actually going to be what happens. What happens is we're going to look for a beta hydrogen that we can take off with my nucleophile as a base. Okay?

So how to find beta protons, just to remind you guys, would be that this is my alpha carbon. The alpha carbon is the one that's directly attached to my alkyl halide, and then a beta carbon is any carbon that's directly attached to the alpha, so this would be beta, this would be beta and this would be beta. All of those are beta carbons. Okay? Because they are carbons directly attached to the alpha. Alright? And then any hydrogen that's directly attached to a beta carbon is considered a beta hydrogen. So what that means is that I have 3 beta hydrogens right here, and we might have other beta hydrogens on those R groups, but the R groups are general, so I don't know how many H's there are or not. So the only ones that I'm given here are these 3. Does that make sense? So those would all be beta hydrogens because they're hydrogens directly coming off of the beta carbon. So, like I said, we're gonna pull off a beta hydrogen instead. Let's go ahead and draw that the nucleophile attacks that hydrogen right there. Okay? Now, is that hydrogen happy with that mechanism? Can I just leave it there? The answer is no. I cannot just leave it there because hydrogen only wants to have one bond and now it has 2. Okay?

So if I make that bond, I'm gonna have to break a bond and this is the interesting part. We're going to take the electrons from the bond, from the carbon to the hydrogen. We're going to give those electrons to that single bond. Basically, the bond in between the alpha and the beta, okay, is going to get a double bond. So alpha double bonded now to beta carbon. Okay? So now I have a double bond there, is that the last arrow. Actually, it can't be because this alpha carbon had 4 bonds already, and now by making a double bond it would have 5 bonds. So if I'm gonna make that bond, then I have to break another bond, and the easiest bond to break is the one for the leaving group. Because remember, the leaving group is gonna be stable after it takes off. Okay? Relatively stable. So let's go ahead and draw our transition state. What our transition state is gonna look like is like this. I'm gonna draw everything that didn't change in the reaction with a solid line. So what that means is that I would have an H in the front and an H in the back that nothing ever changed. I would also have an R in the front and an R in the back that never changed. Okay? So those are the things that, during the course of the reaction, nothing's happening to them. But what is changing is that a bond is being broken and destroyed at the same time between the H and between the leaving group. So the reason I drew it with partial bond is because this is a one-step reaction. Right? So what that means is everything is happening at the same time. The H bond is being broken, the double bond is being made and leaving group is leaving all at the same time. Awesome. Okay.

Rank reactivity toward E2

So, really quick exercise, guys. This is super easy. Go ahead and look at these four alkyl halides and tell me which of them would be the best at E2, then just draw a spectrum of which one would be the best and which one would be the worst. So go for it. Just pause the video and try to figure that out. Alright. So, hopefully what you recognized is that once again, I had the same degrees of alkyl halides, and the best one is just going to be the tertiary. Okay? The reason is because the tertiary is the most backed off, so it's the one that has the worst SN2. So you could say here worst or worst backside, and then over here was the best backside. Okay. Why am I talking about elimination here? Why am I thinking about backside? Because remember, that backside attack and elimination are going to compete with each other. So that means if I have a really good backside like zero and primary, these are going to favor SN2 because there's a really easy backside to hit. Okay? If I have a really bad backside, like maybe secondary and tertiary, then these are going to favor E2. Does that make sense? Because then the backside's really hard to hit. Now, there is unfortunately, it's not as clear-cut as this because it turns out that secondary is kind of right in the middle. Secondary, depending on the conditions, it can do SN2, like we've done before. We did some secondary SN2s, but it can also be forced to do E2. Alright? So that's why for right now, I don't want you to think too much or too hard about it. Later on, I'm going to give you guys a flowchart that explains everything, every condition that you need to watch for. But for right now, I just want you to know that as you go up the spectrum of substitution, you're going to favor E2. As you go down that spectrum, you're going to favor SN2 and the secondary is just kind of like a messy one that we're going to have to dissect later. Alright? Cool. Awesome. So let's go ahead and move on to the next topic.

Here’s what students ask on this topic:

What is the E2 elimination mechanism in organic chemistry?

The E2 elimination mechanism involves a strong nucleophile interacting with an inaccessible leaving group, leading to the elimination of a beta hydrogen and the formation of a double bond. This concerted reaction occurs in one step, requiring the beta hydrogen and the leaving group to be anti-coplanar (180 degrees apart) for optimal orbital overlap. The rate of the E2 reaction is bimolecular, depending on both the nucleophile and the alkyl halide. The process converts two sigma bonds into one pi bond, exemplifying the elimination process.

How does the E2 mechanism differ from the SN2 mechanism?

While both E2 and SN2 mechanisms require a strong nucleophile, they differ in their approach to the leaving group. In the SN2 mechanism, the nucleophile performs a backside attack on an accessible leaving group, leading to substitution. In contrast, the E2 mechanism involves a strong nucleophile interacting with an inaccessible leaving group, favoring beta elimination. The E2 reaction is concerted, requiring the beta hydrogen and leaving group to be anti-coplanar, whereas SN2 involves a direct backside attack without such geometric constraints.

What factors influence the rate of an E2 reaction?

The rate of an E2 reaction is influenced by the concentration of both the nucleophile and the alkyl halide, making it a bimolecular reaction. The strength of the nucleophile, typically negatively charged, and the degree of substitution of the alkyl halide also play crucial roles. Highly substituted alkyl halides favor E2 over SN2 due to steric hindrance. Additionally, the anti-coplanar arrangement of the beta hydrogen and leaving group is essential for the reaction to proceed efficiently.

Why is the anti-coplanar arrangement important in E2 reactions?

The anti-coplanar arrangement is crucial in E2 reactions because it ensures optimal orbital overlap, facilitating the simultaneous breaking of the C-H and C-X bonds and the formation of the C=C double bond. This 180-degree alignment minimizes steric hindrance and allows for the most efficient transition state, leading to a successful elimination reaction. Without this specific geometric arrangement, the reaction would not proceed effectively.

What is the stereochemistry of the E2 elimination mechanism?

The stereochemistry of the E2 elimination mechanism is characterized by the requirement for the beta hydrogen and the leaving group to be anti-coplanar, meaning they must be 180 degrees apart. This specific arrangement, also known as anti-periplanar, ensures that the reaction proceeds efficiently by allowing optimal orbital overlap. This geometric constraint is essential for the concerted nature of the E2 mechanism, where bond breaking and formation occur simultaneously.