E2 Mechanism

Okay. So I want to talk about a mechanism that competes directly with the S and two substitution and that's the E two elimination. So if I were to sum the entire reaction up in one sentence, what I would say is this, I would say an E two reaction happens when a strong nuclear file reacts with an inaccessible leaving group. Okay, I'm just gonna stop right there for a second. Okay, So if you remember back from what we learned about SN two, there's a little bit of a similarity there. Can you guys tell me what condition is similar to the sN two? The nuclear file? Remember that in in essence to mechanism you needed a strong nuclear file to start that backside attack. Okay, same thing with you two. We also want a strong nuclear file 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 very accessible backside? Well, for E two, we prefer an inaccessible leaving group. What that means is that these molecules are gonna be bad, generally bad at doing a backside attack. So they're gonna prefer to do something else instead. All right, so what is that other thing? What they're gonna do is beta elimination. Okay, so elimination of a beta proton. We're gonna talk about that in a second. All in one step. Okay, so 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 gonna 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 gonna start at the negatively charged nuclear file. Just like it did for SM2 because this is a strong nuclear file. So it's going to initiate the contact first. All right. So, I'm coming over here and you'll notice that I have this nuclear file that wants to do it sees this alcohol highlight. There is a very strong di pole there. There's a partial positive right here. And this nuclear file 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 nuclear file 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 where you're going to notice is that this is actually a tertiary alcohol highlight. Okay, do you remember what I said about tertiary alcohol highlights? 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 nuclear files frustrated, It's like, well, I'm a strong nuclear file. I want to do backside attack, but I can't So what am I gonna do? Well, instead it says, okay, instead of being acting like a nuclear file and donating my electrons maybe I can act more like a base. And the way that that bases act is that they are proton except ear's 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 gonna go ahead and erase this arrow. And that's not actually gonna be what happens. What happens is we're gonna look for a beta hydrogen that we can take off with my new profile 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 um are alive. And then a beta, 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 three beta hydrogen right here and we might have other beta hydrogen on those r groups. But the r groups are general. So I don't know how many there are or not? So the only ones that I'm given here are these three. Does that make sense? So those would all be beta hydrogen because their hydrogen is directly coming off of the beta proton. I mean coming off the beta carbon. So like I said, we're going to pull off a beta hydrogen instead. Let's go ahead and draw that. The nuclear file 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 to okay, so if I make that bond I'm gonna have to break a bond and this is the interesting part, we're gonna take the electrons from the bond from the carbon to the hydrogen, we're gonna 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 four bonds already. And now by making a double bond it would have five bonds. So if I'm gonna make that bond then 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 a transition state, What our transition state is gonna look like is like this, I'm going to 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 like 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 age 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's happening at the same time, the H bond is being broken, the double bonds being made and the leaving group is leaving all at the same time, awesome. Okay, now there is there are too many bonds here, so there'll be a negative charge distributed throughout. I'm just gonna write the negative on the outside that just shows that the entire thing is negatively charged. Okay, because you have one too many bonds and now what I want to do is show you one more unique thing about the E. Two mechanism in particular. Okay, this only has to do with E. two. And what it is is that if you were to take a Newman projection of this transition state. Now I know it's been a really long time since we talked about Newman's. So try to try to unburied that information and I had already like had already like buried it whatever. So try to think about a new projection. Remember that? That was a way to visualize single bonds. So here's my eyeball and if I were looking down the center of that bond, what would I see? Well what I would see is that on the top I have 22 H. S. Okay. And what do I have coming off the bottom? The bottom? What I would have is a partial bond to an H. Okay then what would I get in the back? What I would get in the back is I have those two are groups right? With single bonds. So our our but then on the front I would have a partial bond to my halogen. Alright so there you have it. That's what the transition state is actually going to look like. Okay and the unique thing about About E. two is that the transition state will always look like this. It's always gonna have that confirmation where my ex. And my H. R. as far apart from each other as possible. Do you guys remember what that confirmation is called? Remember that's 180° apart And 180° apart equals anti. Okay, so it turns out that whenever you have an E. Two elimination because of what's favored the way that it's favored, It's only gonna react once the X. And the H. R. Perfectly anti to each other or 100 and 80 degrees apart. And later on I'm gonna teach you guys how that's really important in predicting products. Okay. But just to let you guys know these two Hs that I drew at the top would not have actually been able to react unless they rotated down into the anti position. So really, even though I said that I had three beta hydrogen in this reaction, I only had one. That was in the proper position to be eliminated. All right, cool. Don't let that get you to confused because like I said, we're going to have an entire section dedicated to this one thing about the anti. Okay, so then what would the product look like? Well, we know that the H gets removed. Okay. So I'm just gonna chop it off even though this isn't part of the transition state, but you're gonna chop off the H. We also know that the X. Gets removed, Right? So what you get left is just a double bond in the middle with Hs and Rs on both sides. That's what my product would look like. It would just be a double bond with two Hs on one side. Okay. And two R. S On the other. Okay. And that is an elimination reaction. What I just did was I took two sigma bonds. This was a sigma and this was a sigma. Okay. I destroyed those bonds and I made a new pi bonds. And that's the definition of elimination. You take two sigma's and you make one pie. Alright, awesome. Other things that I would get are just my leaving group and then my nuclear file with an H. On it. All right, easy.

what I want to do is like we did with the substitution reactions, go through a set of facts and see if you guys can, you know, see if you guys can figure them out or not. Alright, so for E two, first of all, what kind of nuclear fall is it going to favor or it's going to favor E too strong or weak? And you're exactly right, it's going to be strong. Just like we used for SN two because remember that you need a negatively charged and I said that negatively charged is strong and neutral is weak. So this would be strong. Okay, how about my leaving group? Would I prefer un substituted or would I prefer highly substituted? The answer is highly substituted. Why? Because check it out if I use un substituted, guess what? That's going to favor SN two because then I can do a backside attack. So maybe around here you guys can just write that? This one would favor SN two but this one's gonna favor E two. Okay, and that's just kind of, it's just a spectrum of the more substituted it is the more SN two you're going to get, the more substituted or blocked off. It is the more you're gonna get, they really directly compete with each other. Okay, the bulkier you can make it, the more the chances are that you're gonna get 100% of the E two. Cool, awesome. How about the reaction coordinate? Would this be a transition state or an intermediate transition state. Okay. And we said that that's why because this is concerted or because it's two step, concerted means one step. So it would be concerted. Okay, now let's talk about rate. Okay, so all the rate information just you know, for E two is gonna be the same as SN two. Why is that because the two stands for by molecular? Okay, so this is E two. That means elimination by molecular by molecular. I should probably write it up here too. Okay, and that means that the rate is going to be contingent on both the nuclear file and the alcohol highlight or and the leaving group. Okay, so Remember that analogy that I had with the arrows and with the targets, it's the same exact thing here. Except that my target has changed a little bit. Remember what the target was before for sn two. Okay, what it was that is that you had your strong nuclear file, that was your arrow and you had your target that represented what it represented the back side. Does that make sense? The only way this changes is that for E two you still have that arrow and you still have a target? Okay. The only thing that changes is that the target represents something else. Now it represents beta hydrogen. Okay, and in the same way the more beta hydrogen is that I have around or the higher concentration of my leaving group, the better the chances are that I'm going to get a collision and then that's going to lead to an elimination. Does that make sense? So it's the same analogy just applies to a slightly different part of the actual highlight. But it's the same thing that if I double the concentration of alkali Halide, that's still going to double the rate of my reaction because I'm doubling the amount of beta hydrogen that are available to be collided into. Does that make sense? Cool then, lastly, what is this box I have for line I have for stereo chemistry. That is what I was just talking to you guys about. The stereo chemistry for E two is very particular because it always needs to be and this is the word anti co planer. Okay, now, just you guys know some professors and even if you read it in another textbook or if you look it up online, what you might find is that your professor or online they might call it anti perry planner. If you see those two words, that's the same thing. Okay, all it means is that your beta hydrogen and your Leaving group? So I'm just gonna say alcohol, hail, I'd have to be 180° apart in order for them to react. Okay, so the beta hydrogen that you're extracting as a base and the leaving group that's leaving have to be 100 and 80 degrees or just not gonna react. Okay, why is that? It has to do with a lot of theory about molecular orbital's and stuff that we're not gonna get into. Can I just tell you guys to just trust me on that one? You can also look it up in your book? Your book has a pretty good explanation of why. All right. But for right now, all we're gonna do here at clutch is we're just gonna memorize that and learn how to recognize it. Okay.

so really quick exercise guys, this super easy. Go ahead and look at these four alcohol hail IEDs and tell me which of them would be the best at e two and then just draw like a spectrum of which would be the best in which would be the worst. So go for it. Just pause the video and try to figure that out. All right? So hopefully what? You recognizes that once again, I had the same degrees of Al Kyohei lights, and the best one is just gonna be the tertiary. Okay, The reason is because the tertiary is the most backed off, So it's the one that has the worst s and two. So you could say here worst or worst backside and then over here was best backside. Okay, wise, I'm 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 in primary, these air gonna favor sn two because there's a really easy backside to hit. Okay, if I have a really bad backside, like maybe secondary and tertiary then these air going to favor e to Does that make sense? Because then the backside is really hard to hit now there is. Unfortunately, it's not as clear cut. Is this because it turns out that secondary is kind of right in the middle secondary? Depending on the conditions it can do s into like we've done before. We did some secondary s and twos, but it can also be forced to do e to 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 gonna give you as a flow chart that explains everything, every condition that you need to watch for. But right now, I just want you to know that as you go up the spectrum of substitution, you're gonna favor E to as you go down that spectrum you're gonna favor S and two and secondary is just kind of like a messy, a messy one that we're gonna have toe dissect later. All right, cool. Awesome. So let's go ahead and move on to the next topic.