Time to discuss the most complicated mechanism of the bunch, E2. It’s an awesome reaction, but there are a few extra details we’re gonna have to keep track of!
Concept: Drawing the E2 Mechanism.10m
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. I'm just going to stop right there for a second.
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. 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? Back side attack. So remember that back side attack was favored when you have this very accessible leaving group or very accessible back side.
Well for E2, we prefer an inaccessible leaving group. What that means is that these molecules are going to be generally bad at doing a back side attack, so they're going to prefer to do something else instead.
So what is that other thing? What they're going to do is beta-elimination, so elimination of a beta-proton. We're going to talk about that in a second. All in one step. 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's 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. 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.
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. So actually don't draw what I just drew yet, I'm just trying to guide you guys through the process.
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. Do you guys remember what I said about tertiary alkyl halides? Do they have a really good back side? No. They have a terrible back side. In fact, it's impossible to get through. Just can not get anywhere close.
So now this nucleophile is frustrated. It's like, “Well, I'm a strong nucleophile. I want to do back side attack, but I can't. So what am I going to do?” Well, instead it says instead of acting like a nucleophile and donating my electrons, maybe I can act more like a base. The way that 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 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. 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 because they are carbons directly attached to the alpha.
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-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 three. Does that make sense? So those would all be beta-hydrogens because they're hydrogens directly coming off of the beta-proton, I mean coming off of 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 nucleophile attacks that hydrogen right there.
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 two. So if I make that bond I'm going to 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 is going to get a double bond. So alpha double bonded now to beta-carbon.
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 going to make that bond, then I'm going to have to break another bond and the easiest bond to break is the one for the leaving group because remember the leaving group is going to be stable after it takes off, relatively stable.
Let's go ahead and draw our transition state. What our transition state is going to 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. 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 bonds is because this is a one-step reaction. 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 the leaving group is leaving all at the same time. Awesome.
Now there are too many bonds here, so there'd be a negative charge distributed throughout. I'm just going to write the negative on the outside that just shows that the entire thing is negatively charged because we have one too many bonds.
And now what I want to do is show you one more unique thing about the E2 mechanism in particular. This only has to do with E2. 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 Newmans, so try to unbury that information. I know you had already buried it or whatever. So try to think about a Newman 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 two H's. And what do I have coming off the bottom? The bottom what I would have is a partial bond to an H. Then what would I get in the back? What I would get in the back is I have those two R groups with single bonds, so R-R, but then on the front I would have a partial bond to my halogen.
So there you have it, that's what the transition state is actually going to look like. And the unique thing about E2 is that the transition state will always look like this. It's always going to have that conformation where my X and my H are as far apart from each other as possible.
Do you guys remember what that conformation is called? Remember that's 180 degrees apart and 180 degrees apart equals anti. So it turns out that whenever you have an E2 elimination because of what's favored, the way that it's favored, it's only going to react once the X and the H are perfectly anti to each other or 180 degrees apart. And later on, I'm going to teach you guys how that's really important in predicting products.
But just to let you guys know, these two H's 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-hydrogens, 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 too confused because, like I said, we're going to have an entire section dedicated to this one thing about the anti.
So then what would the product look like? Well, we know that the H gets removed. So I'm just going to chop it off, even though this isn't part of the transition state, but we're going to chop off the H. We also know that the X gets removed. So what you get left is just a double bond in the middle with H's and R's on both sides. So that's what my product would look like. It would just be a double bond with two H's on one side and two R's on the other. That is an elimination reaction.
What I just did was I took two sigma bonds, this was a sigma and this was a sigma, I destroyed those bonds and I made a new pi bond. And that's the definition of elimination. You take two sigmas and you make one pi. Awesome.
Other things that I would get are just my leaving group and then my nucleophile with an H on it. Easy.
Summary: A negatively charged nucleophile reacts with an inaccessible leaving group to produce beta-elimination in one-step.
Concept: Understanding the properties of E2.5m
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 E2 first of all what kind of nucleophile is it going to favor? It is going to favorite E2, strong or weak? And you're exactly right it's going to be strong just like we use for SN2 because remember that you need a negatively charge and I said that negatively charged is strong and neutral is weak so this would be strong, OK? How about my leaving group? Would I prefer unsubstituted or would I prefer highly substituted? The answer is highly substituted, why? Because check it out if I use unsubstituted guess what that's going to favor? SN2 because then I can do a backside attack so maybe around here you guys can just write that this one would favor SN2 but this one is going to favor E2, OK? And that's just kind of it's just a spectrum of the more unsubstituted it is the more SN2 you're going to get the more substituted or blocked off it is the more E2 you're going to get they really directly compete with each other, OK? The bulkier you can make it the more the chances are they're going to get 100 percent of E2, cool? Awesome, how about the reaction co-ordinate? Would this be a transition state or an intermediate? Transition state, OK? And we said that's why because this is concerted or because it's two step, concerted means one step so it would be concerted, OK? Now let's talk about rate, OK? So all of the rate information just so you know for E2 is going to be the same as SN2, why is that? Because the two stands for bi-molecular, OK? So this is E2 so this means elimination bi-molecular...Bi-molecular I should probably write it up here E2, OK? And that means that the rate is going to be contingent on both the nucleophile and Alkyl Halide and the leaving group, OK? 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 SN2 what it was that is that you had your strong nucleophile that was your arrow and you had your target that represented what? It represented the backside, does that make sense? The only way this changes is that for E2 you still have the arrow and you still have a target, OK? The only thing that changes that the target represent something else now it represents beta hydrogens, OK? And in the same way the more beta hydrogens 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 Alkyl Halide but it's the same thing that if I double the concentration of Alkyl Halide that's still going to double the rate of my reaction because I'm doubling the amount of beta hydrogens that are available to be collided into, does that make sense? Cool, then lastly what does this box I have for Stereo chemistry? That is what I was just talking to you guys about, the stereo chemistry for E2 is very particular because it always needs to be and this is the word anti-coplanar, OK? Now just us know some professors and even if you read it in the other textbook or if you look it up online what you might find is that your professor or online they might call it anti-periplanar if you see those two words that's the same thing, OK? All it means is that your beta hydrogen and your leaving group so I'm just going to say Alkyl Halide have to be 180 degrees apart in order for them to react, OK? So the beta hydrogen that you're extracting as a base and the leaving group that's leaving have to be a 180 degrees it's just not going to react, OK? Why is that? It has to do with a lot of theory about molecular orbitals and stuff that we're not going to 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, alright? But for right now all we're going to do here at Clutch is we're just going to memorize that and learn how to recognize it, OK?
Properties of E2 reactions:
Example: Rank the following alkyl halides in order of reactivity toward E2 reaction.3m
For the following E2 reaction, draw the transition state that leads to the predominant product(s), indicating geometry and stereochemistry with wedges and dashes. We will be paying particular attention to the geometry of your transition state. No need to show any arrows for moving electrons here, we are just interested in the transistion state and product structures. In your transition state structure, be sure to show bonds being made or broken as dotted lines
Which of the following shows the first step for an E2 reaction?
Follow the curved arrows to predict the product(s) of the following chemical reactions
How many atoms and electrons are directly involved in the bond-making and bond-breaking of the first step in an E2 reaction?
(A) five atoms, six electrons
(B) four atoms, six electrons
(C) three atoms, six electrons
(D) five atoms, four electrons
(E) four atoms, four electrons
(F) three atoms, four electrons
Which of the following shows a mechanism for an E2 elimination?
Which of the following is the correct mechanism for the elimination reaction of 2-bromo-2,3-dimethylbutane with methoxide?