E2 - Anti-Coplanar Requirement

Alright, guys and I want to talk about a really important topic that Onley applies to the E two mechanism. That's called the Antico Plainer requirement. So as I told you guys already, e two reactions are going to require an antique a plainer arrangement between the leaving group and the beta hydrogen in order to go to completion. Okay. And that's because the orbital's need toe overlap in a certain way in order to make a new pi bon, which is that double bond that you get the end. Okay, so that's the first thing that we need to know now. Not only are we gonna have to look at how many different beta hydrogen we have, but now we're gonna have to look at an extra level of complexity, which is how many of these beta hydrogen can be in the anti position or are in the anti position. So that means that we're going to require two steps to figure out the amount of products that we have. First, we're gonna look a beta hydrogen. And then after we figured out the number of data hydrogen, we're gonna figure out, Are they antique? A plane or not? Okay. On top of that, there's one more thing I should know. Which is that when you have a leaving group in a beta hydrogen on a cyclo hexane, that's actually going to form a chair, Right? Remember that cycle? Hexane is usually are in the chair confirmation. And when you're dealing with an elimination on a chair instead of calling it anti complainer, we're actually gonna call it a die axial requirement instead of this is the same thing as Antico. Plainer. Okay. And why is that? Why do I say co plainer? Why do I say die axial? Because the Onley way that the leaving group and the beta proton can be anti to each other is if they're on adjacent axial positions. The reason is because think about the equatorial positions, the actual positions go like this. The equatorial positions go like this, right? Let me see. I'm doing this all wrong, But let's say that the actual positions are like this. The equatorial positions to do this, that's not an anti arrangement that's actually like a gouache or something like that. Okay, so in order for the elimination to occur, you're gonna need to rotate a chair to the actual position first, even though that's the less stable position and that actually something to do with it as well. Even though this is less stable. I need to rotate it like this in order to make my reaction happen. Because I need my groups to be anti not. Gosh. Okay, so that looks like I was doing a really weird dance, So I hope you guys enjoyed that. Um, we're gonna do here is a really quick practice. Not a lot of drawing. In fact, I don't want you to draw anything yet. All we're analyzing is would these e two reactions happen or not? Notice that I have a strong nuclear file, okay. And I have either a secondary or a tertiary alcohol. Haley, remember that? I said secondary and tertiary can do et too, because I have a bad back side or not that great. So I want you guys to figure out, first of all, like, how many beta hydrogen is you have? Okay. How maney. Beta hydrogen, Different data. Hydrogen. Would you have? And then once you figure that out determined, would they be anti complainer or not? In order to make the reaction occur? So this is two steps. First of all, do the same thing that we did for the beta hydrogen exercise. Figure out how many how many different ones we have. But then on top of that, figure out how many of those are actually anti complainer? Okay, and that's going to be the number of possible products for E two. Alright, so go ahead and try it with the first one and then I'll explain it.

All right. So for this first one, how many different beta hydrogen did you guys find? Actually, it should have only been one. Because this is my Alfa. And if you'll notice my Betas are all exactly the same. They all have hydrogen, and they're all metal group, so it doesn't matter which one I pick. All right, so let's just erase the bottom two and look at the top. OK, now, the next question, this is the part you guys aren't good at yet. Could one of the ant will one of the hydrogen sbi anti? Okay. And I'm going to give you guys a rule of thumb. Any time that you have hydrogen on a metal group like this, always assume that at least one of them, that one of them will be anti. Why? Because this could rotate as much as it wants. So what that means is that at least one hydrogen will always be in the position where it's faced opposite to the direction of the Brahmins. To notice that the Brahmins facing that direction this is the hydrogen that would be eliminated. Ok, do we need to actually figure that out? Not really. You could just assume that if it's a method, then it will have an anti complainer arrangement. Okay. So I could eliminate my nuclear fall. I'm not gonna draw it yet, but my nuclear fall could eliminate from there and make a double bond and kick out my bro. Mean, just like I showed you guys when I did the e two, and that would be in an anti arrangement. Okay, It would be anti, by the way, if we were looking from destruction right here and looking down, you would see that the Brahman is facing straight up and the H is facing straight down. All right, so the answer is that this would give us one product. Okay, Cool. So and and obviously it would react. So let's go on to the next one.

So how many different beta hydrogen is did you guys find? And it turns out that for this one, there should have just been one, because this is my Alfa and my Alfa is connected to two beta carbons. But what we're gonna find is that one of those beta carbons doesn't have any hydrogen on it, so it can't count. So that means I only have one side that's available. And that's why it happens to be a metal. So can one of the hydrogen to be anti absolutely. Like I said for these, if it's metal, you assume that it will work. Just to show you guys I would have one h in the front. I would have one each in the back, and I would have one h going straight down. And the basically the h that's going towards the front would be the one that I could remove because the front one, if you draw your thing correctly, if you draw your name correctly, if it's going the opposite direction, that's going to be the one that is gonna be anti to it. Okay, so I know it might not look like it there, but it's Actually, it actually would work. Okay, so we would only have one area that would form a double bond. That'll be between this Alfa and this beta. So once again, I would get one product. Okay, Cool. So let's move on to the next question.

All right. So how maney beta hydrogen did you find here? Well, the answer is actually two different ones. The reason is because this is my Alfa and these were my Betas. Let's say this is my Blue Beta, and this is my green beta. Okay, Do both of these have hydrogen on them? Yes, they dio if you'll notice we have for the green one. We have a hydrogen coming straight off the front like this. Hoops. Let's do that again. Okay. Why is it going on the dash? Because the alcohol is on the wet. Why is it on the wedge? Because the alcohol is on the dash. Okay. And then what about the blue one? The blue one has a metal group face towards the back. So that means I must have a H facing towards the front. Pretty easy. So far right now. Are these equivalent or non equivalent these air non equivalent? Because the blue direction would eliminate towards, um, Ethel. The green direction would eliminate towards the Ohh. Okay. Is that making sense so far? So I should get. So far, I have two different types of beta hydrogen. That was the first level that's what you're supposed to do just by recognizing beta hydrogen. Now, let's go into the second level, which is just specifically for E two. What does it to require Antico plainer. So are any of these two hydrogen actually anti to my chlorine? And the answer is no. Okay, the answer is that this would be no reaction. Crazy, right? There would be no reaction. And why is that? Because I have two different hydrogen is. But both of them are what we would call sin to my chlorine. Or also what you would call in terms of new and projections we call them eclipsed. Okay, now, I know that I just pulled the word sin like a clips. Okay. Where the hell did sin come from? Sin and anti are words that are used on many times opposite to each other to mean that sin would be sis and anti would be trans or facing different directions. Okay, so if you ever hear me say the word sin, that's literally a synonym persists. Okay, that we'll talk about later, but also in terms of noon projections. If I were to draw a Newman projection of this bond that way that would be eclipse. Because you'll notice that my chlorine facing up my hydrogen also facing up. That's really bad. I need them to be faced anti not zero degrees, 180 degrees. So it's no direction because none of these hydrogen zehr anti. And this is where professors get you This whole chapter Probably the hardest type of question you're going to get is a une to elimination and having to figure out which ones are Antico player and which ones aren't. So I hope that by approaching this systematically, you guys will get into the habit of always looking at How many different data hydrants do we have? And then secondly, how many of them are actually Antico player? Okay, cool. So let's move on to this last one. Now, we do need to give you a preface for this last one. This last one that you're gonna notice is that the flooring is equatorial. Okay. Now, even though that might be more stable in terms of the equatorial preference, it cannot react in this confirmation. So what you're gonna need to do first is you're gonna need to draw a chair flip of this of this molecule. You have to draw it so that all the positions are reversed. Remember that when you do a chair flip, Axl's become equatorial equatorial become axles. So you have to draw all that and then see how maney anti complaint or hydrogen is. You have beta hydrogen. You have. All right, I know it sounds like a little bit of a bitch, but go ahead and try your best, and then I'll show you an easy way to how to do it after you've given it or try. Okay, so go for it.

The answer for this one was that only one product was possible even though we had more than one beta hydrogen. Alright. And that had to do with Antico plan or what we call the dye actual requirement when you're talking about chairs. Okay, Now, at the beginning, I told you guys as a hint, draw the chair, the chair flip, and then figure it out. And even though technically to be totally safe, you should draw the entire flip, meaning that you draw the opposite looking chair and you draw all the positions the same. I have a little cheap for that. Sometimes just gonna work even though it's not technically technically correct, which is, instead of drawing the perfect chair flip instead, just flip all the positions on the same molecule. So what we're gonna do is we're just gonna draw that this f goes here, we're gonna draw that this ch three goes here and we're gonna draw that. This ch three, I guess, goes out here. Okay, So notice that. And if you want, just go ahead and scratch out everything else. So scratch that out. Scratch that out. Scratch that out. Okay. I'm just trying to think of the fastest way. If you write a test and have to figure this out, that'll be the fastest way to figure it out. Okay, so now I look at my positions and I say, Okay, all of my ax sales have become editorials on my equatorial became axles. How maney beta hydrogen is Do I have? Well, this is my Alfa. Okay, So that means that I have to beta carbons. I have. This is my beta carbon, and this is a beta carbon. Okay, How many of those have hydrogen? Actually, both of them have hydrogen. I'm gonna have ah hydrogen facing equatorial here. So where the old metal group Waas and I drew that equatorial a little bit wrong. Let me actually draw it a little better. Should just be going like that, Okay. And then I'm also gonna have a hydrogen axial where the other one was, so I should have a hydrogen facing down like that. Okay, I know this color is difficult to tell, so I'm gonna put a circle around the hydrogen so you guys can tell where it is. Okay, so now my question is, are those two exactly the same. Okay. And the answer is no, because they're in different positions. Okay. Could they both eliminated any to? No, actually, no. The only one that could would be this one right down here. Okay. And the reason is because that one is axial. And this one is Axl. My leaving group is now Axel, remember, we put the leaving group there on purpose because it has to be in the actual position. So now, as you'll see, if I were to draw a Newman projection of that, I would have one going down when going up. It would be perfect. Okay, how about the equatorial over here? That one is not die Axle. It's not in an actual position, so this one would not work. So the answer is I could only get one elimination product. All right, so we're not going to draw it out. But we will just say that you would get one product. Alright, Cool. So I hope that exercise helped you guys to see how important Antico Planner is and also how tricky it can be. It's one of those things. You just have toe always be looking out for it. All right. And now let's just move on to the next video