Hey guys. In this video, we're going to dive deep into a type of pericyclic reaction called a thermal cycloaddition. So, thermal cycloadditions are pericyclic reactions in which 2 pi bonds are destroyed, by the way, we know that it's 2 because it's a cycloaddition, and they always destroy 2 pi bonds, after a heat-activated cyclic mechanism. So, we know that all pericyclic mechanisms are cyclic, but this happens to be a thermally controlled one, not a photochemically controlled pericyclic reaction. It turns out that there's already a very popular example of this type of reaction in your organic chemistry textbook, and that's called a reaction. Now, you may have not studied the reaction yet, or maybe you just got finished studying it, but this reaction is an example of a thermal cycloaddition. So here's just the general kind of rundown of what you would get. You would have 3 pi bonds reacting under heat to form a new cyclic product that only has 1 pi bond. Okay. That's the basics of how it works. Now, very quickly, I do want to show you guys the mechanism. The mechanism would have to be a cyclic, concerted reaction where all of your double bonds are forming new bonds within the system to make a new ring. So, for example, this double bond right here would form a new bond in between the 2, then that's basically going to form a new bond between the diene and the alkene. Then this double bond would come down and form a new double bond here, and then finally, this double bond would come around and attack this one. Now, actually, even though I showed that it's making a bond, typically, this first mechanism arrow is actually drawn to that carbon, just so people understand that it's actually going to form a new link to that carbon, okay? And what you would get as a result is a new cyclic adduct, where now you have these 2 molecules that came together and made a new ring. Okay? Now that's the basics of the mechanism, but to really understand what's driving a cycloaddition, a thermal cycloaddition, you need to go back to frontier molecular orbital theory. We need to understand what frontier orbitals are. We need to understand how HOMO and LUMO are interacting for these 2 molecules. And it turns out that it's a very kind of simple rule, which is that in a cycloaddition, the HOMO from 1 molecule, which I'm calling HOMO a, must fill the LUMO from molecule b. So that means that the HOMO from 1 is going to kick up electrons to fill the LUMO of the other molecule, and that is how they're going to join together and make a new ring. Okay. According to frontier molecular orbital theory, this bonding interaction is the strongest when the symmetry and energy levels between your molecular orbitals match closely. So, what does that mean in terms of symmetry? Symmetry has to do with the lobes. Remember, when you have molecular orbitals, you have lobes facing in different directions, right? So, you want to make sure that the terminal lobes of your HOMO match the terminal lobes of your LUMO, so that there can be good bonding overlap between them. That's what we call symmetry, and I'm going to show you more of an example, but for right now, just know that it has to do with what direction the lobes are facing. You want them facing similar directions, okay? And then, what do we need by energy? The energy also has to match because you want your HOMO and your LUMO to be close in energy, and not far apart. If they're very far apart in energy, one is much higher than the other, it's difficult to make a strong bond. So you're trying to find the HOMO that's closest to the other's LUMO, and that's where you're going to find a really strong interaction. So, just to kind of summarize what I just said, in order for a cycloaddition to take place, you're looking at 2 things. 1, the reaction must be what's called symmetry allowed. Symmetry allowed has to do with the orbitals matching up properly. Okay. Symmetry allowed, not disallowed. Disallowed is also called just, you know, forbidden. This is another term your textbook may use, symmetry forbidden or just forbidden. It needs to be symmetry allowed, okay. And the second one is that you want to try to minimize your gap, which is what I was just talking about in terms of energy levels. The gap gets bigger the further apart those HOMO and LUMO orbitals are in energy. So you want to make sure that you're trying to minimize that gap as much as possible. Cool? Awesome, guys. So now it's time to actually get into the molecular orbital theory of a cycloaddition. I hope you're excited. So here's our diene. Let's start off with our diene. You should know how to draw the molecular orbitals for a diene at this point. We've gone over that in other videos. And what we know about a diene is that it has 4 pi electrons. So that means that psi 1 should get 2 electrons, and I've already gone ahead and labeled our HOMO and our LUMO orbitals for the diene. Once again, a is just to signify that we're dealing with this molecule a. I'm calling it a for this reaction. Okay. Now, in heat, we're going to be exposing that diene to another conjugated molecule. In this case, it's the simplest conjugated molecule which is an alkene, but it doesn't have to be just an alkene. It could be another diene; that would be fine, just something else that's conjugated that has HOMO LUMO orbitals. Okay? So in this case, we should know how to fill in the orbitals for this. How many pi electrons are there? 2. So that means that I should fill in 1₂. And I've also filled in that this is the HOMO for molecule B and on top, we have the LUMO for molecule B. So what did we say about what we're trying to accomplish in a cycloaddition? What we're trying to do is we're trying to take the electrons from HOMO a and use them to fill the orbital of LUMO B. Okay? So that's what the heat energy is going to do. You might say, well, Johnny, why would these electrons raise in energy? Well, that's why you need to use heat.
Thermal Cycloaddition Reactions - Online Tutor, Practice Problems & Exam Prep
Thermal Cycloaddition reactions are pericyclic reactions in which 2 pi bonds are destroyed after a heat-activated cyclic mechanism.
MO Theory of Thermal Cycloadditions
Video transcript
Determining Favorability of HOMOb to LUMOa
Video transcript
Now let's look at the favorability of going the other way around, trying to make a bonding interaction between HOMO B and LUMO A. Okay? So, this also means that we're going to have to redraw our molecular orbitals because they have changed. HOMO B is pretty easy, we already drew this. Notice that it is this guy right here. So I'm just going to bring that down. HOMO B would look like this. Cool. Now LUMO A, we didn't draw because LUMO A is actually side three of the other molecule, so that means we have to draw it from scratch. Let's go ahead and draw it right here. So much room for improvement. That's not the best. So how do we draw this? Remember that for side three, this would continue to face down so it would be dark at the bottom, and remember that this gives my last orbital two chances to flip. So it would have been down, then on side two it flipped up, and then on side three it flips back down. And then finally remember that I need how many nodes? Two nodes because I need one increasing node with every MO, so that means that I would have a node here and a node here, meaning that I should have orbitals, the phases facing up in the middle and that is my MO diagram for side three. This is what is interacting with HOMO B and I'm going to bring this down. Let's bring that down into this area. So what I have is down, up, up, down. Awesome. Okay. So now we're ready to determine and to make some decisions about if this is going to be favorable or not.
So first of all, let's look at the symmetry of this reaction. Is this a symmetry-allowed or a symmetry-disallowed process? And guys, what I see is that actually this is symmetry allowed again because what I have is that this one matches this one, right? So there could be a bond there and then this one, oops, I want to use a different color, this terminal end matches this terminal end, right? So terminal end matches this terminal end. Once again, this is symmetry allowed. Bonds can't form here. What I could do is I could form a bond here and a bond there. Cool? So we know that bonds could form, so this is symmetry allowed.
Let's go to the next one. Is this the smallest HOMO LUMO gap possible? And what we see here is that I actually went ahead and I brought down the same exact distances from my original LCAO diagram into here. So notice that this is purposeful. The reason that the HOMO and the LUMO look a little bit further apart in this one than they do on the first one is because they are further apart. If you actually look at them, they are drawn much further apart and that's the way that it works. Usually, there's going to be one set that's closer and one set that's further away. The set that's further away isn't going to work as well. So this would get a big X in the smallest HOMO LUMO gap. Is it possible to make this one work? Yes. But then you would need to try to decrease, try to do different things to the molecules to try to decrease the gap between them so that the bonding interaction would be favorable. So, several of you, if I wouldn't have gone through this exercise, many of you would have probably thought, well Johnny, why couldn't it just go the other way? And this is why because it's not just about being symmetry allowed, it's also about having the smallest HOMO-LUMO gap possible.
So, guys now you know how to find the favorable reaction of a cycloaddition. Remember that this is true for any conjugated molecules A and B, not just for a four-membered chain and a two-membered chain. It could go up to many atoms in the conjugated system and now you have the tools to figure out if it's favored or not. Okay? So we're done with this video, let's move on to the next one.
Who's ready for some practice? See if you can predict the correct answer.
Use FMOT to predict the product of following cycloaddition reaction.
Use FMOT to predict the product of following cycloaddition reaction.
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