Thermal Cycloaddition reactions are pericyclic reactions in which 2 pi bonds are destroyed after a heat-activated cyclic mechanism.
Concept: MO Theory of Thermal Cycloadditions14m
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 two pi bonds are destroyed, by the way, we know that it's two because the cycloaddition, they always destroy two pi bonds, after a heat-activated cyclic mechanism. So, we know that all pericyclic mechanisms are cyclic but this happens to be a thermal controlled one, not a photochemically controlled pericyclic reaction, so 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 diels-alder reaction. Now, you may have not studied the diels-alder 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 run down of what you would get, you would have three pi bonds reacting under heat to form a new cyclic product that only has one 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, it would 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 single bond in between the two then that's basically going to form a new bond between the diene and the alkyne 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 just have two 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, we 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 two molecules and it turns out that it's a very kind of simple rule, which is that in a cycloaddition the HOMO from one molecule, which I'm calling HOMO a must fill the LUMO from molecule B so that means that the HOMO from one 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? Now, according to frontier molecular orbital theory, this bonding interaction is the strongest, when symmetry and energy levels between your molecular orbitals match closely. So, what that means is what do we talk, what do we talked was symmetry, symmetry has to do with the lobes. Remember, that you have whenever you have molecular orbitals you have lobes facing in different directions, right? So, what you want to do is you want to make sure that the terminal lobes of your HOMO match the terminal lobes of your LUMO so that there can be a good bonding overlap between them, that's what we call symmetry and I'm going to show you more, 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 we do match energy, the energy also has to match because you want your HOMO and your LUMO to be close in energy and not far in energy, if they're very far and energy one is much higher than the other, it's difficult to make a strong bond. So, what you're trying to do is you're trying to find the HOMO that's closest to the other LUMO and that's where you're going to find a really strong interaction, okay? So, just to kind of summarize what I just said, in order for a cycloaddition to take place you're looking at two things, one the reaction must be what's called symmetry allowed, symmetry allowed has to do the orbitals matching up properly, okay? Symmetry allowed not disallowed, disallowed is also called just so 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 HOMO-LUMO 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 what a diene is that it has four pi electrons so that means that psi one should get two electrons psi two should get two electrons and I've already gone ahead and labeled our HOMO and our LUMO orbitals for the diene, okay? 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 sign to another conjugated molecule, in this sense, in this case it's the simplest conjugated molecule, which is an alkene but it doesn't just have to be 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, two, so that means that I should fill in one and two and I've already 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? that's why you need to use heat, heat is going to be able to allow those electrons to jump into a higher energy state, which is the LUMO, for the other molecule, okay? So, just so you know in order to predict that this is going to be either symmetry allowed or symmetry disallowed, we are going to have to draw what the molecular orbitals look like, okay?
Now, there are some shortcuts, I'll show you later, but right now we definitely want to draw those molecular orbitals. So, if you don't mind, let's go ahead and draw them right next to the ones that are important, which are, we should draw what this one looks like, we should draw what this one looks like and then we should draw these two over here, right? Cool. Awesome. So, what would the orbitals look like for the first diene, right? We'll, remember that the way I'd like to do it is I always put my shadings on the bottom to start off. So, I would do this fill in, fill in, fill in, fill in, cool? And for the next one, we know that but if you guys have been practicing drawing molecular orbitals you should know that there's going to be, that this one's not going to move, this one's going to go up and then we're going to one node in the middle, which means that this gets shaded down and this gets shaded up, cool? So, this is what my HOMO a looks like and this is what we're going to be looking at to figure out symmetry, okay? Now, let's look at HOMO b and LUMO b, HOMO b would just be this and LUMO would be this, so the two orbitals I'm going to be looking at to determine symmetry are going to be which ones, I'm going to be looking at this orbital here, this molecular orbital here and I'm also going to be looking at this molecular orbital here, okay? And as I go into the bottom part here, where we're going to be analyzing a little bit closely, these are the orbitals I'm going to be bringing down, okay? So, let's go ahead and look at this little box that I've created to help us think about how this reaction is going to take place. So, we know that the HOMO a comes from side to my diene and what it looks like is above, let's go ahead and bring that down and fill it in, so it looks like this, cool? We know that LUMO b from my double bond, from my alkene looks like this, let's go ahead and bring that down here. Cool. Awesome guys. So, now we have to figure out, we have to determine if this is going to be symmetry allowed or if it's disallowed, if it's allowed we give it a checkmark, if it's disallowed, we'll give it an x, okay? So, how can we tell okay? Well, guys just so you know a new single bond can only form when you have two of the same phase come together, when you have two of the same phase lobes overlapping that's when you can form a new single bond, okay? So, could we form a new single bond the way this is drawn? yes, we could because notice that the terminal ends of both of my HOMO and LUMO they match. Notice that this terminal end looks the same as this one, right? And notice that this terminal end looks the same as this one, okay? This is what we call orbital symmetry, where there's going to be a possibility for a strong overlap between them because the fact that they can match up. So, what that means is that you could form a new single bond here and you could form a new single bond here, cool? So, this would be some symmetry allowed because of the way they can overlap, okay? Now, I do want to make a note for you guys later, when we actually get to drawing reactions, when I get to drawing the mechanism and maybe drawing stereochemistry, usually we don't draw it in this way with the HOMO on top and the LUMO on the bottom, actually most textbooks and professors do it the other way with the HOMO on the top and the LUMO on the bottom but this doesn't have to do with, it doesn't change your answer it just is just a different way of looking at it, okay? Later on it actually is more convenient to do it that way because then you can look at stereochemistry better but the reasons we're doing it here, in this case, is because I'm trying to show how the HOMO is, how the LUMO is higher energy and the HOMO is lower in energy and I want to compare that but later on once you get into like actually drawing thermal cycloadditions will probably start putting the HOMO on the top and the LUMO on the bottom so that we can draw the stereochemistry better, okay? But just letting you know that for right now this works just great, since we're just talking with the theory right now. Cool. So, the question is this the smallest HOMO-LUMO gap pulse possible and you might say, Johnny, what are you talking about? Well, because this is the thing. Notice that there are actually two HOMO's and two LUMO's on this molecule, I mean in this reaction, there's HOMO a, which is reacting with LUMO b, right? But there's also the possibility that HOMO b could react with LUMO a, right? There's no reason. there's like, there's no principle that say that that can't happen, it's not impossible, it is possible that HOMO b could kick electrons into LUMO a, okay? But the question is, which one of the gaps is smaller, is that is the gap for a, for a to b smaller, or is b to a smaller? and it turns out that if you were to measure it out, this is actually the smallest HOMO-LUMO gap. So, then this would be the preferred, the preferred mechanism that would take place, okay? So, guys so now we've just shown that this cycloaddition is possible due to symmetry and due to similar energy levels between your HOMO and your LUMO. In the next video we're going to try to see how favored HOMO b and LUMO a would be to react with each other.
Example: Determining Favorability of HOMOb to LUMOa5m
Who's ready for some practice? See if you can predict the correct answer.
Problem: Use FMOT to predict the product of following cycloaddition reaction.4m
Problem: Use FMOT to predict the product of following cycloaddition reaction.6m
Explain why maleic anhydride reacts rapidly with 1,3-butadiene but does not react at all with ethene under thermal conditions.
Provide the π molecular orbitals for 1,3-butadiene. Give them in order of highest energy (top) to lowest energy (bottom). Label each π molecular orbital (π1, etc.) appropriately. Show how many electrons exist in each π molecular orbital in the ground state of 1,3-butadiene; use arrows appropriately to represent the electrons.
For the 2π + 2π cycloaddition, which orbital matching describes the reaction?
In the 4π + 4π cycloaddition show below, is the reaction favorable or unfavorable? Which two molecular orbitals do you need to look at to determine this?
1) Not favorable because A and B do not interact well with each other.
2) Favorable because A and B interact well with each other.
3) Not favorable because B and C do not interact well with each other.
4) Favorable because B and C interact well with each other.
5) Favorable because C and D interact well with each other.