Thermal Electrocyclic Reactions - Online Tutor, Practice Problems & Exam Prep

MO Theory of Thermal Electrocyclics

Now it's time to learn about a new type of pericyclic reaction called thermal electrocyclic reactions. So guys, thermal electrocyclic reactions are just going to be pericyclic reactions in which 1 π bond is destroyed, and that's because any electrocyclic reaction is going to involve one π bond changing from the reactants to the products. And that happens through a heat-activated cyclic mechanism because we know that it's thermal, so it's going to be heat activated. Cool? So, what do we need to know about this reaction? Well, one thing that you should keep in mind is that this reaction is always intramolecular. So, it's not going to involve 2 different molecules; it's just going to be 1 molecule reacting with itself in a mechanism to make a new ring. Okay? It's usually a ring-forming reaction. Notice that what would happen here, in this specific example, we're starting off with 3 π bonds in the reactant, and then after heat is applied, we end up with 2 π bonds. So this just goes ahead and verifies that this is electrocyclic because remember that in an electrocyclic reaction, you're always going to lose 1 π bond, okay, or 1 π bond is going to change from the reactants to the products. Cool. Also, notice that here we're applying heat. Now, I do want to quickly go over the mechanism really quick and just show you guys what would happen. So, what would happen is that it's concerted and cyclic, so it's all going to happen at the same time. You don't really need to know where to start or end from, but what we do need to know is that like we need to make that new single bond, right? So, the way I would start it is I would take the electrons from this double bond and make a new single bond, and then that means that this needs to break and put its electrons there, which means that this one needs to break and push its electrons there, and there you have it, you would have your new ring and now we're missing 1 π bond. Okay? So guys, it turns out that all conjugated polyenes are capable of doing these intramolecular electrocyclic reactions. So, it's not like a specific type that's possible. This could happen with any polyene, however many π bonds long, it could happen. However, the stereochemistry is variable, meaning that you're not always going to get the same product. Many times, you can immediately predict what ring you're going to get, but to know where the substituents are going to be, we need to think one level deeper and we're going to have to think about frontier molecular orbital theory. Right? We're going to need to think about HOMO and LUMO orbitals and figure out how that plays into this. So, how do you determine the stereochemistry of the product? The way you do it is by looking at the HOMO (Highest Occupied Molecular Orbital) of the molecule. Okay? And it turns out that the HOMO orbital is capable of cyclizing in one of 2 ways, either in a conrotatory direction, okay, a conrotatory direction or in a disrotatory direction. Okay. So when we look at electrocyclic reactions, we're going to be looking at intramolecular reactions, and we're going to try to figure out, did this molecule, did the molecular orbital have to rotate, conrotatory or disrotatory, in order to form a new sigma bond? And I know I just kind of did some hand motions, but here I'm going to show you exactly what I mean. I have some examples drawn out for you already. So, to make a new single bond, we're going to need 2 of the same phase lobes to overlap. Okay. So, what that means is that if we're trying to make a new bond, let's say with this 2 with this 4 π system, this 4 π conjugated system here that I have on the left, what I would need is I would need like, for example, the positive here to overlap with the positive here. Okay. Notice that they're on opposite sides though. So the only way that that can happen is if both orbitals are rotating conrotatory. So what that means is that this one's rotating to the right and this one is also rotating to the right. Okay. Because if they both rotate the same direction, what's going to end up happening is that the positives are going to overlap. Does that make sense? That's what conrotatory means. That you're it's kind of like they're rotating the same direction, but what's actually happening is that you're bringing 2 opposites together. Okay? So orbitals rotate the same direction, that's conrotatory. But what about a 6 π system or a 3 double bond system? If we were to try to make a ring out of this one, which is actually the example that we had at the top, that would actually be disrotatory because in that case, when you draw the HOMO orbital, what you wind up finding is that your orbitals are perfectly symmetrical rotation happens rotation happens when they rotate in opposite directions, one is clockwise and one is counterclockwise, to create that same type of overlap that would lead to a single bond. Okay. Now why is this important? It sounds like no matter what, we can get an electrocyclic reaction to happen thermally. No matter what, it's either going to rotate conrotatory or disrotatory. But what matters are the substituents, because if you have any substituents located on the terminal ends, let's say here or here, let's say that I put an alcohol here, right? How do you know if that alcohol is going to face up or down after the new mechanism has taken place? Well, that's why you have to look at the rotation type. The only way you can predict it is to know the rotation type. So the focus of our electrocyclic problems is not actually going to be on forming the ring because that's the easy part. We're always going to assume that the ring can form. The focus is going to be on the stereochemistry because if you can predict the stereochemistry accurately, that means you understand the molecular orbital rotation that's occurring. Okay? So in the next video we're going to do an example including stereochemistry.

Predicting Electrocyclic Products

Predict the product in the following electrocyclic reaction. Label the reaction as either conrotatory or disrotatory. So guys, I just want to emphasize here that we know a reaction is going to take place no matter what. We know that we're going to get a mechanism that looks something like this, where let's say this double bond makes a single bond here and then this double bond goes here. So what we're going to expect to get is some kind of square, with a double bond here and with methyl groups here. Does that make sense so far? Because we're making a new single bond here. But what is the stereochemistry of these methyl groups? Are they cis or trans? You can't just draw this and get the right answer. We have to actually provide the stereochemistry, the whole product, not just ignoring stereochemistry. So, to figure out if these are facing up or down, we're going to have to look at the molecular orbitals. So, where do you guys think is a good place to start? We're starting from scratch here. What do you think we should do first?

So, probably a good place to start is let's start drawing our molecular orbitals of a diene. So that means I have to draw 4 of 4, right? One, two, three, four. It's not bad. I'm going to use my copy-paste feature, and you guys can always pause the video if you need to. Cool. Awesome. So now we can go ahead and draw in our molecular orbital names. So we know this is ψ₁, ψ₂, ψ₃, and ψ₄. And then we can shade these in. You guys should be pretty good at shading in dienes by now. Why don't we just do this very quickly? It would be that all the bottom is filled, it would be that there's one node on the second one, so then this is what it looks like because there's one node, and the last one has to flip. For 3 and 4, the first ones don't change, for 3 and 4, the back one keeps flipping, so then this would go down then this would go up. Cool. And then finally, we need 2 nodes and 3 nodes. That means that these 2 go up here and then this one goes up and down. Cool.

So, those are our MOs. We should then fill them in with the number of pi electrons we have, which is 4. Cool. Awesome. So now what do we do? What's the next step? This was important. We did need to do this, but what's the next step? Guys, we need to analyze the HOMO. So, remember that we learned in frontier molecular orbital theory, we learned about HOMO and LUMO and you learn how to identify both. Well, it turns out that we don't need to worry about the LUMO for electrocyclic. Electrocyclic only focuses on the HOMO because it's just the HOMO interacting with itself. So that's the good thing that it's the in terms of the frontier orbitals, there's less to worry about. It's just the HOMO. So the HOMO happens to be this orbital here and this is the one we have to figure out if it's going to be conrotatory or disrotatory.

Okay. Now, the way that we tend, that we typically do this is to actually try to do it in 3D, so we can figure out what it's going to look like at the end. So what I'm going to try to do is I'm going to try to draw it like this with the orbitals facing up so we can see how they interact with each other. Okay. Something like that. So let's go ahead and draw in our orbitals. We're going to draw 1, 2, 3, 4. Does that make sense? So what I'm trying to do is I'm trying to actually keep it similar to that one, but now I'm going to be involving what we know about these orbitals here. Let's go ahead and shade them in. Let's say that, this side right here is the beginning of my MO on the other side. So that means that what it would look like is shade the bottom, shade the bottom, shade the top, and shade the top. Cool? Awesome.

Now don't forget we have our substituents that we need to include. So, where are our substituents going? Well, they should go on the inside. So what that means is that I'm going to draw these as methyls, I should draw a substituent going in here and a substituent going in here. Does that make sense so far? This has to do with these methyl groups right here and this is how we're going to determine the stereochemistry. Awesome, guys.

So now, we have to determine if it's going to be conrotatory or disrotatory. What do you guys think? So it should actually be conrotatory, right, because notice that my lobes are on opposite ends. So what that means is they both have to rotate in the same direction. That means this one, let's say it has to rotate clockwise, this one also needs to rotate clockwise. Or if you picked counterclockwise, they should both rotate counterclockwise.

So what that means for my product is that what it's actually going to look like is like this. So I have my new square, my new cyclobutene, I know that I'm going to form a double bond here and then we have to look at with this rotation, where would these groups go, where would the methyl groups go. So what I would see is that, let's start off with the first one, the first one at the bottom. It was, it's going into the page but after I rotate it, it's going to rotate down, right? It's gonna go down because of the fact that, that thing is rotating clockwise. The way that I drew it, it's rotating clockwise, so that means it's going to rotate down. So I would expect this methyl group to go down like this. I hope that's making sense so far.

Now, the other one that's, that's facing this way, that's actually coming out of the page. Like in fact, one way to write this could be that you would write this one into the page and this one out of the page because one is going into the page, one's going out of the page, right? So then that one, when it rotates, it's actually gonna face up, right, because it's rotating clockwise. So then this methyl group would go up, which means that my product is actually going to be a trans dimethyl, where this one faces up and this one faces down.

Now, guys, it turns out that this is actually an enantiomer, this is not a meso compound, so we should actually draw the other enantiomer as well, meaning that we actually get 2 products for this reaction. We get the 2 different trans products that are possible which are this one and this one. And you might be saying well Johnny, how would you get the other one? That would be if you had rotated counterclockwise, then you would get the other one. Cool? Awesome, guys. So I guess now we just have to label it and the rotation turned out to be conrotatory.

Cool? So guys, we did it. Great job. We got our products. Just so you know, it's not always going to be 2 products. If they had been facing the same direction, that would then be a meso compound because it has a plane of symmetry and then you just get one product. But since there was asymmetry here, we had to draw both products as our answer. So, remember guys, when you're drawing electrocyclic, you can't just draw them on the plane, you need to include stereochemistry because that's really what your professor, what your homework is going to care about. Making the ring is the easy part. The hard part is using HOMO-LUMO frontier orbitals to figure out the stereochemistry. Great job. Let's move on to the next video.

Here’s what students ask on this topic:

What are thermal electrocyclic reactions?

Thermal electrocyclic reactions are a type of pericyclic reaction where one π bond is lost, typically resulting in the formation of a ring. These reactions are heat-activated and occur through a concerted, cyclic mechanism. They are always intramolecular, meaning they involve a single molecule reacting with itself. The stereochemistry of the product depends on the rotation of the Highest Occupied Molecular Orbital (HOMO), which can be conrotatory (same direction) or disrotatory (opposite directions). Understanding this rotation is crucial for predicting the orientation of substituents in the final product.

How do you determine the stereochemistry in thermal electrocyclic reactions?

The stereochemistry in thermal electrocyclic reactions is determined by the rotation of the Highest Occupied Molecular Orbital (HOMO). The HOMO can cyclize in one of two ways: conrotatory (both orbitals rotate in the same direction) or disrotatory (orbitals rotate in opposite directions). The type of rotation depends on the number of π electrons involved. For example, a 4π system typically undergoes conrotatory rotation, while a 6π system undergoes disrotatory rotation. This rotation affects the orientation of substituents on the terminal ends of the molecule, making it crucial for predicting the final stereochemistry.

What is the difference between conrotatory and disrotatory rotation?

Conrotatory and disrotatory rotations refer to the ways in which the orbitals rotate during a thermal electrocyclic reaction. In conrotatory rotation, both orbitals rotate in the same direction (either both clockwise or both counterclockwise), bringing like phases together to form a new bond. In disrotatory rotation, the orbitals rotate in opposite directions (one clockwise and one counterclockwise), also bringing like phases together. The type of rotation depends on the number of π electrons in the system: 4π systems typically undergo conrotatory rotation, while 6π systems undergo disrotatory rotation.

Why are thermal electrocyclic reactions always intramolecular?

Thermal electrocyclic reactions are always intramolecular because they involve a single molecule reacting with itself to form a new ring. This intramolecular nature is due to the concerted, cyclic mechanism of the reaction, where the electrons move in a synchronized manner to form a new σ bond while losing a π bond. The reaction does not involve two different molecules, making it inherently intramolecular. This characteristic is crucial for the formation of cyclic structures and the specific stereochemical outcomes observed in these reactions.

How does the number of π electrons affect the type of rotation in thermal electrocyclic reactions?

The number of π electrons in a system determines whether the rotation in a thermal electrocyclic reaction will be conrotatory or disrotatory. For systems with 4n π electrons (where n is an integer), the rotation is typically conrotatory, meaning both orbitals rotate in the same direction. For systems with 4n+2 π electrons, the rotation is disrotatory, meaning the orbitals rotate in opposite directions. This distinction is crucial for predicting the stereochemistry of the product, as the type of rotation affects the orientation of substituents on the terminal ends of the molecule.