So in this video, I'm going to introduce you to another form of cycloaddition called photochemical cycloaddition. So a photochemical cycloaddition is a pericyclic reaction in which 2 pi bonds are destroyed, just like any cycloaddition, but it only occurs after a light activated cyclic mechanism. So it's not heat activated, it's light activated. We're still getting a cycloaddition, but it's happening through the energy that's given through light and as you're going to see it actually has very different effects on the types of products that you can get. So here's an example, I have 2 alkenes reacting under light and what I'm getting is a cyclobutane, right? So how many double bonds did we start with? We started off with 2. How many double bonds are we ending up with? 0. So we can confirm that this is a cycloaddition because I'm losing 2 bonds in the process. I also want to briefly show you the mechanism. The mechanism would be cyclic and concerted and it would make those new single bonds. So it would be something like this where this double bond comes in and attacks that carbon to make a new single bond and then this one does the same thing. Cool. There's no, you don't really know where it started and where it ended because it's all happening at the same time and it's all cyclic. At the end, you get a cyclobutane. Cool. So that is the basics of photochemical cycloaddition, that's all you would need to be able to draw the mechanism, but how do you know if the product is actually favored? Well for that, we're going to need to once again lean on frontier molecular orbital theory, which will give us the tools that are required to really predict if these reactions can happen or not. So in any cycloaddition, regardless of whether it's thermal, chemical, or photochemical, a HOMO must fill a LUMO. Okay? And what I usually try to do is make 1 molecule a, 1 molecule b, so HOMO a fills LUMO b. Now according to frontier molecular orbital theory, the bonding is going to be the strongest when orbital symmetry matches and when orbital energy matches. So, once again just to reiterate, orbital symmetry means that your orbitals are lining up nicely to make new single bonds and energy means that there isn't a huge gap between the HOMO and LUMO that you're choosing to interact with each other. So, we're going to be trying to make both of these things happen with a photochemical reaction. Okay. Now something that is new information here for photochemical versus thermal cycloaddition is that we have to remember what does light do to conjugated systems. Remember that light is able to be absorbed by the conjugated system and the conjugated system can turn that radiation energy into kinetic energy and shoot up an electron to a higher energy state. Okay? So light excites those ground state electrons in the conjugated system to a higher energy state. Essentially what we're going to have, what's going to happen is that a bonding Psi is going to transfer electrons to an antibonding Psi. That's usually what happens and that means that your HOMO and LUMO orbitals are going to change based on the light, okay? Based on light your HOMO and LUMO orbitals are going to change, versus what would have happened in in a thermal cycloaddition. Cool? So let's go ahead and look at what would happen. Once again, I'm just going to be using the example that we had above. I have an alkene on the left that I'll call alkene a, and I have an alkene on the right that I'm going to call alkene b. And just so you know, I can tell that alkene a is getting cut off just a little bit, but if you print in your PDF then you should see it just a double bond, just like the one on top. Cool. Awesome guys. So how would we fill in the electrons for the alkene A? What we know is that psi 1 would get 2 electrons and psi 2 would get 0 electrons, right? Everyone's cool with that so far? So that means that before light has reacted with anything, before radiation energy like is involved, we just have a HOMO A and then we also have B, okay. B, would do the same thing, it would have basically 2 electrons in Psi 1 and then nothing in Psi 2. Okay. So notice that I've already labeled HOMO and LUMO. I've labeled that for a, I have that the bottom one side 1 is HOMO a and that for b, LUMO B is psi 2. Okay? But we haven't taken the light energy into account yet, the radiation energy. So what happens when I involve electromagnetic radiation or light photons? What's going to happen is that one electron from my HOMO a, for my Psi 1, is going to get kicked up to Psi 2. So what it's actually going to look like afterwards is like this. 1, 2. Isn't that crazy? So that means that now this is no longer my HOMO. This is now my HOMO A. So what that means is that the identity of my orbitals just completely change which is going to have massive implications on the symmetry. Now the symmetry changes because light radiation is involved. Okay? So what that means is that in order to predict if this reaction is going to be favored or not, I'm going to have to compare what the molecular orbital looks like for HOMO a here versus LUMO b here. So what that means is that we should draw out what these orbitals look like, let's draw them here. So I have 2 orbitals here, 2 orbitals here, and then I'm going to do the same thing over here, this will be b. 2 orbitals here, 2 orbitals here. Cool. So what do we know about filling molecular orbitals? The bottom one should be completely shaded at the bottom and the top one should have the first one remaining unchanged and the last one changing. So this one should go up. Okay? And the orbitals that we're going to be, that we really care about with this reaction because we're using photochemical energy, it's not this one down here. It's actually going to be this one because electrons got kicked up to that orbital and this one. And these are going to be the orbitals that we're going to bring down to our analysis where we figure out if this is symmetry allowed or disallowed. So let's go ahead and do that now. Let's go ahead and shade in HOMO A, which HOMO A is now this one right here, and let's also shade in LUMO B. LUMO B is this one here. Okay. Now guys, the reason that I have HOMO a and LUMO b, 1 on top of the other, has to do with where the HOMO orbital started. But what we actually know is that the gap between HOMO a and LUMO b is almost 0 because they are almost at the same exact height, right? In fact, they could be at identically same heights because they are both the same molecule. So for sure, I'm just going to say this right now, this is definitely the smallest HOMO LUMO gap possible because, that they are at the same exact energy, but it even turns out that this is the only combination of HOMO LUMO that works because notice that the other molecule no longer has a LUMO at all. So molecule a doesn't have a LUMO. So if I wanted to go from HOMO B to LUMO A, that just doesn't exist anymore because we've already kicked up our electrons to the highest energy state orbital possible. So that means that I would not even have to really analyze this, but just to reinforce symmetry. Symmetry. So symmetry, is this a symmetry allowed or a symmetry disallowed process? Our orbitals match, Our orbitals match guys and this is awesome because you may remember that when we were talking about thermal cycle additions, 2 pi and 2 pi didn't work. 2 pi and 2 pi would be symmetry disallowed and would lead to no reaction, But when we use light, since we're kicking electrons up to side 2, what that means is that it does work. So that means that this is a symmetry allowed process because we're using light. Isn't that cool? So guys, that's it for this video. Hopefully, you guys now have an understanding of what a photochemical cycloaddition is and in the next video, I'm going to introduce you to a summary that of of the things that we learned in thermal chemical and in photochemical cycloadditions so that we can put it all together and we can predict very easily when a reaction will be favored or when a reaction will not be favored. So let's go ahead and move to the next video.
Photochemical Cycloaddition Reactions - Video Tutorials & Practice Problems
Photochemical Cycloaddition reactions are pericyclic reactions in which 2 pi bonds are destroyed after a light-activated cyclic mechanism.
MO Theory of Photochemical Cycloadditions
Video transcript
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Cycloadditions Summary Chart
Video transcript
So now that we know about both thermal and photochemical cycloadditions, I just wanted to make a little summary for you guys so that you could easily and quickly tell when a reaction is going to be symmetry allowed or symmetry disallowed. And if you use this, you can always know with certainty. Now before I get into it, I do have this one disclaimer at the top, which is that this only works if you're assuming suprafacial interactions, okay? So there's this concept of antarafacial and suprafacial that your professor is going to mention, your book is going to mention, examples are going to mention, but I think it's really silly. And that's because all of the examples that you are actually able to do in your class, in an introduction to organic chemistry 1 and 2 and 3 class, whatever, sometimes 3, you're only going to deal with superficial interactions because what superficial means is that basically, like let's just look at this for example, what suprafacial means is that you only have interactions happening on one side of each MO. So this is one molecular orbital and then it's only reacting with this side of the molecular orbital. This is called suprafacial because basically both of these can just overlap easily and then they can interact right. Antarafacial is this other thing that happens with much bigger, with much bigger molecules where you would actually have, let's say one side is interacting here but it's interacting with the top side of this one and with the bottom side of this one. That's antarafacial. Yeah, antarafacial. And antarafacial could happen, it's basically like almost the idea of cis versus trans. Cis can happen with all molecules. Trans, you would need to be a really big molecule before that can happen. So imagine that your MOs need to be so long that they actually twist at the end so that the top part of your MO and the bottom part of your MO can link together and link to other orbitals and share electrons with other orbitals. So basically, if you're dealing with rings that are 8 membered or smaller, a ring is equal to 8 membered or smaller, you can only do suprafacial. And the only way you could do antarafacial is if it's over 9, 9 or more. So that's why in my videos I'm not even going to talk about suprafacial or antarafacial because I'm already assuming that we're going to be only doing suprafacial because the examples that your professors could give you could really only be suprafacial, okay. That being said, these rules apply to superficial interactions. If you start talking about antarafacial, then you would have to mix all this up because now you're literally twisting your MOs. But as long as you're assuming that it's superficial interactions, then this summary is going to work great. And what the summary says, now let's get to the summary. What it says is that all you have to do is count up all the pi electrons in both polyenes, both the polyene a and polyene b. If all of the pi electrons equal a multiple of 4n, which should just be any multiple any multiple of n that's an integer. So that would be 4, 8, 12, 16, etc. If your total number of pi electrons equals a multiple of 4 n, that can only, that photochemical conditions okay. If all of your pi electrons equal add up to a multiple 4+2, which should be the number 6, 10, 14, etc. Then you can only do a cycloaddition in thermal conditions. So a great example of this would be the Diels-Alder reaction, which I showed you guys as an example of a thermal cycloaddition, but remember what did it look like? Remember that it was a diene on one side and an alkene on the other. There were 3 double bonds total. Those 3 double bonds equal 6 pi- electrons. So according to this rule, 6 electrons can only cycloadd in thermal conditions. Diels-Alder is a reaction, so we know that it's favored. In the same way, if we were to do a, let's say, a 2 pi plus 2 pi Cycloaddition, remember that and you're trying to do it thermal. 2 plus 2 equals what number? 4. Can a 4 n number of pi electrons do thermal cycloaddition? No. You need photochemical for that. So that's why just with this little chart you can easily tell what's going to be thermal and what's going to be photochemical. Okay. One last thing which is another way to think about this, like I think the first way is the easiest one, but this is just even another way to think about it, another way to very quickly determine if a reaction is favored or not. You can look at the Psi orbital difference and what this has to do with guys is it has to do with the numbers of the Psi orbitals that are interacting as HOMO and LUMO. Okay? So, for example, remember that, homo A always has to fill LUMO B or vice versa, right? So what you would do is you would actually look at the number of that Psi orbital and take the differences between them. Let's look at an example up here of the one that we just did which was photochemical. Remember that originally before we excited electrons using light, the original homo that we were going to use was this one down here, and the original lumo that we were going to use was this one up here. So notice that what are the numbers of the psi orbitals that we are reacting with. We're reacting with psi 4 and psi 2. So if we were to take the difference between those, so let's just do that very quickly, psi 2 minus psi 1, what we get is a number of 1. Now the number 1 is odd. So what does that tell us if the number is odd? That means that the only way it's going to react is if you use photochemical energy. So that's another way to think about it, that even from the beginning, once I saw that one was 1 and one was 2, I knew this isn't going to work unless we use light. Now on the other hand, if you go back to our example when we did thermal cycloaddition, we actually did 2 different combinations where homo A reacted with lumo B and then homo B reacted with lumo A. And in both of those situations we ended up with even Psi orbital differences. Let me show you. So for the first one, if you go back and look at it, what we actually did was we subtracted psi 2 from psi 2. The number we got as a result is 0, and 0 is even. So would that work according to a thermal cycloaddition? Yes, because even happens with thermal. Then when we did homo B with lumo A, the difference in the psi orbitals was minus Psi 1, which gave us a difference of 2, which is also even, which told us once again that that reaction would be favored to go thermal. Okay. This is just another way to even verify that instead of having to draw your MO's and figure out if they're symmetrical or not, you can just do this math very quickly and if you can just look at your Psi orbitals, you can already know if this is going to be symmetry allowed or symmetry disallowed. But again, an even faster, faster way than that would be just count up your pi electrons. And just by counting up your pi electrons and looking at the activator, you could automatically know if this is symmetry allowed or symmetry disallowed. These tricks could save you time on your exam because it's just very quick shortcuts to know if something's going to be favored or not. Guys, thanks for watching. I really hope that this summary helped. Let's move to the next video.
- Cycloadditions Summary:
![Content](https://cdn.clutchprep.com/guide_visuals/inline_images/IH3QNxcyrN8supYNXkdjPQ.png)
Use FMOT to predict the mechanism and products for the following cycloaddition. If no product is favored, write “symmetry-disallowed” in place of the product.
Use the cycloaddition summary rules to verify that you have come to the correct conclusion.
Had enough of Cycloadditions? If so don't worry because we will now move on to a new kind of pericyclic reaction:Electrocyclic reactions.
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