Photochemical Cycloaddition Reactions

So in this video I'm going to introduce you guys 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 implications on the types of products that you can get. So, here's an example, I have two alkenes reacting under lights and what I'm getting is a cyclobutane, right? So, how many double bonds did we start with? we start off with two, how many double bonds are we ending up with? 0. So, we can confirm that this is a cycloaddition because I'm losing two 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 failing, at the end you get a cyclobutane, cool? So, that is the basics of photochemical cycloaddition, that's you know, that's all you would need to be able to draw the mechanism but how do you know 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 one molecule a 1, which will b. So, HOMO a fills LUMO b. Now, according to FMOT, the bonding interaction is going to be the strongest when orbital symmetry matches and when the 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 HOMO-LUMO 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's in light due 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 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 a thermal cycle addition, 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 on the left, that I'll alkene A and I have an alkene on the right that, That I'm going to call alkene B and just you know I can tell that alkyne a is getting cut off just a little bit but if you printed 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 one would get two electrons and psi two 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 two electrons in psi one 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 one 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 one is going to get kicked up to psi 2. So, 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, okay? 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 it or not, I'm going to have to compare what the molecular orbital looks like for HOMO A here, versus LUMO B here. So, if that means is that we should draw out what these orbitals look like, let's draw them here. So, I have two orbitals here, two orbitals here and then I'm going to do the same thing over here, this will be B, two orbitals here, two orbitals here, cool? So, what do we know about filling molecular orbitals, the bottom ones should be completely shaded at the bottom and the top ones 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 direction because we're using photo chemical 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 leave the orbitals that we're going to bring down to our analysis where we figure out if this is a 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 reasons that I have GOMO A and LUMO B, one 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 and LUMO B is almost 0 because they are almost at the same exact height, right? In fact they could be at identically the same height because they're 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 they're are 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 picked 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 that there's actually very little to no energy difference between these orbitals, which is a great thing. Wonderful. Now, we've to look at symmetry. So symmetry, is this a symmetry allowed or 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 cycloadditions 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 lights, since we're ticking electrons up to psi 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 guy now have a understanding of what a, you know, a photochemical cycloaddition is and in the next video I'm going to introduce you guys to a summary that of the things that we learned in thermal chemical and in photochemical cycloaddition 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.

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 antrafacial 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 the examples that you are actually able to do in your class in a introduction to organic chemistry one, two and three class whatever, sometimes three, you're only going to deal with suprafacial 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? antrafacial is this other thing that happens with much bigger, with much bigger molecules where you would actually have, let's say, one is interacting here, but it's interacting with the top side of this one and with the bottom side of this one, that's interfacial, yes antrafacial, and antrafacial could happen if 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 MO's needs to be so long that they actually twist at the end. So, that's 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 eight membered or smaller, equal ring is, equal is eight member or smaller, you can only do suprafacial and the only way you could do antrafacial is if it's over nine, nine or more. So, that's why in my videos I'm not even going to talk about suprafacial or antrafacial 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 suprafacial interactions, if you start talking about antrafacial then you would have to mix all this up because now you're literally twisting your Mo's but as long as you're assuming that it's superficial interactions then this summary is going to work great, and what's the summary says.

Now, let's get to the summary, what it says, is that always 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 equals a multiple of four n, which would just be 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 for n that can only that, a cycle addition can only occur under photochemical conditions, okay? It's all of your pi electrons equal and add up to a multiple form plus two, which would be the number 6, 10, 14 etcetera 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 cycle addition but remember what did it look like. Remember, that it was a diene on one side and an alkene on the other, there were three double bonds total, those three double bonds equals six pi electrons, right? So, according to this rule six pi electrons can only excite will add in thermal condition, diels-alder is a thermal 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 is 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 size of it'll difference and what this has to do with guys 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 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 with 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 1 for HOMO A and psi 2 for LUMO B. So, if we were to take the difference between those, 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 it so that one was one and one was two I know this isn't going to work unless we use light, okay? Now, on the other hand, if you go back to our example when we did Thermal cycloaddition, we actually did two 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 site 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 cycle addition? yes because even happens with thermal, then when we did HOMO B with LUMO A, the difference in the psi orbitals was, psi 3 minus psi 1, which gave us a difference of 2, which is also even, which told us once again that that reaction will be favored to go thermal, okay? So, there's just another way to even verify that instead of having to draw your Mo 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 this is going to be symmetry allowed or symmetry disallowed but again and 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 can 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 on to the next video.