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.