Show that the [4 + 2] Diels–Alder reaction is photochemically ...

Question

Show that the [4 + 2] Diels–Alder reaction is photochemically forbidden.

Accepted Answer

Hey everyone. Let's do this problem. It says six plus four cyclo addition reaction is photo chemically forbidden. Show why this is true. So let's look at what happens during photochemical cyclo addition. So we have one electron that's going to be in the highest bonding orbital. And when it reacts with light, it's going to get excited and jump up to the first anti bonding orbital. Okay, so that's what's happening during photochemical cyclo addition. So now when you are combining the two compounds, we need to overlap the excited molecules. Oops, excited molecules, homo right? It's new anti bonding orbital with the loom Oh, of the other molecule. So if this overlap has all bonding interactions, then that is allowed. But if there are not all bonding interactions, even just one of them, then it will be forbidden. Okay, and if you had a problem that asks you is photochemical psycho edition allowed? Or is thermal cycle edition allowed? Remember that for the total number of pi electrons We follow four in for photochemical Reactions in four n plus two for thermal reactions. So we have six plus four here, which equals 10. Which means thermal cycle edition will be allowed. But photochemical is not. Right. So, that just confirms everything that this problem is telling us. Okay, so let's look at our 64 edition. If we have six pi electrons, that means we're going to have three double bonds that will look something like this. 123 and we'll have the atomic orbital's six of them. 123456. And they'll each have one electron, Right, two electrons per double bond. And now drawing our molecular orbital's from this will have three bonding orbital's that are lower energy and three anti bonding orbital's at our our higher energy. And labeling from lowest energy to highest energy pi one pi two, pi three pi four star pi five star in pi six star. Okay, so the lower energy half our bonding, the higher energy half our anti bonding. So now we follow our regular electron configuration rules to fill in our new molecular orbital's with these electrons. So starting with pi 12 electrons, two electrons and pi two and two electrons in pi three gives us six. So that makes pi three the homo for this compound. Right? But when we apply that light, we're going to excite one of these electrons Up to Pi four star. Okay, so it's no longer in pi three, it has moved to pi four star, which is going to make Pie four star the new homo. All right. So what do the orbital's look like for this level? Let's draw our lobes and I'm going to start with Pie one and just draw the first and last one. So that I know how to draw pi four star. Right, I'm going off with johnny seven steps for drawing molecular orbital's. So if we draw pi one and there's other orbital's here. I'm just not drawing them all because we don't need them. They're all going to be in the same phase? We'll have the bottom shaded. So for the first orbital we always keep that one the same going up with all the energy levels. And for the last one we're going to alternate the phase. So this one's down. That means this one will be up, this one will be down, which means this one will be up. Okay. And now we need to add our nodes. So pI four will have three nodes. Right? We start with 0123 and they have to be symmetrical. So we'll have one, 2, 3 nodes and now we flip flop the phases across the nodes. So we'll have these two filled on top, These two filled on the bottom and then the last one is on top. So that is the homo molecular orbital's for our six pi electron compound. So let's draw for four. So we're going to have a dying 1234 atomic orbital's each have one electron and it's going to be pretty much all the same thing. Except now we just have two bonding orbital's into anti bonding orbital. So pi one pi two pi three star and pi four star filling in our electrons into these mOS 1234. So that means where is our loom? Oh, lowest unoccupied molecular orbital, that would be pi three star, right, There are no electrons there. So pi three star is our loom. Oh, why does that look so ugly. L. U. M O. So let's draw our orbital's for this one. I'm going to do pretty much the same thing as I did in the first one. Starting with Pie one. So if we fill these both on top, the first orbital will be also filled on top. The second Pie too would be filled on the bottom. The last orbital. So the top Orbital would be filled for Pi three star. And how many nodes do we have in pi three star? Well we would have 012. So two nodes we will add them symmetrically like so Which means the Middle two lobes will both be filled in the bottom. Okay so we have our homo star in our loom. Oh so now we need to show the overlapping of these orbital's right, that's what's happening for cyclo edition. So actually let me not scroll too far. So let's draw our homo on top. So we're starting down. Let me draw the orbital's okay, there's our six And there's our four. So drawing our four, we have up down down up up down down up and for our home, oh we have down up up down down up so down right just pretending it's flat across right down up up down down, Up. Okay we're almost there. So this is our home, oh pi four star and this is our loom. Oh Which is Pi three star. So let's look at these bonding interactions. So we see here that the phases are lining up, right? That's what we want. That's a bonding interaction. But here we see that the opposite phases are closest to each other. Is that a bonding interaction? No. So this one anti bonding interaction is going to make this forbidden, Right? So that is one anti bonding interaction and that is going to make a the correct answer for this problem. That's it for this one. I hope this video is helpful and thanks for watching.

Show that the [4 + 2] Diels–Alder reaction is photochemically forbidden.

Key Concepts

Diels–Alder Reaction

Orbital Symmetry

Photochemical Reactions

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