 Okay, so let's try this reaction, let's do the mechanism for it. So this is MCPBA reaction, remember whenever you see MCPBA, that's going to take an alkene and make an epoxide out of it, okay? MCPBA is this molecule, metachloroperoxybenzoic acid, so we could go over the structure and the name together, okay? So let's just go for it. So meta, do you guys know your ortho-metaphera? Do you guys remember that from last semester, vaguely from last semester? Yeah. Okay, so where are they located relative to each other? Across from the molecule from here. What's ortho? Ortho both on one side, right? They're next to each other. Next to each other. And then meta is when they're one away from each other. Okay, so meta-chloro, so that would be meta-chloro, right? So meta-chloro, peroxybenzoic acid, so the benzoic acid is, benzoic acid would be that structure there, but meta-chloro-benzoic acid would be that thing. The peroxy means that you've got a peroxide here. So that is two oxygens on each other. Peroxides will oxidate alkenes, okay? This one's cool because you can dissolve it into like organic solvents, okay? So if you've got something that'll dissolve into an organic solvent, not an aqueous solvent, you know, you can use this as opposed to just hydrogen peroxide. But anyways, so what's going to happen here is there's going to be a concerted mechanism going on, right, where you're going to have kind of a, kind of a cyclic transition state here. So what I like to do is I like to, here, I'm going to erase all of this part, too. What I like to do is actually draw this in a different confirmation, so take the proton from the acidic proton and draw it like that. So you can kind of see we've got a one, two, three, four, five-membered ring there, okay? So we're going to, you know, do an intermolecular acid-base reaction as a matter of fact, okay? So this proton is actually going to be deprotonated by this. And it's going to be this oxygen here that's going to be the epoxide oxygen, okay? So you can think of this mechanism starting a number of ways, probably the best way to think about starting it is to, I don't know, take these electrons, and again, like I said, you can start it a number of ways. But a good way to do it is to take these electrons, move them down there, okay? And that will induce these electrons to move and take that proton, okay? But let's draw it in a way that makes it obvious that it's doing that. So we'll take these electrons here and take that proton away like that, is that okay? So remember, it's this oxygen here that's going to epoxidate this alkene. So this bond is going to be broken there, of course. It's going to attack that carbon right there. And then this bond, the alkene, of course, this alkene bond is now broken, right? It's going to go and attack that oxygen like that. And then that's going to make these electrons go like that, okay? So again, if you wanted to, you could take these two arrows and consolidate it and do that, okay? If that makes sense, okay? But it's good to, you know, do it step by step because there's a lot of arrows, right? So anyways, this carbon is sp2, of course. This one, too, is sp2. So trigonal planar, right? So flat, flat. So it can attack from either side. So you're going to get the, where's mine? It's all this up and down. You're going to get the two enantiomers here. One attacking from, you know, this side, our side, and one attacking from the opposite side of the board. So let's just, we'll just draw the one attacking from this side, okay? So what I like to do is kind of just turn it to where it would be my, like, thinking about it attacking from the top, okay? So when we do that, we can, why don't we just pretend it's still an alkene and then attack from the top? So got the methyl group coming there, the methyl group, and then the methyl group like that, right? So if that's all happening, what's happening here, right, is that this bond is breaking. And we're adding that oxygen to it, right, like that, okay? Does that make sense from that way, okay? So again, if you wanted to look at it like this, you could think of it as the peroxides coming out and both of those are actually going back. So you're going to get this out of the peroxide, the epoxide, plus the enantiomers here. Does that make sense? Does that make sense? Are there any questions about that one? Is there any questions about that?