 Okay. So remember when we took an alcohol and we made an ether out of it, do you guys remember that? What do we call that? What was the name of that reaction? So we recorded one earlier. What was it? Yeah, the Williamson ether synthesis. Okay, so what are we doing in this one? Right? What do we have as our starting material over here? This is an alcohol, right? I mean, do you guys remember what this functionality is? You have an alcohol and a halogen in the same thing, in the same molecule? Do you remember? Yeah, this is a halohydrate, okay? But we got an alcohol effectively going to an ether, so this is a Williamson ether synthesis. But it's an intramolecular Williamson ether synthesis, okay? So remember, sodium hydroxide, when we put that into water, oh, we're going to do the mechanism, by the way, so I'm going to raise some everything. So, when we put that into water, the sodium just becomes a spectator ion, and the hydroxide becomes the portion that's going to be the active, in this case, Bronsted Viggs. We've got that hydroxide looking for an acidic proton. Hopefully you see the acidic proton right there. What are we going to show? The first step of the mechanism. So we make the alkoxide, and remember, when we ask ourselves, what's faster, an intermolecular reaction, or an intramolecular reaction? So remember, what's the difference? Inter means what? Intermolecular. Like an interstate is interstate across the state lines, right? So yeah, it's between two molecules. An intramolecular reaction is what? Yeah, reacting on itself. So which one's going to react faster? So what was the example I said? It was like if Geno was over there and got mad at me and wanted to punch me in the face, well, I could just be like, oh, I'll do it for you. I'll do it much faster. She would have to get up, walk over here, find me. So that's what's going to happen here. So instead of another molecule coming and reacting with this chloride, or this reacting with another molecule, it's just going to react with itself here. So hopefully you see that this arrangement is such that these two substituents are on opposite sides of the ring. Remember when we named it, it was a trance, right? One, two. So since this is on the opposite side of the ring, is that the antibonding orbital is right there? So we can do an SN2 reaction really quickly here. So what we'll do is attack right there, SN2. Could we group like that? And we'll make those epoxides. Well, in this case, it should be apparent that it would be on the same side. But remember, you can't have trance three-membered rings. So they always have to be on the same side. So that'll be the last mechanism we record today. If you want us to record another one, let me know and maybe I'll do it at the beginning of next class.