 Okay. So, let's do this problem. So, can anybody tell me what the first intermediate is here? What is that thing called? That's not good. Well, you're talking phenol, is that what you're talking about? Yeah, but this is not a phenol, right? Phenol has the OH here, not there, okay? But what are you talking about now? The hydrogen's going to be gone because the hydrogens are coming. You've got a deprotonane, right? So that kind of, there's this, right? It's telling you that hydrogen is going away, so there's hydrogen gas being formed. So how can hydrogen be gas-reformed? Remember, sodium hydride breaks up into, remember the sodium is always the spectator ion, right? The H minus, do you guys remember what that's called? Do you guys remember what the name of H minus is? Hydride. That's why it's called sodium hydride. The mechanism, deprotonane, the alcohol. Again, this is not a phenol, it's just an alcohol, primary alcohol. So that's the intermediate. Hopefully that makes sense. Okay, what's going to happen here? That's going to attack the oxides. Okay, what is it going to attack the oxygen here? No, bottom carbon. One of these two bottom carbons, right? Why? Because of the ring strain. Because of the ring strain, right? It doesn't matter which carbon it's going to attack. Why would it not care? They're the same, right? They're essentially the same. Like if you looked at this in the NMR, you'd only see one carbon, right? So remember this is a good, what type of thing? Good nucleophile, right? So this thing is a good electrophile due to that ring strain, okay? So these three member cyclic ethers are called what? Aboxa, right? So can I erase this first part so I can put the mechanism of that argument? So again, this wouldn't be a mechanism type problem, but for me it's easier to think about things mechanistically. So again, the spectator ion floats away. What we'll do is put that in. So we've got this oxide, like we said. So because of ring strain, there's this delta plus there, right? That makes it a electrophile. The nucleophile will attack it like that. Opening up that ring due to the ring strain. And I didn't give myself enough room for that box. But anyways, we've got to do the second step before we put anything in. So when we do that, what do we get? What do we get? How many carbons are we adding to this side of it? Two. Two plus oxygen. So one, two, oxygen. And oxygen doesn't have a charge on it? Negative charge. So the sodium is still associated there. Okay, does that make sense? Okay, so that's what we got. But we haven't put our acid in there yet. Right? What's going to happen when we do that? It's a protonate den. Of course, yes. Remember, what are the fastest reactions in chemistry? Acid-base reactions. Okay, so whenever there's a base, is there a base in this? In this solution now? Right? What is it? It's this thing. Here's the base, right? De-protonate the acid. Like that. So in that box that was there, I would put this thing that I'm about to draw. And I'm going to erase this part of the mechanism. Okay, so I'll let you guys draw a little book. Or actually, can I erase this part of the mechanism here? Of course. The next step, Billy? This would have been what was in that box. It's going to cover that hydrogen. So the next step is going to be what? You guys remember the structure of this thing? OTS. Of hydrogen? Of hydrogen. 5 carbons. So what is this thing good for? What is it called? What did you say, Kendra? Yeah, it's a non-nucleophilicase, right? Okay, so what is the base do? De-protonate. De-protonate things. Is there an acid-ic hydrogen that it can deprotonate? Yes. Where is it? No. Come on, guys. Where is the acid-ic hydrogen? On the alcohol, of course. Of course. You know what I mean? I'm going to try to doubt myself. Kendra, you're wrong, you're wrong. No, Kendra. De-protonation. What? The blood was rough. So we've got to be alkoxide again, and then we'll have them. That would be complex with the sulfonol, right? What was it, amesolate? Yeah, amesolate. It has a lot of carbon. But what do we do with it? It's committed to the ionic acid. It has de-protonation. Because it's... Because is this... Is this nucleophilic? No. Yes, this is nucleophilic. That's what it is. If it comes to class more often. Like that? So what are we doing here? Why are we doing this stuff? Is this alcohol here? Why are we... What are we trying to do here? Create a group. Yeah, a good leaving group. Is an alcohol a good leaving group? No. So we've got to make it... We've got to invert its reactivity, right? We make it from a bad leaving group to a good leaving group. In order to do that, we make amesolate. Well, we've got to take it out of solution, you know? And you can't have salt and bring it out of solution. That's what I'm talking about. So... Now we've got... And then what happens? It goes down, picks up that chlorine. Chlorine is a good leaving group. That's awesome. And then what do we get? OTS. Yeah, whatever was supposed to be in the box. So... So let's just remember... Like a draw. Or a rim to draw. Can I erase this top part of the paradigm deprotonation? Remember, this is all going to be on YouTube too, so... Go and check it out if you didn't get it all right now. Okay, so we inverted the reactivity. This is now a good leaving group, right? So it was a bad leaving group when it was an alcohol over here. So, since it's a good leaving group, this is going to be a what type of thing. So this is going to be a nucleophile, electrophile reaction. Which is this going to be? The nucleophile or electrophile? The other one's going to be the electrophile. So then that must be the electrophile. If you see anything like that, it's a good electrophile. Okay? So it's these, you know, tosylates, mesylates, these types of things. Okay? So what's going to happen? It's just things that are very good. SN2 reaction. So when we do that, we're going to do the final product. It's the final product. Of course, you're going to have the conjugate base of the mesylate ion. Okay, is everybody cool with that one? Any questions on it? So that would be a good synthesis problem. So something like that will be on the exam. I could only imagine. What's that? In the step on the top right, why does it hit the sulfur? Why does it hit the sulfur? That's why it has to be. Yeah, so somebody else tells me. Well, the sulfur... The oxygen is coming from the sulfur. Yeah, exactly, those oxygens are very electronegative. And in fact, if you remember what sulfones look like, right? Remember the resonant structure that they prefer to have, right? In the sulfone, the sulfur is what charge? You guys remember? The resonant structure that they prefer, they're positively charged. The sulfur is much bigger than oxygen. So they can't really have that double bond very well. We draw it like this because it's easier for us to look at. Okay, everybody cool with that?