 Let's try to do this one together. It asks, write a complete mechanism for the following reaction. Show all intermediate structures, formal charges, and electron flow using the curved arrow convention. So for me, I always, when I'm doing my mechanisms, like to think about them, of course, stepwise, like you're supposed to with mechanisms. But one thing that oftentimes students don't like to do, and they really should, to keep the count of all the electrons, is just to put all the lone pair electrons first off. So that's what I'm going to do, put my lone pair electrons on my VR, just to remind myself so I can get those formal charges much easier. The other thing I want you to realize, and you probably already do, especially since we talked a little bit about it, is that this is saying that these are actually two separate reactions. So one of the reactions is happening after the other reaction. So in actuality, the KOH is there by itself, reacting with the starting material. And then after that reaction happens, then we have the HBR. And then that reacts with whatever intermediate we've made in that. Because of course, if we put the KOH and the HBR together in there, they're just going to react with each other, because that's a strong base and a strong acid. Does that make sense? So don't even think about doing that. The last thing I want you to remember is you're not going to get to this in one step. So this is a serious sequence of going through arrows and intermediates and stuff like that. So if you're trying to do all these arrows in one, that's totally going to be wrong. So don't do that. So what I want you to do now is if we've taken the product of memory, what we're going to do is erase this stuff. And then what do you say? Go through the mentions. So right here on the product. And then OK, so the first thing that we're going to be reacting the starting material with is potassium hydroxide. So what do we know about potassium hydroxide? Anybody tell me? It's a strong base. Strong base. OK, what part of the potassium hydroxide is the strong base part of it? The o-h. O-h. OK, so what does that mean about the k? The k is just a what? Spectator ion. Spectator ion. So when we have the spectator ion in the mechanism, do we even really have to put it in there? No. No, so don't even, if it messes with you, don't put it in there, OK? Because it's just spectator. It's not going to do anything in the mechanism. In fact, I don't like to put it in there. So just erase it and just put the o-h minus. Because that's the important part, OK? What do we say o-h minus was? It's a base. A strong base, right? So what do bases do? Or they react with a deep bromine stuff, right? Yes, and that's what you were trying to say, right? So where is the acidic proton ionist? It's going to be one of the alpha protons to where in the bromines, OK? So more than likely, we're going to want to form the most stable product. Because this is going to be a E2 reaction, right? So the most stable product is the Zatsev product, right? Remember we talked about that yesterday, OK? So which one would be more stable if we make the alkene there or the alkene there? That's the right one there. Why would that one be more stable? More? More hydrogen. Substitution. It's not hydrogen, though. It's more what? Substitutive. Substituents, but carbon atoms, OK? Carbon. So anyway, so let's talk about the mechanism. So it's that hydrogen there that we're going to need programming, OK? So what are we going to show? The arrow is starting from where? The one pair on the oxygen. Yeah, the one pair on the oxygen. Remember arrows always start from long pairs, double bonds, OK? So one arrow, two arrows, three arrows, OK? This is an E2 reaction. We've got a strong base, OK? And we've got an alkyl halide. That's going to go E2. So that's one step, and we get to that intermediate that we were talking about, OK? So that's the end of the first step. Now we're going to, if you want to, you can also put your H2O just to confirm that that happened, and then your Br minus. So now it says to do what? Add HBr, right? So what are we adding that HBr to? We're just adding it to this, OK? This Br minus isn't going to do anything in that reaction. This H2O, it's not going to do anything in that reaction. So they confuse you just for a reason. So like that, HBr, but we're not just going to have to say HBr, right? We're going to put a bond in between. And on the Br, we're going to show our lone parallel electron, like that, OK? So what kind of reaction is this going to be? So an addition. An addition, if you remember specifically, it's called a hydrogromination reaction, right? And so what we're going to do is what kind of molecule is HBr? It's a strong mass. Strong mass. So what happens to strong masses? They get deprotonated. They get deprotonated, OK? So the base is clearly going to be the alkene. Which carbon on the alkene is going to take the hydrogen away? The bottom carbon or the top carbon? Bottom carbon. Why is that? Because it's a more carbonic competition. Yeah, so you're going to do more carbonic competition on this. So the more carbonic competition, remember, that says the hydrogen goes to the more side of the more hydrogen, OK? That's what we're going to show. And I really like to show the arrow actually going through there, just for y'all's sake. So like that, are we cool with what we've done there? Try to draw that product on your own and see. That's just going to be, it's not going to be the final product. It's going to be the product of that reaction or an intermediate. And see if you get the same thing I get. So remember, we have a hydrogen here. Right now we have two hydrogens there. But they also have our Br that have the three lone pairs plus another lone pair. So now it's a Br minus. I like that. OK? Does that make sense? Yes. So we've got the carbocation intermediate, tertiary carbocation that's stable. It's not going to do any like hydride shift or anything like that. So the hopefully obvious mechanism is the nucleophile attacks the electrophile. And that's going to give us our final product. If I remember correctly, is that molecule. OK? So that was essentially a synthesis problem, right? You did a two step synthesis, but you showed the mechanism of the synthesis. So it's kind of a combination problem. Is there any questions on this particular problem? I know there was a lot of stuff that we went through. We cool? Yes.