 Okay, so let's do another addition reaction. Hopefully you guys can see that this is going to be an addition reaction because we've got an alkene first off and we've got Br2 in there. Okay, so we just had those two things. Hopefully you would expect a dihalide to be formed, right? Do you guys remember us doing that? Okay. So in fact, the bromonium ion intermediate will be formed and a dihalide will be formed in this reaction but it won't be the dibromide, okay? It'll have a bromine on it and a chlorine. So it'll be a monobromide or monochloride, okay? So it will still be a dihalide though if you understand those terms, okay? So if we look at this, we've got bromine, sodium chloride, and excess. So whenever you see that excess in parentheses, remember it means that word excess, okay? And CCl4, carbon tetrachloride, that's just your solvent. So that means don't be trying to use this in the reaction, okay? So what I'm going to do, remember, we're going to write this mechanism out and when we write mechanisms out, we want to erase all of those reagents and write them out in their Lewis structures, okay? So I'm going to erase all that stuff. I'm not going to worry about the carbon tetrachloride anymore and the other thing I'm going to do is actually only worry about the reagent that's going to be reacting for first. I know that there's sodium chloride in there. It's going to be reacting in the second step so I'm going to save it for the second step so I'm not going to write it right now. Is that okay for you? Okay. So now what am I going to do? It's going to be the same reaction as we've learned before, the same reaction to get to that bromomium ion intermediate. So can somebody help me with that reaction? What's going to happen? Anyone? To the bromine. Uh-huh, from through that carbon, remember, okay? And then what's going to happen? The electrons are going to first going to push those. These ones? Yeah. Very good. And then is that it? It's going to be what? One more arrow? Yeah. You want to say it? To that carbon there, right? So remember we're going to get one of these ring structures, okay? So it's going to be a three-member ring. And this happens, this bromine is electrophilic because this is a very good leaving, okay? It's very stable on its own, the bromide. So let's draw the intermediate that we get, and then we'll... So what's the hybridization of these two carbons? Try again. What is it? SP2. SP2, right? Okay, so if they're SP2, what is the geometry around them? What is their planar, right? So that means that this bromine can attach where and when? Front side and back side. Yeah, from the front face or the back face, okay? So does that... That means we're going to have two intermediates here. Is everybody cool with that? Yeah. So we'll come from the back face. So now, remember, we all... in the... making the dibromide, we would still be R minus it, right? So we still have that, because that was formed in this step. But what did we say? We had sodium chloride in excess. So there's going to be more Cl minus ions. Is everybody okay with that? So, and there's just one of these relative to, I don't know, say, just for example, 100 chloride ions. The chloride ions are going to be more likely to attack. In fact, it's going to be even probably a higher percentage of chloride relative to bromine. So these actually, bromide ions, get kind of washed out by the chloride ions. Okay, so we're going to actually just erase them and forget about them. Is that okay? Is everybody okay with doing that? So remember, organic chemists don't like to balance reactions very much, okay? So I'm just going to have to move with that. So we've got a chloride ion and we'll write another one down here. So we remember that that's what we're using. And what's that going to do? The same reaction as the bromide did in the last addition that we saw, okay? So here, what kind of hybridization do we have, guys? SP3. SP3, right? So this thing is going to want to attack from the backside. Is everybody okay with that? Not to mention, it's being held, the bromine, bromonium, is being held on to by this carbon that's not being attacked, okay? So that's going to flip that bromine to the front face. In fact, you're going to get a product ratio that's different than 50-50 on this particular reaction due to the substitution pattern of these two carbons, okay? So this carbon here is bonded to two carbons. Does everybody see that? This one's bonded to three, okay? So if you think about the partial charge density or a kind of resonance structure, and I know I told you resonance structures were not going to break single bonds, but I really want you to think about it in this sense, okay? So if we look at just around that center, compare that carbocation there to that carbocation there, right? There's a hydrogen there. Is everybody okay with that? So this is secondary carbocation and that's a tertiary carbocation. Is everybody okay with what I'm saying? So do you remember which one's more stable? If we only had these two things, what would you say? Which one's more stable? The top or the bottom? The top one. Okay, because it's tertiary. Okay, since that's more stable, that's going to be the one that's going to be formed in the bigger amount, okay? Because this carbon here doesn't like to give, it likes to hold on to that bromine harder than this carbon does. Does that make sense? Since this carbon doesn't want that positive charge, it wants to hold on to that bromine, this carbon doesn't mind having it as much as this carbon does, so it doesn't hold on to that bromine as tightly. Is everybody okay with that? Is that good? Okay, so you want to think about this kind of analysis of it. So, what we're saying now is this chlorine can attack like that, knocking that BR out. So that's going to attack that carbon there. So let's call it carbon one. So that'll be attacking carbon one. When we do that, where's the chlorine going to be? In the front or in the back? How's it going to be in the back, guys? Yeah, well, this is SP3, so what kind of... So the bond you just broke was in the front, and it's going to get inversion of configuration. Yeah, we're going to get that inversion of configuration. Okay, so if you're having trouble with that, remember always, so that would be attacking carbon one. So I'm going to erase these mechanistic arrows, is everybody okay with that? And I'm going to actually draw the dash this next time. Or if we have another color of N, let's see. So we'll say carbon two here. Is everybody okay with just labeling that one? Okay. Okay. So I want you guys to predict the product and then look up here and see if you've got it up. So we're going to call this product A and this one product B. From this reaction here, just this one that we're looking at, which one would be formed in the higher percentage? A. Y, A. I was going to do this for me to see if it's going to go a little smaller. Can I interfere with the metal group? Remember what we talked about? Tertiary. Yeah, it was related back to those pseudo-carbocad ions that we were talking about, okay? So the tertiary is more stable, so this one wants to hold on to a problem more. So from this reaction here, right, A would be major, and B would be minor. Now we've got to see what this reaction turns out. Okay, so here, we're going to show this chlorine attacking, that carbon, and that electron is going to that problem. I'm just going to call this carbon three, so we'll say reaction three here. Okay, so what product would we predict? Everybody write it down and I'll write it down, and we'll see if we get the same thing. Is this the same as A or B? Does anybody know? It's not the same, right? So let's call this thing C, because we've got more than two products now. So let's erase this mechanism, and we'll attack carbon four, and we'll do a reaction four, like that. I'm just going to do it up here. So when we do that, we're going to get this other product. No, so this is D. So from this reaction here, which would be the major product? B or C? D, right? Why? Because of the chlorins attached to them? Because of that pseudo-carbohydrate. That's what you're going to say. Sorry for any of that. Then I say the major. That's the major product, and of course this would be the minor product. Okay, so out of C or A, would either one of them be favored? Since they're both major, we could say, well potentially one of them is in a greater concentration than the other, so it would be the major, major product. Would either one of them be favored? No. If you're saying no, so not many people are responding, but shaking their heads, yes, no, you're saying no, you're right, okay? So there would be no preference of A or C, okay, because there's nothing on the rest of the ring to kind of get in the way sterically. A and C would be the two majors, and B and D would be the two minors. That's all you can say, okay? And you would have equal concentrations of A and C and equal concentrations of B and D, or you would assume. So what's the relationship between these two ions? Yes, they're anion tumors. What about between A and C? What's the relationship? Isomers. Okay, are isomers, but they are what type of isomers? Stereo isomers and anion tumors. Yes, they're anion tumors, and C are anion tumors. They don't have the same connectivity, right? They're structural isomers. So C and D are structural isomers, right? All right, what other ones? A and B, right? Have we done all of them? A and D? What is A and D? Yeah, they're still structural. And then B and C, structural as well. Let's kill this video before we run out of juice on our battery. Any questions really quick?