 Anyways, back to chemistry. So let's show what we were doing. In fact, the very last reaction that we did yesterday was the addition of halogens to alkene, right? So do you guys remember what the name of the type of compound we're going to make over here is? No, no, we haven't done that reaction. Well, that's going to be the intermediate, the Ramonium Ionium Intermediate, and then we're going to make a dihalide, okay? So it's an alkyl halide, but with two halogens. So remember, if you're going to do the mechanism, which I'm sure everybody wants me to do, right? Everybody wants me to do the mechanism, okay? So when you're going to do the mechanism, you have to erase this part or rewrite it, okay? So if you've got pencil or a whiteboard erase and show the Lewis structure, what's all the lone pair electrons, if you do that, it'll help you so much. And you can't show all your arrows happening at once, like a lot of you guys like to do on your exams, okay? You have to show them stepwise, the actual steps that the arrows are performing, okay? So remember, do you guys remember how many arrows are in this first step? Yeah, so this is one of the more involved mechanistic steps where we have three arrows, okay? The most involved that you've seen. So first thing, this alkene sees that electrophilic bromine there. Now remember, it's electrophilic because this bromide ion that's going to be formed is very big, doesn't mind to have that negative charge, so it's very stable on its own. That's going to kick that bromide ion off with those extra electrons, so it's going to have a negative charge, right? And then these electrons, because remember, bromine is big, especially relative to carbon, carbon's very small. So bromine occupies a lot of space, all of these halogens do that, do this reaction. So when it gets into close proximity of these two carbons, it's going to make two bonds to, or one bond to each of those carbons, two bonds off that bromine. So how did we depict that, remember? So now draw the intermediate, or the product of that reaction, which is the intermediate of our overall reaction. The product of that reaction stuff, are you guys? So remember, this is sp2, sp2, right? We have two sp2 centers here. We're bonding four things to those carbons who make sp3 centers up, okay? So is this going to be the only intermediate that we form? Okay, yeah, we'll have the R minus as well. What else will we have? Yeah, on the back, right? So we call it the anti-merge. Really well, yeah, it is like. These two intermediates, what do we call these again? Anantumers. Anantumers, but the type of intermediate they are, are. The bromonium ions, okay? And like you guys said, they're in the anti-merge. Is this where we stop this reaction? Did we want to get bromonium ions as our products? What did we want again? Di-highlights, right? So we're going to have to put another what on this thing. Di-highlights? Yeah, another halogen, okay? So we got a Br-minus out here. What is it going to do? Well remember, this bromine is electronegative relative to the carbon atom, okay? So don't let that positive charge trick you there, okay? The electrons are going to flow to that positive charge, but the bromine's not going to attack that bromine, because if that happens, then we just go back here. Does that make sense, okay? So in fact, this carbon here and this carbon here are very electro-positive, okay? So they have a delta plus charge on them. So this bromine can attack that carbon there, or it can attack that carbon that attacks this carbon and knocks these electrons to that bromine and it attacks this carbon and knocks these electrons to that bromine, okay? So these are two different reactions. Normally since, normally you wouldn't want to write them both on the same structure. I just want to show you on the same structure that each one of them can happen, okay? So let's just right now erase one of these and show what happens in those solid ones. You can choose on, say, an exam or a quiz. Well you're going to have to draw them all if I said draw the mechanism. But the thing is, is that would not make you draw this whole mechanism. Mostly it'd be like, what are these arrows showing? Or show this in an antenna or something like that. Show the mechanism to this bromonium line. You're not going to have to do the whole mechanism. It would be too much for you guys to do on a test. What the products are of this reaction though, and you should be able to do the mechanism on your own. So it's not like you shouldn't be able to do it. But to grade you on each step is a little, okay? So does everybody agree that would be the product from that solid reaction mechanistic arrow? Why would the bromine up here be in the back? Inherit everything? Yeah, so that's like an SN2 reaction, right? So notice there's an SN2 reaction within the addition reaction. So let's show the attack with the dash lines that we were talking about earlier. Is everybody okay with that? So I'm going to erase these reaction arrows, just so we're not showing the mechanism on the same substrate. We're going to get here, do this. Well, so can you guys see what the relationship between these two molecules is? Do you have a question? So we can start there. So one was the back phase attack and one was the front phase attack? No, they're both SN2 reactions, so they have to be what? It's not back phase for SN2, what do we call it? Back side attack. Back side attack, right? So we get the inversion of configuration. When we're talking about the face attack, it's like the SN2 centers. Okay, this is SB2 centers, not SB3 centers. So what kind of carbon is this here? What's the hybridization? Look up here. Yeah, look up here. What's the hybridization of that carbon? SP3, right? So we don't have front phase and back phase on those SB3s. We only have SN2 reactions, okay? If we're going to have them simultaneously. So we've got this over here. What's the relationship between, so what kind of reaction is this then? That would just be, I know it's early guys, come on. SN2? SN2. We just said that. Okay, what's the relationship between those two products of that reaction? They're in action. Because all of the stereo centers are opposite of each other. If you can't tell, do your R&S configurations. Okay, so let's do the attack at these ones, okay? So with this immediate, we're going to attack there. Knock that out. So if we do that, do we get either one of these products? So let's call this one A, and let's call this one B. If we attack that, show those magnetic arrows. Do we get either one of those products? Do we get A or B? We get B. We get A, yeah, we get A. Why would you get A? A. Yeah, so there, all right, so the bromine's going to be pointing which way? This bromine here. Well, towards us or away from us? Towards us. Towards us, right, why? Why isn't the methyl group pointing towards us? Do we have to go over it again? So what kind of reaction is happening here? SN2. And what happens in an SN2 reaction? Conversion of configuration. Conversion of configuration. So why isn't the methyl group pointing towards us anymore? Because we had an inversion of configuration. Don't make it harder than what it is. We know all these rules, okay? So here, if we do that, what do we get? We get that. So this goes to A. So this goes to A. Well, the one that we had before the solid bond before goes to A, and let's erase this. And should have dashed lines here? Do we get one of those products up there, A or B? Do you see that? Why would we get B there? Where is this bromine going to go? How about that? Yeah, it's going to stay at the back. Why does it stay at the back? Does it get detached from that carbon at all? No, no, so the carbon's still hanging on to it from the back here, okay? Does that make sense? So let's just show, that's another S and 2 reaction. We're going to have that inversion of configuration on that series. So notice we only get two products from these two intermediates. You might think we would get four products, okay? Oh, we don't. But you have to do all of this to know which products you're going to get. Okay, so what would be the ratio of each of these products? What's the percentage? 50-50. 50-50. Why is that? In any terms? Well, no. Is there S and 2 reactions? Is that racemic? It is racemic mixture, yes. That means 50-50. Why? Why is it racemic? Because of these intermediates here, right? So since there's no preference to attack this carbon or this carbon on this intermediate or this carbon or this carbon on this intermediate, there's nothing really standing in the way. You're going to have a 50-50 or racemic mixture in any tumors, okay? Any questions on this?