 Okay, so the question we had is kind of what was the difference between elimination reaction and an addition reaction, so what we're going to do is do two different videos, this one being the addition video and we're going to compare that with the elimination video that we've done or that we're about to do after this one, okay? And this one we're going to add HBR to the alkene. So if you have any questions, whenever, feel free to ask them. So whenever you see this, of course, remember you don't have to keep it in that form. What I would do is expand it to its Lewis structure so you can actually see the bonds in the lone pair so it'll be easier for you to write the mechanism for this reaction, okay? So let's go ahead and do that right now. So that makes sense, hopefully. And then remember, for me I like to show the arrows going through the actual atom. That's going to be taking the proton in this case or adding to the electrophilic center or whatnot, okay? So let's show that. So here we've got an acid, okay? So if we have an acid, we should be predicting an acid-base reaction. And that's what we have here. The base here is going to be this set of pi electrons, right? And remember, well, let's just show the mechanism first. And when we show this mechanism, what I'm saying here is this carbon atom is actually deprotonating that hydrogen atom, right? That proton. So what we make is this carbocation on here and the very stable bromide ion. So remember, over here we've got two hydrogens. Do you want me to show them the explicitly show the hydrogen? Okay, so we've got two hydrogens there and one hydrogen there, right? So if that carbon is taking that proton, if it's deprotonating and if it's being the base, it's going to have a full set of protons on it. So we're going to make a full methyl group here. So if we want to, we can show all of our hydrogens here. Does that make sense? And we didn't do anything to the amount of hydrogens on this carbon, right? We just took the electrons away from that. So let's just go ahead and show that hydrogen. It's got that positive charge. So that's a carbon cation, right? So the carbon with a positive charge. SP2 center, that means this can attack from either side, okay? So we're going to get a set of enantiomers in this reaction, okay? You may ask, well, why didn't it deprotonate, why didn't this carbon deprotonate this acid instead of this carbon, right? Why did we make this carbocation instead of this carbocation, which was not born? Do you remember what we called this? When the hydrogen adds to the carbon with the least amount of the hydrogens? Do you remember what it's called? Is that Russian guy's name? Markovnikov. Markovnikov, yeah. So what we're showing here is Markovnikov addition, okay? So here we're showing the Markovnikov carbocation. This one's not formed. Well, why is that? Well, if you look, this is a primary carbocation down here, right? This carbon is only attached to one other carbon. Here you see it's a what kind of carbocation? Secondary, right? And do you remember which one's more stable? The secondary, that's why it's formed. Remember in chemistry, you want to form the more stable things, okay? So that's why this intermediate is formed. So this is the intermediate in the reaction. Like if you were drawing the reaction diagram, remember those? So now what we're going to do is show the attack of the bromide ion, okay? Like that. So it can attack. So this is planar. It's trigonal planar, right? This is the sp2 center. So you can attack from what we call the back face or the front face. Remember on those sp2 centers. So hopefully you can see we have a methyl group behind your gen and a cyclohexane ring, right? Those are three different groups. And if we're adding this bromine, that's going to be a fourth different group. So we're going to make a stereo symptom there, okay? And there's no reason why the bromide would prefer to choose the back face over the front face, okay? So we're going to get what we call a racemic mixture. Remember that. So it's a 50-50 percentage of the two stereo isomers, the two and anti-mers. So let's just show that. So what we're showing here is we'll say the back face attack. And what that means is the bromine is coming from the backside of the board to attack that center, okay? So when we show that, if that happens, that hydrogen is going to be pushed towards us. Does that make sense? So the bromine is going to be back. So that would be the mole. Without all the hydrogens. If you wanted to put the hydrogens in, we could. The one hydrogens that's on that SP2 center that's nailed the stereo center would be pointed towards us, okay? So that would be from the back face attack. Let's show the other one. So I'm going to erase some of this. Is it all right if I erase this bottom part? So I'm going to show it on the same carbocation here. So that would be coming from the front face. But it helps, let's erase this other mechanism just so we don't think they happen at the same time, okay? So the front face can be attacked. Again, because this is a trigonal planar center. So if that happens, where's the hydrogen going to go? To the back, right? So the bromine, if we're just going to show this without the hydrogens, is going to be pointed towards. To the front, yeah, to the towards us. So we're going to show a wedge. And again, if you prefer to draw it like that, that's fine. If you want to draw your hydrogen in, just to make sure you can tell the difference. I know you were an introductory organic chemistry student, so it's good to remember all of this stuff. And again, like what we were saying, there's no preference for it to be taken from the front face or from the back face. So what we'll find is that you're going to get 50% of this isomer and 50% of this isomer. We call these stereoisomers. And in particular, since all of their stereocenters are opposite of all of these guys' stereocenters, all one of them, right? We call them an anterior one. So that would be like what we say, Markovnikov addition to an alkene. So Markovnikov of addition of HBr to an alkene. So we can show the opposite reaction, without to show the elimination of HBr from, we'll say, one of these molecules, whichever one we prefer. To give an alkene. So does this make sense? Does anybody have any questions on this one? I know. It's early morning, we just walked in. It's okay if there's none.