 Okay, so let's do this problem now. It says write the stereochemistry of the product resulting from the following reactions and is the product optically active. And the one we're doing says the addition of HBR and ether to two-butene, okay? So ether in this case is just the solvent, okay? You have to have some sort of, what do you say? Like swimming pool for the molecules to swim around in reactant, okay? So that's kind of what the ether is. Gives them a place for it to react. So you just put that on the bottom of the, and you just think, okay, that's not a reactant. I'm not gonna even worry about it. Okay, HBR, we're gonna put on the top of the arrow because that's the reactants, okay? And that's where we put them, okay? Two-butene, it doesn't tell us if it's this or trans. It doesn't matter, okay? Which one do you want? Cistern. Cistern. Cistern, okay, let's do Cistern. Okay, so what is the stereochemistry and is it optically active, okay? So in order to know how to do this, we should do the mechanism of the reaction. It'll help a lot more, okay? So when we do the mechanism, that's the step-by-step breaking and informing of each of the bonds, okay? So I'm gonna erase this stuff and put only the active ingredients into the reaction mechanism, okay? So two-butene and hydrobic acid. So notice I'm drawing that bond there because I'm gonna be breaking that bond, okay? So remember the first step, when you have a strong acid, right? If something's gonna deprotonate that strong acid, okay? Since the other thing is not the acid, I hope you assume that's the base, okay? So anything with extra electrons can act as a base. That's what two-butene is. So what's going to happen? It's going to deprotonate like that. Remember to draw your arrows in the appropriate fashion. Don't draw them the other way, okay? Remember it's the electrons that do stuff, okay? So when those electrons go and grab that hydrogen, now we have that hydrogen on that carbon atom there. That carbon atom already had a hydrogen on it, right? So that's how it works. So you have to remember where those hydrogens are, okay? So now when we add this hydrogen, we're actually gonna have two hydrogens on that carbon. Okay. We're gonna have a Br minus over here because this Br took those electrons away and we're gonna have a plus charge, or a carbon cation on this carbon here. Here, let's put that hydrogen there. Right there because it doesn't have its full octet. Does that make sense what we've done so far? So now this, and I'm here, is sp2 hybridized, okay? So what is the bond angle if it's sp2 hybridized? Say it out loud, 120, right? 120 degrees. So that means that this is trigonal planar, okay? So here we have an electrophile that's trigonal planar and a nucleophile over here, okay? So since this is trigonal planar, the electrophile can be attacked from either side equivalent, okay? So I can attack it from the backside or from the front side because neither one is sterically encumbered, okay? So there is this sp2 business where the reaction is going to be performed in two different ways, okay? The bromine is going to attack, can attack from the front like that, okay? So from the front we get one product, if it attacks from the back, we get a different product, okay? So from the front, hopefully we're still on camera here. Slightly off, so just move it a little bit. Bromine is going to be hatched from the back. If it attacks from the front, the bromine is going to be wedged, 15, 50 ratio. There's no reason that the bromine would attack from the front or the back, okay? Any more than it would attack from the other way. 50% mixture and a 50% mixture of the two and A tumors, okay? So we call this a racemic mixture. We have a racemic mixture. The mixture itself is not optically active, okay? But these two here are really optically active. So if we, and we can tell because they have a stereocenter, each one of them is, okay? So we label them one, two, three, likewise. So that's the R and that's the X. So if you need to move to yourself, that's the S, okay? So since R is 50%, S is 50%, they rotate equivalently in direct, equivalently but in opposite directions, they're going to cancel each other. So it's because we have this racemic mixture that is not optically active, but each one of these as an individual is optically active. Okay, did you have a question? No, okay, do you have a question? Does this have anything to do with SM1? This is an addition reaction, uh-huh. So SM1 will be eventual, okay? So you're a little ahead of the game right now, okay? Yeah, but this is hydrobromination of an alkene. That's what you call this reaction, okay? Or an addition reaction. Okay, any other questions on this one? Okay, one.