 Okay, so we just did addition of HVR to an alkene, and in fact what we did was we made this as one of our products, you recall. So now we're going to show the opposite reaction, which is the elimination of HVR from this alkyl halide to make that alkene that we started with. So if you remember when we did this reaction, to get this product, we actually had a racium mixture. And what we'll find is that the starting material from the last reaction isn't going to be the only product of this reaction, so just keep that in mind. So when we look at this, these reagents and solvents that are over and under the reaction arrow, hopefully you can see that they're very similar, right? This is OETNA, and this is ETOH, right? So the H has been replaced with an Na in this particular molecule relative to the solvent here, does that make sense? Okay, so that's really important, so we're reporting that you don't want to do this on human. That's really important for a number of reasons, but for right now we're just going to focus on the fact that it's a very strong base, okay? So remember what we like to do. We like to erase this stuff and draw out the Lewis structures of the reacting molecules, okay? So let's erase that. Remember Na plus is the spectator ion, so we've got the negative OET being our strong base. So remember, what are we trying to do here? We're trying to eliminate, okay? And we're going to want to deprotonate one of the alpha protons that's on a carbon, one carbon away from that function, because those are acidic protons. So what we'll find is if we look, we've got one, two alpha carbons, okay? So in fact what we can do is deprotonate either that proton or one of these three protons, okay? And in fact the product mixture is going to show that we deprotonated one of those in one of the products and one of the other ones in the other product, okay? But one of them is going to be favored. In fact the one that's favored is the, like the major product or the state set product, remember we called it that, is going to be the starting material of the reaction we used, or we started with on the video before, okay? So this is kind of like going backwards and forwards from the same reaction to the same product, from that product back to the starting material, okay? So what we'll do is show the mechanism of the minor product first, okay? So remember in these E2 reactions, this is an E2, okay, that's recording. So the minor product is going to come from deprotonation of the less substituted carbons. So we'll deprotonate any one of those protons would have been fine with the deprotonation, they're all acquittal ones, okay? So when we do that those electrons are going to flow down to in between those two carbons and of course this carbon cannot have more than four bonds. So it's going to kick out that Br minus ion. So what we're going to do is get that thing here. Actually was that the starting material? I can't remember who that was the starting material of the reaction. That was the starting material. So yeah, the minor product, sorry about that, the minor product of that last reaction is, or the minor product of this reaction is the starting material of that. So this is the minor product here. Why is it the minor product? Well, let's see what the major product is first, okay? So what I'm going to do is erase these mechanistic arrows and I'm just going to show this reaction again. But this time it's not going to deprotonate from the less substituted carbons, deprotonate from the more substituted carbons. That's going to make those electrons going between those two carbons there. And that's going to kick out that Br minus there. So when we do that, we're not going to get this product. Let's keep that reaction as we said was the minor product. We're going to get this product down here. What do you think the double bond is going to be? Okay. So if you oftentimes introduce your organic students who want to also draw this as one of the products, hopefully you guys can see this and this are the same substance, okay? So don't draw both because you get some points. So let's erase this. And we'll determine why this is the major product. And that's the minor product. Can you tell me? Can anybody tell me? That's a rule. Mm-hmm. Yeah, so this is... That's a rule, right? He says that the more substituted alkene is the more stable one. Therefore it's going to be produced as the major product. So substitution talks about how many substituents that aren't hydrogen that are around that alkene, right? So here we've got one, two, three carbons. Here we've got one carbon. So this one's more substituted, so that's the major product. Okay. Pretty good for, well, earlier than eight is when we started. So I'm real proud of you guys. I'm even tired. Thanks for the question.