 So this is the, I believe this is the last one, number five on the quiz that I think a few people wanted me to go over. So the question says, for the following E2 reaction, so that's a helpful hint. Remember E2 is a one-step reaction, okay? So propose a structure for the major product and mechanism for the minor product in the reaction. Okay, so I'm going to go ahead and just propose that structure for the major product. Probably you all know what it is. The mechanism seems to be harder for this one for everybody. The major product, of course, this is the exocyclic double bond that's going to be the minor product because it's only one to dye substitute. You can get a more substituted product from this. So remember an E2 reaction takes an alkyl halide, gives you an alkene out of it. So that would be, thank you, put the double bond there or there. That would be the same product. So the major product, of course, is going to be if you want to use a metal. Like that. So if you drew the double bond up here, like I think a couple of you did, that would be fine too. Okay, it's the same product. That's the major product. Okay, so now it reminds us that it's an E2 reaction and it wants us now to identify what the mechanism for the minor product is. So can I erase the major product? Everybody got that now, right? I can erase that part. And I'm going to draw out this methyl group. In fact, I'm not going to draw the C but I'm going to draw the H's there. And I'll remind us that there are three H's. Eventually you'll only, it'll start drawing one H. So now remember we reacted that with potassium hydroxide to an ethanol, okay? So the potassium hydroxide is going to be the base for this reaction. Remember an E2, that's an elimination reaction, so you need a base. But the potassium, that's just a spectator ion and it doesn't do anything. All your spectator ions, just like always, like we've said a number of times, you just don't even put those in the reaction anymore. You just get rid of them, okay? This is a mechanism. You're just showing what bonds are being formed and what bonds are being broken. So anyways, what's going to happen? You start your arrows always at your lone pairs or your pi bonds. You don't have pi bonds here so lone pairs are going to start. So where are we going to start from there? It's going to deprotonate the hydrogen and these electrons are going to make a double one there. Is that a double one? There, and that's going to kick off that chloride, okay? And ta-da, that's the step. That's it, E2. So it's just a one-step mechanism, but you of course have to actually draw structures that exist. You have to draw arrows that go to the right way, not the wrong way, okay? And you've got to draw all the arrows in the right steps, okay? So for this one, there's actually more arrows than what we like. You know, we've been seeing a number of reactions with two arrows. This one's an elimination, it's got the three arrows. It just got to recognize that, okay? So for those of us who didn't get that mechanism, write it down. I expect you to be able to do it next time we see it. Okay, any questions on that? Okay, wonderful.