 Okay, so let's do another interesting reaction, so this one's an E1CB reaction, or it's going to be. So our substrate here, you can see this is a tertiary alkyl halide, and we've got a strong base. So normally we would think, okay, tertiary, that's going to go through SN1, but we've got a strong base here, so that we would think, oh SN2, but tertiaries can go through SN2. Okay, so what's going to happen here, in fact this, either one of those is going to be an E1CB. And like we said, the CV stands for conjugate base, okay? So the reason being is because the conjugate base itself is going to be the thing that's doing most of the reactivity. Okay, so I'm going to erase that, the ethanol on that was just the solder. So the important part was the OH-, so we're about to do the mechanism here. Okay, so remember, protons alpha to a carbonyl are acidic, so this is a very strong base, so what's going to happen? So we're going to deprotonate that proton there. Those electrons are going to be donated into that double bond, the forming double bond there, making these electrons go up to the oxygen atom like that. Okay, so you've seen reaction mechanisms like this before, so don't let it freak you out. You've got that. Okay, so hopefully everybody got that as their organic product, if you didn't. Make sure you are able to before Thursday. Okay, but that's not our product, right? We've got to get rid of that alkyl halide or the halide. So what's going to happen now is these electrons are going to donate back down, so reform that carbonyl. That's going to mean this double bond has to move. Okay, that double bond is actually going to move to right there. Once that happens, of course, this bromine is a very good leaving group, so what do you think it's going to do on that? Yeah, pop off or leave. So a good leaving group is going to leave, and that's going to be through and off. So, E1, 2. So this one's called an E1 stevee. Are there any questions about this one? Questions? Hold now.