 Okay, so let's look at this satellite anion reacting with this bromocyclohexane. So from what we've been talking about, potentially, you would think that the product that would be formed would be the SN2 product. So what would be the SN2 product? Why don't you guys write that down in your paper right now? So that would be the product that we were just talking about. Remember, in that case, we were reacting it with a primary alkyl halide. This time, we're reacting it with a secondary alkyl halide. Because the secondary alkyl halides have more steric ingredients, more sterics than primaries. That makes sense, right, everybody? Does that make sense right now? Yes. So what? OK, so what you would expect from the SN2 reaction that's not formed would be that product there. Hopefully, you guys all wrote that product. If you didn't, make sure you're being able to by next time. What's actually formed is the E2 product, the elimination product. Why? Because secondary alkyl halides are more sterically hindered. So what you'll find is that instead of just doing the backside attack and letting the leaving group leave, you'll do the acid base reaction. So remember, where did this come from? This came from a terminal alkyne, right? So we can make a terminal alkyne back from it. It's a pretty strong base, remember we were talking about. So when we have that bromine there, that makes that hydrogen even more acidic. So what will happen is that hydrogen will be deprotonated. The electrons there will go into the ring and form a double bond. And those electrons there will bounce out and form the bromide on anion. So what you'll find is the expected SN2 product is informed in this reaction, but the elimination product is formed. And that's, again, due to the fact that this is the secondary alkyl halide as opposed to a primary alkyl halide. Any questions on that one? Wait, the last one. No, that's all right. Oh, OK, sure.