 Okay, so this one from the other day says, write a radical mechanism to form the most stable monochlorination product for the following compound. And then it says methylcyclobutane. So I want you to draw the mechanism for this reaction. So remember, methylcyclobutane is that compound there. If you couldn't figure that out, make sure you can do that. And I want you to do the reaction by chlorine plus light to get the product. Okay, so what's the product here? Does anybody think of that? Where should the chlorine be? The word squared null. Yup, that's right. But you didn't know that? Make sure you're on your game tomorrow or whenever we're taking the quit. Okay, so that should be the product. And those of you who know how to do this, well know that that should be the product. Okay, so the mechanism, that's different. Okay, that's just showing what the product is. Let's draw the mechanism. But this is kind of like what you were doing. So, to erase all of that business, we know where we're going. And then we know where we're starting. So, I like to remind myself that there's a hydrogen there. Remember, the most stable hydrogen is the most substituted hydrogen. The most stable radical is the most substituted radical. So, if we look at the substitution patterns of these potential radicals, we'll see that's the primary, secondary, secondary, secondary tertiary. So clearly it's going to go on the tertiary curve. So the first step, of course, is that the chlorine atoms break. And if you're not drawing fishhook arrows at this point in time, then you're not drawing it properly, okay? So you have to draw your fishhook arrows. Remember, those show the motion of one electron. The first step of the reaction, of course, is making the two chlorine atoms, or if you prefer to call them radicals, which they are. They're actually chlorine atoms. So now one of those high-energy chlorine atoms is going to come and extract that hydrogen in there. Because, remember, we want to make the most stable radical. So it's going to happen. And then the two fishhook arrows go like that. And then the other one go like that. And we were saying that tertiary radical, most stable, since it's going to be the most stable, that's where the monochlorination is going to go. Yeah, we have another chlorine atom, but that's probably doing something else. So it's probably another chlorine molecule at this point in time. If you use the chlorine atom, I wouldn't mark you wrong, but you're probably not doing it correctly. So that's going to give you the product there, and then another propagation step so that other chlorine could keep propagating the reaction. So that gives us the monochlorinated product. Like I was saying, we still have another chlorine atom. Go ahead and keep propagating like you would expect. Any questions on that one? Pretty straightforward, right? Wonderful.