 So there's this mechanism on the board. This is a Friedel-Krabs isolation, right? So when we think about this, before we draw the mechanism, what do we want to do to predict the product, right? So can anybody predict what the major product is like? Yeah, it's going to be a pair of positions, right? So the acetate's going to go on the pair of positions, right? So that's the... Yeah, so that's the major product, it's the para. And so we want to draw the mechanism here. So just keep in mind where we're starting and keep in mind where we're going so we can get there, okay? If we don't remember this structure, we won't be able to push electrons to get there, okay? So has everybody written this stuff down? Okay, so, yes, do you have a question? Thanks for the... At first you just... That's what I would do, yes, definitely. Because you don't know how to get there if you don't know what the answer is, right? Yeah, no problem. So the first thing we're going to do is react the acyl chloride, right? With the aluminum chloride. So you've got these electrons here. They're going to react with the Lewis acid there, the aluminum, like that. These are your formal charges as well. Now that aluminum has a negative charge, and chlorine has a positive charge, okay? So just, if you're having trouble with it, go back and watch this video and count them all up, okay? So now what happens, right? We're going to have these electrons come down like that and then kick those electrons down to that chlorine, okay? So what are we going to get, okay? A really electrophilic carbon there. Okay, you can see that positive charge on the oxygen, okay? So that's where the, what do you say, the nucleophile then comes in, okay? So all of this has to happen before we're going to even react with this benzene derivative, okay? So can I erase all of this stuff up here? You got all that? Okay, yeah, so I guess we can draw that too. So we still have this, right? And then these electrons just go back like that. So when that happens, we now have reformed our catalyst, aluminum chloride, and we have the Cl minus. Are we good? So now this is going to react with this, okay? So I'm just going to write this in an easier structure to work with. It's the same compound, of course. Okay, we good? So remember, the methoxy group here with its electrons, alpha to the benzene ring, is going to be an electron donating group, right? So that's going to, we're going to predict that the product would be either in the what positions? Do you remember? Ortho or para, okay? But since we've got a substituent here, and it's asking for the major product, we would expect the para to be major because it's further away from the other substituent, right? Steric hindrance is going to make the para be more favorite. Is that okay? So what happens here? This carbon remembers very electrophilic, okay? So that's why we can use the benzene ring here as a nucleophile, okay? So what's going to happen is these electrons are going to come down here. That's going to make those electrons move there. And when that happens, can you see? I'll get out of the way in a second. These electrons are going to attack that electrophilic carbon like that, and those electrons go up to the oxygen and relieve that positive charge. So I want you to attempt to draw what you think the products will be, okay? Try to do them before I do. In fact, let me erase this stuff over here. Everybody's got this, right? I'll take that as a yes. Are you trying to draw this one on your own? Try to draw the products. Does that make sense? Yeah. You see the arrows being pushed? Okay, good. Remember your Cl minus that we made before? Are you asked about? We're going to use that as the base here to deprotonate that now acidic proton. Remember, we want to have aromaticity in that benzene ring, okay? And this is not aromatic right now. So we're going to deprotonate that proton there to regain the aromaticity. So again, try it on your own before I do it. If you're watching the video, right now would be a good time to pause it any of these times before I do a set and then draw it yourself. Just like now. Show me what the products are. In fact, you should know because this is the major product that we had predicted that we would get. Are we good? Yes. So you can see we make HCl as well. So you'll see the pH of this reaction go down as the reaction progresses. Any questions on what we've done here? Make sense? Okay, good.