 Okay, so let's try this problem. It's actually a mechanism problem. It says add curved arrows to the following polar reaction mechanism to indicate the flow of electrons in the reaction shim. Okay, so what does it mean polar reaction? Can anybody tell me? Or what kind of arrows do you use in a polar reaction? Yeah, double-headed, right? That's what that two signifies, right? Double-headed arrows. Is everybody clear with that? Okay, cool. Quiet this morning, yes. Double-headed arrow, yeah. Radical, it's always the what kind of arrow guys? Fish on, yeah, or single-headed arrow, okay? Well, so yeah, that's going to be a point where you're going to, sometimes I'll tell you, and sometimes you're going to have to figure it out on your own, okay? But since we're just learning, most of the time right now, I'll be telling you, okay? So like this one said polar, so what kind are we going to use? Again, double-headed, okay? So what was the first thing we said we wanted to do when we ever whenever we see a mechanism problem? Loan pairs, yeah, put in the lone pairs, okay? So around bromine, right? There's some lone pairs. Stop me if I do something that freaks you out, okay? Oxygen, right? Do we have lone pairs? Good job guys. Okay, so did everybody draw those lone pairs in there? Problem? Okay, good. And then you might as well put your lone pairs over on this side too, because again, what is the arrows showing the motion of what? Electrons, right? And what are lone pairs? Electrons, right? So it's good to identify these things, lone pairs, bonds. And remember the arrows themselves are showing where the bonds are being made and being formed, okay? That's like the motion of the electrons to form or break that bond, okay? So in other words, what did we see? In this reaction, right, we see this part of the molecule, the phenyl ring, right? Over here, okay? But what don't we see attached to it by this methylene group? The what? The bromine, right? Okay, so what do we have to show in that case? An arrow, right? Yeah, because that bond's being broken, okay? But it's not going to break by itself, okay? So we gotta show another arrow first. That's the obvious arrow. When I look around at everybody's reactions, that's the one that everybody gets pretty easily, okay? But the first arrow, where are we showing a bond being made? Can anybody tell me? Yeah, not the bromine, right? Oh, I mean just, yeah, the oxygen. The oxygen and the carbon, the bromine's attached to it, right? Does everybody see that, right? So what's going to happen is there's going to be a bond being made between those two atoms and that making of the bond is going to subsequently make that bromine leave, okay? We call that bromine a good leaving group, okay? So that's kind of a good name to remember to kind of identify what it does, okay? So since we've already figured out what's happening in the reaction, all we need to do is draw our polar arrows or our curved arrows, okay? So how do we do those? Remember, we always have to start our arrows from the electrons, right? The lone pairs or bonds, okay? So what did we say? There's going to be a bond from here to here, right? Everybody agrees? So that's what we're going to do. Draw that arrow like that. No problem, right? No problem now that Professor Geith is drawn in, right? Everybody gets the second arrow very easily, but that first arrow is very difficult for people to get. So I want you to just look. This arrow represents that bond right there that we made, okay? The other thing about arrows you got to draw on the right way. You draw them backwards, that's not correct, okay? It's just as not correct as driving your car on the wrong lane and on the highway, okay? That's going to really send you into trouble, okay? So don't do it. So just to beat a dead horse, are we done with this thing yet? No, we have to show what other part of the reaction happening. The bromine leaving, right? So where do we start our arrow from? Always, did we say? The what? Electrons, right? Okay, so we have to start from the electrons. Look at this bromine. How many lone pairs does it have around it here? Well, three lone pairs, right? Okay, so six electrons total. And how many does it have around it here? Four, right? So what have we done? It has more electrons, right? Two more electrons, so that is right for putting a polar arrow. Does everybody understand? Okay, so where are we going to start it from? The bond, and where's it going to go to? The bromine, right? It's all right, you guys can talk about it like that. Is everybody okay with what I did? Do you see what's happened, right? So what are we showing with this bond here? Err, this arrow here, I'm giving away the answers. Again, what are we showing? The bond being formed, right? And what are we showing with this one? Bond being broken. Does that agree with what we've seen over here? Okay, cool. Any questions on that one? This is probably a dumb question. I don't see a double arrow. Isn't a double arrow, if there's an arrow on both sides? What do you mean double arrow? What do you mean double arrow? That's a good question for the mechanism quiz that we're about to take, right? I agree. What does it mean? I'm not sure. Well, look at the arrows that I'm showing. What does it mean? It's a polar arrow, right? There's another way of saying it. The double-headed arrow, right? We're not talking about resonance arrows. We're not talking about this. Okay? We're talking about this. Okay? That's a double-headed arrow. So what does a fish hook arrow look like, or a single-headed arrow look like? It looks exactly like that thing except for what? Okay? So that's a very good question and very important to know. Okay. Yes. So you need to start studying this. Okay. Any other questions? Would that be what type of reaction? Yeah, that's a good question. What type of reaction did we say? Radical. Radical reaction. Good job. Okay. So is this a radical reaction? No. What's one way I can identify that this is not a radical reaction, especially after we've done it? Because of the double-headed arrows. Double-headed arrows. Oh my gosh. Okay. Easy way to identify. No wonder we're getting a lot of hundreds in this class. Okay? Any other questions? Wonderful.