Aromatic halogenation

In this video, we're going to look at the halogenation of benzene. And we'll start with bromination. So here's a benzene ring. And to this, we're going to add some bromine. And our catalyst will be aluminum bromide. And you could've used FeBr3 instead of AlBr3. That's fine. And the end result is substitution of a bromine atom for an aromatic proton on your ring. Let's look at the mechanism for this electrophilic aromatic substitution reaction. And if I look at the aluminum bromide catalyst, and I can see there are six electrons around the aluminum atoms. So here's two, and four, and then a total of six. And so because of aluminum's position on the periodic table, it can actually accept two more electrons. So the aluminum bromide is going to function as an electron pair acceptor, which is the definition for a Lewis acid. The bromine is going to function as a Lewis base. It's going to be an electron pair donor. So we could think about this lone pair of electrons in here being donated to the aluminum and a bond forming between that bromine and that aluminum. So let's go ahead and draw the result of that Lewis acid base reaction. And so now this bromine is bonded to this aluminum. This aluminum is still bonded to these other bromines here. I'm not going to draw in those lone pairs of electrons around those bromines just to save some time. Let's follow those electrons. So the electrons in magenta, those are the ones that were donated to the aluminum and forming of this bond between the bromine and the aluminum. That would give the aluminum a negative 1 formal charge. And this bromine would get a plus 1 formal charge, like that. Now technically, this is the complex that's going to react with our benzene ring in our mechanism. But it's kind of hard to see the electrophile in this complex. So let me just go ahead and show you what you can think about the electrophile being. And then we'll come back to this complex in the mechanism with benzene. So if these electrons in here moved off onto the bromine on the right, the bromine on the left will have lost a bond. So it would now only be surrounded by three lone pairs of electrons, giving it a plus 1 formal charge. And it simplifies things to think about this as being the electrophile in your mechanism for electrophilic aromatic substitution, even though, technically, it's going to be this top complex here that's going to react with our benzene ring. So if we form Br+ over here on the right, we would have this bromine now still bonded to this aluminum. And it would have three lone pairs of electrons around it now. So let me go ahead and highlight these electrons in here in red. I'm saying they're going to kick off onto that bromine there like that. And the aluminum, of course, is still bonded to these three other bromines. And it still has a negative 1 formal charge like that. So we've generated our electrophile. Or this is one way of simplifying it, to think about the Br+ as being our electrophile. And so that could react with our benzene ring. So we come back to our benzene ring here, and we think about Br+ as being the electrophile. And so, remember, electrophile means loving electrons. And so it is attracted to electrons. And of course, electrons are negatively charged. And they're attracted to positive things. So you could think about these pi electrons in your benzene ring as functioning as a nucleophile. And so we get a nucleophile attacking an electrophile here. And so let's go ahead and show the results of that nucleophilic attacks. So we have our ring. We have our pi electrons in our ring. And you can show the bromine adding to either one of these carbons. It doesn't really matter. Since they are equivalent, I'm going to show the bromine adding to the top carbon there. So the top carbon already has a hydrogen on it. And we're going to say that these electrons in here add on to the bromines. Let me go and highlight the electrons that we're talking about. So these Pi electrons in here function as a nucleophile, form a bond with that bromine like that. We took a bond away from this carbon down here. So we're actually going to get a plus 1 formal charge at that carbon. So we make a carbocation. Now remember, technically, it's actually this complex up here that's reacting with benzene. So we could think about it better mechanisms, or a more accurate mechanism, as being these electrons in magenta going up to here, attacking that bromine, and then the electrons in red here, once again, kicking off onto this bromine, forming this complex over here. And so that's the more accurate way of thinking about it. For me, it just simplifies things to think about Br+ as being an electrophile. So now that we've formed a carbocation here, we can actually draw some resonant structures. And so if I took these pi electrons and moved them into here, let's go ahead and draw the resulting resonant structure for that. So I would have my ring. I would have these pi electrons. I would have a hydrogen and a bromine still attached to my ring. And I would move does pi electrons over to here. So let me go ahead and highlight those. So these pi electrons moved over to here, took a bond away from this carbon this time. And so that's the one that's going to get a plus 1 formal charge like that. We could draw another resonant structure. So these pi electrons up here could move down there. And let's go ahead and show the results of that. So once again, we have our ring. We have a hydrogen, a bromine, these pi electrons are still there, and we get some pi electrons moving over to here. So let me highlight those now. So these pi electrons right here move over to here. Took a bond away from this top carbon, and so that top carbon is going to get a plus 1 formal charge. And so we have three resonant structures that we could draw for this carbocation. And remember, the actual ion is a hybrid of our three resonant structures. And we could think about that hybrid as being a sigma complex. So in electrophilic aromatic substitution, the last step of the mechanism is deprotonation of your sigma complex to reform your aromatic ring. And so we could think about this complex right here. It's going to function as a base. And I'm going to say that these electrons in here could take this proton. And then it these electrons would move in to here to reform our aromatic ring. So let me go ahead and draw the product, which is bromobenzene. And since we have a lot of arrows going on there, let me go and highlight some electrons, so we can follow them. So let me go ahead and make these electrons in here blue. So these electrons are going to move in here to reform our aromatic ring. And these electrons in here, you could think about this complex as functioning as a base. And so those electrons pick up that proton. And so you'd form HBr as another product. So we'd have HBr here, and I'll highlight those electrons in green like that. And then, of course, we would also reform our catalyst. So AlBr3, our catalyst has been reformed in this reaction. And so that's the mechanism for the bromination of benzene. If you wanted to think about adding other halogen onto your benzene ring, let's go ahead and look at chlorination here. So if we started with a benzene ring, and we wanted to put a chlorine on our benzene ring, we would add some Cl2. And our catalyst would be, you could use AlCl3, aluminum chloride, or you could use FeCl3. It doesn't really matter which catalyst that you use here. The end result, of course, would be to substitute in a chlorine for one of the protons, one of the aromatic protons on your ring here. And so our product would be-- let me see if I can draw this in here. So we would have our benzene ring. And we would have a chlorine on our benzene ring to form a chlorobenzene. And if it helps you to think about HCl as being another product formed in this reaction, it is. And so these are just two mechanisms for the halogenation of benzene.