EAS: Halogenation Mechanism

Let's dive into the exact mechanism of EAS halogenation. EAS bromination and chlorination are both going to have to complex with a Lewis acid catalyst before any reaction can take place. Remember that in our general reaction, you need that Lewis acid catalyst in order to go anywhere. We're going to be using that to start off this reaction.
In our very first step for our mechanism, before the benzene can get involved at all, we need to complex our diatomic halogen with the Lewis acid catalyst. That happens through the bromine sharing some of its electrons with the Lewis acids. It's missing electrons. It's a great electron pair acceptor.
Now what this is going to do is it's going to from a complex that's very electrophilic. Let's see what it's going to look like. It's going to be a bromine attached to a bromine, attached to an iron, which is attached to three bromines. I'm just going to put Br3. Is that fine? You guys know that that stands for all three bromines. Awesome.
Now are there any formal charges included in this molecule? Yeah, well, now the iron was neutral before. It has an extra bond. That's going to get a negative charge. Bromine, as we know, likes to have seven valence electrons, right now it only has six because it has two lone pairs. Two bonds, two lone pairs, that's going to get a positive charge. It's missing a valence electron. This is our active electrophile. This is our active electrophile. This is the one that reacts with benzene.
What's going to happen in this mechanism is that my benzene is the nucleophile. Now it's going to attack the electrophile. What might seem a little bit weird is that you would think that would go straight for the positive charge because usually negatives attack positives. But actually, let's just bring down the benzene here. What's going to happen is that the benzene is not going to attack the positive, it's going to attack the bromine next to the positive. Why? Because if it can attack that one and remove it, then this bromine can donate its electrons to the one that's missing some, which is the actual positively charged bromine.
Now, going down, what this is going to cause is an interruption of aromaticity. This is going to be our intermediate. So now what we're going to have is one double bond here, one double bond here. We had an H before, but now we've also got a bromine. Here we also have an H, but it's missing it's fourth bond, so that's going to be where our carbocation goes and this specific carbocation is called what? This is our arenium ion or sigma complex. This is going to be our sigma complex.
As we know that sigma complex has resonance structure, so let's just draw those really quick. We know that this double bond can resonate to three different positions. It's going to move over. HBr. And what we're going to end up with is three resonance structures that are stabilizing that intermediate. As you guys recall, this is the slow step of the reaction, making the intermediate.
All right. Now we want to do the elimination step. That was addition. Let's do the elimination step and end this reaction. What do you guys think is going to be the nucleophile that reacts with my arenium cation? Good, it's going to be the FeBr4 that's negatively charged. We've still got this FeBr3 that has an extra Br on it. And the whole thing is negatively charged. So how is that going to react? Well, we can use the electrons from the bond, from the extra bond to eliminate the hydrogen.
Now, to me personally, the mechanism that makes the most sense and you're going to see this a lot in this chapter is what I think makes sense is the bromine grabs its electrons. Says, I'm taking them back. And then it gives its electrons to the H. Actually, don't draw this. Use you eraser really quick. Then it gives its electrons to the H. To me, if I were writing your textbook, that's the way I would have drawn it because it makes the most sense that it takes it's electrons and then it gives them to the H.
But the way that textbooks usually write this mechanism is all in one shot. They'll write that the electrons just go straight for the H. But it's the same thing. This notation of showing that the electrons go from that bond to the H, literally just means that the bromine is taking it's electrons and giving them out to that H and now making a bond with the H.
Now we know that hydrogen can't make two bonds, so if we make that bond, we would then break this bond and reform the aromatic compound.
That was just a note to say guys that if you ever see me in the next few mechanisms drawing straight from a bond, that means that you can just think of it as that two arrow mechanism instead.
So what we're going to get now is we're going to get, since I reacted with this resonance structure, I'll draw it like this, bromine here. Plus we're going to get what else? We're going to get FeBr3. Notice that this is why it's called a Lewis acid catalyst because we regenerated it at the end. And we're going to get HBr. Excellent guys.
If you're wondering if the resonance structure matters, no it doesn't, guys. You could have drawn that resonance structure however because it's constantly – it's a resonance structure, you can draw it however you want. You could just draw it as a circle if you want. But that is our product. And that's it. So let's go ahead and move on to the next mechanism.