Let's dive into the exact mechanism of EAS halogenation. EAS bromination and chlorination both require complex formation 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 proceed. We're going to use that to start off this reaction. In our very first step, before the benzene can get involved at all, we need to complex our diatomic halogen with the Lewis acid catalyst. This happens through the bromine sharing some of its electrons with the Lewis acid that it's missing electrons. It's a great electron pair acceptor. Now, this is going to form 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 what that stands for, all three bromines. Awesome. Now, are there any formal charges included in this molecule? Yes. 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, and two lone pairs. That's going to get a positive charge. It's missing a valence electron. So, 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 it 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. Now what we are 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 its 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 structures. 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. 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. Now we want to do the elimination step. That was an 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. 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 that the bromine grabs its electrons, says, "Hey, I'm taking them back." And then it gives its electrons to the H. Actually, don't draw this. Use your eraser really quick. And 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 its 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 its electrons and giving them now to that H and now making a bond with the H. We know that hydrogen can't make two bonds. 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 get 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. You could have drawn that resonance structure. However, because it's constantly changing. It's constantly in a resonance structure. You can draw however you want. You could just draw it as a circle if you want. But that is our product and that's it. Let's go ahead and move on to the next mechanism.
19. Reactions of Aromatics: EAS and Beyond
EAS:Halogenation Mechanism
19. Reactions of Aromatics: EAS and Beyond
EAS:Halogenation Mechanism - Online Tutor, Practice Problems & Exam Prep
EAS Bromination and Chlorination both require complexing with a Lewis Acid Catalyst before the reaction can begin.
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EAS Halogenation
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PRACTICE PROBLEMS AND ACTIVITIES (12)
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- Phenol reacts with three equivalents of bromine in CCl4 (in the dark) to give a product of formula C6H3OBr3. W...
- Propose a mechanism for the bromination of ethoxybenzene to give o- and p-bromoethoxybenzene.
- Predict the major products of the following reactions. (a) toluene + excess Cl2 (heat, pressure)
- What products are obtained from the reaction of the following compounds with one equivalent of Br2, using FeBr...
- What is the major product(s) of each of the following reactions?a. bromination of p-methylbenzoic acid
- What products are obtained from the reaction of the following compounds with one equivalent of Br2, using FeBr...
- What is the major product(s) of each of the following reactions?b. chlorination of o-benzenedicarboxylic acid
- Why isn’t FeBr3 used as a catalyst in the first step of the synthesis of 1,3,5-tribromobenzene?