EAS:Halogenation Mechanism - Online Tutor, Practice Problems & Exam Prep

Electrophilic aromatic substitution (EAS) halogenation involves the formation of a complex between a diatomic halogen and a Lewis acid catalyst, creating an electrophile. The benzene acts as a nucleophile, attacking the electrophile to form a sigma complex, which undergoes resonance stabilization. The elimination step regenerates the Lewis acid and produces HBr, restoring aromaticity. This mechanism highlights the importance of resonance structures and the role of Lewis acids in facilitating reactions, essential for understanding aromatic chemistry and electrophilic additions.

EAS Halogenation

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.

Here’s what students ask on this topic:

What is the role of a Lewis acid catalyst in EAS halogenation?

In EAS halogenation, a Lewis acid catalyst, such as FeBr₃ or AlCl₃, plays a crucial role by complexing with the diatomic halogen (Br₂ or Cl₂). This complexation makes the halogen more electrophilic, facilitating its reaction with the benzene ring. The Lewis acid accepts an electron pair from the halogen, creating a highly electrophilic species that can be attacked by the nucleophilic benzene. This step is essential for the reaction to proceed, as it activates the halogen for the subsequent steps in the mechanism.

How does benzene act as a nucleophile in EAS halogenation?

In EAS halogenation, benzene acts as a nucleophile by using its π-electrons to attack the electrophilic halogen complex. The benzene ring, rich in electron density, targets the halogen atom adjacent to the positively charged halogen in the complex. This attack disrupts the aromaticity temporarily, forming a sigma complex (arenium ion). The intermediate sigma complex is stabilized by resonance, allowing the reaction to proceed to the elimination step, where aromaticity is restored.

What is a sigma complex in the context of EAS halogenation?

A sigma complex, also known as an arenium ion, is an intermediate formed during the EAS halogenation mechanism. When benzene attacks the electrophilic halogen complex, it temporarily loses its aromaticity, resulting in a carbocation intermediate. This intermediate, characterized by a positive charge on one of the carbon atoms, is called a sigma complex. The sigma complex is stabilized by resonance, where the positive charge can be delocalized over different positions on the benzene ring. This stabilization is crucial for the reaction to proceed to the elimination step, restoring aromaticity.

Why is resonance important in the EAS halogenation mechanism?

Resonance is important in the EAS halogenation mechanism because it stabilizes the sigma complex (arenium ion) intermediate. When benzene attacks the electrophilic halogen complex, it forms a carbocation intermediate, disrupting aromaticity. The positive charge on the sigma complex can be delocalized over different positions on the benzene ring through resonance structures. This delocalization reduces the energy of the intermediate, making the reaction more favorable. Resonance stabilization is essential for the intermediate to survive long enough to undergo the elimination step, which restores aromaticity and completes the reaction.

What are the products of EAS halogenation?

The products of EAS halogenation are a halogenated benzene, hydrogen halide (HBr or HCl), and the regenerated Lewis acid catalyst. For example, in the bromination of benzene using FeBr₃ as the catalyst, the products are bromobenzene (C₆H₅Br), hydrogen bromide (HBr), and FeBr₃. The overall reaction can be summarized as:

$C_{6\mathrm{H}6}+{\mathrm{Br}}_2\to C_{6\mathrm{H}5\mathrm{Br}}+H_{\mathrm{Br}}$