Allylic Bromination - Online Tutor, Practice Problems & Exam Prep

Allylic halogenation involves the halogenation of compounds with allylic positions, where resonance stabilizes the resulting radicals. The mechanism includes initiation with diatomic halogens, followed by propagation that generates multiple products due to resonance. For allylic chlorination, heat (400°C) is commonly used, while N-bromosuccinimide (NBS) is preferred for allylic bromination to minimize unwanted reactions. Understanding the regioselectivity and potential for multiple products is crucial in predicting outcomes in these reactions.

The general mechanism of Allylic Halogenation.

So far we've been learning about radical halogenation mechanisms and how the radical is always going to be attracted to the spot that has the most r groups. Okay? So the tertiary is going to be more stable than primary. But what we also know is that according to the radical stability trend, allylic radicals are the most stable. Okay? So now I want to throw that extra complexity into the mix. What happens if there are allylic positions available to be halogenated? And that brings us to allylic halogenation. So once again, I just want to show you guys, just in case you forgot, that the allylic and benzylic radicals are more stable than primary, secondary, or tertiary. So what that means is if I have one of these two possibilities, this is going to be my preferred spot, not those 3. Okay? Now, what's interesting about these is that these radicals can resonate. Okay? So what that means is that resonance is going to have to play a central role in this mechanism. We can't just ignore resonance. We're actually going to have to embrace it for this mechanism. So I just want to show you guys what this could look like.

So basically, I've got a diatomic halogen. Let's say this is Br₂. Okay? So it's selective. Right? And we've got a double bond. Okay? Well, usually we would just say secondary or primary. We're going to go for the secondary spot. But in this case, I actually have an allylic position. That allylic position is right here. Okay? Because that's the one next to a double bond. So that's actually going to be the position that's the most stable and that's going to be the one that I want to react with. Now let's go ahead and draw the whole mechanism out and I'm going to show you guys how things are a little bit different now.

So first of all, my initiation step is really easy. We're just going to use Br₂. And where do those first arrows go? You would just make radicals. There and there. So now what I'm going to get is Br radical plus Br radical. Alright. So that's it for our initiation. We have our target radical. Now what we have is the propagation step. In the propagation step, I'm going to want to pull off the hydrogen that's going to give me the most stable radical, so obviously it should be this one. Right? So let's go ahead and react that. I would use 3 arrows. I would do 1 into the middle of nowhere, 1 into the middle of nowhere, and then one right there. Okay? So what I wind up getting is I wind up getting a radical that looks like this. Okay? Now usually, and also we would get HBr, right? Okay. Now usually what would happen is that my propagation phase would continue and I would have to regenerate the Br radical by reacting with Br₂. Okay, so that's normal. That's what we're expecting. But wait, there's a little difference. This radical can now resonate. So what that means is that I can't actually just continue from here. I need to draw the resonance structure of this radical before it can continue. So what that means is that the resonance structure actually looks like this. Okay? And what that means is that now my Br that's going to continue in the propagation stage can attach both there and there. So it's going to be able to attack both of those carbons because the radical is hybridized between those 2. Remember that resonance structures, it's not an equilibrium. It's a mathematical like statistical average of where that radical is. So what that means is that I can go ahead and draw my mechanism with the Br here and I can draw it from the first one. Just to keep things easy, I'll just do this, there, there, and there. Okay?

And what I wind up getting is I'm going to wind up getting a Br here plus Br radical. Okay? But I'm also going to get something else. I'm going to get a mixture of products because the radical on the end could have also attacked. So what that means is that I'm also going to get a product that looks like this. Okay? And this is one of the complications of allylic halogenation. If your double bond isn't perfectly symmetrical, you're going to wind up getting a combination of products. And really, we don't deal with major and minor in this section. We're just going to say, hey, we're going to get both of them. Okay? So it kinda sucks. Alright?

So then finally, what would the termination step look like? Well, there are a lot of terminations. We could get a Br and a Br. Okay? We could get just any of these radicals. I mean, any of these that plus might be r. Okay. Sorry. I'm just going to draw this and this. Okay. And that would give us our product that looked like this. This. And we would also get the other one, the other radical that would look like this. The radical here plus Br. And that would give us the other looking product. Okay? Now, I know that there was another one that was the 2 r groups. That is almost never seen in here. But if you wanted to include it, you could just add like R radical R. Okay? And just say that that's going to give you RR. Okay? But honestly, like for these reactions, usually professors just neglect this because it's assumed that you're not really going to get much of this at all. Okay? What is important is that now you can detect when you're going to get just one product or when you're going to get a mixture. And when you're going to get a mixture is when you're doing an allylic halogenation. If it was just tertiary or secondary, you would only get 1. But since it's allylic, that means it can resonate. Okay? In this case, you would get 2 different structures. The way to know how many structures you're going to get is just to draw the resonant structures yourself. Okay? So now what I want to do is go more into depth on allylic halogenation with the specific reagents that we're going to use.

Mechanism of Allylic Bromination.

For an allylic bromination, usually, we want low trace amounts of bromine. We don't want too much bromine or we might wind up getting an actual halogenation on the double bond, which would be bad. So we use NBS. We usually use NBS with a combination of light or heat. It doesn't matter. But in this case, it would be the same exact thing. I would wind up getting a bromine here and I would get oops. How am I trying to draw this? There we go. And I would get a bromine right there. Okay? So really these are the same exact reaction, just the reagents are a little tiny bit different. That's it. Cool. So now what I want you guys to do is take this problem, try to solve it yourself. Be sure to think about the whole mechanism before you predict your products. I don't need the full mechanism as long as you can give me all the products. That's what I'm looking for. So go for it and then I'll solve it.

Predict the product(s) of the following reaction.

So the first thing we need to do here is we need to figure out how many allylic positions have hydrogens. And in this case, we actually have 2. Okay? This position is allylic and it has at least 1 hydrogen, so it can be pulled off. Okay? This position is allylic and it has 1 hydrogen. I actually have a 3rd allylic position. This position is also allylic because it's next to a double bond. But notice that it doesn't have any hydrogens on it, so this one isn't going to count. Okay? So what that means is that I have 2 positions that would make really, really good radicals. Okay? And let me just erase the hydrogens really quick. What that means is that I would have a red position here that would make a really good radical and I have a blue position here that would also make a really good radical. So to figure out what my products are going to look like, I actually need to resonate both of these radicals and see what they look like afterwards to see how many products I'm going to get. So let's go ahead and draw them. We know that for sure at least we're going to have 2 products right now, but maybe we'll have more. Let's go ahead and draw all of these. So I'm going to draw the red one first. The red one would look like this. So I have a double bond here and my radical there. How could that resonate? Well, I could do the whole three arrow thing. 1, 2, 3. And that would give me a resonance structure that looks like this. So now what I would have is my double bond is here and my radical is there. Is that a new radical? Is that a new position that didn't have a radical on it before? Yes. So what that means is that I can also get a halogen there. Okay? Now, let's look at the blue one. The blue one, let's draw it out. So that was the red. That's the end of the red. Let's look at the blue one. So for the blue one, notice that I have a radical at the bottom. So it's its own position. And if I use my 3 arrows again, I would do 1, 2, 3. And what I would wind up getting is a new radical actually that's going to look like this where now my double bond is here and my radical is there. Okay? So how many different places do I have radicals in? Well, it turns out that I have radicals in 4 different places. I have radicals here, here, here, and there. All those four locations at some point have a radical that's going to resonate there. Since these positions aren't symmetrical, that means I'm going to get bromines at all these positions. So this is going to be a really bad mixture of products. So what I would expect to get is I would expect to get basically a bromine here. Okay? With my double bond there. So that's one of them. Let's draw the other resonance structure. So I'd also expect to get a bromine here with a double bond there. That's another product. What I would also expect to get is a lot. Okay? This is like the worst case scenario where you have like a comp

Here’s what students ask on this topic:

What is allylic bromination and how does it differ from other types of halogenation?

Allylic bromination is a specific type of halogenation where a bromine atom is introduced at an allylic position, which is the carbon atom adjacent to a double bond. This process is unique because the resulting allylic radical is stabilized by resonance, making it more stable than primary, secondary, or tertiary radicals. Unlike other halogenations, allylic bromination often uses N-bromosuccinimide (NBS) in the presence of light or heat to avoid unwanted reactions on the double bond. The mechanism involves initiation, propagation, and termination steps, with resonance playing a crucial role in determining the final products.

Why is N-bromosuccinimide (NBS) used in allylic bromination?

N-bromosuccinimide (NBS) is used in allylic bromination because it provides a controlled source of bromine radicals, minimizing the risk of bromination occurring directly on the double bond. NBS, in the presence of light or heat, generates bromine radicals that selectively react with the allylic hydrogen atoms, forming stable allylic radicals. This selectivity helps in achieving the desired allylic bromination without over-brominating the molecule, which can lead to unwanted side reactions and products.

What are the common conditions for allylic chlorination?

Allylic chlorination typically requires the use of chlorine gas (Cl₂) and high temperatures, around 400°C. The high temperature helps in generating chlorine radicals, which then abstract an allylic hydrogen atom to form a stable allylic radical. This radical can resonate, leading to multiple possible products. Recognizing these conditions is crucial for predicting the outcomes of allylic chlorination reactions.

How does resonance affect the products of allylic halogenation?

Resonance significantly affects the products of allylic halogenation by stabilizing the allylic radical intermediate. The radical can delocalize over multiple positions, leading to the formation of different resonance structures. As a result, the halogen can attach to multiple positions, producing a mixture of products. This is particularly important in unsymmetrical molecules, where the radical can resonate to different extents, leading to a variety of halogenated products.

What is the mechanism of allylic bromination?

The mechanism of allylic bromination involves three main steps: initiation, propagation, and termination. In the initiation step, bromine (Br₂) is split into two bromine radicals. During propagation, a bromine radical abstracts an allylic hydrogen, forming an allylic radical and HBr. This allylic radical can resonate, creating multiple resonance structures. The allylic radical then reacts with another Br₂ molecule, forming the final brominated product and regenerating the bromine radical. Termination involves the combination of radicals to form stable molecules, ending the reaction.