The presence of radicals in some familiar looking addition reactions can completely change the product.
1
concept
The general mechanism of Allylic Halogenation.
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So far, we've been learning about radical college nation mechanisms and how the radicals always gonna be attracted to the spot that has the most are groups. Okay, so the tertiary is gonna be more stable than primary. But what we also know is that according to the radical stability trend ah, Lilic radicals are the most stable. Okay, so now I want to throw that extra complexity into the mix. What happens if there are a Lilic positions available to be how originated? And that brings us to a little ecology nation. So once again, I just want to show you guys, just in case you forgot that the A Lilic and Bends Ilic 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 three. Okay, now, what's interesting about these is that that these radicals radicals can resonate. OK, so what that means is that resonance is gonna have to play a central role in this mechanism. We can't just ignore residents are actually gonna 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 diatonic halogen. Let's say this is BR two. Okay, so it's selective, right? And we've got a double bond. Okay, Well, usually we would just say secondary of primary. We're gonna go for the secondary spot, but in this case, I actually having a little position that a little position is right here. Okay, so that's the one next to a double bond. So that's actually gonna be the position that's the most stable. And that's gonna be the one that I want to react with. Now, let's go ahead and draw the whole mechanism out. I'm gonna show you guys how things are a little bit different now. Okay, so first of all, my initiation step is really easy. They're just gonna use br to. And where does those first arrows go? You just make radicals. They're in there. So now what I'm gonna get is B r radical plus b r article. Alright, so that's it for our initiation. We have our target radical. Now we have is the propagation step in the propagation step. I'm gonna wanna 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 three arrows. I would do one into the middle of nowhere, one 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 HPR, 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 br two. 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 residents structure actually looks like this, Okay? And what that means is that now might be our that's gonna continue in the propagation stage, can attach both there, and they're so it's gonna be able to attack both of those carbons because the radical is hybridized between those two. Remember that resonant structures? It's not an equilibrium. It's a mathematical lake. 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. I can draw 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 gonna wind up getting a br right here, plus b r radical. Okay, but I'm also gonna get something else. I'm gonna 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 a little ecology nation. If you double bond isn't perfectly symmetrical, you're gonna 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 gonna get both of them. Okay? So it kind of sucks. All right. So then finally what would determination step look like? Well, there were a lot of terminations, but we could get a B r and B R So you see? Yes. Okay, we could get just any of these radicals. I mean, any of these that plus yard. Okay, sir, I'm just gonna draw this in this, okay? And that would give us our product that looked like this. And we would also get the other one, the other radical that would look like this. The radical here plus PR. And they would give us the other looking product. Okay, Now, I know that there was another one. That was the two are groups that is almost never seen in here. But if you wanted to include it, you could just add, like, our radical are okay. And just say that that's going to give you are are okay. But honestly, like for these reactions, usually professors just neglect this because it's assumed that you're not really gonna get much of this at all. Okay, What is important is that now you can detect when you're gonna get just one product or when you're going to get a mixture on. When you're gonna get a mixture is when you're doing in a Lilic halogen nation. If it was just tertiary, your secondary. You would only get one. But since it's a little like that means it can resonate. OK, And in this case, you would get two different structures. The way to know how many structures you're going to get is just too. Draw the resident structures yourself. Okay, so now what I want to do is go more into depth on a little ecology nation with the specific regions that we're gonna use.
Remember our friendly addition reaction halogenation? Notice that you achieve a vicinal dihalide in this reaction.

Now we see this reaction. Note that the only difference is the presence of a radical initiator.

This one added condition will lead to the formation of a mixture of allylic alkyl halides. Here’s the full mechanism:

2
concept
The products of Allylic Chlorination.
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So now that we know the mechanism, predicting these products is gonna be easy. The number one thing I'm looking for here that I want you guys to know is thes re agents. So it happens to be that when you're doing in a little chlorination, 400 degrees Celsius is just the most common heat that's use. You should recognize that when you see seal to 400 degrees and a double bond and you have in a little position available, we're gonna wind up getting our you know, we're gonna getting in a little college nation with multiple products because I could get the radical here. Or I could also resonate the radical Thio here. Okay, so I'm gonna wind up getting both products. One looks like this, and one looks like this. Okay, easy stuff. You guys should be aware of that


3
concept
Mechanism of Allylic Bromination.
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for an Olympic Romine ation. Usually you want low trace amounts of bro. Mean we don't want too much, bro mean or we might wind up getting like an actual holiday nation on the double bond, which would be bad. So we use NBS and we usually use NBS of the combination of light or heat. Doesn't matter. But in this case would be the same exact thing. I would wind up getting a bro me in here, and I would get soups. How much training? Jealous. There we go. And I would get a bro me right there. Okay. So really, these are the same exact reaction. Just the re agents 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 we'll solve it.
Predict the product(s) of the following reaction.

4
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Predict the product(s) of the following reaction.
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So the first thing you need to do here is we need to figure out how maney a Lilic positions have hydrogen, and in this case, we actually have to. Okay, this position is a lulic, and it has a least one hydrogen so it could be pulled off. Okay, this position is a Lilic, and it has one hydrogen. I actually have a third. A little position. This position is also a Lilic because it's next to a double bond, but notice that it doesn't have any Hydrogen is on it, so this one isn't going to count. Okay, So what that means is that I have two positions that would make really, really good radicals. Okay. And let me just erase the ages really quick. What that means is that I would have a red position here that make a really good radical and have a blue position here that would also make a really good radical. So to figure out what my products are gonna look like, I actually need to resonate both of these radicals and see what they look like afterwards to see how many products are gonna get. So let's go ahead and draw them. We know that for sure. At least we're gonna have to products right now. But maybe we'll have more. Let's go ahead and draw all of these. So I'm gonna 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. One, 23 And that would give me a resident 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 could also get a ah 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 the bottom. Okay, so it's it's own position. And if I use my three arrows again, I'll do one, 23 and what I would wind up getting is a new radical. Actually, that's gonna look like this. Where now my Dole 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 four different places. I have radicals here, here, here. And they're all those four locations at some point, Have a radical could is going to resonate there. Since these positions aren't symmetrical, That means I'm gonna get bro means at all these positions. So it's gonna be a really bad mixture of products. So what I would expect to get is I would expect to get basically a Roman here, Okay? With my double bond there, that's one of them. Let's draw the other residents structure. So it also expect to get a bro mean here with a dull bond there. That's another product. What? I would also expect to get its a lot. Okay, this is like the worst case scenario where you have, like, a completely asymmetrical double bond. I would also expect to get April me in there with a double bond there and obviously the methyl group. And then finally I would also expect to get a bro mean. Oops, that's right. Where in my face is let me take myself out of camera really quick. I would also expect to get a Brahmin there with a double bond right here. Okay, notice that none of these products look the same. So I'm gonna just put pluses with all of these thes air, all products, and we don't have a way of telling which one's major. Which ones? Minor. We just know it's gonna be a mixture. Okay, so I'm just gonna go ahead and add myself again. This is one of the complications with a Lilic halogen nation that you get mixtures of combinations of products often. So you need to make sure if you're gonna do this to make sure you use a really symmetrical double bond or really small molecule because something like this that is just messed up. You don't want to get four different products. This is like a worst case scenario, so you can use this reaction, but it's advised to use it on Lee on small, symmetrical molecules. All right, so hopefully that makes sense. Let's go ahead and move on.
Additional resources for Allylic Bromination
PRACTICE PROBLEMS AND ACTIVITIES (5)
- The light-initiated reaction of 2,3-dimethylbut-2-ene with N-bromosuccinimide (NBS) gives two products: 2,3-d...
- The light-initiated reaction of 2,3-dimethylbut-2-ene with N-bromosuccinimide (NBS) gives two products: b. Th...
- Using 1,2-dimethylcyclohexene as your starting material, show how you would synthesize the following compounds...
- When N-bromosuccinimide is added to hex-1-ene in CCl4 and a sunlamp is shone on the mixture, three products re...
- Predict the products of the following reactions. (c) 2-methylpropene + NBS, light