Radical Selectivity - Online Tutor, Practice Problems & Exam Prep

Radical halogenation reactions vary in selectivity based on the halogen used. Fluorination is highly exothermic and nonselective, while chlorination is less exothermic but still produces a mixture of products. Bromination is more selective, favoring the most stable radical intermediates, typically tertiary carbons. Iodination is endothermic and nonspontaneous. The Hammond postulate explains that the transition state resembles the higher energy side of the reaction, influencing selectivity. Bromination's late transition state allows for better stabilization, making it the preferred method for selective halogenation.

Radical selectivity:Alcoholics Anonymous Version

Hey everyone, so up to this point we've learned that radical halogenation prefers to occur at the most stable radical cation intermediate. In most cases, that's the tertiary position. But in this video, we're going to learn that not all halogens are created equally and, in fact, some halogens at times choose a spot that's not always the best spot. Now, in order to think of this in everyday terms, just imagine you've been invited to the greatest steakhouse in the world where it has all types of meat selections, whatever you could think of. Now, here we're going to attach this idea of our steakhouse to the different types of meat entrees you select and we're going to say that these meat entrees represent different types of halogenation reactions. And the more exothermic your halogenation reaction, then the more non-selective it is. It basically halogenates anywhere on your organic compound and therefore the more unhealthy it is. Again, playing along with this whole idea of food selection. Now, this video is just taking a look at qualitative radical selectivity and we're going to go into greater detail later on and try to connect the idea of exothermic reactions and selectivity. Now, first of all, what exactly is radical selectivity? Well, selectivity is defined as the ability for radicals to only halogenate the carbons with the most stable radical intermediates.

Again, we're going to learn that not all halogens play by the rules when it comes to this idea. Alright. So here we have 4 individuals that have been invited to this awesome steakhouse, the greatest in the world. The first one is not really picky in terms of food. They'll eat whatever. They are not selective whatsoever, and they choose a typical hot dog. Okay. We're going to say that this hot dog, which represents a highly exothermic reaction, also represents fluorination. Here, fluorination has an enthalpy value that is negative 432, so really negative ∆H. So it's very nonselective, very unhealthy because we know hot dogs are made up of who knows what tons of different parts of animals. And we're going to say here, that fluorination, because it's not very selective, because this diner doesn't really care what they're eating, it is not useful in terms of, radical halogenation. So when we say fluorination, we have F2 and we have ultraviolet light, we're going to say it's a no-go. It's too exothermic, too uncertain. We don't know what it's going to do. It's so negative, in fact, that in some cases we can deem that as being an explosive reaction. So we want to stay clear away from fluorination, and let's move on to our second person.

So here, the second person, is not willing to eat whatever. They're kind of stuck in their ways. They only want to select something that they've eaten before. So they're gonna play it safe. They're gonna choose meatloaf and other dishes that they're familiar with. If we try to connect this to chlorination, we're going to say that the overall enthalpy value of chlorination is negative 101. Now, here it's still exothermic because it's negative, but it's not as negative as fluorination. Because it's exothermic, it's still a spontaneous reaction which is good. But the issue is it's still just a little bit too negative. As a result of this, chlorination, we're gonna say the only useful radical chlorinations are reactions with a single type of hydrogen. Meaning that if I choose an organic compound that has different types of hydrogens: primary, secondary, tertiary, chlorination will give me a mixture of products. It's not very selective. It'll do a little bit of here, a little bit of there. Just like our patron at this restaurant, they're only going to stick to the foods that they know. And they've eaten quite a few things.

Now we have our 3rd person. This person really knows their different types of meat selections, and they know exactly what they want. They want the most expensive. They want the highest quality cut in terms of this. Here, they want maybe Wagyu beef or something, something they see that as being the most expensive. So this, we relate to bromination. Bromination has an enthalpy value that's negative 26. So it's still exothermic, not quite as exothermic as fluorination and chlorination, But since it's exothermic, it's still spontaneous. We're going to say here that bromination is the only useful method for selectively halogening alkanes because you can control what it'll brominate. So when it comes to bromination, if we have a mixture of tertiary, secondary, primary hydrogens, bromination always goes for the most stable radical cation intermediate that's presented. In most cases, tertiary is more stable than secondary, which is more stable than primary. Right? So we can count on bromine to brominate what we want it to do. So here we have Br2 HV, and depending on what we have it'll brominate in the place that we want.

And then finally, we have our 4th person. So we've come to the greatest steakhouse, greatest meat selection in the world, but then our 4th person says, I don't want any meat. I don't wanna get involved in any of this. So here, we're gonna relate this to iodination. Iodination has an enthalpy value that is positive 53. Because its enthalpy is positive, it's an endothermic reaction. Remember, endothermic reactions are nonspontaneous and it would require our investment of energy for it to go. So you'd have to basically, push this person to try anything. But again, they don't really want to do that. They don't want to get involved in any of this meat selection. Just like iodination doesn't want to get involved with any type of radical halogenation. So here again, it's nonspontaneous so it doesn't even want to get involved.

Now, finally we're going to say here, chiral products are always, racemized. So here we're talking about racemic mixtures. Now, what exactly does that mean? Well, if we think about it, this has to do with if our starting material belongs to a chiral center or not. So let's say that we had this organic compound and here we have a methyl group, and then here we have our hydrogen, here we have ethyl, and here we have propyl. And we did bromination. So bromination will want to replace this hydrogen here. Now that bromine can either come in from the front or the back of the molecule in order to replace that hydrogen. So because it has those two options we have 2 possible products. One where the bromine would be dashed, and one where it would be wedged. And it would just switch positions with the methyl. And you will get 50% of this one, and 50% of this one. So you create a racemic mixture. We have, 50% of both of these products. So that's what we mean by chiral products are always racemized. So again, we're looking at this qualitatively. We're trying to relate the whole type of selectivity associated with the halogen based on the preference of different patrons within the greatest steakhouse in the world. Right. So again, we're going to go into greater detail in terms of this from a qualitative to quantitative analysis later on. For now, this is just helping us to relate how the different halogens react when it comes to radical halogenation.

Predict the Product.

So was number 2 synthetically useful? Yes, it was. Okay. The reason this was synthetically useful is because I've got 2 different types of carbons. I've got the carbons in the middle and the carbons on the outside. Okay. Now, even though I have 2 different types of carbons, the middle ones are like quaternary. The blue ones are primary because they're all attached to just one carbon. Okay? But my question is how many types of hydrogens do we have? Because it's not about carbons, it's about hydrogens. We only have one type because the red carbons, even though they're different, they don't have any hydrogens on them, so they don't really count. So what that means is that my product is simply going to be taking any of these methyl groups, you pick whichever one you want, and just stick a chlorine on it. Okay? And that would be my product. It doesn't matter which methyl group you pick because it's all the same. You just have to rotate it to look like the same molecule. Okay? So what that means is that this is a synthetically useful reaction. I could use bromine if I wanted to. Bromine would give me the same thing. I would just have bromine instead of chlorine. But chlorine works as well. So it turns out that this is one of those molecules that actually works well with chlorination if I want to use it. Okay? That doesn't mean you have to use chlorination. You could also use bromination. You get the same thing. But I'm just saying it's an option now. Alright? Whereas for the first one, that wasn't really a good option at all. All right? Cool. So let me know if that makes sense. Let's move on.

Predict the Product.

How many products would we yield with this 1? With b? And the answer was 2. Okay? The reason it's 2 is because in this case I have 2 tertiary positions. Okay? So this is an example where maybe this isn't such a great reaction. Okay? Because now I'm going to have to wind up getting molecules that are very structurally similar. I'm going to have to separate them. It's going to suck. But there's no better option. Okay? Chlorine would have been way worse. So this is just the best that we can do. Could I use iodine? No. Because that wouldn't even be exothermic. It wouldn't give you a chain reaction. Okay? Cool. So I hope that made sense, guys. Notice that these problems come a lot easier once you're not drawing the full mechanism, which you don't have to for every question as long as you know how to do it in general. Okay? Cool. So let's go ahead and move on.

Using the Hammond Postulate to describe radical chlorination.

Now that you guys know the funny way to memorize which halogens are going to be more selective and which ones are less selective, I want to teach you quickly the rigorous actual reason why bromine is so much more selective than chlorination. And it doesn't really technically have to do with how exothermic it is. In fact, it has to do with the Hammond postulate. Okay? So, remember what we learned about the Hammond postulate? It is a way that describes what transition states look like and the Hammond postulate is really the reason why bromination is so much more selective than chlorination. Let me show you guys that now. Once again, my definition of selectivity hasn't changed, but it turns out that I'm going to use the Hammond postulate to explain why bromination is so much more selective. So first of all, I just want to show you the reaction diagram or the energy diagram for radical chlorination. Let's go ahead and look at the x-axis first and just look at the coordinates. Notice that all of these things should look familiar. Basically, this is the first step of my propagation phase. Okay, this is all propagation here. I'm just going to put "prop." Okay. So, I've got my propagation phase. We would expect that that takes energy because you're breaking a bond. Right? So that's going to be this—basically this activation energy here. Okay? That's going to be the activation energy required to do that first step. Okay? Then what that's going to make is an intermediate that looks like that. Okay? That's a radical. That's my intermediate. And notice that here I put "rate determining step." Remember that your slow step, the one that makes the intermediate, is always going to be your rate-determining step. Does that make sense so far? Cool. So then what happens is that in the next part of my reaction diagram, there's like a story. Then what happens? Well, remember my propagation phase isn't done yet. I need to react my... well, in this case, this would be a termination step here. Let me just say here this will be termination. In my termination step, what I get is now two radicals colliding to make a new single bond. Now, just so you guys know, technically, this might come up conceptually; the activation energy for two radicals to collide and make a new bond is always going to be exactly zero. This is just a definition that you guys should know; the activation energy of two radicals colliding is always zero. It doesn't take energy for that to happen. So that's why there is no hump here. Okay? Usually, we would expect to see two humps, but there's no hump there because it doesn't take any energy for that to happen. Okay? Now notice here that my overall enthalpy at the end is negative -101 like I told you guys. But what's really important that I want you guys to know is what the transition state looks like. Okay? Now, I'm not going to go through this just yet. I'm just going to let you guys know. Pay attention to what that transition state looks like. Hammond's postulate. And now we're going to look at bromination. See how bromination is different.

Using the Hammond Postulate to describe radical bromination.

So for bromination, notice that my energy diagram is a little different. I get the same first step. Same first step. And what I notice is that I get my same intermediate. Now, there's just one major difference between my intermediate here and my intermediate up there. Notice that the energy level for my rate-determining step here is actually going to be endothermic. What that means is that I'm actually having to put in energy in order to become this radical. Now, look at the energy difference for my chlorination. For chlorination, it was exothermic. It actually released energy to make that radical. Isn't that crazy? So for chlorination, it was exothermic to make that radical. For bromination, it was endothermic.

What's the significance here? Remember, what did Hammond's postulate say? It said that basically Hammond's postulate said that your intermediate is going to look most like the side of your reaction or the side of your reaction that has the highest energy level. So, in this case, what that means is that the transition state for chlorination and bromination is going to look radically different. For chlorination, the thing with the highest energy, I'm just going to use green as the highest energy, is going to be the starting molecule. So what that means is that my transition state looks a lot like my starting molecule. Notice that my bond between the H and the stage 2 is really short. That means that the intermediate looks a lot like the original alkane. Does that make sense so far? So my intermediate really looks like the original.

Now what we notice is that for radical bromination, my highest energy state is actually the radical. So what that means is that my intermediate looks a lot more like the radical than the alkane. Notice that in this case, my H is really, really far away, meaning that this CH2 almost already has a radical on it. Is that making sense so far? So, really, the biggest difference is the way that the intermediate looks.

So why is that important? Well, because if you think about it, if the alkane if the transition state looks a lot like an alkane, then the energy savings of making a more stable intermediate isn't going to matter that much. Notice that I have here that this is the primary. This is this energy savings that I get as a primary radical. But if you make a secondary radical or a tertiary radical, you save even more energy. Right? So you would think, okay, it really wants to be tertiary so that it can make it dip all the way down here and be more stable. But no, chlorine doesn't care. Because chlorine doesn't really look like the radical once it's in its transition state. Really, it looks like it's an alkane, and the alkane was always stable to begin with. So it doesn't really matter whether it's primary, secondary, or tertiary because it looks like the alkane anyway. So it doesn't really need to have a stable intermediate.

Now let's look at bromination. For bromination, what we notice is that it actually looks a lot like the radical already in the transition state. So, if I can make it secondary or if I can make it tertiary, even better, that's going to make my transition state a lot easier to form. It's going to make my transition state a lot more stable, so the rate-determining step can happen faster. So basically, that's the whole point. Bromination cares a whole lot more about making the graph look like this. Because the intermediate according to Hammond's postulate looks a lot like the radical.

So what's this difference? Well, the difference is that remember that if you had a transition state that looked like the original starting product, you called that an early transition state. And notice that over here, what we wind up getting is a late transition state because late means that it looks more like the second step. Once you have a late transition state, you're going to care a lot more about how stable that transition state is or how stable that intermediate is because it's going to stabilize your transition state as well.

So what I'm trying to say is that my simplification of the exothermic rule works. But technically, the way that we explain selectivity is through Hammond's postulate. Now, for some of you guys, this is just going to be theoretically just helps you know more about this class. Some professors are going to be really picky about it and actually want you to be able to describe this. So it's really up to you guys. If you didn't totally get this, it's not the end of the world. It's probably not going to be on your test. But I just wanted you guys to know it because I like to be thorough and I like you guys to know everything. So let me know if you have questions. But if not, let's go ahead and move on.

Here’s what students ask on this topic:

What is radical selectivity in halogenation reactions?

Radical selectivity in halogenation reactions refers to the preference of radicals to halogenate the carbon atoms that form the most stable radical intermediates. This stability is typically highest at tertiary carbons, followed by secondary and primary carbons. Different halogens exhibit varying degrees of selectivity. For example, bromination is highly selective and prefers tertiary carbons, while chlorination is less selective and can produce a mixture of products. Fluorination is highly nonselective and often too exothermic to be useful, whereas iodination is endothermic and nonspontaneous, making it generally ineffective for radical halogenation.

Why is bromination more selective than chlorination?

Bromination is more selective than chlorination due to the Hammond postulate, which states that the transition state of a reaction resembles the side with the higher energy. In bromination, the transition state is closer in energy to the radical intermediate, making it a late transition state. This allows the reaction to favor the formation of the most stable radical, typically a tertiary carbon. In contrast, chlorination has an early transition state that resembles the starting material more closely, making the stability of the radical intermediate less significant. This results in less selectivity for chlorination compared to bromination.

How does the Hammond postulate explain the selectivity of halogenation reactions?

The Hammond postulate explains the selectivity of halogenation reactions by describing the nature of the transition state. According to the postulate, the transition state resembles the side of the reaction with the higher energy. In bromination, the transition state is closer to the radical intermediate, making it a late transition state. This allows the reaction to favor the formation of the most stable radical, typically a tertiary carbon. In chlorination, the transition state resembles the starting material more closely, making the stability of the radical intermediate less significant. This results in less selectivity for chlorination compared to bromination.

What are the enthalpy values for fluorination, chlorination, bromination, and iodination?

The enthalpy values for halogenation reactions vary significantly among different halogens. Fluorination has a highly exothermic enthalpy value of approximately -432 kJ/mol, making it very nonselective and often explosive. Chlorination has an enthalpy value of around -101 kJ/mol, making it exothermic but less selective. Bromination has a less exothermic enthalpy value of about -26 kJ/mol, allowing for high selectivity. Iodination, on the other hand, is endothermic with an enthalpy value of +53 kJ/mol, making it nonspontaneous and generally ineffective for radical halogenation.

Why is fluorination considered too exothermic for practical use in radical halogenation?

Fluorination is considered too exothermic for practical use in radical halogenation because its enthalpy value is approximately -432 kJ/mol. This high exothermicity makes the reaction highly nonselective and difficult to control, often leading to explosive outcomes. The extreme energy release can cause the reaction to proceed uncontrollably, halogenating multiple positions on the organic compound and producing a mixture of products. Due to these risks and the lack of selectivity, fluorination is generally avoided in radical halogenation processes.