Free Radical Halogenation - Online Tutor, Practice Problems & Exam Prep

Alkanes, while stable and largely unreactive, can undergo radical reactions, particularly with halogens, transforming into alkyl halides. This process involves three key steps: initiation, where radicals are generated; propagation, where radicals react with alkanes to form new radicals and alkyl halides; and termination, where radicals combine to form stable products. The radical chain reaction allows for complete conversion of alkanes, making alkyl halides valuable intermediates in organic synthesis, enabling further reactions such as substitution and elimination.

The one reaction that alkanes will actually undergo.

Back when I taught you guys about functional groups, I told you that alkanes actually don't count as a functional group. Okay? Even though they're super abundant, they're everywhere. And the reason that I said that is because it's true. A functional group implies function, that they actually do something, and alkanes really don't react with much at all. Alkanes just come from underground. You dig them up in petroleum. That's what oil is. It's alkanes, and you can't really react with it much. They're super stable. All you can do is blow them up. Okay? You can put them in your car and combust them, but you can't really react with them a whole lot. So, they seem kind of worthless at first glance, but it turns out that there is one thing that they actually can undergo, and that is that they can undergo a radical reaction. Okay? Because radicals are very high energy. So they're going to be able to react with something that's seemingly unreactive, which is alkanes. So I want to show you guys the mechanism by which they do that. So as I just said, alkanes are the backbone of organic molecules, but they're almost completely unreactive. That's why they last for millions of years underground because they don't react with anything. Okay? But there is one thing that they can do in the presence of radicals, and they can add halogens. Okay? So here I have an unreactive hydrocarbon. Like I said, that's from the dinosaurs. It didn't do anything that whole time. Now, I bring it up to the science lab, and I react it with a radical reaction and, lo and behold, I get a halogen on that alkane. Now, what's cool about that is that now I can do a bunch of other types of reactions to it. This is now called a functionalized hydrocarbon. Why? Because now I have a functional group, an alkyl halide. Once you have an alkyl halide, that's the gateway towards organic synthesis because now, guess what? I can do a bunch of stuff so that I can do substitution reactions, elimination reactions, addition reactions, all kinds of stuff because I first added that halogen. Okay? So what I'm going to show you right now is really the first step of all organic synthesis. Okay?

Radical Chain Reaction Mechanism.

So let's just go ahead and talk about it. It turns out that radicals are so high energy that once they react with something, they're going to keep trying to give away that high energy intermediate. Okay? And what winds up happening is that it's like a game of hot potato where no one wants to have the hot potato, so they keep passing it along and it forms what's called a radical chain reaction. Okay? Now the chain reaction, it actually does mean that. It means that once you start it, it actually can't end until it's fully reacted. Until you've fully reacted with all of your alkane. That is useful for us because remember, alkanes aren't that great to begin with, so if we can react them completely, that's a useful reaction as an organic chemist. So let's go ahead and see how this works. Our first step is going to be the initiation step. The initiation step is where I get that first radical because I can't play a game of hot potato without the hot potato itself, so I have to create that first radical. Now, what you notice is that this mechanism is broken down into 3 different steps. And we're actually going to need to write all 3 of these steps. In fact, it's smart that you actually write the words if you do have to draw this mechanism for a test, that you write these three words: Initiation, propagation, and termination. Okay? So let's look at the initiation step. And let's say that we're just using the easiest radical initiator, which is X₂. Okay? Let's use X₂ over heat. Now, what I taught you guys is that in the initiation step, what we're going to wind up getting is electrons from 2 electrons, 1 on each side, jumping onto each X. So what I'm going to wind up getting is X· + X·. Okay? That's the end of my initiation step. Really all I need for the bare minimum of my initiation step to work, all I need is one radical. In this case, I have 2, so I'm great. Okay? Now that I have that radical in place, that radical is free to react with other molecules. Okay? And it turns out that it reacts really well with alkanes. Now, for the sake of a really simple mechanism, let's just use the simplest alkane possible, which is methane. Methane just being a 1 carbon hydrocarbon. Okay? CH₄. So now I've got CH₄ and I'm reacting that with X·. Okay? This X· hates itself right now. It's super high energy, super unstable. It's saying, how can I get rid of this hot potato? Can I get rid of it? And then it sees all these electrons in the methane and it's thinking, maybe I can take one of the electrons from one of those carbon-hydrogen bonds. And that's exactly what it does. So it turns out that radicals are going to react with hydrogens and alkanes. And the way we draw these arrows is just so you know, radical reactions are always going to have 3 arrows. So I'm going to draw 1 fishhook into the middle of nowhere. Okay? Then I'm going to draw another fishhook from the CH bond meeting that one. Okay? What that's implying is that now there's going to be a new bond between the H and the X that's going to form from those 2 electrons. So that's looking great, but I still have one electron left over. Notice that the bond between the CH had 2 electrons. So where do you think that last electron goes? It goes onto the C. Okay? It goes onto the carbon backbone. So what that's going to do is it's going to give me a structure that now looks like this: C···HH·. Now notice that I'm drawing the geometry differently because now this would be trigonal planar, right? So you should draw it with like a triangle and that would be + HX. Making sense so far? So notice that the reason this is called a propagation step is that propagation means like I'm reproducing myself. Propagating. And notice that the radical just reproduced itself. Now, it has kind of moved through my medium and now I've got a radical on a new species. Okay? Well, it turns out that your propagation step isn't done yet because you're not done with the propagation step until you fully reproduce yourself. 100%. So what that means is not only do I need to have a radical at the end, I need to have the same exact radical that I started with. So if I start off with an X·, I need to end off with an X·. So what that means is what could I react my C· with to generate that original X· that we had at the beginning? Can you think of anything? It turns out the easiest thing to do is just to react it with another X-X, diatomic halogen. Now, you might be wondering, well, why isn't this already radicals? Because we just did that in the first step. We made it radical. Well, it turns out that not all of the diatomic halogen is going to cleave at the same time. So some of it's going to do the initiation step, but some of it isn't going to be hit by enough light or enough heat to split yet. So what that means is that the one I'm reacting with here hasn't really cleaved yet. Okay? So this is one that's just waiting around for enough energy to finally do that homolytic cleavage. But wait, before the light can even get to it, another radical just did. So instead, we're going to propagate to the X-X. And the way we draw these arrows is once again, 3 arrows. So I'm going to take the radical that always starts it. I'm going to put that one out into the middle of nowhere. Okay? Between the C and then the X. So then I'm going to take one electron from this bond and make it go there. This>

Explaining the following problem.

Now I want you guys to practice the general mechanism for radical halogenation on your own. And I want you guys to notice that this alkane that I'm reacting with has carbons of different stabilities. Okay? I want you guys just to assume that we're going to react with the most stable carbon in this case. So you're going to have to think back to what I talked about with radical stability to figure out which of those hydrogens to pull off in the radical halogenation. I think you guys can get this, though, so I'm just going to let you guys loose on your own, try to draw all 3 steps, and then I'll give you guys the answer. So go for it.

Show the entire chain reaction mechanism.

We know we need to start with the initiation step here. But even before we get to that, I want to ask you guys which hydrogen did you react the radical with, enoyalkane? Basically, there's actually only one choice that made sense, and you should have reacted it with the H right here. The reason is that this H belongs to the only tertiary carbon on this molecule. Now notice that there are no allylic sites here, so I don't worry about the resonance thing. I just worry about which hopefully this will be a learning experience for you. So now let's go ahead and draw the 3 steps. My first step is going to be initiation. I'll just draw it up here to make more room. Okay. So my initiation step is really easy. We're using XX again, so I'm just going to draw like that and like that. I'm going to do this. And what I'm going to wind up getting is 2 X radicals. Cool?

So now let's go into propagation. Now this is the part where it actually matters which hydrogen I used. And I just showed you guys why we're going to use that hydrogen there. I'm going to redraw my alkane and I'm going to draw the hydrogen sticking off this way this time just to make it easier to pull it off. I'm going to react that with x radical. And what that winds up giving me is 3 arrows. 1 here, 1 here, and 1 there. Okay? So what I wind up getting is a radical that looks like this. Okay? That radical plus HX. Okay? Now, what can that radical react with? Well, it in order to fully propagate and reproduce itself, it's going to have to react with another XX. So I'm going to do this, that, and that. And what that's going to give me is it's going to give me an alkyl halide. Notice that this is now a tertiary alkyl halide because I reacted it at a tertiary position, and I'm going to get that final radical. Cool. So now we're just going to end off with the termination step. And with the termination step, we basically had 3 different possibilities. We had X terminating with X. That would be my first product, and that would give me basically that would give me XX. Then we had another possibility, which was now my alkane radical terminating with a halogen radical. And what that would give me is another equivalent of a tertiary alkyl halide. And then, lastly, I had the third possibility, which would be I have basically 2 R groups colliding with each other. Okay? And that would give me a small amount of this kind of random-looking thing, which is going to be, that and then a single bond. And that single bond is attached to basically 2 more methyls and something like that. Okay? So anyway, I know that's ugly. Okay? But the whole point is that I can change the one. In the end of the day, I'm not going to get a whole lot of 1, I'm not going to get a whole lot of 3, but I am going to get a lot of 2. So my final product would be this guy. And then plus, I would get, obviously, a lot of HX as a byproduct. Okay? Because that's going to pretty much be forming all the time. Alright?

Here’s what students ask on this topic:

What is the mechanism of free radical halogenation?

The mechanism of free radical halogenation involves three key steps: initiation, propagation, and termination. In the initiation step, radicals are generated, typically by homolytic cleavage of a diatomic halogen (X₂) under heat or light, forming two halogen radicals. During the propagation step, these radicals react with alkanes, forming new radicals and alkyl halides. For example, a halogen radical abstracts a hydrogen atom from methane (CH₄), creating a methyl radical (CH₃•) and HX. The methyl radical then reacts with another X₂ molecule, forming CH₃X and another halogen radical. Finally, in the termination step, radicals combine to form stable products, effectively ending the chain reaction.

Why are alkanes generally unreactive, and how does free radical halogenation change that?

Alkanes are generally unreactive due to their strong C-H and C-C bonds and lack of functional groups, making them stable and resistant to many chemical reactions. However, free radical halogenation changes this by introducing high-energy radicals that can react with the alkanes. The process involves generating radicals that can abstract hydrogen atoms from alkanes, forming alkyl radicals. These alkyl radicals then react with halogen molecules to form alkyl halides, which are much more reactive and can undergo various organic reactions such as substitution and elimination.

What are the three steps in the free radical halogenation mechanism?

The three steps in the free radical halogenation mechanism are:

1. Initiation: Radicals are generated, usually by the homolytic cleavage of a diatomic halogen (X₂) under heat or light, forming two halogen radicals.

2. Propagation: The halogen radicals react with alkanes to form new radicals and alkyl halides. For example, a halogen radical abstracts a hydrogen atom from an alkane, creating an alkyl radical and HX. The alkyl radical then reacts with another X₂ molecule, forming an alkyl halide and another halogen radical.

3. Termination: Radicals combine to form stable products, effectively ending the chain reaction. This can involve the combination of two halogen radicals, an alkyl radical with a halogen radical, or two alkyl radicals.

What is the significance of alkyl halides in organic synthesis?

Alkyl halides are significant in organic synthesis because they serve as versatile intermediates for various chemical reactions. Once an alkane is converted into an alkyl halide through free radical halogenation, it can undergo numerous reactions such as nucleophilic substitution, elimination, and addition reactions. These transformations allow for the synthesis of a wide range of organic compounds, making alkyl halides essential building blocks in the field of organic chemistry.

How does the termination step in free radical halogenation work?

The termination step in free radical halogenation occurs when radicals combine to form stable products, effectively ending the chain reaction. This happens when the concentration of radicals becomes high enough that they are more likely to collide with each other than with unreacted species. Common termination reactions include the combination of two halogen radicals to form X₂, an alkyl radical with a halogen radical to form an alkyl halide, or two alkyl radicals to form a larger alkane. These reactions remove radicals from the system, preventing further propagation.