Lecture 014: Free Radical Reaction Part 2

Lecture 014: Free Radical Reaction Part 2

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Jean-Claude Bradley: Okay, let's get started. So we're going to continue on Chapter 4.


[student chatter, settling in]

Jean-Claude Bradley: All right, so this is the reaction we're looking at, free radical halogenation. What we did in the last class was basically look at the steps that were involved in this reaction so we know now that this does not happen in one step, it happens in multiple steps. Let me just reiterate those because we're going to need them to redraw the energy diagram.


So the first step is that we're going to take bromine and then in the presence of light we're going to generate two bromine radicals. In the second step, the bromine radical will react with the CH4. Then in the third step, the methyl radical reacts with the bromine. Then finally, we have three possible termination steps. I'm not going to list all of them here, just one of them. Termination being the combination of any two radicals.


Ok, now the key thing with this is if we want to understand the energy diagram, is we have to look at how the bulk of the products are formed. That means that the initiation step is a very very small part of this whole process, it just happens for one out of every 10, 000 or 100, 000 molecules. The termination step is also an extremely minor part of any products being formed. So the only two reactions that we are going to look at in the energy diagram will be propagation steps, steps two and three.


In the energy diagram we will ignore one and four. Does anyone have a question with that or is that clear?


Okay, so lets start off then, we're going to have two steps. The first step will be the first propagation step. It will be the bromine radical plus methane. So we're going to start off here, and I'm doing step two only here: bromine radical plus CH4. And the end result of step two will be the formation of HBr and CH3 radical. Okay, so it's going to look a little strange, the way it is here, because I'm going to have products then I'm going to start at the same place then I'm going to do the second propagation step. So in the second propagation step what do we have? We have methane radical plus Br2. And that is going to go to CH3Br and a bromine radical.


So what I've done basically here is I've filled in a little bit of what was missing in the middle. What we had done initially was we determined from the left hand side to the right hand side was an 8kcal/mol drop, so that's still true but now we see that in between there we have another step which is actually a higher energy than where we're ending up. Let me draw this energy curve. It will look something like that. This energy curve is two steps. The first part is propagation step two the second part is propagation step three. If you remember a little bit about how these energy diagrams work, we have multiple components here.


The first thing that I'm interested in is this distance here, what's that called? Activation energy. That is the energy that's required to go over the top of the energy hill to make it to the products. The higher the energy of activation, the slower the reaction is going to be.


So the first arrow is energy of activation for step two. Now the enery of activation for step three is going to be much smaller, it will be on the other side of this.


Knowing where one of the energy of activation is much much higher than the other that tells us that the rate of the entire reaction is going to be dependent not on both of these steps but is only going to be dependent on the slow step. The step that has the higher energy of activation is going to be the slow one, which is also called the rate limiting step.


What we can tell from this diagram is that step two is the rate limiting step. That's going to be really important when we look at some additional reactions that have multiple steps because if you know what the rate limiting step is, then you can control the rate of the whole reaction. Because when you do this reaction, you don't actually usually measure the intermediates. You don't measure the methyl radical, you don't measure what's happening in the first point. What you are measuring is basically the rate of the production of your two products.


Greater production of CH3Br, HBr, and if you know the mechanism though, then you can start to control that reaction. You can change conditions, you can speed it up or slow it down. But the key concept here is that no matter what you do to the second step in terms of making it faster or even a little bit slower will have no effect whatsoever on the rate of the entire reaction. Okay, so that's the concept of rate limiting step. Any questions?


Okay, there's one more part of this diagram that we need to point out. And that's at the very top of these energy hills we have a structure that's called what? You remember what this is called? Transition state. Which is not a molecule, it's basically a configuration of energy and geometry that is the highest possible energy, the most unstable part of this entire curve. So this is called a transition state. We have one for step two and we have another one for step three. The other thing you have to learn how to do is to estimate what is the geometry of the transition state going to be. That's one little trick you're going to have to use in this reaction and in some additional ones we will do later.


In order to predict transition state, what you need to understand is that atoms are going to move from left-hand side to right-hand side and they don't move by jumping around. They move in a continuous way. You can guess what that transition state is going to look like by just using common sense and thinking what's in between what's completely on the right and what's completely on the left.


Let me use number two as an example. We start with Br radical and CH4. We start off like this and start off on the left with that. And on the right what I'm going to do is keep the position of all the atoms the same that don't move. So if you look carefully at what the products are like the only thing that changes here is I move one of the hydrogens over from the right to the left. In the end I move that hydrogen over and on the CH3 group I now have a radical. If we want to figure out what the transition state looks like in between those two, you basically take an intermediate view.


A transition state will be drawn with square brackets around it. The bromine on the left is uncharged, the bromine on the right is uncharged, so the bromine in the transition state is likely not going to be charged, that should be your first guess. So I would put a bromine. And on the left there are zero bonds between the bromine and the hydrogen and on the right there is one bond between the bromine and the hydrogen. So we can guess that in between those two there's going to be a partial bond between the bromine and the hydrogen.


So I'm going to put the hydrogen between those two but there's something else I'm going to have to worry about before I put the hydrogen. There's a full bond between the carbon and hydrogen on the left and there's zero bond between the carbon and hydrogen on the right so I will again have a partial bond between the carbon and the hydrogen. So when I'm done here I will put that hydrogen. The other Carbon hydrogen bonds are unchanged.


There is no change in charge there's no change in bonding between the left and the right so I know I can just leave those things intact like this. So the only thing I haven't put here is the hydrogen and partial bonds. If we don't know the energies of the reaction, we could take a guess and just put the hydrogen in between the carbon and the bromine. That would be a good guess. We don't know exactly where it's going to be so if we don't know the energies, then we'll just put it randomly in the middle. However in this case we do know the energies. Let me go back to this diagram.


We know that there is a huge energy of activation, goes to a top, then it has just a little bit more to go before it goes to the products. So based on that, would you put the hydrogen closer to the bromine or closer to the carbon?

Student: Carbon.

Jean-Claude Bradley: You think closer to the carbon? Does the transition state look more like the products or more like the reactants? Let's go back. Just on a picture level, the height of this peak, is it closer to the reactants or the products? Definitely the products. So an easy way to think about this is the transition state is going to look almost like the products. And what do the products have? They have a full bond between the bromine and the hydrogen and have no bond between the hydrogen and the carbon.


So the hydrogen has migrated from the right to the left. In the transition state it's almost completely connected to the bromine. So when we put the hydrogen, now that we know the energies, we'll put it a lot closer to the bromine than the carbon. My little dash here is a partial bond. The hydrogen is closer to the bromine because the transition state is more product-like.


The transition state is more product-like just because it's closer in energy. This is called Hammond's postulate. Let me write that out. Hammond's postulate basically says that the transition state will look like the species that it is closest to in energy. Hammond's postulate is one of those things that if you try to learn it by looking at the definition you're just going to confuse yourself. It's much easier to understand by looking at the picture first.


So just to make sure everybody's on the same page here, any reaction you can do this with, but you have to know the energies to be able to apply Hammond's postulate. If you don't know the energies, you just can't apply it and that's why you would just guess and put everything in the middle cause you don't know. When I talk about energetically closer, I'm just simply looking at is it closer to go down the right-hand side of this hill or the left-hand side.

Student: [xx]

Jean-Claude Bradley: Well, I'm telling you that. I have to tell you what the energies are for you to be able to do this, but yeah, if I didn't tell you what the energies were you really couldn't do the problem. Anything else on that? Any questions?


So I will let you do step three as an exercise. See if you can draw the transition state for step three.


[phone rings, silence while students work on problem]

Jean-Claude Bradley: We looked at how to brominate methane. All we've really looked at is that you know this reaction is kind of general, because we called it the halogenation of alkenes, we didn't call it the bromination of alkenes so you know that you can probably do it with other halogens and you know that you can probably do it with a lot of other alkenes. But the one thing you haven't learned yet is how you can predict the products you're going to get. So let's take a look at a simple example, which is propane.


So if you treat propane with bromine in the presence of light, and if you're not used to this way of writing, if I put a reagent on the arrow, it's the equivalent of saying propane plus Br2; it's just more convenient sometimes to do it that way. So if I react it with bromine in the presence of light what we know is that we're going to swap a hydrogen for a bromine and generate HBr. So, the key question now is how many different products can we make from propane. Because there's more than one place that I can put the bromine on here.


Let me see what the answer is. Three? Let's start with the first one. Where's the first place I could put the bromine? At the ends? ok so let's start with that. Where would be another place? On the middle carbon. Anyplace else? The reason we spent so much time going through sigma bonds, and all these compounds are the same; this is the reason, because now we have to apply this stuff. So I gave you a couple of the tricks to find out if compounds are the same or not.One thing you can do is just name it. One is 1-bromo propane and the other is 2-bromo propane. If you draw anything else, you'll come up with either 1-bromo propane or 2-bromo propane; however you draw it.


Okay, so, that's the first thing you can do here is determine how many products you'll have; if you attempt the quiz here you'll see that that's a common question. Now, we also have HBr that is resulting from this. So HBr will be kind of a side product here. When I ask you how many products you get I'm only interested in the organic products. Be very clear about that. Propane will give you two products only. Two possible products.


Okay, so if we try to predict the ratios of these two, there's a couple of approaches that we could take. What would be an argument? Could you argue that you'd get more of one than other? Is there any argument that you can make?

Student: It depends on the stability of the radical.

Jean-Claude Bradley: We didn't learn that yet; so, you're right, it depends on the stability of the radical. Let's say you didn't know that. Just based on the information that we have so far, how could you guess rationally?

Student: [xx]

Jean-Claude Bradley: There might be a hindrance issue, so that the central carbon would be more hindered. Let's say though that there's no hindrance issue whatsoever. It's purely a statistical effect. Let's say that the probability of substituting any hydrogen in this molecule is exactly the same. What would be the ratio that I would expect of those two compounds? Fifty-fifty? It's not fifty-fifty.

Student: Wouldn't it be the first one because [xx]

Jean-Claude Bradley: Right, so how many chances do I have? Well, I've got six hydrogens that would give you 1-bromo propane, I only have two hydrogens that would give me 2-bromo propanes, so I would expect to get this in a ratio of three to one just from a statistical expectation. I expect three to one, just by counting the hydrogens.


When I do the reaction I find that that's not at all what I get. I get the second one at over 99%. Just using common sense and just the tools we have looked at so far we did not have enough information to do this problem properly. So we need to understand some additional properties of these intermediates, and as you said, it's going to be related to the radical stability between these.


Let's think about what's happening in these two competing reactions. So let's go back to making at 1-bromo propane and we're going to step through the steps of the free radical halogenation and see what's happening exactly.


[phone rings]


You should become very familiar with this mechanism. We're going to go through free radical halogenation. The first step is always going to be the same, don't forget that you need light to do this first step. Now the bromine radical is going to approach the propane. And it has a choice. This is where the decision gets made as to what product you're going to make.


If it goes on one of the six end hydrogens, I'm going to get HBr and an n-propyl radical. And then that n-propyl radical will react with bromine. And then we're going to have termination steps and we can have three different termination steps; I'm not going to bother to write all three of them, we don't need that really to understand the problem so here's just one of the three.


Again, the key thing to remember here is what happens at the second step. Once we're at the second step, that propyl radical is not going to change, it's not going to move around.


[phone rings]


So that means if it's at the tip, it's going to stay there and always go to step three. So we can think that the outcome, the product is decided at step two and is decided because we made an n-propyl radical.


I could redraw this whole thing to show you how to get to the isopropyl radical but it would be the same thing. At step two, I would have the isopropyl radical and that would commit you to making 2-bromo butane. So, if we're not getting a statistical ratio here, then clearly it's got to be an energy difference. As was stated earlier, that reason is that the n-propyl radical is much less stable than the isopropyl radical.


So let me draw that in the next slide here. That's the comparison you need to make whenever you do these problems. You figure out the radicals that are involved and you determine their relative stabilities. So this will be the isopropyl radical, will be more stable between these two.


How do we know that one radical is more stable than the other? What's the key difference between these two?

Student: [xx]

Jean-Claude Bradley: What is the type of carbon?

Student: [xx]

Jean-Claude Bradley: That's the key thing, exactly. This is a secondary radical and the one on the left is primary. So you don't have to memorize that isopropyl is more stable, what you need to understand that will work for any instance is that there is an order of stability for radicals.


And that radical stability will be - tertiary is more stable than secondary is more stable than primary which is more stable than methyl radical, which is even beyond primary. It doesn't have any carbons connected to it.


We're going to stop our analysis at that point, you don't need to understand further why is it that those radicals are more stable, the key thing that you need to understand as base knowledge is that here's your order. Okay?


All right. So we're going to do a bunch of examples where you're going to be comparing -- you're going to have a molecule, you're going to have to compare different sites on a molecule and try to predict which is going to be your major product and which is going to be your minor product. Let's go back to this problem that we did with propane and try to do it in a little bit quicker fashion here.


So the first step is to determine how many different products you have then the second step is to determine your major product.


You don't have to go through the mechanism every time, that was just to explain to you why this happens; this is secondary alkyl halide and this is primary. The secondary will be your major product and the primary will be your minor product. So in this chapter the definition of the major product will be the one you get the most of and its going to be related to the order of the carbon. So if you have two products that are both secondary carbons, you would have two major products and everything else will be a minor product.


The other thing you'll notice is that I've been using bromine consistently through this. That's not an accident. Basically, this will work for bromine but it doesn't work so well for other halogens. Let's pick again the example of propane. This is what we just looked at. Do the bromination, we saw that we get above 99% of the secondary and less than 1% of the primary.


If we try to do the same thing with chlorine, the first point is that we do in fact get all the same products, we'll get the 2-chloro propane and the 1-chloro propane. But in this case, the ratio is going to be very far from an overwhelming amount of the secondary bromine. And the reason is that the reactivity of the chlorine radical is much higher than the bromine radical.


So what's happening is that when the bromine radical is coming along and is about to grab a hydrogen off of the propane, it's not too reactive so it has chances to bounce off the propane many times until it reaches just the right angle, just the right energies, and it's going to make the more thermodynamically stable radical. When the chlorine radical is coming along, it is so reactive that it's going to grab pretty much the first thing that it bumps in to.


And because we have more hydrogens that are primary than hydrogens that are secondary, its going to skew the results a lot. Basically, you cannot predict major products and minor products for chlorine, so remember that when you're doing problems. However, you still get the reaction. Ok so that would be the case for fluorine and chlorine. Those would be way too reactive to be selective.


The problem with iodine is that most of the time it doesn't react; it's not even reactive enough to even do the reaction so that's why bromine is just in that sweet spot. It's just reactive enough to do all the reactions that we want but it's not so reactive that it's not selective.


Let me write out here an order of reactivity. So fluorine radical would be most reactive, followed by chlorine, bromine, and iodine radical, where the iodine is not reactive enough, and what follows from that is the selectivity. The selectivity will be: bromine is more selective than chlorine which is more selective than fluorine. I'm leaving the iodine out of it because if it doesn't react then selectivity is meaningless.


Those are the two concepts that you need to add to the rest of this theory.


Last thing I want to touch at in this chapter is make sure everything is clear about these radicals. Okay, so a radical is not charged, that's a very common mistake that I see. The radical is not charged. Why is it not charged? Go back to the basic theory of the first week. Carbon normally has four electrons for charges, there are exactly four electrons here. This does not satisfy the octet rule, which is one of the reasons it's really unstable but it does exist. It exists fleetingly, but it still exists long enough to do the reaction. But remember that it's not charged.


You can have a CH3 with a charge; if I have a lone pair of electrons on that carbon, I would have five electrons for charges, carbon normally has four, so this would have a charge of -1, and that's called a carbanion. Anytime you have a carbon with a negative charge on it it's a carbanion. The only other possibility is if I have no electrons on that carbon left, now I have three electrons for charges, and so this is going to be a carbocation.


Okay, so make sure you understand the differences and why it is that we don't put charges when we do free radical halogenation mechanism.


So this is a pretty good place to stop, we're going to go right into the problems next class.


[general noise of students packing up and leaving]

Transcription by CastingWords