Friday, May 19, 2006

Lecture 013: Free Radical Reaction Intro

Lecture 013: Free Radical Reaction Intro
Jean-Claude Bradley: We are going to take this entire chapter to look at one single reaction, and that's something that we looked at briefly, the halogenation of alkanes. When we looked at the reactions of alkanes, this was one of the three. As we will see, as you move on through organic chemistry, you'll see that reactions fall under certain types. This particular type is called a free radical chain reaction; you'll understand why it has all those terms in its name after we're done here.

This is something that's specific to halogenation of alkanes, but it's also general. Once you learn the free radical chain reaction, then you can understand a lot of other chemistry that does not seem related at first, but actually is.

I'll just remind you of the example that I likely gave, which was if you take methane and reaction it with a halogen, let's say bromine, and shine light on it, you're basically going to swap the hydrogen for the bromine, I'm going to put the bromine on the CH3 group and we end up with HBr from that.

That's useful on one level; you do know what the product is from methane reacting with bromine. Because we covered this in the alkane chapter you could probably guess that propane will also reaction with bromine, but you really have no means of predicting for different products that you would get. For methane, there is only one product; I can only put the bromine at one place. For longer alkanes, there's all kinds of different places I can put the bromine. Without knowing the mechanism, you really don't have any predicting power to be able to even guess, rationally, as to what kind of products you're going to get.

By looking at the mechanism, it's pretty powerful because you'll be able to predict how the free radical halogenation will work, not only on molecules you haven't seen before, but on molecules that have never been made. That's the whole point of organic chemistry, to make stuff that hasn't been made before and predict how it's going to behave.

Let's start to analyze this, based on some stuff that you already have seen, which is let's look at the energy of this reaction, at least get a big picture view of it. To look at this we're going to consider bond association energy, which is the energy it takes to break a bond. There are two ways hat we can break a bond, and we have to be specific about which one we are going to use. If I bring a bond, the bond has two electrons and here are two ways that I can split those up. I can either electron on each hydrogen, so each hydrogen ends up with one electron. To do that, I'm drawing these arrows that have not a full arrowhead, these are half arrows. That means movement of one electron.

What we saw previously with the curved arrow formalism was the movement of two electrons. If I wanted to show that in the bottom example, all I would do is use a full arrow. That would basically tell me to take the two electrons in that bond and put them on one of the hydrogens and leave no electrons for the other hydrogen. These are two ways that I can disassociate; does anybody remember what these are called?

Student: Homolytic and heterolytic.

Jean-Claude Bradley: Right. Homolytic, homo means the same, so they get the same number of electrons. Heterolytic, different. The one that we're going to be concerned with, to do these calculations, is going to be the homolytic. On page 134, there was a whole list of bond association energies that are relevant to this chapter, that's what I'm going to use to do the following calculations.

In order to look at that, which is going to tell us how much energy, if energy is being released or absorbed by the reaction. We are going to draw only the bonds that are being broken or formed in that first reaction. I'm going to draw a CH4 like this, CH3-bond-H. Then another bond that is broken is a bond between the two bromines. All I need to do to figure out the energies of this reaction, is I'm going to write down the energies for the homolytic dissociation. For the CH bond in CH4, we have 104 kilocalories per mole. For the bromine, we have 46. For the CH3-Br, 70 and then 88. The higher the number, the stronger the bond is. You can see here very quickly, the bond between the two bromine atoms is a lot weaker than the CH bond in methane, which is partially why bromine is so reactive.

To calculate the energy of the whole reaction, we're basically going to sum up the bonds that are broken, minus the bonds that are formed. That's just basically (104+46)-(70+88), those are all kilocalories per mole, we end up with -8 kilocalories per mole. What does that tell you about the reaction?

Student: It's giving off energy.

Jean-Claude Bradley: Right. It's giving off energy when you have a negative number. It's not a huge amount of energy, but still, it's going to be definitely liberating heat as you do this reaction.

The first thing that it tells us, if we look at the energy diagram, if we start off with CH4 and bromine on the left, we end up with CH3-Br and HBr on the right. The only thing that we really know is that we are going to have an eight kilocalorie per mole difference between the starting material and the products. What we don't know is the landscape between those two, what actually happens energetically between those two. That's what we're going to look at now. We're going to fill in this diagram as we do the mechanism.

You want to think of this in a logical way. What happens is we're going to take a container and fill it up half with CH4 and half with bromine. The first thing is that we know that without light, nothing happens. I could have bromine and methane coexist and nothing will happen. Obviously, light is going to be involved in the first step of this mechanism. The very first thing that happens, methane is pretty inert to light, but bromine is photosensitive, so the bromine is going to dissociate, like that. That requires light. H nu is traditionally the symbol you use for light in organic chemistry.

Again, all of the electrons, because we're going to use these arrows extensively through the rest of the class. The half arrow tells you it's moving only one electron. Whenever you break a bond, using only half arrows, you're going to need two of them every single time, because I have to remove two electrons to break a bond. The first thing that this tells us is that we now have something new in the mixture, and that is a bromine atom. When I write Br dot (Br.) what I actually mean by that - I'm only going to do this once because it's a pain to draw it like this every time - is the full correct Lewis structure for a bromine atom is a bromine with seven electrons around it. It's neutral; bromine only has seven electrons. Now that we've done the first couple of weeks, you know what it should look like. We are going to abbreviate that by just drawing Br dot, but there are three lone pairs there that exist.

We're going to follow this down logically, to a generated bromine radical. There are only two things that can happen, the bromine radical can reaction with either Br2 or it can reaction with CH4. It probably does reaction with Br2, but it will just break up again into another Br2 and a bromine radical. That's not something that we care about, what we care about is the reaction between the bromine radical and the methane. You don't really even have to memorize this, if you just think about it logically, about what gets added.

Now, I have something new and there's only one logical thing it can do and that is to reaction with the methane. It's going to take three arrows to do this. The bromine radical is going to form a bond with the hydrogen. We're going to make HBr at this point. In order to form a bond between the bromine and the H, it's going to take me two electrons. Bromine only has one electron to donate, so the other electron has to come from somewhere. The only place that it can come from is between the CH bonds. When I'm making a bond, I'm taking two electrons and bringing them together in space. I'm going to depict that by putting one half arrow in space, between the Br and the H, and the other one will come from the bond.

If you think about the meanings of the arrows, you're not going to make mistakes. If you just look at this, it should be obvious that this is wrong because I can never just remove one electron from a bond. What would I have, where is the other electron? That electron has to go somewhere. Where is that going to go? It has to go on a carbon. It can't go on a hydrogen because the hydrogen is fully bonded now to the bromine. It's not going to go off into space. You have to put it someplace and the only place it can go is on the carbon. I'm going to, again, start the arrow on the middle of the bond, and end it up on the carbon atom itself. That will give me BrH and CH3 radical, so this is CH3 that's uncharged. How do you know it's uncharged? Go back to the basics that we learned in the first week; there are four electrons around the carbon for charges. Carbon only has four electrons, so it's not charged.

With this step, we have now generated two new species. We now have HBr and we have CH3 radical. Now, there are a number of things that can happen here. First thing you'll notice is that we generated the HBr, which was one of the two products of the reaction. That's actually how that HBr forms.

The next product that we have to make is the CH3-Br. That's going to be a reaction between the methyl radical and some more of the bromine. The flow of the arrows is going to be completely identical to number two. I'm going to take the radical and put it into space between the C and the Br. I take the other electron from between the two bromines, because we're going to break that bond and have it join the other one. Again, if I leave it like that, I have to show what I'm doing with the other electron in that bond, so I have to put it on the other bromine, just like I did in number two. There are lots of arrows here, but it's the same pattern over and over again.

I've just now created CH3-Br and a bromine radical. I already have the bromine radical in here, so the only thing new is the CH3Br. This is actually an interesting reaction because I now have a bromine radical. That bromine radical can go back into step two and react with another methane molecule. I have here basically, the way that it's drawn, an endless cycle. For every step two that I don't I make a bromine radical, for every step three I do, I can go back and do step two again. Steps two and three keep going in a cycle. It doesn't go on forever, for one, I have a limited amount of material here. Presumably, after all the methane and bromine is gone, then the reaction is over. More practically, this will only go around ten thousand or a hundred thousand times before it stops. Anybody have an idea why it might stop?

Student: Because the reactants will eventually run out?

Jean-Claude Bradley: No, I'm saying that way before I run out of reactants.

Student: Some bromine might [unintelligible] and Br2.

Jean-Claude Bradley: Right. As long as I use up the bromine in step three and step two, I can keep going until I run out of reagents. However, if I trap that bromine radical, then it's over, and I have to start over. I have to go back to step one and generate more bromine radicals. That step, where you take two bromine radicals, or generally any radicals, in steps two and three, if you combine them, you now have a termination step. That's why it's not infinite.

I'm going to draw steps four, five and six. Step four could be the reaction of two bromine radicals. Step five could be a bromine radical and a methyl radical. The third possibility is I could have two methyl radicals combining. These are very straightforwardly just two half arrows because that's all I have. I just have two single electrons; they come together in space somewhere and they create a bond.

Basically, we have all the steps of the mechanism, there are six total steps, but they're divided up into categories of reactions. The last three are the termination steps, because they stop steps two and three. Steps two and three are called propagation steps and number one is called initiation. You'll notice here, light is only needed in the first step. If you were to flash a brief pulse of light, what would happen is you would generate the bromine radicals, then you get steps two and three, like I said, ten thousand or a hundred thousand times, it would go over, and then it would be over.

The thing you have to understand about the relative importance of each one of these steps. Why is it that steps two and three can go on for tens of thousands of times? That has to do with the probability of two radicals finding each other. After I do step three, I go back to what's in the mixture. I have bromine radical, and around it I've got 50 percent methane and 50 percent Br2. The probability is it's going to find a methane molecule way before it has a chance of finding another bromine radical. It's all a concentration thing. That's why the termination steps are so rare.

Consequently, when we look at the energies of these things, a common misunderstanding when doing this is that step five produces one of the products, CH3Br, but it does so in an insignificant amount. You're only going to have one part in every hundred thousand methyl-bromide molecules made this way. It simply does not come into our calculations of the energies. It's important that you understand that because it really is a big distinction when we do the energies. The terminations are just aberrant behavior in the system.

What we are going to do next time is we're going to look at the energies of those steps, and then make predictions.


Post a Comment

<< Home