Freshman Organic Chemistry II

CHEM 125b - Lecture 2 - Peculiar Rate Laws, Bond Dissociation Energies, and Relative Reactivities

Chapter 1. Processes with Fractional or Negative Kinetic Orders [00:00:00]

Professor Michael McBride: Today’s the day I should have welcomed you to sunny New Haven I guess.

So we’re studying reaction mechanisms this semester, but we’d taken a little diversion to look at reaction order–how the dependence of rate on concentration gives information about mechanism. So this is the kinetic analog of the law of mass action, the exponents we talked about on concentration last semester. And we saw last time that we would have rate laws, the d(Product)/dt is some function of concentration raised to some power. And we looked last time at zero order and first order and second order. 

Now we’re going to look at some complex reactions and how they can get weird orders like fractional orders, or even inverse orders.

So a fractional order: Suppose the rate is proportional to some rate constant times the concentration of  A times the concentration of Bto the ¼ power. How can that be? How can you have 1/4 of a molecule participate in a reaction?

Well, this all depends on, when a reactive species in equilibrium with other species, what is the dominant species? Because when you add more material to a solution, what you increase is the dominant species by how much you add. If you double it, you have twice as much of the dominant species. If you triple it, you have three times as much of the dominant species. But suppose what’s reacting is not the dominant species, but is related to the dominant species by some equilibrium constant.

So, for example, suppose the dominant species were a tetramer, four units. But the reactive species were a monomer, a single B unit. So you have the equilibrium between B₄ and 4B. Now if B₄ is dominant and B is reactive, then if you double the concentration–or actually let’s say you increase the concentration by a factor of 2, 4, 6, 8–you increase the concentration by a factor of 8. How much do you increase the concentration of B₄? It’s practically all the stuff, it’s the dominant species, so you increase it by a factor of 8. By how much do you increase the concentration of the monomer, B, that’s in equilibrium with it? 

So if the dominant species is proportional to how much you add, that’s how much you think you have. But the minor species tags along following the law of mass action. So we actually, last semester, talked about a tetramer–that methyllithium gives this distorted cubic tetramer. Remember it has four ether molecules associated with it if possible.

But if you put in a lot more ether then it rips the tetramer apart, because all the oxygens want to be sharing their electrons with lithium. So excess ether rips it apart, first to give a dimer with two more methyl ethers per dimer unit. And ultimately to a monomer with another methyl ether per monomer unit. So each monomer now has three dimethyl ethers. 

So if you have plenty of dimethyl ethers, then the equilibrium shifts to the right, and most of the stuff is monomer. And if you increase the material you put in by a factor of 8, you’ll increase the monomer by a factor of 8. But that’s not what happened when you don’t have the excess ether available. Then the equilibrium lies to the right, and when you ladle 8 times as much material in–you don’t ladle it because it’s very reactive with air and water–but if you increase the concentration by a factor of 4, this is what you increase by a factor of 4. And how do you increase that? Well, you have this equilibrium constant where the monomer is to the fourth power, so the monomer to the fourth is proportional to the tetramer. Or you could write the same thing, the monomer is proportional to the tetramer to the one-fourth power. 2, 4, 8–16 was what I was supposed to say, right?

So if I increase the tetramer by a factor of 16, I increase the monomer by 16^1/4: by a factor of 2. So it would only double the rate. So reaction of monomer in a hydrocarbon solvent turns out to be 1/4 order in how much reagent you add. But the reaction order proves then that the monomer is reactive, but the tetramer is dominant in the hydrocarbon. But if you have excess ether, then the reaction shifts to the left, and if you add 16 times as much stuff, you increase the concentration of the monomer by a factor of 16.  Then it’s first order in how much stuff you added, and the reaction becomes first order.

So that’s one example of a weird kinetic rate law, a weird dependence on concentration, fractional. Suppose you had a 1/6-order reaction? What would you infer? Elisa? 

Student: That it’s a hexamer?

Professor Michael McBride: Right. That the dominant species is a hexamer, but the reactive species is a monomer. Good.

Now here are two enantiomers, an R and an S configuration of this funny molecule that’s a little bit related to amino acids. Now it has what are called lpha-hydrogens–those are hydrogens attached to a carbon adjacent to a carbonyl group. So it turns out that a base can pull off that hydrogen, leaving an anion on that carbon, which then is stabilized by π* of the carbonyl. So it’s easy to pull off that H and put it back on again.

So that means that if you put base in with these two molecules they’ll interconvert easily. Now that would lead to racemization. If you started with all one hand, you’d have a reaction going one way which has a certain rate constant times the base, and a reaction going the other way which is the same rate constant, because they’re mirror images, times the concentration of base.

Now the rate, how much goes from one side to the other, is the rate constant times the concentration. So if you have a higher concentration of one than of the other, its rate will be faster. So as it says here, that since they’re identical rate constants for the two enantiomers, that guarantees a faster rate for the major concentration, the major enantiomer going to the minor enantiomer, until they equilibrate, they come to 50/50, and then the rates will be equal.

So that’s a guaranteed mechanism for racemization. You have an -hydrogen at the chiral center, the stereogenic center. You put base in, pull it off, put it on, and you’ve lost all the effort you’ve put in to make the thing a single enantiomer in the beginning.

Incidentally, how do you make something a single enantiomer? When you generate something, typically you make both right and left equally, with equal probability. But what you have to do, remember as we talked about last semester—Pasteur, one of Pasteur’s methods—is to react it with something pre-existing that’s only one enantiomer in some reversible reaction. So you can make the pair of enantiomers into diastereomers, separate them, perhaps by crystallization, then take them back to what you want. That’s how you get one to begin with. It depends on having something pre-existing that’s a single hand. 

Now these two molecules turn out to crystallize, and they crystallize in crystals that are composed all of one enantiomer or all of the other. So if you let such a mixture crystallize, it would give 50% of one crystal, 50% of the other crystal–it would be racemic. Except if you grind it.

So there was this very interesting paper two years ago by a group in the Netherlands–Noorduin was the lead author–about grinding a suspension of these crystals. That is there are these right- and left-handed crystals of that compound. When they’re in solution they can interconvert, because there’s base in there. So the molecule comes off the crystal, goes into solution, racemizes, and then goes back on one crystal or the other.

So the experiments lasted some time, you see, up to 50 or 60 or 70 days. And they looked in the solid phase of the crystals at what the enantiomer excess was–how much more of one crystal there was than the other. And they started with a little bit of excess of either the S, those are the blue points, or the R, those are the red points. And I notice we have these front lights on. I better turn those off to get better definition on the slides.

So we have red or blue. It could be a very tiny excess, like 0.04%. Now what would you expect from what I’ve been telling you if you had base in the solution and you start grinding to encourage the crystals to dissolve more rapidly? What would you expect to happen in time? Suppose you started with the one that’s 17% excess of S? What would happen in time? Come on, be naive. Megan, what do you think would happen?

Student: Less enantiomeric excess?

Professor Michael McBride: Can’t hear very well.

Student: Less enantiomeric excess?

Professor Michael McBride Yeah, it would go to zero in enantiomer excess, it would go to 50/50. So you’d expect a curve that would look like this. It would go along, depending on how fast it is, it would become zero. This is what actually happened.

So it DE-racemized. It became a single enantiomer. Whatever was in excess at the beginning, all the crystals became that.

So this is very different from what we were talking about before where the rate law guarantees that whatever’s in excess will go more rapidly to the minor than the minor to the major, and it’s guaranteed to come out 50/50. This one goes exactly the opposite. So this was a great puzzle, what caused this to happen.

It could be that there’s faster conversion of the minor crystals. So that if you have, say, more R than S, it’s faster to go from S to R than from R to S. What would that say about the rate law? If the less concentration you have, the fewer crystals, the faster it goes. It’s an inverse rate law. That certainly sounds weird. But there’s another interesting thing. The lines have a curious shape. Do you see what’s funny about, as a reaction goes along, having this kind of shape or this kind of shape? Remember, we looked last time at a first-order reaction or a second-order reaction, what it looked like. You started with a given starting material and what happened in time? It went to product. It went to zero concentration. But how do it go? It went like this, because the more of it there was, the faster it was going–half-life, half-life, half-life. This seems to be backwards. And you see it both in the fact that it goes all to one, to the major one, and in the fact that you have these funny curves.

Now the explanation is that when you’re grinding you make tiny, tiny pieces that are guaranteed to dissolve. So grinding accelerates dissolution by making very tiny crystals below the critical size that we talked about last semester. So they’ll dissolve, they won’t grow. Except they can be rescued from dissolution by combining with another crystal of the same kind. So the crystal could grow not by adding one molecule at a time, but by two small crystals coming together.

Now that’s going to be second order in how many of these little pieces there are. If two things are coming together, two A’s are coming together, that’s second order in A. So the more stuff there is the faster it avoids dissolution. So really it’s a second-order process of avoiding dissolution.

But that means that if you’re looking at the dissolution, which allows stuff to go from one to the other, that’s faster if you have less stuff, because you don’t save it. So the majority solid dissolves more slowly than the minority solid does, because it could be saved by a second-order process.

But the key to understanding this really weird phenomenon was the kinetics, to see that the lines had these funny shapes. That’s what clued people in to the idea that it was this coalescence of little shards of crystals that made the thing to do this. And this could be of interest. The people who do this, the reason they’re motivated to do it, other than they just think it’s cool, is people have always wondered how it is that in the biosphere things had a single hand, that all amino acids are L-amino acids and all sugars are D-sugars, or most sugars at least. How did things get resolved in the first place when there wasn’t anything pre-resolved to do this separation we talked about?

But this can start at 50/50. If you have a lot of particles, it’s never exactly 50/50–there’ll be a little bit more of one than the other. So one of them will be saved from dissolution a little more than the other, and ultimately that one will build up autocatalytically–the more it goes, the faster it goes. So that seems to be the explanation of this phenomenon that could, indeed, have something to do with the origin of single handedness in the biosphere. But the key to understanding it was the kinetics, the rate law. 

So we’ve seen the guidance that rate laws can provide for understanding reaction mechanisms–zeroth order, that’s when the catalyst is saturated. First order is the normal business. Second order when two things have to get together. Fractional order when things have to come apart. And negative order, in this case, although you could argue that it’s actually second order, but it appears to be negative order, that the minor stuff goes more rapidly.

Chapter 2. Problems in Understanding Relative Bond Dissociation Energies [00:17:28] 

So now back to what we were talking about, the reaction we’re talking about, which is bond dissociation to generate free radicals. The reason we’re talking about that is try to figure out how you can predict rate constants from things we already know. 

So we’re studying free-radical substitutions for this purpose because a) they’re simple. There’s minimal influence from the solvent, so you can just look at the molecules that are doing the reaction and try to understand the rates. 

They also turn out to be very important, these free-radical reactions. They’re very important in atmospheric chemistry, so there’s a lot of work done on them in, say, Boulder, Colorado, where the Atmospheric Administration and the Joint Institute for Laboratory Astrophysics are very interested in these kinds of things. They’re important in combustion, you make free radicals. Also in oxidation reactions that we’ll talk about more later.

And in terms of strictly organic chemistry, they provide great examples of selectivity. And that’s a pervasive theme, as I say here, in synthesis and in biochemistry. Almost always with any kind of complicated molecule, the problem is not what can react–there are always lots of different ways of reacting. The question is how you can be selective and do one thing rather than another. That’s often the problem in organic synthesis–is being selective. And the free-radical reactions are a very simple example of showing something about selectivity, as we’ll see later in the lecture. 

OK. So last time we were looking at this table of experimental bond dissociation energies, bond enthalpies as they’re called. There are a zillion of them and they’re known very well. But what we would like to understand is why. Why do they have these values? So you don’t have to memorize the whole table. That is, do you just have to suck it up and memorize this, or can you understand this kind of lore–what the rate constant should be, what the energy should be for the strength of these bonds that you’re trying to break?

Well, it’s very difficult to predict exactly what these energies should be. But often you can predict relative energies. If you’d know one, you can predict another one. So most often in exams, questions I’d ask on exams, it’s not a question of knowing the precise rate, but if you knew the rate for this, what would the rate for this be? And you notice that it has to do with selectivity, when you have two processes that are analogous to one another that can go on, how much better should one be than the other? That’s much easier, to get the relative values than to get the absolute values.

We started last time looking at this, in the case of the bonds to methyl from the various halogens. Remember, that it has to do with overlap? That as you make the halogens bigger, the overlap isn’t so good. And also that has to do with energy match between the orbitals–we talked about this last lecture.

And we also talked about other cases where the radicals were unusual. That, remember, had to do with starting material, how strong the bond is. But then others had to do with how stable the radicals were. Like the hybridization of the C-H and the resonance–the SOMO mixing with  and p*  [this should be shown as pi and pi* ] we talked about that last time–has to do with the radical, whereas the hybridization that explains the strong ones has to do with the starting material, how strong the C-H bonds are. So we’ve done that. 

Now let’s look at a more subtle example, which is as we go from one alkyl radical to another, breaking hydrogens off from carbon, but from just different alkanes. And the first thing we notice there is that they’re not very different. You can explain this by saying that the carbon-carbon bond is more sensitive to hybridization, its energy, than the carbon-hydrogen bond is.

Now when you go from a tetrahedral carbon, the starting material, you pull off an H and get a radical, which is trigonal. You go from sp³ to sp² bonds. And the sp² bonds, as we saw last semester, overlap better and give stronger bonds. That’s fine. That’ll be true, they’ll be stronger if they’re bonded to hydrogen; they’ll be stronger if they’re bonded to carbon. 

But it turns out that carbon-carbon bonds are more sensitive to the hybridization than carbon-H is. So if when you’re pulling the hydrogen off from your tetrahedral carbon there are other carbons attached to that carbon, then in the process of going to the radical you’ll get more extra stability than you would have if there had been hydrogens there, because the carbons profit from the change in hybridization more than the hydrogens would have done.

So that’s what we see here. Methyl, which has three hydrogens is hard to form the radical–104.9. But it gets easier and easier and easier as you go to more and more carbon-carbon bonds rather than hydrogen-carbon bonds on the radical you’re making. So that’s one explanation: that the carbon-carbon is more sensitive. And that’s to say that that’s why the radical is more stable. It has to do with the stability of the radical.

The other possibility is analogous to what we saw over here in the allyl and benzyl cases. It was where the SOMO mixes with π and π* on the adjacent carbon. So we’re going to pull a hydrogen off here, make a radical, a p-orbital here. If we had a π system here that could overlap with that we mix it with π, mix with π*, and get stabilization. We looked at that last time.

Now, of course, if you have a hydrogen here attached to the carbon from which the hydrogen is being lost, it’s orthogonal; it can’t overlap with the p-orbital here. But if you have a carbon here with a hydrogen on it, then that σ and σ* bond of C-H can overlap with the p-orbital. So you can get the same kind of mixing that you get with  and *, except σ and σ*. So they’re much further away in energy–the energy match is much poorer with this p-orbital.

So you don’t expect it to be a very big phenomenon, and the changes here, indeed, are not nearly as big as they were for benzyl and allyl. But those are the explanations that are standard for explaining why more substituted radicals are more stable. That’s what people say, so it’s easier to break the bond. That is, they say that it has to do with this radical stability. Notice, we’re assuming that the bond dissociation energy difference, as you go across from methyl to ethyl to isopropyl to t-butyl is due to differences in radical stabilities, not the differences in the starting material energy. Remember, it could be either.

The standard explanations are in terms of the radical. And what I’m going to show you now is that that’s not so obvious. Even though I think it would be interesting for you to look in these various textbooks that you have at this subject about free-radical halogenation–every one of them will have a chapter about that. And see what they say about what the explanation is. Some of them won’t give any explanation at all, they’ll just say more substituted radicals are more stable. But those which give an explanation will give one of these. But I want to show you that it’s not so obvious that that’s true.

If it were true, we can test it by checking more examples. Instead of looking only at hydrogen breaking away to give those different radicals, we could break other things away. There’s not enough data for fluorine, but chlorine, bromine, iodine, methyl, ethyl, isopropyl, t-butyl all can be broken away from these radicals, and they all should show the same phenomenon, if it has to do with the stability of the radicals rather than the starting material. So that’s what we’re going to do is expand the scope of what we’re looking at using that table.

So here is R along here and X. We’re going to break X away from R and leave the R radical.. Now what connects B? X can be hydrogen, that’s this. Those are the data we just looked at, and we explained it in terms of different stability of the radicals that are being left behind when you pull the hydrogen away. And here are all the different alkyls–methyl, ethyl, isopropyl, t-butyl. Breaking chlorine away is the green one, about like the alkyls. Breaking bromine away is the red one, and iodine.

Now what this shows is that bonds to iodine are much weaker than bonds to hydrogen. Iodine forms weaker bonds than bromine, than chlorine, than hydrogen. Fine. So hydrogen is stronger than R-C. R-C is about the same as R-Cl, but R-Cl is stronger than R-Br, and R-I is the weakest, whatever the radical you’re breaking away.

Now there are some obvious differences here that the phenyl, when phenyl is the group that’s being broken away in the left column here, it’s unusually hard to break away. And we saw why that was last time. They have good overlap because you have sp² C-X bonds that have to be broken. They are the strongest–whatever X you’re breaking away that’s the strongest bond. And on the far right we see that if you have allyl and benzyl you get those resonant stabilized radicals. And no matter what X you’re breaking away that’s the best.

What we want to do is now focus on these ones in the middle and see if we can understand these ones where the various methyl, ethyl, isopropyl and t-butyl are the radicals being left behind, and see whether the data as a whole echoes what we saw in hydrogen, that more substituted radicals are more stable. So all these should be the same. That is, it’s going to be different depending on which atom we’re pulling away, but the relative stability of the radicals as we go across a given color should be the same, if that’s the explanation.

So here’s when we’re breaking H off. And now I’ve blown up the scale so we can see the variation. So we’re going to compare it. These are relative bond dissociation energies, taking zero to be methyl. So we see that it’s 4 kilocalories easier to break hydrogen away from ethyl. Another 2 and a fraction to get isopropyl. And another couple to get t-butyl. So people do say, your books I’ll guarantee you will say, that those radicals as they’re more substituted are more stable. 

But if it’s true, it should be the same whenever we’re breaking away from the radicals. So what they tend to say, for example, the Jones and Flemming book that I mentioned, on page 479, sp² σ bonds to carbon are preferentially stabilized the more substituted radicals. The C-C overlap is more sensitive to hybridization than C-H. That’s what I just told you. And they also say on page 478-479, there’s probably a bit of stabilization from SOMO overlap with σ and σ*. That’s what I said last time.

Now let’s see. So if this trend is due to radical stabilization by these two phenomena, then whatever X we break off we should show the same trend. So here’s when we break t-butyl, when we break t-butyl away from the various radicals. That is, break the bond between t-butyl and methyl, between t-butyl and ethyl, between t-butyl and isopropyl, or between two t-butyls. And indeed it looks pretty much the same. Within the error probably, these are the same trend. So that looks good.

But look at the others. If you break these others off they don’t show–they show it’s always going down, but not nearly as dramatically. So it can’t be just the difference in stability of methyl, ethyl, isopropyl and t-butyl. There has to be something in the starting material that’s contributing to this. It makes a difference what it is you’re breaking away. 

Now, t-butyl is a very special compound, because it’s so crowded. So if you put two t-butyls together, it’s very crowded. And that could make the starting material unstable and relatively easy to break the bond. But we can calculate how much instability there is due to that by using molecular mechanics, as we mentioned at the end last semester.

So we can calculate strain energy in the starting material using molecular mechanics. This is the amount of strain energy you count in two t-butyls (t-butyl-t-butyl), t-butyl, in t-butyl with isopropyl, in t-butyl with ethyl, and in t-butyl with methyl. That’s how much strain energy there is in the starting material.

Let’s just look at the one there to see where it comes from. So here are two t-butyls bonded to one another with idealized bond lengths and angles. And we calculate by molecular mechanics, if we have normal C-C and C-H bond distances, normal angles at carbons, how much strain will there be due to things running into one another. Because you’ll notice if you make a space-filling model of that you can see down the middle its crunch region.

Now how big is that? The van der Waals energy calculated for this from these things running together is 26.9 kilocalories per mole–pretty high. Now, of course, the molecule doesn’t have that shape. It adjusts, as we mentioned last semester, to minimize the energy.

So what it does is adjust to that position. So it drops the van der Waals energy by 16.8 kilocalories per mole, down to only 5.2. That’s how it does it, by moving the two apart. But it’s at the expense of stretching the bonds and bending the bonds a little bit to get those hydrogens apart from one another. So although it started with 109º bond angles at 1.52 Å, it changed like this–you can see how it moved to get the hydrogens apart, but there’s an expense to pay.

So even though we’ve gotten down to only 5.2 kilocalories per mole of van der Waals strain, there’s now an increase to 4.8 kilocalories per mole from stretching and bending. So the total strain in two t-butyls bonded together is 12.2–that is 4.8 plus 5.2 plus 2.2 of torsional energy that’s in there. So that’s where that number comes from. So that’s the 12.2, and you can do similar calculations to get these other values.

Now notice something interesting. Those are strained in the starting material. But notice that the range of that from 2.3 to 12.2 is 9.9. So that’s bigger, in fact, than the whole range of this difference attributed to differences in the radicals. That is, in the case of t-butyl the starting material for t-butyl-t-butyl is strained enough to explain the entire difference of why it’s easy to break t-butyl-t-butyl–not the radicals, but the starting material.

Now we can look at the case of methyl, how much strain was there in the starting material when methyl is attached to these various Rs. And what you see is the same thing, that the amount of strain can explain the range of difference in bond association energies without saying the radicals are different at all.

So for X equals alkyl, methyl, ethyl, isopropyl, t-butyl, for the X that’s breaking away being methyl, ethyl, isopropyl or t-butyl, almost all of the change is due to strain energy in the starting material. But not in the case of H-R, because in the case of H-R there’s very little strain in the starting material. So there’s something different going on with H-R. It’s not typical.

So as far as I know, no one really understands this, why it is that way.

Now here we’ve done another plot where we correct for the R-X strain. That is, subtract out that strain in the starting material and then see how different they would be. And you see there’s hardly any difference at all as you go from methyl, ethyl, isopropyl, t-butyl radicals.

So this is an alternative to the hypothesis that radical stabilization by substitution of methyls for hydrogens–more substituted radicals are more stable, which I guarantee you is what your books will say, whatever book you’re looking at.

So the intrinsic C-C bond strength corrected for strain is practically insensitive to the substitution of the radicals. But if you have C-H bonds they are, indeed, weakened by alkylation of the carbon, but no one understands it. And on the other hand, if you have chlorines–chlorine or bromine, they go the other way. And again, nobody understands it. No one I know at least understands this, but the textbooks seem to be wrong. So a word to the wise, as you look at your different textbooks.

Now that’s what we’re going to say about bond energies.

Chapter 3. Predicting Relative Rate Constants – The Hammond Postulate [00:37:34]

Now how can we predict activation energies, when you’re not just breaking a bond, but making a new bond as you break the old bond? What energy should that have? Can we use the energies of stable structures, ones that we understand to the extent we–it’s clear that we don’t really understand things perfectly—but can we infer the energies of transition states so we’re able to predict reactivity? For example, might exothermic reactions be faster than endothermic ones? 

Now let me take this–someplace I dropped my prop. Is there a green thing over there? OK, you’ll have to imagine my green thing that I thought I brought along. I had a packing strip that comes around big shipments. So it’s a piece of wire. So imagine that there’s a wire between my two hands and I’m holding it so that it goes like this. So it’s like a reaction coordinate. What I’m showing you is an endothermic reaction. You start here, you go over a barrier, then you get to product. Or if I lower the product, it now looks like this. You have that in your mind’s eye?

So I can go back and forth, knowing the energy difference between starting material and product, I have some idea what the activation energy might look like. Of course, this is really nonsense. Activation energy isn’t a packing strip like that, but we can think of it that way. And if that were so, then you notice that an endothermic reaction would have a high transition state. But an exothermic reaction would have a low transition state. 

So we could go from a knowledge of the starting material and the products to infer something about what the activation energy was, as long as we’re looking at reactions that are very, very similar to one another. So they would have the same kind of strip.

So we’re going to see if we can use that to understand something, to predict things, about free-radical make and break. So we’re not going to try to do absolute energies like this–that’s way too hard. But we’ll hope we can at least do relative energies if we look at analogous reactions.

So how can we predict activation energy? It’s no easy task a priori, especially when interaction with the solvent is important. But often you can say something sensible about relative values of the activation energy or the free energy of activation. It’s a question of compared to what–you want to compare two things that are very close to one another.

Now this is related to what’s called the Hammond Postulate. George Hammond, who incidentally was also a student of Professor Bartlett, who’s your chemical grandfather as we talked about last time–in 1955 formulated something that has become called the Hammond Postulate. What he actually said was, “If two states, for example, a transition state and an unstable intermediate, occur consecutively during a reaction process and have nearly the same energy content, their interconversion will involve only a small reorganization of molecular structures.” 

I found my prop here.

So here’s an exothermic reaction. Here’s an endothermic reaction. Now notice that the transition state and the product are very near one another in energy in this case. But they’re also near one another in structure, from here to here. Whereas if it were an exothermic reaction, the transition state is very different from the product, it’s like the starting material, and it’s very near in structure to the starting material. It’s a little bit crummy, but I think you got the idea.

So whether that’s what you would read into this necessarily, I don’t know. But what is clear is that this stimulated organic chemists to begin thinking about transition states and to try to generalize plausibly about reaction coordinates–how you might infer from starting material and product energies, the kind of molecules we know about, to these transition states, when one bond is being made and another broken, for example, which is very difficult to reason about.

So here’s another picture of the same thing. The more exothermic a reaction the more similar the transition state to the starting material, both in energy and in structure. So here’s the starting material, here’s the transition state. When it’s very exothermic it’s close to the starting material in energy, it’s close to the starting material in structure.

If it’s not very exothermic, then the transition state is sort of halfway along, and considerably above the starting material in energy. And if it’s endothermic, it’s near the product in energy and in structure, so that the reaction is then–and the reaction is slow because the transition state is so much higher.

Now we can be fairly confident of this, at least if we’re looking at very similar reactions. For example, the ones we were talking about. If you have an X radical come up and take H away from R, but we vary X or we vary R to make the reaction more or less exothermic, then we might be able to use this kind of reasoning to get from one to another. Knowing one we can predict another.

Now there’s likely a continuum between starting material and product with respect to factors that influence stability. For example, resonance stabilization. Remember the allyl radical and the benzyl radical are especially stable because they have resonance stabilization, the SOMO mixing with π and π*.

So if we’re talking relative to just a regular old alkyl group, methyl, so we got here methane goes to methyl radical. So we make those, by definition, the same, because we’re going to compare relative. What happens if we have benzyl with hydrogen on it and the benzyl radical. And we see that the benzyl radical is more stable because when you get here you have this resonance stabilization. And this idea that there’s a continuum is that the further the transition state is along, the more it will experience that special stabilization.

So if the transition state, when this bond is partly broken because X is taking H away, if it were a late transition state like that, an endothermic process, then it should have a late–if it were a late transition state it should have pretty much stabilization by resonance. But if it’s an exothermic reaction and it’s a very early transition state, then it shouldn’t have much of this. So the idea is that you’re going to have a steady increase of this. It might not be quite a straight line, but at least it’s steadily going down.

So the rate of the slower reaction should be more sensitive to the overall difference in free energy. That is, if you have a very fast reaction and you move this up and down, change the exothermicity, ∆G, by quite a bit, you hardly move that at all. But if it’s an endothermic reaction and you move this up and down, this’ll move up and down almost as much. So the reaction becomes less sensitive to the ∆G when you have an exothermic reaction, because it happens so near the starting material.

So this gives rise to what’s called the Reactivity-Selectivity Principle. So here’s an exothermic reaction, here’s an endothermic reaction. It could be pulling H off of certain R by chlorine or by bromine. So it could look like this. So this would be more reactive and that would be less reactive. Higher activation energy.

But consider a similar pair of reactions where you’re pulling the H away from a different R, an R whose bond is easier to break. So the radical here is more stable, at least compared to the starting material–it may, in fact, be the starting materials that are different as we were talking about. So you have a certain difference here, it’s the same difference here–you’re going from the same starting material to the same radical, it’s just that in this case you’re using bromine to pull it off, in this case you’re using chlorine. 

So notice that in this stage it’s not selective because no matter which of the compounds, the blue RH or the red RH, no matter which H you’re pulling off with the chlorine, they have almost exactly the same energy–activation energy, the same rate. But here in the case of bromine it’s going to be more selective because you’re going to have a bigger difference, because it’s further along.

And I’ll stop now and we’ll pick up here next time on this Reactivity-Selectivity Principle–the idea that the less reactive it is the more selective it will be.

Thanks.

[end of transcript]

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