Freshman Organic Chemistry II

CHEM 125b - Lecture 1 - Mechanism: How Energies and Kinetic Order Influence Reaction Rates

Chapter 1. Introduction and Outline of the Course [00:00:00]

Professor Michael McBride: Welcome back. You look healthy and tanned. I hope you’re rested and ready to go again. So welcome back to sunny New Haven.

We’re lucky this semester to have holdover of all the people that are going to help you–the graduate TAs, Jon and Phillip, and the peer tutors, Eva, Jack, Julia. We’re going to have the PowerPoints, timed PowerPoints available on the website after the lecture and the audio as usual.

We’re not going to have a textbook this semester either. But last year I used the Jones and Fleming–this textbook here. So the order I’ll follow is sort of more or less the order that’s there. If you need to buy a textbook, that’s one that would be appropriate. But the one I gave you in the fall will be fine, too. The topics we are doing are generic.

So as your remember the first semester we studied bonding and molecular structure. Then toward the end we did some sort of thermodynamics, which is what connects structure to energy, which is what determines what chemistry will happen. So this semester we’re going to talk about that chemistry, about reaction mechanisms, and about using these reactions to do synthesis. And we’ll also do some spectroscopy around the middle of the semester.

So just a brief excursion through what we’re going to be doing. The first two quarters of the semester are going to be on how mechanisms are discovered and understood in terms of structure and energy–what we talked about last time. So first we’ll look, naturally enough, at the simplest reactions. There can’t be anything much simpler than just cleaving a bond. But remember there’s also the possibility of making a bond at the same time you break a bond; and we introduced these ideas last semester, of course.

So we’re going to start with free-radical substitution, one of the earlier reactions to be studied, as you remember, in the 1830s in France. But we’re going to use it to talk about the concepts of reactivity and selectivity–when you can get different products, how do you control or predict which ones you’ll get?

Then the nice thing about free radicals is they’re not charged, so they’re relatively insensitive to solvent. So you can just think about the molecules themselves reacting, whether it’s in the gas phase or solution, more or less the same thing. But it’s quite different when ions are involved either as starting materials or products or both, because then solvent effects are very important. So as we go from free radicals, we’re then going to do solvent effects.

Then go on to nucleophilic substitution reactions, which almost always involve ions either as starting materials or products or both. And going to use that as an example of showing how people go about “proving” a mechanism. Then we’ll have the first exam on February 2.

Then we’ll go on to electrophilic addition reactions to alkenes and alkynes. You’ll recognize that these are concepts we introduced last semester, but this time we’re going to focus on how you know about how they work. And we’re going to talk there, especially about the role of nucleophiles. You usually call these additions electrophilic, but there’s quite often a very important nucleophilic component as well.

Then we’ll get onto polymers and their properties for just a lecture or two. And then conjugation, aromaticity and pericyclic reactions, which you’ll know what they are when we get to them. Then we’ll have the second exam. And then we’ll have an interlude about spectroscopy and go on to more focus on synthesis.

So first, spectroscopy for structure and also for studying dynamics of molecules and reactions: so ultraviolet-visible spectroscopy; electronic; IR spectroscopy–vibration, which we’ve talked about a good deal already; magnetic resonance imaging and nuclear magnetic resonance. Then on to aromatic substitution, and finally carbonyl chemistry and the concepts of oxidation and reduction. Then we’ll have the third hour exam.

And the last quarter of the course then will be more having to do with carbonyl chemistry; acid derivatives; substitution taking place at carbonyl group; reactivity adjacent to the carbonyl group; the classical condensation reactions; then something about carbohydrates and Fischer’s classical proof of which isomer is which of glucose.

And finally we’ll talk about some complex synthesis. We’ll be talking about synthetic ideas all through this, of both unnatural and natural products. Then we’ll have the final. So that’s how long you have to endure.

Chapter 2. Energy and the Reaction Coordinate [00:04:05]

So at the end of last semester we were talking about energy and how it related to things–that free energy determines what can happen at equilibrium–and remember it was all statistics, and the statistics filtered down to give us this handy relationship at room temperature that an equilibrium constant is 10^–3/4 of the difference in free energy. Or if you’re not worried about entropy, the difference in enthalpy, in the heat of the molecules. So that’s both energy and entropy the Gibbs free energy. 

But there’s a completely independent question, which is “Will it happen?” You know, graphite is more stable than diamond as a form of carbon. But the advertisers tell us that a diamond is forever–just because it’s not as stable doesn’t mean it’ll convert in any reasonable time. So the questions of kinetics are just as important as the questions of equilibrium, or at least almost as important.

And we saw last time that we could approach this by studying lots of trajectories, trying to calculate how all the atoms move in going from one arrangement to another one. But that really provides too much detail. In fact, over the last couple of weeks we’ve had interviews from people who are applying for a faculty position here, and a number of them are talking about fancy new ways of actually looking at one molecule at a time as it reacts. So that you actually could do something getting closer to trajectories, but really that’s more detailed than we want.

We want to summarize things statistically to know what’s going to happen in a flask. So we have collective concepts, enthalpy and entropy, and then we can have a reaction coordinate diagram–remember we rolled a marble on that thing last time, on that potential energy surface. Or we could slice along the potential energy surface and look at the starting material, the transition state and the products. Or we could simplify things to notice that this is just a sequence of three species: the starting material, which is molecules that we know about, the transition state, which we don’t know about, and the products, which we know about.

So our challenge is going to be to try to figure out something about the transition state given what we know about the starting material and the products, because its energy relative to those others is going to determine rates. So we have then free energies with just these three species, rather than trying to look at a detailed trajectory.

OK. Now. So free energy determines what can happen at equilibrium, but also how fast it happens in kinetics, and for that purpose we need to know what the energy is, either the free energy or the heat, at least, of the transition state. OK. And remember we talked about this last time that in Eyring’s transition state theory, the rate constant for the reaction per second is 10¹³ times that sort of equilibrium constant between the starting material and the transition state. So again, we can use at room temperature that 3/4∆G when we express it in kilocalories per mole.

Chapter 3. Bond Strength and the Mechanism of Free-Radical Substitution [00:07:31]

So we want to use energies to predict equilibria and also to predict rates first for the very simplest one-step reactions. And no reaction is conceptually simpler than just breaking a bond in the gas phase to give atoms or free radicals. So we need to know the energy for that.

Now, in the textbooks that I’ve handed out to you there are tables like this one that give bond dissociation energies. How much energy does is actually take to break a bond? This particular one is from a text we used to use in the course called Streitwieser and Heathcock. That was in 1993, but as of 2003 there’s a new set. And you can see that these don’t change very much in time–98 became 99, 111 became 113. Some of them are a little bigger–81 became 85, and so on.

And these values I refer to as Ellison’s. I mentioned him last semester–there’s Barney Ellison. He and his friends compiled these new values, and he’s going to come and talk to you about how he did this in April some time. He can’t get out of Boulder now–he’s all snowed in out there, in Colorado. 

So this is his table of molecular bond dissociation energies for losing H from something. R-H becomes R plus H atom—R radical plus H atom. So these are experimental bond enthalpies. And as you can see, some of them are known to very high precision, like H₂ is known to six significant figures. Most of them, of course, aren’t known that well, but most of the ones in this table are known pretty darn well, plenty accurately enough for our purposes.

OK. Let’s see if we can understand some of these, so that we don’t have to memorize the table. Let’s look at the bonds from H to halogen. You’ll notice as we go down to larger halogens from fluorine, chlorine, bromine, iodine, we go from 136 to 103 to 87 to 71–quite a difference in the energy involved. So the idea is that larger halogens don’t overlap as well with the hydrogens, so they don’t make as strong a bond at their normal bond distances. And there’s less electron transfer to the halogen.

So whereas HF looks like that and the two electrons are lowered quite a bit, in HI the net amount of lowering of the electrons in forming the bond isn’t nearly as much. So that makes sense to us. So less electron stabilization means a weaker bond.

Now here’s the Table 2, which is bond enthalpies for various atoms or groups attached to various hydrocarbon radicals. Previously we looked at H attached to halogen, now we’re looking at methyl attached to halogen and you see there’s a very similar trend. It’s strongest for fluorine, weaker for chlorine, weaker still for bromine, and weakest for iodine. So many of the same features.

But let’s try to understand different radicals. So we’re going to look at the bonding of H to methyl, ethyl, isopropyl and t-butyl. And the first thing you notice about this is that they’re all almost the same. As we went down the halogens it varied by 60 kilocalories per mole. But as we go across these different methyl, ethyl, isopropyl, t-butyls, they’re all 100 kilocalories per mole, plus or minus 5. We’re going to look next lecture at this in some more detail and see if we can understand why they vary at all. But they don’t vary very much. 

On the other hand, some hydrocarbon radicals have substantially different ones, like vinyl and allyl, phenyl and benzyl. The vinyl and phenyl are much stronger, 10 kilocalories stronger. The benzyl and allyl are about 10 kilocalories weaker. So let’s see if, based on what we did last semester, we can understand why this might be so.

Are these unusual bond dissociation energies to be explained as unusual bonds, that is the bonds are unusually strong or weak? Or is it the radicals are unusually strong or weak? This is compared to what? We have the starting material and go to the products where the bond is broken; we go uphill in energy by these amounts. The question is: are things unusual because the starting bond is weak or unusually strong? Or is it because the radical is unusually stable or unusually unstable? Compared to what?

So let’s look at a couple of these cases and see if we can understand. First let’s look at vinyl, which is the H attached to a carbon that’s double bonded. Now if we look at the radical, we see that we have a SOMO, so there’s the possibility for interacting with something and changing its energy. But notice that the HOMOs and LUMOs that it might interact with are π and π*. And there’s no special stabilization because they’re perpendicular to one another, they’re orthogonal, there’s no overlap. So there’s no reason that having this double bond should make that radical unusually stable. And notice that, it, in fact, is hard to break as if this were unstable. But it’s neither stable or unstable, it’s just what it is. 

On the other hand, if we look at the starting material where we have this C-H bond, notice that as compared to these others, which are 100 plus or minus 5, that this one is made from an sp² carbon, right? So very good overlap. So in this case, it’s that the bond is unusual. The bond is unusually strong; the radical is nothing special. It’s very hard to break. So it’s hard, it’s 111.

The same thing is true in the phenyl radical where, again, the C-H bond we’re talking about is attached to a double bonded carbon. Here it’s part of a benzene ring. But again, the unusual energy orbitals in the ring are perpendicular; they don’t overlap with the singly occupied orbital we’re talking about. Therefore, nothing special here. But again, it’s an sp² carbon to hydrogen bond, and again, it’s unusually strong, 113 kilocalories per mole. So these are unusual because the starting material has an unusually strong bond.

Now let’s look at these others, allyl and benzyl, where it’s an unusually easy bond to break. Now we look at the allyl radical where we’ve broken H off this carbon, and now what’s different about that as compared to the ones above? We broke the H off, we got a p orbital that has the single occupancy. Anybody got an idea about whether it’s going to be usual or unusual? Yeah, Sebastian?

Student: π*.

Professor Michael McBride: Pardon me?

Student: It can mix with the π*.

Professor Michael McBride: Now it can mix with the π* because it’s going in and out of the plane. So now it overlaps with the π and π*. Now what’s that going to do?

Well here’s the singly occupied orbital, the red one here. Here’s π*, vacant. They’ll mix. That would suggest that the single electron would go down in energy. That would be good. However, we should also think about the fact that there’s an unusually high HOMO associated with the blue orbitals there, the π orbital. So we could consider instead the π. And now we see that that would shift this electron up, but these two electrons would go down. Again you would win – more down than up.

Now. So this is a little schizophrenic on the part of the singly occupied orbital. Does it move up or down? The answer is “No.” One of them pushes it down, the other pushes it up, it stays the same place it started. But the others go up and down, the ones that came from the π and the π*. So you get net stabilization due to this pair of electrons going down. So we have this special “allylic,” it’s called, stabilization from mixing the SOMO with the π and π* orbitals adjacent to one another. So this radical is unusually stable.

On the other hand, in the starting material, the bond was just a regular old sp³ C-H bond, nothing special there. So as compared to here, where the starting material was unusually strong, here the product is unusually stable. So that was easy to break, 89 kilocalories per mole. And the same is true for the benzyl radical where the p orbital on the adjacent carbon, again can overlap with the π system of the benzene ring. So you get special stability and it only takes 90 kilocalories per mole.

OK. So we can understand these special cases of bond strengths, of bond enthalpies.

Now, we spoke last semester about the halogenation of alkanes. So we can use these bond dissociation energies to do some calculations about the possibility of doing halogenation of alkanes. So let’s just take as an example methane, some various dihalogen molecules, and we can trade partners, a double displacement sort of reaction, as they called it in the early days, and make CH₃X and HX.

So we break the red bonds and form the green bonds. So we have a cost to pay in breaking bonds, we have a return from making the bonds, and we’re going to see whether we get a net profit from trying to run this operation. 

So we’re going to do it for fluorine, chlorine, bromine and iodine. Now, of course, in all those cases the C-H bond is the same, it’s 105 kilocalories per mole.

Now, but the halogen-halogen bonds are different. It’s fairly weak for fluorine, strong for chlorine, but then weaker again for bromine, and back to the start for diiodine. What does that tell you that as you go down the rows of the periodic table, you don’t get some monotonic up or down in the bond strength, but it goes up and down? What do you infer from that kind of thing when you get curves?

Student: There must be two factors.

Professor Michael McBride: There must be at least two factors involved in this. And those factors are probably the overlap of the σ bond, which is best for fluorine-fluorine, but the interaction of unshared pairs which is worst for fluorine. So you have two things going on. OK. Anyhow, we can add those two together to see what it’s going to cost us to break those two bonds. And there are the values: they’re in the range of 150 kilocalories per mole. 

But then, when we do the reaction, we’re going to profit by making the bonds on the right, so that we have various values for the carbon-X bond that we got out of that table we just showed. And for the H-X bond, and there’s the return we get.

Now, are these going to be favorable reactions? Well, in the case of fluorine it’s wildly favorable. It’s favorable by 109 kilocalories per mole. But it’s only 19 kilocalories per mole for chlorine, only 9 for bromine, and it’s 12 kilocalories UPHILL in that case of iodine. So you’re not going to make a profit if you try to set up a factory to convert methane and iodine to methyl iodide and HI. In fact, you’d want to run at the opposite direction. Right? But the others, at least overall, are favorable. So we know from equilibrium now that for three of the halogens this reaction should be favorable. And for one of them it should be unfavorable.

But how about the rate? How fast will it be? Now if this were the mechanism, first you break two bonds, then you change partners and re-attach, the activation energy for getting up to the transition state for this where both bonds are broken is going to be this cost. Right? So is break-two-bonds-then-make-two a plausible mechanism? Could we get a rate that’s reasonable on the basis of that? If you know the activation energy you have to get to, how do you go about calculating the rate? Debby?

Student: 10¹³.

Professor Michael McBride: Right, it’s 10¹³ per second, times 10 to the–

Student: Negative ¾. 

Professor Michael McBride: 10^–3/4 of that energy. OK. So let’s think about it then in this case. Suppose we take one of these that’s about 140. So at room temperature, 300K, which is where your approximation works, the 3/4, we find that it’s 10¹³ per second times 10^–106, 3/4 of one of these numbers. Right? So it’ll be 10^–93 per second. That’s a very unfavorable number. So forget that. There’s no way a mechanism like that could work at room temperature.

What might you be able to do to realize such a reaction? Yeah, Amy?

Student: Heat it.

Professor Michael McBride: Higher temperature. So suppose you go to 3,000K instead of 300K? Now instead of 3/4 it’s 3/40. So if we go to 3,000K, then it’s 10¹³ times 10^–10.6, which would be 250 times a second. Right? So that would be accessible. Except that there probably is something else that could happen when you’re at that very high temperature. So fundamentally this is not a reasonable mechanism.

And we already know a better way to go about it from looking at these potential energy surfaces for transferring an atom between two other atoms. Remember we start with H plus H₂, and then we can move around on this surface and have, for collinear three Hs, we can get the energy differences between two of the Hs and the other two Hs. 

So there’s the product we want to go to where a hydrogen has been transferred from the right to the left. And one way to do it, the way we’ve been talking about, is to go by way of three separate atoms. Which means we go up there, way up in energy, very slow, and once we get there then zing, we go down. But it’s very hard to get up to this plateau. 

On the other hand, as we knew from the marble, the way it actually works is to transfer a hydrogen between two rather than to break a bond first and then make a bond. So make it as you break it that way and that’s going to be much faster. So this kind of displacement reaction is the way we want to do it–to make a new bond as we break the old one. 

OK. That’s how we’re going to do it, and we already talked about this last semester. First you take chlorine and you break its bond. That’s going to require some energy. But once you have the free radical, then you can transfer a hydrogen atom making a bond as you break a bond. So going through that pass rather than over the big plateau.

And now the product of that is HCl, one of the products you want. But the other one is another free radical made without having to spontaneously break a bond. And now it can react with Cl₂ to give the other product we want and the chlorine atom, and then we go back to the beginning in this free-radical chain, and this is what we talked about last semester.

OK, another way of writing the same thing is to say we could have the starting materials here to the top left and bottom right, the halogen and the alkane, and we could have the X atom that we got somehow.  It could be by heat breaking the relatively weak bond, it could be by light, it could be by some chemical reaction that generates a free radical, but somehow we get a radical. Then it can react in a hydrogen atom transfer to give one of the products, the green XH here, and another radical, and it could react with the halogen to give the other product and go back where we started.

So what we have is a cyclic machine that just cranks round and round.  It preserves the radicalness of the situation. You don’t have to break any new bonds spontaneously – only to get the first radical.  We’re going to talk about this some more later on

But to look now at how we’re going to rearrange things – if you want to have a mechanism for a reasonable rate, what you’re going to do is trade two of these columns. Right? Instead of breaking two and then making two, we’ll trade those two columns so that on the left we make and break; on the right in Step 2 we make and break.

So now when we look at how much each of these takes, Step 1 now in the case of fluorine is actually favorable. Right? Step 2 is wildly favorable in the case of fluorine and overall it’s very favorable, as we knew. But now you can see that these others, although overall favorable, one of the steps is unfavorable, but not drastically unfavorable–only 2 or 10 or 20 kilocalories per mole. That kind of barrier you could get over; you can’t get over the barriers of 50 kilocalories or more. 

So now you have two steps, each of them a very simple make-as-you-break atom transfer. And, of course, here the first one is so high that you’re not going to be able to get over it, and even if you could you wouldn’t give the product, the reaction would run backwards.

OK. So we know the energies of the starting material and products for each of these steps, but we don’t really know the activation energy–how much you have to come up before you come down. And that’s what we’re going to be starting to look at next lecture, is how could we predict the activation energy for a simple one-step reaction if we know how exothermic or endothermic it is.

Chapter 4. Complex Reactions and Kinetic Order [00:27:23]

But even if we could predict the rate of Step 1 or Step 2, there’s still a problem, because the overall process is not just one step, it’s two steps. And how would we know the overall rate if there are two reaction steps going on? This is a little more complicated. So we’re going to take a little digression now to learn to cope with complex reactions that involve several steps. So this is a digression on reaction order and complex reaction.

Now the reaction order is the kinetic, the rate, analog of the law of mass action that we talked about at the end last time–remember, it was just statistical, how likely is it that there are going to be two things next to one another? The same thing, getting next to one another is the same problem, if things are going to react with one another in getting to the transition state as well.

So there’s going to be a dependence of rate on concentration, the same way that equilibrium depends on concentrations, as we saw last semester. Moreover, this dependence of the rate on concentration can give us information about what kind of mechanism the reaction might be undergoing. So that’s why we’re particularly interested in it. 

Now just to look at some simple ideas first. The rate is how much per second, say, or per minute or hour a day or whatever, how much per second. So we could have a faucet, we turn it on and the water comes out a certain rate per second. Now what would happen if you had two faucets? Well, if things were simple the rate would double. You know, if you think carefully about it, it won’t quite do that because there might be a restriction in how much water can get to the two faucets. But in a simple point of view if you double the number of faucets, you double the rate.

There’s another way of doubling the rate, which is to use a bigger faucet, one that’s twice as big, twice the cross-sectional area.

Now in chemistry we can look at how many moles or molecules we get per second. And if we have one beaker and then have another beaker, or a single beaker with twice the volume, we double the rate. So that will double the rate. But chemists are clever. They can not only change the volume, they can also change the concentration in the same volume, and that’s the question we have. Obviously you’d expect it to get faster if you have more stuff, but how much faster? Will it double?

So this has to do with what are called rate laws. What that exponent is going to be that says if you double the concentration of a reagent, how much will the rate change, the number of molecules you get per second? So the rate is the increase in product with time. Then there’s a rate law that’s some constant times the concentration of something. And maybe several concentrations, but that question mark is the exponent–what power do we raise it to? That’s what we mean in talking about the kinetic order. If it’s to the zeroth power, it’s a zeroth order; first power, first order; second power, second order and so on. But those exponents will have to do with what the mechanism of the reaction is. So the only way to discover it is by experiment.

So let’s first look at some simple one-step reactions, and then we’re going to look at complicated reactions. So in a simple one-step reaction you could have zero order–this sounds weird, doesn’t it? That the rate doesn’t depend on how much material you have. You double the amount of material, double the concentration of material, and it doesn’t change the rate.

How can that be? Well here’s a picture that helps illustrate it for sheep getting from one field to the other. Now in this situation, if the shepherd had more sheep, would the rate increase? Would you get sheep from one field to the other faster? No, they’re going as fast as they can no matter how many sheep are lined up. Because it’s saturated. It can’t go any faster because of how wide he’s opening the gate.

So there are real cases like this in chemistry where you have a catalyst that’s involved in converting the starting material to the product. When the catalyst is working as fast as it possibly can, increasing the starting material isn’t going to increase the rate of the reaction anymore. The gate can’t open any wider.

And that’s often the case with enzymes. So you have the substrate, the starting material, which interacts with the catalyst in order to get a product. But if the catalyst is rate-limiting, if it’s saturated, you can’t go any faster. So the rate then is proportional to how much catalyst you have, of course. If you doubled the amount of catalyst under those situations you’d double the rate. But the substrate is raised to the zero power, it doesn’t make any difference.

So the rate is actually just proportional to the catalyst, independent of the substrate. But if doing the experiment you didn’t realize that there was a catalyst in there doing it, you just thought it was happening spontaneously, then you would say it’s zero order. So that’s a zero order reaction, zero order kinetics.

But that works only when the sheep are really crowded here. If the sheep were all over the fields and some coming through, if the concentration were lower, then having more sheep would allow them to get through faster. You’re not saturating the catalyst. But it becomes first order in substrate when you have very low concentration of substrate, but at some higher concentration it becomes zero order. So it’s always, in principle, possible to lower the concentration until it becomes first order. But, in fact, under many conditions it would be zero order.

So that’s an example of zero order. Now, first order says the rate is proportional to the first power of the amount of reagent you’re using. And that’s, of course, very reasonable–it’s like the water coming through the taps or something like that.

Now if you have first-order kinetics, the amount of product you get in time looks like this. It’s an exponential approach to the 100%. Or this particular one is drawn with a certain rate constant. If the rate constant were faster it would rise more quickly. If the rate constant were lower it would rise more slowly, but always it would go, in this case, to 100%.

You could, instead of monitoring the product, you could look at the disappearance of starting material, which is then an exponential decay, just the same thing upside-down.

So exponential decay means you have a constant, so-called half-life. Now I chose 0.69 here for a reason. It’s because the half-life, the time it takes for half the material to go away, is 0.69 divided by whatever k is. So if I chose k to be 0.69, the half-life is 1. So after 1 second, half of the material remains. After another 1 second, a quarter, then an eighth, then a sixteenth and so on. And that’s the necessary–you can prove it easily by mathematics for a first-order reaction. There is a repeating half-life.

Now, suppose you have a more complex situation where it’s first-order kinetics but it’s reversible. So the starting material at a certain rate goes to product, but the product can also come back to starting material, suppose at a much lower rate. So once you reach equilibrium the two rates balance, so things cease changing.

In fact, you know, in the nineteenth century, equilibrium situations were initially discussed not as equilibria, but as balanced rates. So at any rate, they’re the same thing. So whatever k₁ is times the concentration of starting material will be equal to whatever k_-1 is times the concentration of product. Or we could divide through by starting material here and k_-1 over here. So k₁/k_-1 is the equilibrium constant.

So now the product doesn’t go to 100% nor the starting material to zero. But if you start with all starting material, you get these exponential behaviors, it’s still exponential. That’s what’s interesting. But it doesn’t approach zero. So here, after 1 second where with that rate constant you would have fallen to half, you don’t fall quite to half, because some stuff is coming back.

But if you go to the ultimate goal of equilibrium here, then you find–notice the equilibrium constant is 3, I chose that to be a third of that. So there’s one quarter of the red material and three quarters of the blue material at equilibrium. But if we draw this at 25%, the value that the starting material actually approaches asymptotically, exponentially, we find now there is a repeating half-life–this is truly an exponential. But the interesting thing is that it’s an exponential decay to the equilibrium mixture, but the half-life is 0.69 divided by the SUM of the two rate constants.

Again, this is easy to demonstrate using calculus, but it’s just sort of cute that you add forward and reverse rate constants together to get how rapidly it approaches equilibrium. So they’re first-order reactions, both just one-way and reversible.

Now, a second-order reaction is proportional to the square of A. That’s obviously true if two A’s have to get together in order to reach the transition state. But it could also be second order if you have an A interacting with a B. In that case you say it’s first order in A and first order in B and second order overall.

But what if B is effectively constant? So as time goes on, A is decreasing but B is not decreasing. How could that possibly be the case? How could you have a reaction where B doesn’t decrease?

Student: B is a catalyst.

Professor Michael McBride: Ah, if B were a catalyst, it doesn’t get consumed. That’s one way. Another way is that the B is in gross excess compared to A. So even though A consumes a B, it doesn’t change its concentration appreciably, because the concentration is so large compared to that of A. So if [B] is much greater than [A] you get that. And in that situation you say that you can neglect [B] by incorporating it into the k. Whatever [B] is, it’s not changing.

For all you know when you’re doing the experiment, you might not even though there’s a catalyst there, but it’s there in always the same amount, you always get the same rate constant. And in a situation like that you say it’s pseudo first order. It behaves as if it’s first order, even though it really depends on something else as well. And so a second-order process could be a pseudo first order rate constant.

For example, suppose B were the solvent. Then it appears that it’s a first-order reaction, even though it turned out that the solvent had to react with the thing in order to do it. So that’s pseudo first order.

Now, if you have a second-order reaction compared to a first-order reaction, that is a regular second-order reaction where two molecules of A have to get together, what you find is that a first-order reaction goes like this. But the second order starts faster and ends slower. So it’s no longer exponential.

Obviously it’s decreasing but not with a half-life, not exponentially, not a repeating half-life, because it gets slower as it goes along, because it depends on the square of the concentration. So it slows faster. It’s not exponential. There is no constant half-life. So that’s how you would know that something is second order. You measure how much in the first increment of time, the second increment, and they’re not exponential.

Now we’ll get on to complex reactions and the idea of the rate-limiting step. So suppose that you go to an intermediate and then it goes to product. But suppose the transition state energies are like this. So it’s very slow to get to the intermediate. But once you’re at the intermediate it rapidly goes to product.

Now this reactive intermediate never builds up, you never have an appreciable amount of it, because even if it were at equilibrium with this there’s very little of it with a big energy difference here. So you might not even know that the intermediate is there. You get there and it immediately goes to product. In that case, who cares really about what rate this is, as long as it’s fast? The rate at which you get product is the rate at which you get this. It never builds up. Every time you get it, it goes immediately to product, or very quickly.

So this first step then, even though it’s a two-step reaction, the first step is the one whose rate makes a difference. If you double the rate of the first step then you’ll double the overall rate of the reaction, because the second step doesn’t make any difference. So that’s called the rate-limiting step.

On the other hand, suppose that that first reaction is fast and the second one is slow. So you have a rapid pre-equilibrium formed between these two–again, not very much of the intermediate. But it reaches an equilibrium concentration compared to this, or almost, and then slowly it goes to product.

Now two things that are in equilibrium with the same thing are in equilibrium with each other. So to the extent that that intermediate is in equilibrium with the starting material, and we treat the rate of going over here as an equilibrium between the intermediate and this transition state, we could also pretend that the transition state is in equilibrium with the starting material, since they’re both in equilibrium with the intermediate. 

So now, who cares that there’s an intermediate, that there’s a first step. We can calculate the rate just on the basis of getting how much of this transition state there is by assuming it’s in equilibrium with that starting material. So now the second step or the second transition state is rate-limiting. All we have to know is how high it is compared to the starting material. We don’t care about the intermediate.

So you can have one step or the other step be rate-limiting. Or if they’re not drastically different, both of them could affect the rate. Many cases one or the other is rate-limiting. But if you want to see how it works out in a complicated case, where you have a starting material that can go reversibly to the intermediate, and then the intermediate goes to product, all with certain rate constants, on the website you can get an Excel program, which is a crummy Excel program. I’m sorry, it executes very slowly. You can change the parameters to make it go a little faster. But you can put in what energy you want for the starting material, the intermediate and the product; and what energy you want for the transition states, one and two, and see how the overall rate compares with what you would have calculated if it had been just the first step that you had to get over, or just the second step that you had to get over.

In this particular case where each of these is about between a kilocalorie and a kilocalorie and a half, the difference between those two and the difference between these two, you get this. So this is the actual rate. Notice the amount of starting material immediately falls quickly–that’s when you’re establishing equilibrium between these two. So starting material is going to intermediate. And then slowly it goes away as the intermediate goes to product.

Here is the intermediate. It immediately builds up to some concentration and then it changes very slowly. It stays fairly stable for any short period of time.

Now, if you didn’t have the first barrier and had only the second one, it was a one-step reaction. This is what you would have calculated for those same energies. So not too bad. So, in fact, the second transition state is pretty much rate-limiting–you get pretty much the same result.

On the other end, if transition state two weren’t there and you only had to get over transition state one, then you’d have that blue one there, if it were the sole barrier. So very far from that. But even in this situation, you don’t make a very big mistake if you just ignored the intermediate and the first transition state, and said it was only transition state two. It’s rate limiting. And if you fiddle with this, you’ll find that if you start changing these differences very much, then one of them or the other one becomes clearly rate-limiting–you get much better agreement. 

Now, I said that this was between a kilocalorie and a kilocalorie and a half, 3/4 of that is about one, so the equilibrium ratio is about 10–or actually, I said 9 here, 9 times as much starting material as intermediate. So once the intermediate reaches its steady sort of equilibrium relationship with the starting material, this very quick rise here, once you get there, then there’s 9 times as much starting material as there is in intermediate. That’s 9 times as much as this. And that persists then, that ratio. When you get to here it’s 9 times as much of one or the other and so on.

Now this is also a factor of 10 between this. But this has to do with rates, remember. So this is how fast the intermediate goes back to starting material is 10 times faster than how fast it goes to product because of those differences in the ∆H of activation. And what that means is that once the intermediate reaches its steady state, its equilibrium relationship with the starting material, it yields product 1/10 as fast as it is formed.

So it forms very rapidly, you can see that here. But now individual molecules are still being formed just as rapidly, or almost as rapidly, but they’re going away just as rapidly as they’re formed. Every one that’s formed at a certain rate goes back to starting material 10 times for every time it goes to product–9 times for every time it goes to product. So 1/10 of the time it goes on to product.

So now the rate of going from starting material to product is however fast it goes over the first one times 1/10, because it’s not changing. It’s reached the so-called steady state. It’s changing very slowly. Individual molecules are going very rapidly. So the rate of an individual molecule to go from starting material to product is how fast it goes to intermediate times 1/10.

So that’s a sequential reaction–the idea of the rate-limiting step.

Now a really interesting concept is fractional order. But I see that we’ve reached the time now, so we’ll have to wait for fractional order until the next lecture. 

[end of transcript]

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