CHEM 125b: Freshman Organic Chemistry II

Lecture 10

 - Cation Intermediates – Alkenes: Formation, Addition, and Stability


Bridged pentavalent carbon structures can be intermediates or transition states of cation rearrangement during SN1 reactions, and short-lived ion pairs explain net stereochemical inversion. The different perspectives of preparative organic chemists and mechanistic organic chemists on reaction yields are illustrated by a study designed to demonstrate that molecular rotation can be rate-limiting in viscous solvents. “Electrophilic” addition to alkenes is the reverse of E2 or E1 reaction, and its mechanisms can be studied by analogous techniques. The NIST Webbook provides thermochemical data to help understand the relative stability of isomeric alkenes.

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Freshman Organic Chemistry II

CHEM 125b - Lecture 10 - Cation Intermediates – Alkenes: Formation, Addition, and Stability

Chapter 1. Rearrangement of Cation Intermediates [00:00:00]

Professor Michael McBride: OK, so we’re going to finish up on nucleophilic substitution, elimination reactions today with an example of practical application of them. And then go on to the reverse reaction of elimination, which is addition to alkenes, so a new functional group. And we’ll be talking about the stability of them today, the stability alkenes.

OK, so last time we got started on the silver-assisted substitution reaction of neopentyl iodide. And notice something interesting in this, if we look at those red crosses that trace the carbon skeleton of the starting material and the product, what do you notice about the carbon skeleton? Does it look like a normal substitution? Ellen, what do you say?

Student: No.

Professor Michael McBride: Why?

Student: There’s an extra carbon.

Professor Michael McBride: There’s not an extra carbon. There are five carbons on the left, and there are five carbons on the right, but they’re not hooked together the same way. So there’s been a rearrangement of the carbon skeleton during this reaction. So how did this happen?

Well, silver comes down and pulls off iodide, which would generate this cation. But in a cation we have a low LUMO, obviously, on the cationic carbon, but there are HOMOs next door. They’re not very high HOMOs. They’re the C-C σ-bonds. But you could imagine that since primary cations are rather unstable, and we’ll give some more evidence on this later on, that you could get a rearrangement in which first, the HOMO next door is shared with the LUMO of the carbon cation. Did you ever see a structure like that before, where the pair of electrons is shared with an adjacent atom? That’s the Y-bond, right? The three-center, two-electron electron bond.

And then it could keep going, which would give the tertiary cation, a carbon cation with three other carbons attached to it, which is much more stable than the primary cation is. In fact, it’s possible that it doesn’t even go that way. That at the same time the iodide is leaving and beginning to develop this unstable primary cation, the methyl next door should shift across, so that you directly get the cationic rearrangement product.

But at any rate, once you have the carbons rearranged like that, it’s a simple matter to have–and notice what that means is, that the [gr[s of the carbon-carbon σ-bond is actually acting as a nucleophile a displacement reaction. The carbon next door is backside attack as the iodide is leaving, but it’s within the molecule that the reaction is occurring. But then you can bring water down of course, and finally lose a proton, which will give the product. But the crucial thing is this rearrangement, because the fact of the rearrangement shows that there must have been a cation intermediate. Without that cation, there would have been no means for the rearrangement to take place in the carbon skeleton. Carbons skeletons don’t normally rearrange. But when you lose electrons and can get these two-electron, three-center bonds, then a carbon could move from one place to the adjacent one. So the fact of the rearrangement shows that this was an SN1 reaction, that there was a cation intermediate. Had there been no cation intermediate, there couldn’t have been a rearrangement.

OK, let’s think just a second about the arrangements–so–we and about this three-center, two-electron bond. So we could imagine bringing a proton down, mixing with the π electrons to form a three-center, two-electron bond. We could also imagine bringing a hydrogen down on one end of an alkene to form a normal kind of cation. And of course it could go on the other end too. You can get from one of these to the next by shifting the hydrogen from left to right. That’s the rearrangement. But the question is, is it a single minimum, or is it a double minimum? Do you have two normal cations which have a transition state they can go across back and forth between them that’s not too high, because it has this three-center, two-electron bond? Or might it be that that bridged structure is in fact the lowest energy structure, and it’s a single minimum?

So you can do simple calculations to check this, and you may or may not believe them, but these particular ones are well supported by experimental evidence that we don’t have time to go into right now. But if you take the open compound, that’s what its HOMO looks like. And you notice that the HOMO, which you would have guessed from the structure on the top is the σCH bond, in fact is mixed with some of the electron density in that HOMO, or HOMO–1 it is, is shared over onto the cationic carbon. It’s distorted; you can see that the electrons are shifting that way. And the LUMO is the other way around, that there’s something vacancy of the orbitals that are involved in the CH bond.

So you could imagine that resonance structure, the one that’s drawn in color just below to the right, but you could also have this structure, which is a proton plus a carbon-carbon double bond, or you could have a resonance structure that has a long, stretched bond. Remember when you draw resonance structures, you don’t move the nuclei. So you don’t draw that hydrogen over on the right, you draw it with a really long bond, because they have to be at the same position. And if you take those three simultaneously into account, it says you can have positive charge either on the right carbon, on the hydrogen, or on the left carbon. So that’s what’s denoted by drawing this Y, that the two electrons are shared among these three centers, and the positive charge, therefore, is shared. And we can see it in the orbitals that that’s the way it works. And of course it makes sense that you should have that. Of course, the CH σ bond isn’t a very high HOMO, so it’s not drastically shifted, not 50/50. And we call that, the stabilization due to that HOMO-LUMO mixing, hyperconjugation, I think we’ve talked about that before.

And incidentally, that’s one of the factors that makes highly substituted carbon cations stable. Because if you have a hydrogen next door, there’s no way that the electrons involved in that bond can mix with the p orbital, the vacant p orbital, because they’re orthogonal. But if you have a carbon and then a bond, the electrons in that bond can overlap. So the more carbons you have attached to this carbon, the more bonds there are whose electrons could be stabilized by mixing with the vacant orbital. So that hyperconjugation is one reason that more substituted cations are more stable.

Now if we go to the bridged structure, which is actually symmetrically bridged, although the program drew that for its own reasons with a different kind of bond, and if we look at the HOMO, you can see that it’s symmetrical. You can see it’s the π orbital, but it’s distorted because the hydrogen orbital is mixed with it. And the LUMO is the antibonding combination, or the LUMO + 1 in this case. So you can see how those orbitals participate. But in fact, in this calculation, it’s a double minimum. The open structure is lower in energy than the bridge structure, so it’s a double minimum. But it’s different, instead of a hydrogen on the top, you have a methyl group on top. Because the methyl has a somewhat bigger orbital than hydrogen does, so it’s better at overlapping two neighbors than hydrogen was. 

So now it turns out to be a single minimum, that’s the lowest energy structure. And again, you can see the HOMO, which is a p orbital on the carbon overlapping with π, and the LUMO, which I put in only because I’ve been drawing these things, but I wouldn’t spend a lot of time worrying about that one. But at any rate it’s a single minimum, so that structure is in fact bridged.

Now here I’ve made the alkene, the C-C double bond, unsymmetrical, such that a carbon cation would be more stable on one side than on the other, one side is more substituted. And in this case, it distorts so that it’s sort of a double minimum, but the one side is very low, so it’s an unsymmetrical cation, and open, not strongly bridged. That’s its structure, and there’s the HOMO, which again is somewhat mixed, but don’t waste time looking at those things. I just put them in because I had them. But anyhow, it’s a sort of double minimum in the sense that it’s not a bridged structure, but it’s unsymmetrical and open.

Chapter 2. Studying SN1 and E1 Mechanisms: Stereochemistry and Rate [00:10:20]

OK now, SN1 and E1, there’s an interesting thing about the stereochemistry of SN1 reactions. If you substitute the chloride here, this tertiary chloride, you could imagine getting either the OH group substituted directly where the chloride was, or the OH group on the opposite face of the chloride. Now what would you expect if it was SN1? What about if the SN1–so the rate determining step is ionization, you get a cation intermediate, what structure do you expect for a carbon cation? We talked about this last semester in XH3. So it’s the same number of electrons as BH3. There’re no electrons in that fourth orbital. What should that orbital be? There are four atomic orbitals on the carbon, or boron in BH3, but you only have electrons in three of them. How should you hybridize it? sp2 to get the strongest bonds, and leave the vacancy in the highest-energy orbital, the p orbital, so it should be planar.

Now if you get a planar intermediate, how much attack should be from one side versus how much on the other? Should be 50/50. But in fact, when you do the experiment, as it says here, you get 40% retention, and 60% inversion. So 80% of it is racemization, that’s what you expect, that is 40% of right, 40% of left handed. But the other 20% is excess inversion. It should have been 100% [racemization] if there was a planar intermediate cation. Can anybody see how you might explain this? How could it be that you indeed form a cation, but it’s not equally attacked from the two faces? Mostly it’s attacked from the face opposite the chloride ion, where the chloride ion came from.

Why should it be mostly opposite where the chloride came from? Yeah, Nathan. 

Student: Steric-steric hindrance.

Professor Michael McBride: Can’t hear very well.

Student: Steric-steric hindrance? 

Professor Michael McBride: Steric hindrance of what?

Student: The electrons in the chlorine atom.

Professor Michael McBride: Right, the chloride, even though it’s broken the bond, for a certain time is still around, blocking access to the side it came from. So the idea to explain the net inversion is that that chloride, for a short time, is still there, even though it’s not bonded. So it temporarily blocks the face that would lead to retention to give the same stereochemistry coming in where the chloride was, and you get more of the water coming in from the opposite face. So this short-lived ion pair gives an explanation for the net inversion. So even though people would usually say that SN1 gives racemization, in truth it doesn’t always do so, although it depends on the solvent. Because the more the solvent can stabilize ions, the faster it can get the ions apart from one another, the less of this affect you would have. So it depends on the polarity of the solvent. That was acetone and water. 

We talked about this elimination reaction; the t-butyl bromide with sodium hydroxide gives a large amount of the alkene. But is that process that gives the alkene E1 or E2? That is, is the rate-determining step formation of the t-butyl cation, and then the hydroxide attacks it? Or is the rate-determining step attack of the hydroxide on a hydrogen, pulling it off at the same time that the chloride leaves? So how could you tell? How could you tell whether it’s E1 or E2? Luke?

Student: Maybe test the rate?

Professor Michael McBride: What about the rate?

Student: See if it’s bimolecular.

Professor Michael McBride: That’s what the names mean. So you could follow the names. E2 means it’s bimolecular, that the rate depends on the concentration of hydroxide. If the rate-limiting step were breaking the bond, no hydroxide involved there, so the rate wouldn’t depend on that. So you can look at the overall rate, not just the percent of alkene, because more hydroxide could mean that, even though in the rate determining step you made the cation, the more hydroxide, the more hydrogens you pull off, compared to water attacking. So if you see the rate, not just the product ratio, but the rate depends on hydroxide, then it’s E2.

OK, another way to see whether the proton is being pulled away in the rate-limiting step is to look for a kinetic isotope effect. Remember that? Hydrogen is faster than deuterium by a factor of up to about six or seven usually, only when you’re breaking the bond in the transition state. If the transition state is forming the ions, and later you break the proton away, then you wouldn’t see a hydrogen being faster than deuterium.

OK, now this was a question I asked on an exam once. “Give an example of the influence of a change in reactant structure on the ratio of SN2 to E2 products. Be as specific and quantitative as you can.” And the simplest-minded answer to such a thing, it would be neat to have a more subtle example, but a really simple one is methyl bromide. How about the amount of SN2 and E2 in that case? Anybody venture a guess? This is in fact a trick question. Wojciech. I can’t hear.

Student: There cannot be an elimination.

Professor Michael McBride: There are not two carbons, you can’t make an alkene, there’s no hydrogen on the adjacent carbon, so obviously that’s going to be all SN2. So that was sort of a trick. But then in this one, we’ve seen from last time that it’s much easier to do with a high concentration of hydroxide, it’s much easier to do SN2 than E2, to the extent–pardon me, to do E2 than SN2, because the SN2 is so hindered. If you get substitution, it’s SN1 in the cases we saw last time and looking at the rates of the various RBr’s.

Chapter 3. Preparative and Mechanistic Perspectives on Competing Reactions [00:17:13]

Now these are really drastic ratios. Now there are two different organic chemical perspectives on situations where you have drastic ratios. One is the perspective of a synthetic organic chemist, who says it’s great to have drastic ratios, because then you have a reliable way to get a really high yield. It’s not going to be 40/60 or something like that, and you have to separate and throw away the stuff you don’t want. You get all the stuff you want, that’s great. But for a physical organic chemist trying to understand mechanism, if you have a really drastic ratio, then the energy of activation is quite different between the two, and you can often get an unambiguous interpretation of what the cause of what’s determining the rate if you can get a really big difference. If you have just a small difference, you can’t be so confident that it’s not just some subtle change in solvation or something.

So for example, the 103 steric retardation of isopropyl via SN2, remember you had substitution at methyl, at ethyl, at isopropyl, they got slower and slower. Isopropyl was 1,000 times slower than methyl. So that’s a pretty big effect, 1,000 times, 103. That meant it must be 4 kilocalories per mole at room temperature difference in those two. So that’s pretty big, that’s not like half a kilocalorie or something like that. So a factor of 1,000 or more often you can be confident in interpreting. Or in the case of t-butyl, remember it jumped way back up again, it was accelerated by 105 compared to isopropyl, by putting that next one in, which showed that there was a new mechanism coming in, the formation of the cation intermediate. 

Now what if the ratio is not drastic? What if the two products that you could get form in about a 50/50 ratio? The synthetic organic chemist says this is deadly, because he’s doing a lot of reactions in a row to try to get some product. If each reaction has a 50% yield, and you do a whole bunch of them in sequence, for example if you did 12 steps getting a 50% yield at each step, you’d have a yield at the end of 1/212, 0.02%. So a synthetic chemist wants a reaction that gives a really, really high yield. Although it might be possible to have a reaction that doesn’t give such a good yield, but to tinker around with it and get the yield up so that it becomes acceptable in a sequence like that.

But a physical organic chemist finds such a situation valuable, because it’s a borderline. There’s a delicate balance between going this way and going this way. And now you can see changing this effect, changing the solvent this way, making it more polar or less polar, putting a methyl group on, taking a methyl group off. You can interpret small effects, so it allows sensitive tests of subtle influences. For example, isotope effects. Remember that a deuterium isotope effect, we said the maximum is about a factor of 7 usually. So that’s not very big in terms of kilocalories per mole. But even though it’s not a very big effect, even though it’s a small thing, you can interpret it, because you know that it must involve breaking a hydrogen away in the rate-limiting step.

Chapter 4. Preparing t-Butylhydrazine to Study Rate-Limiting Motion [00:20:51]

And finally in our discussion of SN2 and E2 eliminations, I’m going to give you a valuable lesson that was learned, he taught himself, by Jo-David Fine, who as you see here was the class of 1972 at Yale College, and was doing undergraduate research during the spring of 1971, so 40 years ago he began this project. And he continued it through the summer when he lived on York Street, and then through the fall, and the next spring as a senior, so this was his research project. Now what was his research project? There was an interesting question he wanted to ask that required preparing this compound. And the question was this: if you have a very viscous solvent, can motion of the pieces be a rate- or product-determining step? 

Usually we think of bond formation or bond breaking being a rate-limiting step. But obviously for two things to react with one another, they have to get together. So that requires motion. And if the medium is sufficiently viscous, and the reaction that follows when they get close together is sufficiently fast, then it could be that the rate-limiting step would be motion, not breaking or making bonds. So that was the question that Jo-David wanted to find the answer to, was whether it was possible that motion would be the rate-limiting step, even very subtle, short range of motion. It’s easy to imagine that long-range motion could be rate limiting. For example, in free radical chain reactions, you have radicals floating around. Once they find one another, bingo they’re gone. That’s the termination reaction that stops the machines in the free-radical chain reactions. So obviously what’s rate limiting is moving them through the solvent. But that’s long-range motion, how about very subtle motion? Can it also be rate limiting?

So this is what he was going to do. He was going to use UV light to knock out a molecule of nitrogen between these two radicals. So the two radicals are almost next to one another, but there’s a nitrogen molecule in between. But if those two radicals are going to react with one another, they have to get the nitrogen out of the way. So how do they get the nitrogen out of way so they could combine? Well in fact, these radicals studied in solution don’t combine. Why would those radicals not combine easily? Any ideas? Radicals, usually two SOMOs, come together, form a bond. Why not in this case? Lauren?

Student: The methyl groups would be too close together.

Professor Michael McBride: Right, it’s a question of steric hindrance; there are too many methyl groups to get in the way. So that’s not what happens. So in fact what happens, instead of that, is that you rotate that t-butyl radical together with the nitrogen. So it’s just a rotation, you don’t have to actually move things by translation, just rotate them to get that geometry shown at the right. And now one radical is close enough to the deuterium on the other, that you can get a deuterium atom transfer, which is very exothermic, easy, and fast. So that would generate then the alkene and the alkane, by deuterium transfer.

Now there’s an alternative reaction for these two radicals. Does anybody see what it is, besides doing the deuterium atom transfer? What other action could you imagine? Debby, do you have any idea?

Student: No.

Professor Michael McBride: Ok, that particular process was called radical pair disproportionation. That’s another way that radicals can kill one another besides coupling. But the second alternative is to rotate the other group, the one with hydrogens, which would give the geometry down on the right. And then you would transfer a hydrogen atom, and that would be very exothermic, easy, fast, and you would get that. So there’s a competition then between transferring hydrogen and transferring deuterium. But what’s the product determining step? What determines which one you do? Well if it’s the hydrogen transfer, then we know that there’ll be a kinetic isotope effect, and you’ll transfer hydrogen more often than deuterium. But if it’s the motion, then the two rotations should be just about equal. There’s not a very big effect from deuterium on the rotation.

So all you have to do to answer this question of whether this very subtle short range motion, just this rotation, is rate limiting, is to make this compound, photolize it to knock out the nitrogen, and see which products you get. Whether the deuterium is in the alkane product or the alkene product. If it’s 50/50, that says that there wasn’t a kinetic isotope effect, hydrogen was not being transferred when the product was being determined, the rotation was limiting. If there is an isotope effect, then that means that the bond breaking, not the motion, was rate limiting. So that’s easy, that’s what it says here. 

So all he had to do then was to prepare this compound. And notice how the compound was pretty carefully devised, not so you can sell it to your neighbor and make a big profit, but to answer this particular question. So this is in the same spirit as the Bartlett and Knox paper, where the idea was to make some molecule that was designed to answer a particular question.

Now often when you’re planning to make something, you draw a scheme that where the arrow doesn’t show what something goes to something else, it shows it backwards, it’s retrosynthesis, that is, what could you prepare this compound from. Well we haven’t talked about the reactions that would get from that starting material on the right to this product, but by the end of the course, they’ll seem trivial to you. They’re easy reactions that would do that.

OK, so if you could prepare this compound, t-butyl hydrazine, then you’d be home free according to what we planned. So you need t-butyl hydrazine. Now where are you going to get t-butyl hydrazine? So it requires having a t-butyl group connected to nitrogen. Well you could buy hydrazine, that’s cheap; you could buy t-butyl chloride, that’s cheap. At that time 40 years ago you could buy a can of it that size for about $5 or something like that. So all you have to do is do a substitution reaction where the nitrogen displaces chloride. Does that sound like a good synthetic plan? Kate, what do you think about doing, say, an SN2 reaction, to have the unshared pair on the nitrogen attack the carbon, displace chloride? Seem OK to you, or not so good? 


Professor Michael McBride: Pardon me?

Student: I think so. Like a HOMO-LUMO interaction.

Professor Michael McBride: Yeah, you could imagine the HOMOs and LUMOs. Any problems with it? Ayesha?

Student: Steric hindrance– 

Professor Michael McBride: Steric hindrance, right? When you have all those methyl groups, it’s hard to get at that carbon, and what would happen instead? If you can’t do the substitution? Elimination. So you’re going to have E2 will be much greater than SN2 here. And in fact, E1 is greater than SN1 too, so you’re going to get this, not that. So you’re going to have to figure some other way of making t-butyl hydrazine.

Well fortunately there was in the literature this paper in the Journal of Organic Chemistry from 1958 about preparation of these hydra–that wasn’t the title of the paper, but to make the stuff in the paper, they needed to prepare hydrazines. 

And they say in this paper that, “we explored most of the synthetic methods that have been used for other alkyl hydrazines. Most failed all together, or gave only negligible yields” when you tried to do it with t-butyl. But there was a seldom-used method that had been discovered by Zerner, and that worked well. And it involved hydrolyzing this CN double bond. So that reacting with water gave from this half a ketone, a ketone that I think some of you have seen, and on the right, the t-butyl hydrazine, the compound you want. So hydrolysis of the immediate [to] benzophenone. So the idea is to break a CN double bond and put water in there to put oxygen on the carbon, the H2 of water on the nitrogen, and you have the amine you want, it’s connected to another nitrogen and then to t-butyl. And it’s very common to change CX double bonds into CO double bonds and H2X, and we’ll discuss this ad nauseam later on.

And notice that this reaction is catalyzed by acid. The way it happens is you put a proton on this nitrogen, and then that resonance structure has the cation on the carbon, and that can add water and get the oxygen on there. But suppose you protonate the wrong nitrogen. If you protonate that nitrogen and then shift these electrons in there, now you have the t-butyl cation, which is going to lose a proton, you have isobutylene, and you’re lost. So it says hot acid caused elimination of the t-butyl group, so you have to be carried out at room temperature, you don’t want to warm it. So fine, do it at room temperature. So this is what Jo-David set out to do, the synthesis starts with this compound, weirdly enough, and you have to react it with a Grignard reagent here, which puts the t-butyl group on the nitrogen, and then you’re going to break that bond, put oxygen on there, H2 on there, and you’re going to have the product you want. That’s the idea, that’s the synthetic scheme.

So in the first place, it turns out that t-butyl Grignard is not the easiest reagent to make, especially in the summer when it’s humid. So it was hard for Jo-David to do this, but finally he got this stuff to add there, and got the stuff that needed to be hydrolyzed, and then he tried to do it. He worked on this from April to October, through the whole summer, and never got it to work. And finally he got very frustrated, so he telephoned Peter A. S. Smith, a professor at the University of Michigan who had published that paper in 1958, and asked him what gave. And this is a page from Jo-David Fine’s notebook from October 1971: “Private communication with Doctor P. A. S. Smith. Most importantly, he stressed that his published method will not give the desired product, instead the t-butyl group is lost. They haven’t been able to reproduce Lakritz’s’ results, nor to make any more. Although he feels that hydrolysis in the desired manner is possible, if hydrolysis is killed very rapidly. He said to, quote, play around with it.” Which of course, Jo-David had been doing from April to October.

So the next comment is the one I really like in his notebook. “For the sake of public courtesy and sophistication, I will reserve my own comments on this subject.” Now in preparing for this lecture, I e-mailed Jo-David Fine, so there’s a happy ending to this rather lachrymose story. He’s now a respected Professor of Dermatology at Vanderbilt, and is the world expert on Epidermolysis Bullosa, which is a skin disease; he’s a dermatologist after all. And his son graduated from Yale in 2000. Here he is, you see, wearing his Yale crew tie. He’s a very nice guy. And he said in a reply e-mail, he’s a fellow of the Royal College of Physicians in London now, as well as a Master of Public Health and an MD. He said, “That project, and my conversation with that pleasant prof at U of Michigan, did teach me more about the rigors and foibles of bench research than anything that I subsequently experienced as a postdoc at NIH. It therefore prepared me for many more happily transient foibles during the 19 years that I ran a bench immunology lab at University of Alabama at Birmingham and University of North Carolina. As I found out the hard way, trying to predict ‘good behavior’ with cell lines, antibodies, and semi-purified proteins was even more subject to random whimsy then when I was trying to work with relatively straightforward purified chemicals in a far more structured laboratory setting as an undergrad.” So he feels that this was very good training for him.

But the important lesson was learned by me the next year at a Gordon Conference in the summer, up in New Hampshire, when I ran into my graduate school friend Steven Nelson, a professor at the University of Wisconsin. He said that the way to make t-butyl hydrazine is to react this with this. Now it’s indeed true that 95% of it is elimination, but that product is a gas, so it just goes away. And the other 5% is the stuff you want. Furthermore, it’s very easily purified; you can crystallize it. So in one day, I made 10 grams of it. This was after Jo-David had graduated. So what’s the lesson here? If you have cheap starting materials, that the yield you get is not a big deal. If you got cheap stuff, and you can easily do it and purify the product, that’s the way to make this stuff.

OK, so there’s a second happy ending, that Jo-David Fine then had a successor the next year when we actually had the compound. He was able to make it, test it, and find out that if you have a fluid solvent, there’s an isotope effect. Hydrogen is transferred more than deuterium, the atom transfer is rate limiting. But in a very viscous solvent at low temperature, the kinetic isotope effect disappeared, it was 50/50 hydrogen to deuterium transfer, because this demonstrated that motion had become rate limiting. So it is possible for very subtle motion to be the rate-limiting step in a reaction. So there were two happy endings to this, Jo-David learned a valuable lesson, and I learned something both from Jo-David and from Steve Nelson, and from this result.

So this is stuff that’ll be included on Friday in the exam, I’m sorry that it’s taken so long to do this. And I’m going to give a review this evening, but the registrar seems to be having trouble getting to work or something, I’m not sure. That’s not surprising today. So we’ll have a room in WLH I’m sure, and I’ll e-mail you where it’s going to be.

Chapter 5. “Electrophilic” Addition to Alkenes [00:36:46]

OK now, the next topic is going to be backwards what we’ve been doing. So what we’ve been doing is substitution and elimination. And the elimination reaction is of course this, where the hydrogen and the leaving group go away, proton and leaving group minus typically. So we have a leaving group. Now the next topic is going to be electrophilic, as it’s called, addition to alkenes. Now notice that this is the reverse of E1 or E2, with some of the names changed. So where we have a proton there, that’s a low LUMO, of course an electrophile, some place where electrons could be easily accommodated. But we can generalize it, it doesn’t have to be a proton, it can be any electrophile, just call it El. And the thing that was a leaving group, we’re going to change the name of, it’s a high HOMO. We’re going to call it a nucleophile, and now with those changed names, we’re going to reverse the reaction, and take the alkene to a substituted alkane.

Now there are a number of questions involved with this, both ones involving mechanism, and ones involving synthetic practicality. One is what can you make? That’s a synthetic question. What can electrophile and nucleophile be? And you’ll see that they could be a wide range of things. Then there’s a mechanistic question, is this a HOMO-LUMO kind of reaction, or does it involve singly occupied orbitals? Is it a free-radical reaction? There’s the question of timing. Do you make both bonds at once, or do you make one bond and then the other? And if so, which one? There’s the question of stereochemistry. Do the electrophile and nucleophile come in on opposite sides of the double bond, or on the same side? And that’s particularly relevant, of course, if they happen to be in the same molecule, then almost certainly they have to come in on the same side. But if you come in stepwise, one first, then the other side can flop around. So you could attack either side of the other one, even if the two things are coming in from the same side. You could attack the opposite face if it was stepwise and it rotated. So that’s going to be an interesting question, whether the product is from syn or anti, we studied that already in elimination, whether elimination was syn or anti.

And then there’s the question of regiochemistry, or as it’s sometimes called orientation, whether the reaction follows Markovnikov’s rule. Now Markovnikov’s rule happens when you have an unsymmetrical case, and you’re adding HX. So you can have X go on the carbon that’s more substituted. And a name that’s sometimes been given to this is the Matthew Rule, from this Bible quotation, Matthew 25:29: “For unto everyone that hath shall be given, and he shall have abundance.” So the carbon that already has two methyls gets another X, an H goes to where the H’s were. Contemporary economics seems often to work that way, that the people that have a lot get a lot, right? So that’s one rule, that’s the Markovnikov rule. But sometimes, it’s anti-Markovnikov; the product puts the hydrogen on the carbon that has more things. So all these are questions we’re now going to address.

Last year we had a guest lecture from Professor Siegel from Zurich who talked with a more synthetic perspective to the class. If you want to read his lecture, you could listen to it and see what he wrote on the board on the web for last year’s website. But anyhow, from a synthetic perspective, the idea is you have an alkene, an unsymmetrical alkene, and you’re going to react it was some AB reagent, and put A on one end and B on the other. And what can A and B be? That is, what products can you make from an alkene? You can imagine putting two H’s on, an H and an X, an H and an OH, two X’s, X and OH, OH and OH, or you could imagine OH and H. Now we already had H and OH, but this is different. It’s a different orientation, like that Markovnikov, anti-Markovnikov, and the same here with these. So in the course of the lecture he went through and showed a number of reagents and filled in this scheme of what happens, and what kind of stereochemistry you have and so on. So that’s the synthetic motivation for studying these things, what compounds can you make from alkenes?

Chapter 6. NIST Webbook and the Stability of Isomeric Alkenes [00:41:41]

But we’re going to take, you won’t be surprised, a little more physical approach, and look first at the thermodynamic stability of some of the alkenes, and then at how they react. Now you wonder where you get this information about stability of compounds. A great place to get it is from this website, which I can get on the other computer here. If you put NIST, which is the National Institute of Science and Technology [correction: National Institute of Standards and Technology] into Google, you get this. And the very first thing you see is Chemistry WebBook. So if you click there, you get the Chemistry WebBook and I think I’ve got that here too. And then you can search in that WebBook by different criteria, and the easy one is formula for our purposes. And let’s suppose we’re interested in butenes. So we have C4H8, and let’s work in kilocalories, not in SI units. So let’s search and see what we can find.

OK, so here’s what I just found, and it’s also on here. So I have 1-butene, 2-butene, in two flavors, Z and E, cis and trans. Then I have 2-methyl-1-propene, an isomer, and cyclobutane. Also methylcyclopropane and 2-butene, generic that is, not separated as to isomers. So let’s just look at 1-butene and see what kind of information is available. So we see there’s all kinds of data available here. Gas phase thermochemistry, condensed phase thermochemistry, phase changes, reaction thermochemistry. That’s particularly interesting, but let’s first look at gas phase thermochemistry.

So here we see the heat of formation in the gas phase is almost 0, it has about the same energy as the elements graphite and hydrogen do. And if you look at the heat of combustion it’s this. And what’s particularly valuable, is you can see how well it’s known, what the error limits are. These data have been evaluated by here ALS, which is Afeefy, Liebman, and Stein. But it also gives where the information came from. So if you click on that, you get that those things came from Prosen, Maron and Rossini.

Now this Rossini is not the guy who wrote The Barber of Seville. It’s Frederick Rossini, who was a great thermodynamicist who measured lots and lots of these things. Notice incidentally that also there’s a lot of data from the Thermodynamic Research Center at Texas A&M in Texas. Now Rossini started at the National Bureau of Standards, which is what’s become NIST when it got renamed, then he went to Notre Dame, and then he went to Rice in Houston. Now some of you are from Houston. Why should somebody who lives in Houston be interested in the energies of hydrocarbons? He measured a zillion hydrocarbons. Why? Cassie?

Student: Oil?

Professor Michael McBride: Oil. This started in 1942. The American Petroleum Institute started a big project to determine thermodynamic properties of all the things that could be in petroleum and refined petroleum and so on. So this was done first at the National Bureau of Standards, then to Mellon Institute in Pittsburgh as it was than, and then he went to Rice. So there’s a method in what they’re doing here. And I’ll be back up. OK so we’ve got that.

If we go down a little further, we could look at reaction thermochemistry. So this is the heat evolved when you do a reaction, so notice that here is 1-butene becomes 2-butene, trans-2-butene, and here are different people who measured it, either by equilibrium constant, or by isothermal calorimetry, and these are the values they obtained. And you can see they’re pretty similar to one another. That the 2-butene, the trans-2-butene, is more stable than 1-butene by about 3 kilocalories per mole.

Now you can go down in this to other reactions. These are different reactions. There’s adding HBr to 1-butene. The one I’m looking for–here’s going to the cis-2-butene. Oh that’s the first one I had. So cis and trans. You see it’s a different energy involved in going to cis and trans. But the one I’m really looking for is a little further down here in the middle here, is adding hydrogen to get butane. So here for that reaction–wait, I’m out of sync with what’s over here. This is changing 1-butene to cis-2-butene, and here are all these values for that. And the next one over here on the web is the one I’m just talking about, 30 kilocalories for that. But you can also do it, and I’m not going to do it over there for the sake of time, you can see how much energy is giving off in adding hydrogen to cis-2-butene, and how much to trans-2-butene.

Now these differences in energy that you get here–wait, here now–all give the same product. Now there’s a big advantage to doing a reaction that’s not very endothermic or exothermic. It’s not like combustion, which is 600 calories per mole. To take one thing to another that involves rather small amount of energy, like this adding hydrogen to the alkene, and to take three different isomers and convert them to the same thing. You see why that’s a nice experiment to do? Yes?

Student: You can make more?

Professor Michael McBride: Pardon?

Student: You could make more of it?

Professor Michael McBride: Well you could make more, but not many people want to make butane. You get that for free, or from the petroleum to begin with. It’s for purposes of measurement that it’s important. Because if you’re measuring a small amount of energy and make a 2% error, it’s not very much. If you’re measuring 600 [kilo]calories and make a 2% error, that’s 12 kilocalories. But here, you see these errors are a part in 200, a part in 300, a part in 300. So now you know all these compounds relative to the same compound, to very high precision, which means you know them relative to one another. So that’s the way to really get the data properly.

So here’s a table from a textbook, I think it’s the Jones textbook which gives these things, but you’ve seen now where it really comes from, from this NIST Web book. Here are data I collected from the NIST Web book for the hexenes. And you’ll notice that a whole bunch of them were done by Professor Wiberg in the 1980’s and 1990’s, the guy who talked to you last year. And this is a problem to do after the exam, so it could be due next Wednesday, say something like that, a week from today. But to try and look at these numbers here, the relative stability of the hexenes, and see whether it’s consistent with what you find for the butenes in terms of the relative energies. That is, are there regularities of the energy between cis and trans double bond? Of the difference between a double bond at the end of the chain, and a double bond in the middle of the chain. Can we make generality that way? So that’s a fun project to look at. But the first thing to do is to get to the exam on Friday, so I’ll talk to you this evening about that. That’s it.

[end of transcript]

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