CHEM 125b: Freshman Organic Chemistry II
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Freshman Organic Chemistry II
CHEM 125b - Lecture 7 - Nucleophilic Substitution Tools - Stereochemistry, Rate Law, Substrate, Nucleophile, Leaving Group
Chapter 1. “Proving” a Mechanism by Imagining and Disproving All the Alternatives [00:00:00]
Professor Michael McBride: We’re talking about nucleophilic substitution reactions. And today we’re going to talk, in the next lecture as well, about mechanistic tools. How you find out what the mechanism is. You can talk about plausible mechanism, but whether that’s really it or not, and almost always there are several, requires experimentation. So we’re going to look at experimental tools you can use to decide what the mechanism of the reaction is.
So last time we saw a whole bunch of examples of nucleophilic substitution reactions, where a nucleophile approaches a compound that has a low LUMO, the substrate. It reacts in a particular solvent to give the product in which the nucleophile has replaced the leaving group. So now we’re interested in what the mechanism of that is. We talked last semester about the possibility of attacking on the carbon of the σ*, and the leaving group leaving with the electrons that were formerly in the R-L bond. That’s certainly plausible; the question is whether it’s true. Because as you’ll see, there are other possibilities as well.
So there are many different mechanisms. In fact they’re not all the same. But many of them fall in the same two classes, as you’ll see. And the one we’re going to talk about first is called nucleophilic substitution, SN. Substitution, nucleophilic, 2 means second order in kinetics, and you’ll see what that means. So SN2, nucleophilic substitution. And we’re interested, as I said in the pragmatic logic of proving a mechanism with experiment and theory.
Now this kind of logic isn’t like the logic they teach you in philosophy or mathematics. Things are rarely 100% proven. But there’s a practical use to know how these things work, because they can help you in designing new reactions, or changing the conditions to make a reaction work better. So you’ll see what it means to prove a mechanism in organic chemistry. And actually, mostly it involves disproving all the alternative mechanisms. You remember Sherlock Holmes in The Adventure of the Beryl Coronet, said “It’s an old maxim of mine, that when you have excluded the impossible, whatever remains, however improbable, must be the truth.” That’s generally the logic of how you prove things in chemistry. But the problem is you have to disprove all the alternative mechanisms, which means you have to imagine all the alternative mechanisms in disproving it. Of course it’s always a possibility that there’s one you haven’t thought of that’s the real McCoy.
Now let’s just look at what has to happen. You have to break the R-L bond, leaving group has to leave, and you have to make the new bond. So there has to be a dissociation and there has to be an association. And you could classify the possible mechanisms in the sequence that’s involved in these two processes.
You could first have dissociation, and then association. And we’ll see that actually happens–that’s called the SN1 reaction. You can have association followed by dissociation, which happens, but not with carbon. And you’ll see also the possibility of a simultaneous or a concerted reaction as you call it, where you make the new bond at the same time you break the old one, so that they come at the same time. That seems to pretty much cover the territory of how you can do it.
So you can have concerted, where they both happen at the same time, association/dissociation, or dissociation followed by association. And we can imagine reaction-coordinate diagrams for these, and geometries. So in the concerted, the nucleophile is coming in and attacking the carbon at the same time the leaving group is leaving, and that’s a transition state.
You can have a pentavalent carbon intermediate. So you first make the new bond, and only subsequently do you break the old bond. Or you could break the old bond, and get a trivalent intermediate, and then subsequently make the new bond. So if we look at the reaction coordinate diagram, the first one just has a transition state, but the other two have intermediates. Either a pentavalent intermediate of pentavalent carbon, or a trivalent carbon intermediate.
As I’ve drawn it there, the first step is rate limiting. We talked about the complex reaction of two steps, where either the first or the second step would be rate limiting. It would be possible to imagine the second step to be rate limiting as well. However, that might not be so common. We will ask what’s normally the case in a reaction. These ones where the second step has a higher barrier are not so common if you have an exothermic process because it’s against the Hammond postulate.
Not that it couldn’t be, because you have the intermediate, and you can go either to the product or back to the starting material. But if the second step is rate limiting, it implies that you have a higher barrier going to the lower energy product, which could be, but it’s not so likely. So it’s much more likely to have these up at the top in a reaction that’s exothermic.
Of course the same reaction run backwards would retrace the same path, and running this backwards would have the second step rate limiting. But generally we’re going to talk about cases where the first step is rate limiting, because we’re interested in reactions that go to give a product, not the ones that come back from the product and the starting material so much.
Chapter 2. Kenyon and Phillips Pinpoint Backside Attack in Nucleophilic Substitution [00:05:59]
Now if you have three different other substituents on the carbon that’s being attacked, other than the leaving group that’s leaving, then you have the possibility of chirality. So that the first transition state and the pentavalent intermediate are chiral. But the trivalent intermediate, if it’s planar, is achiral. And that means that there are stereochemical implications. Because in the first two cases the nucleophile comes on the opposite face from where the leaving group is leaving. But in the third case when you have the trivalent intermediate it could do that, but it could also come in from the other side. So you would get two enantiomers. So the stereochemical consequences, the stereochemistry of the product, whether it’s inverted from the starting material, or whether it’s racemic, you attack both sides, will tell you something about the mechanism.
I suppose if we want to be complete in our thinking about this, we could also imagine that there would be a frontside attack where the nucleophile would come in on the same face that the leaving group is leaving, either in a concerted process, or in a process with a pentavalent intermediate, where nucleophile and leaving group are on the same side. That’s a possibility, which isn’t shown here. But whichever it is, whether it’s the same side or the opposite side, the first two cases are going to give a single chirality to the product, where the trivalent intermediate will give both.
There are tools for testing, that is for excluding mechanisms. They can involve stereochemistry, rate law, rate constant, and structure (X-ray work and quantum mechanics). In the particular book that you might be using, the Jones book, these are in the sections indicated there, and I’ll indicate those as we go along. But these things will be covered in all the texts, so you’ll see a good treatment almost anywhere.
We’re first going to look at stereochemistry, what we were just talking about. So when you have nucleophilic substitution, where the nucleophile comes in place of the leaving group, it could either do exactly that–it could REplace the leaving group, so the nucleophile is on the right in the product, the same way that the leaving group was on the right in the starting material, so they just change positions. Or it could be a DISplacement, where the nucleophile comes in on the opposite side, and the things that make it chiral are inverted. You could imagine either way.
If you were naive and in the nineteenth century, which do you think you would think more likely? That it would actually replace the thing that it was exchanging with, or that it would come in and invert everything? Which do you think you would have thought was more likely? How many would think it more likely if you’re going to replace something, you replace it? How many think it would displace it? That’s exactly what they thought in the 19th century, that it should be replacement not displacement.
So when it was found in 1898 that there was an example, where it inverted the configuration, Emil Fischer, the guy who invented the Fischer projection after all, who was a big expert on stereochemistry at the turn of the century, said this was “the most astounding discovery in stereochemistry since the groundbreaking work of van ‘t Hoff,” which was a quarter century earlier.
So it was really astounding to these people that it could invert the stereochemistry. They thought this weird case that showed that, was just that–a weird case. But the normal case must be replacement. So the goal then was to find out which one was normal. The first really good evidence on this, and a beautiful experiment, was done in England by Kenyon and Phillips in 1923. And it involved this alcohol, and they resolved it, and got a single enantiomer, which rotated light +33º, to the right.
Now they want to do a substitution reaction, replace the OH by something else, and see whether the configuration is the same or whether it’s been inverted. Notice incidentally that if in doing the process you do two steps, you’ll get the same ultimate product, whether it’s two inversions, or two retentions of configuration. They’ll both give the same product. So you have to have just one step that involves the nucleophilic substitution reaction.
So now in the first place, OH–, as we’ll see later in the lecture is a crummy leaving group. You can’t undergo the replacement reaction easily with OH. So what they did first, was to react it with this chlorosulfonic acid. And this is indeed a nucleophilic substitution, where O attacks sulfur this time, not carbon, and chloride leaves. So now you have a product where the group attached to the chiral carbon, is not OH but O with sulfur on it.
Notice incidentally that nothing happened to the chiral carbon at this stage. The reaction was at the oxygen. So whatever configuration of the carbon you had in the starting material, you have in the product. This is not an interesting reaction stereochemically. It’s just the way to get there to be a group that’ll leave. And notice that this mechanism is A/D. It’s first association, attack the sulfur, and then dissociation, lose the chloride. And the reason you could do that is that there’s a vacant d orbital on the sulphur. So there’s a way that you can form a new bond without breaking the old one. You don’t have to attack σ*.
And once you have this product, that now is a good leaving group, because if you break the bond between carbon and oxygen, now the anion you get is not OH–, but an anion analogous to the anion of sulfuric acid, OSO2. Now that has a rotation of +31º, and the starting material was +33º, so that confirms what we were saying, that the configuration hasn’t changed at the chiral carbon. Am I right? Is that a good inference? That it’s + at the product and + at the starting material? Does that prove that the carbon didn’t turn inside out? How many think it proved it, or at least supported it? How many think it had nothing to do with it? Right. Because they’re different compounds and who knows how they’re going to rotate light, so that doesn’t prove anything.
Now we’re going to do the nucleophilic substitution reaction where the anion is acetate. And it’s going to attack the carbon, and break the bond from carbon. So the nucleophile attacks, the leaving group leaves. This is the reaction we’re interested in, whether it inverts the configuration. And we find out that we get that product, and that’s the nucleophilic substitution at saturated carbon. And the rotation of that product is –7º. So it was an inversion right? It shows that the carbon turned inside out, because it’s opposite the starting material, right? I hear you say–wrong! Because it has nothing to do with it. Same as the first step. So we don’t really know. That proves nothing.
The only way we’re going to prove is to get a product that’s exactly the same as the starting material, except for the possibility of configuration. Then we’ll know if it’s the mirror image, but we can’t use any derivative. So what we needed to do is change that OCOCH3 into OH. And you can do that. You can bring in hydroxide, attack the carbonyl carbon, which generates this intermediate, and then that can do this reaction, breaking the CO bond and generate that anion.
Now there’s another product, of course. The top right of the molecule is this. It’s got OHCCH3, and then that becomes a double bond. So it’s acetic acid. Incidentally, notice this is a substitution reaction too. The reaction across the top was a substitution. The reaction coming down was a substitution. The reaction going at the bottom is a substitution. But the top was a substitution at sulfur. The coming down was the one we’re interested in, substitution at a saturated carbon. The one going across is substitution at a carbonyl carbon. That one. So OH–comes in, and this anion comes off. But again it’s possible to do that by association first, and then dissociation, because there’s that vacant p* orbital that you can go in without breaking anything away.
So we’re not interested in the one on the top and the one on the bottom, we’re interested in that one that comes down the side. That’s the reaction were interested in. Nucleophilic substitution at saturated carbon.
Now notice the product. Remember that was an acid and this was an anion. But of course this is a stronger acid than hydroxide, so the proton will be transferred in this direction. And now we have a product which is the same as the starting material, except possibly for its configuration. So the question is, “Is it the same as the starting material?” Is the normal course of the reaction a replacement or a displacement? And the rotation is –32, where it was +33 before. That’s experimental error in the difference. So it’s a displacement. It inverted the configuration. And so we know that in this kind of reaction, backside attack on the carbon, the kind we anticipated, although nobody could have thought about it in 1923, on the basis of a σ* orbital being attacked where it’s big.
So backside attack in nucleophilic substitution at saturated carbon. And notice in this scheme that’s the only case where you broke a bond to that carbon. That’s the beauty and the design of this scheme. Now there’s another nice feature in the design of this scheme. It looks a little clunky, actually to have to do all those things. And you might wonder, “Why not avoid all that stuff going around the bottom by just doing hydroxide attacking this sulfonate ester?” Because if you brought hydroxide in there you’d get this product, and you’d do it without all this stuff.
But there was a reason in the experimental design they did that. Because if you use hydroxide, a much stronger base than acetate is, then it turns out it attacks that hydrogen, and pulls it off, when the leaving group leaves. So you generate a double bond there. So you don’t get the product. And of course that stuff is achiral, not that you really care. But you don’t do the reaction if you use hydroxide, so they had to use acetate. So this was a very clever experimental design, and proved that inversion is the normal course, backside attack on the carbon.
The trivalent intermediate could have been attacked from either face, and would have given racemic product. So we can be confident that in this case that’s not the mechanism. That doesn’t mean there couldn’t be a case where it’s the mechanism, but it’s not the mechanism in this kind of case, and things that are related to it.
So stereochemistry is a good tool, and it shows inversion. Which means that it’s not a concerted frontside attack, not a pentavalent intermediate where the nucleophile came in on the same side the leaving group was leaving. And it also means that you can’t have had the trivalent intermediate.
Chapter 3. Using Kinetics to Study Mechanisms – Rate Law [00:18:53]
Now let’s look at the rate law, and what it can tell us. How the rate depends on concentration. So this reaction I’m showing here is ethoxide anion attacking ethyl bromide. So the ethoxide replaces bromide as the leaving group. And we’re looking at the rate of forming the product or losing the starting material as a function of how much the concentration of ethoxide is. And we’re going to fix the concentration of ethyl bromide. It’s assumed that ethyl bromide, which is the subject of this whole thing, that rate’s going to be proportional to it. The question is, is the rate proportional to ethoxide?
So you plot the rate as a function of ethoxide, and you get a line that looks like that. So it’s a second order rate law. It’s proportional to how much ethyl bromide you have, and it’s proportional how much ethoxide you have. So both of those things must be in the rate-determining transition state. Both the ethoxide and the ethyl bromide.
Now there’s something funny about this plot. Do you see what it is? It’s nice and linear, so it’s first order in ethoxide. But what else? Ayesha?
Student: It doesn’t start from zero.
Professor Michael McBride: It doesn’t start at zero. If you don’t have any ethoxide, you get the reaction. How can that be? It’s because, although it’s second order–
Wait a second. What just happened? Did I press the wrong button? This was just to say about the second order. Sorry I got out of whack here.
So the initial rate-limiting dissociation in the dissociation / association would give a rate independent of nucleophile, this one. Because the first step doesn’t involve the nucleophile. Only the second step. So if the rate-limiting step depends on nucleophile, it can’t be this mechanism. So we could exclude it on this basis, as well as on the stereochemistry.
Back to this, we have the problem that Ayesha brought up. How can you get the product if there’s no ethoxide? So there must be a reaction that’s going on all the time, whether you have ethoxide or not. And then another reaction that depends on ethoxide. What’s that first reaction? It doesn’t depend on the concentration of ethoxide, so it seems to be a first order process. It’s constant.
First order could be dissociation followed by association, so there could be some of that going on all the time. And then a little bit of this where the ethoxide attacks, getting more and more as you have a higher concentration of ethoxide. But that’s not what it is. This is done in ethanol solvent. So the unshared pair on the oxygen of the solvent can attack, the same way the unshared pair on ethoxide can. Of course the ethoxide has a much higher HOMO because of its negative charge, so it will react faster. But there’s a lot of the ethanol there compared to how much ethoxide, especially here at the beginning when you don’t have any ethoxide at all.
So actually that second process is a pseudo-first-order reaction. It depends on the concentration of ethanol. But ethanol solvent’s concentration is constant. So it appears to be first order. But it’s actually a pseudo-first-order process where the first-order rate constant is kpseudo[EtOH], which is not changing.
So now we have these two possible reactions, ethoxide or ethanol, unshared pair, attacking the σ* of ethyl bromide. We can tell at this point here, when it’s half and half, half the mechanism is pseudo-first order, half the rate is the second order. Then we know that whatever the difference in these rate constants is, is made up by the difference in concentration, so that they have the same rate. That is, there’s enough ethoxide, a small amount of ethoxide with a much higher rate constant, so it has the same overall rate as the pseudo-first-order rate constant does.
And from that you can figure out that the second-order rate constant is 20,000 times faster than first-order rate constant. So this is 20,000 times faster than that. Now does that make sense to you? Yeah, it does. Because you’d expect the much higher HOMO of the anion to attack more rapidly. How much more rapidly? Do you have any way of measuring how reactive these two things should be? These two HOMOs? Any way of measuring how high the relative height of those two HOMOs, and their ability to form a bond. Hint: it’s what we talked about in the last lecture. Debby?
Professor Michael McBride: pKa. How different are they in attacking a proton. So, the analogy to that is attacking protons. And that’s an equilibrium that’s easy to measure by the way we did last time. We find out that their pKa’s differ by 17.5 powers of ten. Here they differ by only a little more than 4 powers of ten. So you might expect it to be even bigger, the difference.
But in fact this is measured in equilibrium. When you’ve completely formed the bond. Whereas in the nucleophilic substitution, we’re only at a transition state. We haven’t completely formed the bond. So you don’t expect it to be as big a difference by the Hammond postulate kind of thing. You’re only part of the way across. Not all the way. So it’s maybe not surprising that seems a plausible rate constant ratio. So that makes sense.
So the rate law then, shows that it can’t be dissociation first, because that wouldn’t depend on how much nucleophile you have there. Nor can it be frontside attack, either concerted or pentavalent intermediate and then lose the leaving group. It has to be from the backside.
Chapter 4. Rate Constant – The Influence of Substrate Structure [00:25:44]
Another thing you can test, is not just how it depends on concentration, but what the rate constant is. In fact we just looked in the last slide at whether these rate constants were plausible for a higher HOMO and a lower HOMO.
So the rate constant will depend, certainly on what the nucleophile is, ethoxide versus ethanol. But it will also depend on what R is, what the leaving group is, what the solvent is. It also may be on the nature of the product. For example whether the product is charged or not. So all of these things will influence the rate constant, and we’re going to look at the whole set of them, one at a time, and see whether these tell us something about the mechanism of the reaction.
First let’s look at the substrate, the R group. Here are different R’s, and the relative rate constant for how fast they get attacked by iodide in acetone solvent at room temperature, when bromide is the leaving group. So this is how the relative rates as we change R. We’ll call ethyl 1. Then, methyl is much faster. But propyl is about the same rate. Isopropyl is ever so much slower. The isobutyl is faster again. t-butyl is very slow. In fact, you don’t know how slow it is. It’s got to be slower than that. Because something else happens other than the reaction we’re talking about. So all you know is it can’t be any faster than that. And neopentyl is really, really slow. Now, do these make sense? Let’s look at it in two ways.
First we could have substituents on the α-carbon, the carbon that has the leaving group on it. And notice here we add one methyl group to that carbon. Here we’ve added two methyl groups to that carbon. Here we’ve added three methyl groups to that carbon. So we can have α-substitution.
But we can also have β-substitution. We can add methyls to this carbon. Here we put one methyl on that carbon. Here we put two methyls on that second carbon. Here we put three methyls on that second carbon. So we can try to classify it as we put extra methyls on a particular carbon. So those are the β-substituents, for example, one, two, or three methyl groups on that β-carbon.
Now notice that as we put a methyl group on the α-carbon, the first carbon, we add a methyl group to it in place of hydrogen. As we add it, it decreases the rate by 145 times. So there’s a higher activation energy, enough to slow it by 145 fold.
Now let’s put a second one on. That slows it down by 128. Another factor of 128. About the same. So the influence of the first methyl group in slowing it down is about the same as the influence of the second methyl group.
How about the third methyl group? Do you see what it does? So that means it’s raising the activation energy by about the same amount each time you do that. So you might think that the third one would do it by that amount, but in fact all you know is that it’s greater than 15 times. You can’t really measure it because something else happens.
Now let’s look at the same thing with respect to the β-hydrogen. The first one slows it down almost not at all. From 1 to 0.82. Only by a factor of 1.2. The second one slows it down by a factor of twenty-three. But how about the third one? Slows it by 3,000 fold! Why does the third one make so much difference, where the first two don’t make much difference of adding those methyl groups?
Well here’s the case in methyl with bromide ready to leave, and we have to attack the backside. Or let us suppose we’re attacking the backside, which seems plausible on the basis of the stereochemistry and the rate constant. That means we’re attacking that LUMO. Now let’s look at that same thing in the case of the α-substitution. There’s that–between the red and the second blue is the anti-bonding node that’s going to break. Here’s the surface potential. The blue area is a good place to put an anion, so that makes sense. So we have methyl, ethyl, isopropyl, and t-butyl. Let’s look at the total electron densities. Notice methyl, the arrow, gets there OK. Ethyl gets there OK. Although it’s bumping into this thing at the back a little bit, but it could lean forward just a little bit to avoid that. Here it’s getting in trouble. That slows it down by a factor of 23.
Oh. Let’s see. I’ve got the wrong one. This was 100, that’s another factor of a hundred–
But this one is a much bigger factor because you just can’t get there now when you have three. So the third one is what really stops it. And this is called steric hindrance. The space gets in the way. So that’s the effect of α-methylation.
Let’s look at the LUMO. It looks pretty complicated. The reason I put it up is not to look at that, but to look at that one orbital in the context of the total electronic density. So there it is at 0.06 contour. And now I’ll put the total electron density there, and look through that to see how much we can see of that LUMO that we need to attack. So you see, you can really see it here. You could see it a little bit here, and a little bit here, but you can’t see it at all from the backside here. So this one is going to be shut down. And there’s the surface potential, and you see even the surface potential isn’t as good here as it is in the other ones for the anion to come in and attack.
Now let’s look at the β-methylation, where we have this carbon, and we start putting one, two or three on that one. And here are the rates remember. One, 0.82, 0.036, and then 0.000012. Why is the third one so bad? Well you see this one can get in there. That one can get in there just as well, because this methyl group, the substituent is out of the way. You can put two of them out of the way, but the third one you can’t get out of the way. Once you have three, there’s no way to avoid it. So that one gets slowed way down. No way to avoid the third β-methyl group. There you can see that you can’t get any place near the really good blue areas. So this is consistent with the idea that you have a transition state that involves backside attack. Not the planar intermediate.
Now there’s a question. Might it be possible to have frontside attack? Might it be just easier to do backside, but you could do frontside attack if the other one weren’t faster? How hard is it to do frontside attack? So is this possible? Or is it possible to form a cation that’s not planar? So could you have it bent, and not planar, so that the stereochemical consequences would not necessarily be inversion. Might it be possible to have a non-planar cation as an intermediate? But remember that BH3, which has the same electrons that CH3+ does. Remember we talked last semester about XH3, that when that’s a cation, or in BH3, you expect it to be planar, but still might not be too bad to have bent.
This was the subject of a second really classic experiment in the twentieth century. The first one I talked about, which I really love, is that Kenyon and Phillips that proves that it must be an inversion. But there’s another one that shows that it’s not possible to have frontside attack. At least it’s very, very hard. And it’s not possible to have a non-planar cation. At least it’s very, very hard.
That was in this paper published in 1939, which if you click there you can download a copy of the paper, by Bartlett and Knox. So this is your chemical grandfather, P.D. Bartlett, who was a young man at that time. He was 32-years-old, and Knox, who was his graduate student, who was almost the same age, because he was an African American, one of the first African Americans to get a Ph.D. in chemistry. And if you click on that asterisk, you can download an account of his life, which is fascinating. He’s the grandson of a slave, and he has a lot of trouble. And that’s why he was old at the time he got his Ph.D. But you can read about it there if you want to.
At any rate, they specifically designed and prepared this molecule to test the mechanistic questions. This is rather different than just taking things, reacting them, messing around, observing what happens, and then trying to interpret it. They specifically designed a molecule that would test these questions about whether you could have frontside attack, whether you could have a non-planar cation. And one that wouldn’t be complicated by other problems. So they thought through very carefully to design it. And then Knox had to prepare it and study it.
Now this is a comment about that paper.
“In 1939 Bartlett and Knox published the account of their work on the bridge-head chloride, apocamphyl chloride.” That’s where X equals Cl. “I believed then, and I believe now, that this was a fantastically influential paper. For thirty years afterwards, no one really accepted any mechanism unless it had been tested out on a bridgehead case. Indeed, the Bartlett-Knox paper shaped the interests and viewpoint of many chemists about the kind of physical organic they wanted to do.”
And that’s a quote from John D. Roberts, who is an Institute Professor at Caltech, and one of the real leaders of organic chemistry through the twentieth century. And he’s still living now. He said that in 1975.
This is the structure of that molecule with balls and sticks. And it became so important, this kind of structure, that people who are in the field know how to draw it quickly. And I’ll show you how to draw it. It has that stick structure. And here’s how you draw it. You first form, or I do anyhow, different people draw it different ways. But you draw this sort of a distorted W. And then you start at the nearest carbon, and draw up and down, and to the right. And then you draw the last bond, and you draw it last, because it goes behind that one. So you make it broken, so it has a sort of a 3D aspect to it. That’s this case.
And notice what it is. It’s a boat cyclohexane. It has a bridge across the top. So it’s a bridged compound. And formally it’s called a bicycloheptane. It’s got seven carbons, the six of the cyclohexane plus the bridge. And you give numbers in brackets in the middle of bicyclo to say how many carbons there are in each bridge. So there’s a two-carbon bridge, between these two, and a one-carbon bridge. And those two positions are called bridgeheads. So this is a bridgehead chloride that we have in the molecule of the apocamphyl chloride that they designed and then prepared to study.
Now here’s a space filling model of it, and there’s where you would have to attack to get in the backside of σ*. So that’s obviously out of the question, that you could get in there. So the molecule was designed so you can’t do backside displacement. The question is if you can’t do backside, can you do frontside displacement? So the attack would have to be frontside. Now notice something else about this structure. That if you were to flatten that carbon, if you were to lose the chloride without attacking it, and generate a flat cation, that would require generating really strongly-strained angles here, and here, and here.
And it was estimated by Bartlett and Knox that that would cost more than 23 kilocalories per mole. Which 3/4 of that is 16. So it would be 10–16 factor in rate. Forget that. So they’ve designed the molecule so it can’t undergo backside displacement, and it can’t give the cation either, lose the chloride to generate the cation. If you were to get a cation, it would have to not be planar, otherwise it would be slowed down by 1016 fold.
So they set up to answer those two questions. But there’s another possible competition. Remember in Kenyon and Phillips, they couldn’t use hydroxide; they had to use acetate, because the reaction would go otherwise if they used the strong base, hydroxide. So might competition from loss of HCl to generate a double bond be a problem here?
Well, although there are hydrogens on the carbon adjacent to the carbon bearing the chloride, there are β-hydrogens. But they’re not in the position you would need in order to get elimination. Notice that if you had hydrogens here, here, and here, then you could do the elimination. This is anti to the leaving group, so you to do that, and make a double bond there. But there are not hydrogens here, or here, or here. So you can’t do the elimination. Those are gauche, not anti to the chloride that going to be leaving.
And even if you could break off H and Cl, this would be the double bond you would generate. Now think about that double bond. Here’s the p orbital on the carbon that used to have the chlorine in it. But here’s the p orbital on the carbon that now has only one hydrogen. What’s wrong? They don’t overlap. There’s horrid overlap. So in fact, there’s a rule called Bredt’s rule that a carbon-carbon double bond cannot originate from such a bridgehead.
Incidentally, I looked up Bredt’s paper yesterday, and it’s interesting that Bredt himself calls it Bredt’s rule. Usually somebody else names it in honor of the inventor, but Bredt didn’t wait around.
Would competition from loss of HCl make it hard to do? No. So here’s the beauty of the design of this molecule. It can’t do backside displacement. If it forms a cation, it has to be a non-planar cation, and it can’t do elimination, so how reactive is it?
Well here’s the part of the paper that discusses that. It says “all attempts to replace the chlorine failed.” You can’t do displacement on this bridgehead chloride. They refluxed it “for twenty-one hours with 30% potassium hydroxide in 80% ethanol.” And they didn’t get any reaction. Notice they said it was a little hard to do because these vigorous conditions attacked the glass, but they didn’t attack this molecule.
So that frontside attack, where you can’t come at the backside, you have to come in from the front, must be at least a million times slower than the typical backside attacks. So probably much, much slower than that.
Now how about the possibility of forming the cation. They said they refluxed it with silver nitrate in aqueous ethanol solution for two days. What’s the idea of the silver? Because silver can attack the unshared pair on chlorine, and form a bond, and break the bond from the R, to leave a cation of R. So instead of pushing by attacking the backside, you pull the chloride off leaving the cation. So that’s a good way to generate cations, except in this case it didn’t work. There was no opalescence. The solution remained clear. Why opalescence? Because if you get this reaction, the sodium chloride, makes a precipitate. If you get just a little bit of it, you can see a little bit of cloudiness there. But they saw none, and in fact as you can read here, “the reaction of silver nitrate with 1-chloroapocamphane is least 2 x 10–10 at about 85 º.”
Now the reaction in this compound, so it’s carbon that has two methyls and an ethyl on it. This one here is at least a billion times slower than that. And when this one was measured, it was done 60 degrees cooler, and without the silver pulling on it. So it’s many, many, many powers of 10 more difficult to form this cation when it’s bent. So it’s very hard to form bent cations. And it’s very hard to do frontside attack. So this was a beautiful experiment where they specifically designed a molecule to answer these questions. And it succeeded, and it became the model for many kinds of mechanistic studies that were done over the next thirty years.
Now there are also interesting cases when the R group is in a ring. So here’s no ring at all, just two methyl groups, then three, four, five, and six membered rings. And we’re going to displace the bromide. Now cyclopentyl bromide is almost exactly the same rate as isopropyl bromide, which we’re taking to be 1 for relative purposes. But cyclobutyl is very much slower, 200 times slower. And cyclopropyl is very, very much slower than that. Now any ideas about why there’s no problem with cyclopentane, but cyclobutyl and cyclopropyl are bad. What do you think about when you compare those rings? Chris?
Professor Michael McBride: Strain. Right, exactly. That one has 109º bond angle, and so does cyclopentane, but the others have very small bond angles. Now wait a second. This strain is in the starting material. If you destabilize the starting material, you should make a reaction faster, not slower. So that in itself seems to be backwards. What the heck is going on?
Let’s think about the transition state for the reaction. How hard does it get from the starting material to the transition state? That’s what determines the rate. So the transition state looks like this. And it wants–it’s sp2 hybridized to make those bonds to the original substituents. So it wants to have 120º bond angle. Beforehand, it wanted to have 109º bond angle. Now it wants the bond angle to be even bigger. So if you have a 60º bond angle and it wants to be 120, that’s worse than having 60 when it wants to be 109. So indeed the starting material is strained. But the transition state is even more strained. So you always have to compare the starting material and the transition state in a situation like this. So the increased strain in the transition state is what slows it down.
Now how about cyclohexane? It has 109º degree bond angle here, but it’s slowed down. It can’t be this kind of thing. It could go to 120 no problem. No problem with making 120º-bond angle here. Can you see what the problem is in the six-member ring? Why is it slower than five? Remember the shape of the five-member ring? It’s almost flat. A little bit puckered. But cyclohexane is the chair, which means that chloride is running into these two CH2 groups right here. In cyclopentane, those other groups would be out here, not up here. And you notice that it’s strained because the line between the leaving group and the nucleophile is bent here. This thing is being pushed back. So there’s a different kind of strain in this case that slows it down by a factor of 200, compared to that one.
Chapter 5. Rate Constant – The Influence of Nucleophile and Leaving Group [00:48:27]
So we’ve seen the rate constant dependence on R. Now we’re going to look at it just really quickly on the nucleophile. So here are a bunch of nucleophiles, and they get faster, and faster, and faster. Now, is it the same with the proton? How about the pKa’s? If we look at this, water as a leaving group [correction: as a nucleophile], fluoride, and hydroxide, you see that indeed, as it gets to be a stronger acid, it’s slow. Is that right? A weak acid is fast. Is that reasonable, or is that backwards?
Student: It seems like it would be backwards.
Professor Michael McBride: A weak acid means it holds on to a proton tightly. But it also is holding onto–attacking a carbon well. So that makes sense. That makes a lot of sense.
So for first row elements, the nucleophilicty attack on σ*, parallels basicity attack on H+. Both require a higher HOMO, which’ll make it more reactive. So far, so good.
Now let’s look at the halides. Fluoride, chloride, bromide, iodide. Notice that as the acid get stronger, it speeds up. Now it’s backwards. The halides are going backwards. The better they are at attacking a proton, the worse they are at attacking carbon. That doesn’t make sense. And we see the same thing here for oxygen and sulfur. Sulfur is a stronger acid, but it’s better at attacking carbon. This seems nuts.
Now this is actually tied into the effect of solvent. Because if you look at the relative rates in water for attacking methyl iodide with these different nucleophiles, you see this which is essentially what we see here. This is 14 times faster, here’s 10,000. This is 160 times faster; here it’s 80 times faster. These are generic numbers, and these are specific ones for a particular case.
But look what happens if you do it in acetone. It turns around. So it’s what you expect in acetone, but it’s backwards in water. What’s special about water? Why are things backwards in water? What’s different about water and acetone?
Student: Water is polar.
Professor Michael McBride: Water forms hydrogen bonds. And when you form hydrogen bond to an anion, you deactivate it. You have to break the hydrogen bond in order for it to attack something else. So you can see that this is sensible in acetone, but it’s backwards in water, because it’s harder to break hydrogen bonds to the smaller ions. So as the ions get big, like iodide, then they’re not very much hydrogen bonded, and even though they’re not so good at attacking protons, they’re in fact better at attacking carbons because you don’t have to break the water away. OK, will continue with this stuff next time. Thanks.
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