CHEM 125b: Freshman Organic Chemistry II
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Freshman Organic Chemistry II
CHEM 125b - Lecture 9 - Pentavalent Carbon? E2, SN1, E1
Chapter 1. Using X-Ray to Investigate the Possibility of a Pentavalent Carbon Intermediate [00:00:00]
Professor Michael McBride: We’re going to finish up on nucleophilic substitution, SN2, where we’re spending so much time looking at how you prove a mechanism. And then look at the elimination that’s second order, and then the substitution and elimination that are first order kinetically.
We’ve looked at proving mechanisms, that is disproving mechanisms by stereochemistry, by rate law, by rate constant, varying all the different components of the reaction. And finally there’s another way, which is to look at structure, which you can do either with quantum mechanics or by X-ray.
The problem, remember, is this really subtle distinction, one that probably doesn’t make much difference. But if you’re really looking carefully at mechanism and want to understand them it’s an interesting question, which is, is there an energy minimum at the top? Is there an intermediate in the concerted SN2 reaction, or is it just a transition state? And of course they blend into one another as the stability of the pentavalent intermediate, if there were one, gets less and less.
The interesting thing is when you have the nucleophile, a leaving group and a carbon in the middle, trivalent carbon, is it actually a pentavalent carbon? Is there some stability associated with that?
You could imagine arranging groups like oxygens, so there are unshared pairs are on the left and right, and a planar carbon cation in the middle and see whether there’s a bond there. How are you going to get them? Remember the non-bonded distances between atoms, when they’re in contact, is about twice as big as bonded distances, right? So if the thing is only marginally stable, what’s going to keep it from flying apart? How will that last long enough for you to look at it? Well, you can put other things in a molecule that hold those pieces in place, like this set of three aromatic rings of the anthracene system.
Now we’ve got the groups up there on top where we want and the question is going to be–Well first, how do we get them there? The thing in the middle started as an ester, right? But we want it to be a cation with a vacant p orbital in the middle. So what you do is react it with a molecule we’ve talked about before, Meerwein’s reagent, which is a way of giving methyl groups. It’s got a great leaving group on it. A dimethyl ether is the leaving group. We can have a substitution reaction and put a methyl group on and now the carbon has a positive charge. Although if you want to be careful about it, not all the positive charge is on carbon because the unshared pairs on the adjacent oxygens, of course, are mixing with that vacant orbital.
We’ve got our trivalent carbon cation in the middle. There we see it. There should be this anion that was left from the original Meerwein salt. And the people who did the X-ray work we’re surprised to find that it was not BF4–, but B2F7–. It had gotten together with another BF3. That’s neither here nor there, but it’s what was there in the crystal.
They have the oxygens next to the central carbon. And remember, calculated when you had a cation with waters on both sides of a trivalent carbon, was that at the symmetrical pentavalent “geometry,” it was a transition state. OK? If you believed in the theory, you wouldn’t expect there to be stability here now. How do you know if there’s stability? How do you know if there are bonds there? Well, you can look in the paper that’s cited down there on the bottom right, and it shows a picture and you can see that indeed there are bonds there. Right? That’s supposed to engender a laugh. I see Chris smiling. Why do you smile?
Student: They just drew the bonds.
Professor Michael McBride: They just drew the bonds, right? How do you know there are bonds there?
Well there’s one interesting thing. This structure is symmetrical, as drawn. And, you notice these funny shapes the atoms have. What they are, are ellipses with an octant cut out of them to show the size of the axes. And one thing you measure in X-ray is how much the atoms vibrate. Those show how much it vibrates in the three principal directions for vibration. If we look at that C19, the central carbon there, notice it’s not elongated. It would be elongated if it vibrated with great amplitude back and forth. It would move a lot in that direction.
So it could be–remember in X-ray, what you see is the average structure over many molecules–it could have been that one of them is like this, another one is like this, another one was like this, another is like this. And the average would appear to be in the middle. But it would have a long displacement, on the average, you see. But that’s not what you see there. It could be like this bell clapper, sometimes there, sometimes there, but on average in the middle. But it’s not because it’s not stretched out that way.
That still didn’t answer the question, whether they’re bonds there. It could be in the middle but no bonds. So let’s look right edge-on at the central carbon. And ask the question, how far apart are those two oxygens? If the interaction is repulsive then they should move apart. If it’s attractive they should move together. Compared to what? Well, the authors compared it with the two carbons they were attached to, which are 5.02 Å apart. And the white is 4.86. So, they’ve been drawn together. The distance is shortened by 0.16 Å, as if they’re being sucked in. That looks like there’s a pentavalent carbon that’s actually attracting those adjacent oxygens. This was published in a distinguished journal, the Journal of the American Chemical Society. And notice that it’s not only that distance but you can see also that the angles are distorted. Instead of those being 120º angles out on the left there, one of them is larger and one of them is 7º smaller. It’s bent in, as if it’s being sucked by the central carbon, so that there’s a bond there.
So pentavalence seemed to be a safe inference, and that was the subject of this paper. But, if they did compared to what, and just had hydrogen in the middle instead of that positive carbon, it turned out they were sucked in even further. So instead of being 4.84, whatever it is, it’s 4.75 Å. So much for that proof that these groups are being sucked in. With nothing in the middle, just the hydrogen, they’re sucked in?
Now, notice that there wasn’t just that positive carbon there. There were these other things, the oxygens, and the methyls. There was some space to that which could have pushed things apart. Now lacking that push apart, they come together closer. From one point of view, that’s reasonable, but why should they come together? It’s not a pull to the central carbon. It seems to be a push. Where is the push coming from? Can anybody see?
What pushes those methoxy groups, the oxygens, toward the center? Helen?
Student: Maybe it’s the steric hindrance or the strain of the two methyls.
Professor Michael McBride: What are the methyls being pushed by?
Professor Michael McBride: It must be some push to bend those angles in. Anybody got an idea? Chris?
Student: Is it with the hydrogens?
Professor Michael McBride: Ah, the hydrogens. Right? They’re eclipsed. And there’s repulsion there that causes them to bend in.
Now, you can put instead of a positive carbon, you can put oxygen in the middle with this extra group on it. And that, you see, does push them further apart. But still they’re bent in a little bit as compared to the carbon. But the central oxygen is only slightly repulsive, compared to the carbon. So the carbon’s not pulling things in at all. It’s pushing them out trying to fight against that hydrogen pushing into the methyl out on the side.
If you have BF3 in the middle, which clearly has the vacant orbital on boron, has very small other things on the boron, so they’re not pushing apart. And those other things are fluorine, so not only are they small, but they’re withdrawing electrons and making that B orbital especially low in energy. Then you can see them genuinely being sucked in, further than in any of the other cases, as compared there to the positive carbon in the middle.
Now, if you try to make the things on the outside more willing to give up their electrons, that is higher HOMOs to make stronger bonds, and make it O– rather than CH3O. So we have the minus charge raising those HOMOs. Then indeed, it forms a bond, but not two bonds. It’s unsymmetrical. One of them is a reasonable bond. The other one is just a non-bonded interaction. In fact it turns out that this is a salt. It’s an anion. So there’s a positive cation nearby, a potassium, which makes it a little bit unsymmetrical. So it’s better to have the minus charge on the right near the potassium cation than on the left. That also helps distort it. Now remember that as compared to the anion that was calculated, which again, was said to be by calculation a transition state.
Here are a whole bunch of things they studied, and I don’t want to spend the time going through this in great detail. But compared to what? Compared to just having a hydrogen in the middle–to see what these distances are. Then in the case where C+ was in the middle–they made the C+ that way–but it’s not such a great C+ because of the vacant orbital on the carbon being mixed with the unshared pairs on the adjacent oxygens. But if you have–if the central atom is bonded to oxygen or sulfur, which has these extra electrons, that seems to use up the vacant orbital. So it’s not so good at bonding and therefore sucking things in. If you have fluorine on boron then, as we said, the electron withdrawal lowers the energy and indeed it does seem to suck it in to give a pentavalent atom in the middle. But if you have higher HOMOs on the neighbor of the boron, giving their electrons to the boron, then that’s a better source of electrons for the boron than the neighboring atoms are. And you can see in these cases.
If you look at this whole bunch of things plotted in this particular way, you can see, for example, there’s boron with two chlorines in the middle and methoxides on both sides. The bar then shows the two different bond distances. So it’s unsymmetrical. One is short, one is long. It’s not a pentavalent carbon. There’s some that are like that on the right, very different and unsymmetrical. Then there’s some on the left that are about equal bond distances, right and left or symmetrical. There’s our reference with hydrogen in the middle. And it’s pressed in to give somewhat short distances by the hydrogen pushing on the methyl as we spoke about before.
If you have boron in the middle it’s not sucked in very much. Unless you have that case with fluorine on the boron, and then it attracts enough that you could make a pentavalent atom in the middle. But most of the borons are tetracoordinate, that is unsymmetrical not a single minimum but a double minimum. If you look at the two cases of carbon, the one with oxygens on each side there on the left, is symmetrical but not short. And the one on the right is short, one of them is short, but the other is long. It’s not symmetrical. There’s no sign from this exercise for a pentavalent intermediate of the SN2 sort being stable. It seems genuinely to be just a maximum on the way across. So it’s a transition state as calculated by quantum mechanics, so that should give us a little more confidence, perhaps, in the reliability of quantum mechanics, which of course, can’t take into account all the neighbors around the molecule.
Chapter 2. The E2 Reaction and Kinetic Isotope Effects [00:14:07]
That’s as far as we’re going to go looking at the mechanism of the SN2. Now we’re going to look at alternative paths of reaction, the elimination reaction, which we already mentioned last semester and also first order nucleophilic substitution and elimination.
First E2, or β-elimination. Now, this is from last semester. You remember, you can attack here with a high HOMO, antibonding here, antibonding here. You break off fluoride. You break off hydrogen, and form the carbon-carbon bond, as shown by the curved arrows here. We said last semester that that’s the E2 elimination mechanism.
What influences the rate? The rate is influenced by the base. That’s what makes it second order. It depends not only on the concentration of this substrate but also on the concentration of hydroxide. The two must be getting together in the rate-limiting step, or before the rate-limiting step in a preliminary equilibrium that gets drawn off. That’s the same thinking as we had on the SN2, that both things are involved in the rate-limiting, or before, the rate-limiting step. It depends on the nature of the leaving group. A better leaving group reacts faster. So that means that the fluoride is leaving in the rate-limiting step, or before the rate-limiting step in a pre-equilibrium.
There’s a hydrogen isotope effect, called a kinetic isotope effect. If you change hydrogen for deuterium, you change the rate, and this has special implication, as you can see here. These others only meant it was at the transition state, or before the transition state, that something was happening. But if this makes a difference it shows that bond is being broken during the transition state. We know the leaving group is leaving, we know the base is coming in, those two things either before or at the transition state, and now we know the hydrogen is leaving at the transition state, and let’s see how that works.
We saw in Erwin Meets Goldilocks that you have a vibrational potential surface. And you’d have also a vibrational potential surface for the hydrogen that’s being transferred after it’s transferred. It’s stuck first place to the carbon, in the second place to the oxygen. But halfway across, when the bond is being broken, it’s between the two so it’s not being bound to either. It’s very low–It’s easy to move back and forth when it’s halfway between.
What implication does this have for the kinetic energy of that hydrogen? Remember, there’s a lowest possible kinetic energy, the one that gives no nodes in the wave. It’ll be here in that case, here in the oxygen case, but when it’s very easy to move you can stretch the wave way out, very little curvature, very little kinetic energy. So you can have much lower minimum kinetic energy when the hydrogen bond is being broken than when it’s either attached to the carbon or attached to the oxygen.
What’s the implication of that? When you do the reaction you have to get to the transition state so you have to put in that much energy.
What’s the implication if we change from hydrogen to deuterium? You remember what happens when you change the mass in Erwin Meets Goldilocks to the lowest energy? If you make them into deuteriums, then you have a wave that looks like this, but the bigger mass means it has lower energy. So, we have lower energy here, lower energy here, and lower energy here, but it can’t be lower than the minimum. There’s hardly any change in the energy here, when you go from hydrogen to deuterium, but there’s a substantial change here and a substantial change here.
If you look at the energy that’s required for the reaction, more energy is required for the deuterium, because it started lower. If you were talking about an equilibrium between here and here, then that difference would cancel out. It’s only in that transition stage, which has the very low potential, that means that there’s something special about the deuterium, or perhaps you should say something NOT special about the deuterium. When it’s bonded, it’s unusually low in energy, but when it’s not tightly bonded, then it has the same energy as hydrogen. Because you have more of a barrier for deuterium, the reaction should be slower. Hydrogen is faster than deuterium if that is the rate-limiting step. If that’s not the rate-limiting step, if it happens before this or after this, then you don’t have the difference between hydrogen and deuterium. It’s only when the hydrogen is being broken away during the transition state, during the rate-determining step, that you can see it. You get that difference in rates, that so-called kinetic isotope effect, only if bond is being weakened in the rate-determining transition state.
Chapter 3. Stereochemistry & Regiochemistry of E2 Elimination [00:19:51]
There’s another interesting implication as there was in SN2, which is stereochemistry. So we have a hydrogen and a leaving group that are going to come off. And if we look at that from the end with the Newman projection, we see them anti to one another. We can imagine pulling the two off and forming a bond by what’s left.
They could’ve been syn to one another in an eclipsed conformation. Which should be better? Should we pull them off anti to one another or syn to one another? Well of course, the top conformation would involve eclipsing. So that would be higher in energy than the anti. And furthermore it turns out, although we don’t have time to talk about it right now, that anti orbitals overlap with one another better than syn orbitals. To me, that seems counterintuitive. You’d think when they’re like this, they’d overlap better than when they’re like that. But in fact they overlap better when they’re anti. By both accounts, one would expect the anti to be the preferred mode for elimination. But is it true? How could you test it experimentally? How could you tell whether they had been pulled off, or were being pulled off, from the same side or from opposite sides? Well, if you look at the title of the slide you use stereochemistry.
Notice then, that if we put R groups on here, then when we lose H and L here these two R’s will be on the same side. But if it’s rotated into this form, then they’ll be on opposite sides of the new double bond that’s being made across the middle. We can start, for syn elimination of this particular compound, where notice this configuration is S and that one also is S. It’s rotated into a conformation where the leaving groups are syn to one another. Of course it could rotate, and they could be anti to one another. And the question is, which one really happens? We can look at the product and find that the two methyls are on the same side of the double bond as they are here, both in back. Whereas here, one’s in front, the other’s in back. This suggests then that the fact that one gets the E isomer of the alkene that it was the anti form that gave rise to it. Stereochemistry showed that it’s anti elimination in this case.
Of course there could have been another reason for anti elimination. It could’ve been that this product is more stable than the one where the phenyl group is near the methyl. How can we discriminate whether it’s just that it’s the preferred conformation and anti is good for the reason shown here, or whether the only reason you get that one is because it’s more stable? Can anybody see how you can test that? You could use a different stereochemistry on the starting material and see if it still gives that one. So maybe E is just more stable than the Z isomer.
Here we can change the configuration on the right to make it R. The methyl is now in back and the hydrogen in front here. So now the syn is the one that should give the E isomer, and the anti should give the Z. In fact, you see it is Z. So no matter which way–it can’t be just because this one is more stable, because if you change it you get that one. It has to be that it favors the anti conformation for the elimination.
There’s anti stereochemistry but nature is not dogmatic about this. If you can’t do the anti, then the syn is still OK. And here’s an example of that taken from the Jones textbook. Notice that in this case, the treatment with base to eliminate hydrogen or deuterium from this side and tosylate from that side, removes the deuterium, leaving two hydrogens. It gives that a 98% yield even though those two are required to be syn to one another because, again, this bicyclic framework, the same one used by Bartlett and Knox, that idea of holding things in place that way, doesn’t allow conformational change around here. So these have to be syn to one another. It loses DOTs not HOTs despite the kinetic isotope effect. Remember hydrogen is eliminated in preference to deuterium, other things being equal.
Other things aren’t equal as you can look here. If you do a sort of semi-Newman projection along this bond you can see the D and OTs, their bonds are in the same plane. So in this rigid, eclipsed case the overlap of σ* here, which is losing its electrons–that is, as the OTs– leaves–this vacant orbital is able to overlap well and stabilize the σ orbitals in this bond, so it takes away those electrons and D+ leaves. So, the eclipsed D is better than H, which would be better if it could be anti. But this framework doesn’t allow it to be anti. There’s not good overlap between the H and the OTs. It’s anticlinal. Notice in this case you don’t have to pay a penalty for having the starting material for the syn elimination eclipsed. It’s required to be eclipsed by the nature of the linkage among the carbons.
That’s a question of stereochemistry, whether you get E or Z isomer. There’s also a question of what’s called regiochemistry. Where does the double bond appear? If the leaving group is here on the second carbon, and you treat it with base to remove H and L, then you can get the double bond either in this position, or in this position. And that can be both cis and trans. I draw only the trans isomer. You could remove a blue hydrogen or a red hydrogen. And this is the kind of thing, where before people knew anything about mechanism, people tried to figure out how you can get a certain product and different pathways. This was just lore: this would do this; this would do this. There were two kinds of rules. One was the rule due the Hofmann. The other was the rule due to Saytzeff. So there was the Saytzeff rule, which said you should take off the blue hydrogen. And there was the Hofmann rule that you should take off the red hydrogen. Some cases did one, and some cases did the other. And it was your business if you were a synthetic chemist to know which ones did which, so you would know what kind of leaving group to choose or what kind of base to choose to get the particular products you were desirous of.
If you have the leaving group of the halogens, iodine, bromine, chlorine, you see they’re Saytzeff. The dominant isomer you get is in the second position not the terminal position. But notice as we go iodine, bromine, chlorine it gets less and less Saytzeff. In fact if you go to fluoride it turns around and it becomes Hofmann in its orientation. The same is true for the leaving group being trimethylamine. The trimethyl ammonium starting material loses this trimethylamine and this proton, so it’s 98% of Hofmann orientation. This big group seems to be Hofmann. Iodide seems to be Saytzeff.
Let’s look a little bit about the energetics that are involved here. This is a ratio of about 4:1. This is about 1:50. The whole range, which is very important, what you’re going to be getting, is only a factor of 200 between 4:1 and 1:50. So 400 means a factor of 102.3, which means that it’s about three kilocalories change in energy that makes it from going 4:1 one way, to 50:1 the other way. A change of just three kilocalories. Remember that a hydrogen bond–not a particularly strong hydrogen bond–is worth three kilocalories. So you can see that subtle things are going to enter into this. It’s not surprising that there are not hard and fast rules, that sometimes it’s Saytzeff and sometimes it’s Hofmann. This is a very subtle thing. It’s important for synthesis, but it’s not a big factor that you could be confident, that you could put your finger on it and say that’s the reason it went that way.
You can have E2 versus SN2. You could have the high HOMO come in, either attack a carbon or take a hydrogen. How well do we understand what does that? Suppose we take t-butyl. To do an SN2 reaction you’d have to attack there, but we’ve seen this picture before and you know that’s sterically hindered. You can’t get in there. Steric hindrance tends to disfavor SN2 processes. But it’s much easier to get to the neighboring hydrogen and pull it off. It’s on the surface of the molecule. Steric hindrance favors E2 over SN2. It does it by disfavoring SN2. If you have a very hindered center, you tend to get elimination, not substitution. That makes perfect sense.
Furthermore, remember when we looked at nucleophilicity, how good things are at attacking carbon, we tried to see whether it would be the same as attacking hydrogen. There wasn’t such good parallelism. Some things were more dramatic in their pKa in attacking hydrogen than they were in attacking carbon. That obviously means that if you take something that’s a fairly weak nucleophile, compared to how basic it is, the pKa, if you look at those which are grossly out of whack, then they could be ones that would prefer attacking the hydrogen. For example, hydroxide is of that case. It’s a strong base but not as good a nucleophile as you would expect for such a strong base. So hydroxide, or alkoxide, tends to do elimination rather than substitution. But always these are a balancing act and it could depend on other factors as well.
Chapter 4. Strategies for Substitution in Organic Synthesis [00:31:46]
I’ve given the title of these next things Synthesis Games because synthesis is for many people the real goal of organic chemistry is how to make new molecules, or make old molecules in better ways. One of the best ways is to follow lore. But it’s also guided by understanding. There has been lots of progress made in the last fifty years because of people having studied mechanism. If something isn’t going right you have some idea what you might change in order to make it, not just looking up in the literature to see what people did before, although that remains a very important procedure. We saw all these different nucleophilic substitution reactions, and the question then if you’re trying to apply somebody else’s work to yours, is how general they are.
For example, the Williamson ether synthesis, how general is it? For example here’s the compound, MTBE. Have you ever heard of that? I bet you’ve seen the word, or the acronym. It’s methyl t-butyl ether. Have you ever seen MTBE? Where did you see it?
Professor Michael McBride: On the gasoline pump. This is an additive to gasoline. So it’s made in enormous quantities. Here, you want to make this ether. Here’s a way to do it. You can start with methanol, treat it with sodium hydride. There is a high HOMO, which can attack the proton and take it away as H2, leaving the sodium salt of methoxide. We have a nucleophile. Now if we want to make this bond here, the one between the oxygen and the tertiary carbon, then we would react it with t-butyl bromide. Are you going to get rich in the petroleum industry, making MTBE that way? Are there any problems? What have we just been talking about? The competition between substitution and elimination. And what are the factors that favor E2 over SN2? Amy?
Student: Steric hindrance.
Professor Michael McBride: Steric hindrance. Right? Crowded here. It’s hard for this oxygen to get at the carbon. And what else? If you have an anion, a nucleophile, that’s even better as a base, taking hydrogen. Here we have the worst of all possible worlds. We have something that wants to take hydrogen, and something that doesn’t want to be attacked at the carbon. So you get elimination. It’s too hindered for SN2, and the strong base favors E2. All you get is the double bond compound. You don’t get methyl t-butyl ether.
Does the Williamson synthesis then not work? Do you just throw up your hands, if you’re a synthetic chemist? We said we wanted to make MTBE. We wanted to make that oxygen-carbon bond. Can you see any alternative? To make this ether? You can make that oxygen-carbon bond. There’s more than one way to skin this kind of cat. That one won’t work. But if we do it the other way around and make this anion, which again would prefer to eliminate rather than substitute. But you attack methyl bromide which can’t eliminate because it doesn’t have a hydrogen on the other carbon because it doesn’t have another carbon. Now you’ve got the MTBE in high yield although nobody makes it that way. The stuff that’s made for petroleum is made a different way, and we’ll talk about that a little later on.
Anyhow, this illustrates that often there are different ways of choosing putting the reagents together. One will work and one won’t work for reasons that you can understand mechanistically.
Here’s another interesting synthetic intermediate which is, if you start with a compound that has OH and chlorine on adjacent atoms and notice it has one in the front and one in the back. You treat this with base. It gives this funny ether, an ether that’s a three-membered ring. It does so in pretty good yield. How does it do it? Well, hydroxide takes off this proton and can you see what happens next? It’s a reaction that’s familiar to you. We want to make this bond. How do we make that second bond, the new bond, in the three-membered ring? It’s an SN2 reaction. This HOMO attacks σ* backside, so it’s an intramolecular SN2 reaction. And chloride leaves to generate this three-membered ring. The three-membered ether is called sometimes epoxide, sometimes oxirane. Some people call it one, some call it the other. In fact, epoxides are very useful synthetic intermediates, because, although RO– is a crummy leaving group, in this case it’s helped by losing the ring strain. So it’s spring-loaded to open up. You can attack here with some nucleophile, break the bond, the O– comes away, and this gives a way of adding C-C-O to a nucleophile. You bring some nucleophile in, and it adds to the nucleophile two carbons and then an oxygen. Synthetic chemists often think, what is it? If I want to get this product, that’s something plus C-C-O. How could I do that? Ethylene oxide. That epoxide can do it. In fact we’re going to talk more about that as a synthetic intermediate later. For example, that’s the stuff from which you make crown ethers because it has, remember, C-C-O, C-C-O, C-C-O, but we’ll talk about that in a few lectures further on.
For synthetic purposes a particularly useful set of nucleophiles is cyanide, acetylide, the thing you get by losing a proton from acetylene, and this other type of anion, which has an α-anion. It’s lost a proton adjacent to carbonyls. We talked about that one before. The reason these are interesting is because they are nucleophiles that form carbon-carbon bonds. Most of the things we’ve talked about replace one functional group by another. A leaving group goes away. The new one comes in. The new one is typically something like oxygen or nitrogen. But these anions have carbon coming in and that means you make a new carbon skeleton. That’s something that’s very important in putting together organic molecules.
Notice that these two both have a pKa of 9. You put the starting material–that is, the protonated forms have a pKa of 9–put them in with base and you get the anion and it does its trick. Acetylide is pKa of 25. You can’t use that in water. It would immediately take the proton away from water. So it’s a base, and it actually tends to do elimination sometimes rather than substitution. So you need a stronger base if you want to pull off a proton and do that. You can do it with sodium amide notice that this is the anion coming from NH3 as the acid. And NH3 then is the product after it’s pulled off this proton. And it is a much, much stronger base, a much weaker acid by nine powers of 10 than the acetylide is. You could make it that way. These could be important in synthesis. We’ll see later on in the course how very important nucleophiles that are carbon are in putting molecules together.
Chapter 5. SN1 and E1 Reactions - Kinetic Evidence for SN1 Substitution [00:41:05]
Now if we look at the rate of the reaction of sodium hydroxide with an alkyl bromide, for different R groups. We’re going to look at a log scale so we can cover five powers of 10 in the rate. And it’s how much of R-Br gets converted to HO-R per minute. We’re going to do it in ethanol/water, 4:1 at 55 degrees. We see 1% about 10–2, of methyl bromide is converted to this ether–or pardon me, to the alcohol–per minute, 1% per minute.
If we make it ethyl bromide, it’s only a tenth of a percent per minute. What would you expect for isopropyl? Looks good. What do you expect for t-butyl? So there’s something funny. Something has changed when we went from methyl, ethyl, isopropyl, and t-butyl.
This was done with 0.01 molar sodium hydroxide getting these rates. It turns out that the rate of these depends on how much hydroxide there is. That doesn’t surprise you, it’s SN2. Depends on how much of the nucleophile there is. In fact, if you looked at the second-order rate constant, which takes into account the concentration of base, then you see these fall on a nice curve and it’s slowed down by crowding, the steric hindrance that we talked about before. But you’d expect t-butyl to be down here someplace. In fact, it’s up there. That’s because that one is not dependent on hydroxide. It’s not SN2. The rate doesn’t depend both on the halide and the hydroxide. It depends only on the halide. At the rate-determining step, the base hasn’t gotten involved yet. Notice that it’s accelerated by crowding. When you get to the most crowded one, that mechanism goes really fast. It can happen because the cation is more stable when it’s more substituted. It’s also accelerated by a polar solvent. If you try a solvent that’s not polar, it’s harder to make ions in it. But the ethanol and water is a reasonably polar solvent.
This is the SN1 reaction, as opposed to SN2. This categorization was formulated, SN1, SN2, in England, in London, in the 1930s by Hughes and Ingold. That rate actually wasn’t measured at that temperature. They measured at a lower temperature and figured out what it would be at this temperature. This is how much is converted to HO-R but some of the starting material is not converted to HO-R. Some of it undergoes elimination. At this stage 19%, about 1/5 of the product, is in fact the elimination product, E2. The base attacks the adjacent hydrogen rather than waiting for it to ionize. And we know it’s E2 because the amount you get depends on how much hydroxide you put in there. The rate of forming the alcohol, the rate of the substitution, doesn’t depend on the hydroxide. But the rate of forming the alkene does depend on the hydroxide. You change the ratio of the products by changing the hydroxide concentration.
Chapter 6. Carbocation Intermediates: Competition and Rearrangement [00:45:09]
Now you can have SN1 and E1 as well. In that case we just talked about, if you put more hydroxide in, the starting material goes away more rapidly because the elimination is dependent on hydroxide. But in the case here where it’s cyanide, that’s the high HOMO, that’s doing the attacking, you find that you get not only the cyanide substitution but you also get substitution by water. The product ratio depends on how much cyanide you put in. That doesn’t surprise you. You put in more cyanide; the formation of this product is faster. But the rate doesn’t change. The rate of destroying the starting material doesn’t change when you do this. Can anybody see how that could be? How could you change the products but not change the rate? Matt?
Student: Because even though you have a different ratio, it could be doing a different mechanism, possibly.
Professor Michael McBride: Let’s try it here. We have this starting material and these two products. We’re going to draw a reaction coordinate diagram to get from the starting material to the products. Here we formed, as in SN1, an intermediate cation. And then it goes on, if there’s cyanide around, to give that product or with water to give that product. What determines the rate of the process? What’s the rate-determining step? The red one, right? That happens before you decide which product you get. If you have more cyanide in there, more of it will go this way. You can change the products without changing the rate, if you have this cation intermediate. So if the product is determined after the rate, by competition for the short-lived cation, that’s evidence that you have a cation intermediate in this substitution.
Here’s silver nitrate helping to pull off the leaving group, iodide. Notice what’s funny in this case.
Whoops, what’s funny is I’ve talked my head off here. There are a few more slides that I think it would be helpful to have to make the exam coherent. Since the exam’s going to be on Friday, why don’t we put a few more of them onto the next and I apologize. I’ll let you go now. We’ll do a few more slides next time for the exam. And there will be a review session on Wednesday.
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