Freshman Organic Chemistry II

CHEM 125b - Lecture 17 - Alkynes. Conjugation in Allylic Intermediates and Dienes

Chapter 1. Addition to Acetylenes: Regio- and Stereochemistry [00:00:00]

Professor Michael McBride: OK. So we’ve been talking about alkenes and their polymerization. And now, we will go on to the triple-bond analogs, acetylenes, for just a little bit, and then on to conjugated systems, starting with allylic intermediates, and then dienes, and then, in general, about linear and cyclic conjugation and aromaticity.

So, we can generalize from what we know about alkenes to acetylenes – that we have the same kinds of reactions, additions of HBr, addition of water, addition of H₂. But all these have a little bit of a different twist. HBr, we already mentioned this but I’ll go over it again, stepwise addition, but Markovnikov. The addition of water, we see the possibility of keto-enol tautomerization and regioselection, as in alkenes. And then in H₂, there’s a question of a stepwise addition. One double bond and then the other, one multiple bond and then the other, and the question of stereo selection. And finally, some stuff about acidity and base-catalyzed isomerization of acetylenes which you’ll see are much more acidic than alkenes are,

First, we talked about this already but just to go over it quickly once again, it’s possible to add HBr to a triple bond like 1-hexene, and you can use a catalyst that helps makes the HBr more acidic, and you can get a single addition to give the hexene– 2-bromo-1-hexene– and you often use an inhibitor so that you don’t get free-radical addition of HBr.

Now we mentioned before it was 100 to 1000 times slower than the comparable ionic addition to the alkene, and the reason is the vinyl cation is not so great an intermediate. And you can see that already in gas-phase ionization of the chlorides. The one that gives the ethyl cation requires only 193 kilocalories per mole to ionize the chloride. But the vinyl compound, to make the vinyl cation, requires another 32 kilocalories per mole. It’s hard to make the vinyl cation. So that is what, of course, what you make when you add the proton to the hexyne.

It’s possible to isolate that in 40% percent yield which seems funny, as we mentioned before, because if it’s not very reactive, wouldn’t the hexene you make be even more reactive, so that it would be impossible to isolate something that was part way along. So, you can go further to 2,2-dibromohexane. Obviously it’s slower, and it gives a second Markovnikov addition, right, which if the bromide slows the addition to an alkene, in order to make it slower than the initial addition to the already slow alkyne, why is the orientation Markovnikov? And we saw the resolution to that riddle a few lectures ago, that the halogens do two things, they withdraw electrons through the sigma bond and, therefore, make it harder to form the cation. But if you form the cation, make it next to the halogen so that at least you can profit from stabilizing the unshared pair on the halogen. OK, we saw that before.

Now, hydration and hydrogenation of alkynes. First hydration, we can do the same kind of thing we did before, the alkoxymercuration, which you might remember forms an unsymmetrical bridged ion, then water can open it up, and you get the addition of Hg and OH, hydroxymercuration. Now remember what we did before when this was an alkene to begin with, and we had a single bond in the product? Nobody wants the mercury compound. Why did people make it? The next step was reduction, remove the mercury, and replace it with H, so that what you added was effectively water. Right? So, they used sodium borohydride to do that. However, it doesn’t wait around for you to do that under these conditions, because you have acid there. And with the double bond it’s possible for the acid to add to it. Now, so you get that cation. What’s good about that cation? Why is that easier to form than it would have been if it had just been a regular alkene? Elisa?

Student: There’s the O there.

Professor Michael McBride:  What does the O have to do with it?

Student: It’s able to protonate it and form the double bond.

Professor Michael McBride:  No. Well, a double bond gets formed ultimately. But, what’s good about the cation when you have the oxygen next door? Brandon, what do you say?

Student: The p-orbital?

Professor Michael McBride:  Yes, it has unshared electrons in the p-orbitals, that could be stabilized by the vacant p-orbital next door. So there’s a resonance structure that makes it unusually stable. A HOMO-LUMO interaction. But, what leaves then is the Hg⁺, OK. So, you get this compound, which is an enol. Notice that it was formed with Markovnikov orientation, as would have been the case if it had started with an alkene, right. But enols can undergo this same acid-catalyzed reaction to get that cation, which is stable as we can see in that resonance structure, and what could that do? Lose a proton.

So it gives a ketone. This got traditionally the name keto-enol tautomerism. “Tauto” means equivalent to, right? So the idea was that the ketone and enol interconverted so easily that it was impossible to isolate one or the other. So that’s a really easy allylic rearrangement from one end of, from one allylic position to the other. And we talked about the ketone-enol equilibrium last semester when we were talking about bond strengths. Just to remind you, this is where we went through it, added up the bonds, and showed that the ketone is much more stable than the enol. I’ll just flip through that, so when you’re reviewing it you can look at it there. OK, so it goes to the ketone. So the product from hydration of an alkyne is a ketone. The alkoxymercuration of an alkyne doesn’t wait around to do the sodium borohydride reduction.

Now, remember, you could do the other thing in an alkene, add in the other regiochemistry by using a borane. For acetylenes, you use a special borane. You get a vinyl borane as the product, and it won’t add a second time if you have those R’ groups sufficiently hindered, so that it blocks access, OK. A way you can do that is with a compound called disiamylborane, which is made with BH₃ and that alkyne. And so you might think about what the R groups are, the R’ groups, that you put on there to make that hindered one with a single H that then adds only once.

So you got that vinyl borane and remember what you did with boranes when we added to a double bond? What was the next thing? Hydroboration and then what?

Incidentally, this reminds me that a week from today we have another exam. And it’ll cover today’s and Wednesday’s lectures. I’ll give a review session Wednesday evening. I’ll tell you, it will probably be 8:00 to 10:00. And so, you have a week to learn this stuff. That’s good news.

Anyhow, it was hydroboration, oxidation was the next thing. Remember you treat it with hydrogen peroxide and alkoxide to make the borane into an alcohol. Right? And then, in this case, we have an enol as above. So, that’ll isomerize to a carbonyl compound. But notice it was anti-Markovnikov, the addition above was Markovnikov the oxygen went to the more substituted center. This one, the trick you use when you do borane, the O goes to the less substituted center. So in that case you get an aldehyde. So, the H₂ on one carbon and the O on the other, add in one sense here and in the opposite sense here. So that’s nice because if you’re a synthetic chemist that gives you the control of which product you want to get. You can use one method to get the Markovnikov ketone and the other to get the anti-Markovnikov aldehyde. So this is hydration with either regiospecificity.

But, also, we could talk about hydrogenation, adding H₂, and here it’s a question of stereospecificity. Up at the top there if we take an alkyne and reduce it with H₂ and a catalyst, as we talked about with an alkene. You can get first this alkene, and you use a special catalyst, this is called… was developed by a guy named Lindlar, and he added lead to the palladium catalyst. Lead is a poison for catalysts. It makes the catalyst less active. Why might you want the catalyst not to be very active? Chris?

Student: So it doesn’t go all the way.

Professor Michael McBride: So it doesn’t go all the way. It’ll do the acetylene, but it won’t do the alkene. So you can stop and get it there. If you use a more reactive catalyst, you can go all the way. So that deactivates the palladium and stops at the alkene stage. And notice the two hydrogens came in from the same side. So it’s a syn addition. For example, you get a 97% of that syn addition if the R is this (CH₂)₃ ester group,.

Now on the other hand, if you use sodium and ammonia, a thing called a dissolving-metal reduction, you get the opposite stereospecificity. It’s an anti addition. Now you might wonder– the syn addition we talked about before, the same thing in an alkene about how the palladium comes in and attacks, then it gets replaced by one hydrogen, by another hydrogen, always front-side attack, so that it comes out to be syn. That one won’t surprise you. But why does this one give anti? The key is that it’s a dissolving-metal reduction. Incidentally, in that case, with n-propyl groups it’s an 80% to 90% yield of that anti addition and I think none of the syn. OK, so it’s a dissolving-metal reduction. So sodium actually dissolves in ammonia to give electrons and sodium plus. So it’s a very interesting ion pair. It is an electron and a sodium ion. Of course, the electron is solvated by ammonias around and about it, and it looks blue, the solutions of electrons.

Now if you have an acetylene, an electron will add to it. What orbital will it go into? What’s the LUMO going to be? Yigit, what do you say?

Student: The π*.

Professor Michael McBride:  Yeah, π*. So notice it puts in– there is already a pair of electrons in that corresponding π orbital. So now you have three electrons among those two p-orbitals.

So, you can write it as one being an anion with two electrons, and the other being a radical with the third electron. So, we have an anion radical as an intermediate. That, of course, is a very strong base the carbon anion. So it can react with the ammonia and pull off a proton. So, we’ve added an electron and a proton. That’s the equivalent of adding a hydrogen atom, although it’s done is two stages. So, we get this vinyl radical, which is bent, right. They’re sp² hybridized at that functional carbon. But they invert easily in the case of a radical, right. So it goes back and forth between the R being on the same side and the opposite side. Now which one do you think is favored? Obviously, the one with the R groups on the opposite side. So at equilibrium that’s the one you have.

And now a second electron comes along and goes into that singly-occupied orbital. So now you’re going to have a pair of electrons there. We will write the minus charge in that sp² hybridized orbital. But, now it’s harder to invert. Remember XH₃, how if you added more electrons it ceased to be planar and became pyramidal. The same idea here. So now it doesn’t flip back and forth rapidly. So, a second proton can be taken away from the ammonia, and we get that product, and the two H’s as are on opposite sides. It’s an anti addition. Why anti? It’s because that isomerization of the radicals favored the one with the Rs far apart from one another. Then it got trapped as the anion in that geometry. And then it went on to give the anti product, as we gave yields of on the previous slide.

Chapter 2. Acidity and Isomerization of Acetylenes [00:14:07]

Now, the final special thing about alkynes that we’re going to talk about is their acidity and how that relates to isomerization of alkynes. So, we looked before at these pK_a values for carbon, nitrogen, oxygen; in the order you would expect them to be. But that if you have an sp² hybridized carbon, or an sphybridized carbon, they get much more acidic. We talked about this before. And in some cases if you have next door… if you’re breaking a CH bond and next door there’s a double bond, you can get extra acidity. OK, that stabilization. OK so anyhow, we’ve looked at those pK_as, and here we can see the energies of these three isomers of an alkyne. So you have the alkyne on the left, the lowest energy is heavily substituted, on the far right is one that’s at the end of the chain, so not having as many carbon-carbon bonds with the special sphybrid, it’s not a stable. And in the middle we have the diene version, the 1,2-diene.

So at equilibrium you would strongly favor the internal acetylene. But how do you establish an equilibrium among these? You have to move hydrogens around. So at equilibrium, given those energies, there’s only 0.1% of the 1, 2-diene and 0.03% of the terminal acetylene. So very little of those.

So, we just talked about the fact that the terminal acetylene could be, rather the internal acetylene could be acidic because of the π bonds next door making the anion more stable. So indeed you could do that, or you could do this one, that was on the previous slide as well. Or you could do the terminal acetylene, which is the easiest one to take off to put the electrons in the low-energy sp hybrid orbital.

Now the pKa over here is 38. So that means that the K_a is 10^-38, alright. Or the energy involved in getting up there’s about 51 kilocalories per mole. Over on the right it’s 33 kilocalories per mole. Still way uphill. So those energies are quite far above the.. that’s what we know from the pKa s. So if we wanted to take off a proton, then put it back on and take off a different one, put it back on, and finally go to the final anion, we would have to go way up in energy. If you have to go up 51 kilocalories per mole, that’s not a very fast process. The rate constant for that would be 10¹³ per second times 10^-38 , or the halftime would be about 10¹⁷ years, which is about 10⁴ times the time that’s elapsed since the big bang. So don’t wait around for that one to happen, OK? So, obviously, you’re not going to be able to isomerize that way by just let it get ionize off a proton.

But, if you put hydroxide in it reacts with the proton to make water and that’s worth 21 kilocalories per mole. Remember, the PKA is 16, 4/3 of 16 is 21. So, 21 kilocalories per mole. That makes it 21 kilocalories easier to make the anion, because you’re not making proton, you’re making water. So that’s going to lower these. Is that going to do the trick to be able to isomerize? Well, now the halftime would be about 30 years at room temperature. But, if you warmed it to 150K it would be about two minutes. So, if you heat it up you could make the isomerization occur. Then no matter which of these you start with it’s going to equilibrate, and this is the lowest energy form, and that’s going to be everything, except there’s going to be 0.1 % here and 0.03% there, OK. So no matter which of these you start with you’re going to get that.

However, suppose you use a really strong base like amide. So now that favors loss of a proton. That is it’ll take the proton by 45 kilocalories per mole. So now we’re going to lower those ions even further, like that. And notice what happens on the far right. It goes all the way down there. That’s now the lowest energy form, if you have NH₂^-. So, instead of having it be all here, which is what you’re going to have in any equilibrium among the neutral forms, this is going to be the lowest-energy form. So everything will be over there, 7.2 kilocalories down, which means that at equilibrium you will have only 0.001% of this stuff, which was always the favored stuff before. So everything is now over here. So the trick is now you neutralize it very rapidly. So the trick to obtain a terminal acetylene, remember that anion is the one that has the anion on the last carbon, that’s why it’s such a stable anion because it’s sp hybridized. So, you equilibrate with an amide base in an amine solvent to form the terminal anion. Then you quench it by adding water to donate a proton to the terminal anion and also to the RNH. So you have only hydroxide now, which is too weak to allow equilibration. Or if you want to, you could just add acid so that you don’t even have OH there, although, that’s gilding the lily. You’ve already made it by not having that really strong base. So you stopped equilibrium, you put the proton on the terminal acetylene, so you can go from the internal, the more stable one usually, to the terminal acetylene using the trick. OK, so that’s all I wanted to say about acetylenes.

Chapter 3. When Does Conjugation Matter?  Allylic Intermediates and Transition States [00:20:32]

And now we’re going on for these next two lectures, and a little bit probably still beyond the exam, to talk about conjugation and aromaticity. So conjugated π systems is what you’re talking about. The name conjugated comes from an interesting root. Conjugated systems are when you have π bonds separated by a single bond. And they are called conjugated because they are yoked together. The word yoke is the same root. Yoke comes from jungere from jugom, which is to join. So they’re joined together. They don’t act independently. And we talked about this already in the first exam when you memorized functional groups. And on the exam I had a thing that had both the carbonyl and an amine, and they were near one another. But it’s not a carbonyl group and an amine group. It’s an amide group. They are conjugated and they behave in their own special way. So now we’re going to talk about conjugation.

So, conjugation is where the localized orbital picture, which remember we’ve talked so much about pairwise two atoms coming together to give a bond and an antibond, or isolated atomic orbitals. That was our intermediate between hydrogen-like atomic orbitals, which we really understood– we had a table that we could draw them out– and the computer MOs, which looked very complicated. But in between were these pairwise things. But sometimes we have to think more deeply when these things are conjugated. They don’t behave that way.

So when does conjugation make a difference? First we’re going to look at experimental evidence, and then we’re going to go into the theory. So a classic example is allylic stabilization. Allyl comes from the word– the same as aioli– or ail in French meeting garlic. There’s a sulfide in garlic that has a double bond, that is carbon double bond, single bond, and then a functional group, a sulfide. So that’s an allylic sulfide came from garlic. So, allyl is these three carbons in a row, double bond, single bond. So, you can get special stabilization when you have functional groups involved, when you have rather intermediates involved. They’re like the cation intermediate. So the energy for pulling off a chloride for primary and secondary are these values in the gas phase. Of course, very hard to separate the ions, but if it’s allylic it’s as good as being secondary. So allylic is very good.

Or if you look at anions. Look at the pK_a of the OH. Here it is 16 we know, if you have a double bond here it falls by six orders of magnitude. And in the case where it’s an oxygen here it falls by another five orders of magnitude. You actually call it an acid. So the anions get much more stable when they’re allylic by about 8 kilocalories per mole, in the case of the enol.

Or free radicals. If you look at the bond dissociation energy of a normal CH bond, 101, but if you make a double bond next door so that this is allylic, it’s 89. It is stabilized by about 13 kilocalories per mole. So there’s stabilization of these reactive intermediates when they’re allylic.

Now let’s look at some more detail about some of these like the cation intermediate. The addition of HX butadiene. We talked about HX, halogen acids, adding to single double bonds. How about adding into two double bonds. OK, so at minus 78°, you can add HBr to butadiene, and you could imagine the H⁺ going on one of the internal carbons to generate that cation, or it could go on the terminal carbon, the other regiochemistry, and generate that cation. Which of those do you think is more likely to get? Which one would you prefer, if you were the proton? Ayesha?

Student: The one that’s down.

Professor Michael McBride:  The lower one. Why?

Student: It’s a secondary cation.

Professor Michael McBride:  First, it is secondary. And it’s also allylic, right, which the first one isn’t, because from this carbon to the double bond there’s an intervening saturated carbon. It’s not two double bonds or two sets of p orbitals separated by a single bond. There’s no overlap between this p orbital and that p orbital. So, the second one is, of course, also that, because those are resonance structures. But as Ayesha tells us, the top one doesn’t look so good. So we’re going to get one of those. But then in a second step we add bromide. If we do that at minus 78°, of course, we generated the bromide when we attacked the proton. So, we could imagine getting those two forms by attacking this intermediate, which is denoted by two resonance structures.

Now, in fact, you get those in the ratio of 80% to 20%. Does that surprise you? What would you expect for the ratio of these two? Here you have a terminal double bond. Here you have an internal bond. Which one is more stable? Do you like to have double bonds more substituted or less substituted? Arvind, do you remember?

Student: More substituted.

Professor Michael McBride:  More substituted. So, this is backwards. It’s 4 to 1 in the wrong direction. Now that means that if you allow it to equilibrate, it’ll go the way Arvind wants it to. But, the initial product formed at really low temperature, at dry ice temperature, is backwards. You can cause, this in fact, can ionize to go back to the cation and then come back, and that’s a root for equilibration between these two. And you can encourage the bromide to come off by putting ferric bromide in, if you want to, it’ll go faster. But, at any rate, that’s equilibrium. So this is called kinetic versus thermodynamic control.

So thermodynamic is equilibrium. But kinetic is what you get first, and those aren’t necessarily the same. It’s this case we’ve seen before where you start and you go up, a certain rate, and then down to a product, or you could go to a more stable product. Hammond would say, you’d go faster to the more stable product, but sometimes, remember, it crosses and it’s easier to get to the less stable product, and that’s the example here. It’s a kinetic control under that low temperature. Now why when it’s a question of rate it’s the first one to be formed. Why do you think you form the top one? Anybody got a theory? Yes.

Student: The positive charge.

Professor Michael McBride:  What about the positive charge?

Student: The electrons on the bromine, the bromide ion, are attracted to the partner. .

Professor Michael McBride:  Right. It looks like this is a secondary versus primary; more of this, so that’s where it attacks. That looks like a good explanation. And that’s what you’ll see in almost all textbooks. So very good, Anurag. You can write a textbook.

Let’s look at the orbitals that are involved in this just for fun, quickly. So here’s a top view down on the four p orbitals to the butadiene, and here’s the side view. So here’s the HOMO - 1. That’s the favorable combination of those four p orbitals. And here’s the HOMO, that’s the unfavorable combination. It’s π on this side, π on this side, but unfavorable between them. That’s the HOMO. The LUMO has two nodes among the four p orbitals. And what do you think the LUMO + 1 looks like? It’s got three nodes. So you’ve got no nodes, one node, two nodes, three nodes for combinations among those p orbitals.

Now, the HOMO is the one you want to attack with the proton. So where should you attack it? Probably, you get the best overlap around in there. Or you could look at how the potential of the surface that we talked about to see where the best place for a proton is, and it’s in the orange region. So that seems to more or less to agree. It’s about there. So that’s where the proton should attack at the terminal carbon. And then it would give the propenyl cation, which is shown here. There are three hydrogen on this carbon here. So that’s the best product then. Now that’s the HOMO - 4, it’s a little bit silly to look at that, but it’s actually a CH bond that is hyperconjugated a little bit with the p. Now we have three p orbitals here, here, here. The one that was here became part of this CH bond. Now, how about the rest of them? Here’s the HOMO it’s π, here’s the LUMO, one node, and here’s the LUMO + 1, two nodes among those three p orbitals. So that’s the propenyl cation.

There’s the LUMO that you’re going to want to attack, if you’re the bromide. So you could attack at either end, not in the middle. And if you look at the surface potential, the worst place for a proton is where it’s very blue here. So that would be the best place for bromide to come. But it’s in the plane of the ring. It’s not a place you can form a bond. If you want to come in from the top where you can form a new bond, it’s a little bit green there, and a little bit blue, even, better there. So the blue place +144 would be the best place to attack, if you want to form a bond. That’s where bromide would tend to be in order to attack. So one would think that this is exactly what Anurag has proposed, that it should attack there.

But here’s a really interesting paper that was published in 1979, where what it’s being added to is not butadiene, but pentadiene, another methyl group, on there. So they add DCl which adds to the end to get the, as one would expect, more substituted allylic cation. But now there’s something interesting about that. That’s symmetrical, except for the fact that there’s a deuterium there. So if a bromide now comes in, or in this case they use chloride, it could attack either here or here. If it attacks here, you get the 1,2-addition, D and Cl. If it attacks here, you get 1,4-addition, D and Cl.

Now, if it had to do with the kind of thing we talked about before, where’s the best place to put a chloride, where do you got the best overlap, and so on. That’s the same here and here. So it should be 50/50, in this case because of the intermediate being symmetrical. When it was butadiene, and we didn’t have this methyl group and it was unsymmetrical when we did the analysis of the previous slide, where would be the best place to attack? But now it’s symmetrical, so we should get 50/50. So here’s the product distribution done in a bunch of different solvents and different temperatures and just look at a couple of these. If you do it at room temperature, at 25°, you get half again as much of this as you do of this. More chlorides adds here than here. And if you do it at low temperature, you get three times as much addition here as here, even though it’s symmetrical. So we’ve got to find some other reason why when we do this, the chloride tends to attack here, rather than here. 1,2-addition rather than 1,4-addition. Can anybody guess a reason for that? Well let’s look at the mechanism. The first thing that happens is we a proton on and generate this cation and also chloride. Now, do you have an idea? Ayesha?

Student: It’s closer.

Professor Michael McBride:  It’s closer. It has to move farther to get to 1, 4 because this backside attack on the hydrogen meant that the chloride has to be out here someplace– closer to this end than to this end. So it appears that rapid collapse of this ion pair that you’ve made, before the ions could move independently and establish true symmetry, that competes, so that the collapse competes with the motion that would make it truly symmetrical. So it looks like it might not have anything to do with where the charge is on the surface or which resonance structure is better. It’s just where the chloride got formed and that attacks really quickly when the ion pair collapses. So you at it you could have written a textbook, but you might not have been right.

Now, allylic transition states also are unusually stable, and we actually mentioned this before. But if you look at an S_N1 reaction, so you’re breaking away to get a cation, the relative rates are, if you take 100 for t-butyl which is very easy to form the t-butyl cation, the secondary is 10⁴ times slower, and the primary just doesn’t happen. It’s way slower than that. But if it’s allylic then it’s 5 times faster than secondary, even though you might think it would be primary. And if it’s allylic with a methyl group on it, to make the cation more stable that you’re forming when you break off chloride, it’s half as fast as a tertiary– t-butyl chloride. If you put two methyl groups there, it’s very much faster, 63 times faster than t-butyl.

And if you put a methyl group here, however, on this carbon… So we put a methyl group here it accelerates enormously. If we put two methyl groups there it really accelerates. [brief confusion] If we put it on the other end it’s just as good. But, this is what I wanted to tell you, if you put the methyl group in the middle it doesn’t help at all. So the methyl group has to be on one end or the other of the cation you’re going to make when you lose the chloride, in order to get that acceleration. If you put in the middle, it doesn’t help.

So again, we could look at where the charge is in the cation, and you have to put the methyl, to be effective in accelerating the S_N1 reaction, it has to be where the charge is going to be. Now, how about S_N2? If we take the rate of ethyl chloride to be 1 reacting with ethoxide, it’s 37 times faster if it’s allylic, and 97 times faster if you put a methyl on, and 560 times faster if you put two methyls on. If you put the methyl in the middle it doesn’t help compared to not having a methyl. But if you put the methyl at this end, remember before when we were doing S_N1 forming the cation we could put it either at this end or this end and got the same rate, now it’s slower here. Any idea why it should be slower in this case for the S_N2?

How about the rate of t-butyl for S_N2? You have t-butyl, three methyls, and you want to do S_N2. It’s very hindered. So this hinders the approach because, remember you’re making a pentavalent carbon in the transition state. So allylic helps, but this time you’re going to get help from the methyl at this end, but not at this and. So we looked at the allylic cation, we looked at the transition states involved in forming the cation, and in the S_N2 process.

Now let’s look at the allylic anion. That’s easily done by looking at the acidity. So we have a pK_a of something like 52, it’s hard to measure for that, but if you have it allylic it’s 43. And that’s helped by 9 powers of 10 or about 11 kilocalories per mole of stabilization for losing that proton at equilibrium. And if you a phenyl ring there, it what’s called benzylic. And now it’s even more so, about 16 kilocalories per mole. So again having these π things where they can overlap with the anion that’s being formed is very favorable. The order of 10 or 15 kilocalories per mole.

Chapter 4. Allylic Radicals and Allylic Bromination [00:38:31]

How about free radicals? Well here’s a very interesting… these figures came out of this paper by Ziegler. This Ziegler is not the Ziegler who teaches CHEM 220. This is the one who invented Ziegler-Natta polymerization and became quite wealthy. So in 1942 working in Germany in the Second World War he was studying this process. He could get a 58% yield of substitution. This bromination is a substitution– pulling off one of these hydrogens and replacing it by bromide. Now for this purpose, the paper that describes this particular compound which is called N-bromosuccinimide. And in ether solvent in 30 minutes using light. What does the fact that he was using light suggest? That it might be a free-radical chain reaction. He was able to get this 58% yield.

Now where does he get the N-bromosuccinimide? So that compound is succinimide. He reacted it with base to make the anion. Does it surprise you that that anion can be formed with hydroxide? Remember, NH has a pK_a of 30 or something like that. And hydroxide is only 16, the pK_a of water. How’s it able to pull off that proton? What’s good about that anion that we formed? Megan, do you have any idea.

Student: The lone pair is stabilized by the carbonyl π*.

Professor Michael McBride:  Right. You have two carbonyl π*s to stabilize the high-energy electrons on this nitrogen. OK, so you can make that anion, the pK_a, incidentally, of that is 9.5. It’s a very strong acid as NH bonds go. And that could do an S_N2 reaction on Br₂ to put bromine on the nitrogen, and make the bromide anion together with the sodium ion that came from hydroxide. So that’s how we got NBS.

But you can take it apart again, if instead of using base, use acid. So if you use HBr, you can protonate this. You can protonate that oxygen, and you could draw a resonance structure here like that. So it’s easy to protonate here. But now you can take the bromide and do an S_N2 reaction on the Br. So you generate this, now neutral, which is an enol that easily isomerizes back to the starting material. So in base you could make this with bromide, and if you add HBr, you come back. And of course also get Br₂ that was what was formed here.

So you can get the Br₂ back again, if you want to, from NBS.

Now back to the reaction. We can imagine doing a reaction of Br₂ with a double bond. What reaction have we talked about Br₂ plus a double bond? It’s the addition, electrophilic addition to a double bond. We talked about this a good deal. No doubt it will be covered on the day exam next Monday. You do this here at room temperature in the dark. You get the bromonium ion first. Remember, how we talked about that and then you get anti addition.

Now, but suppose you want to do substitution. That’s what Ziegler did. He didn’t add bromine to the double bond; he replaced an allylic hydrogen with bromine. So those substitutions are typically free-radical substitutions. So you could have this same compound, but I show the hydrogens, because if you have ways to get a free radical, like a bromine atom, it can pull off, get HBr, add this.– but remember, we use bromine in here, not chlorine, because it’s slow and selective. So if you use chlorine, it takes not only this hydrogen but also this hydrogen almost as rapidly; it’s not selective. But bromine is selective; it’s uphill. Then you can react it with Br₂ and replace where the hydrogen is with bromine.

So you’ve done allylic bromination. And, of course, this allylic intermediate was resonant in this form. You could have put the bromine either here or here. So you also get that, which is the same material in the case shown, but if you had a substituent, it would be in a different position. If you had something on the ring substituted here it would be a different compound. And it also gives a bromine atom. Why is that relevant? What happens to the bromine atom? Matt?

Student: It gets reused in the free-radical chain.

Professor Michael McBride:  Right, goes back to the start. It is a chain reaction. So how do you control what you’re going to get? Are you going to get bromine addition or allylic substitution? Can you think of any way of controlling which of these. You put bromine in, which of these things is going to operate? Is there any handle you have on it? Matt?

Student: You can use photoinitiation to get the radical.

Professor Michael McBride:  If you use, noticed this was done in the dark. If we can’t get the radical, we won’t do this one. So one idea is to keep it in the dark. Don’t use the initiator. So that’ll allow us to do this one, that dark one. How do we get this one to happen? Well you might say we use light or use an initiator. But this can still happen if you have the light on. You can still get that. It doesn’t require dark. It’s just this one requires light. OK, so this one would also be going on

But there’s another way, which is to minimize the concentration of Br₂. Now, why does that have anything to do with it? This is a really interesting thing about the kinetics of the top reaction. It’s second order in Br₂. That looks really weird. Why do you need two molecules of bromine in order to do this? The reason is that when you get the Br⁺ coming on, Br^- is coming off from Br₂. So you have to get Br^-. And, in fact, Br^- is stabilized by reacting with Br₂ to give Br₃^-. So, in fact, what happens is that the second Br₂ helps the bromide leave from Br⁺ in a non-polar solvent like this. So it’s second order. So that means if you have a really low concentration of Br₂, this one gets really slow, because it’s a bad leaving group. It’s like making the Br^- leave by putting Br₂ in is like making OH^- leave by putting a proton on it. Make it a better leaving group,

So if we minimize the concentration of Br₂ we’ll slow this one down, but that reaction with Br₂ is very fast. Remember, the first step of this is endothermic, is slow, but that’s step is really fast. So we can tolerate a low concentration of Br for this purpose, but how are you going to do it. Because suppose you put in just a teeny amount of Br₂ there. OK, this start in, but quickly you’ve exhausted the Br₂. now the radical chain stops, and that’s the end of the line. Now you could say I’ll very carefully run in Br₂, very slowly just to balance out. But you put a drop in the bromine concentration is higher in that drop, locally, right, so you’re going to get this reaction, because of the local high concentration of Br₂.

But this is where Ziegler comes in with his clever NBS molecule. This looks like it’s tedious to impossible to maintain a really low concentration of Br₂ while it’s being consumed all the time. But notice that this reaction also yielded HBr, whenever that happens. And if we have N-bromosuccinimide plus HBr that’s the reaction we looked out on the previous slide that destroys N-bromosuccinimide– takes it back to succinimide. And you remember what the other product was? Br₂. So isn’t this neat? Whenever a Br₂ molecule gets used up, a new Br₂ molecule is created by this reaction. So it automatically maintains the minimum possible concentration of Br₂. Every time you use it, you get a new one. You don’t have to sit there with your burette, trying to run it in very slowly. So that’s the trick with n-bromosuccinimide.

Chapter 5. Modest Stabilization of Conjugated Dienes [00:47:59]

Now, so we’ve looked here at a lot of different allylic stabilizations. But dienes can also be conjugated, and how much is that worth? OK, we can look at the heat of combustion of this one, where they’re not conjugated, and this one, where they are across a single bond, these isomers of one another. And you can see that the conjugation seems be worth about 7 kilocalories per mole. Or down here it seems to be worth about 8 kilocalories per mole. But these numbers, although they’re claimed to be very precise, are big numbers. And I think that precision might be a little bit overstated. If you look at the heat of hydrogenation, the heat that is given off when you add H₂, then you can see that these are all 30.

I’m going to let you go after you make two guesses, OK. Now guess what each, this double bond, this double bond, this double bond, they all give off 30 kilocalories per mole of heat. How much should this one give off? When you hydrogenate both double bonds. Leen?

Student: 60.

Professor Michael McBride:  Right. How about this one? Natalie?

Student: 60.                                                                                                    

Professor Michael McBride:  How about this one?

Student: Something bigger.  Much more. 62.

Student: Oh. Much less.

Professor Michael McBride:  So conjugation. This one is more stable. It doesn’t give off as much heat. And it seems to be stable by about four kilocalories per mole. OK, now you can go.

[end of transcript] 

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