CHEM 125b: Freshman Organic Chemistry II
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Freshman Organic Chemistry II
CHEM 125b - Lecture 11 - Carbocations and the Mechanism of Electrophilic Addition to Alkenes and Alkynes
Chapter 1. Alkene Thermochemistry [00:00:00]
Professor Michael McBride: So the second part of the last lecture introduced the idea of alkenes, and we were talking about the stability of alkenes, cis versus trans, and the effect of substitution. We’ll go on with that today, and then go on to addition mechanisms, which is a characteristic reactivity of alkenes.
So if we look at the stabilities of these C6-alkene isomers that we talked about last time, we showed how you can get the numbers for those from the NIST website. It’s also possible to get them from this book by Pedley. That’s the one that Professor Wiberg, who’s a big expert on this, recommended to me as a good source. So here’s the heat of formation of the straight chain with a double bond in the number one position. What does heat of formation, what’s that the heat relative to? The energy of this molecule compared to what? The heat of formation. Rehearse from last semester. Mary?
Student: Is it the heat of atomization to–
Professor Michael McBride: It’s not the heat of atomization. How do you measure such things? How do you measure the heat of a molecule?
Student: You have to heat it up–
Professor Michael McBride: Can’t hear too well.
Student: You have to heat it up until it’s really, really hot.
Professor Michael McBride: That was the way that you got the heat of formation of carbon atoms from graphite, remember? But that only needed to be done once, to get where carbon atom is relative to graphite. Then you need to know the energy of molecules relative to graphite. How do you do that?
Student: You want to break the bonds in graphite and then break the bonds–
Professor Michael McBride: But that’s a very difficult thing to do, to break the bonds in graphite. Matt, do you have an idea?
Student: Do you combust graphite?
Professor Michael McBride: You burn them. You compare them to CO2. So you take your molecule and burn it, we talked about this last semester. You burn graphite. You also burn H2, so you can get a hydrocarbon. What’s the energy of your compound, compared to the amount of energy that was in hydrogen and carbon of graphite. So anyhow, the heat of formation is –10 or a little bit more kilocalories per mole. That means it’s that much more stable than the elements would be in their defined standard state. Now if you look at it for the cis-3-hexene, the double bond beginning on the third carbon, it’s a little bit more stable. And that doesn’t surprise you, because we’ve talked about having more carbons as opposed to hydrogens on a double-bonded carbon makes the compound more stable. That is, that the CH bonds don’t profit as much from the change in hybridization as C-C bonds seem to do.
Now if you have the trans isomer, it’s still more stable by about 1.5 kilocalories. I don’t know whether you had a chance to look at the problems yet. Probably not. They were for Wednesday. Just look at the generalities. But you find that the trans isomer is about 1.5 kilocalories more stable than the cis. Can you think of any explanation for that? Amy?
Student: Less repulsion from the–in cis there’s repulsion from the hydrogens that are on the same side.
Professor Michael McBride: The groups run into one another, it’s a steric effect when they’re cis. That’s worth about one and a half kilocalories per mole. OK, now if you make three carbon-carbon bonds from the double bond–incidentally there’s a table that has those in it too in the Jones book. If you put three carbons on the double bond, then it’s more stable still. And obviously if you put four, it’s a little bit more stable still. Now if we come back and look at correcting for strain, which was what we attributed the difference between cis and trans to, we find that there’s 3.6 kilocalories of molecular mechanics strain in the cis, and 2.5 in the trans. So it’s what you said, Amy, that the molecular mechanics explains it, in terms of a steric effect. So the average then, for those two, is about there.
And now if we make strain corrections for the other molecules as well, calculate the strain by molecular mechanics, there’s 2.8 kilocalories of strain calculated there, 2.6 and 7.7 in the one that has four carbons on the double bond. And now you notice something interesting about the trend. Once you’ve corrected for strain, what do you notice? It’s almost exactly linear according to the number of C-C bonds. So you gain in stability, about 1.5 kilocalories per mole every time you substitute a C-H with a C-C, keeping the same number of carbons and hydrogens in the molecule, discussing isomers of one another. Because of course it makes no sense to compare heats of formation of molecules that are not isomers of one another. You’re comparing apples with oranges. So that could be due to this hybridization. It could, in part, be due to hyperconjugation, that we talked about before, and we’ll talk a little bit more about it later.
Chapter 2. Alkene Addition Mechanisms [00:06:19]
Now let’s think about the thermochemistry involved in alkene/HCl reactions. So here’s HCl plus an alkene, to give the addition product, where H is added to one carbon, and Cl to the other. So notice we’ve converted a single bond and a double bond on the left into three single bonds on the right. Is that likely to be favorable or unfavorable? What’s the energy of a double bond compared to two single bonds? That’s what we’ve done here. You remember from our table we looked at in the first semester? Is the second bond of a double bond worth as much as the first? More? Less?
Professor Michael McBride: It’s worth less. So if we look back at that table that we looked at last semester and put actual numbers to these bonds that we’ve drawn here, we see that the hydrogen chloride bond is worth 103 kilocalories per mole. The double bond, carbon-carbon double bond, is a 146. So there’s 249 that we’re losing in the starting material. So how much do we gain in the product making the three single bonds? So there’s 104, and 83, and 81. So 268. In other words, it’s profitable by 19 kilocalories per mole to do this addition. And we talked about this before, when we were talking about chain reaction addition of HCl, or HBr, of HI, and so on. All of them were exothermic by about that much. So it’s favorable by about 19 kilocalories per mole.
Now we could also look more specifically at these compounds by using that data that’s in the table that we accessed last time from National Institute of Science and Technology [correction: National Institute of Standards and Technology], to look at the actual heats of formation of these particular molecules in the gas phase. That is, before we were just looking at average bond energies to get an order of magnitude idea of what’s going on, or better than an order of magnitude. But these now will give us the precise values. And where we said it was 19 before, now we see that, for this specific set of compounds, it’s 14.5. But we were about right. It was a good guess before.
So this reaction should go to the right. But we’ve already talked about a reaction that goes to the left. What reaction would that be, the one that starts on the right and goes to the left? That’s the elimination reaction. An E2 reaction goes from right to left. What’s the difference? If this equilibrium lies to the right, how could we have spent all this time in the last quarter of the course talking about reactions going the other way? Isn’t that uphill in energy? What reagent do you use to do the elimination?
Professor Michael McBride: Strong base, right? Like hydroxide. So let’s see what happens if we put hydroxide into the mix. So in acid we get addition. And it’s favorable by 14 kilocalories per mole, in this particular case. But if we put hydroxide in there, now what will it react with? Well, we talked about how it pulls off the proton at the same time chloride leaves in the E2 reaction. But if all we’re doing is studying thermochemistry, we could look at where it is in the product, in the balance of the two sides of the equation. So what it does on the left is take the proton away from HCl, to give water and chloride.
Now HCl is a much stronger acid than water is. So that’s a very favorable process, to take the proton away from HCl, and to put it on hydroxide. And in fact, we could find out how different it is by looking up in that same table the proton affinities of OH– and chloride, and we find they differ by 19 kilocalories per mole. So adding hydroxide stabilizes the left side of this equation by 19 kilocalories per mole, and now it’s favorable to go the other way by four and a half kilocalories per mole. So in base, you do elimination. You go the other way.
Now are these really the reverse of one another? Do we just run the machine backwards to get elimination, or to get addition, as compared to the elimination we talked about before? Or is it a different mechanism because the conditions are different? The elimination was done in the presence of strong base. The addition is done in the presence of acid. So you might not be able to run the machine backwards. So incidentally, here’s a problem for you to look at for Wednesday too. Use the pKa values for HCl and H2O to estimate–here I looked it up, actually, the proton affinities in NIST, and got 19 kilocalories per mole. But you could have done the same thing out of your head, or out of a simple table. Does anybody remember what the pKa of HCl is?
Student: Maybe nine?
Professor Michael McBride: The exam was Friday, I realize that. You can scrub that now. Anybody remember about what it is? I think it’s –8. Isn’t it? [correction: -3] How about the pKa of water?
Professor Michael McBride: Right, –16 [correction: positive 16]. So from that difference, you can calculate what the energy of that reaction would be. So try it from that difference and see if you can do it. That’s a good exercise for you. But at any rate, we have the question of the addition of the mechanism. Is it the reverse of the E2 or E1 mechanism? So we’re going to talk about addition to alkenes, and we’ve already done that to a certain extent. We talked about the SOMO mechanism, the free radical chain addition of HBr, and you remember what that is, that the bromine atom adds to give the more stable radical, which then attacks HBr, taking off the hydrogen atom, to give the product. It gives that product, and there’s a reference in the textbook, it gives that product, not the one above. And it doesn’t give that one. It’s regioselective. And remember we called this anti-Markovnikov. Markovnikov had the idea, as we talked about last time, that the carbon that has more substituents gets more. So if you had two methyl groups already on one of the double-bonded carbons, it would be the one that gets the bromine, that the hydrogen goes to the other one. But this is anti-Markovnikov, it does the opposite. So we’ve already talked about that. And the key feature in the regioselection is getting the more stable carbon radical.
But it’s also conceivable that there could be a HOMO-LUMO mechanism, a concerted one, where the new group comes up to the double bond, and the partners just exchange like this. Suppose you had H2 to react with, with the carbon-carbon double bond, a very exothermic reaction. So you could just shift electron pairs like that, and perform the addition. Now let’s look at the orbitals that would be involved in that. What’s the HOMO of the C-C double bond? Ellen?
Professor Michael McBride: π. And the LUMO is π*. So let’s try to bring the LUMO and the HOMO of the H2 down on top. Look good to mix HOMO and LUMO? What do you call that situation? Rahul?
Student: It’s out of phase?
Professor Michael McBride: They’re out of phase. In fact, there’s a specific word to talk about things that don’t overlap for this symmetry kind of reason. Remember what it is? They’re orthogonal, Anurag says. So there’s no mixing of HOMO of one with LUMO of the other. Of course you could shift them so that they mix. But then there’s nothing to be gained, because on the left, you’re mixing filled with filled. That’s repulsive. And on the right, the orbitals are empty. There’s nothing that has whatever energies are available for those orbitals. So you can’t do this reaction. You can’t just bring AB single bond down on top of a double bond and change partners, because the symmetry is wrong of the orbitals. They won’t mix that way. You can’t draw those two arrows.
However, it turns out that if you use certain catalysts, in particular metals like platinum, they make it possible to do this, because the platinum gets involved. It gets its orbitals involved, and there can be favorable overlap. And we’ll be talking about that pretty soon. You see it in the Jones book; it’s in chapter 10.
So that would be a concerted reaction, where you make and break bonds at the same time. But it’s also possible to do it stepwise. That’s why the reactions are called electrophilic, because you have an acid, for example HBr, and the HOMO attacks the σ* LUMO, attacks the proton to give off bromide. And then the bromide can attack the intermediate cation. So it’s a two-step reaction. So the initial reaction is the reaction of an electrophile, something that wants the electrons. A low LUMO, the σ* of HBr is the reagent that reacts with the double bond. And notice that that one works for any hydrogen halide. Remember all the hydrogen halides were exothermic for their addition to the double bond. But only HBr would work by the top mechanism here, the SOMO, because both steps were exothermic to compete with termination. Whereas any of the hydrogen halides will do this stepwise electrophilic reaction. And notice, in this case, that it does follow Markovnikov’s rule. It’s not that one. It’s the Markovnikov, where bromide goes to the more substituted carbon, and it goes there because that carbon is more stable in the inter… that carbon cation is more stable as an intermediate. So that’s stuff that mostly we’ve mentioned before.
Now let’s think about the orbitals and where they should get attacked. So here’s ethylene, an alkene, and we can look at the surface potential to see the energy of a proton on the van der Waals surface. So a proton likes to be up above where the double bond is, and it doesn’t like to be around where the hydrogens are. So this electrostatic feature of the surface of the molecule is important in positioning the fragments before they begin to react. But new bonding requires mixing orbitals, not just positioning things electrostatically. So we should look at the HOMO and the LUMO. So the LUMO is this π* orbital and the HOMO is the π orbital. So the addition, as we’ve said, to alkenes is electrophilic, it goes after the π electrons, the high HOMO. So the proton, with its low vacant orbital, electrostatically likes to be up on top. That’s the best place to be, if you’re on the surface of the molecule. And when it’s there, it can overlap with this π orbital.
So in the various textbooks you have, you could read about this electrophilic addition to alkenes. I guarantee you there are all these sections in there. One about hydrogen halide addition by R+, by the cation intermediate, what we just talked about on the previous slide. There’ll be discussion of the regiochemistry, which end of the H+ adds to, to give the more stable cation intermediate. And there’ll be discussion of hydration, where instead of the X– being the thing that adds to the H+, [correction: C+] it’s the solvent water, the unshared pair on oxygen can attack the cation as well, and then lose a proton, so that it’s OH that goes on instead X–. So you can be reading that in your texts. And this, just to show an example, is from the lecture that I mentioned was given by Professor Siegel last year, where he talks about how you can do this Markovnikov addition, chloride going to the more substituted, OH going to the more substituted, Br going to the more substituted, when it’s clean, that is when you don’t have free radical initiators there. So all of these form by proton first going here to get this more stable cation intermediate. And in this case, with dilute sulfuric acid, it was water that came on to do this. And that mechanism is shown here.
Incidentally, on the exam, remember the last question was to draw a very careful curved arrows and everything. I put this up, in part, to give you fellow feeling, because we graded that very rigorously. When you draw a curved arrow to denote electron pair shifts, where do you start the arrow?
Student: Where the electrons are.
Professor Michael McBride: Where the electrons are in the starting material. And where do you end the point of the arrow?
Student: [INTERPOSING VOICES]
Professor Michael McBride: Where those electrons are in the product. Sometimes they will be on an atom, become an anion, or becoming an unshared pair. Sometimes they will be between two atoms, because you’re forming a new bond. And if you want to be thinking really clearly about it, you should draw those arrows clearly. That’s why I asked it that way in the question. But notice that Professor Siegel here, in drawing some of these arrows, drew it going on to the hydrogen. He drew the X– going on to the carbon, where I would expect you to draw this curved arrow ending between X and the carbon. And to position those two things that are going to react with one another in your drawing, such that they’re close enough to one another that it makes sense to end the arrow between them, not to have it off here and the arrow terminating in space, just because it’s halfway between. So that’s a lesson that I hope will become clear. So this shows the case of water intervening.
Chapter 3. Understanding Carbocation Stabilities [00:22:41]
So there are sections of the book about the addition of hydrogen halides, about regioselectivity, about the intermediate cations that are involved, about cation stability, and about cation rearrangements. And I’ll talk about a few of those things here, but you can review them in your book as well. First about cation stability. I took this from the text we used a couple of years ago by Loudon. So he shows primary, secondary, and tertiary cations, how many carbons are substituted on them, and says of the stability of carbocations, that tertiary is more stable than secondary is more stable than primary. This of course is the explanation for Markovnikov orientation. Now we can look at the rationale for that, rehearse it once again. Hyperconjugation says it’s better to have a carbon on the carbon cation with its hydrogen, or whatever else it’s bonded to, than to have just the hydrogen. And the reason it’s better is that there’s some kind of HOMO that’s in position to have its electrons stabilized by that unusually low-energy LUMO, the one associated with the positive charge, the vacant orbital on carbon. And in the case of hydrogen, there are no electrons to be stabilized, those are orthogonal to this. But in this case, the σ electrons that are in the bond, even though they aren’t particularly high, can interact with a really unusually low orbital, and get some stabilization. So we’ve talked about this before, this hyperconjugation. And it could be denoted by writing a resonance structure, which has a double bond between the carbons and an H+, or R+ if it’s an R group out there.
The other rationale for this order of cation stability is bond energies. And we’ve talked about this several times, changing CH to CC bonds when you have this especially advantageous sp2 hybridization of the carbon. But those are theory, and it’s different to know what the evidence is that they actually have that order, tertiary more stable than primary. So the first part is the rationalization. Now we have to say where do we actually get the evidence that their stability is that way? And several people on the exam, when asked for evidence, provided theory, rationale like this, rather than actual numbers, where the evidence comes from.
So to get evidence, you have to compare something. So you could compare products or transition states, that is you could look at reactivity. If you have a very unstable cation, you would expect it to be more reactive. So that would be one way that you could do it. Another way would be compare with a starting alkene. If it’s easier to put a proton on, that would indicate that the cation you’re leading to is more stable. The cation is uphill in energy typically, but if it’s a more stable cation, it should be easier to form. So the reactions, where you protonate the alkene, should be easier. Or in the reactions we were just talking about, the SN1 substitutions, the ease of losing the anion, in order to generate the leaving group that is to say, in order to generate the carbon cation should be easier when the cation is more stable. So we could do either of those. Or we could directly compare the cations with one another, to see which one is more stable.
Now here’s a table taken from the textbook, which gives a table of the heats of formation of carbon cations in the gas phase. So they’re very hard to make in the gas phase. And it says, “secondary vinyl cations are more stable than primary vinyl cations.” Now what’s the basis for that here? In the table, what’s the basis for saying that secondary vinyl cations are more stable than primary vinyl cations? Well we look up there and we see a primary vinyl cation is, in fact, the highest energy of any of these. And the secondary vinyl cation is also pretty high. It’s substituted; it’s secondary, which should make it better. And indeed, it is better. So it says it’s better by 54 kilocalories per mole, in the gas phase.
Now there’s a problem with using this table in this way. And the problem is compared to what? Is this really saying that a primary vinyl cation is more stable than a secondary vinyl cation? Compared to what? These heats of formation, what are they compared to? So this is the energy of the vinyl cation compared to what? What’s the heat of formation? Helen?
Student: Carbon dioxide and [UNINTELLIGIBLE]
Professor Michael McBride: That’s the heat of combustion. It could have been heat of combustion. We would add something to these, to get heat of combustion. Actually what’s given is the heat of formation.
Student: Compared to the carbon and hydrogen?
Professor Michael McBride: Yeah, compared to carbon and graphite and hydrogen and H2 gas, at standard temperature and pressure and so on. That is compared to the elements, in their standard states. But notice that there are different numbers of atoms in these things. They’re not isomers of one another. So it’s not that you can convert one to the other and say that this heat difference shown here takes one to the other. You’re comparing apples and oranges. They’re just not the same thing. If you want to compare with heats of formation, they have to be things that are isomers of one another, where you know their energy relative to one another, rather than relative to something completely different, different numbers of atoms.
So this particular table doesn’t help you at all directly in figuring out, except that there might be some cases here, like this one. Notice that this is C4H9, and this one is also C4H9. So here’s a primary and a secondary, which differ by 20 kilocalories per mole in the gas phase. And that really does show that secondary is more stable than primary. But this table definitely doesn’t show that secondary vinyl cations are more stable than primary vinyl cations. So you have to think carefully about compared to what. Here you’re comparing to different numbers of atoms in their standard states, not relative to each other, nor relative to their respective starting materials or products, as we talked about last time. So this table, in general, is irrelevant to the question of cation stability in the sense that we were talking about it in the previous slide.
Now on the other hand, in Loudon here, Table 4.2, gives the heats of formation of isomeric butyl cations. So the same atoms, they’re being compared to the same thing, so differences between them now do tell us the difference between primary, secondary, and tertiary cations. They got it from the same numbers, heats of formation. Although, in fact, you’ll find that these numbers are slightly different from those in the previous table. Neither book says what the source of these numbers is. These NIST values have been pretty carefully gone through. So if push comes to shove, I often look at those, or ask Professor Wiberg what the best value is. But here you see that the relative energy of the tertiary butyl cation is taken as zero, the most stable. And now the others are 16 or 32 or 37 kilocalories less stable. So tertiary is more stable than secondary is more stable than primary. And here are two different primaries.
Now if we look at those two, we see that going from primary to secondary is a difference of 21 kilocalories per mole. 37 to 16, primary is less stable by 21 kilocalories per mole. Now remember that the reason for this could be things like hyperconjugation, or the number of carbons bonded to the sp2 carbon, the same factors that we talked about in the case of alkenes, making more substituted alkenes more stable. Here more substituted cations are more stable. But the difference is much bigger here. 21 kilocalories per mole versus two and a half kilocalories per mole, from the slope of that plot we made for the alkenes at the beginning of the lecture. So it’s ever so much more important for cations than it is for alkenes, whatever the factor is that’s doing this.
So it could be hyperconjugation. Would you expect that to be more stable, more important for the cations shown here, or for the alkenes, where the carbon is a carbon-carbon double bond? You have the same factor, that the carbon-carbon double bond here would have a low LUMO. This has that p orbital on carbon. That has a π* orbital, a low LUMO, and the same deal here. Would you expect the effect of that mixing, HOMO-LUMO mixing in hyperconjugation, to be the same as when you have a double bond? If one stabilization is bigger than the other, which? So here we have the electrons in a σ bond. And here we have a vacant orbital. In the case of the cation, we have just a vacant p orbital. In the case of the double bond, we have π* orbital, plus/minus. Which one’s going to give more stabilization of these electrons? What determines it? What determines how much stabilization you get out of mixing orbitals? Megan?
Professor Michael McBride: Overlap is one thing. So it’s a p orbital on carbon in either case here. In one case, that p orbital is also associated with another p orbital. So each of them is 1/21/2. So it’s smaller in one case. So there indeed would be less overlap in the case of the carbon-carbon double bond, rather than just a full-fledged p orbital. So that’s one factor. How about another factor, Megan?
Professor Michael McBride: How about the energy match?
Student: When you have a positive charge it raises the energy.
Professor Michael McBride: Think about that carefully. Say it again.
Student: It lowers the energy?
Professor Michael McBride: The positive charge makes the energy fabulously low for the cation. The π* is low, compared to σ*. But it’s not really all that low, compared to the vacant orbital. So both on the basis of overlap, and on the basis of energy match, you expect a much bigger factor for the cation than you do for the alkene. And in fact, that’s what you see. It’s worth a heck of a lot more. But the bond energy factor might be more or less similar between the two, because in both cases, you’re changing C-H to C-C single bonds. But this is the one that should make a really big difference.
But there’s another factor as well, actually, which is that these are measured in the gas phase. So we have this positively charged carbon in the gas phase. We saw, when we talked about the ionization of water, how much effect solvation has on the ability to form ions. So here we’re trying to form an ion on this carbon. Can you see another difference among these, about stabilization by this positive charge, other than hyperconjugation and bond energy? Nathan?
Student: The top one’s more polarizable.
Professor Michael McBride: Right. The other stuff in the molecule, the other three carbon atoms and their hydrogens, are polarizable. But remember, there’s a very strong distance dependence. You have to be really close to get a lot of advantage out of that. So with t-butyl, you have all the other atoms really close to the positive charge to be stabilized by polarization. Here you have that one much further away. Here you have two of them significantly further away. And here you have this one significantly further away, and this one off in left field. So we can see that there’s going to be “intramolecular solvation,” you might call it, but nobody else calls it that. But this effect from polarizability in the gas phase is going to be very important. But if you go into solution, it’s just the reverse. Because the primary one up here has the positive charge very exposed. So neighboring molecules could get very close to that. So in the gas phase, it’s the atoms within the molecule that are solvating the positive charge, and their electrons being stabilized. But if you go into regular solution, it turns around.
So these differences in the gas phase are much bigger than the effects that you would actually see in a solution reaction, where the primary cations are more stabilized by the surrounding molecules. So for example, if you do SN1 reaction, t-butyl bromide is only about 5 kilocalories per mole easier to ionize than isopropyl bromide. That we saw in that–remember where methyl, ethyl, isopropyl, t-butyl. t-Butyl was unusually easy to substitute, because it went by SN1. But if you looked at the isopropyl, it also was going partly by SN1. And it turns out that it’s about 5 kilocalories harder for isopropyl. But here the difference between t-butyl and the isopropyl–the equivalent of isopropyl–is 16 kilocalories per mole, or even more. So it’s very strongly attenuated in solution.
People often talk about gas phase being intrinsic values for things you can measure, like the heats of formation of these things, that that’s the ideal. The reason it’s ideal is you can compare it with calculation. But it’s not an ideal reference if what you’re interested in is predicting reactions in solution, where that solvation is going to make a lot of difference.
Chapter 4. Skeletal Rearrangement of Carbocations [00:39:01]
Now here’s addition of HCl to an alkene. So is this Markovnikov or anti-Markovnikov addition? Natalie, do you remember what Markovnikov and anti-Markovnikov means?
Student: Your anion goes to the more substituted–
Professor Michael McBride: Does the anion, the non-hydrogen thing, go to the more substituted carbon in this case? Yeah, it did. And why? Of course, what we would expect is that it would be attacked, and this would give the more substituted cation, rather than attacking the central carbon, and having the primary cation as the intermediate. Then chloride goes over and forms a new bond, and you have the product. So that’s Markovnikov addition. However that’s only 17% of the product. The 83% of the product is that. Can anybody see what’s funny about that product? Is that the anti-Markovnikov addition? Natalie?
Student: No, it’s also just as equally substituted–
Professor Michael McBride: I can’t hear very well.
Student: That carbon is equally substituted as in the first case.
Professor Michael McBride: Ah, the substitution of the carbons is different. What has happened? We saw this before. Chris?
Student: A methyl shift.
Professor Michael McBride: A methyl has shifted. It’s rearranged. In fact, what shifts is a methyl group, but it’s a methyl with its electrons. It’s a methide shift when you’re talking about cations, as you will see here. So it’s shifted a methyl across, so what’s happened is that those electrons–it’s like hyperconjugation, that those σ electrons are stabilized by the vacant orbital on the carbon. But they’re stabilized so much that the methyl with its electrons, with those bonding electrons, the methide, shifts across. If you want to have fun, sometimes when you draw a rearrangement, you draw a little loop in the arrow, isn’t that cute? So the methide has shifted across to give this more stable cation, this was secondary, this one is tertiary, and then that gives 83% of the product.
So what this says is that the rearrangement, the methide shift, competes with reaction, with the collapse of the ions, chloride attacking the positive carbon. So both things happen, and they have similar rates. Now suppose you had a lot of some other nucleophile there, like water. Now the unshared pair of water could fill the role that chloride filled here, then lose protons, so you got H2O+, then it loses a proton. So you can get this alcohol, or you can get the rearranged alcohol. So the fact of this rearrangement is, of course, additional evidence that you had the cation intermediate as it was in SN1 reactions, where the rearranged skeleton showed that there’d been a cation intermediate.
So when you do acid-catalyzed hydration of a double bond, adding H and OH, you have to be aware that there’s the possibility of rearrangement. If the skeleton is one where you would form one cation, but then a methide shift, or a hydride shift, could give a more stable one.
Chapter 5. Stepwise Addition to Alkynes – Competing Influences of Halogen [00:43:16]
Now let’s look at the influence of halogen in these cations. So here we start with acetylene and add HCl. So we’ve started to give the more substituted vinyl cation. Remember the table, it said primary vinyl cations are less stable than secondary vinyl cations, even though the data was not the appropriate data to support that, it’s absolutely true that vinyl cations are very hard to make. So this is not such an easy process. But then it can add chloride, so you have this product, which is the Markovnikov product. The proton added to the terminal carbon to make the more stable secondary, rather than adding to the central carbon to make the less stable primary cation, so the chloride ends up there. And this is perfectly reasonable.
But it went by way of this unstable vinyl cation. So it turns out that this addition, this kind of electrophilic addition to alkenes, is 100 to 1,000 times slower than it–to alkynes is 100 to 1,000 times slower than to alkenes. It’s much harder to add to triple bonds than to double bonds, because you get these unfavorable vinyl cation intermediates. Well fine. But now we have a double bond. We have Markovnikov regiochemistry, but if you have excess HCl, you can add it again, and get a second Markovnikov addition.
Now suppose you didn’t have an excess of HCl. Suppose you just had one equivalent of HCl. What product would you expect to get? Jack?
Student: Still that.
Professor Michael McBride: Pardon?
Student: Still that.
Professor Michael McBride: Why still that?
Student: Because the double bond is easier to attack.
Professor Michael McBride: Ah! The double bond is easier to attack than the triple bond. So if you didn’t have enough HCl, still this would be much more reactive than this. After you form any of that, it will be reacting 100 to 1,000 times faster per molecule than that did. So you’re still going to get this product. Jack’s right.
Under certain conditions, these are the yield you got. It was 56% and 44%. I suspect that if the reaction had been carried on longer, you would do what Jack did, well you’d go all the way there. But under certain conditions at least, the reaction was interrupted, and that was the product distribution.
Now it went obviously by way of this cation intermediate, the second step, protonated to give this substituted cation. Now we have a question: is the halogen, as a substituent, favorable or unfavorable for purposes of forming the cation?
Jack, help us out here. Was this second step what you expected to be in terms of rate?
Professor Michael McBride: Is it unusually fast or unusually slow compared to what you expected?
Professor Michael McBride: Fast? Slow? What did you say before? This should react 100 to 1,000–being a double bond–should react 100 to 1,000 times faster than this. Did it react 100 to 1,000 faster?
Professor Michael McBride: No. It didn’t react any faster at all, or roughly the same. So for some reason this was slowed down by some several powers of 10. What could’ve slowed it down?
Student: The chlorine could’ve stabilized the cation.
Professor Michael McBride: Because of the chlorine, it could be hard to get a cation. Would that make sense to you? Why?
Student: Because the chlorine has electrons that can overlap.
Professor Michael McBride: The question is how hard is it going to be to form this cation, which we did by putting a proton on that carbon. The alkene carbon gets a proton gets that cation. So is it harder? Sebastian, what do you say? What effect do you think the chlorine had?
Student: The electron induction pulls the electron away from–
Professor Michael McBride: The σ bond is electron withdrawing. The electrons around the carbon are lower in energy because they’ve been sucked away by the chloride, so it’s harder to form the cation. That would explain why we get this ratio here. That this second step, even though it’s an alkene and should be enormously faster is not any faster, because this chlorine is slowing it down by pulling electrons away. It can’t provide electrons to the proton. Now Sebastian, keep going for us. How about the regiochemistry? So the second step is slow, that’s what we were just talking about. How about the regiochemistry of the second step? That is, is it Markovnikov or anti-Markovnikov? That is, the proton could have added here, to give this cation, or it could have added to the central carbon to give the other cation. Which did it do? Did the new chlorine go on the more substituted or the less substituted?
Student: On the more substituted.
Professor Michael McBride: On the more substituted. So the chlorine slows it down, but it’s still Markovnikov. This cation is better to get then the other cation.
Student: It reacts better with the chlorine.
Professor Michael McBride: Now you might say, aha. The choice here is between a secondary, whatever the chlorine is, and a primary. And primaries are bad to get, you don’t want to protonate the center one, because it would be a primary cation. But notice here with HBr, when it’s symmetrical. Still the two bromines go to the same place. So that it looks like the chlorine is welcoming the formation of a cation. Or at least the bromine, in this case, is welcoming formation of that cation, rather than that cation in adding the second proton. So the answer is yes. Is it favorable or unfavorable? It’s both. It’s hard to make the cation because of this σ-electron withdrawal. But if you form the cation, where do you want to form it? Next to the chlorine. Why? Chris, do you have an idea? If you’re forced to make a cation, say in this case when you put that second bromine in, you could’ve made the cation either here or here. Both of them with respect to the carbon would be secondary, but this one, the one you actually form, where the bromide comes in has a bromine on it. It’s hard to form it because the σ bonds are electron withdrawing. But if you form it, form it next to the bromine. Why?
Student: It’s polarizable, so it makes it even more effective.
Professor Michael McBride: Polarizability is a good point. That wasn’t what I was thinking of, but it’s a good point. What else?
Student: Lone pairs on the halogen.
Professor Michael McBride: Lone pairs on the halogen can be stabilized by overlapping, so you get resonance stabilization. So if you’re going to make it, make it where you can get help in the π system. So that’s really, I think, interesting about the halogens. That they could be deactivating, but still want to have the cation next to them, because of the different role of the σ electrons, and the π electrons. OK, thanks for sticking around here.
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