CHEM 125b: Freshman Organic Chemistry II
|Transcript||Audio||Low Bandwidth Video||High Bandwidth Video|
Freshman Organic Chemistry II
CHEM 125b - Lecture 3 - Rate and Selectivity in Radical-Chain Reactions
Chapter 1. The Reactivity-Selectivity Principle [00:00:00]
Professor Michael McBride: So we were talking last time a little bit about the reactivity-selectivity principle. The idea is that if you’re talking about two similar reactions, two similar reagents, the one that’s more reactive will be less selective in which product it gives.
So here we have the example of two radicals, a chlorine or bromine atom, reacting with propane. Remember, the chain reaction is that first the halogen atom pulls off a hydrogen, generates a radical, which then takes on a new halogen. So wherever that first hydrogen gets pulled off, that’s where the halogen will end up in the product.
But in the case of propane there are two different hydrogens that you can take off–the one in the middle or the one at the end. And it’s easier to take off the one in the middle, because the radical is more stable, the secondary radical, the internal one rather than the primary radical. We talked about that last time–the extent to which that’s true. But whether it’s differences in the starting material or differences in the product, however you slice it, it’s easier to take off the hydrogen from the secondary position than the primary position.
So the overall heat of the reaction as you see here is better by 2.5 kilocalories for generating the secondary radical, as opposed to generating the primary radical. And that’s true whether chlorine is doing the abstracting or bromine is doing the abstracting. The same difference overall in the products. But the actual change in energy is much more positive in the case of bromine. It’s harder to do than it is in chlorine. So chlorine is more reactive, and bromine is less reactive.
But notice that the difference in the activation energies is more dramatic in the case of bromine than it is in the case of chlorine. The earlier the transition comes, the less difference there’s going to be. So chlorine, the more reactive reagent, is less selective. And bromine, the less reactive agent, is more selective.
So this is called the reactivity-selectivity principle, and it only can apply to the extent that the transition states are mirroring what the products do and that the two reactions are very, very similar to one another, and there’s not something else that makes the reaction transition state come much earlier or much later with a much different structure.
So in the case of chlorine, the relative rate of forming secondary versus primary is a factor of 4. But in the case of bromine it’s almost 10 times as large, the selectivity. It’s a factor of 35. That means in the first case, the difference in energies of activation, the Δ ΔG‡–ΔG‡ is the activation energy. But the difference in the activation energy is 0.8 kilocalories per mole, about 1.
Whereas with the later transition state the difference is very much bigger, 4.5. So typically more reactive means less selective. So if you were a chemist who wanted to prepare a certain product that required selectivity, you might choose to use a less reactive reagent rather than a more reactive one, so you’d get a higher yield of the material you wanted, even if it was a little more difficult to do, if it took a little longer.
Now let’s look at this a little more carefully, the fact that the difference in energies of activation in the case of bromine taking off a primary or a secondary hydrogen, the difference in activation is 4.5 kilocalories per mole. We just went through why it was reasonable that it should be large. But should it be that large? Notice that the difference when you get all the way to products is 2.5 kilocalories per mole.
Now we had the idea that relative to a standard, say methane, the energies for taking off a secondary hydrogen should be–either the starting material is less stable or the product is more stable, compared to our standard methane in this case. But at any rate, there’s an increase in stability relative to the normal case, up at the top here. Relative to that, there’s an increase in stability as you go from starting material to this free-radical intermediate.
Now there’s a question where the transition state comes on that. If it’s early you don’t see much of that increased stability. If it comes late you see more of that stability difference.
However, you see what’s nuts here? What we see is this. There’s more difference at the transition state than there even is when you get all the way to product.
So how can the transition states be more different in energy than the products are? This might not be a straight line, but at least you’d expect it to be monotonic–not go down and then up again. There’s something we’re missing here.
The trouble we’re having is understanding the energy of transition states. Regular molecules we’re more or less comfortable with, but transition states are tougher. So one way to look at transition states is molecular orbital theory. Another one is with resonance theory. And the problem is that you don’t know which of the R or Br the hydrogen is bonded to when it’s being transferred.
You could write one resonance structure in which it’s bonded to R and bromine is an atom. Or you could write another resonance structure where it’s bonded to Br and R is the free radical. And it’s exactly poised between those two at the transition state–that’s the definition of the transition state that you’re about halfway there.
But what’s special about the transition state if there could be another resonance structure. A third resonance structure, which has Br–, because bromine, after all, is electronegative. And R could be plus.
So at the transition state here we could have cation character to the R, so that will be helpful at the transition state. A third resonance structure here stabilizing things in resonance theory. But the R cation is irrelevant when it’s R-H over here, or when it’s an R radical. Once the radical gets away it only has a certain number of electrons. It’s not R+ anymore.
So as we’ll see next week, substituting a methyl for hydrogen that is going from a primary to a secondary cation stabilizes the cation a lot. So this is unusually stable. But this is important, this resonance structure, only at the transition state–not in the starting material and not in the product. So we make a mistake in this case if we assume that the character of the transition state is intermediate between starting material and product for this one-step reaction. There are actually special factors that apply only in the transition state. And we’ll see when we talk about pericyclic reactions that this is very important.
So sometimes factors involved in stabilizing transition states could be different from those involved in stabilizing either the starting materials or the products. And in these cases we can’t easily rationalize the activation energy changes on the basis of the overall exothermicity or endothermicity of the reaction step, because there’s something special going on at the transition state. You hope that’s not always the case. Most cases you could do thinking that more exothermic reactions will be faster.
But you have to be careful in some cases. And as I said, we’ll see dramatic examples where the transition state has something special about it when we get to discussing pericyclic reactions in the third quarter of the course.
Now here are two problems for Wednesday that have to do with free-radical substitution, and you can work in groups if you wish. Remember, we won’t have a class on Monday, and there won’t be discussion section in the evening on Monday either.
So here’s problem one. Suppose, as we’ve just been talking about, monochlorination of propane gives n-propyl chloride and isopropyl chloride in ratio 43:57. Note I say monochlorination, because, of course, these products can also have a hydrogen abstracted and you can get dichlorination, trichlorination and so on. So if you want to get these ratios accurately, you do just a little bit of reaction and see the first products before you get subsequent substitutions on the initial products. You do just a little conversion and see what the ratio is. And in the case of propane it’s 43:57, primary and secondary with the hydrogen places.
So if that, and if monochlorination of 2-methylpropane gives–it’s called isobutane also–gives t-butyl chloride and 1-chloro-2-methylpropane, 36% and 64% yields. Notice this is a little test also on knowing what the structures are of compounds that have these names.
Then predict the products for monochlorination of 2-methylbutane. So you’re going to get relative reactivity of primary and secondary from one, relative reactivity of primary and tertiary from another, and then you’ll use the relative reactivity of secondary and tertiary to solve the problem.
Now you’ll find in the textbooks that you have–every textbook I’m confident has this kind of problem in it about selectivity in free-radical chlorination. It probably has the same examples. So consult your textbook on free radical halogenation to be sure you’re taking the statistics of hydrogen numbers properly into account. That’s the first problem.
The second problem is solve the puzzle that’s going to be coming up on slide 18, so hold on and we’ll get to that one.
Chapter 2. Radical-Chain Addition of HBr to Alkenes and Its “Regiochemistry” [00:11:11]
So we’ve been talking about free radical substitution–take off a hydrogen and substitute it with a halogen. But there’s another free-radical chain reaction that involves halogens, and that’s addition to alkenes. So the starting materials are an alkene and hydrogen halide. And somehow you get a radical to start it off, say a halogen radical. And instead of pulling off a hydrogen it adds to the double bond. So it gives that intermediate carbon radical.
And then that one is the one that does an abstraction of hydrogen, this time from H-X. So you get a product where X and H have added to the carbon-carbon double bond. So free radical addition to the alkenes, and it’s a chain reaction. Once you get a radical it can go round and round without getting rid of radicals.
Let’s look at the energetics that are involved in this. A curious thing about this reaction is it doesn’t work with all X-H’s. Now let’s see why. So we’ll analyze the energies of the bonds–we’re using here just average bond energies.
So in the starting material we have a carbon-carbon double bond that’s worth 146. And we have the X radical, and also there’s X-H in the starting material. And in the product we’ve added the X-H to the alkene making a new carbon-carbon bond that’s worth 83 kilocalories per mole, and a carbon-hydrogen bond that’s worth 99. And we’ve converted X-H to X•.
So if we look at the average bond energies involving fluorine, for R-X, the product, we have 116 that’s going to go in here. And a starting material we’re going to have H-X, hydrogen-fluoride, 135 that’s going to go there.
So we’re going to add these together, these bond energies. Here it’s 146 for the double bond, and 135 for the X-H in the case of fluorine. And in the product, it’s going to be 83 and 99 and 116 for the carbon-fluorine bond. So we add them together and we find out this is favorable by 17 kilocalories per mole to do the overall transformation. So it’s going to be exothermic in the case of HF.
How about in the case of the other halogen atoms? We could use the corresponding values for their bonds, and we get that they’re almost exactly the same. So all these additions would be the same exothermicity. So you might think they’d all go at about the same rate. But they don’t at all. Only HBr works. Why?
Overall, the thermodynamics is fine for the transformation. But the question is mechanism and activation energies. And remember, there are two steps in this reaction. First the X adds to the alkene, then the new carbon radical abstracts a hydrogen. So we have to look at the individual steps to get what the activation energies are.
So we first have to go to this intermediate, which has a single bond worth 83 kilocalories per mole, instead of the double bond worth 146. So that’s going to be fabulously exothermic, this first step. So very exothermic should mean very fast, so that’s going to be no problem. That’s in the case of fluorine, the blue one. In the case of chlorine it’s still exothermic–that should be not as fast but easy still. Bromine still exothermic. But iodine is endothermic. So that reaction is going to be very slow.
So we can understand now why iodine doesn’t work. But why not fluorine and why not chlorine?
Student: Because the second part isn’t favored…
Professor Michael McBride: Ah, Derek, you’re right. It’s the second step that’s hard for them. Because having this intermediate we now have to go here by abstracting the X-H bond. And the X-H bond we have to break in the case of fluorine is fabulously strong, 135 kilocalories per mole. So that reaction is going to be way endothermic and it’ll be very slow. So the cyclic mechanism will cough on the second step for that one. And in the case of chlorine the second step is also a little bit endothermic so it’s not very fast.
But in the case of bromine it’s exothermic again. In the case of iodine it would be exothermic, but the first step is where the machine coughed. So only bromine is set up to do both of these relatively rapidly. Now you might say, OK, we’ll do the others, we’ll just wait longer. It won’t be as fast because one or the other the steps will be slow. We’ll just wait.
But you can’t do that with free-radical reactions. Only HBr works fast enough in both steps, and you need both steps to be fast.
Now we’re going to get to discussing what would happen otherwise, if both steps aren’t fast, in just a minute. But first I want to comment on one other aspect of addition to the alkenes. What’s called regiospecificity. I don’t particularly like that word, but it has to do with when you add HX to a double bond there are two regions they could go to–the H could go here and the X here or vice versa.
So when we add HX we could imagine X going to this carbon or H going to this carbon. And it turns out that for any halogen that can undergo addition, there’s a principle called “Markovnikov orientation,” after a Russian chemist at the latter part of the nineteenth century. And he discovered this in 1870, just about the time that people knew about structures, of course, that what you tended to get was the more substituted halide, that is the X went to the carbon between these two that has more substituents on it. And it’s understood in terms of initial addition of H+. This was figured out in the 1930s.
And it’s because of this ionic mechanism that we’ll discuss in three weeks that Markovnikov orientation works. We’ll talk about that when we come to it.
But what’s important is when you do this HBr addition by the free radical mechanism, the cyclic mechanism, you get the anti-Markovnikov product. It goes exactly the other way–the bromine goes on the less substituted carbon.
So this was quite a mystery for people from 1870 on, why HBr was sometimes a bad actor. Sometimes it would give the Markovnikov product, the expected one that all the other hydrogen halides would give. But sometimes it gave the other. It tended not to give mixtures, it would give all one or all the other.
But in the 1930s, 1933, it was found out that this happened when the samples contained little bits of peroxide. Now let’s see why that is. It’s because it’s an initiator of radical chains. Remember you needed some way to have a radical to begin this whole process.
So if you have R-O-O-R, a peroxide, and HBr and the alkene, the first thing that happens in this mechanism is that you break the O-O bond to get two R-O radicals. Now you have the radical that could do a hydrogen abstraction. So it abstracts hydrogen from HBr, and now you’re into the machine–you have X• that could add to the alkene and get one or the other of those two radicals.
Now if you’ve got the radical on the left you get Markovnikov orientation, because H would come here the next time it abstracts H from HBr. But, in fact, this is the one you get because the secondary radical is more stable. So when you’re getting radicals you get that one rather than the primary radical.
Chapter 3. Rates of Radical-Chain Halogenation: Rate Laws for Catalytic Cycles [00:20:23]
So that explains then the regiospecificity of the special addition of HBr, but only if you can get free radicals. Now why can’t you just do it waiting longer? This has to do with the rate law when you have a catalytic cycle. So we’re going to discuss it in this case, but it’s important in all sorts of cases where there’s catalysis in inorganic chemistry, in biochemistry. When there’s catalysis and you have these cycles, the rate laws you get are a little bit different from what you might expect.
So there are two steps in this reaction. The first step has a certain rate constant and, of course, the rate then multiplies that rate constant by concentrations to get a rate. And the two reagents are R-H and X•. Then there’s the second step which has some other rate constant, k2, and its starting materials are, in this case, that dihalogen and R•.
Now a couple of years ago when we were having a visiting lecture by Barry Sharpless, who won the Nobel Prize in catalysis–you can look at his lecture if you want to–but one thing he said caught my imagination. He said, “This is a real democracy, catalysis.” What did he mean by it being a democracy? It’s because when this thing is going around, these two rates have to be equal. Whatever k1 times those is has to equal k2 times those. And the concentration of the free radicals will adjust the ratio of the two to make those two equal.
What would happen if the first one were really fast and the second one were slow? Then you would convert X• to R•. So you’d have a big R•, which would favor the second one, and the small X• on the first one, and soon they’d become equal and then the thing can go around at whatever rate they allow. But both steps are involved. These two things become equal. So that’s what he meant by a democracy that they were equal.
Now you can see an example of that if you look down on the Garden State Parkway here. Notice that for the cars going north there’s a lot of difference between these lanes of traffic and these lanes of traffic. And the same is true here. There’s a big difference between those two. And the reason is that this is cash and this is E-ZPass.
So if we look just at the cash lanes, you notice that there’s a low rate constant here. You’ve got lots of starting material building up its concentration trying to get through paying. But once they get through they can accelerate and go on fine. So here you have a high rate constant, here you have a low rate constant. But the number of cars per minute, the fluxes here and here, are the same. The cars that come here also end up here.
So the point is that here you have high concentration, here you have low concentration. But the rates are the same. The small rate constant paying cash times the high concentration is the same as the big rate constant once you speed it up times the low concentration. So it’s the same as you go around in the cycle. The rates of the two have to be the same, even though the rate constants are different. So the concentrations adjust.
And you’ll notice, as Sharpless said in that lecture, if there’s a slow step, like this one, then there’s 99.9% of the titanium is what he was calling about–remember we talked about that last time. That’s what he was talking about. “They’re stuck before this big mountain that goes way up like Mount Everest,” he said. He’s a colorful speaker. If you get that barrier down then the rate goes way up. And if you get them all at the same height then you’re really rolling.
So notice that’s what happens in the E-ZPass lane, that they go at the same rate because you’ve gotten that first big rate constant up, the barrier down, for that first step.
Now let’s apply that principle back to this catalytic cycle. So we had from the previous slide that these two had to be equal when the machine is running. Now suppose k1 is small and k2 is large. So now how does the rate, the amount of material you produce per minute, depend on the concentration of R-H?
Well, we can look here. We could rearrange this to say that the ratio of X• to R• is k2[X2]/k1[RH]. That we get by just dividing through. So we know the ratio of these two. But the total of the two, there’s a certain amount of radical there. As you go around the cycle it could be either in the form of X• or in the form of R•. So the ratio can change, but the total amount is limited.
So let us suppose that X• is dominant. Suppose there’s 98:2, the ratio of X• to R•. Now let’s double the amount of R-H. We’re leaving the radicals the same but we’re doubling the amount of R-H. So we’re going to double this, double this, leave that the same, these are fixed constants, but we’re doubling this. So what’s going to happen? The ratio’s going to change. That’s what we’ve just been talking about.
So instead of 98:2, we have to get a ratio that’s only half that big. We have 100 radicals to deal with. They’re in the form of 98 X• and 2 R•. If we double R-H, up here–we’re interested in what that exponent is–we double R-H, we’re going to halve that ratio. How will we do it? We still have 100 radicals.
But instead of 98:2, if we assume that the sum is constant, we’re going to have 96:4. Everybody see that? This ratio is half that ratio, very close. But how much did that change the concentration of X•? Notice it doubled [R•]. But it hardly changed [X•] at all in relative terms. It just went to 96 from 98. So X• is still the same concentration.
So the rate at which you go through here, which is the same as the rate you go through here, is now the same here. This one has been doubled, and this one is practically the same–it’s 96 instead of 98. So what happened to the rate approximately? Matt?
Professor Michael McBride: The rate doubled. So who’s surprised that when you double the starting material here you double the rate? That is that the question mark there is about 2. I see no hands. What do you expect?
So this rate grows by that now, 1.96 fold, practically 2. So it’s first order in R-H. Did I say 2 before? It’s first order in R-H, no one’s surprised. Of course, the rate is also proportional to k1.
Now how about if we change [X2]? What’s its exponent? So let’s do the same analysis. Now we’re going to double X2. So we’ll double X2 and now we’re going to have to double the ratio of X• to R•. Remember we’re starting from 98 to 2. What do we do in order to double that ratio? We have to go 99 to 1. Everybody see that?
That’s now going to make a difference because if we look at this rate, which is the same as that rate, it’s the same k2, the same–pardon me, we doubled X2. But R• is now only half as big as it was before.
So we doubled X2, but we halved R•. So we didn’t accelerate it at all. So it’s zero order in X2. So it depends on the concentration of this starting material, the one that goes into the slow step, but it doesn’t depend on the concentration of this starting material, the one that goes into the fast step. So probably if I’d asked you, my guess is if–we should have done this as a democracy, as Sharpless said. And I should have asked you before what you think that exponent is and what you think that exponent is. My guess is you’d have said they’re both 1. But in fact that one’s zero. The rate is also, therefore, if it’s insensitive to X2 it’s also insensitive to k2.
So in summary, when one step of a cycle is much slower than the others, the rate of cycling is pretty insensitive to the rate constant and concentration of the incoming reagent for the fast step. Because concentration of the minor form of recycling reagent adjusts to compensate, and fractional changes in concentrations of dominant radical are much more modest. So the rate’s obviously going to be proportional to the amount of radical. There’s no surprises–if you were running 13 machines it’s going to be 13 times faster than running just one machine.
But that’s the other factor that we were going to get to a few minutes ago. That we’re going to have initiation and termination, and in my fancying up the slides I forgot to animate that properly, sorry.
So initiation typically involves breaking a weak bond with light, heat or adding an electron, or sometimes taking away an electron. So heat, you can take a weak bond, and with heat, break it to get two radicals–the oxygen-oxygen bond is unusually weak, as we’re going to be discussing not long from now. Or you can use light–we talked about this last semester already–take an electron that’s in σ, put it in σ* and the chlorines will come apart.
This one has an energy of 30 kilocalories per mole. What are the energies of other σ bonds we’ve been talking about, more or less? C-C or C-H, for example, order of magnitude. 100, right? 3 times as strong. So this is an unusually weak bond, so you can do it with heat. And the reason it’s weak is because you have repulsion among these unshared pairs of electrons that you don’t have in the other cases. We’ll talk about that in just a minute as well.
Or you can put an electron into σ* by a reduction reaction and get OH–, but the free radical OH, which can do a hydrogen abstraction and start the chain.
But even if you’re generating radicals, you could lose by radicals being destroyed. Remember we said why not just wait longer and let this machine go around, even if it’s going slower? The reason is that’s not all the radicals can do, abstracting hydrogens or adding to double bonds. If radicals find each other the SOMO-SOMO reaction is really easy. It’s all downhill in energy. So you have to be sure they don’t find partners, and that process is called termination of a free-radical chain reaction.
So you have initiation that generates the radicals that allow the machine to work. Propagation, that’s the cycling of the machine. And termination, which destroys the machines. Remember the New Hampshire license plate, it says “Live Free or Die.” If the radicals aren’t free from one another, bingo they’re gone. So that’s termination.
So now how about the kinetic order in the amount of initiator? You might think if you initiate the chains twice as fast–if you have twice as many machines operating–then the reaction should be twice as fast. The overall rate should be twice as fast. But that isn’t true, as you’ll see here.
So the initiation, indeed, depends on how much initiator you have, and it has a certain rate constant for initiation to give those radicals. And the rate of forming radicals is, therefore, first order. It’s proportional to the concentration of the precursor.
A termination has two radicals coming together. It could be any radicals. It could be the X•’s coming together, it could be R•’s coming together, or it could be an R• coming together with an X•. All of those reactions are a radical plus a radical–it will terminate. And the rate of destroying radicals is, of course, second order in how many radicals there are, because two of them have to come together. And at steady state, once the thing is cranking along, the rate of initiation, which is this, here, and the rate of termination must be the same, because we’re in the steady state, we’re not changing the number of machines.
So once we get to steady state then those two will be proportional to one another, or the amount of R• is proportional to [peroxide]1/2 because of that termination. So, in fact, if the rate of the reaction is proportional to how many machines you have, [R•], is proportional to the square root of the amount of initiator there, because the destruction is second order.
So that’s an interesting additional rate law, that when you see that half order, in this case it means that there’s a chain reaction going on and termination is involved. So the answer is one-half order.
Chapter 4. Ionic Control of Radical Regiospecificity [00:36:19]
Now we’re going to shift from free-radical reactions toward ionic reactions by considering free-radical reactions that involve ions. The reason this is important, to make this transition, is because ionic reactions are so different because they’re dependent on solvent in ways that radical reactions aren’t.
So here’s another substitution reaction where now instead of a halogen, instead of having Cl-Cl here, we have a nitrogen-chlorine bond. So we have nitrogen radicals that can pull off hydrogen to generate an amine, the N-H bond of an amine, and give a free radical, just as we did before, which then can pull chlorine off here to make R-Cl. So we can do a free radical halogenation, but instead of using a halogen we use a chloramine.
Now this bond is about 100 kilocalories per mole, this one is a weak bond, about 46 kilocalories per mole. Again, like chlorine-chlorine would be a weak bond. And those product ones are strong bonds.
So the first step is a little bit exothermic, and the last step is–pardon me. The first step goes from a strong bond to a weak bond, it’s endothermic. So that’s slow. The second one goes from a weak bond to a strong bond, that’s exothermic. So that one’s fast. The top one is the slow step. But that means that this will be the dominant species between these two. You’ll have a lot of this so that this rate can be equal to this rate–not much R• at any given time.
This is then what happens. We have the free radical, which–pardon me, two free radicals–can come together in this way that one radical takes the hydrogen from this isopropyl group, the hydrogen here, from the other one. So that’s the termination of the reaction. That’s what’s going to kill it. So that’s slowing the reaction down by destroying machines.
So if we could retard this process then we can speed up the reaction, if you don’t have termination anymore. So this is termination by hydrogen atom transfer–get rid of the radicals, and stop the machine.
Now what happens if you add acid? Now you protonate the unshared pair here in the amine. And charge will keep–remember, most of the radicals are in this form. But the charge, if they’re both charged, it’ll keep them apart. And if they’re apart then they don’t terminate. So you can slow down the termination by protonating these free radicals.
Now let’s look at these results from 1971 on using this reagent. This is one step in the reaction, the chloramine, to substitute, to halogenate, a long chain alcohol. So what I graphed here is what percent of isomers you get for putting chlorine on this carbon or on this carbon or this carbon or this carbon or this carbon or this carbon or this carbon. And you can see that you don’t get more than 20% of any of them, and you get a real dog’s breakfast mixture here of isomers–this is useless. You get a very low yield of whatever it was you wanted, and you have to do a lot of work to purify it.
If you do it in 50% sulfuric acid it’s pretty much the same story. If you do it in 60% sulfuric acid it’s a little bit better. You get more of this. But if you do it in 70% sulfuric acid you get almost all of this material–chlorination specifically at that position. So isn’t that a great mechanism? By changing the solvent, by making it a strong acid, you can make it selective for this one. See the advantage of that, obvious in synthesis. Why? Why when you change from–notice 50% and 60% didn’t make much difference. A little difference. But going to 70% really zoomed up on that one.
So why does changing the acid concentration influence the selectivity? Notice that those amine radicals are not really strong bases. It takes a very strong acid to get most of this stuff protonated. To add this proton on there takes a really strong acid. When you get to 70% it’s almost–pardon me. I said exactly the wrong thing, and I apologize. Forget what I just said for the last 20 seconds.
The amine is a reasonable base. It’s easily protonated. Even in weak acid this is protonated. But there’s another base in the system that’s hard to protonate. And what’s that? Sebastian?
Student: The alcohol.
Professor Michael McBride: The alcohol could be protonated. But you need a really strong acid to protonate the alcohol.
So here the ROH is half protonated. Now if this molecule is protonated it’s hard to get the two protonated things together, so the reaction slows down. The reaction still proceeds, but it goes with the material that’s not protonated, and it’s only half protonated.
But at 70% it’s fully protonated, everything is like that. Now the reaction will slow down. So it won’t be as fast. But what? What’s the advantage? It’s hard for the charges to get together. So they’ll stay apart to the extent they can. So what? Ayesha?
Student: The amine would not react…
Professor Michael McBride: It won’t react here near the other positive charge. It’ll react out at the end. It’ll get a secondary hydrogen, not the primary one–we’ve talked about that already. But it gets this one, not even this one. Isn’t that a cute idea–to use the charges to make the thing selective. But it only happens if you can make both reagents be charged. This one’s easy to charge. But this one only when you get in very strong sulfuric acid. I’ve always admired that experiment. So it comes down there in order to do its reaction.
Now what I haven’t shown is reacting there. And this is going to be related to your next problem. So if you use 30% sulfuric acid, then you get a completely different product. Remember I went 50%, 60%, 70%. But if you use 30% sulfuric acid, then you get that product. This carbon becomes an ester or an acid.
So this carbon gets oxidized–the two H’s are removed and oxygen’s put on. And that probably got formed from an aldehyde. The first thing that probably happened was that you changed this CH2OH to the HC=O. And mostly that’s formed not by a radical–it could be formed by the radical-chain reaction. But probably it’s not. It’s probably formed by a HOMO-LUMO reaction, an oxidation reaction, involving this reagent.
Remember the original reagent we’re using–the thing you put in is chloramine, NCl, and at 30% acid it’s protonated. So it’s probably a reaction between this stuff and this. It could be the free-radical reaction, but it’s probably a HOMO-LUMO reaction that’s able to convert this to that. And this is that puzzle that’s your other problem for a week from today, since we’re not having class on Monday—pardon me, next Wednesday.
The puzzle is to propose mechanisms to form the aldehyde involving two steps. A substitution, but also an elimination reaction, reaction with a base. So now that substitution reaction could either be the radical substitution, this mechanism going like that. Or it could be a HOMO-LUMO reaction.
So I’ve said you could get together and work on this in groups. So I think you should be able to do both these, both the free-radical substitution and the HOMO-LUMO, that together with elimination, removing HX, gives rise to the aldehyde. So see if you can figure out why at 30% sulfuric acid you get a completely different product.
I think that’s enough for now. We’ll go on to this problem next time.
[end of transcript]Back to Top
|mp3||mov [100MB]||mov [500MB]|