CHEM 125b: Freshman Organic Chemistry II

Lecture 6

 - Brønsted Acidity and the Generality of Nucleophilic Substitution


The coincidentally substantial extent of ionic dissociation of water provides an example of Brønsted acidity, or nucleophilic substitution at hydrogen. Relative pKa values are insensitive enough to solvent that they provide insight on the role of energy-match, overlap, and resonance in ionic dissociation. The titration of alanine in water illustrates the experimental determination of pKa values and the phenomenon of buffering. The limited pKa scale in water can be extended dramatically by titration in other solvents, providing one of the best ways to measure many “effects” in organic chemistry. A wide range of important organic reactions discovered in the 19th century and many biochemical reactions can be understood under the rubric of nucleophilic substitution.

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Freshman Organic Chemistry II

CHEM 125b - Lecture 6 - Brønsted Acidity and the Generality of Nucleophilic Substitution

Chapter 1. Solvent Influence on Ionic Dissociation [00:00:00]

Professor Michael McBride: So as you know, we’re getting on to ionic reactions, and started by talking about how important the solvent is. We started last time looking at the example of the disassociation of water into ions to give H+ and OH, both, of course, solvated. We’re interested in the important role that the solvent plays.

We saw last time that we could talk about water in the gas phase. Dissociating into radicals, we know the bond dissociation energy. Then transferring an electron. And most importantly, then getting the two ions apart from one another, which is going to cost 332 kilocalories per mole just to get them apart from being like 1 Å to infinitely apart.

Altogether those processes cost 392 kilocalories per mole. So that would be absolutely out of the question. In fact, even once you start in the gas phase it’s 386, so that would be 10–290. Impossible. But the solvent makes it possible–you know that water dissociates.

So when you put another water molecule, which is a dipole, next to the OH, it makes it more stable. We talked about that last time or two times ago how you get hydrogen bonding between a hydroxide anion and a water. So it’s more, in that case, more than just a dipole. Remember it was that double minimum that, in fact, the lowest energy level had the hydrogen in the middle. Anyhow, that’s not fabulously strong; it’s not like a covalent bond, but it’s 28 kilocalories per mole, which is nothing to sneeze at.

If you put another water on that you get another 18 kilocalories per mole. And if you put another and another and another and another and they line up, you get stability by 106 kilocalories per mole for solvating the hydroxide ion, to put it in water. So that’s worth more than most covalent bonds.

Now you also have the proton that could be solvated. That one actually forms a bond–you can form a bond between the unshared pair on the water that’s coming up and the H+. That bond is, in fact, a very strong bond. It’s 164 kilocalories per mole–about twice as strong as most covalent bond. Because it’s not only a bond, it also gets the positive charge in close proximity to a bunch of other electrons. So it polarizes them. That’s a little bit like solvation within the molecule.

So you get 164 just for the first water onto a proton. Then, of course, that looks rather–it’s charged and it’s not very much bigger than the OH. So it’s not going to surprise you that you can put more waters onto that. And from doing a whole bunch of waters to put it into liquid water, you get another 100 kilocalories per mole on top of what you got forming H3O+.

So if we sum all those together, 164, 106 and 100–notice that those are similar to one another–we get 370 kilocalories per mole. That means we get the energy of forming the proton and the hydroxide if they’re solvated down to within only 21 and a half kilocalories of the water in water.

So that means the pKa is 15.8, you know. But it has enormously to do with solvation, as you can see here. It’s, in fact, very delicately balanced. It’s a small difference, this 22 kilocalories per mole, of very large numbers, things that total 392 kilocalories per mole. So if any of these steps had been off by a few percent, the dissociation of water would be very different.

So this is a sign that you have to be careful, because whenever you’re dealing with small differences that come from very large effects in the first place, it could go any place. In the case of water it happens to be that water will dissociate with a dissociation constant of [correction: 10^16]16. But it has to be regarded mostly as an accident because of this important role of solvation.

Chapter 2. Brønsted Acidity as Nucleophilic Substitution at Hydrogen [00:04:45] 

Now we’re going to generalize into Brønsted acidity–other things giving up protons. Or you could call it substitution at hydrogen. Here’s HF–and we talked about this last semester ad infinitum, of how we could make and break a bond at the same time by bringing up, say, water, and get H3O+ and F. That is, HF acting as an acid, hydrofluoric acid. So that’s Brønsted acidity, giving up a proton, but it could be thought of as substitution at hydrogen–something new comes in, the old thing leaves, fluoride in this case. 

Now this happens, of course, in a solvent. Just as in the case of water, HF isn’t going to disassociate to H+ and F in the gas phase. It’s the stabilization of those ions by water that make this accessible. 

So is it going to be like water where you have these enormous things and there’s no predicting what will happen, as to how strong the acid will be? To a certain extent that’s true. But fortunately, the solvation energies of analogous compounds are similar enough that we can often make reasonably accurate predictions–or at least if not predict, at least once we’ve seen the results we could rationalize them reasonably–of the relative acidities in terms of molecular structures.

That is, if we can cancel out, because they’re very similar, all these solvation things, then we can look at the molecules themselves and try to understand why one is more acidic than another. That’s what we’d like to be able to do because it’s much simpler to think about just the molecule than to think about how all the solvents would be arranged around any particular molecule or ion.

So let’s try that. Now as background for talking about acid you’ve, in your AP chemistry, looked at pH and pKa, and there’s the question why should organic chemists bother about pH and pKa, which you’ve already done in your AP. It seems like a topic for general chemistry, not organic chemistry. But there are several reasons why it’s important in organic chemistry, and I’m sure that all those various textbooks you have have a chapter, a section, about Brønsted acidity.

Because in the first place, whether a molecule is ionized or not is important for predicting reactivity–that is, the availability of HOMOs and LUMOs–how high, how low they’ll be depends on whether it’s an ion. Also things like conformation, the color, the proximity to other species, mobility, particularly if you’re in an electric field, is a big function of whether there’s a charge. So you want to know whether molecules are ionized or not.

One of the simplest ways of making ions is dissociating a proton. And because the ease with which a species reacts with a proton might predict how readily it reacts with other LUMOs. So often we’re not specifically interested in how easily a proton would react with some high HOMO, but we are interested in how someone else would react. And if we’re comparing two high HOMOs to see which one’s more reactive, it wouldn’t be surprising if the one that’s more reactive with a proton is also more reactive with the molecule we’re interested in.

So if we can easily measure reactivity with a proton, the reverse of which is acidity, if we can easily measure that equilibrium, then we have a scale to talk about how low LUMOs are, how reactive LUMOs might be. So that’s why acidity is particularly important in organic chemistry, because it could predict organic reactions. That is, reaction with the LUMO σ*CX or π*CO might parallel the reactivity with a proton.

Now we know that the acid dissociation constant is the product of [H+] and [B] over [HB] where B is some base or HB an acid. And if we take the logs of both sides and rearrange a little bit, we see that the pH is the pKa minus the log of the ratio–how much of the stuff is ionized and how much of the stuff is in the acidic form. If that ratio is 1, if it’s half ionized, then the log of 1 is zero so we forget that. And then pH is equal to pKa

So we have an acid with a certain pKa. If the pH is that value, then you have 50:50 ionized and unionized form. Now you often use things like that as indicators. It might be that the ionic form is colored or has a different color from the protonated form. Then if you see that color or not and can measure its intensity, you can tell how much is ionized. So you could measure that ratio. But you can’t measure it over a really wide range, because suppose you could measure it when there’s 5% of the colored species or up to 95%, but beyond that the colors aren’t distinguishable. That would be fairly reasonable. 

So if you go from 95:5 to 5:95 that covers just this ratio, that covers just 2.5 pH units. So if you want to use an indicator to measure the pH of something, you can only measure it with that indicator over a very narrow range by measuring the ratio of how much colored to uncolored stuff there is. Beyond that you have to get some other indicator and then some other indicator and some other indicator. So you could have a ladder of indicators that allows you to cover a wide range of pH. And then you could bootstrap that way with overlapping indicators to get a wide coverage.

So with a known pKa, you measure the pH by measuring the ratio of the ions–ionized to the unionized form.

Chapter 3. Understanding Relative pKa in Terms of Energy Match and Overlap [00:11:09]

Now let’s look at factors that influence the acidity, because they may also be the ones that influence reactivity and other organic reactions. Now pKa normally is defined in water. We just saw that the dissociation constant for water in water, that is, losing the red proton there, is 15.7. Now if you tried to get the pH higher than that, what would happen?

If you had water and you made its pH 20, let’s say, or 18 instead of some place in this range, what would happen? You’d have the strong base in there. That base would pull protons off the water, and you wouldn’t have any water anymore. So it wouldn’t be anything like the same solvent if you tried to get higher pH than that. And the same thing is true if you look at the dissociation constant of H3O+. At a pH of –1.7 it would be 50% protonated. But if water is 50% protonated it’s not water anymore.

So if you’re within this range between, say, zero and 15 or something like that, then you have water and you can talk about things. But if you tried to get out of that, you can’t use water anymore because all you’re doing is making all the water into ions. But in that range you can work.

So in that range there are a number of interesting things, like ammonia can be protonated. The ammonium ion has a pKa of 9.2. So at pH 9.2, half of ammonia is protonated, the ammonium ion, half of it is NH3

Does it surprise you–now let’s get this straight. So NH4 acting as an acid, NH4+ is 9.2. H3O+ is –1.7. Which one gives up its proton more easily? H3O+ or NH4+? Which is a stronger acid? Which one would take the proton from the other one? Debby, do you have an idea?

Student: I guess H3O+ is a stronger acid.

Professor Michael McBride: Right. So it would protonate NH3. So if you had H3O+ and NH3, it would be downhill in energy to transfer the proton to the NH3. Why? Why is it better to put the proton onto NH3 than onto H2O? What is it about NH3? That makes that a stronger bond.

So you have a proton plus something that has an unshared pair, they come together in the form of a bond. How much energy are you going to get out of that bond? The proton is the same in all cases. What difference is there in the unshared pair between nitrogen and oxygen that means that nitrogen will hold the proton more strongly. Sebastian?

Student: It’s a higher HOMO because nitrogen has a lower nuclear charge.

Professor Michael McBride: Right. Lower nuclear charge, higher HOMO, stronger bond. Better energy match. So we understand that difference here of about 11 pH units.

Here’s FH, 3.2, hydrofluoric acid. Not nearly as good an acid as H3O+. But it’s a lot better acid than H-O-H than water by 12.5 powers of 10. The reason is the same. That oxygen has a higher HOMO and holds the proton more strongly. So we can understand that. 

H2S is 7.0. So now that seems a little funny compared with water. But notice the bond association energies. Actually, it’s reasonable–sulfur is not as electronegative as oxygen. Its HOMO is lower. Pardon me, let me get this straight. You always have to think these things through. I realize I’m rattling it off quite fast, and it’ll take a little while looking this over in your own room to get it straight in your head.

But it’s interesting that if you look at the bond dissociation energies, sulfur is between oxygen and fluorine. So it’s more than just how strong the bond is that determined sulfur to be in the middle. So we’re going to talk a little bit about that.

So here are hydrogen attached to various elements, and we’re going to look at the pKa’s. Now if we go across the top row you see that the ease of heterolysis, how strong an acid it is, increases to the right. So it gets easier to break the bond into ions as you go to the right. But if you look at the bond strength, the bonds get stronger as you go to the right. So that ease of homolysis goes the other way. But this is exactly what we talked about a time or two ago when we looked at the energies of two things and said that homolysis and heterolysis are opposite one another, whether you take the two electrons or whether you put them both here in heterolysis, or whether you put one here and one up there in homolysis. So that’s the same thing we talked about before.

So the energy mismatch, the electronegativity difference makes it easier to do heterolysis, but harder to do homolysis. But notice it’s not the same when you go down the periodic table. Now the red and the black arrows are in the opposite direction. Fluoride is the most electronegative, but it’s the weakest acid among the hydrogen halides. And in that case, it is the bond strength that’s making the difference. And that’s what made the sulfur more acidic than oxygen is because of the weaker bond strength. So that has to do with decrease of overlap; going across had to do with the energy match. So we understand those things.

Now we can learn, by knowing pKa values, some of the things that influence the energies of these bonds. So here’s an O-H bond in acetic acid. Of course, it’s called acetic acid–water is not called aqueic acid or something like that. So this is much more acidic–4.8 versus 15.7. So it’s 11 powers of 10 stronger as an acid. Why? Is it because the bond is intrinsically stronger? They’re both OH [sigma]s bonds. The bonds are essentially the same. But the anions are different–and we’ve talked about this before. What is it that makes the anion that you’re going to make here unusually stable compared to H-O?

Student: Resonance.

Professor Michael McBride: Right, there’s going to be resonance that’ll make it stable. Or HOMO-LUMO mixing, if we wanted not to use the language of resonance. So the low LUMO of π* stabilizes the electrons we’re putting on oxygen.

Now if you make chloroacetic acid, then it’s another 2 powers of 10 stronger. Why would that be sensible? Anybody got an idea about that? Ellen?

Student: Electron induction by the chlorine.

Professor Michael McBride: Right. It’s an inductive effect. The chlorine is electronegative, it withdraws electrons from this carbon, which withdraws electrons from that carbon, which withdraws electrons from that oxygen, and makes that a better place to put a negative charge. So it’s worth the factor of 100. So we have some measure here. Because these equilibrium constants are easy to measure for reversible acid dissociation, it gives us a scale that’s easily accessible of how important something like inductive effect can be. 

Or if we look at this one where we’re breaking a C-H bond. Now normal carbon-hydrogen bonds aren’t acidic, but this one is rather acidic. It’s as acidic as the ammonium ion. Why? Because there are two π* vacant orbitals next door that can stabilize the high HOMO when you make carbon minus by losing that one. So again, we have a quantitative measure of how important this might be.

Now here’s an amino acid, alanine, in the form that’s protonated here on nitrogen, and its pKa, it’s even a little bit stronger acid than the chloroacetic acid. Why would you say so? Nitrogen is not more electronegative than chlorine–not withdrawing more electrons than chlorine is. Or is it? Lauren?

Student: It’s got the positive charge.

Professor Michael McBride: Ah, the positive charge. That makes it very electronegative.

Chapter 4. Measuring the pKa Values of Alanine [00:21:18]

Let’s look at that one in a little more detail. And it also shows how you determine these pKa’s. So you can titrate alanine. You start with this protonated form in acidic medium at pH 0, say, and it’s got that red proton that could be lost. And then we’re going to start running in hydroxide until we’ve put in two equivalents of hydroxide. When we add the first equivalent, we’ll take off that red proton. And then we’ll take the one off the nitrogen when we add a second equivalent.

But let’s look at how the pH changes as we add the base and pull off protons. So it looks like this. It changes very slowly in there and is said to be buffered. And then it changes rapidly again. Now why is that? Why isn’t it just a straight line as you add more and more hydroxide and watch the pH change?

If you want to change the ratio by 9-fold, you could change it from 3:1 to 1:3. So that’s a 9-fold change in the ratio. And remember, the ratio is measuring the change in pH. So if you change the ratio by 9, say, approximately 10, the log of 10 is 1, so it would change the pH of the solution by 1. So adding half an equivalent of base, going from here to here, changes it by about 1 pKa unit, 2 to 3.

Now suppose we want to change the pH from 3 to 4, how much do we have to add? If we change the ratio by a factor of 9–again, I use 9 just because it’s convenient, 3 to 1 to 1 to 3. If we want to change it by a factor of 9 again and go up by another pKa unit, pH unit, logarithmic unit, then we only have to go from 1 to 3 to 3 to 100. That’s another factor of 9 change. But it only takes 0.22 equivalents to do that.

Then if we want to change it by another log unit, it takes only 0.03 equivalents. So during this region when the ratio is some place close to 50:50, it takes a lot of conversion in order to change the ratio very much. But when you’re out at the end and you only have 1% or a couple percent of something, then it doesn’t take much change to change the ratio enormously. So that’s why you get this behavior as a buffer, where if you put a lot of that stuff in, and you have roughly 50:50 of the ionized and unionized form, you can do a lot of adding acid or base and it doesn’t change the pH significantly.

But if you’re out in this region where you only have a few percent of one or the other, then it changes drastically. Now if we keep going, we take off this proton, and again it’s slow and buffered and then it goes up again.

Now, how can you find the pKa of the compound, having done one of these titrations where you run in and measure the pH as you’re going along? Remember, the pKa is the same as the pH when what? When the ratio is 50:50. So that means if we take it where the ratio is 50:50, that’s the first pKa, the pKa for this proton, the red one here. So the pK1 is 2.35.

Now it’s reasonable that the proximity of the positive charge to the carboxyl group that’s going to be losing a proton, having the plus charge here–what Lauren was telling us about–is going to make it, here it is, 300 times easier to remove the H+ from this cation rather than acetic acid, which was 4.5 in its pKa. So we went from 4.5 to about 2.5. Well, a little bit more than 2.5. We went 300 times better because of that positive charge, stabilizing the anion we’re going to get here by having these two charges near one another. So that’s reasonable.

Now we take off the second proton, and we find that that’s pKa 9.87. Now how would you expect that to compare with taking a proton away from a normal amine? Here we’re taking the proton away from an ammonium ion to generate an amine, but now we have a negative charge that’s making this more stable, be harder to pull away.

So at first glance you’d say it would be at about the same. If this was easier by a factor of 300 because this is so stable, then this one, taking that second proton off, should be harder than 300 compared to a normal amine because this thing is so stable. So with a proximity to the negative charge should make it about 300 times harder to remove proton from this so-called zwitterion–the thing that has both positive and negative charge–than from a normal amine, like say ethylamine.

Now ethylamine is pKa 10.6. It’s actually 5 times easier to take the proton away from alanine. From the zwitterion of alanine it’s easier to take it away. The same way this one was easier as compared to a normal ammonium ion. That seems absolutely backwards—that it should be 300 times easier. Why? Why is it easier rather than harder to take it away from here rather than here?

So you’ll notice that it’s a question of compared to what. Is this a good model for this when it has a proton on it and isn’t charged? Not really, because this has this carbon with a bunch of oxygens in it. And this one, the ethylamine doesn’t. So maybe the electron withdrawing effect of those oxygens is making it easier to give up a proton here, stabilizing the anion. The same way chlorine did. Remember, in chloroacetic acid, it was withdrawing electrons and making it easier. So maybe we should compare this, not with ethylamine, but with something that has the oxygens but doesn’t have the charge.

So we can do that by making it an ester instead of an acid so it doesn’t have the charge on it. Its pKa is 7.3. So now, in fact, it is harder to take it away, and it’s about 400 times, about what we expected if we have the right comparison.

So having these charges changes the pKa in a reasonable way. So apparently the CO2 group, lacking charge, is sufficiently electron withdrawing to destabilize that cation more than the negative charge stabilizes it.

Chapter 5. Factors that Influence pKa over an Expanded Range [00:29:09] 

Now notice here we’ve changed the scale–now we’re going from 10 to 50. We’re going to very, very weak acids. And, of course, we can’t do that in water because water has a pKa of 16, and if we try to get more basic than that we just convert water into hydroxide–it’s not water anymore. So we’re going to have to use other solvents to do that. But if we use these other solvents, and again, use indicators with boot straps to go on up.

Then we get that a hydrogen attached to nitrogen. Now we’re not talking about an ammonium ion. That’s what we were just talking about, taking a proton away from RNH3+. It’s not that anymore. It’s taking the hydrogen right away from nitrogen without the charge. So it’s much harder to take it away than it is to take it away from water, which is what you expect because the energy match was better here.

And if we go to carbon then it’s way, way up there, 52. Now in your various books you’ll have different values of that. It might vary by as much as 3 or even 4 pKa units, because you have to go from one solvent to another here and use these indicators to know what the equivalent of pH would be. And it depends on which ones you use to do that. But anyhow, it’s way up there. 

So these values are approximate because the equilibrium for bases stronger than hydroxide can’t be measured in water–you have to bootstrap by comparing these acid base pairs in other solvents. So that one’s a bad E match for O-H, a better E match for N-H, and a great E match for C-H, so it’s holding the bonds stronger together and making it harder to form the anions.

Now here’s another C-H bond, but it’s significantly more acidic. The hydrogen attached to a double-bonded carbon. Anybody got any ideas why that would be easier to break? Can anybody think of an idea why it might be harder to break a proton away, to break a bond to hydrogen when it’s attached to a double bonded carbon? Ellen?

Student: sp2?

Professor Michael McBride: Ah, it’s sp2 hybridized, so the bond’s going to be stronger that you’re going to be breaking. But, in fact, it’s easier to ionize it. So it’s not because of the bond strength–that would go the other direction. It must be the ion that’s unusually stable. Can anybody think about why the ion might be stable in this double-bonded carbon, the unshared pair? Anurag?

Student: The π* of the ion will stabilize the charge.

Professor Michael McBride: Ah, you could get resonance. The π* stabilizes it, but thanks for biting on that, because it doesn’t work. Do you see why? So here’re the two hydrogens on the carbon, double bond here on that carbon. Why is it that when you pull this proton off and generate a pair of electrons here, why isn’t it stabilized by the π*

Student: They’re orthogonal.

Professor Michael McBride: They’re orthogonal. So you’re not going to get it that way. But this orbital is sp2, not sp3, as it is in the other one, the one that’s 52. So it’s a lower energy orbital for the electrons to be in, so it’s a more stable anion. That’s good. 

Now here’s an acetylene that loses its proton, and it’s 20 powers of 10 better at losing the proton. Why? Anurag, let’s come back to you on that one.

Student: The resonance on the ion? 

Professor Michael McBride: No, it’s still not resonance because–so here’s the carbon, here’s the hydrogen coming out at you. There’s a π orbital that’s this way, and a π orbital that’s this way in the acetylene, but both of them are orthogonal to that. What’s good about it?

Student: It’s sp hybridized.

Professor Michael McBride: It’s sp hybridized. So here we go from sp3 to sp2 to sp, and it gets more and more–so sp3 anion, sp anion–no π overlap, and an sp2 anion–no π overlap. And Anurag, I’m going to do you one more because you’re being so good to me here. So we’re going to take it away from this hydrogen, which is not on the triple bond–this one the hydrogen was on the triple bond. Now we’re taking it away from a carbon next to the triple bond. Or here we’re taking it away from a carbon that’s on a double bond. But it’s also next to a double bond. Now why do you say it’s good? 

Student: Resonance.

Professor Michael McBride: Resonance–now you got it right. Because when it’s not one that’s on this, either a double bond or a triple bond, but it’s on the carbon next to it, now it can point this way, and be like that, and overlap and get the resonance stabilization, so you were right.

So this is a HOMO-π* overlap finally. And there we see it.

Now there are fabulous tables of people who have measured these things. So, for example, this is a website from a course at Harvard, Chem 206, which has six pages–this is the first of six pages–so all different pKa’s of many, many different compounds losing their protons. There’s another compilation at another website here, which I think is even bigger.

So these are interesting tables to look at to see if you can figure out–look at molecules that are very similar to one another–so solvation will be more or less the same and you don’t have to worry about that–and see what it is about the change from one molecule to the next. How can you explain in terms like we’ve been talking about–hybridization, resonance, electronegativity. That is, the HOMOs and LUMOs. How can you explain these trends seen over similar compounds?

So here’s the problem for Wednesday. List the factors that help determine the pKa for an acid. Then choose a set of several related acids from one of these pKa tables, or your text will have some, too, I’m sure. And explain what they teach about the relative importance of these factors–which is more important, resonance or hybridization, as we were talking about here.

So just practice yourself by choosing a set of three or four, or whatever number seems to be there, of related compounds where you see structural changes, and see if you can rationalize why they have the pKa’s they have. Then once you’ve done that in the privacy of your own room, explain your conclusions to at least one other class member and decide together how unambiguous your lesson is. Did you think it through, or was there something you forgot or did you get something backwards? So that’s your assignment for Wednesday, and feel free to consult the problems in textbook–they’ll have problems about pKa’s, and the references at the end of the tables if you want to. There’s a hint that that would make a good exam question for you to explain something like that.

Chapter 6. The Generality of Nucleophilic Substitution [00:37:06] 

Now, nucleophilic substitution and β-elimination. So we’ve been talking about this Brønsted acidity in order to understand reactivity and more typical organic reactions. So let’s see if we could use these ideas in understanding nucleophilic substitution and β-elimination. In the Jones textbook that’s chapter seven, but there will be corresponding chapters in all of the books that you have. We also started talking about this in lecture 16, as you’ll remember, where we talked about acid-base reactions, and about SN2 substitution, and that they really were the same reaction in terms of the orbitals that were involved.

In fact, the β-elimination or E2 elimination, remember we talked about at the same time in lecture 16 where the base takes off this proton, you generate the double bond and lose the leaving group there.

Now these reactions weren’t discovered by people who understood mechanisms and applied them to see what they could make in the laboratory. They were discovered by people fiddling around with things and seeing what would happen. And it’s only 100 years or 150 later that people began to really understand what was going on. 

So, for example, Alexander William Williamson in Britain in 1852 discovered that treating this alkoxide with ethyl bromide could do a double displacement reaction where you exchange partners and the ethyl goes on the oxygen and the sodium goes with the bromide.

Finkelstein, who was a chemist in the Netherlands in 1910, found that sodium iodide reacting with R-Cl could give R-I, so you could exchange iodine in place of chlorine and get sodium chloride.

Now all these things have their own little bit of lore about them. The reason that reaction works so well is that sodium iodide is soluble in acetone, the solvent that’s used. But sodium chloride is not soluble in acetone. So when you mix the two, you could imagine either of them replacing the other one. But the equilibrium is driven toward the products by the fact that sodium chloride precipitates. So all of these reactions have their own twists about them that make them interesting and variable.

But there’s generality, too. For example, he could also do this trick with an alkyl bromide instead of alkyl iodide. So if you could only do something in one particular case, no one cares very much about it. But if it’s general, then people can apply it to their own problems and it becomes much more interesting.

So generalization is the name of the game here. And both of those reactions, the Williamson ether synthesis, and the two flavors of the Finkelstein reaction, both of them involve exchanging ions. You start with one pair of ions and you end with a different pair of ions–it’s one of these double decomposition reactions.

Then in Russia, Menshutkin in 1890 discovered that you could react triethylamine with an alkyl iodide and generate the salt–tetravalent ammonium salt as an iodide. Notice that this seems to be different from the first one, because the first one just exchanged ions; this one actually creates ions where there weren’t ions in the first place.

Or Hans Meerwein working right after the Second World War in Germany discovered that you could use this thing called Meerwein’s reagent, which is a special kind of oxygen–oxygen plus because it has three things attached to it. Trimethyloxonium. Fluoroborate is the anion that he used. This could react with ROto give R-O-Me–that is, it’s a methylating agent, it puts methyl onto oxygen.

But notice in this one you actually destroy ions. You start with four ions and you end with two ions. So at the top we have exchange of ions, then we have creating ions, now we have destroying ions.

And then there’s solvolysis, where the reagent actually is the solvent–ethyl can react with the carbon displacing bromide as HBr. So that solvolysis means breaking apart, that’s the “lysis,” by the solvent. 

So all these seem to be–if you had a list of reactions you’d have to memorize, we have all these different reactions–the Williamson ether synthesis, the Finkelstein reaction, the Menshutkin reaction, the Meerwein reagent, solvolysis reactions and so on. But the great thing is that they’re all the same. Once you understand the mechanism, they’re all nucleophilic substitution, which is the subject of this part of the course, because all of them have some unusually high HOMO, which reacts with an unusually low LUMO, which is the σ* orbital. So it breaks the σ* as the new bond is formed with the HOMO. 

So if we’re talking about the generality of nucleophilic substitution, we should generalize over variety and all the different components of the reaction. Just to give you an example of how many different kinds of nucleophilic substitutions there are, consider adenine, which has lots of different unshared pairs on the nitrogen. And ribose, a sugar.

Now here you have a HOMO on the nitrogen and a σ* LUMO of the carbon-oxygen. And you can substitute the nitrogen for the oxygen. Actually, as we’ll see shortly, it’s not quite the same where you push the O-H out. But anyhow that’s a substitution at carbon, which loses water then, and makes that N-H [correction: N-C] bond. And that stuff is called adenosine. So the adenine, plus the “-ose” comes from ribose, so that’s adenosine.

So we have adenosine. And here’s methionine, an amino acid. And it has a HOMO, which is an unshared pair on the sulfur there. And it has an O-H that could be a leaving group. So you can substitute SR2 for O-H at carbon and do another nucleophilic substitution, and you get this thing that’s called S-adenosylmethionine. It’s methionine that has this adenosyl on it. And that is this stuff you may have heard of, SAMe, S-adenosylmethionine–you can buy it in vitamin stores. But it’s very popular with nature as well.

So if you have an amino acid or something that has a nitrogen with an unshared pair, like the nitrogen in arginine, then you can do a nucleophilic substitution reaction where you substitute NHR2 for SR2. So it attacks that carbon, the leaving group leaves, so here’s the third nucleophilic substitution reaction. And it ends up with a methylated amino acid. Or that arginine could’ve been part of a protein and you’d have a methylated protein.

Then you could have another one, where a base comes along and takes the proton away. But that’s nothing but a substitution at hydrogen, and leave the electron pair on nitrogen.

And this is biological methylation, which I know practically nothing about, except for the fact that our daughter, who was a student in this course 20 years ago, is now a professor of biology at Bowdoin College. She works on biological methylation, on protein modification by methyltransferases. I talked to her on the phone the night before last to check this out. She says yes, indeed, this is the compound that’s used in practically all biological methylation reactions of proteins. 

So these nucleophilic substitutions have broad generality and are very important. And what we’re going to be studying then is how the different components influence the rate of these reactions. So we have the nucleophile, the high HOMO that’s coming, we could vary that. We have the substrate, what various R groups we could have. Included in the substrate is, of course, whatever’s going to be leaving–the leaving group, we could vary that and see how it affects the reaction. It takes place in some solvent, that’s going to make a difference. And, of course, there’s some product being formed. 

So all of these things can vary. And now, although we looked schematically early on in the middle of the semester–last semester we looked at this idea of the HOMO attacking the σ* and the leaving group leave. Now we’re going to look at the details of how, as you vary these things, the reaction becomes more or less easy or takes some completely different course. 

So that’s going to be the project we begin next lecture.

[end of transcript] 

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