CHEM 125b: Freshman Organic Chemistry II
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Freshman Organic Chemistry II
CHEM 125b - Lecture 32 - Measuring Bond Energies: Guest Lecture by Prof. G. Barney Ellison
Chapter 1. Diatomic Bond Dissociation Energy from Spectroscopy [00:00:00]
Professor J. Michael McBride: OK, well, today we have a guest lecture. Certainly the most valuable tool in organic chemistry is the covalent bond for thinking about things. We talked about this a lot last semester and, in fact last semester was all about understanding bonds. So, at the end, remember, we got into energies and the question about bond energies, where they come from, and are bond energies real? And the real kinds of energies, remember, are bond dissociation– not average bond energies, which are tables that you can sort of add together to get the heat of formation of a compound– but the actual energy that it takes to break a bond in a molecule.
And we spoke about this table of bond dissociation energies. The best values as of 2003, and the fact that they have been updated through the work of many people, but one of the main compilers was Barney Ellison, who was a graduate student in this department something like 40 years ago. He worked, in fact, with Professor Wiberg, who spoke with you. And so I told you we would try to have him speak with you about where these things really come from, how you know what bond dissociation energies are. And there he is, sitting over there sleeping on me. So he’s going to come and tell you about that. Thank you.
Professor G. Barney Ellison: OK, here. I’d like to see if I can describe to you how you could measure the bond energies of complex molecules. And to start out, let me just remind you what we’re talking about. Let me talk about something like methane. I’m going to need a larger stick. So if you have methane, and you want to measure the bond energy, I literally want to reach in and measure the energy it would take to grab that hydrogen and literally just rip it out of there. And when you can see it does that, the molecule is going to have a change from a tetrahedral geometry to a flat geometry. And there’s a potential curve that I’ve written as a Morse oscillator. And there’s a zero-point energy, and you’re going to have to measure the energy to go from the ground state up to the dissociation limit.
So I thought I’d try to tell you, just in a simple way, if you took a molecule, and you’d like to try to see if you could measure the different bonds in the molecule, how could the specific bond energies be measured? And let me consider something like methanol. I’d like to take methanol, and I’d like to show you how you measure the bond energy to break the OH bond, then how you’d break the CH bond, or how you’d break the CO bond. And I don’t want to calculate this. I actually want to measure it.
Chapter 2. O-H BDE from Acidity in the Flowing Afterglow [00:01:28]
So one way you can do this is to use a gas-phase-acidity-electronegativity cycle. This is a negative-ion cycle that you’ll take this molecule, say methanol, and if you treat this with a base such as fluoride ion, the fluoride ion is actually going to attack the methanol. And when it does this, it’s going to pull a proton off. And if it does that, you’ll form a methoxide ion. The fluoride ion will never pull the proton off the carbon, because the negative ion that you get has the electron sitting on the carbon, and the oxygen is more electronegative, so you’ll always get the top ion.
So if you could do that, what you’d like to think about is, how can I measure an equilibrium between these two species? And if I have the equilibrium, I can get the enthalpy, and if I can get the enthalpy, I can get the acidity of this.
So this is an experiment that a bunch of us do in Boulder. The world’s expert in these gas-phase acidity measurements is my colleague Ronnie Bierbaum, Professor Doctor Veronica Marie Bierbaum. And she’s very good at this. They have a flow tube, and this is the experiment you want to do.
So you would like to measure the reaction of fluoride ion pulling a proton off methanol to get the methoxide ion. So what you do is… here, you can’t get a lecture bottle of fluoride ion, because they’re charged. So what you’ve got to do is take a gas, and you make a plasma. You boil electrons off a filament. And as the electrons interact with the HF, the electrons get captured into a σ* orbital, and the molecule detonates and gives you an F- and hydrogen atoms.
So the fluoride ions are injected into this flow tube. And there’s methanol in here. And what happens is, as the reaction proceeds, the methanol has a proton removed from it. And you’re literally watching the end of the flow tube with a quadrupole mass spec. The quadrupole mass spec monitors the negative ions that have m/Z 31, and you’ll watch them growing in.
So as you watch the methoxide ion going in, you can measure the rate constant, k1, that goes this way. So if you do that, what you’re after is the equilibrium constant. So to get the equilibrium constant, one way to do this is that you would measure the reverse rate.
So if you did that, to measure such a reaction, now you’d just turn the reaction… you do the reverse of this. You now have to make a vapor of methoxide anions. An easy way to do this is to have electrons interact with methyl nitrite. The methyl nitrite then disintegrates, and it forms methoxide anion and nitric oxide, NO. And the methoxide ion then goes down the flow tube. And if the methoxide ion interacts with the HF, now what you’ll see is F- growin in. And you know it’s F-, because Jesus only made a finite number of ions that have m/Z 19, and fluoride is it. There’s no other.
So it turns out, when you do that, you’re able to monitor this. So if you’ve measured k1, and you’ve measured k-1, that is going to give you the equilibrium constant. And this is the difference in acidity between methanol and HF. But the acidity of HF is well-known. It’s 370.424 kcal per mole with this uncertainty. And so it turns out that these two numbers give you the equilibrium constant, and this is going to give you the gas-phase acidity. So by making the equilibrium constant, Professor Bierbaum has measured the acidity of this species.
The way to think about this now, to use acidity, here’s what the gas-phase acidity is. Remember, she really has this number in the gas phase. You know, in the gas phase, these molecules are never going to dissociate into ions, because you’ve got to deal with Coulomb’s law. So you’re talking about maybe 300, 400 kcal per mole. So it’s a whopping amount of energy.
So the way to think about this is, to get that ion, you have to break the OH bond. Then you have to somehow measure the electron affinity. That gives you the energy of the negative ion here. And the ionization energy of a hydrogen atom is how you get the proton.
And here, this is the easiest way, I think, about this. This acidity– you want to get those ions. The first thing you’ve got to do is reach in with a pair of scissors and cut this bond. So you’ve got to pay that energy. You’ve also got to give me the ionization energy of hydrogen atom. You’ve got to pay me that, because that gives you the proton. And then you take the electron and put it back on the alkoxy radical, and that’s the electron affinity.
So it turns out– Here, this is what we’re after. We want this bond energy. And this will be the energy for that bond only. And so the electron affinity for hydrogen atom is known to some pornographic detail. You couldn’t imagine what’s been done to this poor atom. This is done to like 80 digits.
So turns out the only thing you’ve got to do is now, you now have to measure the electron affinity of the radical. So how are you going to do that?
So this is an experiment that’s done actually here by Professor Mark Johnson at Yale University, and it’s done a bit by people in Boulder, Colorado. Here’s what you do. You literally take a vapor of these negative ions. And if you take a vapor of the negative ions, you make a fast beam. You cross this with a laser. I know the energy of the laser precisely. So what you do is, when the laser interacts with the negative ions, you’ll eject electrons, and you catch them in a hemispherical analyzer. You literally measure the kinetic energy. So this is the kinetic energy.
So here. This is easy. If you know what the energy going in is, and you know what the energy coming out is, you’ve measured those two things, and the difference is the binding energy.
So here’s the negative ion. When the electromagnetic wave comes blazing in and strikes it, what happens is, here’s the laser energy. So the molecule literally interacts with the laser, and an electron’s knocked off, and what you’re going to do, is you’re going to measure the red arrow. This is the kinetic energy of the photoelectron. So if you measure what comes out, then the difference is the binding energy.
So it turns out that this is the essential experiment you have to do. There’s a bunch of details you’ve got to pay attention to. If the product radical is vibrating, then the photoelectron kinetic energy will be smaller than the measured binding energy.
And these are just a bunch of words. Let me just show you what you do. This is it. Jesus, we did this like thirty years ago. Christ, I’m an old guy. When you do this, what you do is you take a– right here.
So what you’re going to do is, you’re going to measure the kinetic energy of the electrons coming out. So we’re making a plot of the photoelectron kinetic energy. Here, let me just show, the laser is 488. It’s a beautiful robin’s egg blue. Extremely bright laser. And 488 nanometers is 2.540 electron volts.
So here. Think about this. This is the photoelectron kinetic energy with electrons coming out. No electrons can come out with more than 2.54 eV, because that’s all I’ve got to put in.
So the first time you see electron counts– this is electron counts– is at about 1 eV. When you actually measure it, it’s measured to be 0.968 eV. And so the way you get the electron affinity, is if you put 2.54 eV in, and you measure 0.968 coming out, then the difference is 1.572, plus or minus 4 milli electron volts. And that’s how you do it.
This is detachment to the ground state of the methoxy dot. This is one vibration in the methoxy dot. This is two vibrations, this is three vibrations. Can you see how cool this is? You can actually take a vibrational spectrum. You’re looking at the vibrations in a radical. Well, that’s not hard– that’s not easy to do. Because you don’t have a lecture bottle of radicals.
So this is a very powerful experiment on the one day of the month when it works. I’ll not kid you. These experiments of Professor Bierbaum’s, you spend most of your time with the hardware, trying to make sure the electronics are stable and what not. And then when you’re running, these people basically are dancing around the machine nude with tambourines, just trying to keep the machine stable long enough to get a spectrum.
But when it works, you’re going to get a number very clean. And you see how precise this is? I mean, this is a very nice measurement.
So we can now do what we’re going to do. If Bierbaum has the acidity, and you know the electron affinity, and you know the ionization energy of hydrogen atom, so here are these numbers. When you put these things in, you get out that the bond energy is about a 105 kcal per mole. And this is the bond energy of methanol in the innergalactic space with nobody else around.
And all of these people who like to think they can calculate everything by using all these quantum mechanical programs, now this is the number they have to get. There is no place to hide. This is the cleanly measured value.
Chapter 3. C-H BDE from Radical Equilibrium [00:12:35]
No experimental system is perfect. So for example, problems you can have. I just showed you how to measure the OH bond. Suppose you want to measure the CH bond. Well, this will never work! Because the negative-cycle will never work. Why? Because when you take any base that you apply to this, the base is always going to take off the most acidic proton. Well, the most acidic proton is one off the OH, because you get the negative on with the electron on the oxygen. That’s good. If you have a base that you try to pull a proton off the carbon, the electron is going to be on the carbon atom, and that’s bad. Oxygen’s more electronegative. It’ll hold it much tighter.
So turns out the negative ions are not going to be a useful way to do this. And of course, if you were going to apply the negative-ion cycle, then you’d have to measure the acidity of this proton, then we’d have to measure the electron affinity of this radical.
Does everybody see that this radical– this is hydroxy methyl radical– is different than methoxy? Because the dot sits on the carbon, not on the oxygen, so it’s a different radical. So if you can’t make the ion, then we can’t do our measurements, and we’re dead.
Not to worry. There’s other techniques you can do. My friend David Gutman does radical kinetics. And so a thing you can do is you can make chlorine atoms. And when you make chlorine atoms in a flow tube, he uses photoionization mass spec to monitor the products. And so he can measure k1, and he can measure k-1. So he gets to equal–
Professor J. Michael McBride: Can you explain what a flow tube is?
Professor G. Barney Ellison: A flow tube is a beautiful device that’s a meter long. It’s my fist, ten centimeters in diameter. And Bierbaum has flowing through this a tremendous flow of helium gas. And so when you make ions, the ions are made up here in the discharge, and then–
Professor J. Michael McBride: What’s the pressure in there?
Professor G. Barney Ellison: The pressure will be about a torr. And so it turns out that it’s mostly helium, and you have a few millitorr of the reagents that you’ve got. You have a quadrupole mass spec. The quadrupole mass spec, of course, can’t operate in a millitorr, so there’s some clever pumping that– you have to sample the output of the flow tube.
So the quadrupole mass spec only looks at the ions. So if you have a reaction that occurs in here, one ion would go away, and another one would come up. And you can monitor this with a quadrupole mass spec.
Remember, the way the ions get downstream is they’re carried along by the helium. And what’s true in a flow tube is, the distance equals the rate times the time. So if you have a shorter distance, you have a shorter reaction time. If you have a longer distance, you have a longer reaction time. And if you measure that variation, you can get a rate constant.
Professor J. Michael McBride: How do you change the distance?
Professor G. Professor Ellison: It turns out that you can change the distance by admitting your reagent through a tiny little tube. And this tube is the cutest little thing you ever saw in your life. It slides on a little wire. And they literally, like a gorilla, they’ll set the ejection tube at a certain space, and then what they’ll do, is they’ll systematically vary this by simply pulling this back. Then they’ll measure the ratios of these rates.
Professor J. Michael McBride: We talked earlier about Robert Hooke making a vacuum pump to lower pressure. So is that the kind, do you have somebody down in the basement pumping on this or how do you get the vacuum?
Professor G. Barney Ellison: No, it turns out–
Professor J. Michael McBride: That’s a lot of helium you’re putting through.
Professor G. Barney Ellison: Sure. OK. So Professor Bierbaum uses a set of pumps which are as big as I am. They’re Roots blowers. These pumps will pump just a staggering quantity of gas, like 500 liters a second, at this pressure. These pumps– I’m just telling you words– one of the uses of these pumps are to pump grain in grain elevators in Kansas. Have you ever been to some dreadful place like Kansas? I mean, there’s 10,000 of these silos. And so the grain is pumped around in these, with these huge pumps. So she has two of these in the basement.
So it’s a very hard experiment to do. So you can also do radicals in this. And let me just tell you Gutman’s experiment. He’s going to measure Kequi and so he’s going to measure the reaction, from the equilibrium constant. From the temperature variation of the equilibrium constant, he’s going to get the reaction energy of this. And the reaction energy of this radical reaction is, you’re going to have to pay the bond energy, this CH bond energy, and you’re going to get the bond energy of HCl back.
I now have to tell you something. The bond energy of HCl is known to be 103 kcal per mole. Remember, I told you the bond energy for methanol is 104. So the chloride atoms, if they hit the hydrogen on the OH group, if they hit it, they’ll just bounce off, because they don’t have enough energy to do this. So they’re only going to pull the hydrogens off the methyl group.
So when you do that, if you do the analysis, you can get that the CH bond on methanol is 96 kcal per mole. And on the days when it works, you can get this very precisely. So those are two of the bond energies that you can get from methanol.
Chapter 4. C-O BDE from Radical Heats of Formation; Potential Errors [00:18:34]
So you might wonder, what’s the CO bond in methanol? In other words, you’re literally going to take the methanol molecule, and you’ll karate-chop it in the center, and you just want to cut it in two. So that means I’ve got to measure the energy of the methyl radicals and the hydroxyl radicals.
So it turns out, this is a straightforward thing to do. Turns out that if you want to look at this, this bond energy, to get that heat of formation… Here, I need to get methyl radicals. So to get methyl radicals, you actually have to measure the bond energy of methane. Because that’s the bond energy of methane is the heat of formation of methyl radicals, heat of formation of hydrogen, minus that of methane.
So photoionization mass spec or reaction kinetics both measure the bond energy of methane to be 104.99 kcal per mole. The bond energy of hydrogen, H atom, comes from the dissociation energy energy of H2..
So there’s classical tables. These guys, Mr. Pedley and his friends, there’s 3,000 of these bond energies that these guys anally have put in this book. So it turns out that if you have these guys, you now use the heat of formation of methyl radical.
I also need to tell you that the bond energy of water gives you the heat of formation of methyl radical– of hydroxyl radical. So turns out if I know hydroxyl radical, and I know methyl radical, then Pedley et al will give you the heat of formation of methanol. So this is the other thing I’m going to need.
So turns out that this bond energy is equal to heat of formation of methyl radical, plus OH radical, minus that of methanol. And so this comes from water, that comes from bond energy of methane, dissociation of methane. And this is Mr. Pedley.
So that’s actually the weakest bond in the molecule. The weakest bond in the molecule is the CO bond. So you can actually work all three of these bonds out by going through a procedure. Me and a whole bunch of other really smart people have measured a large number of these molecules, and these are the bond energies. There’s all sorts of complicated heats of formation that you can use.
Let me just talk here at the end, for the last part of this, about trying to compare these things. Professor McBride is typically going on like a broken record. We always try to say, how do you know what you’ve done? Which means to say– Here. I’m not kidding you. It’s very easy to make a mistake. If you get any of these calibrations wrong, or any of these little cycles are done in an incorrect manner, it’s very easy for you to have an uncertainty.
Indeed, let me go back. I need to make one point I wasn’t smart enough to do, where I had this acidity. Here.
This bond energy that we’re going to measure—OK, we’re going to get this through a cycle. And to do this cycle, we have to do two difficult experiments. Bierbaum has to measure the acidity. It’s a very large number. That’s 380 kcal per mole. And the electron affinity is known. The ionization energy of hydrogen atoms is known. But you have to measure the electron affinity for this radical.
So this number comes from a cycle where you’re adding and subtracting large numbers. And you’re smart guys. You know this. If you have a number that you’re after, and there’s a cycle that you go through, and you’re taking differences of large numbers, any fault, any mistake that’s made here, that error is going to be in this bond energy, or any error that comes in this measurement.
There’s a famous chemist who’s English, whose name is Colson. And Colson used to say, procedure is like weighing a captain on the ship by weighing the ship with and without the captain.
Here. So you have an aircraft carrier. OK. You’re a smart guy. You get a pair of scales. And you know you have to measure this aircraft carrier, the weight of this very precisely. And then you put a little guy on top, and then the weight’s going to change just a titch. And your measurements have to be so precise you can pick this up. So on the days when it works, you can really get these bond energies. But these are not easy experiments to do.
Chapter 5. Interpreting BDEs [00:23:20]
Let me talk here about these bond energies. We’ve done several of these things. You might guess that if you have an alcohol– I did methanol, this is the OH bond in methanol– you might hope that the OH bond in ethanol would be the same. And it’s exactly the same within our uncertainty. If you go to– this is isopropyl alcohol, this is t-butyl alcohol. t-butyl alcohol is a little bit larger, but you can see, amongst friends, it’s all about 104, 105 kcal per mole.
When you go to measure something like phenol– this is an OH bond in phenol, but this is only like 86 kcal per mole. And you go, Christ, what’s wrong here?
Well, this is no regular alcohol. Here, I drew this. If you break the bond in methanol, you get a hydrogen atom, and you get methoxy dot. And the dot is stuck on the oxygen atom. But if you pull a proton off phenol, now that dot is not on the oxygen. It’s actually delocalized around. It interacts with the π system.
And McBride’s probably drawn you some of these resonance forms. Or you can talk about the interaction of this extra electron with the π cloud.
This is a catastrophic change. What this is, like 20 kcal per mole. That’s a volt. So it’s a lot easier for you to remove a hydrogen atom from phenol than it is from any alcohol.
We’ve also done a bunch of peroxides. Remember, here’s hydrogen peroxide, here’s methyl hydroperoxide, here’s ethyl hydroperoxide. These, again, look like OH bonds.
Here. You’re also smart people. You know that these peroxides, they’re very reactive compounds. The reason my daughter’s a blonde is because of this. I mean, you can buy a bottle of this, and only—Here, let’s see.
There’s a difference between chemistry and, say, other subjects, in that these molecules are real. When we go to do this… Here. My daughter would get… You could buy a bottle of peroxide in the pharmacy. It’s 5%. So it’s 5% hydrogen peroxide, and 95% water. So we have to make a beam of these things. So there’s clever ways that if you can have these compounds, and you can actually distill them, then you can make 99% hydrogen peroxide. The hydrogen peroxide, of course, is extremely reactive. Why is it reactive? It’s because the two oxygens are held together by a single bond, and you have all these electrons that are crammed right on top of each other. So the bond energy of the O-O bond is actually quite weak. So it’s a very good oxidizer.
And so turns out, I’ve actually… we had a bottle of about 99% percent of this stuff. And I’m talking about this with my students. And I’m not very skillful. And so it turns they maybe had about half a milliliter of this. As long as hydrogen peroxide is in glass or plastic, it’s fine. Nothing’s going to happen. You can handle it safely. But it will oxidize any carbon that it gets to.
So alas, I dropped it. And when I dropped it, it bounced once on the floor, and then it broke. And when it broke, when the glass came down and broke, it went, fuff! And there’s a big orange spot around. And what happened was, all the dirt and stuff on the floor was oxidized immediately! Just gone.
So if you have a solution of– Here, you think I’m kidding. You have a 100% solution of something like this, you take a little bunny rabbit with the ears and sort of lower this thing in– this thing is just simply torn to pieces. Then you pull this thing out, there’s just bones. I mean, this is tremendously reactive! So all of these things, you have to handle ‘em, and you’ve got to get them into your spectrometer.
But OK. So when you do this, you see the OH bonds– see the dots here on this? This is 88 kcal per mole. This is a lot different than these guys.
And of course, the one that you’re really interested in is water. The bond energy for water is 118. And that goes to OH dot and H atom. This is an important number. And remember, water covers 70% of the earth’s surface.
Here, you’re making a list of what are the most stable compounds that God made. It’s N2 in the air. 80% of the air we’re breathing is N2. N2 has got a triple bond. It’s indestructible. Very, very stable. The sand on beaches. I mean, the silicon dioxide, the sand, is just like a rock. I mean, it is a rock. And then the other thing is water. 70% of the earth’s surface. And the reason it’s so stable is, if you want to break this apart to radicals, you’ve got to give me 118 kcal per mole. That’s a whopping amount of energy.
Here. And so you notice, if you just replace this with an ethyl group, OK? So you make ethanol. So the bond energy’s going to drop by almost a volt to 104 kcal per mole.
Here. Imagine you have a swimming pool. So it’s a swimming pool. Johnny Depp is talking to somebody next to the swimming pool. Only instead of the swimming pool, you have water, you have ethanol. It’s vodka in there. So Johnny Depp takes a drag on a cigarette, and as he’s talking to Angelina Jolie, he flicks the thing over. And the split second this hits the pool, there’s a flame like 180 meters high. Because the ethanol ignites! It burns. If it’s in water, the cigarette goes out.
So this bond energy– if it’s a real stable compound, this has consequences.
OK. The last thing I’ll tell you is… Here. These are the bond energies I just told you for methanol. This is the OH bond. This is the CH bond. The weakest bond is the CO bond. These are these numbers in kcal per mole. It’s always interesting. If you knew what these bond energies are, what are the bond energies for the radicals? So if I reach in and cut a bond here and I make methoxy dot. If I make methoxy dot, now the bond energy for the CH bond—Here, I can use the meter stick– the CH bond, which used to be 96 kcal per mole, drops to 22.
You can see why. The minute you cut the bond here, the hydrogen atom leaves, but this dot couples with that dot, and you’ve got formaldehyde. Nice, stable compound. And the CO bond here is about 90 kcal per mole. This is about the same thing it is for the CO bond in methanol.
You go to the other bond, which is hydroxy methyl. You go to this species. Again, if you cut the bond off the OH bond, you get a dot here, and the other dot then combines back here with formaldehyde. So now this is 30 kcal per mole, to be contrasted with 106.
What’s interesting is the CO bond, which used to be 92, goes up. I don’t know why, but it does. And not by a little bit. This is almost a volt. So there’s all sorts of interesting patterns you can have in these bond energies. And you literally can have… you can look at that not only these alcohols, but all sorts of alkyl peroxides and a variety of these things.
OK. Here. I’m going to finish early, you can see.
Professor J. Michael McBride: There’ll be questions.
Professor G. Barney Ellison: So here there’s another thing I want to show you. And that is, when you talk about what a bond strength is, it turns out that an early way that people tried to measure bond strengths would be that you’d consider a molecule like methane– I need my big stick here– and you take methane, you can burn it, and you can actually measure the energy to take this apart to a carbon atom and four hydrogens.
So Chupka described to you how you would measure the energies of individual carbon atoms. You can actually work out that the bond, to break all four bonds in methane is going to cost you 497.5 kcal per mole. Since there’s four bonds in methane, you’d cross yourself, think of Jesus, and then just divide by four. And if you did that, that number divided by four is 99 kcal per mole. And so you figure, that must be sort of the average bond energy.
If you go through these cycles, like I just told you, you can actually measure individually. The first bond is 104.99 plus or minus 0.03 kcal per mole. And that’s the first bond energy. That gives you the absolute heat of formation of a methyl radical. You can now take methyl radical and measure the second bond. Notice the bond energy goes up from 105 to 110. Then the third one is 101, and the last one is 90 kcal per mole. And these guys give you, individually, the heats of formation of all of these species.
Notice that not a single one of these bond energies equals that average. Here, it’s close, but the uncertainties are such that it’s not quite. And so using these average bond energies, you’ve got to be careful. The other thing I’ll also tell you, if you actually take the sum of these numbers and add them up, it’s 397.5. It’s exactly this. And this is a testament to the guys who did these measurements. These are very hard experiments to do, and that means that each one of these four numbers is right. Because you know, this thing has to be right. The four of these guys have to add to these things.
So here. If the sum of these things is this, that means the first law of thermodynamics really works. So this is a very interesting set of experiments to do.
OK. So here. I’m going to give you people–
Professor J. Michael McBride: Let’s have some questions.
Professor G. Barney Ellison: OK. So I’m allowed to show my last little picture, then. Here. I’m in Boulder, Colorado, and we’re not as sophisticated as you people here are in New Haven.
So this is me. Carl Lineberger is the captain of the negative ions. The smartest guy in the world about doing negative-ion spectroscopy. Chuck DePuy here is a Yale PhD, is a member of the National Academy of Sciences. He’s an organic chemist in Boulder. Eldon Ferguson is the inventor of the flowing afterglow. It was at NOAA. The National Oceanic and Atmospheric Administration in Boulder.
This is Steve Leone, who’s one of the most famous chemical physicists in the United States. He’s now at Berkeley. He’s the head of the Chemical Dynamics Beamline out there. Zdenek Herman was a postdoc here in the 1960s, and he’s at the Academy of Sciences in Prague. And this is Roni Bierbaum, who’s the master of negative-ion chemistry.
Chapter 6. Questions: Hot Bands and Resonance Stabilization [00:35:43]
So I’ve given you about ten minutes early, so I’m happy to answer any questions that you people may have.
Professor J. Michael McBride: OK. Do you have any questions or comments? OK, I have one.
So let’s go back to your– how do we get out of here?
Professor G. Barney Ellison: Oh, you hit a B?
Professor J. Michael McBride: But I want to go back to the beginning. Let’s, for what it’s worth– here we go.
That, hold up that spectrum you showed that had vibrational things in here.
Professor G. Barney Ellison: That’s right. Yes, there it is.
Professor J. Michael McBride: Now, do people understand how this was working? You put the laser energy in, and you’re measuring this difference, right?
Professor G. Barney Ellison: That’s right.
Professor J. Michael McBride: But sometimes some of the energy went into vibration of the molecule. So the energy that came out was smaller, to the extent that vibration actually went into the molecule. So here, you say, is when no vibrational energy goes into the product. So that’s how you know what the lowest energy is.
But here’s a peak here.
Professor G. Barney Ellison: Right. So that comes from the fact that you have a vibrationally excited negative ion. The negative ion, instead of being in its ground vibrational state, has a little bit of energy. And it will actually be in the CO stretch, is where this will be.
So if there’s energy in the negative ion, then, to go back to my diagram– if you have hot bands, in other words, if some of the molecules have vibrational energy in this, then the molecules aren’t down here, v equal to zero, but they’re actually populated in higher states here. So that means that those electrons will take less kinetic energy to get off.
And you can always spot them, because they’ll have a different frequency. Here. You’re going to have a set of frequencies here, which will be a mode in the neutral molecule. This is the methoxy radical, and this is that of the negative ion. And so this is a vibration in the negative ion. And they’ll be different, because the electrons are different in these things.
So you can guess, when you get to a molecule which gets to be more complex, like phenoxy radical– I talked to you about that– phenoxy radical has a lot more atoms. So analyzing this steadily gets more and more difficult. But we like to do that. That’s what God put us on this earth for.
You can also guess– I’m interested right now in sugars, because I’m interested in how biomass decomposes. You know what glucose looks like? Glucose is a molecule that has many different hydroxyls on it, and hydroxyls are all slightly different.
So now, instead of making a negative ion like in methoxy– I mean, how hard can this be? You’ve only got one OH. What happens if you add ethylene glycol, or if you take glycerol– ethylene glycol has two hydroxyls, glycerol has three, threose is going to have four, ribose is going to have five. So now you’re going to have to begin to tell which OH this came from.
So this is going to be a bitch. But you if you can do this, you would be able to take these molecules, and you’ll be able to break all the bonds in the order that you’d like to do it. And that’s very interesting to do.
Professor J. Michael McBride: So you were very kind to speak about putting electrons in the σ* orbitals. You don’t believe in σ* orbitals. Professor Ellison is not a big fan of molecular orbitals. I think. Is that true?
Professor G. Barney Ellison: Yes. Yes, that is true.
Look, I do valence bond work. This is a matter of taste. There’s different ways to view these things. And in the end, the only thing that counts is what you measure. Here, if you have a model that can account for the patterns that you’ve measured, and you can predict what the next values are going to be, you understand this. But it’s the data is what survives. The rest of it is all a matter of opinion. He likes σ* orbitals, I don’t.
Professor J. Michael McBride: You were talking about the stability of the phenoxy radical, in terms of its resonant stabilization. So I think these people could help you out with that. Or at least point out something interesting. Would you say that this radical is stable because you can draw resonance structures that have the electrons over here.
Professor G. Barney Ellison: Yes, that’s right. Correct.
Professor J. Michael McBride: But how about if you say that in the starting phenol, that’s the bond that you’re going to break in here, right? Here I’ve got a pair of electrons that I can draw resonance structures for. So it looks to me like it’s worth more here than it is here.
Professor G. Barney Ellison: Listen, dream on, big boy. Look—Here. How do I push this thing up?
Professor J. Michael McBride: OK, we’ll get that.
Professor G. Barney Ellison: Here.
So this electron is going to get delocalized over the whole molecule. If you try to do this with an electron pair, what you’re going to have is– OK. So if I do that, I’m pushing charges around. Well, I don’t like that. I don’t like the fact that the oxygen is going to have– you’re starting to suck electrons– if the electrons migrate down, the oxygen is going to become positively charged. So jeez, I just would never do that.
But here. But what’s true is that bond energy, honest to God, really and truly, is 86 kcal per mole. That is a fact. And this is a flippant way to have anticipated this.
Contrast that with, say, methoxy dot– where’s that electron going to go? It’s really stapled to the oxygen atom. There’s nothing it can do. It can’t go anywhere. So when you break that bond, you get hydrogen atom. You get the dot that’s just stuck right there. As you pull this bond off, the dot now is able to be delocalized and spread around the whole ring, and this bond energy, then, in phenol, this bond energy that’s going to be 86 kcal per mole– this thing lets you measure the absolute heat of formation, 298, of this thing. This low bond energy is going to mean that this heat of formation is going to be low.
So what that means is, if we’re trying to study lignin, and lignin are trees– you take a chainsaw and you’re cutting trees down, that’s very tough material. All these compounds are long, three-dimensional polymers of aryl alkyl ethers.
So if you know the heat of formation of the phenoxy radical, this bond turns out to be 62 kcal per mole. And if you do this in this, it’s like 90. And that’s a volt and a half.
So when you heat this, all these molecules break apart, and the first thing they do is they form phenoxy radicals. And this triggers the decomposition of these things.
So it turns out, I don’t like these things in the neutral molecule being delocalized. Because I think you have to separate charge, and I don’t like that.
Professor J. Michael McBride: There are actually people back there who are smart at this kind of stuff. Do you guys buy that?
Professor G. Barney Ellison: You.
Student: What I’m thinking is, if you have that kind of species, you’d think there’d be a spin observable.
Professor G. Barney Ellison: Oh, indeed there is. If you look at the EPR spectrum of this radical, it is delocalized all over the place. And here, you’re an expert in this. You know about all these hyperfine couplings and whatnot.
Professor J. Michael McBride: Yes, there’s coupling from the protons around here. It’s like NMR. So if the electron spin gets out on these protons, then it interacts with the nucleus, even if the molecule’s tumbling, you don’t have to worry about the anisotropy. So it’s no doubt it gets delocalized, but why doesn’t this one? So Barney says it’s because he doesn’t like to separate charge in resonance structures. But that’s not– is this Coulomb’s law, or what? Anybody got an idea back there?
Professor G. Barney Ellison: Well, here. What do you mean, is this Coulomb’s law? Coulomb’s law is what it is. So here. If resonance forms like this are important, then that means that I think, as a consequence of this, you have to have charge separation. And I think these are high-energy resonance forms. And if they’re high-energy resonance forms, they don’t contribute to the superposition in the wave function, is the way I say that.
Professor J. Michael McBride: That’s because you believe in resonance.
Professor G. Barney Ellison: Because I’m a valence-bond guy. But what’s true is this 86 kcal per mole. Nothing can be done about that. OK.
Professor J. Michael McBride: OK. I think we can thank Professor Ellison.
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