Freshman Organic Chemistry II

CHEM 125b - Lecture 20 - Electronic and Vibrational Spectroscopy

Chapter 1. Electronic Spectroscopy: Atomic Absorption and Time Dependence [00:00:00]

Professor McBride: OK, so we’re ready for a new phase here, to talk about spectroscopy, the interaction of light with matter and what it can tell us about molecules. We’ll talk today about electronic and infrared spectroscopy, and in particular, with infrared spectroscopy, about normal modes and the mixing and independence of vibrations.

So the purposes we use spectroscopy for are to determine structure and to study dynamics. So we use electronic spectroscopy that’s visible and UV, where it’s motion of electrons that’s being excited by the light. Or vibrational, where it’s motion of the nuclei, the atoms, being excited by the light. Or NMR, where it’s a magnetic interaction with light that’s involved. 

Spectroscopy comes from the word specter. In 1605 it was used, according to the Oxford English Dictionary to be “straunge sights, visions and apparitions,” so that’s like the specters of ghosts, right? Or Newton used it in the 1670s to be “Sunbeams…passing through a glass prism…exhibited there a spectrum of divers colors,” so this kind of ghost. So that’s where the word comes from.

Now Atom in a Box, which we used last semester and is shown over on this screen, can be used to show a number of things. It can show spectral transitions for the H atom. Let’s get this here. And if you click “spectrum” here, you get– where is it now? I think it’s off the screen. I enlarged things here. But anyhow, it gives you what the spectral transitions are, what energies are involved in going from one– oh, I think it’s because I didn’t get “simplified” at work here. What’s going on? Why isn’t it simplifying? My guess is because I blew this up to be on the screen. It’s not showing. Anyhow, we’ve done that before and seen that you can see the energy levels of different things, different atomic levels and the wavelength that’s involved.

We can see the static shift in electron density, from mixing 2s with 2p orbitals. Maybe I’m going to have to bring this back down again, and see if it works now. I don’t know why it’s not working. Well, fortunately, the most important stuff I’m going to have is (A) displaying there, and (B) also on the PowerPoint, so you can do this on your own time.

OK, you can see the shift, as we saw last semester, from mixing 2s with 2p. Remember, that’ll shift the electrons from one side to the other of the nucleus, depending on the hybridization. We can see the oscillation of electron density that comes from mixing orbitals with different energy, and that’s what’s involved here. You’ll notice over here, we’ve got a mixture of the 1s with a 2p orbital. What we showed before was fixed in time, and it would shift it up or down. But if you look at it as time goes on, you see that it actually goes back and forth between being down and being up, being down and being up.

So the electrons, when you mix 1s and 2p, oscillate up and down in time. And that’s because of a change in relative phase with time. You add s with p, and then subtract s with p, and it goes one way and then the other. And I’m going to illustrate that. This is a feature of time-dependent quantum mechanics, where the phase of the wave function changes at a rate proportional to its energy. And I’ll go through this just a little bit on the next slide. We’re not going to spend too much time on it, because it’s not really the business of this class. But it’s so neat that I thought you’d like to know about it.

But anyhow, when you have the electron shifting up and down like that, it’s like an antenna. It gives off light. In fact, it can absorb light, too. Antennas can either transmit or receive. So that’s what allows electrons in atoms to interact with light as an oscillating dipole. You can also do breathing by mixing 1s with 2s, but that doesn’t interact with light, because it doesn’t generate a dipole. And I’m going to illustrate that.

So here’s 1s, mixing 1s and 2p. So as we see here, if 2p is both up and down, but with opposite phases, 1s is a single phase, a single sign. If you add the two, it reinforces above. OK, and we’ve seen that before. But if you make it time dependent– that’s what’s being shown over there– then what happens in time, is that it goes up and down. And the oscillation frequency, how rapidly it goes up and down, that is what color of light would interact with it, what frequency of light, has to do with the energy difference between 1s and 2p.

So here, this is just a footnote. You’re not deeply responsible for this, but I wanted you to see it. So I’ll just read it to you. A time-dependent wave function looks just like the ψ’s we’ve been talking about, functions of position. Except that it’s also got something to do with time, obviously. It’s multiplied by e^iωwt. And you know e^ix is cos x + i sin x. Have you done that? I think you may have seen that.

So you have a cosine and a sine of ωt, one is real and one is imaginary. It’s a complex number that you’re multiplying the wave function by. So i is √(–1). ω is the energy in frequency units. That’s how rapidly things are changing in time, because it’s cos ωt. So as time increases, the cosine goes up between plus and minus one. And how fast it goes up and down depends on how big ω is, because ωt is what you’re taking the cosine of. So the higher frequency ω is, the faster it’ll go up and down, the higher energy ω is.

And now that’s an interesting question, because it depends on what– the magnitude of the energy for any given system– depends on where you define 0 to be. If you to define 0 to be separated particles, that’s one energy. If you define it to be the ground state, that’s a different energy. And this would suggest that the rate of oscillation depends on what you call 0. That seems nuts, right? It can’t depend on you, what the rate of oscillation is. And the reason is, you don’t care what the rate of oscillation is. And I’ll show you why. 

Because when you square the wave function to get the probability density, at any given time, you multiply ψ times e^iωwt, and ψ times e^-iωwt. That’s how you do squaring with complex numbers, multiply by the complex conjugate. And that gives ψ². All the stuff about time cancels out. So you don’t really care. So why do you have it in there in the first place? If you’re mixing two different states that have different energy, then they oscillate at different frequencies.

So really, you don’t care how fast either one is oscillating alone. You only care what the difference in frequencies is. So now it doesn’t depend on what you call 0, because the one state you’re taking and the other state you’re taking have the same difference, whatever 0 is. That’s where you make a difference. So it says, when a problem involves actually mixing two states of different energy, one considers a wave function of the form, e^iωw1t, plus a different psi, the other wave function, times e^iωw2t.

Now, if ω₁ and ω₂ are different, this means that the two spatial functions cycle in and out of phase with one another. If at a certain time they add at a time 1/2 divided by the difference in frequency later, they subtract. So 1s + 2p_z will become 1s – 2p_z. So if this is the energy here, the oscillation frequency of 2p_z, then 1s will have a lower oscillation frequency, by that energy difference between them. And it’ll look like that. So here, they’re in phase with one another, the cosine part, they’re adding. But in the middle there, they’re out of phase, and it’s subtracting. And then it comes back into phase again at the far right.

So as time goes along, it comes s + p, s – p, s + p, s – p, and the electrons go up and down. So if you mix these two orbitals– so an electric field that’s oscillating up and then down and then up and then down favors this at the first time, then plus, then minus, then plus, then minus. So the light interacts with the electrons and pushes them up and down, mixing 1s with 2p. Now, if you do that for a while, and then turn the light off, it may end up being in the s + p, when it started in the s – p, or vice versa. So you can get an absorption, or indeed, an emission of light, the interaction with light and things, because of this time dependence.

But again this is not really our business, and you’ll see that later. So this is the source of the oscillation we observe, when superimposing functions of different n using Atom in a Box, different energies. So that thing going up and down over there. And note, just as a footnote to a footnote, this is different from what we were doing when we made hybrids. We mixed s with p before. Usually we were mixing 2s with 2p, which have very similar energies, right? But in those cases, we weren’t trying to mix two real states, that’s why I said “actually” up there.

There we were just trying to get in a reasonable form for a wave function of some particular energy. But when light is involved, you actually mix states of different energies, so it’s a different situation. So we’ve got this oscillation up and down, which interacts with light. It can generate light. It can absorb light. Now, the 1s to 2p transition is said to be allowed, because it happens with light. But there are other mixtures that don’t generate light. For example, if you mix 1s with 2s. If you add the two, it reinforces where they’re blue, and subtracts when they’re red. That makes a big circle, a big sphere. But if you subtract them, make the one on the left red and add it to the other one, then they’ll reinforce in the middle and cancel on the outside, and it’ll be inside. 

So if you mix those two, you get this dependence with time. I think it’s going to do it. Whoops, I didn’t wait long enough. But what happens, and you can do it yourself, it goes [makes expanding and contracting sounds]. It’s breathing. But that’s not separating the average position of the nucleus and the electron. It’s not going up and down to make a vector, a light. It’s just expanding and contracting. 

So that “symmetrical breathing” electron-density deformation has no oscillator strength. It doesn’t interact with light’s magnetic field. So that transition is said to be “forbidden.” You can’t cause it to occur with light.

Chapter 2. Organic Chromophores [00:12:58]

Now, what does this have to do with organic molecules and the things we were just talking about, which were polyenes. You remember a bunch of bunch of double bonds in a row. Well, suppose you have C=X double bond, and X is some heteroatom, like oxygen or nitrogen, it’s something that has an unshared pair of electrons. So it has n electrons. But it also has a π* orbital, a vacant orbital.

Now, can you mix those two? What would happen if you mixed those two orbitals? You could add them together, and they’d reinforce in the top-right and cancel on the bottom-right. And you get the electrons to move up. And if you changed the sign, they’d move down, if you subtracted one from the other. So here’s a way that light gets a handle on the electrons that are an unshared pair, and can make them go into π*, by mixing the σ, unshared-pair electrons, with π*.

Now, that’s neat, because as it goes up and down you absorb light, and the large energy gap between the unshared pair and π* means that this transition occurs at very high frequency. Because remember, the frequency of cycling in and out of phase is the difference in the energies of the two things. So if they’re fairly far apart, it’ll be a very high frequency of light, not in the visible. So it’s in the ultraviolet.

How could you make those two orbitals closer in energy? The unshared pair you couldn’t do much about, except maybe changing one atom what is involved, oxygen, nitrogen, sulfur, things like that. How could you change the energy of the LUMO in the π system? Anybody got an idea? So if you put more double bonds conjugated with it– remember, the LUMO comes down, the HOMO comes up.

So you can bring it closer to the energy of the unshared pair. If we put more and more double bonds in there, then the LUMO is going to look like this. It approaches the energy of an isolated p orbital, so the light will get redder and redder that interacts. And this makes a lot of differences as the π* approaches the energy of the 2p orbital.

So for example, this molecule, which has an unshared pair on nitrogen here, which could mix with π* of this really long conjugated system. That’s the imine of retinaldehyde. That’s what’s in rhodopsin. That’s the light sensitive stuff in your eye. So when light comes in, it excites this transition, and it triggers then to your brain and so on, because the cis-isomer changes to the trans-isomer at this double bond. And that’s how you see.

Incidentally– that compound, I just put this in as another footnote; let’s not spend time on it– but this is the actual report of Lindlar’s catalyst. Remember, the poisoned catalyst that’ll reduce triple bonds to double bonds, but won’t reduce double bonds. And he said it was in the context of vitamin A synthesis. “During work on the synthesis of vitamin A, a palladium-lead catalyst was developed, which one could hydrogenate a triple bond without attacking double bonds already present.” 

But the things he was working on we’re things like this. Really long conjugated things, like b-carotenyne, which he made, which had a triple bond in the middle, as you see. And then he could react it with his special catalyst and get the cis addition of two hydrogens to give the β-carotene. And he says the “hydrogens are added in cis arrangement” to one another. But then he goes on to say that he could, with heat and light he was able quantitatively to convert the cis isomer into the trans isomer, which is what he wanted to have for β-carotene.

OK, now notice, this was C₄₀H₅₆, this big long thing. And you know why it’s that kind of number, because it’s made up of these isoprene units that we talked about in the last quarter, polymerization of isoprene to give these natural products. OK, that’s just a footnote to last semester, it’s made from isopentenyl pyrophosphate. Now, retinal is the stuff that’s involved in making this visual pigment. And β-carotene is that, and obviously, it’s called β-carotene because that’s what makes carrots that way. And that’s why you eat carrots to make your eyes good, right? Bugs Bunny has good vision because of that. 

Now, this related compound is called isozeaxanthin. And it turns out to be the stuff that’s in the feathers of the scarlet tanager. Now, what’s weird about this, it’s what makes the scarlet tanager yellow-green, right? What’s wrong with that? It’s supposed to be a scarlet tanager. But this is actually late in the fall. If you looked a little bit earlier in the fall, the scarlet tanager looks like this. So it’s got a little bit of red in it.

But notice that this OH out on the right, which has an unshared pair that could go into the π system, that oxygen is not part of the π system. You can’t generate any overlap by that, because the oxygen is not where the π system is. It’s not conjugated. The OH unshared pair is isolated, so you can’t do this thing of taking an unshared pair and putting it into the LUMO, the π LUMO. But if you oxidize it to make that, then you get it conjugated, and that’s the scarlet tanager in the summer. So it’s exactly this changing of unshared pairs into the π* orbital with a really long conjugated chain that gives this most vivid of our birds’ colors. And these come in East Rock Park about a month or month and a half from now, so you can go see them yourselves. They knock your eyes out.

Chapter 3. Infrared Spectra, Hooke’s Law, and Vibrational Frequency [00:19:38]

OK, so we’re talking about spectroscopy. A spectrum is usually presented as a graph. And when you see a graph you wonder what the axes are. That’s the first thing. And depending on what point of view you’re taking about spectroscopy, how you’re understanding it, there are different ways of labeling the graph. For example the horizontal and the vertical coordinates here in this particular graph are called wave number and transmittance, but the meaning of the axes, with respect to an experiment, is color, which light you’re talking about along the horizontal axis, and vertically, what the light intensity is.

For example, how much light of a certain color gets absorbed by your sample. Or on the other hand, how much light doesn’t get absorbed and gets through the sample. And in fact, that’s what’s plotted here is transmittance, how much light gets through rather than how much gets absorbed.

OK, so that’s one way of looking at it, just experimentally. But another way is to look at it in terms of quantum mechanics. The horizontal axis is the molecular-energy gap. What are the orbitals that are involved? How far apart are they in energy? What does that have to do with the frequency of light? You remember, the rate of this oscillation back and forth has to do with the energy difference between the two states you’re talking about.

And if you’re talking about quantum mechanics, the vertical axis is how much overlap you generate. So when that oxygen had the π system on it, in the scarlet tanager, then you generated a lot of overlap by mixing the n with the π. But if they were far apart and the light couldn’t induce overlap, then you didn’t get the fancy color. You got just the yellow-green color.

Or you can look at it in terms of classical mechanics, which is the way we usually talk about infrared spectroscopy, where the horizontal axis is some actual vibration frequency of the atoms in a molecule. And the vertical axis is how much “handle” the light has on that particular vibration. For example, if you’re vibrating H₂, you don’t change any electronics by doing that. You don’t generate a dipole. It doesn’t interact with light. But if you vibrate HCl, then you get a big separation of positive and negative charge as you do that, and it does interact with light. Light has a good handle on it.

So it’s molecular vibration frequency though– of course, a molecule of H₂ or HCl is just two atoms. But normally, the atoms whose bond you’re interested in vibrating is part of a whole molecule, and then things get more complicated, as we will see. So infrared spectroscopy uses light for a number of purposes. One is to fingerprint molecules. And I think you’ve done this in lab, to see whether your spectrum corresponds with a known spectrum. Or even if you don’t have an authentic sample, you may have a table of where functional groups, even in different molecules, if it’s the same functional group there maybe some line that’s characteristic. So you can identify functional groups and molecules that are otherwise unknown.

And you can also use it to use the molecular dynamics that’s involved to study bonding, to see how strong bonds are, whether atoms are linked by springs, for example. And that’s what I’m going to concentrate on in the lecture, to use it to understand molecules. You you’ve used it in lab to identify molecules, to identify functional groups, but we’re going to use this last thing to see what it tells us about how molecules hold together.

Now, just a quick excursion into physics or math, to see what makes vibrations sinusoidal. When things vibrate we often think of them following a sine wave. But what is the condition under which they do actually vibrate according to a sine wave in time?

So if it’s vibrating according to a sine wave, then the displacement is some height times the sin ωt. That’s how fast it’s going up and down. That describes the displacement. So that’s the frequency, and t is the time, and h is the 1/2 amplitude. It can go from +1 to –1 as the sine goes back and forth. OK, and that says the velocity, the change in position with time, is the derivative of that, so the cosine. And the acceleration is the second derivative, so it’s that. And from that then, if the motion is sinusoidal, as we said at the beginning, then the acceleration divided by the displacement is minus the frequency squared.

Now, Newton said that acceleration is force over mass. And if things are obeying Hooke’s Law, then the force is proportional to the displacement. And that means we can substitute in Newton with the force, –fx, and we get that a/x is –f/m. And a/x, we already knew, is –ω². And that means that the frequency squared is how strong the spring is, if it’s Hooke’s law, the force constant divided by the mass. Or, if we take the square root, the frequency is the square root of the force constant divided by the mass.

So the frequency then is interesting. It’s constant. It doesn’t depend on the amplitude of the vibration. If you have a system that’s obeying Hooke’s law, it’ll have the same frequency no matter what the amplitude. And that turned out to be very important. Incidentally, one of the texts we’re involved in says the opposite, but it’s not true. OK, so why is that so important? Hooke’s law. Hooke’s book, Of Spring in 1678, says vibrations, if you’re obeying Hooke’s law– this is where he set out Hooke’s law—“shall be of equal duration whether they be greater or less.” Why did he care that the frequency be always the same no matter how big the displacement is? Can you see why that would make a difference, practically? Ellen? 

Student: For clocks. 

Professor McBride: For clocks. Or more particularly, for watches, because you know, about 90 years or 80 years later, John Harrison made this marine chronometer, which allowed to solve the problem of longitude. So ships could tell how far they were from Greenwich by having accurate time. And if you look inside this watch, it looks like that. There’s a spring vibrating like that. And it’s really, really clever.

I’ll show you one other thing about it. Notice here, there’s a little rod that goes out here under the spring. It’s right here. And it goes out until it makes a little fork there around the spring. Now, that little leaf is bimetal, so it’ll bend with temperature. So it changes the length of the spring as temperature changes to correct for the temperature changing the stiffness of the spring. Isn’t that clever? OK, that’s obviously beside the point, but it’s really a neat feature of Harrison’s H4 chronometer.

So anyhow, when Hooke’s law applies, the frequency is proportional to the square root of the force constant divided by the mass. So m is the mass. And this means you could use this thing to measure the frequency. You could measure a mass that way. And that’s what a quartz crystal microbalance does. You can have a quartz crystal that will vibrate. I mean, the quartz in my Timex watch does that, right? It vibrates as a given– it keeps very good time. So the quartz crystal microbalance, if you change its mass by absorbing something on top of the quartz crystal, it’ll change its frequency, and you can measure it.

In fact, it’s so sensitive, that if you coat it with platinum, and then let ethylene in to absorb on the platinum, it’ll get heavier and the frequency will slow down. But if you then admit hydrogen, the platinum catalyzes the reaction and the ethane goes off, and you can measure that change of one monolayer of ethylene on platinum by using a quartz crystal microbalance. That’s just a neat feature.

So if we’re talking not about what’s going on the surface of a quartz crystal, but about molecules vibrating, then, for atoms, the force constant should be how stiff the bond is. A single bond should be easier to stretch than a double bond, than a triple bond. But are they really that? Is the force constant really proportional to 1, 2, 3? Is a double bond twice as strong as a single bond, for this purpose of how rapidly things vibrate? Or a triple bond three times as strong? So this is something we can check by using infrared spectroscopy.

Now, with respect to the mass, things get more complicated, because it’s not just one atom that’s moving. Usually it would be at least two atoms that are moving. And then what do you use for the mass? It turns out, that in a diatomic, what you use is called the reduced mass. It’s the product of the two masses divided by their sum. And we don’t need to go into why that is. But what that means is that that effective mass– use the Greek µ, the Greek m, µ instead of m– that’s dominated by the smaller mass. Because suppose m₂ is really small. Then the denominator, for practical purposes, is m₁. m₂ has nothing to do with it. But the m₁ in the denominator then cancels out the m₁ in the numerator, and all you have is m₂. So it’s the smaller thing that counts. You’d usually think it would be the larger thing.

OK, so let’s look at what that would be. For example, for hydrogen and carbon that we’re vibrating, it would be 1 × 12 / (1 + 12). Almost 1, right? The 12’s almost cancel. If it’s carbon-carbon then it’s going to be 1/2 of carbon, 6. If it’s carbon-oxygen it’s going to be 6.9. If it’s carbon-chlorine it’s 8.9. So even when it gets much heavier, it doesn’t make much difference. It’s mostly pretty much like other carbon bonds.

So what that means is that bonds involving hydrogen have really, really different frequencies from those involving not hydrogen, because it’s so much lighter. 1 to 6 here, or more. So it stands apart, the mass of hydrogen.

Now, how about the bonds then? So C-H then, is proportional to the square root of a single bond over 0.9, that’s reduced mass. And that turns out to be around 3000 wave numbers, or 10¹⁴ Hz. So that’s how fast C-H bonds vibrate.

A C-O bond has a heavier mass, 6.9 versus 0.9. If it’s a single bond, that’s going to be about 1100 wave numbers, or 3 × 10¹³ Hz. And that, remember, is what you put in front in Eyring to get a rate, it’s 10¹³ times that equilibrium constant to the transition state. That’s because of this. That’s how fast atoms like carbon and oxygen vibrate. OK, so a C double bond, though, if a double bond is twice as high as a single bond, should be 2, the square root of 2/6.9, and that’s about 1500. And that’s about right. And a triple bond is higher still, with a 3 for the force constant, about 1900.

So you can see that if you have a spectrum, where 3000 is on the left, and 1000 is toward the right, then you’re going to have hydrogens up here, hydrogens stretching about 3000, and then you’re going to get triple bonds, double bonds, single bonds involving other atoms, not hydrogen.

Chapter 4. Why IR is Complicated: Coupled Oscillators and Normal Modes [00:33:09]

But it’s not that simple, because a single atom is connected, not to just one other atom, but others, and they couple with each other. So this, our study of coupled oscillators is going to illustrate several things. The complexity of what’s really involved in vibration, and why physical chemists thought at the beginning that it was going to be impossible to interpret anything in IR spectroscopy. And then the idea of normal mode analysis, and what a normal mode is, and phase of mixing, and the possibility of things being independent, so you can think about a functional group having a particular thing, not being all mixed up with others.

So for this purpose we’re going to look at some oscillators, at this particular thing here. Let’s see if I can get this, get the lights I need here. OK, so here are some things that’ll oscillate. So if I take this, it vibrates at a certain frequency. It’s a spring that’s being bent back and forth, so it’s Hooke’s law. And that’s what I’ve shown here. It’s a simple oscillator. I bend it. I let it go. It flops back and forth at a given rate. And we know what that rate’s going to be. The square of the frequency is going to be whatever the force constant– how stiff the spring is– divided by the mass of that thing up on top, and of the spring itself, of course. OK, that’s fine.

Now, suppose we put a spring in there as well. But let me hold this one fixed, or hold this one fixed. And I’m going to do this then. Now, can you tell me whether that vibration should be faster or slower than it was when I didn’t have the spring there? What’s the difference? The difference is that now we have a stronger spring. Everybody see that? Because we’ve added this spring to however hard it is to bend that. So it’s going to be a higher frequency. And we could time it, and figure it out. We don’t have time to do that, but it’s faster.

If it’s coupled to a frozen partner like that– I’m holding the partner– then I let it go and it stretches back and forth, and it has a stronger spring. It’s the main spring plus this little one, plus s. We could couple that in, so that’s easy enough. Now we’re going to look at it when I’m not holding it. It’s going to be in-phase coupling. So they’re going to be moving the same direction. So I move them both out here, and I let them go, and they go back and forth. Now, what should that frequency be? Is this spring being stretched? No, so it’s just as it was originally. So if we do this and do that, and let go, then it goes back and forth, not stretching the spring. Or if you look at the thing as a whole, there are two weights and two springs being bent, so it’s 2f/2m, but it’s the same as we started with. 

Now, the interesting one is to do out-of-phase coupling, where they don’t go in parallel but antiparallel to one another. Now, what about this situation here now? Is that spring being stretched? In fact, it’s being stretched twice as much as it was before, because the middle is staying in place, and both things are stretching it. So if we do that, if we do this one, out-of-phase coupling– and notice that the center stays fixed– now as it does that, we have a much stronger spring. It’s 2s in the middle now. I’m sorry you can’t see this very well, but you can look at what we review. So it’s 2s in the middle, so now frequency is higher.

Now, so when the two were isolated from one another, that is, when we held one fixed and just looked at one vibrate, all we had was that little extra spring in the middle, then they would both vibrate at the same frequency, isolated. Can you see it there? I think you can. Actually, let me just put it on the other screen and then you’ll be able to see it. Yeah, OK, good. But once we let them both move and they couple their motion, then there’s an out-of-phase and an in-phase. One is higher frequency. One doesn’t stretch the spring, the other stretches it twice as much. Does this remind you of anything you’ve seen before? That diagram? It’s the same thing as orbitals mixing, to give an in-phase and an out-of-phase. The math is just the same. 

So this is called, either of these, the in-phase or the out-of-phase, is called a normal mode. The meaning of it is that, in that pattern of vibration, all the atoms are oscillating at the same frequency in a normal mode. It’s a lower frequency here and a higher frequency here, but both atoms are moving in phase with one another, you know, at the same oscillation frequency. Now we’re going to do another one, in which– so they move back and forth at the same amplitude, the same frequency, or when they started the same. So that’s out-of-phase and this one was in-phase. 

Now, notice that if I added those two together, those two displacements, those two vibration patterns, I would get something where one moved and the other didn’t. So I could do that. And what I’m saying is, that that is the sum of the in-phase motion and the out-of-phase motion. So if I sum them, I get this one to move and these cancel. So when I let go now, I’m actually exciting two normal modes, the in-phase and the out-of-phase. Does everybody see what I’m doing? Because what I’ve done here is the sum of the two of them. So now when I let go, something neat happens.

Students:  That is cool.

Professor McBride: So what’s happening?

Students: [murmurs] 

Professor McBride: It’s going back and forth between one vibrating and the other vibrating. Why? Because those two modes have different frequencies, and they get in step and then out of step with one another, and when they’re in step with one another, this one is moving, that’s how we started it, and not this one, and when it’s one minus the other it’s this. And because they have different frequencies, they’re cycling in and out of phase with one another. I think that’s just so wonderful. [LAUGHTER]

So that’s exactly what we were looking at at the beginning, where things went in and out of phase with one another, where it was 1s going to 2p. What we started, when we moved just one of them and held the other one fixed, was a superposition of two normal modes of different frequency. And then the vibration “beats” in and out of phase. So now, what happens, if one of them’s heavy and the other one’s– so then you can’t talk about just one of them vibrating, when they’re coupled this way– but suppose I made one of them heavier than the other, so it vibrates normally at a very different frequency. If I can get this on there…

And now let me move this one. Let me move this one and not this one. And what should happen– what we saw before was the vibration goes from there, to there, to there, to there– but now look what happens. I actually moved that one a little bit at the beginning. This one’s not moving nearly as much as that one is. This one keeps moving. Remember, the other one went dead still and then the other one and then the other. Now it’s staying localized mostly. And if I did this one, it stays mostly here. Very little comes over there. So if the two things have very different frequencies, then they don’t couple very much. They stay mostly independent, if they have different independent… different individual frequencies. 

And that’s exactly like the wave functions mixing. If we have one that’s low energy and one that’s high frequency, and we couple them, you’ll get one that’s in phase and one that’s out of phase, but they won’t be very different from one another. It’s an exact analogue of the energy-match and overlap thing, where the energy-match is now how different the frequencies are, or the energy-mismatch, and the overlap is how strong this spring is between them.

So if I make the spring really strong, then even though they have different frequencies, they’ll be more mixed with one another. Now this, it’s weird, but this one is moving more. But if I make it very weak coupling, like that, then they really stay independent of one another. So it’s a really neat physical demonstration of what we saw last semester.

OK, that’s enough of that for now. If I can get this off. So you can come play with that afterwards if you want to.

So vibration remains localized when the coupling is weak, compared to the energy mismatch. These illustrate that. So if you have a general molecule that has N atoms in it, there are 3N independent geometric parameters, x, y, and z for all the N atoms. 

But it takes three numbers to say where the center of the molecule is. And it takes another three numbers to tell how much you rotated around this axis, how much you rotated around this axis, how much you rotated around this axis, so that’s three more numbers to fix the orientation. So you have 3N – 6 coordinates that involve internal vibration. But it’s not just two atoms that are vibrating, it’s the whole set of atoms that are vibrating. And in a normal mode, they all vibrate at the same frequency.

But there are going to be 3N – 6 normal modes, and this is where physical chemists threw up their hands, and said, it’s just so complicated how all these springs are interacting with each other that we’re never going to understand it. So forget trying to interpret. OK for a fingerprint, but forget trying to interpret it. So it sounds hopelessly complex, though it’s good for a fingerprint.

But remember, to get mixing, you have to have frequency match and some coupling mechanism. So if you have some of these bonds that vibrate without similar frequencies to anything in the neighborhood, or not coupled to things in the neighborhood, then they’ll appear independently, and you can hope to identify a particular functional group.

Chapter 5. The Normal Modes of n-Butane [00:45:38]

So if they’re isolated, they’re independent, but if they’re coupled, then you see complication. So it’s again this question of energy-match and overlap. Now let’s look at the butane, C₄H₁₀. It’s got 42 degrees of freedom, 42 of these, minus the 3 translations and the 3 rotations, so there are 36 vibrations, 36 normal modes that are involved. If you just had the four carbons, there’d be three stretches, two bends and a twist, the torsion around the middle. With 10 C-H’s, you have 10 stretches, and 20 bends or twists. But it’s not that simple, because they get mixed up, according to frequency-match and coupling, into 36 complicated normal modes.

Let’s see if we can understand anything about these normal modes. So here’s a spectrum of a hydrocarbon, it’s C₈H₁₈. You’ve seen a spectrum like this, I bet, in lab. Let’s see if we can understand what any of the peaks are. First notice, that there are peaks at the far left, around 3000, and the others are all 1500 or below. What do you think’s up at the top, high frequencies?

Students: C-H

Professor McBride: C-H’s, right. Because the H is so light. OK, so there are 72 normal modes here, but not all of them are IR active with this C₈H₁₈. So let’s look at that particular peak there. There are actually several there but, anyhow, one of them is this. So there we have a molecule and here we distort it. Now, tell me what’s happening in the distortion? What’s mostly changing? 

What’s mostly changing?

Student: C-H bonds.

Professor McBride: Yeah, its C-H bonds are stretching and shrinking, right? And notice the way they’re doing it. Focus on the two central CH₂’s. The hydrogens are moving up and down, up and down. So the charge is moving up and down, up and down. So that’s going to interact then with light, with an electric field of light that’s pointing in that way. So light can help push those four central H’s up and down. That particular vibration, that particular normal mode, interacts with light, and you get absorption of light at that position. When it’s up, it pushes it one way. When the field is down, it pushes it the other way, and we interact with light. So that’s a C-H stretch. 

Now, how about this one? Here we have– or one of the peaks under there at least– here it’s this and then that. What’s happening now? What bonds are changing? What kinds of bonds? Is that C-C or C-H? 

Students: C-H. 

Professor McBride: It’s C-H. And notice how they’re doing. They’re all going in or out together. So that’s going to do it. What direction of the light’s electric vector will excite this? It’s going to be perpendicular to the screen, to push them all in, all out. So it’s a different polarization of the light that’s going to do this one. It’s going to be the light that goes that way or goes that way.

Now there are a lot of others that we won’t go into. One of them is this. Look at this now. What’s happening now? Again, we’re stretching C-H bonds. How strongly is this going to interact with light? Notice, for every H that goes one way, there’s another H that goes exactly the opposite way, so those cancel out and there’s no interaction with light. So that sort of breathing, where they’re all going out and in at the same time doesn’t interact. You don’t see that in the IR. 

So in fact, half of the 10 C-H stretching normal modes have no handle on them. They don’t change the dipole moment. So they don’t absorb light. And that’s where we’re going to have to stop now and go on to the other peaks next time.

[end of transcript]  

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