Subtitles section Play video Print subtitles >> Bond character is somewhere between that of a single bond and that of a double bond. In fact, it's sort of about 30% double bond and 70% single bond that makeup this picture. So as a result, you have slow rotation about this bond and I'll say slow which means if I put a star next to this ethyl group here, I'll just put sort of a star here to remind us it's special, you do have a dynamic equilibrium where you swap positions right. So in this case, the start ethyl group assists to the carbonyl. In this case, the start ethyl group is trans to the carbonyl but this equilibrium is slow on what we'll say the NMR time scale. What I really want to do in today's talk is to give us more of a feeling of what's slow, what's fast both in time and also in energy. So when I saw this is about 30% double bond character, I want you to get a feeling for what that means in kilocalories per mole and how I know that number. Then I also want us to get a feeling for what happens as we cross from the slow regime to the fast regime so we'll get sort of one equation, a couple of calibration points on energy and time out of today's talk. Now, if you think about this, let's take a simpler situation. The case that people invariably use for didactic purposes is something with singlets. The case that sort of was the classic was dimethylformamide just because it's easy to think about and easy to simulate. So here with methyl groups, of course, you have singlets and just like the ethyl groups for one of them is more down field; the one that's CIS to the carbonyl is more down field; the one that's trans from the carbonyl is more up field. You have the same with the methyl groups here. I just want to, we'll call these HA and HB. So you have some equilibrium here we can call this KA and KB. In the case of a perfectly symmetrical molecule, the rate constant is going to be the same but if these were two groups say of disparate size like a tert butyl group and a methyl group, then the rate one way would be faster than the other because you would have an equilibrium constant that wasn't 1. In other words, you'd have an equilibrium constant where if you had a bulkier group, say the bulkier group would spend more time here and the less bulky group would spend more time here or vice versa. All right. So let's just imagine a little thought experiment for this situation. So, the situation we just saw was one where you have 2 singlets so this is just my old drawing of an H1 NMR spectrum. At some condition where you're slow, you're going to see 2 sets of peaks. For the same window if it were very fast, you'd see 1 peak and I'm just drawing the same spectral window and sort of plotting this out so I guess I'll kind of put these on the scale, on the same scale. So we can say this is fast and, of course, what's the best way to take something that's slow and make it fast in the laboratory? >> Heat it up. >> Heat it up. So this is cold and this is hot. All right somewhere between that point of cold and hot you hit a middle point which is called medium, yep, medium, okay. You hit a middle point, medium, where you're at what's called coalescence. Now at coalescence what's happening is each of these peaks is getting broader and broader until they merge. This is the uncertainty principle at work. Remember we talked about line width and I said that if you were able to measure the velocity for infinitely long, in other words, if there were never any spin flip, any relaxation, any swapping of population between alpha and beta states, your lines would be infinitely sharp, but I said when we were talking about the uncertainty principle your lines are typically about a hertz wide or a little less than hertz wide because your relaxation time is on the order of a couple of seconds. In other words, you cannot get your lines infinitely sharp because you're only literally measuring the velocity for finite amount of time. As you heat things up, what's happening is you're flipping faster and faster and so your lines are broadening out. So, what's really happening at coalescence so let's just go back to the equation I presented by the uncertainty principle, which is if you look at your line, in theory there's some exact position of the line. In other words in theory if you could make that measurement infinitely, your line would have this position but what you're getting is signal out here because you're not measuring that line with infinite time. You're not able to because of relaxation and so you have a certain width and that's the value when we talked about the uncertainty principle we called Delta nu, right? Delta nu basically is the half line width at half height. In other words, it is the level where you're sort of within these arrow bars so you're plus or minus Delta nu of that theoretical central value and we said from the uncertainty principle Delta nu times the time that's the lifetime so it's not really a half life it's almost like a half life, it's an eth life, you know, 1 over 2.3 instead of 1 over 2.0. Delta nu times time is equal to 1 over root 2 pi. In other words, if you're able to make that measurement for a second, then Delta nu is going to be .22 hertz or the line width at half height, the full line width is going to be .44 hertz. If on the other hand you're only able to have that lifetime be say a tenth of a second, so I'll say t equals .1 second, now we get to Delta nu is equal to 2.2 hertz. In other words, now that line has become 4.4 hertz wide at half height. [ Pause ] So what you were really seeing at coalescence when I have this sketch of this broadened hump like this, what you're really seeing is 2 fat lawrenciums [phonetic] that are adding up underneath there. So let me make this sort of with dotted lines to show a fat lawrencium like so and if the separation of the lines here, right, if this separation here if we call this Delta nu lines for want of a better term, in other words, the separation of the lines in hertz, at this point at coalescence then now each of these lines is fattened out so that its Delta nu, not the Delta nu of lines, but this distance here, this Delta nu is now, so if this is what we call Delta nu of lines that this Delta nu is now half of the Delta nu of lines. Does that make sense? In other words, each of the lines is broadened out so that it's half width at half height is halfway across and that is when your lines are going to be coalesced where you're no longer going to see a distinct line on the left, line on the right. If they're broadened anymore, they're going to be merged together and eventually you just have a single peak and you're at this situation here, but right now when they're broadened out they're broadened out to a point where they have merged together and so at that point Delta nu of the lines the separation of the lines times tau, which is now going to be our lifetime at coalescence is equal to 2 over root 2 pi. This is just the equation that we have over there except now because we have the difference each of these is fattened out halfway. If we have 2 of them, it's going to now be 2. So, what this boils down to then is a simple equation that tau when you just work this out is equal to 0.450 over the Delta nu of the lines. In other words, the lifetime at coalescence. [ Writing on board ] Is equal to .450 divided by the separation of the lines at a lower temperature. [ Writing on board ] Does that make sense? [Inaudible question] The .54 is simply what happens if I take in my calculator 2 divided by root 2 divided by 3.1415 and then I put that in the numerator and put the Delta nu lines in the denominator. [Inaudible question] Well, you mean the 2? Well, I'm saying because here our Delta nu is half of the separation. [Inaudible question] Within what you can measure it's exactly. Let me show you. Maybe the best way is for me to show you how things look as you vary here. So, basically if you go any, coming to, if it was more, you'd start to pull in and you'd start to pull together. If it's less, you'll see a dimple in the middle and let me show you what this can best be pictured as and this is just a simulation, this is from a chapter undynamic NMR spectroscopy in a book, let's see which book? This may be a book on dynamic NMR spectroscopy. So this is an old, just an old drawing of a simulation of what you would expect and it's really based on dimethylformamide and it's actually probably based on why DMF on a 60 megahertz spectrometer or something like this. So their simulation is as follows, and the reason I say it's a 60 megahertz spectrometer is the lines in this simulation are Delta nu lines is equal to 20 hertz. In other words, out of 60 megahertz NMR spectrometer that would be about 3/10 of a PPM, which is pretty reasonable. Now, on a 500 megahertz spectrometer that would be .04 PPM. So anyway for their little simulation, they're saying imagine that you have 2 lines, those 2 singlets, and they're separated by 20 hertz. Imagine that you have a T2, that's a relaxation time, of .5 seconds. In other words, imagine that the native lifetime for this molecule due to relaxation was half a second. In other words, that your lines are about .9 hertz width, full width at half height, not half width but full width at half height. So that's how your normal spectrum would look. Now, imagine that you start to heat this sample up so that you have rotation between the 2. So you have the 2 flipping back and forth. So, imagine here, for example, that K for, you know, our equilibrium this is our like dimethylformamide spectrum where we can call this A and B or star. [ Pause ] Imagine now that our rate constant, oh, yeah, imagine that our rate constant was 5.0 per second. If your rate constant is 5 per second, then your lifetime is 1 over K, right? So your lifetime at this point is 200 milliseconds, it's .2 seconds. [ Pause ] So your lines have met because they're not staying in the CIS or trans state as long. You can think of this as we started, here we're told and here we're starting to heat the sample up. Here they actually have a very slow K, K is equal to in this case .1 per second. In other words, it's a 10-second lifetime. Do not swap at any appreciable rate. As you heat up the sample in this simulation, they go to K equals 5 and K equals 10 per second so your lifetime is now .1 seconds and finally you get to a point and you notice so here you are and now your line width is still less than half the distance between them. So you still see this dimple here, this is it. Now you can kind of see right at this point now they are coalesced together. So, right at K is equal to 44.4 per second they're now coalesced together. Then as you heat the sample up more, as you get hotter and hotter, as you get faster and faster, now we get to K equals 100 so now your lifetime is 10 milliseconds and then you get to K equals 500 so your lifetime is now 2 milliseconds and finally you get to K equals 10,000. So now your lifetime is a tenth of a millisecond. [ Pause ] [Inaudible question] That's at relaxation time. That's your native line width. So what they're saying, of course, is all lines have a native line width. Let us pretend that we had a native line width of about a hertz. [Inaudible question] You could just as well have it be T1. Either T1 or T2 is going to contribute to line width. I don't know why they chose T2. It's completely, it's completely arbitrary because whether your magnetization is spreading out in the X, Y plane so you're no longer able to get signal or your magnetization is returning to the Z axis you still have line width. In the case of small molecules, typically T1 is the predominant relaxation pathway in the case of very large so strychnine, for example. In the case of very large molecules like proteins, T2 is the predominant relaxation mechanism and this depends in part on how fast the molecules tumble and how viscous the solvent is. Other thoughts and questions? [Inaudible question] It's flat, yeah, it's basically this is, right, this is perfect coalescence. [ Pause ] All right so this is simulated data for sort of a textbook example and now what I want to do is show you a real example, show you how to get a rate out of this and then we're going to translate that into a free energy activation. So, okay, so the case that I'll show you which is kind of cool, this is just one I pulled from my own experience. [ Pause ] It's a sort of neat molecule because we're going to see that there's actually 2 different things going on here. So, the molecule is diethyl ortho toluamide and I'll show you the spectrum of it here. I have another handout for you. [ Pause ] Yeah, one of the great things about being in graduate school if/else a lot of times you get to observe stuff that's cool and beautiful and relates to your classes and this just happened to be something I noted back when I was in graduate school and it's like, oh, this is cool; I'll keep this as an example. There's just something I was doing on the synthetic methods project just working out a method. So I had my sample of diethyl toluamide, diethyl ortho toluamide and I started to warm it up. I noticed, I was curious because there was some broadness to the peaks here. So you had your two ethyl peaks, this is your methyl peak. So these are your CH2s, these are your CH3s, and I was just curious about what was going on. So this was a sample in DMSO D6. DMSO has a very high boiling point so you can heat it up to a high temperature. Deuterochloroform boils at I think 66 degrees. So if you were to try to heat it up to 160 in NMR tube, the NMR tube would if you're lucky just blow the top off the tube if you're not lucky exploding the probe. Either way you'd have a very angry department at UB because you would trash the NMR spectrometer and do serious damage. So, as you warm it up, the CH2s coalesce and 110 is really perfect for the coalescence temperature of the CH2. Now the CH2s are pretty far apart. The CH2s are at this, this one is at 3.45. This is a 300 megahertz NMR spectrometer and the other one is at 3.03. The methyls are a little closer together so their Delta nu lines, the separation of the lines is smaller so they actually will coalescence even with smaller rotation. So the further you are apart the faster you have to spin in order to have the 2 lines coalescence into 1. If 2 lines are very close together, you only have to spin it slowly, only have to have rotations slowly, to get coalescent. If 2 lines are very far apart, you have to have rotation very quickly. So we're already coalesced. I'll write coalesced here to indicate that it's in the past tense and here at 100 degrees we're not yet coalesced. So somewhere here at about 105 I would say would have be coalesce if I had bothered to do that experiment, but let's for a moment focus on these 2 methyls and I want us to figure out the rate here and then we're going to translate that rate into an energy. So for the CH2s the Delta nu lines, the separation of the lines, is equal to 3.45 minus 3.03 times it's a 300 megahertz spectrometer so that's 126 hertz. So, now literally it's plug and chug in this equation. Tau, the lifetime at coalesce. [ Writing on board ] Which is 110 degrees for this particular set of resonances, the lifetime at coalesce is just equal to .450 divided by 126 which is equal to 3.6 times 10 to the negative 3 seconds. I didn't measure it at 105, but I think it's about 105 is the coalesce temperature. So for the CH3s, the CH3s were separated by 60 hertz and so in that case tau so the lifetime at their coalesce 105 degrees for them let's say is going to be about 7.5 times 10 to the next 3 seconds. In other words, about 7.5 milliseconds. To put it in terms of that other example, at room temperature rotation about this amide bond is slow on the NMR timescale. In other words, it's on the order of let's say seconds or hundreds of milliseconds so we see 1 ethyl peak, we see another ethyl peak. As we warm it up, it rotates faster and faster and faster as it gets warmer and warmer and warmer. By 105 degrees it's spinning around with a lifetime of 7.5 milliseconds. The methyls being close together have coalesced and we heated up a little more to 110 degrees the methylenes being further apart now they have it spinning at 3.6 millisecond lifetime. The methylenes have coalesced and by the time I heated up to 150 or 160 the peaks are now relatively sharp and usually by that point it's pretty hard to get good shims so we probably would see a quartet and a triplet there if I could shim the spectrometer better. [ Pause ] All right I want to translate this lifetime into a free energy. So one of the take home messages here from this example is below a millisecond let's say is fast on the NMR time scale and, you know, above 10 milliseconds or 100 milliseconds is slow on the NMR time scale. Let's now see how that relates to free energy of activation. So, what I want to do is translate our K to Delta G double dagger [phonetic] and one came from transition state theory you have the Eyring equation which basically deals with the Boltzmann population of molecules that are able to cross an energy barrier and the Eyring equation is that the rate constant is equal to kappa which is the transmission coefficient which is generally taken as 1 times the Boltzmann constant times the temperature over Planck's constant times E to the negative Delta G double dagger over RT where you have the gas constant with temperature. What this is, the way the Eyring equation is derived is basically you're setting up an equilibrium between molecules in the ground state molecules in the transition state and then assuming that half the molecules go over the transition state at that point and, of course, K here at coalesce K is equal to 1 over tau so we're going to use our 100 degrees K is equal to 1 over tau, 1 over our lifetime. So what I want to do is figure out our free energy here. So add 110 degrees Celsius if I now just plug into this thing I get 1 over 3.6 times 10 to the negative 3 is equal to 1.35 times 10 to the 23rd, 10 to the negative 23rd and at 110 we're at 383 Kelvin and we divide by Planck's Constant, 6.63 times 10 to the negative 34th and for the sake of taking on our math for a moment I'll just keep this as E to the negative Delta G double dagger over RT. [ Pause ] All right if I just continue to work through my equation, I get 3.484 times 10 to the negative 11th equals e to the negative Delta G double dagger over RT and if I work through that's Delta G double dagger is equal to negative 1.987 times 10 to the negative third times 383 times the natural log of 3.484 times 10 to the negative 11th and when all is said and done number I get is 18.3 kilocalories. All right. So what is that saying? That's saying coming back to this point I raised at the beginning of class what's the degree of double bond character in an amide? Well, you've got an 18 kilocalorie per mole barrier to rotation. If that were a single bond, you'd have essentially no barrier to rotation. If that were a pi bond, you could say that would be maybe, you know, 60, 70 kilocalories per mole. In other words, like in ethylene the pi bond is 60, 65 kilocalories per mole so I look at that and say oh, that's about 30%, 25-30%, of a pi bond. In other words, if I had ethylene, well, if I had, if I had CIS 2 butene, no matter how hot I heated it in the NMR spectrometer, I'd never see isomerization between CIS and trans 2 butene, the energy barrier to rotate about a real pi bond is so high you just don't get that from thermal energy, but with this partial pi bond, I have 18 kilocalorie per mole barrier. All right. There's something else that's really cool that's embedded in this spectrum. So take a look at the spectrum at 30 degrees Celsius and you'll notice even at this point one of our methylenes is a little board. You see that? There's actually, this is a cool molecule. There are actually 2 dynamic processes that are going on here. So I figured at the time I was just curious but now, of course, I'm using this as an example and it's a fun example because it actually ties into some cool concepts in stereochemistry. So anyway I figured I want to cool the same down and take a look at it. Now, DSMO freezes just a little below room temperature so you can't do super high temperature NMR in chloroform, you can't do super low temperature NMR in DMSO. There are chlorinated solvents you can use like 1, 1, 2, 2 dichloroethane can up to very high temperatures and down to very low temperatures, but DSMO is common so I use that and chloroform is common. You can use methylene chloride if you needed to go to a lower temperature. So I took the NMR spectrum in chloroform to see what the heck was going on. It's really beautiful. So, at room temperature, which happened to be that day 22 Celsius now we were just not quite almost at coalesce. [ Pause ] And you notice as you cool it down now we're at 10 degrees, well, my Xerox didn't come out well. By the time you're at 0 degrees you notice that CH2 is resolving itself into 2 sets of peaks. By the time you've gone down to negative 40, you can see these, these happen to be doublet of quartets and by the time you're down to negative 40 degrees Celsius you can see that even the other methylene which started as a quartet now has a more complex splitting pattern. So this is really kind of cool. So there's a second dynamic process with a coalesce temperature that's just a hair above 22 in chloroform. Remember, it's a different solvent so you have slightly different rate constants. So we'll say that I didn't measure it exactly but we'll say that coalesce is probably at about 25 in chloroform, 298 Kelvin. So, what I want to do now is to play with this process figure out what's going on and then look at the energies that are involved and get us calibrated on energy. [ Pause ] All right. So, this system happens to be way cool. So, because you have these 2 substituents and the methyl group, your methyl group is not going to be coplanar. In other words. [ Pause ] You have a situation where the tolumide ring is rotated out of planarity from the amide group. They are orthogonal or close to orthogonal to each other. This is a situation where you have what's called axial chirality. Stereochemistry is cool. It's the same thing you have in allene. If you have 1, 3 dimethylallene, there are 2 antimeres of it. So if you have simplistic example you can come up with. [ Writing on board ] This molecule is chiral. All right so we have an equilibrium here of 2, atropisomeric rotamers. [ Writing on board ] And these 2 atropisomeric rotamers are antimeres. [ Writing on board ] Which means your CH2s are diastereotropic. [ Writing on board ] Now the CH2 that's next to the carbonyl has very low magnetic anisotropy so you see that at very low temperature you do see something that's other than a simple quartet, but this one the 2 protons, the pro R and the pro S have a high degree of magnetic anisotropy. The one that's on the same side as the ring. So we actually have separation there and in this case the Delta nu as I cooled it down remember we could see the 2 different lines here and so here one of these lines is at 3.94 and the other line is at 3.30 and so the Delta nu lines then is going to now be for those 2 lines for the methylenes it's going to be 3.94 minus 3.30 times 300 is 192 hertz. So our tau, the lifetime at coalesce, is 0.450 and that's divided by 192 is 2.3 times 10 to the negative 3 seconds at I don't know if it's 298 or 295K. I said, I don't know. That's close to coalesce. If it's at 295K, I'll plug in Delta G double dagger is equal to 13.7 kilocalories per mole. All right so take a minute to think about this. We've got 2 processes in this molecule. One that's almost invisible at room temperature because, well, that's invisible slightly above room temperature, which is the rotation about this bond, and the other that is the rotation about this bond. The rotation about this bond has an 18 kilocalorie per mole, 18 point what did I saw 3 kilocalorie per mole barrier, so it doesn't become fast until you heat the thing up hot, like 110 degrees or 105 degrees. The rotation about his bond is kind of medium scale at room temperature. You cool it down, it becomes slow. You heat it up it becomes fast and is invisible. So that's kind of cool. Now I want to give you 1 caveat because every student, I had to say student, every new person's first excuse when they see something in the NMR that they don't understand when they see 2 sets of peaks is say hinder rotation. All right I'm going to make a very sweeping generalization. The hindered rotations that you see are only going to involve things where you have an SP2 atom connected to an SP2 atom. Chances are if it's an SP2 atom connected to an SP3 atom or an SP3 atom to an SP3 atom it's going to be fast. So hindered rotation generally only for SP2 to SP2. So here we have an SP2 hybridized benzene connected to an SP2 hybridized carbonyl. When you have some extra stereic hindrance it's slow to rotate. Here you have an SP2 connected with partial double bond character, you have slow rotation. [Inaudible question] It's pretty darn flat. It's actually that nitrogen really is SP2. What I mean specifically is I can think of no simple bonding situation where you have SP3 atoms connected, single SP3 atoms, where anything is slow without some SP2 intervention. Cyclohexane ring flip where you have 2 sets of eclipsing, 10 kilocalories per mole, which is still fast on the NMR time scale at room temperature. Cyclohexane is coupled together. That one if you cool to negative 80 degrees, does become, can become slow. Actually let me use this as a chance, did that sort of answer your question? >> Yeah, yeah. >> Other questions? >> How did that [inaudible] change? >> You rotate, you spin so basically the benzene ring is like this here with a methyl group pointing out and it spins back and forth, but it has to bang in doing so the methyl group has to bang past the carbonyl. There's enough steric hindrance there that it can't do it rapidly. [Inaudible question] It's okay. It's not going to be perfectly perpendicular. It'll be at about a 60-degree angle and it'll nicely rock back and forth but to cross to the other atropisomer. That's where it's hard. If you want to make a model of it, this is a great one to use pinmol [phonetic]. You can easily make a model in pinmol and you'll see how they sit and how they sit and how they bang. [Inaudible question] It'll put it, you still have axial chirality as long as you have a barrier. [Inaudible question] Oh, did I, oh, ah. Oh. My goodness. Oh, thank you, yes, no I meant to have this going back, yeah. There you go. Okay, yeah, yeah. Methyl back, methyl forward. >> Does it matter though actually? >> They are 2 antimeres. Yeah, so. [Inaudible question] So, okay, my thumb is the methyl group. Methyl out, flip, rotate methyl back. All right. Thank you, thank you, thank you. All right. I want to show you, the last thing I want to do is give you 2 take home messages and let me start with the message and then I'll go and show you my thought on this. My thought is the take home message is the NMR time scale, you know, I like to have simple things in my mind as ways to keep things, is let's say less than 1 millisecond is fast, about 1 to 10 milliseconds is intermediate and greater than 10 milliseconds is slow. These are obviously sweeping generalizations because they're going to depend on separation of lines and they're going to depend on field strength of the spectrometer, but let me show you my thinking on this. If we imagine a Delta nu of lines and it's 50 hertz, I'm going to give us 2 scenarios. Let's start with a scenario where we're 50 hertz and let's say what that is in PPM at 500 I'll just take 500 megahertz because that's sort of a typical modern spectrometer so that's going to be .1 PPM. So you see 2 lines at .1 PPM separation and the tau, the lifetime at coalesce is .009 seconds. In other words, it's 10 milliseconds or K is equal to 111 per second. So in other words, with 2 lines that are close together if your process is occurring on the order of 1 millisecond, it's going to be fast. If it's occurring on the order of 100 milliseconds, it's slow. If our separation of lines was 500 hertz, that's pretty far apart. That's 1 PPM but we saw half a PPM over in that example. If it's 1 PPM, the lifetime at coalesce would be .009 seconds. In other words, 111, 1,111 per second. In other words, if the lines are further apart, if it's spinning around, you know, many times per millisecond, it's fast. If it's spinning around once every 10 milliseconds, it's slow. So that's how I calibrate myself. Let me give you my other calibration that I like to keep in my head. So the other calibration I like to keep in my head is typical NMR energies. [ Writing on board ] Is going to be say 10 to 20 kilocalories per mole. In other words, a process that's 15 kilocalories per mole kind of teeters between slow and fast at room temperature. A process that's 20 kilocalories per mole is slow at room temperature but it's going to be fast at a very hot temperature . A process that's 10 kilocalories per mole will be slow at very low temperature but fast at room temperature and so let me just show you my thinking on this. So if I take, if I go ahead and look at 1 over K and I'm just going to make a little table of 1 over K in seconds as a function of Delta G double dagger and temperature and so this is me windowing myself there's your take home message on top but let me show you my window in my cell. So, imagine we consider energies of 10, 15, 20 and 25 kilocalorie per mole Delta G double dagger and then consider from the, from the rate equation we consider temperatures and I'm just going to window at 3 temperatures, negative 50 degrees C, which is kind of cold, 298K, 25C, which is kind of room temperature, and 373K, 100C which is kind of hot. All right then if I simply calculate from the calculate the rate that applies the lifetime in seconds is 1 millisecond for a process with a 10 kilocalorie barrier at negative 50. That process has a lifetime of 3.5 times 10 to the negative 6 seconds at room temperature and 9.3 times 10 to the negative 8th seconds at high temperature. In other words for a 10 kilocalorie barrier process at negative 50, we teeter between slow and fast. So that's sort of our intermediate, our coalesce temperature. By this point we're fast and by this point we're very fast. So 10 kilocalorie barrier cyclohexane ring flip slow and cold. Okay, if we go to 15, 108 seconds is the lifetime at negative 50. In other words, it's slow at negative 50, but that 15 kilocalorie barrier process becomes milliseconds, 16 milliseconds, at room temperature. In other words, 15 kilocalories becomes intermediate at room temperature. That was like that rotation about the benzene bond. It's fast by the time we're hot, it's 7.9 times 10 to the negative 5th seconds lifetime. Twenty kilocalories per mole, 8.6 times 10 to the 6 lifetime that's forever. Seventy-five at room temperature, 75 seconds is slow, but you get to hot and it's 68 milliseconds and now we get into the intermediate regime in high temperature. That was like our amide bond rotation; 18.3 kilocalories we had to heat it up to 110. Finally by the time we get to 24 kilocalories, 6.9 times 10 to the 11 seconds 3.5 times 10 to the 5th seconds and 57 seconds. In other words, by the time we have a 25 kilocalorie per mole barrier even at high temperature, you are still slow. All right so that's where I window myself and I say 10 degrees slow at low temperature, 15 degrees, you know, intermediate at room temperature, 20 degrees intermediate at high temperature. All right. Our midterm will be next time. We have an in-class part on Friday and then an open book part on Saturday. So a closed book part on Friday. ------------------------------858acd1bce6f--
B1 delta nu temperature lifetime nmr hertz Chem 203. Organic Spectroscopy. Lecture 18. Dynamic Effects in NMR Spectroscopy 81 3 Cheng-Hong Liu posted on 2015/02/02 More Share Save Report Video vocabulary