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  • >> 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--

>> Bond character is somewhere between that of a single bond

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B1 中級

化學203.有機光譜學。第18講.NMR光譜學中的動態效應 (Chem 203. Organic Spectroscopy. Lecture 18. Dynamic Effects in NMR Spectroscopy)

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    Cheng-Hong Liu 發佈於 2021 年 01 月 14 日
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