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  • >> All right.

  • Well, I will leave these and get started then, I guess.

  • All right, so I think I want to finish

  • up what I'll call basic NMR spectroscopy today.

  • In other words, things that all sort of fall at the level

  • of basic interpretation of structures, and in turn things

  • that we'll get on the mid-term.

  • I think probably where I'll pick up next time is going

  • to be introducing 2D NMR and then our next homework set,

  • not Monday's, but the one after that,

  • we'll start using 2D NMR spectroscopy.

  • So what I wanted to do today was to talk more

  • about carbon 13 chemical shifts.

  • And I gave you, when we talked last time, I gave you this sort

  • of general information that just like proton NMR, carbon NMR,

  • aliphatics tend to be upfield, aromatics tend to be downfield,

  • things that are next to electron withdrawing groups,

  • particularly oxygen, tend to be mid -- we'll call it mid field,

  • sort of in that 50 to 70 range.

  • I also indicated -- and we said

  • that the range is a lot, lot bigger.

  • It's about 20 times bigger in ppm for C13 NMR.

  • In other words, an aldehyde CH is at roughly 10 ppm whereas

  • in carbon NMR an aldehyde carbonyl is at roughly 200 ppm.

  • So it's sort of like 20 times bigger range.

  • Now, C13 shifts have a bigger range.

  • And there's also more richness.

  • In other words, when we talked

  • about proton NMR, it was pretty easy.

  • And we were able to come up with some really simple, you know,

  • back of the envelope calculations

  • where you could typically peg the chemical shift to within,

  • you know, a few tenths of a ppm.

  • We talked about if you're next to an ester, you know, let's --

  • or if you're next to an oxygen, figure you're going to be

  • about 2 ppm downfield of where you'd normally be.

  • If you're next to a benzene ring or a double bond or a carbonyl,

  • figure you're about 1 ppm further downfield.

  • And I gave you several ways of thinking about this.

  • And you should all be able to pretty much estimate things.

  • We talked about the effects of alpha substituents and said

  • that alpha substituents have a really big effect being next

  • to a CH next to a halogen next to a nitrogen next

  • to an oxygen shifts you down, you know,

  • a couple of tenths -- a couple of ppm.

  • We talked about beta effects.

  • And we said that they're smaller.

  • Beta effects in proton NMR shift you down by like .2 to .5 ppm.

  • C13 NMR is a little bit more subtle and a little bit --

  • I won't say less predictable, because we're going

  • to see it's actually quite predictable.

  • But the factors, there are more factors.

  • For example, gamma effects as well

  • as beta effects tend to be big.

  • And there are some really interesting steric effects.

  • Now, along with this richness comes tremendous power.

  • Because this means that carbon NMR also can give you tons

  • of rich information about structure

  • and can be really useful for figuring out structures,

  • confirming structures, disproving structures,

  • and at the end I'll show you a beautiful example

  • of fraudulent work that was disproven

  • by Professor Rychnovsky's laboratory.

  • And also going ahead and basically having a tool

  • that can get you a lot more information than meets the eye.

  • So, I want to show you some of the factors that contribute

  • to these sorts of general ranges,

  • and particularly now, perturb them.

  • And let's start with something pretty simple, inductive effects

  • and resonance effects.

  • [ Writing on Board ]

  • And electron density, of course, plays a huge role

  • in chemical shift because electrons contribute

  • to the shielding or de-shielding

  • if they're absent, of various nuclei.

  • So, if you have substituents that increase

  • or decrease the electron density in a carbon, you're going

  • to shift that carbon upfield or downfield.

  • Let me show you what I mean.

  • We'll start with a simple benzene example.

  • Now, the easiest way to, in your mind,

  • think about chemical shifts is to start

  • with a base value and then perturb it.

  • So a great way to think about benzene is the normal position

  • for benzene is 128.5 ppm.

  • And then if you put substituents on it,

  • you perturb things in a rational way.

  • So let me show you what I mean.

  • If we put a methoxy on the benzene,

  • the oxygen is electron withdrawing

  • through the sigma bond.

  • And so the carbon attached

  • to the oxygen shifts substantially downfield.

  • So you go to 160 ppm.

  • In other words, you shift downfield by 30 ppm,

  • more than 30 ppm by putting that oxygen there.

  • Now, what's interesting, then, is the ortho carbons end

  • up having by resonance extra electron density at them.

  • In other words, the two ortho carbons you can push your

  • arrows, and you see they're electron rich.

  • It's the same reason why electrophilic aromatic

  • substitution occurs ortho and para

  • when you have a methoxy group on there.

  • So, the ortho carbons appear at 114 ppm in methoxybenzene.

  • You don't get a big effect at the meta carbons,

  • which makes sense because you have inductive effects

  • that are now quite removed.

  • So it's very small.

  • And resonance doesn't pump

  • up the electron density at the meta carbon.

  • You go to the para carbon,

  • and now you also see an upfield shifting, although it'll be

  • at a smaller upfield shifting of 121 ppm -- at 121 ppm.

  • Now, I'll come in a moment

  • to empirical additivity relationships,

  • but one way to think about this is to think about it

  • if you have a methoxy group or an alkoxy group on a benzene,

  • that it shifts the ortho protons upfield

  • by about 14 or 14 1/2 ppm.

  • And if you have a methoxy group,

  • it shifts the para protons upfield by about 7 1/2 ppm.

  • And if you have a methoxy group or an alkoxy group,

  • it shifts the meta protons downfield

  • by just a fraction of a ppm.

  • And what we'll see in a moment is that you can add

  • up all these effects and then calculate

  • for different aromatics,

  • the effects of different substituents.

  • All right, let's take a look at some other examples

  • of electron -- of inductive and resonance effects.

  • So, let's take an alkene.

  • And I'll give you cyclohexene as a base value.

  • In cyclohexene, your alkene is at 128 -- 127.4 ppm.

  • Let's compare that to cyclohexanone.

  • And in cyclohexanone, we see a very big effect

  • at the beta position and just a little effect,

  • just a little inductive effect at the alpha position.

  • The beta position is 151 ppm

  • and the alpha position is just shifted downfield by a hair.

  • And that makes sense because you look at this and you say okay,

  • now you can think of a resonance structure

  • in which you're electron deficient, right?

  • We all know that enones are Michael acceptors.

  • That nucleophiles like to attack at the beta position.

  • Are you doing this with orbitals in Van Vranken's class now?

  • >> Yes.

  • >> Yeah, so you know

  • about frontier orbitals and electron density.

  • And so you see, the effect is actually very substantial,

  • right?

  • Both of these carbons, they're symmetrical, they're at 127.

  • You go more than 20 ppm downfield

  • by decreasing the electron density.

  • These effects can be absolutely humongous.

  • And one of the things that I've tried to emphasize

  • when I've talked about these ranges here are these are

  • general ranges.

  • These are not carved in stone.

  • And so you already see, for example,

  • that this one inductive oxygen here brings us even outside

  • of this very generic range here.

  • Let me show you just how huge the effects can be.

  • So, ketene acetal is a good example, right?

  • Alkenes are normally like 110 to 150 ppm.

  • But if hugely perturb the electron density,

  • you can have huge effects.

  • So it probably doesn't surprise you too much if I tell you

  • that you now, by having two methoxys

  • on an alkene can shift it downfield to 169.7.

  • But what's really huge is you look at the position here,

  • the beta position on the alkene, and now we're

  • so electronic rich, this thing is so nucleophilic

  • at this position, there's so much electron density

  • at that position, that we're at 45.5 ppm.

  • You look at a spectrum of that

  • and you wouldn't even know it's an alkene because you'd say, oh,

  • that's got to be aliphatic.

  • That's got to be somewhere over here.

  • And we've just pumped up the electron density hugely.

  • The most radical example I know off the top

  • of my head is this sort of push me, pull you system here,

  • where you have two electron-donating groups

  • and then two electron-withdrawing groups,

  • two cyano groups.

  • And so this alkene, you go to 39.1 ppm.

  • And then this carbon here is at 171.

  • And 171, you'd say all right, well, it's really downfield.

  • But you'd say it makes sense.

  • You've got these beta electron withdrawing groups.

  • But you look, 39.1, who would of thunk that that is an alkene.

  • [ Erasing Board ]

  • >> Can I ask you a question?

  • >> Yeah.

  • >> What is that letter to the -- CW, or CN?

  • >> CN.

  • >> Okay.

  • >> So these are two nitrile groups.

  • These are two cyano groups, CN groups.

  • >> Okay.

  • >> All right.

  • So, substituents have substantial effects whether

  • they're alpha, whether they're beta, whether they're gamma.

  • And you can really see this.

  • I'll give you -- we're going to walk through this

  • and I'll give you some examples.

  • So first let's talk about alpha alkyl substitution.

  • And if I want to give you a general principle, in general,

  • alpha alkyl substitution leads to more downfield shifting.

  • So if we, again, take our benzene example,

  • and remember we said that benzene was at 128.5.

  • If we put a methyl group on it and make it

  • into toluene, now we go to 137.

  • The point is we shift downfield by about 9 ppm by putting

  • on a methyl group onto benzene.

  • Let's take a look at alkyl systems.

  • So, we'll look at propane.

  • And we'll look at the central methylene

  • of the central carbon of propane.

  • You put on an alkyl group onto propane and you get isobutane.

  • And now you're down at 25 ppm.

  • And so you notice, it's kind of about the same.

  • Here we moved down by about 8 or 9 ppm.

  • And here we, again, moved down by about 9 ppm.

  • So in other words, we're talking on the order of, eh,

  • 10 ppm or thereabouts.

  • All right, you put on another alkyl substituent

  • and the effect isn't as dramatic.

  • But again, you move further downfield.

  • Now we're at 28 ppm.

  • Now, what's nice about this is these ideas are generalizable.

  • There is really science here to it.

  • And this is the point, that you can take little bits

  • of knowledge and generalize and build

  • up in your mind what's going on.

  • So let me compare us, say, to ethanol.

  • If we start at 58 for ethanol, and now we envision going

  • to isopropanol, what would you predict your carbon

  • to be at here?

  • >> 67.

  • >> 67 would be a very good prediction, because we say,

  • okay, we take 16, we add 9, we get to 25.

  • You add 9, you get to 67.

  • And that's a very good guess.

  • 64 is the answer.

  • All right.

  • Now imagine we go further and we go to tert-butanol.

  • Now what do you think, for tert-butanol?

  • >> 66.

  • >> 66?

  • >> I don't think it's going to change.

  • >> You don't think it's going to change.

  • Okay. Other guesses on that this estimates?

  • >> 67.

  • >> 67. All right.

  • And the answer is 69.

  • No, you're doing good.

  • You don't have to peg it to the nearest ppm.

  • [ Inaudible Audience Response ]

  • Oh, okay. The Price is Right approach.

  • The Chemical Shift is Right.

  • Do you think we have a show here?

  • Maybe really late night TV.

  • [ Laughter ]

  • >> Let's call up Bob Barker [inaudible].

  • >> Ooh.

  • Maybe we can do a Webcast of this.

  • Try it out in a test market.

  • All right.

  • So in the case of proton NMR, we also saw that shifting,

  • adding an alkyl group shifts you downfield, right?

  • We said a methyl group is typically

  • at about .9 parts per million.

  • And methylene group we said 1.3 to 1.5.

  • So in other words, you're shifting down 1.3 --

  • 1.2 to 1.4 I think I said.

  • So in other words, we're shifting downfield,

  • you know, to about .5 ppm.

  • A methine is down at like 1 1/2 or 2 ppm.

  • These are the baseline values.

  • So in other words, we go a few tenths of a ppm more.

  • And so it kind of makes sense.

  • If you get half a ppm for going --

  • adding one alkyl group and you know, another couple of tenths

  • of a ppm for adding another alkyl group,

  • it kind of makes sense here with a scale 20 times bigger

  • that we get about 9 ppm or 10 ppm for adding one alkyl group

  • and about 5 ppm for adding another alkyl group.

  • With the case of proton NMR,

  • beta alkyl groups don't make a huge effect in chemical shift.

  • In the case of C13 alkyl groups, though, they do.

  • Beta substitution ends up also leading --

  • also leading down to further downfield shifting.

  • Not quite as dramatic.

  • But let's take an example.

  • And what I'll do is I'll take butane as an example

  • and we'll look at this carbon here.

  • And then we're going to put substituents over here.

  • So they'll be beta to it.

  • In other words, instead of directly

  • on it, there'll be one over.

  • So if we add one methyl group here,

  • so I'll go from butane to 2-methyl butane.

  • Now this carbon goes 7 ppm more downfield.

  • We go to 32 ppm.

  • And then if I add another one,

  • we go another 5 ppm more downfield, we go to 37.

  • Not as big, but still pretty darn big.

  • Okay, so that's beta substitution.

  • Let's look at gamma substitution.

  • Now, gamma substitution is interesting

  • because gamma substitution leads to upfield shifting,

  • and it's really a steric effect.

  • So, with gamma substitution,

  • what basically you're doing is now getting interactions

  • that are by having steric repulsion,

  • pushing more electron density onto the carbon

  • and shifting you upfield.

  • So let me show you what I mean.

  • So we'll still keep to this same type of general structure.

  • But now we'll consider this carbon here.

  • And so if you look at the carbon here,

  • and you have hydrogens banging into it,

  • you basically will have hydrogens banging together.

  • This is going to lead to more electron density on the carbon.

  • In other words, electrons of the CH bond repelling electrons,

  • push electron density into the carbon.

  • And since you don't want to have steric repulsion,

  • this occurs most pronouncedly

  • when the molecule cannot avoid steric repulsion.

  • Let me show you what I mean.

  • In other words, if we start with butane, and we start over here,

  • the methyl group of butane is 13 ppm.

  • If I add one methyl substituent gamma, so in other words

  • if I attached it directly to this carbon, we'd be alpha.

  • If I attach the methyl substituent here we'd be beta.

  • If I attached the methyl substituent here we'd be gamma.

  • If I add one methyl substituent,

  • that carbon shifts from 13 to 12.

  • And you'd say, ah, that's not a big deal.

  • That's not a huge effect.

  • And part of it is this methyl group can adopt a rotational

  • isomer, a rotamer, where it keeps out of the way.

  • But, if I add one more methyl group, then no matter what,

  • we're going to be staring that other methyl into the face.

  • So if I go like that and go to 2, 2-dimethylbutane,

  • now there's no way to avoid that steric repulsion.

  • And we shift over to 9.

  • So now we've shifted upfield by about 4 ppm relative

  • to our initial compound.

  • So even this very remote substituent can make a

  • big difference.

  • You can really see this

  • in conformationally constrained systems like cyclohexane.

  • So if I take cyclohexane and I consider the carbons.

  • All the carbons, of course, in cyclohexane are the same.

  • And they're all at 27 parts per million.

  • If I now add a methyl group gamma to this particular carbon.

  • So remember, if I attached it here, it would be alpha.

  • If I attach it here it would be beta.

  • If I attach it here we're gamma.

  • If I add one methyl group, that methyl group keeps

  • out of the way because it keeps equatorial.

  • And so we don't see any change.

  • But if I add a second methyl group gamma,

  • then one methyl group has to, has to, has to be axial.

  • And there's no way to avoid 1, 3-diaxial interactions.

  • And so now that carbon shifts up to 21 ppm.

  • And that's a pronounced effect.

  • And I'll show you the implications

  • of that in just a moment.

  • All right.

  • So these are all of the sorts of biggie effects in C13 NMR.

  • And now I'll show you the last biggie effect,

  • and that's heavy atom effect.

  • [ Writing on Board ]

  • All right.

  • So let's take a look at what happens

  • when you make various halomethanes.

  • So again, it's very helpful to have a baseline value.

  • Methane occurs at negative 2.3 parts per million.

  • So we're going to compare the halomethanes to methane itself.

  • And so we'll look at monohalomethanes, CH3X,

  • dihalomethanes, CH2X2, trihalomethanes, CHX3,

  • and tetrahalomethanes, CX4.

  • And we'll start with X being chlorine.

  • And so, no great surprises here.

  • You put a chlorine onto methane, you get chloromethane.

  • And you go downfield.

  • It's 25 ppm.

  • You add another chlorine, you go further downfield, it's 54 ppm.

  • Chloroform?

  • Where does that show up at?

  • >> 78.

  • >> 77. 77 is chloroform.

  • So you go further downfield.

  • You put yet another chlorine on there,

  • and you go further downfield.

  • And so you look and you say, hey,

  • what am I wasting your time for?

  • This doesn't seem like any sort of surprise.

  • Well, if we move down the periodic table,

  • the atoms get bigger.

  • And they start to have d orbitals, and they start

  • to have their orbitals extend outward.

  • And then we see something very surprising.

  • You put on a bromine and you'd say 10.

  • And you'd say, well that's not so surprising.

  • Bromine is less electronegative than chlorine.

  • It's going to shift you less further downfield.

  • You add two bromines, you go to dibromomethane, and you say 21.

  • And you say, okay, that doesn't sound so surprising.

  • We're moving further downfield.

  • Three gets a little bit weird,

  • because we start to move upfield.

  • And by the time we're at four, we're at negative 29.

  • And then iodine gets downright funky, so iodine --

  • iodomethane is at negative 21.

  • Diiodomethane is at negative 54.

  • Triiodomethane is at negative 140.

  • And tetraiodomethane is at negative 292.

  • And what's happening is the orbitals are extending so far

  • out with these heavy atoms,

  • as you move down the periodic table,

  • that their electron clouds are actually enveloping that carbon

  • and shielding it from the applied magnetic field.

  • And as you can see, in some cases shielding it

  • very dramatically.

  • Now, the practical implication for this becomes,

  • if you go ahead and do a reaction to, say,

  • make an alkyl iodide in synthesis,

  • and you don't go ahead and look upfield,

  • you may miss your carbon and say, oh my god,

  • I can't see my carbon.

  • There's a parameter in the NMR called sweep width

  • that you adjust.

  • That sweep width is the spectral window that you look at.

  • If you have a compound that you expect to be very far downfield

  • or very far upfield, this is true in the proton NMR as well,

  • if you have a compound that you expect to be very far downfield,

  • or very far upfield, you can just increase your sweep width,

  • the field width, by whatever number of hertz you like.

  • You type in the parameter SW, get a bigger spectral width

  • and then you can see, for example,

  • your carbon that's attached to iodine.

  • Same thing with protons attached to iodine.

  • I once took a spectrum of HI.

  • I was doing a de-protection reaction in an NMR tube.

  • And the HI, the peak was at negative 10 ppm

  • in the proton NMR, which was absolutely wild.

  • >> Why, like in the bromine one,

  • that at first it puts it more downfield.

  • And then when you get three bromines,

  • then all of a sudden now it pushes it back upfield?

  • >> I think what it is --

  • so the question is why are two bringing it more downfield.

  • Obviously, you have the inductive effect pulling away.

  • I think by the time you get to three you just have

  • that carbon completely surrounded

  • by the electron cloud.

  • So now you just have electron clouds sticking

  • out in all directions.

  • And in four, it's just all around it.

  • These are, in part, relativistic effects.

  • When you're doing MO,

  • when you're doing molecular orbital calculations

  • on heavy atoms, you actually need

  • to have special terms taking

  • into account relativistic effects of the orbitals.

  • All right, so one of the take home messages

  • from this is there are a lot of factors that go

  • into carbon 13 chemical shift.

  • And for this reason, carbon 13 chemical shifts are very rich

  • and very valuable.

  • And being able to predict them can lead you to lots

  • of useful information on structure that can be extracted,

  • sometimes can give you a problem-solving tool

  • that you might not otherwise have.

  • And so what I'd like to do now is to show you some

  • of these problem-solving tools and their applications.

  • So we'll talk a little bit now

  • about C13 chemical shift prediction.

  • And we're going to talk

  • about three different ways of doing this.

  • One is what is called empirical additivity relationships.

  • That's the simplest.

  • [ Writing on Board ]

  • The idea there is we're going to go and say okay,

  • the effect of an ortho substituent is this

  • if it's an oxygen.

  • The effect of a meta substituent is this if it's a chlorine.

  • The effect of an alpha substituent is this.

  • The effect of a beta substituent is this.

  • The effect of a gamma substituent is this.

  • And we're going to add up all of those effects.

  • If you use ChemDraw and you have one

  • of the more advanced versions of it

  • that does chemical shift prediction,

  • that's exactly what it's doing.

  • Those calculations also for all sorts of systems, aliphatic

  • and aromatic systems, are shown in your book

  • in your structure determination, the orange book

  • that you have for the course.

  • You know, the supplemental book.

  • And you can do it for all sorts of system.

  • And they do it to a limited extent in Silverstein.

  • And I'll show you that just for aromatic systems.

  • Unfortunately, the ChemDoodle program,

  • which tries to do chemical shift prediction,

  • thus far they're still very much in development.

  • But they've thus far got it wrong.

  • There is also a small CD, Windows only,

  • that goes with the orange book in the course

  • with the supplemental book.

  • That has some limited chemical shift prediction based all

  • on empirical additivity relationship.

  • Another way of doing this is based on databases

  • where you come up and rather than simply saying, all right,

  • we're going to have alpha effects and beta effects

  • and gamma effects, you have a training set.

  • And then from that training set,

  • or that set of reference compounds,

  • you find parameters that fit them.

  • And then you can extrapolate.

  • So you first interpolate to encompass all

  • of the compounds in the set.

  • And then you can extrapolate to encompass other compounds.

  • And so there are generalized databases.

  • And then there are also specific ones.

  • So for example, for various stereochemical problems.

  • And the third way of doing this is going to be

  • by electronic structure calculations.

  • And we're not going to do this in the course,

  • but I'll show you an example at the end.

  • Basically, since carbon 13 chemical shifts come

  • from electron density,

  • if you can calculate the molecular orbitals

  • at an appropriate level of theory to figure

  • out how the electrons are distributed in your molecule,

  • then you can figure out the electron density

  • around any carbon, and hence, its shielding

  • from the applied magnetic field.

  • All right.

  • Let's take a look at these various methods.

  • And I'll start you with a simple page right out of Silverstein.

  • And Pretch has a -- the orange book,

  • Pretch, has a similar table.

  • Ooh, that doesn't look good.

  • Ah.

  • Helps to plug it in.

  • All right.

  • So this should be on your handout.

  • And as I said, this is just one of the many tables in Pretch.

  • But a tool is so much more useful

  • when you actually have used it and know how to use your tool.

  • All right.

  • So this is a table of empirical additivity relationships

  • for substituents on a benzene ring.

  • And literally what was done to make this table was,

  • as we said, benzene's 128.5.

  • So somebody took a spectrum of phenol

  • and said the carbon that's directly attached to the OH

  • of phenol is shifted downfield by 26.6 ppm --

  • or 20, what did I say?

  • Yeah, 26.6 ppm.

  • The carbon that's ortho is shifted upfield by 12.7.

  • The carbon that's meta is shifted downfield by 1.6.

  • And the carbon that's para is shifted downfield by 7.3 ppm.

  • All right.

  • let's take a look at the following compound,

  • and then I'll give you two spectra.

  • And I'll show you how we'll use these.

  • So, let us imagine for a moment that we were trying

  • to distinguish the following compound from other isomers.

  • So let us suppose for a moment that we were trying

  • to distinguish, say, this compound, 2, 4-diochlorophenol

  • from this isomer, 3, 4-dichlorophenol.

  • That would be a tough one to do based on coupling patterns

  • because -- in the proton NMR.

  • Because if you look at the proton NMR,

  • you would say we would expect to have one proton

  • that is split by an ortho coupling.

  • So we would expect it to be a doublet.

  • And its J value would be somewhere on the order

  • of about 7 or 8 hertz.

  • We would see another proton.

  • What would you expect this proton here to appear as?

  • As a doublet of doublets with J values of what?

  • >> 7-ish and it's --

  • >> Metacoupling?

  • >> 3?

  • >> 3?

  • >> 3. So we'd expect this to be a doublet of doublets

  • with a big J of about 7-ish and a small J of 3.

  • And this one we would expect to be?

  • >> 3 [inaudible].

  • >> 3 as a doublet.

  • And you would expect this exact same pattern of coupling here.

  • A doublet of doublets with a big J and a small J. A doublet

  • with a big J and doublet with a small J.

  • So we wouldn't be able to tell them apart.

  • But let's look at this compound here and use these principles.

  • If we want to calculate the chemical shift for,

  • let's just take -- I'll just do three of the carbons here.

  • So if we were to want

  • to calculate the chemical shift here, we would start

  • with the base value of 128.5.

  • And then we would say, okay, we have an ortho alkoxy

  • and we'd look in our lookup table

  • and we'd say -- or an ortho hydroxy.

  • And we'd subtract 12.7.

  • So minus 12.7.

  • Now, we would then look at the effects --

  • can we see from this angle?

  • Should I move these calculations over?

  • Let me move this over.

  • 128.5 is our base value.

  • Then we have an ortho hydroxy, and so we'd take away 12.7.

  • Now we're going to look at the effects

  • of chlorine for this compound.

  • And so we have two meta chlorines.

  • So we'll go to our lookup table and we'll look

  • at the effect of a chlorine.

  • So I'll draw a line here.

  • And so, now we have two meta chlorines.

  • So we have a meta chlorine over here.

  • And that's going to have an effect of plus 1.0

  • and we're going to have --

  • I guess I should make my numbers line

  • up if I'm going to run a tally.

  • And we have another meta chlorine,

  • and that effect is plus 1.0.

  • So we tally all of those up and we predict 117.8.

  • And now I can do -- so that's for this carbon here.

  • Now I can do this next carbon.

  • So you help me out with this calculation for this carbon.

  • So what do we use as our base value?

  • >> 128.5?

  • >> 128.5. And then what do we do for the oxygen?

  • [ Inaudible Audience Responses ]

  • >> One -- so we add.

  • So the oxygen is meta OH, so we add plus 1.6, okay?

  • And then what do we do next?

  • James?

  • >> Add .2.

  • >> .2 for?

  • >> The meta chloro?

  • >> For?

  • >> Sorry, the ortho chloro.

  • >> Ortho chloro, so we do O chloro is -- you said plus .2?

  • >> Plus .2.

  • >> Plus .2.

  • And then somebody else, what do we do last?

  • >> Subtract 2 for the para?

  • >> Subtract 2 for a para chloro.

  • And so we tally all of that up and we predict 128.3.

  • All right.

  • I'm not going to do any more.

  • We could do it for all six of them.

  • But, let's flip to the next page in your handout.

  • And I grabbed from the Aldrich Library of Spectra the proton

  • and carbon spectra of the two compounds.

  • And we're just going to look at the carbon spectrum.

  • And the thing that jumped out at me in the carbon spectrum

  • for a difference that was really cool was in the carbon spectrum

  • of the first compound.

  • Remember the quats are small and the CH's are big due

  • to relaxation, differences in relaxation time

  • and nuclear Overhauser effect.

  • And what jumped out at me immediately was

  • that here we have two of the CH's close to 130.

  • And one of them -- so two basically downfield of 120,

  • and one upfield, so at about 115.

  • And what jumped out at me here was the dramatic difference

  • where here we have one at about 150, about 130.

  • And then two of them at the range between 110 and 120.

  • And of course, we could do this for the quats.

  • There are dramatic -- there are substantial differences as well.

  • You can see, here are your quats.

  • So all I did in preparing

  • for today's discussion was I just calculated

  • for the two compounds.

  • And so we did 118.5 and -- let's see, was that right?

  • 100 -- wait.

  • Oh, what did I do?

  • I did my -- yeah.

  • I don't know how I managed to tally things up wrong here.

  • Wait, 100 and -- let's see, 128, 129, 130.

  • 118. -- wait, 118.

  • -- what was it.

  • >> No, that's right.

  • It'd be 119.2.

  • Oh, sorry, that's minus.

  • I'm sorry, I got lost.

  • >> Wait a second.

  • Anyone have a calculator?

  • >> You were right.

  • [inaudible].

  • >> Was it 128.5?

  • Wait. 117.8.

  • Okay. So I get -- I don't know what I was writing down here.

  • 117.8 and 128.3.

  • And over here -- I think -- what did end up doing?

  • 130.7. And then over here 115.2 and 117.5

  • and I think here 131.7.

  • All right.

  • So the point of this is even though it might be very hard

  • looking at the proton NMR to assign which isomer we have,

  • by looking at the carbon NMR,

  • you can say oh wait a second, okay.

  • The first isomer that has the 130.7,

  • 128.3 and 117 matches this spectrum.

  • And we could do this with the quats as well.

  • So this spectrum corresponds to the first isomer.

  • The second isomer that has the 115.7,

  • 117.5 and 131 matches this spectrum.

  • And so imagine for a moment

  • that you're doing an electrophilic-aromatic

  • bromination of phenol, you want

  • to determine whether you got the product you expect.

  • You know you have a dichlorophenol.

  • You don't know regiochemistry.

  • You can very quickly check yourself.

  • And the good news is these types of relationships generally good

  • to let's say plus, plus or minus let's say 3

  • to 5 ppm on the average.

  • So the point is that you can,

  • with very little difficulty, tell things apart.

  • All right.

  • There are many, many more sophisticated ways

  • of doing this.

  • As I said, you can do this for aliphatic compounds as well.

  • With aliphatic compounds you've got to add up alpha effects,

  • beta effects, gamma effects.

  • There's even a little bit of delta and epsilon effects.

  • So you have to look at the whole molecule.

  • There's steric effects.

  • You can do this easily by pencil and paper, but it's even easier

  • to do it with computer software.

  • And as I said, even ChemDraw

  • in the more sophisticated versions incorporates this.

  • There are other empirical additivity relationships

  • that are out there.

  • But in addition to that, there are databases

  • that have training sets.

  • So for example, this is just one poster I pulled that's a

  • database approach.

  • And what they were looking at was how well their system

  • that was trained on a training set would predict the C13 NMR

  • of taxol.

  • And in general, their values were good to plus

  • or minus 1 ppm on the average.

  • And so for example, they weren't able to distinguish --

  • here's another one that would be hard to do by NMR.

  • They were able to use their database approach.

  • And here we have two tautomers.

  • And so if you synthesized this molecule, you might not know

  • which tautomer you have, whether you have the molecule

  • with the carbonyl or the molecule with the hydroxy.

  • And it's not so easy that you can take an IR spectrum

  • because by that point, you end

  • up with very strong stretches associated with the CN bonds

  • of the pyridine tautomer.

  • And so the IR spectrum is not clearly going

  • to scream carbonyl at you.

  • The C13 NMR, the difference

  • between what you would expect here

  • and what you would expect here

  • for the carbonyl is small enough.

  • You can't just look at say obviously I have the pyridone

  • or obviously I have the other compound.

  • But, in fact, they can go ahead and calculate

  • to within a couple of ppm.

  • So they were able to experimentally measure the value

  • and then say, okay, what's the deviation?

  • And here they're getting an average deviation

  • of about 2 ppm.

  • In other words, some of the shifts will be off by 3 ppm,

  • some of the shifts will be off by 1 ppm.

  • But on the average, they'll be good to within about 2 ppm.

  • All right.

  • And so that's an example of a database approach.

  • And in general, their method's good to about 3 ppm.

  • Many times you encounter specific problems.

  • You were working on a research project

  • that involves a specific class of molecules.

  • For example, the Rychnovsky group, here, and the Kishi group

  • at Harvard are very interested in 1, 3-diols.

  • 1, 3-diols are important classes of natural product.

  • They come from biosynthetic pathways that give rise

  • to molecules of many different structures.

  • And the big issue becomes determining stereochemistry.

  • The Rychnovsky group noticed after the fact --

  • they were working with 1, 3-diol acetonides.

  • And the question was do they have a syn-diol

  • or an anti-diol, right?

  • So a syn-diol would be a 1,

  • 3-diol where the two OH's are like so.

  • And an anti-diol would be one in which they are like so.

  • And the beautiful thing is if you're a clever graduate student

  • or a clever professor and you keep your eyes open and look

  • at your data, you say hey, we've got a pattern here.

  • And that pattern tells us something.

  • And that pattern means something.

  • And when they went back and looked at all

  • of their C13 NMR spectra, of the 1, 3-diol acetonides

  • that they had prepared, they saw something very interesting.

  • The methyl groups of the 1, 3-diol acetonides,

  • when the diols were syn, appeared split at about

  • into two sets of diastereotopic peaks.

  • One peak at about 20 ppm, the other peak at about 30 ppm.

  • And when they had the syn-diol acetonides,

  • both of the methyl groups appeared at about 25 ppm.

  • In other words, you could now take a diol

  • of unknown stereochemistry, make the acetonides and just

  • by looking at the C13 NMR spectrum, say oh,

  • that's a syn-diol or that's an anti-diol.

  • It's hard to do that otherwise.

  • That's extremely valuable.

  • Why does this occur?

  • It's that gamma effect again.

  • When you make the syn-diol, you end up with a chair.

  • And in that chair, one of your methyls is axial and one

  • of your methyls is equatorial.

  • And remember, the axial one gets steric compression.

  • So it is shifted upfield.

  • The equatorial one doesn't get that steric compression,

  • so it's shifted downfield.

  • On the other hand, when you make the anti-diol acetonide,

  • then you can go ahead and both your methyl groups are sort

  • of in the middle.

  • Neither of them really has -- you get a twist conformation.

  • Neither of them has unusual effect.

  • And they end up in the middle.

  • So just by a quick C13 NMR with this bit of knowledge in hand,

  • you can go ahead and assign the stereochemistry.

  • The Kishi group at Harvard has taken this type

  • of approach with their diols.

  • And this is a really, really cool use of synthesis.

  • So they have diols where you have alternating hydroxy

  • and methyl groups.

  • These are compounds that occur in a variety

  • of different natural products.

  • And so they wanted to be able

  • to tell the stereochemistry of these patterns.

  • So they went ahead and they said all right, we've got lots

  • of natural products with this type of unit

  • where you have a diol, a methyl --

  • an alcohol, a methyl, an alcohol, a methyl.

  • And so what they did was make a training set

  • of all the possible stereoisomers.

  • So they made ones in which they were all syn,

  • in which one is anti, in which two are anti

  • in different positions, and so forth.

  • And they found that there were characteristic, that each carbon

  • in their training set -- so they used this molecule

  • as the training set --

  • each molecule in the training set had an average chemical

  • shift of a certain position.

  • And so then they took the average value and they said

  • if you compare it to the average value, you have a fingerprint

  • where certain ones are shifted upfield,

  • certain ones are shifted downfield.

  • And each of the possible stereoisomers has their

  • own pattern.

  • So that means you can then take an unknown compound based

  • on this training set and the database

  • and take an unknown compound and say, ah!

  • The pattern matches this stereochemistry.

  • All right.

  • Last example I want to talk about.

  • And this was really cool.

  • So this is -- this is all calculating based

  • on known information.

  • We started with the simplest thing.

  • You just took phenol and you said okay, what's the effect

  • of an ortho substituent?

  • What's the effect of a meta substituent?

  • What's the effect of a para substituent?

  • Took chlorine, took chlorobenzene, did the same.

  • And then say you could predict another compound.

  • These diols are also examples where you have a training set.

  • But for really funky and unusual compounds

  • that have very unusual rings and strained rings,

  • there may not be a good model out there.

  • And there was a really cool compound that was synthesized --

  • end quote -- by a single researcher in San Diego

  • who published this on his own in "Angewandte" with no co-authors

  • and no university affiliation.

  • His name was James J. La Clair.

  • Still is James J. La Clair.

  • And he published this synthesis that looked really cool

  • because the molecule he said he synthesized was this,

  • hexacyclinol.

  • One that Phil Baron, I think, would appreciate

  • as a challenging target.

  • And like a student who maybe didn't quite get an A in Syn.

  • 1, his synthesis had some stuff that was funny about it

  • in addition to him being the sole author.

  • There were steps that just didn't seem to make sense.

  • And so the buzz in the community was something wasn't right here.

  • Now, nobody wants to tell their entire lab, hey guys!

  • Let's reproduce James J. La Clair's experiments

  • and see if it really works.

  • That's several graduate students

  • and dissertation's worth of work.

  • That's not exactly a great job to be doing.

  • So, Professor Rychnovsky used electronic structure

  • calculations to see how well he could first predict the chemical

  • shifts of a known molecule.

  • So he took a known compound, elisapterosin,

  • and did these electronic structure calculations

  • and plotted a graph of the actual chemical shift

  • versus the observed chemical shifts.

  • And as you could see, most of the chemical shifts calculated,

  • the calculated ones and the observed ones matched

  • within a couple of tenths of a ppm.

  • So in other words, here was have a match of plus

  • or minus 2 ppm on the average.

  • All right.

  • So confident, confident of his methodology,

  • he then took the structure of hexacyclinol --

  • and I'll put this in quotes here as you'll see in a moment --

  • and found a terrible match.

  • The vertical scale on this graph is from 0

  • to 5 parts per million.

  • You look at the scale on this graph,

  • it's from 0 to 25 parts per million.

  • The deviations on the average are within about 7 ppm.

  • It's many of the shifts are way off.

  • So that's the average.

  • Now, he then went ahead and thought,

  • this guy got the structure wrong.

  • And he had a hypothesis of what --

  • so not only did he think that the synthesis was a fraud,

  • the thought that the structure wasn't even right

  • because it didn't match.

  • So he reinterpreted the published structure

  • of the molecule, the one from the Journal of Natural Products

  • where they had isolated this molecule,

  • and thought it was this structure,

  • reinterpreted the data,

  • and considering biosynthetic factors,

  • came up with what he believed the correct structure to be was.

  • And there the structure matched.

  • And he did two different conformers of it.

  • And there, for this structure, matches within plus

  • or minus 2 ppm average.

  • By 2 ppm average, I mean RMS deviation,

  • root-mean-square deviation.

  • So on the average within 2 ppm.

  • So, the answer became, not only did James J. La Clair --

  • what would is say?

  • Cheat, forge a synthesis of the molecule,

  • but he forged a synthesis of the wrong molecule.

  • And it was an impossible synthesis.

  • Anyway, it was pretty cool

  • and the Rychnovsky group was really happy for a while.

  • >> Who is this guy?

  • And what is he, exactly?

  • Is he known?

  • >> A very unusual character.

  • So the other thing that was interesting, and it's all

  • out there, is if you look at his spectrum of hexacyclinol,

  • hexacyclinol is a terpene, the has methyl singlets in it,

  • which are big, tall peaks.

  • If you look at his spectrum of his synthetic hexacyclinol,

  • the methyl groups are missing C13 satellites,

  • those little side peaks spaced 125 ppm, 125 hertz around there

  • at .5 percent of the height,

  • which for a 20 centimeter height peak you can actually see

  • in a spectrum.

  • So he fabricated this synthesis

  • and his spectrum out of whole cloth.

  • [ Multiple Speakers ]

  • >> I assume this person had his PhD [inaudible].

  • >> It wasn't part of his PhD -- if you go and you earn your PhD,

  • you're a legitimate, and then later

  • on as an independent scientist,

  • working at the Xenobe Research Institute,

  • your own research institute, publish absolute nonsense

  • in the literature, yeah, I don't know.

  • >> How did he get published?

  • >> Yeah, how did he published?

  • >> Ah, interesting question.

  • A very controversial story. ------------------------------dc9c2cfd829b--

>> All right.

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