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  • >> All right, today we're going to talk

  • about two different things.

  • We talked about proton chemical shifts and today I want

  • to talk briefly about C 13 NMR shift.

  • We'll talk more about them later on but then we're going

  • to start, we're going to spend a reasonable amount

  • of time talking about spin-spin coupling and in order

  • to understand this we really have to understand the concept

  • of chemical equivalence which ties into concepts of symmetry

  • and stereochemistry and confirmational analysis

  • and it's really beautiful, chemical equivalence

  • and so we'll be talking about chemical equivalence

  • and spin-spin coupling.

  • We're actually going to be spending a good deal of time

  • because there's a lot to understand

  • and as you can already see from the problems people are saying,

  • hey what's going on here?

  • And these problems these very simple molecules have all sorts

  • of cool issues of spin-spin coupling and all sorts

  • of cool issues of stereo chemistry

  • so we're actually going to spend a number of lectures on them.

  • Next time we're going to develop a concept called magnetic

  • equivalents which is an amplification

  • on chemical equivalents but it's too much to take in on one

  • and then we're going to spend a couple of times talking

  • about details of spin-spin coupling.

  • All right, what do I want to say

  • about carbon NMR spectroscopy first of all?

  • All right, if proton NMR was difficult

  • because you have a very small population between your alpha

  • and beta states carbon NMR is even worse and first you know

  • that carbon 12 doesn't have an NMR spectrum.

  • It's not active.

  • It doesn't have a magnetic dipole

  • and we only have one percent

  • or more specifically 1.1 percent C 13, so most of your molecules,

  • your small molecules don't contain any C 13.

  • For small molecules some of them contain one C 13.

  • Now things get worse.

  • The magnetogyric ratio for C 13 is only a quarter of that

  • of the magnetogyric ratio for a proton.

  • Now remember what the implications are of that.

  • That means that you get roughly a quarter, I'll use tilde

  • to say approximately a quarter of the Boltzmann difference,

  • say difference in Boltzmann distribution in alpha

  • and beta states, so already we've got fewer nuclei

  • that can flip up.

  • To put it another way a 500 megahertz spectrometer gives you

  • a carbon spectrum at 127.5 megahertz, a hair over a quarter

  • because the magnetogyric ratio is a hair over a quarter.

  • Things get worse than that.

  • Your dipole is only a quarter as strong.

  • Guess what?

  • If you want to generate an electric current

  • like a generator you want a big hunking magnet

  • to spin in a coil.

  • Proton is a little magnet but a carbon is a tiny magnet

  • because it's got a quarter of the magnetic dipole.

  • So you're damned again and then you further get damned

  • because the procession rate is also a quarter

  • since for a proton at that same 117,500 gauss magnet you're

  • processing at 500 times per second.

  • For carbon you're only processing at 125 million times,

  • did I say thousand, million times per second.

  • So that also gives you a quarter as much electricity in the coil.

  • So you've got 1.1 percent and a quarter of a quarter

  • of a quarter which means you're only 1/5,800th as sensitive.

  • Question.

  • >> Is the procession, are you talking

  • about the limiter frequency?

  • >> The limiter frequency, yeah.

  • So if you take a magnet and just spin it in a coil,

  • if you spin it faster you get more voltage.

  • If you spin it twice as fast you get twice as much voltage.

  • If you take a magnet that's twice as big you get twice

  • as much voltage and so for all of these reasons and you've got

  • in addition to a smaller magnet you've got fewer of them

  • because you've got, even

  • if you give a 90 degree pulse you get only a quarter

  • of the magnetic dipole from having only a quarter

  • as many nuclei going down into the X, Y plane.

  • >> Is that a quarter of all or is

  • that a quarter of what's protons?

  • >> It's a quarter-- compared to protons.

  • So carbon is much less work, much less sensitive technique

  • than proton NMR spectroscopy.

  • Now there are few redeeming features

  • so one thing that's redeeming is we typically do proton

  • decoupling, so normally a carbon would be split by all

  • of the protons so for example,

  • the carbon in ethanol would be split into a quartet

  • in the carbon and the methyl group of ethanol, would be split

  • into a quartet by the three hydrogens that are attached

  • to it and then it would be further split by the hydrogens

  • over on the methylene carbon.

  • But what we do is we irradiate the proton

  • so all the carbon you're going to see,

  • virtually all the carbon you're going

  • to see is called proton-decoupled carbon.

  • That flips the spins of protons rapidly

  • which means the carbon doesn't see them as spin up or spin

  • down so the carbons appear as singlet.

  • Well that's good because that means all your carbon signal is

  • gathered in one peak so that gives you more intensity.

  • Now the other thing is when you do that, so that leads

  • to singlets which is good, sharper,

  • bigger and the other thing it leads

  • to is what's called the Nuclear Overhauser effect

  • and we'll talk more about this as a technique

  • but the basic principle of the Nuclear Overhauser effect is

  • that by perturbing the alpha and beta states

  • of the protons you end up enhancing the difference

  • in Boltzmann population between the alpha and beta states

  • of the carbon, so that too gives you a bigger signal,

  • so all of this leads to a better signal

  • than you would otherwise get in a proton non-decoupled carbon.

  • All right anyway, suffice it to say now days it's easy

  • to collect a carbon NMR spectrum.

  • It typically will take more sample

  • so you can collect the proton NMR sample on strychnine

  • and use a milligram of material or even tenths of a milligram.

  • For a carbon you might want to put 30 megs

  • in the NMR tube if you have a chance.

  • You could do it at ten.

  • You could do it at a milligram but it takes a lot more time.

  • And remember if you have one milligram

  • versus ten milligrams it's going to take a hundred times as long

  • to collect the same signal to noise ratio which means

  • if you're in your research lab

  • and you have some sample it actually makes sense

  • if you're trying to collect a carbon spectrum

  • to weigh your sample or at least be cognizant of how much you put

  • in your NMR tube because you want

  • to get your spectrum quickly and you want

  • to get good signal to noise ratio.

  • It also makes sense when you're filling your NMR tube

  • to only fill it with the appropriate amount for the coil.

  • The coil on our [inaudible] is three and a half centimeters

  • or requires a three and a half centimeter high sample.

  • That's a half a mil, so if you dissolve your sample don't

  • dissolve it in a mil and a half.

  • Don't try to be clever and dissolve it in .3 mils

  • because then you miss up the shims on the spectrum

  • because you get flux lines at the end of the sample.

  • So anyway that's the way to get good spectrum.

  • All right, I want to talk about where the peaks show

  • up so carbon NMR spectrum,

  • the carbon NMR spectrum has a big range typically

  • from about zero to about 200 or 200

  • and change parts per million, 220, 240 parts per million.

  • Aliphatics show up at about 10 to 40 so on that big range

  • of 200 or so ppm that's in the up field region.

  • Carbons next to an electron withdrawing atoms show

  • up down field but it's not quite as pronounced so carbon next

  • to a halogen you might even see it in this range,

  • carbon next to a nitrogen.

  • It's going to be sort of at the end of that range

  • but by the time you're next

  • to an oxygen is an electron withdrawing group I'd say 50

  • to 70 for a carbon next to one oxygen

  • so that's sort of stands out.

  • Alkines aren't that common but remember I said there's

  • about two and a half parts per million,

  • 2.2 parts per million maybe for a typical alkine CH.

  • For an alkine carbon it's about 70 to 80 ppms

  • so that kind of stands out.

  • All right, alkenes and aromatics whereas

  • in proton NMR the alkenes show up a little bit more up field,

  • 5 to 6 and the aromatics a little more down field 7 to 8,

  • alkenes and aromatics are all sort of lumped

  • in at about 110 to 150.

  • Beyond about-- and of course these are all typical values.

  • If you put in oxygen on an aromatic like anisole or phenol

  • where you put an oxygen directly on a carbon that carbon might be

  • at 160 parts per million.

  • If you have a very electron donor--

  • we'll talk more about specific shifts but you could easily

  • if you have high electron donations say an ortho carbon,

  • you could easily be a little below, 115 ppm.

  • All right carbonyls, esters

  • and carboxylic acids you get a little bit of a resonance affect

  • so they show up a little bit less far down field than some

  • of the others let's say about 170 to 180 parts per million.

  • Aldehyde show up further down field, let's say about 190

  • to 200 ppm and ketones I'll say are RCOR prime for ketones,

  • let's say about 205 to about 220 ppm.

  • Bless you.

  • I just want to give you one little handout.

  • By the way, if you ever miss your handouts

  • or misplace them I put them up on the web

  • with the video part course so you can always go ahead

  • and download the handouts.

  • All right so just as I had my-- does anyone else need one?

  • Just as I had my little pigeon drawing of my take of what

  • to look at in reading a proton NMR spectrum I have my little

  • pigeon drawing of what to look

  • at when reading a C 13 NMR spectrum.

  • In other words, this type of region is aliphatics.

  • Here you have a carbon next

  • to a significantly withdrawing group like an oxygen.

  • Here's your alkene aromatics.

  • Here's your esters and acids then you go

  • to aldehydes and ketones.

  • So to a large extent it's sort of like H 1 NMR

  • but with maybe a factor of 20 on the scale.

  • In other words most of what you're going to see

  • on a proton NMR is from zero to ten.

  • Most of what you're going to see

  • in carbon NMR is from zero to 200.

  • Carboxylic acid will be further out.

  • Ketones will be further out but that kind

  • of gives you my read on things.

  • All right this should give you the basics to start

  • to use carbon NMR in helping to analyze some

  • of the homework sets that we're going to get.

  • So one thing that you can get is a reading of things

  • like carbonyl groups in there and another thing you'll be able

  • to do is count up and see how many different types of atoms

  • because whereas in proton NMR you may have overlapping

  • resonances most often because the carbon spectrum is more

  • dispersed and peaks typically show

  • up as singlets most often you will be able to see one peak

  • for each type of carbon.

  • All right I want to talk now about spin-spin coupling.

  • And if I had to give you a very, very general way of thinking

  • about it the way I'll describe it and we're going to amplify

  • on this in today's talk, the way that I would describe it is

  • that protons that peaks are split

  • by adjacent protons that are different.

  • We'll later on amplify on the concept of adjacent talking

  • about two bond coupling, three bond coupling

  • and long-range coupling.

  • But today what I'd like to amplify on is the concept

  • of same and specifically different.

  • So let's start with an example that's very, very,

  • very intuitive, very, very, freshman,

  • sophomore rather chemistry.

  • Let's take the H1 NMR spectrum of chloro ethane

  • and so my little pigeon sketch of this spectrum

  • of chloro ethane would look something like this.

  • We have a one to three to three to one triplet somewhere

  • down field of three ppm or just a hair down field of three ppm

  • and then we have, I'm sorry, one to three to three to one quartet

  • and then somewhere just hair down field of one we have a one

  • to two to one triplet.

  • The triplet comes of course from the CH3 and I can say

  • that in this particular example CH3s are split

  • by the adjacent CH2s and I can say because the three hydrogens,

  • well let's talk about the CH2s.

  • The CH2s are split by the CH3.

  • The three protons of the CH3 group are the same

  • so the CH3 does not split each other

  • so I'll say the CH3 protons do not split each other.

  • And then I'll say in this particular case the CH2s do not

  • split each other.

  • In other words, in particular case the two hydrogens

  • of the methylene group are the same

  • so they don't split each other but more

  • in general I'll give this as an exception,

  • so if your compound has a chiral center in it

  • or the protons are otherwise diastereotopic then they will

  • split each other.

  • And so this except is something that we're going to be playing

  • with in today's lecture.

  • [ Silence ]

  • All right, before we come to this very important concept

  • of diastereotopicity let's tackle this basic notion

  • of the one to two to one triplet and the one to three

  • to three to one quartet.

  • So we have a triplet and it's one to two to one.

  • It comes from the CH3 and it sees the CH2

  • and the two protons act as little magnets in each molecule

  • and those little magnets can be spin up or one can be spin up

  • and one can be spin down or they both can be spin

  • down so you get three lines.

  • There's one way for them both to be spin up.

  • There are two ways for them

  • to be one spin up and one spin down.

  • And there's one way for them to be spin down and so

  • that gives rise to our one to two to one triplet.

  • For our quartet we have a one to three to three to one ratio.

  • The quartet of course comes from the CH2.

  • The CH2 sees the CH3 and all of their protons can be spin up

  • and some of the molecules there are or two can be spin up

  • and one can be spin down and there are three different ways

  • that that can occur or two can be spin down and one can be spin

  • up and again there are three ways that can occur

  • or they all can be spin down.

  • And we can generalize this idea to say if there are N equal

  • and I'm going to underline equal couplings lead

  • to N plus 1 lines.

  • So if there are four equal couplings you get five lines,

  • those five lines it follows Pascals triangle.

  • You can work out the statistics yourself or you can say

  • that those five lines end up being in a one to four to six

  • to four to one ratio and so forth you basically add the ones

  • above or you work out the statistics.

  • So for example, if we'd go

  • to isopropyl chloride now the CH here is split equally

  • by three methyl groups, so the CH appears as septet

  • and the ratio of the lines are 1 to 6 to 15

  • to 20 to 15 to 6 to 1.

  • And because the quartet, because the septet is going

  • to be very small compared to triplets and quartets

  • in a molecule unless you look hard you might not see these

  • lines on the outside, and in fact by the time you get

  • up to great big multiplets like septets and so forth,

  • octets often people will just report them as a multiplet

  • in reporting an NMR spectrum.

  • All right let's take a look at a case of what I mean by couplings

  • that may or may not be equal.

  • So imagine the methine group in [inaudible]

  • and specifically this methine group.

  • So this methine group is split by two methyls groups

  • and it's also split by this methine group and so depending

  • on whether the couplings are equal,

  • that is whether the coupling constants are equal it may be an

  • octet or it may be more complicated and I can tell you

  • from experience in this case it would be more complicated.

  • It would be in this case a doublet of septets

  • which you would probably report is a multiplet.

  • The OH of alcohols and I know had has already come

  • up on the homework.

  • The OH of the alcohol may or may not couple so the hydrogen

  • of isopropanol, the methine of isopropanol might appear

  • as a septet or it might be more complicated.

  • In order to couple this hydrogen has

  • to stick around in one place.

  • It has to stay attached to that proton

  • for a time that's roughly one over the coupling constant.

  • In other words, it has to stay attached

  • since coupling constants are typically on the order of oh,

  • about 7 hertz it has to stay attached for hundreds

  • of milliseconds or longer.

  • If that hydrogen exchanges on the order of tens

  • of milliseconds this methine is not going to see it

  • as either spin up or spin down.

  • It will see an average and you won't get coupling.

  • If this hydrogen doesn't exchange then we'll be

  • in the situation here and it will see it either as spin up

  • or as spin down and you will have splitting either

  • as an octet or as a doublet of septets or septet

  • of doublets depending on the relative Js

  • but we'll get into that later.

  • Typically, primary alcohols often exchange rapidly

  • so an alcohol at the end of a CH2 chain like you may have

  • in the homework problem usually will exchange rapidly,

  • secondary alcohols sometimes do, sometime don't.

  • They're more spherically hindered.

  • The exchange, the rate of exchange is going to depend

  • on a number of factors including the concentration of the sample

  • because the molecules can exchange by colliding.

  • It will depend on the amount of water in the sample.

  • There's invariably adventitious water in your NMR tube

  • about 20 millimolar concentration, 10, 20, 30,

  • millimolar concentration

  • and chloroform undergoes photo oxidation

  • to give hydrochloric acid or DCL in the case of CDCL 3

  • and that will rapidly promote exchange so if I want

  • to minimize exchange I'd typically will pass my

  • chloroform through alumina, not my sample but my chloroform

  • through alumina that I've dried over a flame and that will take

  • out the acid and minimize the amount of water in the air.

  • Okay, so alcohols are kind of a wild card on J coupling.

  • [ Inaudible student comment ]

  • Do you still see the proton?

  • In general, if the proton is exchanging

  • with other molecules you will see it and it will be there.

  • If the proton is exchanging with water and you have a little bit

  • of water on the sample you will see it and the water peak

  • at the weighted average of the two.

  • If it is exchanging with water and there's a lot

  • of water then it will become part of the water peak.

  • Also there are two different time scales.

  • One is the time scale for coupling.

  • The other is the time scale for being able to see a peak.

  • Remember I talked about the uncertainty principle before?

  • So in order to see coupling your exchange rate has to be,

  • you have to be able to make the observation

  • such that your line width is on the ordinary

  • of better than 7 hertz.

  • In order to not have swapping with water

  • which is very far away so water is typically 1.6 parts per

  • million in chloroform, remember I said an alcohol can be 1

  • to 5 parts per million, let's take 4 ppm ?

  • Those are hundreds or even thousands of hertz away.

  • A 4 ppm difference is two thousand hertz.

  • So in order to not have exchange with water

  • where you can see the alcohol peak it has to stay attached

  • for tens of milliseconds rather than hundreds of milliseconds.

  • Carboxylic acids are real bad boys in this regard

  • because they do exchange rapidly which is one

  • of the reasons the peak typically broadens

  • out in chloroform.

  • Alcohols often will stay attached.

  • Secondary amines are a big pain in the ass.

  • Secondary amines are almost impossible to see.

  • You have base catalyzed exchange mechanisms.

  • Primary amines tend to be a little bit better behaved.

  • Aromatic amines are well behaved.

  • So I'm talking aliphatic amines but if you're working

  • on an alkaloid project or some other project

  • and you have a secondary amine you really are in trouble.

  • >> So you said earlier the hydrogen

  • in the alcohol may or may not couple.

  • If that doesn't couple you can also assume that the proton bond

  • in the alcohol will not couple as well.

  • >> You mean yes.

  • >> If one doesn't couple the other one will.

  • >> Absolutely, coupling is mutual.

  • Now you could envision a circumstance

  • where you had J coupling but quadripolar broadening

  • so you could have a hydrogen on a nitrogen

  • that could be broadened to a point

  • where it was a broad singlet but the hydrogen

  • on the carbon could be split by it

  • and there the broadening would be quadripolar

  • but yes typically, so again let's talk in rules rather

  • than exceptions now, typically it's going to be seen both ways.

  • Sometimes what will happen is you may for example,

  • if you have a very complex coupling pattern,

  • if you do have coupling you may just see this guy

  • as a complex multiplet and then see this guy as a doublet.

  • Why? Because this guy is splitting this guy but it's

  • such complex splitting that you can't actually discern what the

  • splitting pattern is.

  • On the other hand it's the same coupling constant

  • but you only have one coupling and you split into a doublet.

  • Other questions?

  • These are good.

  • There are real important.

  • This is really understanding the real stuff you're going

  • to see in the trenches.

  • >> What about the methyl groups?

  • >> The methyl groups won't couple to this

  • because it's far away.

  • In general, coupling is going to be through two

  • or three bonds although on the case of double bonds

  • and triple bonds you can get four bond couplings.

  • There's a very good appendix in the back of your book.

  • I forget whether it's appendix A or appendix F

  • at the back of Silverstein.

  • What was the appendix?

  • Somebody looked it up.

  • Anyway, that is a very good place to start but in general,

  • unless you have long-range coupling,

  • we'll get to that in the lecture too.

  • F, unless you have long-range coupling

  • in general coupling is going to be through two or three bonds.

  • All right, let us now tackle the concept of chemical equivalents.

  • Chemical equivalents is the first level of sameness

  • that I was talking about.

  • Remember I said that protons only couple to other protons

  • if they are different?

  • Two protons are the same are chemically equivalent

  • if they exchange by a symmetry operant, can be exchanged

  • by a symmetry operation or rapid process.

  • [ Silence ]

  • For example, rapid rotation about a bond.

  • All right the other caveat and this ties

  • into something I said last time.

  • Virtually all, you could say all rotations about SP 3,

  • SP 3 carbon bonds are rapid at room temperature.

  • [ Silence ]

  • I'll say virtually always rapid at room temperature.

  • That's a hint.

  • If you see two hydrogens splitting each other

  • and they're both on SP 3 carbons attached

  • to each other chances are you need to think deeper

  • and it's not, oh, there's some slow process.

  • All right, one corollary to this is chemically equivalent protons

  • have the same chemical shift.

  • So this thing here is kind of like the golden rule

  • of splitting, the basic do unto others

  • as you want others to do unto you.

  • If you keep this in mind you're going to be very, very well set

  • on all of the details.

  • Let's talk about specific way of thinking

  • about what's the same and what's different.

  • One way that I like to do it,

  • you can do it by a couple of ways.

  • If you're good with symmetry you can just go ahead

  • and say all right do we have a symmetry operation

  • that interchanges these two hydrogens say

  • in diphenylmethane?

  • You can say major change by reflection

  • and therefore they are chemically equivalent

  • but we're going to get into some cases

  • that least the first time you see them are tricky

  • and another way to do it is just

  • to perform a little thought experiment and say all right,

  • what is their topological relationship to each other?

  • And the way you do this thought experiment is you replace one

  • proton by a deuterium and the other proton by a deuterium

  • and then you ask yourself what is the relationship

  • between these two molecules and in the case

  • of this particular thought experiment these two are the

  • same and since they're the same topologically we call those two

  • protons homotopic and homotopic protons are

  • chemically equivalent.

  • Let's try another molecule.

  • Obviously, obviously, obviously these are trivial.

  • So let's take instead of diphenylmethane,

  • let's take ethyl benzene

  • and again we'll consider the methylene group.

  • The two hydrogen groups of the methylene group still exchange

  • by reflection so they meet the criterion there.

  • They're chemically equivalent.

  • Yeah.

  • [ Inaudible student comment ]

  • . You can't tell them apart

  • so they have the same chemical shift and two protons

  • that are chemically equivalent are the same

  • and don't split each other

  • so chemically equivalent protons don't split each other.

  • Now later on we're going to talk

  • about a concept called magnetic equivalents

  • and what you'll see is

  • that although chemically equivalent protons,

  • well actually I'm going to hold off on this statement here.

  • Right now I'm just going to leave it as there are two types

  • of sameness and chemically equivalent protons don't split

  • each other if they're also magnetically equivalent

  • but we're going to get to that later.

  • For now the simple rule is chemically equivalent protons

  • are the same in a certain way.

  • All right let's try our little gadonk experiment here.

  • We replace one by a deuterium and we replace the other

  • by a deuterium and now these two molecules are they the same?

  • No. What are they?

  • Enantiomers, so now they are enantiomers

  • which means topologically the protons are enantiotopic

  • and enantiotopic protons are also chemically equivalent.

  • All right so now let's come to a molecule

  • where we have a chiral center in the molecule

  • and since we've been building on this idea

  • of phenyl groups I'm going

  • to give us the molecule phenylalanine.

  • It doesn't matter whether I draw it racemic

  • which means we have two enantiomers

  • or whether I have one molecule.

  • Okay, these two protons are now no longer exchangeable

  • by reflection and no matter how rapidly you rotate they still

  • are not the same, so these two protons now,

  • the methylene protons are not chemically equivalent.

  • And let's try our thought experiment here again

  • because sometimes things can get more complicated

  • than you might anticipate and so we'll take one,

  • replace it with a deuterium and we'll take the other

  • and replace it by a deuterium

  • and we'll ask ourselves what the relationship

  • of those two structures is to each other?

  • What are they?

  • >> Diastereomers.

  • >> They're diastereomers so we say that the two hydrogens

  • of the methylene group are diastereotopic

  • and therefore they're not chemically equivalent.

  • [ Silence ]

  • All right, so here's a spectrum of phenylalanine

  • so we can see what's going on.

  • This is a spectrum taken in D2 O with DCL to dissolve it.

  • Deutero-HCl to dissolve it, so all of the OHs

  • and the NHs have exchange with deuterium

  • and are exchanging rapidly and for all intents

  • and purposes don't J couple.

  • So we see a peak for our HOD and then we see the rest

  • of it are carbons attached to hydrogen.

  • We see our phenyl group over here

  • in the seven to eight range.

  • We see our proton connected to the alpha carbon

  • over here just a hair down field at about a 4,

  • about 4.3 parts per million and then we see our beta protons,

  • the ones that are next to the phenyl

  • and we see them as two peaks.

  • Each of the beta protons appears and these are expansions here.

  • Each of the beta protons appears and these are expansions here.

  • Each of the beta protons appears at distinct positions.

  • They're not the same.

  • They're not chemically equivalent.

  • They could be coincident meaning appearing at the same position

  • but they are not and so they appear at different position

  • and what's more each of them is a doublet of doublets or DD.

  • Why? Because each of the protons splits the other one

  • and is split by the alpha.

  • So we have both a J2 HH.

  • That's a two bond coupling since you go one two bond to get

  • from these two beta protons to each other

  • and we have a three bond coupling

  • between the alpha proton in each of the beta protons.

  • And the alpha proton is also a doublet of doublets

  • so each proton is a doublet of doublets.

  • The alpha proton is a doublet of doublets because it's split

  • by the two beta protons and it's split

  • with different coupling constants.

  • The beta protons are coupling to the alpha proton probably

  • with about 6 and 8 hertz coupling constants

  • and we'll be talking more about these in a moment

  • or in a future lecture and the beta protons are coupling

  • to each other with about a 13 or 14 hertz coupling constant.

  • All right, let's talk about why we have

  • to keep our heads attached to us.

  • So I'm going to give us another example

  • and the example I'll give us is a molecule called acetal.

  • Acetal is acid aldehyde diethylacetal.

  • [ Silence ]

  • All right, so what do we have here?

  • We have a couple of ethyl groups and we have a methyl group.

  • The CH3s of the ethyl group appear here

  • as a triplet so this is OCH2CH3.

  • That looks pretty reasonable.

  • We have another methyl group it appears as a doublet.

  • That's our CHCH3.

  • That looks pretty reasonable.

  • We have our methine.

  • That's the one that's attached to two oxygens

  • so it's shifted down field.

  • It's at 4.6 parts per million.

  • It's a quartet because it's split by the methyl group

  • and finally we come to the CH2s

  • and the CH2s show up as two peaks.

  • And it takes a moment to wrap your head around this

  • and to wrap your head around this what you have to realize is

  • that these two protons are not chemically equivalent.

  • You can think of this a couple of ways.

  • They don't exchange by symmetry operation.

  • If you reflect through the plane you don't get the same thing.

  • Another way to think about it and I think the easiest way

  • to keep your head on straight is to think about it rigorously.

  • You imagine replacing one of them with a deuterium,

  • replacing the other of them with a deuterium and the molecules

  • that you get are diastereomers.

  • In other words these two protons are diastereotopic.

  • >> When you talk about an IR how you get

  • like an internal alkene sometimes you don't see the

  • stretching because there's like that pseudo symmetry.

  • Do you see the same thing with this

  • that if you have a stereo isomer like a stereo center

  • but it's like really far away?

  • >> Absolutely, great question.

  • Just because two protons are topologically diastereotopic

  • doesn't mean that they won't be coincidental and in general,

  • the further that they are

  • from the stereo center the more likely they are to be coincident

  • and behave as if they are equivalent.

  • Here we see them as well- defined peaks

  • at very different positions.

  • They're both doublets of quartets.

  • That's a DQ.

  • We'll talk more about the splitting pattern

  • about their being split with one big geminal coupling

  • of about 10 hertz, one two-bond coupling

  • and with three vicinal couplings, three,

  • three bond couplings of about 7 hertz.

  • All right, last thing I want to show you,

  • one last handout while worry on the concept

  • of chemical equivalents.

  • [ Silence ]

  • So here we have the molecule 3 methyl 2 butanol and what I want

  • to point out to you is

  • in 3 methyl 2 butanol the two methyls are not the same.

  • They're diastereotopic.

  • You can think of this a couple of ways,

  • one way is to think about,

  • remember the OH constitutes a chiral center

  • and you can do a little thought experiment and say all right,

  • I'll envision replacing one of the methyl groups

  • or the other methyl group with a deutero methyl group

  • or the other way is just to think about the topology

  • but when you look at it you say all right,

  • how many peaks do we see in the C 13 NMR?

  • One, two, three, four, five?

  • There are five types of carbon

  • because there are five non-chemically

  • equivalent carbons.

  • If you look at the proton NMR it's a little confusing at first

  • until you think about what's going on.

  • It's not hard to realize that this methyl group that's beta

  • to an oxygen gives are rise to this doublet.

  • Now we have two more methyl groups in the molecule and each

  • of those methyl groups is next to a hydrogen.

  • Each of those methyl groups appears as a doublet

  • but the doublets are very close.

  • In other words we have a doublet and we have another doublet

  • and so it looks like a triplet.

  • [ Inaudible student comment ]

  • Well you can discern it if you look carefully

  • because it's not one to three, one to two to one

  • but it's confusing so it's two doublets

  • but let me tell you a secret.

  • This is a 300 megahertz spectrum.

  • Remember I said the coupling constant is fixed in hertz.

  • The position is fixed ppm so my two doublets would look

  • like this at 300 and it would look like a triplet if I went

  • to 500 megahertz the two doublets would now separate

  • and I'd see a doublet and a doublet

  • and that would be one way you could proof it.

  • All right, I think that's what I want to say for today

  • about chemical equivalents.

  • We'll pick up next time talking about magnetic equivalents. ------------------------------23065aa15093--

>> All right, today we're going to talk

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B2 中高級

化學203.有機光譜學。第10講。13C NMR化學位移。 (Chem 203. Organic Spectroscopy. Lecture 10. 13C NMR Chemical Shifts.)

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