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  • >> I want to spend the next 3 lectures, 3 classes,

  • talking really closely about first order coupling

  • and the reason is that there's so much to be gained

  • by deeply understanding NMR spectra.

  • As I said, a lot of what one is going

  • to be doing is asking specific questions about stereochemistry

  • and being able to ask those questions is intimately linked

  • to understanding what's going on.

  • Also just in general for solving structures,

  • being able to read spectra, really read them at a level

  • that goes beyond the level of sophomore organic chemistry,

  • involves intimately understanding [inaudible].

  • So we're going to take a relatively slow path

  • through this.

  • In fact, we're going through the midterm exam only have 1-d

  • spectra on our exam so that we really focus

  • on understanding things.

  • So I want to start by kind of making the bridge

  • between last time's lecture where we talked

  • about magnetic equivalents and we talked

  • about non-first order systems and so last time was sort

  • of the bad and today is going to be the good.

  • So the bad is that I said a lot of the rules that you learned

  • in simple sophomore organic chemistry really

  • were oversimplifications.

  • There are very few systems that truly behave in the way

  • that you learned they should behave.

  • These are the first order systems.

  • So first order systems are anything

  • like AX systems, AMX systems, A2MX.

  • In other words, anything where coupled protons,

  • protons within a spin system are far apart in chemical shift

  • and if you do have 2 protons that are chemically equivalent

  • like we have in A2MX system

  • that those protons are both chemically equivalent

  • and magnetically equivalent.

  • We divided this and separated it from non-first order systems.

  • [ Writing on board ]

  • And these are systems in which you either have magnetically

  • inequivalent protons that are chemically equivalent

  • or you have protons that are similar in chemical shifts.

  • For example, a non-magnetically equivalent protons we saw,

  • for example, A, A prime, X, X prime systems and we talked

  • about just how ugly those systems could be.

  • Those were like the phthalate [phonetic] system

  • where I said no matter how far apart,

  • no matter how high a magnetic field you look

  • at dioctyl phthalate or ortho dichlorobenzene is never going

  • to get better than this complex pattern of lines

  • and then I said we have other systems like AB systems

  • where the protons are similar in chemical shift and ones

  • that are related to this, for example, ABX systems.

  • The good news about many of these types of systems is

  • that many of these non-first order systems behave very much

  • like first order and that you can start to apply some type

  • of simple, rational understanding to them,

  • which is more than I can say for an AA prime system,

  • XX prime system or an AA prime BB prime system.

  • Now sometimes these systems will look like first order,

  • which is great because sometimes you can analyze these types

  • of systems as first order and many times you can,

  • but what I tried to show you last time was how there are ones

  • that simply defy simple reduction.

  • >> So what do you use X or Bs based on the distance

  • or the separation of chemical shift not the actual distance

  • between them?

  • >> So let me show you exactly and let's take the AB system

  • because I think this is a great starting point

  • and what's nice is the AB system is going to be an archetype

  • for many sorts of systems

  • that although they're not first order we can apply first order

  • analysis to and we can start to see the distortions that occur.

  • So, a pure AX system is one in which you have a doublet

  • so it's 2 hydrogens that are J coupled.

  • Again, that's going to be the whole spin system

  • so I'll just put on XX and YY to represent some other nuclei

  • that aren't going to couple and not, of course,

  • something with a hydrogen on it where it's J coupling.

  • So you would have a doublet and then a big, big span between it

  • and then another doublet.

  • This little squiggly is just a break, break in the spectra.

  • If those 2 doublets are far apart in chemical shift,

  • then you're going to see them each as a simple 1 to 1 doublet.

  • Now as the distance between them becomes smaller.

  • In other words, either you have different substitutents

  • that instead of having them be very far apart they're closer

  • together in PPM or you simply went

  • to a lower field spectrometer, now you start

  • to see a distortion that we would call an AB pattern

  • where the inner line, and so now instead

  • of saying these are effectively very,

  • very far apart now I'm saying they are far apart like so.

  • In other words, this means, you know,

  • 1 here and 1 way over there.

  • Okay, now the typical way

  • in which one characterizes this is the distance

  • between these line sis the J value, the distance

  • between these doublets

  • and technically one takes not the dead center of the doublet

  • but the weighted average because technically

  • with a multiplet the position of the multiplet is not

  • at its average but at its weighted average.

  • In other words, since this line is a little bit bigger we take

  • the center as just a hair over.

  • It's the weighted average.

  • In other words, if this line is 4 times as, if these lines are

  • in a 4 to 3 ratio and they're separated by .07 PPM, we'd say,

  • all right, you're .4 of the way over there; just a little hair.

  • So, if we call this distance Delta nu, typically if Delta nu

  • over J is much, much greater than 10.

  • [ Writing on board ]

  • We're in the situation like this and if Delta nu over J is less

  • than or equal to 10 and those are approximations,

  • then we're sort of into this AB situation.

  • By Delta nu I mean the difference in position in hertz.

  • So in other words, let's say the center of this line was

  • at 7.30 PPM and the center of this line was at 7.10 PPM

  • and let's just say here that our J value is let's say R,

  • what will work out?

  • What will work out well?

  • Let's say that our J value equals 17 hertz.

  • Now, imagine for a moment you're

  • on a very low field spectrometer.

  • Imagine you're on a 100 megahertz spectrometer what's

  • Delta nu at that point?

  • [ Pause ]

  • 730 hertz.

  • Everyone agree?

  • >> Delta [inaudible] 20 hertz.

  • >> Delta, 20 hertz.

  • So at 20 hertz these guys would be hugely close together.

  • In fact, we'd have a situation that looked.

  • [ Pause ]

  • Like this.

  • At this point Delta nu actually will be just a hair further

  • apart because it's the weighted average.

  • I'm going to shift it over just a hair.

  • I'll make the outer lines just a little bit bigger.

  • This would be a situation where Delta nu over J is very small

  • where Delta nu is about 20 hertz and J is about 17 hertz.

  • If we had the same system at 500 megahertz,

  • what would the difference in,

  • what would Delta nu be for 500 megahertz?

  • [ Pause ]

  • A hundred hertz, right?

  • So at 500 hertz, 500 megahertz, Delta nu is equal to 100 hertz.

  • So you'll look at this situation

  • and at 500 megahertz you'd be more like this

  • and 100 megahertz you'd be like this.

  • So this is your AB pattern

  • and if they were ever closer they'd be like what I sketched

  • out before where the inner line would be huge

  • and the outer line would be very tiny.

  • What? That might, it would be like a 60 megahertz spectrometer

  • like one of the freshmen or sophomore and we actually have

  • like a 100 or maybe it's 60 in the sophomore lab it would be

  • like this or imagine the situation that instead

  • of having substituents that put these apart at .2 PPM,

  • imagine they were separated by .1 PPM, but the main thing

  • to keep in mind is for any given doublet no matter what the

  • center of this peak whether I looked at it

  • at a 500 megahertz spectrometer or at a 100 megahertz the center

  • of this peak is going to be 7.30 and the center of this peak,

  • again, weighted average center is 7.10.

  • Now 17 hertz is more characteristic

  • of a trans alteen [phonetic],

  • which was actually what I was doing when I was drawing this.

  • For something like this we'd be more

  • like about 7 hertz for a J value.

  • Thoughts or questions at this point?

  • [ Inaudible question ]

  • Okay, will the center move, so, if you improve the equipment?

  • So here we've gone from this is our 100 megahertz,

  • this is our 500 megahertz and the point is the center

  • of this peak for this whatever hypothetical compound this is,

  • the center of this peak is always at 7.3 PPM whether I'm

  • at 500 megahertz or 100 megahertz, but the distance

  • between the peaks because the number

  • of hertz per PPM is much smaller at 100 than at 500, the distance

  • between the peaks here is very far, it's 100 hertz apart

  • or relatively far, and over here it's only 20 hertz apart.

  • [ Inaudible question ]

  • The inner one?

  • The closer they are together the more they tent into each other

  • and that really is the difference between the AX.

  • [ Inaudible question ]

  • The center is related to the ratio of the bigger.

  • [ Inaudible question ]

  • Absolutely.

  • Absolutely.

  • Well, here the bigger one is at 7.29 PPM or 7.29 PPM

  • and the smaller one is at 7.31 PPM

  • and by here we've got these 2 lines and one of them is

  • at 7.2 PPM and the other is at 7.

  • whatever the number is.

  • Now what's valuable about looking at AB pattern

  • and understand it is it really becomes an archetype

  • for all sorts of systems that behave very near to first order.

  • So we were talking before about phenylalanine

  • and I guess the example I gave when we were talking

  • about spin systems was acetyl phenylalanine methyl amide.

  • [ Pause ]

  • And I pointed out that we had 1, so this is like a spectrum

  • in chloroform solution so I'll say NCDCL3, and we decided

  • that we had 1 spin system over here and the multiplicity

  • of this proton of the NH is a doublet because it's split

  • by 1 coupling partner.

  • Each of these protons they're non-chemically equivalent

  • so they split each other but they're going to be similar

  • in chemical shift, they're similar in the environment

  • so they'll both be at about 2 and a half, 3 parts per million.

  • Why do I say about 3 parts per million?

  • Well, they're off of a phenyl group

  • so if we were methyl group off

  • of a phenyl I'd say 2 parts per million.

  • It's an ethylene so that pushes it to like 2

  • and a half parts per million.

  • They're beta to a couple of electron withdrawing groups,

  • they're beta to a nitrogen, they're beta to a carbonyl,

  • so then it's going to shift them downfield

  • by about another half a PPM.

  • So we'd expect them to both be at about 3 parts per million

  • but probably not to be on top of each other.

  • So each of these is going to show up at as a DD

  • and that DD is going to be part of what looks

  • like an ABX pattern because this is part of an ABMX system.

  • M is something that's far apart from either A and B and C

  • and so forth and X. So we have 1 proton that's going

  • to be way downfield and nitrogen protons are typically

  • at about 7 parts per million.

  • One proton that's going to be moderately downfield

  • because it's next to an electron withdrawing group and it's alpha

  • to a carbonyl and beta to a phenyl group so this is going

  • to be about 4 and a half parts per million.

  • Then these guys that are both going to be close

  • to 3 parts per million.

  • So we have far apart from this and H far apart from alpha

  • and the alpha is far apart from the beta.

  • So this guy here is going to be split by 3 different protons.

  • So he's going to be a DDD if all of the Js are different or a TD,

  • and we'll talk more about these or DT,

  • if 2 of the Js are the same.

  • Or a quartet if all 3 Js are the same within the limits

  • of experimental error.

  • [ Pause ]

  • So the one I really want to draw our attention

  • to then is these 2 hydrogens here because now this type

  • of AB pattern really can serve as an archetype

  • for more complex patterns that are non-first order

  • but are close to first order.

  • So an ABX pattern is something where you have the AB pattern

  • in which each line is further split.

  • So imagine this type of pattern here with some level

  • of separation, but now with each

  • of these 2 lines split into a doublet.

  • So what you see is line line, line line, and then line line,

  • line line for these 2 protons.

  • [ Pause ]

  • This is for the ABX system.

  • So this is what we're seeing right at about 3 PPM.

  • [ Writing on board ]

  • Which of them is which?

  • [ Inaudible question ]

  • Okay, so stereochemically one of these protons is pro R

  • and the other proton is pro S. They're diastereotopic protons.

  • If we can go ahead and so, for example, if I replace this

  • with a deuterium in the thought experiment, then the ranking

  • of the carbons, the ranking of the 4 constituents

  • on here becomes highest rank, next rank, next ranked and so

  • that would become an S center.

  • So this is a pro S proton and this proton is pro R.

  • If I knew the geometry here, for example,

  • if I knew the phenyl group preferred to point in one way

  • or another or I could by Nuclear Overhauser Effect experiments

  • and the like detect certain proximities,

  • then I would be able to get an experimental correlation

  • or predicted correlation based on say proximity

  • to anisotropic groups of 1 proton and 1 peak.

  • So right now I don't know which is which,

  • but with additional experiments --

  • this is obviously just a sketch --

  • but with additional experiments in context, yes, you can figure

  • out with diastereotopic proton is which.

  • Other thoughts and questions?

  • [ Inaudible question ]

  • Do they ever look the same height?

  • Great question.

  • So, right now I've made a sketch for a situation

  • in which these are relatively near to each other.

  • In other words, maybe they're separated by a tenth of a PPM.

  • So if they're separated by a tenth of a PPM,

  • the JAB here is going to be about 14 hertz.

  • At 500 megahertz that would be a separation if they're separated

  • by a tenth of a PPM about 50 hertz.

  • So Delta nu over J would be about 50 to 14 about 3,

  • but if they were very, very far apart, if something held these

  • in very different magnetic environments,

  • then you would see the outer lines getting bigger.

  • [ Writing on board ]

  • And the peaks becoming more like a regular doublet of doublets.

  • So if they were further apart, it would look more like this,

  • these guys would be bigger.

  • They would be tenting into each other less.

  • Other questions?

  • These are really important and this is one

  • of the reasons I'm going really slow over this.

  • [ Inaudible question ]

  • Coupling, coupling is always going to be mutual.

  • So if we call this, let's name our peaks 1, 2, 3 and 4

  • and let's name our peaks 1 prime,

  • 2 prime, 3 prime and 4 prime.

  • So the JAB is going to be 1 minus 3 and 2 minus 4.

  • They will within the limits of experimental error be the same

  • and JAB here will be the same within the limits

  • of experimental error.

  • It will be 1 prime minus 3 prime and 2 prime minus 4 prime.

  • In this case since it's an ABX pattern, the coupling

  • with the other proton so the coupling

  • with the remote partner, we'll call it AX, JAX equals 1 minus 2

  • and 3 minus 4 and, again, those will be the same

  • within experimental error.

  • So let's say for a moment this is 14 hertz that's also

  • 14 hertz.

  • Let's say for a moment that this is 6 hertz,

  • actually it looks the way I've draw it it looks more

  • like about 9 hertz so let's say this is 9 hertz, that's going

  • to be 9 hertz within the limits of experimental error.

  • Here this distance will also be 14 hertz as will that distance.

  • This distance the way I've drawn it looks like it's

  • about 12 hertz and that looks like it's about 12 hertz.

  • [ Inaudible question ]

  • Ah, where?

  • Oh, no, no, no.

  • One minus 2 will not equal to 2 minus 3.

  • So over here JBX equals 1 prime minus 2 prime

  • and 3 prime minus 4 prime and, yeah,

  • if you look at this pattern

  • and you draw a splitting tree diagram, we split into a doublet

  • so that's our big J and then each

  • of those lines is further split with a small j

  • and so you get this pattern of line line, line line,

  • and if I call my lines 1, 2, 3 and 4, 1 minus 2 is the small j,

  • 3 minus 4 is the small j, 1 minus 3 is the big J,

  • 2 minus 4 is the big J.

  • [ Inaudible question ]

  • From the AB coupling, the geminal coupling

  • in this particular case, and the source of the small j is

  • from the coupling to, in this case, this nucleus over here.

  • [ Inaudible question ]

  • That's going -- beautiful question --

  • that's going to depend on the geometrical relationship

  • on the Karplus curve

  • so typically you will not have exactly the same coupling

  • between 1 of the 2 protons, let's say the pro S

  • and this proton versus the other proton and this proton.

  • So rather than my, let me put up some real data.

  • You'll see exactly the same thing

  • but at least it will be a nice, nice chance for you

  • to have a real spectrum and I know I passed this out before,

  • but we didn't look as deeply at this spectrum.

  • Now let's look at it with a fresh pair of eyes.

  • Let's look at it more deeply.

  • So I didn't use this exactly compound.

  • I just grabbed this right off of the Aldridge [phonetic] webpage

  • and remember you can go to sial.com, www.sial.com,

  • to get yourself lots of spectra.

  • It's a great way to check out your ideas

  • and your understanding of things.

  • So here we see a real compound.

  • I've shown you this before.

  • This is phenylalanine in D20.

  • [ Pause ]

  • So unlike the example that I, the hypothetical example I gave

  • in chloroform, this one doesn't have an amide here it has an

  • amine here and it has a carboxylic acid.

  • In D20, those exchange and so this becomes ND2

  • and you essentially see no coupling and don't see it.

  • Actually there's DCL here so this really becomes ND3 plus

  • and you don't see coupling.

  • This becomes D. So this system here what remains

  • of the 2 methylenes and the methine,

  • so this is an ABX system and so you see coupling here

  • and this is your phenyls and this is your HOD

  • and this is your alpha proton and these are your beta protons.

  • So this is a very real example of what I've sketched out

  • and you'll notice the distance between these 2 lines does,

  • indeed, match the distance between those 2 lines.

  • In other words, the 1 to 3

  • or 2 to 4 distance does match the 1 prime to 3 prime

  • or 2 prime to 4 prime distance, and you'll also notice

  • that the 2 couplings with the alpha proton are a little bit

  • different, a little bit different from each other.

  • So it looks like if I had to eyeball it here

  • that our coupling here, this distance between 1 and 2 or 3

  • and 4 is about 6 hertz and the distance here, the distance

  • between 1 prime and 2 prime or 3 prime

  • and 4 prime is about 8 hertz.

  • So you'll notice now the alpha proton is split into a doublet

  • of doublets so each of these is a DD ABX pattern,

  • DD ABX pattern, and then you'll notice

  • that our alpha proton is a DD

  • and the doublet reflects the 2 different couplings.

  • In other words, the distance between lines 1 and 2 or 3

  • and 4 corresponds to this coupling,

  • to this 6 hertz coupling, and the distance between lines 1

  • and 3 and 2 and 4 corresponds to this coupling

  • to the 8 hertz coupling.

  • [ Pause ]

  • Thoughts or questions?

  • [ Inaudible question ]

  • Well, great question.

  • So the question is if the alpha proton is a doublet

  • of doublets shouldn't it be leaning a lot more?

  • You notice these guys are really tenting into each other

  • and this one is just barely tenting in.

  • So now look this is a 300 megahertz spectrum.

  • So the distance between these 2 looks like it's

  • about a tenth of a PPM.

  • So they're separated by about maybe it's two-tenths.

  • So, the distance between these 2 is about 60 hertz

  • because it's 300 megahertz so it's 300 hertz per PPM.

  • So they're separated by about 60 hertz and the J value is

  • about 14 hertz so that's the case where Delta nu over 12 is

  • about 4 or 5 whereas here the difference

  • between the alpha proton and the beta protons is about 1 hertz,

  • about 1 PPM, about 300 hertz, and the J value is 6 and 8.

  • So this is a case remember I said the big difference

  • between the AB and AX type of situation this is a case

  • where Delta nu over J is very big, 300 versus 6

  • or 8 is a factor of well over 10.

  • So they're effectively far apart

  • and you get very little tenting inward.

  • Other thoughts?

  • [ Inaudible question ]

  • So great question.

  • So this is in D20 the most hydrogens

  • on heteroatoms exchange and so most hydrogens on heteroatoms.

  • As a matter of fact, I will say I can give you exceptions

  • but I will say hydrogens on nitrogen, oxygen are replaced

  • with D. So they get replaced with deuterium.

  • Deuterium shows up in completely,

  • shows up not 500 megahertz but at about 80 hertz, 80 megahertz.

  • So they don't show up in the same spectrum

  • and the J values are so small that for intents

  • and purposes you don't see coupling plus they're exchanging

  • very quickly.

  • [ Inaudible question ]

  • Even without DCL they will exchange.

  • Because of DCL the amine is protenated

  • so the amine is an ammonium group and because

  • of DCL it dissolves whereas phenylalanine

  • in just pure water wouldn't be nearly as soluble.

  • All right so I've started to hint that different types

  • of coupling relationships have different coupling constants

  • and what I'd like to do at this point is to talk

  • about typical coupling constants and see how we can use them

  • to enhance our understanding.

  • So, this coupling constants generally

  • if you needed just one number to keep

  • in your head you could keep 7 hertz or 6, 6 to 8 hertz.

  • Let's say we're talking SP3 to SP3 right now

  • and if I needed a number, actually, I'm going to put this

  • as a general CH to CH.

  • I'll show you double bonds in a second but if you need 1 number

  • to keep in your head, 6-8 hertz or 7 hertz is a great number

  • to keep in your head without confirmational preference.

  • I'll say without a confirmational biases.

  • What do I mean by a confirmational bias?

  • Well, if there's a strong, strongly held relationship,

  • for example, if 2 hydrogens are locked

  • in an antiperiplanar relationship

  • so here we have 180-degree dihedral angle,

  • now our J value is going to be bigger.

  • It's going to be about 8 to 10 hertz.

  • So, for example, if you have 2 axial protons.

  • So I'll put this as JAX, AX, that would be a typical example

  • for 8-10 hertz where you're locked

  • into an axial relationship.

  • If you have something locked into an equatorial relationship

  • where now you have axial equatorial or equatorial,

  • now you're talking about a 60-degree dihedral angle

  • and so a typical JAX equatorial

  • or J equatorial equatorial is on the order of 2.

  • I'll put little tildas, little squiggles here just to indicate

  • that that's approximate 2 to 3 hertz.

  • [ Inaudible question ]

  • This is based exactly off of the Karplus curve.

  • So a general way of thinking about coupling is

  • that coupling comes from interaction of the nuclei

  • with electrons in the bond that polarize the next bond,

  • that polarize the next bond.

  • If at one extreme you have 180,

  • a 90-degree dihedral relationship

  • between those 2 bonds, you get no overlap of orbitals.

  • If at the other extreme you have 180, you get very good overlap

  • and antiperiplanar relationship

  • and at 0 you also get a good overlap

  • and so you see very large coupling constants at 180

  • or at 0 and very small at 90 or 60.

  • So, Karplus relationship is basically a relationship

  • between theta, your dihedral angle, and J,

  • and as sort of a general relationship would be if we go

  • from 0 to 90 to 180 that we go at 0 it's about 8 hertz.

  • We have kind of a cosine wave going down at 90 to a minimum

  • and up to about 10 at 180 degrees.

  • [ Pause ]

  • Now this is sort of for general kind of plain vanilla,

  • it's modulated, the coupling constants are modulated

  • by electronegativity and hybridization.

  • In general, electronegative substituents lead

  • to a smaller coupling constant.

  • In general, if you've got an SP2,

  • SP2 bond between the 2 atoms like a double bond,

  • you have bigger J values.

  • So let me show you a couple of examples.

  • So as I said, first number to keep in your head is

  • about 7 hertz, but now if you want to think

  • about some oddball situations, you can think

  • of like an aldehyde where now you have an electronegative

  • oxygen and you have an SP2 carbon here

  • and aldehydes are very funny in that your coupling constant is

  • on the order of 2 to 3 hertz so that's sort of unusual.

  • Alkenes, it's good the keep.

  • These are numbers that really are worth keeping

  • at your fingertips.

  • For a CIS alkene we're talking typically on the order of say 7

  • to 12 hertz with let's say 10 hertz being typical.

  • For a trans alkene we're talking maybe 12

  • to 18 hertz or 14 to 18 hertz.

  • Let's say 14 to 18 hertz with maybe 17 hertz being typical.

  • So these are all examples of vicinal couplings.

  • One more example falling right in that sort

  • of general 7 hertz range.

  • Let's take on a benzene as an example.

  • On a benzene as an example, we're talking maybe

  • for ortho coupling maybe 6 to10 hertz

  • with maybe 8 hertz or 7 hertz typical.

  • [ Pause ]

  • Two bond couplings tend to show more variation

  • than 3 bond couplings.

  • So, for example, if you have 2 carbons on a methylene group,

  • on an SP3 hybridized carbon with different substituents

  • on the carbon you can see anything from say, oh,

  • 5 to 20 hertz depending on the electronegativity

  • if it's just sort of carbons on here maybe 14 hertz is typical.

  • [ Pause ]

  • If you have an SP2 carbon on a double bond,

  • you're talking maybe 0 to 2 hertz.

  • Maybe 1 hertz being typical.

  • [ Pause ]

  • So all of these are examples of vicinal and geminal coupling.

  • In other words, 2-bond and 3-bond coupling.

  • These are J2 HH couplings, these are J3 HH.

  • If you have anything more than that, if you have greater than

  • or equal to 4 bond coupling,

  • we're talking long-range coupling.

  • So, for example, J4 HH would be an example

  • of long-range coupling.

  • Normally in saturated systems, you don't see coupling,

  • but if you have a system

  • where you have certain geometrical relationships,

  • then you may see it.

  • So, we're talking about say a carbon not with its neighbor

  • but with a carbon 1 over.

  • So where does this come up?

  • Usually if you have intervening double bonds.

  • So, for example, [inaudible] systems.

  • We've talked a little bit about this in discussion section.

  • Typically let's say 0 to 3 hertz depending

  • on the geometrical relationship.

  • Metacoupling on a benzene same type of thing.

  • Let's say 1 to 3 hertz.

  • [ Pause ]

  • The only real situation that you can actually see a visible

  • splitting where you have just SP3 carbons is

  • if you have a locked relationship what is sometimes

  • called W coupling.

  • So usually you need a locked relationship

  • where you have a series of antiperiplanar bonds.

  • This occurs, for example, in the norbornene ring system.

  • They call it W coupling because you have a W-like relationship.

  • So say these 2 hydrogens

  • on a norbornene you can see you have the series

  • of antiperiplanar relationships

  • that make a W. Norbornene actually is loaded

  • with W coupling so you have another W relationship

  • across the ring like this

  • and there's even a third geometrical W relationship

  • hiding in the molecule like so.

  • [ Pause ]

  • So I want to pass out, there's 1 table that's really useful

  • in your book, and I want to pass it out just

  • because the stuff that's in the back of your book is

  • so much better when it's passed in front of your eyes rather

  • than simply waiting there in the back

  • of the book undiscovered and uncared about.

  • So this is the Appendix F I mentioned before and I just want

  • to show you how many good things are hiding

  • in this 1 little appendix here.

  • So everything we've talked about

  • and more is hiding in this appendix.

  • So we have our [inaudible] coupling and this is all going

  • to come up on homeworks.

  • If you're wondering what your typical [inaudible] couplings

  • are, you'll find answers here.

  • If you're wondering what happens if you have a double bond next

  • to a double gone, you'll find answers here.

  • If you're wondering, we've already seen

  • in the homework we've already seen coupling across acetylenes

  • and so how many bond couplings is that?

  • [ Inaudible question ]

  • Four-bond coupling, right?

  • So 1, 2, 3, 4 bond coupling, but if you ask as many people do

  • when they come up with some of the homework assignments here,

  • well, can you couple further?

  • Can you get 5 bond coupling?

  • The answer is right here waiting,

  • waiting for you to read about it.

  • Right over here typically you see a little coupling

  • across there.

  • Coupling on pyridines is going to come up won the homework

  • and already I think we see that very soon.

  • Now, all of this is given in more detail in Prech [phonetic].

  • So Prech has wonderful examples for pyridines, for thiophenes,

  • for all these types of systems off of real compounds and off

  • of typical examples, but here distilled

  • into really 1 little appendix and really 1 page

  • of the appendix is so much different good stuff that's

  • going to help you out with some of the problems

  • that you're working on.

  • So, okay, I think this is where I'd like to wrap it up today.

  • We're going to talk more about first order splitting next time

  • and we're going to walk through some examples of doublets

  • of doublets and triplets of doublets and doublets

  • of triplets and doublets of doublets of doublets

  • and that'll prepare you

  • for a workbook assignment that comes later on. ------------------------------bb79b27d2c62--

>> I want to spend the next 3 lectures, 3 classes,

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化學203.有機光譜學。第十二講.系統中的耦合分析 (Chem 203. Organic Spectroscopy. Lecture 12. Coupling Analysis in Systems)

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