and get good at recognizing and identifying the multiplets and extracting coupling constants. So the things that we're talking about and the question that was raised, is that coupling really 700 hertz?

What makes it 700 hertz? So let's talk about factors that J depends upon and I've tried to sort of group out maybe five factors and give us some points. So let's say one is the magnetogyric ratios

of the nuclei involved and so remember how when we started talking about magnetogyric ratio, I gave you two examples. We talked about the C13 satellites of chloroform

and proteo chloroform, right? So when you see the H1 NMR of CDCL3, of course you don't see any of the deuterium chloroform that's in there but you see that .2% or .1% of proteo chloroform

and you see a peak that's very big at 7.26 but then symmetrically disposed about it you see these two satellites and so this is your CHCL3 or more specifically your C12 HCL3 peak

but then your satellites come from your 1%, 1.1% of C13 CDCL3, CHCL3 and the spacing of those two lines, the distance here between these two is 208 hertz

and that distance happens to be 1.6 is 1 over 6.5 the distance that we see in the C13 NMR of CDCL3, right?

In CDCL3 you do indeed see the C13, the 1% C13 and you see this 1 to 1 to 1 triplet centered at 77 PPM and the distance between the lines here is 32 hertz and that ratio of 208 to 32 is equal to 1 over 6.5

which is the magnetogyric ratio of proton over the magnetogyric ratio of a deuteron meaning

that when everything else is equal if you have a magnetogyric ratio nucleus with a magnetogyric ratio that's big you get big coupling if you have a nucleus with a magnetogyric ratio that's small you get small coupling. For heteronuclei we said that fluorine

for example has a big magnetogyric ratio so fluorine is going to have big coupling. We said that carbon has a relatively small magnetogyric ratio so carbon in general is going to have small coupling so the gamma of the nuclei is important.

The number of bonds, so generally, generally, generally we see coupling through 1 or 2 or 3 bonds sometimes 4 or 5 that's called long range coupling. So if we have some system W, X, Y, Z and we look

at if we're talking about coupling through 1 bond, coupling through 2 bonds or coupling through 3 bonds and I need to make some wildly general generalizations I could say our J1s are on the order and you can just sort of look at the appendix I handed out and if you want

to keep a number in your head I'd say let's say 30 to 300 hertz kind of 1 bond couplings are big. They're on the order of 100 hertz give or take a factor of half a log unit. We saw you can have a 700 hertz coupling in the phosphorus but generally yeah like 100 hertz would be a good number

to keep in mind. J2s are generally on the order of 0 to 20 hertz and again I'll give you examples that fall out of there but if you want to keep a number in mind you know 10 hertz, 20 hertz, something like that and J3s are generally on the order of 0 to 20 hertz.

So in other words 1 bond coupling is huge, 2 bond coupling, 3 bond coupling is smaller. All right other factors involved geometry particularly in 3 bond couplings where good overlap leads to big coupling.

Remember coupling is carried by electrons in bonds where a nucleus feels another nucleus

through polarizing the pair of electrons making up the bond so that the one with its spin up is nearer to the one with its nearer to the nucleus that spin down and so that travels along the line and if you have good orbital overlap for example an anti periplanar relationship or a synperiplanar relationship you'll have a big

coupling constant. If you have bad overlap for example a 90 degree dihedral angle you'll have essentially no coupling constant, no 0 coupling constant and if you're sort of at a low angle like a gauche angle, something close to 90 degrees, 60 or 120,

you have still relatively poor overlap so your coupling constants are small. That was what we were talking about the other day with cyclohexane where I said yeah for an axial coupling on the order of 10 hertz, 8 to 10 hertz. For axial equatorial, so that's axial axial.

For axial equatorial or equatorial equatorial where your dihedral angle is about 60 degrees, yeah 2 or 3 hertz. All right so geometry also matters. Hybridization and a general rule is more S character leads

to bigger J's and other factors, electronegativity.

Hybridization is like why in a trans alkene we see a J of 17 hertz typically versus in a diaxial interaction

on a axial axial coupling on a cyclohexane which also has 180 degree dihedral angle just like a trans alkene we only see a coupling constant on the order of 10 hertz. In other words you have more S character in an SP2 bond in a bond involving an SP2 carbon than a bond involving an SP3 carbon

so you have a bigger coupling constant. So another factor let's say, I'll just say other factors such as electronegativity.

So for example a hydrogen that's next to an oxygen that's on a carbon with an oxygen will have slightly smaller J values particularly if there are 2 oxygens on there. All right let's, let me give you some typical J values

and then we'll look at some examples and I don't want, I don't want us to have a, I mean these sort of generalizations that I just put up are pretty useful but unless you're doing a project with particular nuclei you may not end

up keeping all the numbers in your head but let's just look at sort of some typical examples and these are actually in the appendix here. So let's take a look at a fluoroalkane. So your J2HF and this is all in the appendix is really,

really big, it's like 44 to 81 hertz so even bigger than a sort of generic 2 bond coupling that I talked about. Fluorine has a large magnetogyric ratio so you've got very big coupling. Your 3 bond HF, your 3 bond coupling is on the order

of 3 to 25 hertz and fluorine is so, so good at coupling, right, so this is 2 bond coupling, this is 3 bond coupling.

Fluorine is so good at coupling that you can even sometimes see a little 4 bond coupling, J4HF is on the order of let's say 0 to 4 hertz.

Part of it's the polarization but part of it is the orbitals that fluorine is using in bonding. So if you have a hydrogen, right, we talked about 2 bond and 3 bond HH coupling, right, so a typical 2 bond HH coupling might be I said 14 hertz sort

of in an unperturbed system and a typical 3 bond HH coupling I said let's use 7 hertz as a typical number so that hydrogen is contributing a 1S orbital but the fluorine is going to be contributing orbitals that are going out further because it's using the second shell in coupling so we're in the 2 SP3 technically

so your coupling is going to be bigger with fluorine so I don't think it's just, it probably involves some electron. Yeah, actually you know it would involve electronegativity as well because it is going to be pulling those electrons in really tight and that's going to give more interaction

of the nuclei with the electrons so yeah I guess it's both of those. Other questions or thoughts on this? When you have a controlling role and an axial of controlling, they're both on the wrong side. Of one over the other?

No, I mean think about your Newman projection of a perfect chair cyclohexane. A perfect chair cyclohexane if I draw it as a Newman projection I'll just draw half of the cyclohexane so this is axial, this is axial,

this is equatorial, this is equatorial and in a perfect chair cyclohexane this dihedral angle is 60 degrees, this dihedral angle is 60 degrees and this dihedral is 60 degrees so it really should be

about the same unless you flatten the ring out or do something to perturb it. Ah, okay, do you ever see coupling that's essentially

through space? One can, so outside of the realm of covalent chemistry that we're seeing you can for example see coupling across hydrogen bonds and other situations where things are held

into close proximity with each other. So for example you can do N15 coupling across DNA for example from one nucleobase to another through the hydrogen bonds but just for a normal ordinary organic molecule where it was

in some confirmation, not that I know of any example. Nothing where the molecules aren't locked together. I mean I suppose you could come up with some example where you take some poor molecule and bang two methyl groups into it so hard that they're banging in at Van der Waals radius

and you might, I don't know the answer, that's the sort of thing that one would try experimentally. Other questions? Benzene's are interesting with fluorobenzene's and you will have, I think it's on this coming Monday's homework set you'll have some

fluorobenzene's and it's a little bit counterintuitive. The problems are pretty easy. They're out of chapter, was it five, four, what's the one that I gave you to read that's the coupling involving other nuclei? Six, okay. Anyway, so keep this in mind. So your ortho coupling, your J3HF is on the order

of 6 to 10 hertz and so that's kind of what you'd expect and what's sort of surprising, I mean in the case of hydrogens you have meta coupling but it's usually

on the order of a couple of hertz, you know, two or three hertz, one to three hertz. In the case of fluorine your meta coupling is bigger than you might otherwise expect. It's on the order of 5 to 6 hertz and then because we know that fluorine is very good at coupling you even see some

para coupling and it's on the order of 2 hertz and these are all kind of approximate. Yeah, I mean the good news on all of this

that appendix F is just such a treasure trove and so the good news is appendix F puts a lot

of these data right at your fingertips and the only reason I'm talking about this is because I think hearing it once is sort of helping you see where to look it up and so forth. Let me just point out also the numbers that I had given you before

and just put this into some context. So when we started talking about NMR and spin active nuclei I mentioned the magnetogyric ratios and just let me put those up here again.

Magnetogyric ratio for a proton is big, magnetogyric ratio for a fluorine is big, magnetogyric ratio for phosphorous 31, I guess I'll even put up F19 and H1 is 10840 and the magnetogyric ratio

for C13 is 6728 so let me tell you what I was talking about when I was saying you can correlate things roughly with magnetogyric ratios.

So for example, JCF, J1CF is pretty darn big. It's at the high end of what you would typically expect for couplings. It's on the order of 200 to 300 hertz. J2CF is also pretty big.

It's at the order of 20 to 40 hertz and J3CF is on the order of 10 hertz and then phosphorous as was already pointed out, one bond phosphorous hydrogen coupling is huge.

It's on the order of 700 hertz. Two bond pH coupling is on the order of 10 hertz and three bond pH coupling is on the order of 20 hertz.

Certainly these aren't numbers to keep in your head. These aren't numbers to keep in your head. These are just sort of numbers to have seen once.

Often J3s are bigger than J2s. In carbon hydrogen coupling, often J3 to hydrogen, often three bond coupling is a little bigger than J2. Sometimes J2 is bigger. Remember, of course, one case you can actually have an antiperiplanar relationship so if you think about it

in J2 you can, in J3 you can get a nice zigzag relationship. In J2 you've got, you know, different relationship.

These electrons aren't as directly overlapped and J2 often depends on hybridization and geometry. I mean striking, striking example. So you look at alkenes, right, and this coupling is on the order of 0 to 2 hertz, the geminal coupling, and then you look at your vicinal couplings and it's

on the order of 17 hertz and even the cis coupling is on the order of 10 hertz. So here's another example where J2 is smaller than J3.

These ones, yeah, carbon fluorine. So we've now just had this tremendous traipse through all of these weird and wild couplings and sort of in the abstract in the sense I've listed some numbers. So let's now look at some real compounds and these are just compounds that I pulled out of Aldrich

and for that matter compounds that you might well or types of groups that you might well encounter and the first thing I'll start with is triethylphosphite and you would encounter this last night. We talked about Horner Wadsworth-Emmons type reagent. So if you made a Horner Wadsworth-Emmons reagent,

you will encounter or a Wittig reagent or a phosphonium precursor to a Wittig reagent, you will suddenly find yourself encountering fluorine proton and fluorine carbon coupling as you get to know your molecule. So let's take a look at an example

and I love these Aldrich spectra because you can just pull all sorts of spectra out and say, all right, what would I get? So I pulled this out of www.sial.com and then I just blew things up. Okay, so let's start with we have a 300 megahertz proton

spectrum, all the Aldrich spectra are 300 megahertz H1 and 75 megahertz C13 spectra. So triethylphosphite, we see something that looks like a triplet for the methyl group and then something

that looks a little more complicated over here. How do we describe this pattern? Quintet would be a good place, so we could call this a quintet or what was that?

Quintet, I like that, so quintet I like even better apparent because I've taken this peak and I picked it up and blown it up here and what do we see?

Little shoulders, so we know it's not a perfect quintet, we don't have four couplings that are exactly the same. Can we recognize a pattern hiding under here? Triplet of triplets, now doublet of quartets, watch this,

so doublet of quartets is a pair of quartets, 1, 3, 3, 1 for this shoulder and then here for this shoulder 1, 3, 3, 1.

Do you see that, 1, 3, 3, 1, 1, 3, 3, 1. So it's a doublet of quintets, quartets, I'm sorry,

so if we want we can extract both J's. The smallest J, the small J, the J associated with the quartet is going to be the distance from the last line to the next to the last line

or the first line to the second line and I handily put a scale on here and I read that distance as 7 hertz and the big J is going

to be the distance from this line to the, from the first line, the second from the last line to the last line or from the first line to the third line and I can't exactly read where that line is but it looks like it's about a hertz more than here

so let's call that 8 hertz. So we could characterize this as a DQ, use Q for quartet, quint for quintet, J equals 8, 7 hertz.

So 7 hertz, that's just our, and a quartet, that's just our coupling to the methyl group, right? Because we have P, O, C, H, H, C, H, H, H. So 7 hertz is

just our J3HH and 8 hertz is our J3PH through 1, 2, 3 bonds.

Those are just approximations. I, yeah, I'm trying to give you sort of general numbers to ballpark it.

So 1 bond couplings on the order of 100 hertz, 2 bond, 3 bond couplings on the order of 10 hertz, you know, sometimes bigger, sometimes smaller. Depends on gamma, depends on other factors, depends on hybridization.

We can do the same thing with carbon here and even though our 2 peaks in the C13 NMR are really small here, they conveniently give us a peak printout up over here so those 2 constitute a pair and those 2 constitute a pair. Each of them is a doublet.

The doublet for this carbon we can deal with as 63.64. I want to calculate the coupling constant minus 63.56. Those are just the 2 values on the top of the table

and PPM is 0.08 PPM and we have 0.08 PPM times 75 hertz per PPM is equal to 6 hertz.

So that's our J1, that's our J2PC and we can figure

out our J3PC from this other one. So we have 16.19 in the upper left hand, upper right hand corner and 16.10 as the position

of the 2 lines and that difference is .09, .09 times 75 is equal to 7 hertz and so that's our J3PC.

Thoughts, questions in the carbon NMR? So in the upper right hand corner they list the 2 line

positions, they're in the proton NMR. You mean how we see the doublet? So okay, I can't do this well with my hands because my hands aren't in 1 to 3 to 3 to 1 ratio but okay.

Imagine my hand is a quartet and if we have, so the distance between my hands is the big J and if the distance was 1 finger's worth we would see a perfect quintet. Oops, this is a bad, oh I've got a thumb there.

Okay, if it were 1, I've got to get rid of my thumbs, chop them off, okay if it were 1 finger's worth we would have 7 hertz spacing between the fingers and then it would be split into 5 lines all equally spaced at 7 hertz apart but they're a little bit further apart

so we see a quartet and another quartet and the distance between lines 1 and 2 is the J of the quartet and the distance between lines 1 and 3 or in this case 1 and 2 and 1 and 3 is the distance of the doublet.

Interesting question, so do you notice these little jaggies right on the edge of the peak? Do you see how the peaks aren't smooth? So remember I was talking about digital resolution so you take a spectrum that's let's say 16,000 points wide

and you have those 16,000 points over a spectral width that's let's say 4,000 hertz here and so your digital resolution is a quarter of a hertz so you think of your spectrum as a smooth curve but what it really is is a series of points that have been splined together

and because we may be missing a point right on top of the peak you end up with something not looking completely symmetrical. That's just an artifact of this. As a matter of fact I would say if I went and told the spectrometer by increasing the acquisition time, told it to have more digital resolution we would probably see

something that was better resolved. You just type in, there are two ways, so you type in number of points and you make the number bigger or you make the acquisition time longer and the number of points in the acquisition time and the spectral width are all intimately linked

so you can basically collect data for 5 seconds instead of 3 seconds to increase your number of points or you can tell it more points. So that would be the J between the methylene and the methyl group.

So let's take a look at a fluorine NMR or rather a fluorinated compound and maybe I'll just show you some highlights. So I grabbed from Aldrich the spectrum

of fluoropentane and so let me just write out the structure here just so you can see it. All right, what do we call that?

It's a doublet of triplets and what's doing the splitting? What's giving the doublet part of the splitting? Which proton is it first of all? Which protons are it?

One alpha to the fluorine and so this is giving a 2 bond JHF and what else is giving the splitting?

The methylene so we have this methylene is coupled to the adjacent methylene and so that's giving the triplet part of the splitting and you can read off the 2 bond JHF. What value do we get here?

Somewhere around 48, 47 so your 2 bond coupling I think I got 47. You notice this multiplet over here? It's not so neat.

Remember I talked about non-first order coupling when you've got things lumped on top of each other you get virtual coupling so the next couple of methylenes are lumped on top of each other and you've got coupling here and coupling to the fluorine so it's not so pretty but if you pull it out and you expand it you can pick out your J3HF even

if you can't quite pick out what the multiplet is if it's just sort of a generic multiplet you can measure that distance there and it looks like it's about 24 hertz.

So our J3HF is 24 hertz. What's that? And that fits into the model. You can do the same thing with the carbons and if you look

at the carbon spectrum here you'll notice a big, big, bigly separated doublet over here. What does that correspond to? J1CF so that's the carbon that's directly alpha

to the fluorine and you've got a J1CF and if I want to calculate the J1CF what do I do and what numbers do I subtract to multiply?

85 minus 83 or more specifically 85.26 minus 83.09 times 75 and that gives us 167 hertz so that's a J1CF coupling

and I won't work through it but you'll notice that your next peak here is split into a doublet. You can see the pair of lines over here and that corresponds to these two.

That's our J2CF and then the next one is split and that corresponds to our J3CF coupling. So you can pull all of this data out of your spectra.

Answer on that blackboard. Take your carbon NMR and go ahead and run it at 75 or I guess for our department we have carbon at 100,

carbon at 125 and carbon at 150. So you could just go ahead and run it at 2 and see if the two lines are now at the same PPM. If each line is at the same PPM their distance in hertz has changed and they correspond to 2 singlets.

If on the other hand their positions in PPM have changed and they've moved toward each other but they're still centered at the same position in PPM and their distance in hertz apart is the same then they corresponded to a doublet. All right, let's have some fun with carbon-carbon

and carbon-hydrogen coupling now, CH coupling.

So SP3, so I'm talking right now about one bond coupling. We'll start with one bond CH coupling. One bond CH coupling is important because all

of your heteronuclear techniques and your dept technique rely on carbon-hydrogen coupling, one bond or in some cases two bond and three bond coupling. So a typical, remember I said how hybridization matters and the more S character you have

for everything else being equal the bigger the J if everything else is equal. All right, so typical J1CH for an SP3 carbon is about 125 hertz. So for example if you look at ethane your value is 124.9 hertz.

If you look at cyclohexane and these are all in your appendix table as well so 123.0 hertz. I'm just pulling out some highlights from one of the appendices I've handed you.

SP2, more S character, 25 percent S character and 6 in SP3, 33 percent in SP2 and so your J1CH goes up proportionally, it's about 160 hertz. So you look at ethylene and it's 156 hertz, 156.2 hertz

and these are all in one of the appendices that I've passed out to you. You look at benzene, it's 159.0 hertz.

All right, now where does this, where else does this become interesting? Where else does it become important? Yeah, beautiful, yes. So C13 NMR you will invariably run it proton decoupled

so you won't see proton coupling and yet you will still see people use the term to refer to a dept spectrum where methylene is referred to as T for a triplet and there's history there

and the history is that before they had dept and other techniques they would run what's called off resonance decoupling, partial decoupling and so you would get a methylene and you would see it as a triplet because it would be blowing out most of the couplings, you wouldn't see the 2 and 3 bond coupling which is really horrendous because peaks

when you talk about 2 and 3 bond coupling you can imagine how split your carbons are like the methyl group in ethanol is split into a triplet of, into a quartet of triplets and the methylene group in ethanol is split into a triplet of quartets.

So these are, you know, you have huge, huge splitting in fully proton coupled carbon but you will still see a methylene referred to as parenthesis T from that old, old thing. So yeah, all the carbon NMR we're going to run is proton decoupled. You do see, remember I said in your C13 satellites

of course you see the reverse because you're not carbon decoupling when you're running a proton NMR and for the most part it doesn't matter because 99 percent of your carbon is carbon 12 but if you look closely on your methyl groups you can see little satellites and on other sharp singlets.

You could indeed. One of the problem with X nucleus, so the question was could you decouple fluorine so you didn't get fluorine coupling and the short answer is yes. One of the problems is the amount of power you put in has

to, depends on your spectral width and so if you have the width of fluorine which is very wide, it's 200 or 300 PPM, you've got to put in a huge amount of power to irradiate all of the fluorines and basically that turns your NMR into a giant microwave

and so you'll cook your sample so it actually isn't so simple. It's easy, protons have a narrow spectral range. It's only 12 PPM. Carbons on the other hand have about 200, 240 PPM so you would have to put in a lot of power to carbon decouple. You could do specific decoupling experiments where you irradiate

at one specific frequency and that's another way you could do it but it's hard to blast an entire wide spectral width. Yeah, but normally it's proton decoupling is, proton's the easy one.

Negative, okay, negative and positive for the most part don't mean anything in terms of what you will observe except in a phase sensitive spectrum but a positive J value means that if one nucleus is spin up the other feels a stronger magnetic field. The other, but because you're going through electrons and you're polarizing the electrons in the first bond

which are polarizing the electrons in the second bond, you may end up with polarization so that if one proton is spin, one nucleus is spin up, you end up with the other feeling a weaker magnetic field so that's what a negative J value means but in terms of the doublet, you'll see the same doublet.

All right, let's try a little bit more and then I want to show you one really, really, really cool example. All right, it kind of makes sense that if you start to change the amount of S character in the CH bonds you're going to end up changing the coupling constant so for example, if you go to cyclobutane you use more P orbitals

in making the carbon-carbon bond framework, you use more P character so you have more S character for the CH bonds so your J1CH is bigger so your J1CH for cyclobutane is 134 hertz instead of 125 hertz. For cyclopropane your J1CH is 161 hertz.

All right, so this is SP3 and SP2. Now you come to SP though and things end up really, really tricky. So SP, your J1CH, now you have twice as much S character.

Your J1CH is twice as big as an SP3. It's 250 hertz. Now that's tricky because a lot of the experiments you do depend upon using an average value so when you do a dept experiment there's a delay

in there that corresponds to 1 over the J1CH coupling. That's how it ends up working and picking out whether something is a methylene or a methine. We'll talk more about it when we talk about complex pulse sequences but if the J value is twice as big everything can get mixed up.

The practical implications are that acetylenes in dept so for example anything where you have a methine on acetylene may be completely messed up.

Opposite I'll say of what is expected. So I'll say dept and later on we're going to talk about HMQC and so because your J's are opposite you may see things

that you do not expect there. All right, the last example I want to give is incredibly wild and I pulled this out because I thought this is fun and I thought this is fun and this is cool

because it gives us every sort of heteronuclear coupling you could imagine and all sorts of bond coupling and I found this in that Aldrich collection of spectra and I thought hey we've got to take a look at this because it's fun and it's cool.

All right, so this is an isotopically labeled Horner Wadsworth-Emmons reagent. It would be something that if you wanted to do an isotopic labeling experiment and put C13 into your molecule in specific places you could put it in

but it's got an absolutely funky proton and carbon spectra. So just to orient ourselves remember the Horner Wadsworth Emmons reagent is a phosphonate ester and you have a regular ester and the typical Horner Wadsworth-Emmons ester. Both of them have ethyls and we won't worry

about the ethyl groups other than to say this is OCH2CH3 from both parts of the molecules and this is the OCH2CH3 but what I would like to worry about is what we see here in the middle.

Now what we see here in the middle is that CH2. Let's think about what's going on here. So that CH2 is being split.

It's being split by the carbons to which the hydrogens are attached which are C13s. It's being split by the phosphorous through a 2 bond pH coupling and it's being split by the C13 of the carbonyl

by a 2 bond CH coupling. So what pattern do you observe with 3 distinct coupling constants? Doublet of doublet of doublets and that's exactly what you see over here. It's a DDD. You can pick out your smallest coupling is this distance.

Your next coupling is this distance and your biggest coupling is this distance over here. So I get that this distance is 7 hertz.

This distance is 22 hertz and this distance which corresponds to the biggest J. It's either the distance between 1 and 4 or 1 and 5 but here it's clearly the distance between 1 and 5.

I get that it's 130 hertz. So it's a DDD, J equals 130, 22 and 7 hertz and that corresponds to a J1P, J1CH equals 130 hertz.

J2PH equals 22 hertz and J2CH is equal to 7 hertz

and that's pretty cool. Now there's even more cool stuff embedded in this

if you look at the C13 NMR. Remember this thing is isotopically labeled so most of the carbons are present only at 1 percent abundance but then that central methylene and the carbonyl are present at 100 percent or 99 percent isotopic abundance. So you get to see something that you never,

never see in regular carbon NMR except in what's called an inadequate experiment which we'll talk about at the very end of the quarter which is carbon-carbon coupling and if you look at the spectrum now we see some neat stuff. So this is the proton decouple. Wrap your head around this.

This is the proton decoupled carbon NMR but, of course, you still see carbon-carbon coupling and carbon-phosphorus coupling. So you'll look at this peak over here and you'll look at the lines over here and that peak is ADD. That's your carbonyl.

So that's your carbonyl. It's a doublet of doublets and doing the same thing we did before of taking 1 minus 2 and 1 minus 3 we can extract our J values. So our DD here for the carbonyl ends up being J is equal to,

let's see, I get 59 hertz and I get 6 hertz for analyzing that DD and if I do the same thing for this carbon here and I look at this carbon

and we can see it over here as this very, very, very, very nice, very pretty doublet of doublets for the CH2 and I do the same thing here and I get 134 hertz

and I get 59 hertz. All right, so let's figure this out because now we have a puzzle problem. We have 3 different J's

and we have 3 different relationships here. What does the 59 hertz have to, have to, have to correspond to, has to correspond to the carbon-carbon coupling because carbon, because coupling is mutual. So your J1CC is equal to 59 hertz.

Now, this guy is a doublet of doublets with 59 hertz and 6 hertz. So what does the 6 hertz correspond? Two bond phosphorous-carbon coupling

so your J2PC is equal to 6 hertz. The last coupling we observe is 134 hertz. What's the 100 for the methylene here? What's that coupling?

134, yeah. That's your 1 bond carbon-phosphorous coupling. That's your, right, 1 bond J1. Whoops, did, yeah, that's your J1PC is equal to 134 hertz.

All right, last thing, stretch your imagination. I don't have a phosphorous NMR, but imagine I had a proton decoupled P31 NMR spectrum.

What would the peak for phosphorous look like? One peak, everyone agree? Splitting pattern. What would its splitting pattern be? A doublet of doublets and what would the J's be?

Six and 134. So if we took a proton decoupled phosphorous NMR, you'd expect to see a pattern that looks, what is this Richard Nixon here? A pattern that looks like this, a doublet of doublets

with a really big J for the 1 bond coupling and a much smaller pattern for the 2 bond PC coupling. All right, well I think I've taken enough of your time today, so that sort of wraps up and that will set you in good stead

to attack this next homework set that has all sorts of cool coupling stuff. You had a question?