I want to talk about two things today. I want to talk about the inadequate technique and then give you some thoughts on homework set 9 and build on some of the stuff that I kind of started to scratch at the blackboard and I worked up something last night

with some models that I thought would be really useful for some of our thinking and tie into some of our conformational analysis. So let's talk about the inadequate technique first. It stands for incredible, it is a stretch, incredible natural abundance double quantum experiment

and what it is is essentially a CC COSY. So it's essentially a CC COSY and what's

so incredibly powerful about this technique is it means that you really can stitch together any carbon skeleton with it and what's inadequate of course about it is the amount of C13. So I'll say very powerful but requires lots

of sample and lots of time.

So by lots of sample I mean like 100 milligrams. In other words, if you're dealing with some rare natural product where you've isolated a couple of milligrams, right now it's essentially out of grasp.

If on the other hand you have some side product in a reaction and you think this is key to optimizing a reaction, maybe not a late stage intermediate in a synthesis but maybe in a methods project and you're trying to suppress something and you can get yourself 100 mgs

and you just can't get a handle on it any other way. This may be a way. So I mean the killer on this thing is because it's a COSY like experiment what you're relying upon is having a C13 next to a C13 and since the probability of having a C13 at either position

in a molecule is 1% the probability of having both of those is 1% of 1%, in other words 0.01% and that really is the killer because it means carbon,

you've done carbon on strychnine, we loaded up the sample tube and you know it's like 40 mgs of strychnine in the sample tube and it's pretty quick but if you're talking about having two C13s next to each other your signal to noise is going

to be 100 times lower. Now remember what I said if you're talking about a particular signal to noise level if you drop your concentration of sample by a factor of 10 you've got to collect data 100 times longer to get the same signal to noise ratio.

If you're C13 if you're dropping to a percent of a percent then you're talking about 100 times less sensitive so it really can be killer. Now there are tricks that get played and they're not fun if you've got a compound that you really want to get back so one of the tricks is that you often can add chromium 3

to increase relaxation. Remember how the quats are always small,

the quat carbons are small because their relaxation time is typically on the order of many seconds and you're pulsing every couple of seconds so most of your magnetization is not returning to the Z axis so you can add a paramagnetic compound, a metal ion that promotes relaxation that's going

to reduce the relaxation time so the magnetization returns the Z axis that will help you out a bit but we're often talking many hours or days of NMR time.

Now I will say this, the cryo problem, we haven't really sat down and tried this thing with the cryo problem, maybe Phil has, maybe somebody has. So the cryo probe instrument is super, super sensitive because you're cooling the coil in the probe and reducing the amount of noise. It's not increasing the amount of signal

but if you reduce the amount of noise you reduce, you increase the signal to noise ratio. The cryo probe is like 5 times or 10 times more sensitive than a traditional probe, let's say 5 times so that's going to buy you a lot because remember I said 5 times more sensitive, 5 times more signal to noise can translate

to cutting your experiment time down by 25 to get the, I'm sorry, no 5 times will cut it down by 5 but that still can help you out. So again I'd urge you to think about this. I want to walk us through one example and then get you started. We have 2 homework problems that have inadequate

and I want to get you started on one of them. So there are 2 variants of the experiment, just different variations in pulse sequence. One of the variants is like a COSY in terms of how it looks.

I'll show you both of them in terms of how your axes look so it's just like carbon on one axis, carbon on the other axis and you just correlate like you would in a COSY. The other is with what's called a double quantum axis

and I'll show you what that is. All right, so I think what I'm going to do is pass

out an example and we'll actually piece together, piece together the whole structure on this thing very, very easily and quickly.

Oh yeah, there are extras over there, just sweep them, sweep them back to your right.

Everyone have one? Great. All right, so unfortunately most, because this experiment really is a tour de force in terms

of instrument and so forth, most of the examples that you find out there are not worthy of the technique. They are didactic examples that sort of are nice but it's like okay, we could have done this without bringing such a powerful technique to bear.

A lot of the examples and handouts I've been giving you are from a book by Koji Nakanishi who's a natural products chemist and basically he put together various nice little sort of two-page handouts in the book on various spectroscopic, various NMR techniques.

They were recorded on a 400 megahertz machine so this is recorded on a 100 megahertz carbon NMR spectrometer so nothing and you know they're kind of old so it's nothing near as powerful as what you have downstairs with the cryo probe. So what we have here is an example of an alcohol

and it happens to have, I'll give you a formula here. It happens to have a formula C10H20O and so you'll have 10 unique resonances

in the carbon spectrum in this molecule and what we're going to do is we're going to stitch together the whole structure really painlessly, really, really effortlessly and really mechanically with this inadequate experiment. So I'm going to number my 10 resonances, 1 to 10 just like we always do.

These letters by the resonances, remember I mentioned before that before the DEPT experiment was developed there was an older

technique called off resonance decoupling. So when you collect your carbon normally it's proton coupled, proton decoupled meaning you're radiating the protons, there's flipping spins very rapidly so you don't see any J coupling to the carbon.

But with full decouple, with full coupling all of your peaks are very heavily split because you've got your 1 bond coupling which splits with like 125, 160 hertz coupling splits your lines into doublets and triplets and quartets.

But you also have on top of that your 2 bond and 3 bond coupling which means every peak is heavily split into many, many lines so it becomes like a quartet of triplets would be say what you'd see for the methyl peak in ethanol and the lines are very small. In the off resonance decoupling technique you're

partially decoupling, you're applying proton energy to flip the spins of the protons but you're not doing so right at the exact right frequency so you're doing so slowly so your multiplets collapse down and you don't see your 2 and 3 bond couplings

and you see just your 1 bond coupling and instead of having like a quartet spread out with the lines 125 hertz apart, they're just like 10 hertz apart. Anyway, that older notation which isn't used anymore has stuck around so a doublet means CH, a triplet means CH2.

You'll still sometimes see this and when people report depth data sometimes they will report it as D and T for no particular reason, D meaning CH, T meaning CH2, Q meaning CH3 of course

and the rest of these are CH3s. So okay, so our molecule contains some methyl groups, it contains some methylene groups. We have this one methine group down here, it's 70 parts per million. I said we're an alcohol so obviously this carbon,

the 70 is right in the range where you'd expect a carbon connected to an oxygen so obviously that's going to be the one connected to the oxygen. All right, the double quantum axis is a diagonal and what you're seeing, each of these pairs corresponds to two carbons that are coupled to each other and you'll notice

that each one appears as two little lines. What are those two little lines? Doublet and from the carbon-carbon splitting,

yes so you're seeing the carbon-carbon splitting. Remember carbon has a much smaller magnetogyric ratio than protons so the carbon-carbon coupling constant, the one bond J1CH is going to be smaller than a typical carbon-proton one bond coupling

so you can kind of see. This was done at 100 megahertz, this is done on 400 megahertz spectrometer so each of these tick marks is 100 hertz and you can see the spacing is maybe like 30 hertz so that's your J1CC coupling.

All right, here's how the double quantum axis works. Each of these coupled peaks, they come in pairs and there's a diagonal that goes between them and so if you look, all of these pairs go right around this diagonal.

I haven't split them exactly down the middle. I tried as close as I could so this pair is centered on the diagonal, this pair is centered on the diagonal, this pair is centered on the diagonal, this one on the diagonal, this one on the diagonal, this one and so on and so forth up the double quantum axis.

This is really important because in inadequate you're pushing the signal to noise level which means you will have noise in your spectrum most likely. This one, as I said, this is a 10-carbon compound in a demonstration experiment so there isn't a lot of signal

to noise in it but if you're doing this technique, you're basically going to be sitting on this spectrometer running like an overnight block, coming in in the morning, maybe doing this on a weekend, hoping when you Fourier transform and process your data that you can get off the machine but maybe you've signed up for all of Saturday, you know, Friday night through all of Saturday with the idea

that if you need more time you can get it and so you will see noise. I mean we see noise in our other spectra but some of that's artifacts rather than noise. You will just see basement noise here but what you want to be doing is saying yes, I can see a pair of peaks and you know where your pair of peaks is going to be.

In other words, it's not going to be a pair here and here, it's not going to be a pair here and here, it's going to be a pair centered around the double quantum axis. All right, so let's start to build up our structure and we can do so very mechanically. I'm just going to trace my pairs

and you can start anywhere. I'll just start here and we'll just sort of get our cross peaks and so this guy, this pair becomes 6 to 10. Okay, so 6 is a methine so I'll draw that as C6H and it's coupled to C10H3 so we're well on our way

to building up our structure. Okay, so let's continue and so I will just select my next pair and so that is 6

to 9 so that's good, C9 is another methyl group so C9H3 and then I'll continue.

We have the next one looks like 5 to 8. I could start to draw, was that right? 5 to 8, yeah, so that's over 5 and that one's over 8 so I could start to draw out another chain but I'm just going to sit tight for a moment

because I'll figure out where to put everything in just a moment. The next one looks like 4 to 7. The next one looks like 4 to 5 so as soon as I can figure out where 4 is I'll be in pretty good shape.

Next one is 2 to 7 and the next one is 2 to 6. Okay, that's good, now I've got a lot of information I can start building.

So 2 is a methine so I will say C2H and let's see, 2 connects I think the other one I got is to 7 which is a methylene so that's C7H2 and it looks

like we have 4 to 7 over here and 4 is another methylene. So C4H2, okay, so I've used up the 2 to 7, the 4 to 7.

Okay, it looks like I have 4 to 5 and 5 is a methine so C5H, okay, what else do I have? I guess I still have 5, 5 goes to 8, I haven't used him up

and 8 is a methyl so C8H3 and let's see, I think now I've used all of those up so let's see what I have next on the list here.

I have 3 to 5 so that's a methylene.

Yeah, isn't it? It just makes you want to run this thing until it takes you a day. Well the great thing with this, okay so I think what was happening on the 500 was people were getting antsy over aural

so if you want to sneak back there now I think trying to leave a few blocks of time but you've got this free NMR time on this course here, right, no one's going to be around Thanksgiving, right? So you want to run an inadequate

on strychnine on the cryoprobe. All right, so let's see where are we going here? Okay, so we've got, that looks like 1 to 3.

What's that? You mean I haven't drawn a straight line here?

Is it? Let's take a look here. It's hard staring into this thing so let me slap a grid on this. They look like, if you look at the grid lines they look like they line up.

I think I've put the grid on the axis here. I think everything's pairing up. They look like they're pretty well lined up. All right, so 1 to 3. Now 1 is that one that has to bear the alcohol. So C1H that has to be the one with the OH on it

and our last one here is 1 to 2 and so the whole structure gets built up.

Now we still don't have issues of stereochemistry answered at this point. So we have 3 stereocenters in the molecule and so if you were trying to figure out what the structure was at this point you'd, of course you can't get the absolute stereochemistry

so you have to get the stereochemistry of 2 relative to the remaining third one and that would be from things like coupling constants and NOEs. In fact this is a cyclohexane ring so it fits in very well to all the models and issues we've dealt with cyclohexane rings.

Anyway, that's largely what there is to the experiment and as I said very, very powerful, very tempting and teasing because of course it is so powerful and yet very frustrating because it isn't something

that you can pull out except in heroic circumstances. So this is the handout from Nakanishi's book on the technique which what I like about his book is he always has just sort of a short little paragraph on the technique

and then an example here. So this particular molecule happens to be the monoterpene menthol and so as I said there are 2 basic flavors of the experiment.

He's demonstrated both of them. I simply took the one on the left here and used that for our example here. So this is the one with the double quantum axis. You can see the pulse sequence. You notice it's all a carbon detected experiment

and then you have the version that's the COSY like version and your COSY like version basically just traces up and over just like you would with a COSY. No. No, absolutely not.

The natural products chemistry these days is often pushing the envelope where, so Phil Cruz is a great example of this and in fact he has a really good NMR book. The only thing is in my opinion it's more oriented toward natural products chemists rather than towards say synthetic

and other sorts of chemists who build molecules. So when you're doing that you're often like collecting 10 kilograms of wet sea sponges, getting a kilogram or 2 kilograms of dry sea sponge, fractionating it by extracting with say methanol and chloroform and then chromatographing and isolating bioactive components

and running various chromatographic techniques and then getting a few milligrams of compounds, sometimes less than a milligram. So one of the frontiers in NMR right now is really small probes and probe volumes so that you can concentrate your sample in a very small space and get enhanced signal to noise ratio

with small samples. In a way a 5 millimeter NMR tube, what's sort of this typical standard is a big sample in the sense you're putting a lot of milliliters in there but it's a very convenient size for preparative chemistry.

It's just that by the time you're dealing with a really late stage intermediate, say in a long total synthesis or a natural product, it may not be enough. So let me get, I want to get you started here on, were there other thoughts or questions on this I think?

Yeah, somebody? Why are the methyls on the isopropyl not equivalent? They're diastereotopic. Whenever you have a stereocenter in the molecule,

an isopropyl group is and a methylene group are going to be diastereotopic. If they're very far from the stereocenters, they may be coincident. They may show up at the same place but they are always topologically diastereotopic. So if I had like a 5 carbon, just a linear chain

with an isopropyl group at the end and a stereocenter way off 5 carbons away, yeah, chances are those 2 methyls would show up at the same position but if it was just a couple of carbons away, probably at least in the carbon NMR which is more sensitive, they'd show up at different positions.

All right, I want to give you just, so one of the things I wanted to do, this last problem set really is very, very beautiful and we get a lot of stuff in it but it's also pretty tough and I wanted to spend just a couple of moments getting you started on several

of the different problems and also talk about some conformational analysis and stereochemistry that's germane to them and actually was germane to the last problem set as well.

All right, so this is one of the problems from the problem set and if you look at this, I want to keep this fun for you. You get a low res mass spec, you're going to be able

to basically infer the molecular formula. It's just a carbon hydrogen oxygen compound. You're going to be able to figure out the molecular formula. There are 9 resonances in the C13 NMR and yet if you look

at it you'll see something interesting about the proton NMR. If you just work your integrals you'll see 2,

1, 1. You can slap, what do you do? You usually measure with a ruler or you can slap a grid on it if you're feeling lazy for measuring your integrals

and you don't want to be bothered drawing lines with a ruler so my preferred way if you can get a hydrogen count is always to add up all your integrals here so if I was adding this up it'd be like 1.7, 1.7, 1.7, 1.7 and then what are we at?

We're at about 9.3 and then divide by the number of hydrogens but if you're being lazy you can say okay. Otherwise you're putting all that weight on 1.7 and if the 1.7 is really 1.5 then you come to a peak that's like 9 hydrogens and that error of 10% is going

to make your 9 hydrogens be 10 hydrogens or 8 hydrogens but if you're pretty good here you probably, probably can get away eyeballing it as 1 and 6, right? So you're probably going to be well on your way

to a molecular formula here. Is that right? Oh yeah, we've got this guy over here so we've got 6, 8, 9, 10, 11 and it looks like we've got something here as well. OHs and the like are always hard to integrate in part

if they're often broad, often the integrals aren't so accurate because phasing of your integral becomes really, really important. This has got to be some sort of OH, some sort of OH over here.

All right, with 6 you might even think about what that means in terms of the structure of the molecule. I'm going to take us on to the inadequate and honestly you can pretty much just by inspection

with the inadequate put the thing together and then you're going to have to use your head a little bit to think about what's going on. All right, so they give you two spectra, the spectrum

and then the expansion of this crowded region here so this is your inadequate and we can go ahead and do 1, 2, 3, 4, 5, 6, 7, 8, 9.

You can identify your double quantum axis in addressing this spectrum.

Should have 10 carbons. What? Did I miscount? This here?

Yeah, so that's what we were seeing with the 6 hydrogen so what is that telling us about the 6 hydrogens? Probably not diastereotopic, probably some sort

of isopropyl group in there. Yeah, so okay, so I want to point out a couple of features. So one feature is, all right, so like here for example, you can pick up 8 and 9 and it's right across the double quantum axis. And here for example, you can pick up,

they've even drawn a line for you, you can pick up 8 and 3 but what can you tell about this and this spot and now I've just circled it so badly that I've circled over it. What can you tell about those two?

What? That can't be, it may be an artifact, it may be noise but they're not a cross peak because they're not spaced on that double quantum axis.

So some type of artifact but not a product of the double quantum axis. You know, it is something off of, so what's interesting here, so this is our carbon doublet, you see that,

that's your CC coupling but here these are singletons, these are spots, they're not doublets. So whatever it is, it is not J1 CC coupling. We have a spot under here, here let me transcribe my line so it's 1, 2, 3, 4, 5, 6, 7 so we have a spot under 6,

a spot under 7, a spot under 1, a spot under 2 but they are not part of our pairs that we need to consider and so then as we continue our double quantum axis here,

you'll see that you can just pick up your pairs. You're going to get your 4 to 7 pair, you'll get your 4 to 5 pair here, you've got a 3 to 6 pair here, you have a 3 to 5 pair below that, a 2 to 7

and a 1 to 6 pair here and then what is it, what's going on over here?

That's got to be a 1 to 2 pair. If 1 and 2 weren't giving a cross peak, you wouldn't see a spot here so even where they're really close we're seeing a 1 to 2 cross peak so that's going to, you're all set

to go ahead and tackle this on your own. I don't want to, you know, I think the answer is it could be an issue. Your Prech book and Silverstein will give you your J2

and J3CC coupling and they're going to be small, they're going to be a lot smaller. We already saw on problem number, on that Santonin problem, that very nice sesquiterpene problem, we saw that darn peak from 4 to 7E which by the way now I have gone ahead

and put a note on the problem. So we have seen in some cases you can get some longer range coupling showing up. It's surprising. So I wouldn't say never but it would surprise me.

Certainly if you look at your little paired couplings, that's got to be an indication that it's 1 bond coupling. I can already tell you, so this molecule has a lot of SP2 carbons in it as well as some SP3 carbons in it. If you had like SP carbons in it you could imagine them

being further, even bigger spacing but I think if you were and that may be what we're observing here. It might be something that's a long range type of coupling. Honestly I don't know but there, you notice we're not seeing this paired lines here so that's kind of an indication. All right, I think this and the IR spectrum are going

to be enough. You can do a little bit with the COSY on it. You can do a little bit. Take a look at the IR spectrum. There's obviously some stuff going on. This molecule has carbon, hydrogen, oxygen.

We obviously have some type of alcohol. We've got some stuff going on here that's interesting either with carbonyls or double bonds or something. So what I'll say is something that's pretty obviously an OH but then something interesting.

You'll figure it out at this point. All right, I think that's all I want to do on that problem to help you get started. Let's see, what else do I want to do? I think the other thing I want to do at this point is to spend some time talking about some of the issues,

issues that I was talking about on the blackboard in discussion section. I worked up an exercise and you're welcome either to follow along or not on your computers on this.

If you were to remove that, no, then that would,

they would at that point become equivalent, yes. However, the most, yeah, so meso compound are going

to be equivalent but remember the most rigorous test is always replace one methyl group with like a C13 or a CD3 group, replace the other methyl group and then ask yourself what's the relationship of those two compounds? Are they diastereomers or are they not diastereomers?

But remember, if those two methyls are interchangeable by a symmetry operation, a mirror plane in the menthol which they are, then they are not diastereotopic. So plane and simple. All right, so I want to talk about some stuff that is directly germane to the homework

and also germane to last week's homework and I want to talk a little bit more about this concept of some very, very basic conformational analysis and this is fundamental to organic chemistry. It's fundamental to Kerry and Sundberg part A as well.

It's fundamental to understanding structure. So I was quipping before that I can draw only two things and everything else I can fake it and that's really kind of true in the sense if you can do two things, you can do basic conformational analysis

on a whole bunch of cyclic, bicyclic and polycyclic ring systems. So if you understand chair cyclohexane, then you can basically understand an envelope cyclopentane. Cycloheptane has a number of different conformers

but you can basically fake cyclohexane. You can start to figure out, okay, if hydrogen is here and here and so forth and you can fake a whole heck of a lot. If you can understand norbornane,

if you can understand bicyclo221 heptane and that really is a poor drawing so let me do a little bit better here.

If you can understand bicycloheptane, then you can fake a lot of other cyclic ring systems because you can go ahead and fake say a bicyclo321 system.

This bridge, you notice that embedded in the bicyclo321 system you have and I guess this is a little bit better drawing which means I probably should be faking this like this actually. So you notice you have a cyclohexane embedded in this system and it's a chair cyclohexane.

You can also flip up if you had a bad steric interaction here you could flip up, that would be a boat cyclohexane. Can you see your boat embedded in there? And you can go ahead and if you want to then you can fake other systems so if you can fake this system then you can fake

like remember we had the pinane diol problem and you can fake the conformational analysis of this as well in your drawing and if you can't see this the first time around the beautiful thing is a few basic models can help you

out and so after thinking about this and thinking that really my own understanding of this comes from sort of playing with models I figured well if something is good for me it's going to be good for you

and so both on the class materials and the assignments I created a webpage that links to some basic models of all of those structures and so I created this simple conformational analysis

of these structures so I've hot linked these all to PSE files, to the pi mole files and I guess what you have to do at that point is you sort of have to open them separately. I've programmed my computer to, let me see if I can turn down the lights here, let's see we have a lights button

so okay that's going to be a little bit better. Yeah, I've programmed my computer to open these like this. Obviously, I mean you know exactly

by this point how you get what probably some of you are teaching your sophomores to draw the relationship between cyclohexane projection and the three dimensional structure of the molecule

so of course you have your axial and your equatorial hydrogens on that but what I was saying about an envelope cyclopentane structure is you can go ahead and basically apply that same conformational analysis

to methyl to cyclopentane and so you see I've sort of continued and done half the chair. Now this brings up an interesting point. Why am I talking about this? Because I'm interested in coupling and being able to identify stereochemistry from coupling and NOE's

and being able to identify NOE's and one thing we already had one problem on the homework where we had a cyclopentane ring system. It was that hydroxyproline one where we had to figure out the stereochemistry from the NOE's and I think it was in part confusing because in the 1D NOE there was non-specific irradiation

of the diastereotopic methylenes. One of the points when I introduced NOE's earlier on and we were talking about distances and I said well you know on a cyclopentane ring basically if you've got a cis relationship

or you've got a trans relationship you know you're going to have some difference in distance but it isn't necessarily if I only saw an NOE here and I said oh these two must be cis to each other because I'm seeing you know 2.8 angstroms will give you a nice strong NOE you'd be absolutely wrong here.

You know ditto if I were looking at say this hydrogen and these two hydrogens and this is one of the reasons I like these canonical structures and I like being able to play with these models is it really helps me think about what we're talking about.

In other words you really are ending up having to make comparisons on cyclopentanes and 1, 2 relationships. In pi mole or from NOE's? Oh in pi mole it's from the wizard and there's a measurement for and it gives you either distances or it's a pull

down menu and you can get dihedrals and so forth. So remember that coupling constant calculator and we did it two ways. One was by taking it on to the modeling facility computers and using macro model where it had a car plus calculator built in and the other way was a website that one linked to the tetrahedron paper on typical J values

with different substituents. So if you just wanted to use the website you could just get your dihedrals, you pull dihedrals there, you click on the four atoms, you go hydrogen, carbon, carbon, hydrogen, you get your dihedral and you put in your substituents on the pull down menu on that tetrahedron calculator or at least

if you're thinking basic car plus curve you say, oh wait a second, 90 degree dihedral virtually no coupling constant, 170 degree dihedral, well remember it's a cosine wave basically, 170 is going to be a big J value. Now one thing that's interesting

with cyclopentanes is for 1, 3 relationships, so let me hide some of these measurements, I'm just going to click on these to turn it off. For 1, 3 relationships you really don't get close transrelationships. SIS ends up, an NOE in a 1, 3 relationship,

so we're going 1, 2, 3 in a 1, 3 relationship, an NOE is pretty darn diagnostic of being SIS because that's 2.7, so here we go in the trans, it's 3.8 and 3.8 angstroms, remember how I said NOE's fall off as distance to the inverse 6 power, 3.8 angstroms is going

to be a very, very small NOE, just a small percentage of what a close NOE would be, so that can be really, really diagnostic. All right, I want to show you at this point a 7 membered ring, 7 membered rings as I said can have different confirmations

but often, often you're going to find a confirmation that's very similar to this canonical cyclohexane, it's my faking it and you notice you have 3 hydrogens

that are basically axial and 3 that are basically axial here, I've put the methyl group equatorial here on the ring and so you can start to see, oh yeah, then all those things I was thinking about like a big J between these hydrogens or a big J between these hydrogens or a close contact between these hydrogens is going

to apply, the other family I thought that was really useful, here's our norbornane structure, these are all minimum energy structures, you know the clean function, it's generated with molecular mechanics with the MMF force field.

So here's the norbornane structure, if you want to go ahead and see the projection that I drew over there what I badly sketched out on the blackboard, that's our basic structure but now you can start to imagine with issues like when we were dealing with the camphor problem

and we were dealing with that issue of, oh yeah, what's the distance of a methyl in the bromo camphor sulfonic acid problem, what's the distance here, I'll even clean this thing up not that we really need to to measure a basic distance,

what's the distance for say a cis distance, right, because that was where we had this issue of is the bromine pointing up or is the bromine pointing down and you'd be able to look and say okay, you know, by golly a cis distance is going to be a nice strong NOE, a trans distance is going

to be a much, much weaker NOE. It also helps you calibrate, so remember we were dealing like with a methylene or a hydrogen off of there and you can start to say, oh yeah, I can begin to think about these types of distances as well. So that's why, I mean this is so integral

to understanding stuff. So I'll give you a bicyclo octane, bicyclo 321 octane structure and so we've got this. This is, you notice you have your cyclohexane embedded in it. This is what I was talking about.

You see your chair cyclohexane embedded in the structure there so it's sort of just like a chair cyclohexane embedded in the structure so you can play with that. Here's the bicycloheptane. This you could use as a, the bicyclo 311 heptane.

This is one that you could use as a template. You did, right? Because you said it was flattened, right? You told me it was flattened. You modeled it or? Yeah, so this is our pinane diol problem. The other canonical structure for,

so basically if you get a few simple structures, as I said you can do so much. The other one that's sort of canonical for understanding conformational analysis in so many natural products is trans-decalin and which is basically just cyclohexane

with another ring extended on and it's chair-chair and cis-decalin is also chair-chair and here I've drawn it the way you'd get from a projection and here's your cis-decalin and it's just a chair

and then we have an axial group and another chair. Let me, last thing I'll do is let me just show you how these fit into a couple of things. So here's our strychnine molecule. You've all modeled this and you notice

on the seven-membered ring, here I'll bring it over. Here you can see the seven-membered ring even though we have an oxygen in it, even though, let's see, that's a little hard to see.

There you go. Even though you have an oxygen embedded in your seven-membered ring, you can see this chair-like structure. Even though you have a double bond in it and oxygen in it, you've got your three sort of axial hydrogen

so that's going to influence your thinking as you start to assign your protons and assign your stereochemistry. The problem that from the previous problem set that I presented as a demonstration where I was talking about and drawing things on the blackboard, so you notice this,

you've got your chair cyclohexane, your angular methyl, your angular methyl was the one that gave the NOEs to these axial protons.

That helped us assign our stereochemistry. You had your NOEs like this hydrogen was one that was a real, I guess this was the real linchpin because this one was the hydrogen that gave the NOE to the methyl, gave the NOE across here, gave the NOE here. All of those are short distances.

All of those are just, you know, two basically Van der Waals contact or thereabouts. So all of those are nice, short, strong NOEs, 2.6, 2.2 angstroms. The other one which was a little messier to see when we irradiated the methyl and we saw the NOE

over to this other hydrogen. Again, that sort of gave us our stereochemistry here. And you notice this structure here with the cyclohexane and the five-membered ring is basically what I'm talking about with the trans-decalin and then your five-membered ring, the conformational analysis is essentially

that envelope cyclopentane. So you understand trans-decalin. You understand envelope cyclopentane's relationship to cyclohexane to chair cyclohexane and you can fake the conformational analysis of this system and then you just sort of have a double bond coming off kind of at the middle position here, equatorial position here

and you have that last ring. So there's a lot of power in being able to model and analyze structures. All right, well I think that's where I'd like to end things up in terms of discussion. Go and attack the homework problems. These are some of the prettiest

and most satisfying problems of the course. On some of them, if you make a model, it's going to be of the actual molecule, it's going to be incredibly enriching when you start to get your stereochemistry and get your diastereotopic assignments and you can start with those PSE templates if you want.

Oh yeah, last thing I'll mention. All right, actually I will skip it. I will leave you on your own for one last point. You will figure it out. You want it?