Lecture 11. Magnetic Equivalence, Spin Systems, and Pople Notation
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00:00
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Transkript: Englisch(automatisch erzeugt)
00:07
Today's lecture is basically going to show you that a lot of the rules that you learned in sophomore organic chemistry like the N plus 1 rule are simplifications and we'll be seeing a lot of examples
00:21
that look really weird because I want to show you where these simplifications break down and then next time we'll get to more sort of the typical rules of splitting. All right, so as I said I wanted to begin with this notion of magnetic equivalence and let me begin with a definition.
00:40
We'll say magnetic equivalence and let's say that two protons are nuclei, in other words normally we're
01:01
talking about proton NMR but of course it could be fluorines or phosphoresis or whatever, are magnetically equivalent
01:21
if they are chemically equivalent and remember two nuclei that were chemically equivalent were nuclei
01:42
that were interchangeable by a symmetry operation or a rapid process like rotation about a bond. So we saw lots of examples of chemically equivalent nuclei by symmetry and then we saw how when you have for example a chiral center methylene group is no longer chemically equivalent.
02:01
So chemically equivalent, so they're a subset of chemically equivalent and they have the same geometrical relationship to all other nuclei in the spin system.
03:08
So this brings up one other concept and that's the question of what is a spin system and a spin system is just a complete set of nuclei and I'll underline complete meaning all
03:28
of the nuclei. So a complete set of nuclei in which members are coupled.
04:04
All right this is, let's start with the easy idea, the spin system and let's just do this by example. So for example if we have ethyl propyl ether CH3, CH2, O, CH2, CH2, CH3 we have in the molecule two spin systems.
04:30
We have one spin system comprising the ethyl group and another spin system comprising the propyl group.
04:48
So in other words the ethyl group is a set of nuclei, obviously we're talking about the protons since for all intents and purposes there are no C13s in this fragment so we have these hydrogens
05:02
and these hydrogens and at least one member is coupled to every other member. They make up a set together. In the propyl group we have the methyl group, the methylene group and the other methylene group and there's coupling among them. In other words the CH3 is coupled to the CH2,
05:21
the hydrogens of the CH2 are coupled to each other, the hydrogens of the CH3 are coupled but that doesn't count. The hydrogens of the CH2 are coupled to the next CH2 and what's important, so each of these is complete meaning it takes in all the coupled nuclei. What's important is we don't have coupling
05:41
between the ethyl spin system and the propyl spin system so they constitute two separate sets. So we can consider the ethyl, we can consider the propyl and there's no interaction between them. Let's try another example. Let's take acetyl phenylalanine
06:03
and we'll take the methyl, the mid. So what are the spin systems in this molecule?
06:47
So you have the two methyls on the end and the benzyl group.
07:28
Okay, for all intents and purposes the benzylic protons are not coupled to the phenyl so what would you do
07:40
for the spin system here? Separately, so I'm going to revise this, okay, so we have the phenyl that's going to be one spin system. What do we do here in the middle?
08:03
Alpha and beta, what about the NH? Now we saw an example where I said you have an D2O and you exchange so it's deuterium there which although it has a little spin for all intents and purposes you can discount but what about this NH?
08:23
Is that going to be J coupled? Remember amides are different than alcohols. Alcohols exchange amides and remember I said alcohols can exchange or cannot. Amides on the laboratory time scale if I throw them in D2O will exchange but on the NMR time scale
08:43
that NH stays there and we're not doing this in D2O so what should I do with this middle part of the molecule? So that all becomes a spin system and then what
09:00
about the very end of the molecule? Okay, good, so these guys interact. Now I'll tell you right now so we have 4 spin systems
09:20
in the molecule, we have the methyl group, we have the NH, the alpha and the beta protons and we have the phenyl group and then we have the methyl amide group so 4 spin systems in the molecule and we can look at each of these separately. I'll tell you there's a miniscule like undetectably small and I'll show you how to see it
09:44
if you squint right later on coupling between these hydrogens and the benzyl group but for all intents and purposes you can say that the phenyl group is not coupled over here. So for all intents and purposes we have this
10:02
as an isolated spin system, the phenyl group, the alpha, beta and NH, the methyl and the methyl amide. Other thoughts?
10:25
Yeah? Yep, yep and so in chloroform solution in D2O these would eventually wash out but in chloroform solution what we'd see for this NH is a doublet, it would be a little bit broad.
10:42
This one is going to have 3 coupling partners in chloroform solution where this hasn't exchanged or in DMSO so we'd see either a doublet of doublet of doublets or a triplet of doublets or a doublet of triplets and we'll talk more about that if you're not familiar with those terms. This NH would appear as a quartet
11:03
and chances are it would be broadened out a little bit, remember I mentioned this nitrogen quadrupolar coupling so a couple of ways this can appear. So it can appear as a 1 to 3 to 1 quartet slightly broadened.
11:22
It can appear just due to this nitrogen quadrupolar broadening as an envelope that encompasses the whole thing or it can appear as something where if you don't see the wings of the quartet and you just see a little dip you might say oh it looks
11:42
like a doublet to me so depending on the quadrupolar broadening from the nitrogen this methyl group in turn is not going to have significant quadrupolar effects, it's going to be split into a nice doublet so this will be a quartet
12:05
or broad quartet or something that looks like a singlet. If it's very broadened the methyl group will be a doublet.
12:37
All right let's now tackle this notion
12:41
of the same geometrical relationship. So let's look at this molecule. Let's take 2, 6 dichloro 1 tert-butyl benzene so as far
13:06
as chemical equivalence goes we have 2 types of protons. We have the proton that's para to the tert-butyl group and the proton that's meta to the tert-butyl group so these 3 constitute a spin system.
13:22
Chlorines don't count their quadrupolar nuclei and essentially not spin active. The tert-butyl group is magnetically isolated, it's its own spin system so we look at this and we say all right we have 2 protons that are the same as far as chemical equivalence, they're interchangeable
13:42
by a symmetry operation and now we ask this geometrical question. Do they have the same relationship to all other nuclei in the spin system? And this hydrogen says oh look I'm ortho to this hydrogen
14:00
and this hydrogen says oh look I'm ortho to it also. So these 2 are magnetically equivalent as well as chemically equivalent.
14:28
Now there's a way of naming systems where you have different types of protons and we'll give a different letter to each type of non-chemically equivalent protons. So for example we'll use letters like A and B and C
14:45
and M and X and Y if you need to. And the general idea is if the protons are close in chemical shift we'll use letters that are right next to each other in the alphabet, A's and B's and C's.
15:02
If they're far apart in chemical shift we'll use letters that are far apart in the alphabet, letters like A's and X or A and M and X. So depending on whether these protons are close in chemical shift to the center proton or whether they're far
15:22
in chemical shift we'll either call this an A to B spin system or an A to X spin system. Now technically only ones
15:42
where they're far apart are truly first order. But even if they're close there are some very regular patterns that you can see. If they're far apart in chemical shift and by far apart what I mean is the separation of the peak centers in hertz is many, many times the coupling constant.
16:02
So like a typical ortho coupling constant is about 7 hertz. So if the peaks are far apart like 10 times as far apart like 70 hertz or 100 hertz or 200 hertz apart then they will end up being A's and X's.
16:20
Now remember at 500 PPM 1 PPM is 500 hertz. So in other words if these guys are about two-tenths of a PPM apart, three-tenths of a PPM apart we would call this an A to X spin system. What we'd expect would be to see a doublet for the two
16:47
on the outside because they're being split by the one in the middle. And a triplet and so I'll just draw a little squiggly to indicate these are far apart in the spectrum.
17:01
And a triplet like so for the center hydrogen. I guess technically the triplet would be shorter than the doublet so I'll make the doublet a little bigger.
17:26
If they're close together and I'm going to actually start in just a moment with the archetypical AB system. If they're closer together what you will see, so if it is indeed A to B what you'll see is a slight
17:42
tenting inward depending on how far. In other words the lines of the doublet instead of being equal in height will become unequal in height with the bigger line toward its J coupling partner. And the lines of the triplet will be similarly distorted
18:04
so that the inner line is a little bigger than the outer line. I always like to think of these as sort of tenting in toward each other and that would be what it would look like as an AB, A to B system.
18:28
Let's try another example and I'll take difluoromethane.
18:40
Now remember fluorine is spin active, spin of a half. It's magnetogyric ratio is about 90 percent of that of a proton so it shows up a million miles away. Whereas your protons are resonating at 500 megahertz, your fluorine is resonating at 470 or 460 megahertz.
19:06
So they are far, far, far away from each other but they're J coupled to each other. So collectively the hydrogens and the fluorines constitute a spin system. If I want to remember my geometry because we're going
19:22
to ask what type of spin system it is, it's a tetrahedral molecule. So the geometrical relationship of this hydrogen to the fluorine is the same as the geometrical relationship of this hydrogen to the fluorine. In other words we would call this spin system an A2X2
19:46
spin system. We have two hydrogens that are chemically equivalent. They're interchangeable by a symmetry operation and two fluorines that are chemically equivalent. They're interchangeable by a symmetry operation reflection
20:01
and the hydrogens if you test everyone has the same geometrical relationship to all other nuclei in the spin system. So this is an A2X2 spin system. No, it's just that they're far away.
20:21
As I said, if these guys are more than a few tenths of a PPM away we would call this X in this case and if there were two of them in some circumstances we'd call them X2. So if you look at the H1NMR spectrum or the F1NMR you would expect the H1NMR spectrum to show up as a triplet.
21:07
Now I want to contrast this example with another example, difluoroethylene.
21:31
How would we characterize that spin system?
21:57
What ABXY?
22:02
Okay, let's start with this issue of chemical equivalents here. So are the two hydrogens chemically equivalent to each other? Okay, so these guys are chemically equivalent and the two fluorines?
22:25
Yeah, okay, chemically equivalent. All right, so we're going to use the same letter but we have a problem now. They're not magnetically equivalent. So we need to introduce another term. When we have hydrogens that are chemically equivalent
22:42
but not magnetically equivalent we'll use or nuclei in general we'll use primes. So what we'll do is we'll call this an AA prime, X, X prime spin system and the big difference is
23:03
that while a system that's an A2X2 system is first order and even these types of systems I'll call pseudo first order, this is a spin system
23:28
that is distinctly not first order. I want to show you the difference between them.
24:20
Okay, so on the top I have and these are very old spectra. These are spectra from a book that were taken at 60 megahertz so they're probably from the 1960s. They're taken on a CW instrument. CW instruments are non-Fourier transform instruments not used
24:40
anymore but these little wiggles are just artifacts of it being a CW instrument so don't worry about that. But the main thing here is the difluoromethane, these are proton spectra is exactly what you would expect.
25:01
It's a triplet. The difluoroethylene you look at and you say what's going on, there is no simple description of this pattern.
25:29
This type of pattern can be calculated. Indeed our NMR spectrometers have software that will calculate it. It can be calculated by computer program, by back of the envelope calculations for simple systems
25:44
but basically defies a simple description. Why are the hydrogen, which one? Well here but it's not the hydrogens. Here it's the fluorines are splitting the hydrogen. So the difluoromethane is a triplet
26:02
because the two hydrogens are split by the two fluorines. In the case of the difluoroethylene the problem is the adage that we use that hydrogens that are the same don't split each other really replies most
26:24
rigorously to hydrogens that are magnetically equivalent but hydrogens that are chemically equivalent but not magnetically equivalent kind of sort of do. There are many circumstances where for all intents
26:41
and purposes you don't see any effect and these are cases that I'll call pseudo first order cases and what I want to show you today is how all the stuff that we learn in sophomore chemistry for coupling really doesn't rigorously apply to lots and lots of common systems.
27:02
Now the first thought when you look at this is okay, well this is, you know, this is difluoroethylene. It's not a common system. So let's take a common system and it's one that you're going
27:24
to see in the course of your graduate career most likely. So this is dioctyl phthalate. It's commonly used as a plasticizer
27:40
in all sorts of plastics. If you go to the store and you buy yourself a water bottle and it says phthalate free that's probably saying or you see plastics listed as phthalate free, that's saying it doesn't have this plasticizer, this compound added this oily compound to make plastics pliable.
28:02
Tigon tubing is great with water. However, because it contains dioctyl phthalate if you use it on your manifold, methylene chloride and THF vapor will dissolve into the tigon, dissolve into the dioctyl phthalate, dilute it and it will dribble into your reaction flask
28:22
and when you work your spectrum up you'll see the spectrum. When you work your compound up you'll see a spectrum for this and you'll say what the heck, what's going on? Now let's take a look at the molecule and figure out what sort of spin system it is. So we have the octyl groups which are separate
28:42
and we have the benzene groups. How do we characterize the benzene here? AA prime, BB prime indeed or if they're far apart
29:07
in chemical shift, AA prime, XX prime. So in other words if these guys are within two tenths
29:21
of a PPM of each other with typical couplings you might call it AA prime, BB prime if they're more than a couple of tenths of a PPM apart. You'd probably call it AA prime, XX prime. What that says is your rules of simple coupling may not apply here
29:41
and so many a graduate student has taken their reaction mixture and gone and seen the following pattern in it and come to their advisor or their group mates and said what the heck is going on? This is dioctyl phthalate and again it defies a simple description.
30:02
This is an AA prime, XX prime system and it doesn't matter how high a field you go to the spectrum of dioctyl phthalate is going to look like hell. The only difference as we change field strength is you see
30:21
how these lines on the inside are a little bigger than these lines on the outside. If we went to a lower field spectrometer they'd be a little more unequal. If we went to a higher field spectrometer they'd be a little bit more equal, but no matter what the spectrum of dioctyl phthalate is going to look like that
30:42
and that should freak you out just a little bit because it says all that stuff you learned in software chemistry kind of sort of applies but kind of sort of doesn't. Nothing. So the neoprene is great for your manifold.
31:03
The one that's even better so that is more permeable to air and so it's okay. The one that's really good is butyl rubber which I think it's a butyl nitrile copolymer.
31:21
Anyway, that's particularly good for manifold lines but stay away from tigon. Tigon is meant for aqueous solutions. All right, well now that I've messed with you a little bit for your own good, really it's for your own good. Now that I've messed with you for a little bit let's take a look and see when the rules
31:41
of software chemistry apply and when they don't necessarily apply rigorously and let's play with the implications of this. All right, so if I look at chloroethane, now my first thought and I'll do a Newman projection on it.
32:07
My first thought is wait a second, well the methyl group we talk about geometrical relationships and this is confusing because this hydrogen is para to this hydrogen and this hydrogen is ortho to these hydrogens but of course there's rapid rotation
32:23
so let's see what happens. So remember I said rapid processes can equal things out. So we'll call these hydrogens 1, 2, 3 and we'll call these guys 4 and 5 and I'll just imagine a series of rotations and so you have 3 equal rotamers
33:11
and although H5 is anti to H2 in the first rotamer, we have the second equal rotamer,
33:22
they're all equivalent, they're all equally populated where H2 goes to H5 and H1 is anti to H5 and this third where H3 is anti so in the end it all evens out and if we want to describe this spin system how would we describe it then
33:50
and which take a guess, we've talked about the chemical shifts of dichloroethane. AX and what specifically would we use for numbers?
34:02
How many hydrogens are there in the methyl group? Three. So what sort of spin system do we, it's an A3X2 spin system and all of these spin systems
34:28
where there are no primes are truly first order and while the Bs will, when you get things that are close it will deviate from first order but if they're on top of each other it gets really messy but if they're a little bit separated you basically can call
34:42
it an AB type of system. All of these spin systems are first order. First order means those simple rules of coupling count up number of different types of neighbors work out perfectly
35:01
so an A2X2 spin system you can trust is going to give you a triplet and another triplet. In the case of dichloromethane the other triplet is at 460 megahertz whereas one triplet is at 500 megahertz but you get two pure triplets and in the case
35:22
of here we expect a triplet and a quartet and everything is hunky dory. Now let's take a look at a different compound. I'll take a look at bromochloromethane and we're going
35:41
to do the same thing. I'll Newman project, I'll put the chlorine on the back and the bromine on the front. We'll call it H1, H2, call these H3 and H4 and I'm just going
36:11
to imagine the rotamers.
36:50
So we have two gauche rotamers and one anti-rotamer
37:05
so we could consider the gauche rotamers as a pair but they're separate from the anti-rotamer. All of them of course are interconverting but let's just look at the anti-rotamer for now because the anti-rotamer isn't equivalent to the two gauche rotamers and you'll see the conundrum that we end up with.
37:23
See the problem we end up with in the anti-rotamer and in this here you can say well H1 and H2 are here's 1, 1 is gauche and then the other gets to be gauche so you can say let's consider them as a pair. Here we say okay well the bromine is anti, these two fine, they have the same relationship.
37:43
They're interchangeable by symmetry so H1 and H2 are symmetrical, H3 and H4 are symmetrical but the problem particular to the anti-rotamer is that when you apply this test
38:00
and say what's their geometrical relationship it's different and so you look at the anti-rotamer and you say this is going to be an AA prime, XX prime or AA prime, BB prime spin system and the problem is this
38:25
is that situation that's truly not first order. Now the good news on this is most of the time
38:46
and I've just given you some of the ugliest examples, most of the time non-first order systems can be approximated as first order systems. In other words you ask your sophomores what the spectrum
39:01
should look like and they'll say oh well they told me the N plus 1 rule, it should be a triplet and a triplet and that's largely true but I'm going to show you cases that break down from that. Now I want to show you the sort of break down from non-first order to let's call it pseudo-first order.
39:28
So in other words if we take a simple AB system or a simple AX system, let's say I do bromine, bromine,
39:41
chlorine, chlorine so this is a simple AX system if they're far apart you'd see a doublet and a doublet and again I'll draw a break to indicate that they're far apart and if they're closer together we see
40:02
what's called an AB pattern where the two tent into each other and if they're very close you'll see an AB pattern with the inner lines very big
40:23
and if they're really close you might even mistake it for a quartet, it's not, it's an AB pattern. Anyway, so most of the time you can get away approximating non-first order systems as sort of pseudo-first order systems.
41:09
In other words often non-first order systems will show behavior that's very much like you would expect with just the notion that inner lines may become a little bit bigger but as we saw
41:22
in our example with di-octyl phthalate you can have some very, very big deviations and so what I want to do now is show you really a catalog of typical deviations because once you see them and once you see when they come up I think you'll be much less freaked
41:40
out by things that occur. So the scary thing about the diagram that I made
42:02
on the right hand board is any time you have a methylene chain technically it is not first order. Every pair of methylenes, every methylene the pair of hydrogens technically they are chemically equivalent
42:22
but not magnetically equivalent. Of course if there's a stereocenter in the molecule they're not chemically equivalent then it's like an A2 and ABX system but in the case of just a plain methylene chain without a stereocenter they are chemically equivalent
42:41
but not magnetically equivalent. Normally you can get away and you say okay they taught me as a sophomore the N plus 1 rule I expect to see a triplet. Normally it works pretty well. I'm going to show you some cases where we see some very, very big deviations and I want to show you where these things come up.
43:00
So these are two different molecules that I've worked with. One of these has a propyl chain connecting an azulene, a very bulky group to a phenyl group and so you look at the protons on this chain and you say okay they all look
43:21
this methylene that looks kind of reasonable looks like a triplet. You could call it an apparent triplet if you liked but the one that's right next to the azulene, this very, very bulky group really ends up looking very, very funny.
43:48
You see this pattern that has what kind of looks like a triplet except in the center it's further split and the one over here in the middle also looks a little funny. It doesn't look quite like a quintet.
44:02
See the thing is with the CH2 chains if you've got a mix of anti and gauche conformers, you've got some anti but also some gauche, basically it averages out enough that it behaves like a first order system. It behaves like you were taught
44:21
it should in sophomore chemistry. However, if it's heavily biased toward the anti-conformer then just as we saw in di-octyl phthalate, this really funny splitting, you see the same thing and this group is very bulky here. So in other words when you're looking at this you have the azulene group
44:42
and then you have the hydrogens and then it is almost completely locked in the anti-conformer and so you really, really end up seeing this. So this ends up being an AA prime, M, M prime, X, X prime spin system.
45:02
That's what it technically is. So technically it's not in first order but we get that effect full force over here. You'll notice these types of patterns come up again and again so here's a very different compound where you still have a tri-propyltrimethylene chain
45:21
but now your bulky group is this TMS group and the hydrogen that's next to your bulky TMS group again gives this exact same pattern, a completely different molecule but exact same pattern of non-first order behavior.
45:40
I would just call this peak a multiplet. So I would call all of these multiplets and I would simply list their range. I guess I'd call this guy an apparent triplet. Exactly. I mean we have to understand this is a non-first order system and that normally we can get,
46:03
often we can get away describing non-first order systems as first order but you really can't always. Let me show you some other non-first order behavior. So okay, so most of the time take bromopropane, another molecule with a chain in it and remember most
46:23
of the time you can get away describing things as first order. So you look at your bromopropane and you'd say, oh that looks pretty good. You have a triplet, you have a sextet, you have a triplet. Beautiful. Just what you would expect from sophomore chemistry.
46:41
Now another sort of breakdown that occurs is when protons end up lumped on top of each other even when they're not chemically equivalent. So you go from bromopropane which has a beautiful triplet to bromobutane and you say okay, that still has a beautiful triplet. We see all of our resonances disperse.
47:04
You go to bromopentane and now these two protons, the protons at the beta and gamma position are starting
47:24
to get very close to each other and your triplet just starts to look a little funny. You can already see there's some tinting in there. That's just your AB behavior. That's perfectly normal. But now you see the methyl is starting to fatten out
47:40
and by the time you get up, so these guys are really, really lumped on top of each other. So by the time you get up say to bromohexane where now these two protons or three protons are really, really, let's see we've got alpha, beta, let's see that's beta, okay.
48:01
So now we end up with these guys really lumped on top of each other and now you notice our triplet really is breaking down and it doesn't look like a clean triplet. You could still call it an apparent triplet but it's much uglier than you would expect
48:23
and this is exactly what I'm talking about for non-first order behavior. You get up to bromooctane and now you see it even more so. So this is what I'm talking about for non-first order due
48:43
to overlap so in other words you have all of these guys in the chain overlapping with each other and often what we will call this is virtual coupling. In other words when hydrogens are overlapping the methyl group
49:02
is coupling to the adjacent methylene but you can say in effect it's also coupling to these others down the chain because they're right on top of each other in chemical shift and you'll see this effect. It's extremely pronounced and you've already seen this before. Any of you who's seen a spectrum
49:22
of THF has seen this behavior. Let me show you. Succinic acid, no problem. You're a singlet. You have four chemically equivalent protons in the chain. All four show up at the same chemical shift. They don't split each other. You go to glutaric acid and you'd say, okay,
49:44
even though remember none of this is a truly first order system, none of these compounds with chains are a first order system. You'd say that doesn't look bad. It looks like what they taught me in sophomore chemistry. I see a triplet for the outer two CH2s and I see a quintet
50:08
for the inner one. Everything should be hunky dory. Then you come to adipic acid and you say what the heck is going on? The problem is this business of virtual coupling
50:23
which is just one way to say it's a non-first order system. When you have these two methylenes right on top of each other, this methylene group looks at it and it says, well, I'm coupled to one but I'm virtually coupled to the other.
50:41
Now these break down into non-simple patterns here and it's reciprocal. So this one looks and it says, well, I'm coupled to this but I have my neighbor and he's coupled to that and we're all lumped together. It's nothing about the length of the chain. It's all about this issue of overlap. So you go up from adipic acid to pomelic acid,
51:04
one more carbon and everything's back to being hunky dory. In other words, you look at your outer methylenes and they each look at their neighbors and even though it's not a true first order system, they say, okay, we're fine. We're not overlapping. It will behave largely like a first order system
51:23
and this is what I'm talking about where you go ahead and you say normally you can get away with this sophomore level analysis. Normally you can get away with treating things as first order or pseudo first order systems but watch out because like this example
51:42
and like our azulene example, we're really playing on thin ice. As I said, these types of patterns come up again and again so tetrahydrofuran has exactly the same principle as adipic acid. You have a methylene chain.
52:02
You have this issue of virtual coupling and so you have this hydrogen is virtually coupled. The one next to the oxygen is virtually coupled to the other two and everything is reciprocal and so you see this pattern here and you see this pattern in other places. So we're very good as human beings at pattern recognition
52:24
and I've shown you three types, four types of patterns today. I've shown you the pattern of an ortho-disubstituted symmetrically disubstituted aromatic and we saw it for phthalic acid. We're di-octyl phthalate but you'd also see it
52:43
for ortho-dichlorobenzene or any other ortho compound. I showed you the distortion pattern for methyls that occurs. I've shown you the pattern that occurs in an extended methylene chain when it's locked in an anti-conformation and I've shown you this pattern
53:01
that you get where you get virtual coupling in the middle of a chain. So keep those in mind because you will see them again in the course of your graduate career. Thank you.