1H NMR: Spin-Splitting

Concept: Concept: Splitting without J-values

13m
Video Transcript

Now we're going to move onto the third important piece of information that you can get from proton NMR and that's going to be what we call the spin split.
First of all, it's just important to recognize all the different terms that this could be called and how they're all the same concept. If you ever hear of spin-spin coupling, or J coupling or a term I mentioned earlier called multiplicity, these are all the same exact concept. Spin splitting it just has to do with the concept of neighboring protons interfering with each other and this will reveal to us the distances between different protons.
Now I have to throw out a big side note here, which is that this topic can be taught either in a really simple way or a really, really complex way. Since I have no way of knowing exactly the way your professor taught it this semester, I'm going to teach it to you both ways. I've actually made two different videos so you can pick the easy one or pick the hard one depending on how your professor explains it and I'm going to help you determine which one to watch. Obviously, if you just want to be thorough, you can definitely watch both of them.
But for right now I'm going to teach this topic without J values. J values are the things that complicates the spin splitting concept a lot. Just so you know I've been teaching organic chemistry for a long time and the explanation without J values, which is the simplest explanation works for about 9 out of 10 of my classrooms. Most likely you're in one of those classrooms where you don't really need to learn a rigorous explanation of J values. That being said, then this session, this one video would really just cover you for spin splitting.
Now if your professor goes deep into J values and starts talking about how to draw a tree diagram, that's where you're going to want to watch the second video. If you suspect that you do need to learn that, go ahead and click on the second video after this one and I'll get more into the specifics of how you know that it's important for your class. Awesome. Let's get back to the lesson.
Basically, if you're not learning J values, then this is a very simple rule. All it says is that adjacent non-equivalent protons will split each other's magnetic response to NMR. Now there's a really simple rule that we use to predict what these splits would look like and that's what we call the n plus one rule. The n plus one rule just basically says that n stands for the number of – so I'm just going to write this here. N is going to stand for the number of non-equivalent – the dyslexia is coming out. Equivalent adjacent protons. Basically, n is going to be equal to this definition here.
It turns out that Pascal's triangle is actually going to do a really good job of predicting what these different splits will look like. Now I know it might have been a really long time, you might have not seen Pascal's triangle since grade school. So I'm just going to go over this really quick just to remind you.
Basically, Pascal's triangle is this weird kind of mathematical revolutionary phenomenon. I don't know if that's the right way to put it. Where it basically is a pattern of numbers that if you add up the two numbers above, then what you're going to get is the number below. 1 plus 1 equals 2.
As you go to multiple layers you start getting bigger and bigger sums. What we can see is if we go down in one of these directions where you basically say the two numbers on the top here, two plus one would equal to three or three plus three would equal to six. So you basically take the sum of the two numbers and that would be the number on the bottom. Then you just so on and so forth.
It turns out that these values are really good at predicting the height of splits. This actually has to do with split heights. As n gets bigger you're going to get more splits that have varying complexity. For example, if you do n plus one and your n is actually equal to zero, basically meaning that you don't have any protons around that are splitting, then what we're going to call that is a singlet. That's going to be the name of that and you're just going to get one peak.
Now let's look at a more complicated example. Let's say that n in your example is actually equal to two. That means that you have two non-equivalent adjacent protons next to your target protons. That's going to be called a triplet. What a triplet predicts, according to Pascal's triangle, is that since we're on the third level of the triangle, you're going to get a single split of size one, a second split of size two and a third split again of size one. That's exactly what is represented right here. Notice that I basically have a one, two, one pattern.
Basically, you can use Pascal's triangle to model what your splits should look like going all the way - let's just do one more example, going all the way to a quintet or a pentet, really, technically, it should be called a quintet. Quintet is the better way to say it. If you have quintet, notice that your peaks – you're going to have five different splits and they're going to be of the size one, four, six, four, one and that's exactly what we see here. That is when n would equal to 4. So we'd have four plus one which equals five, which would be our quintet.
I hope that's making sense in terms of the shapes. Now you just have to actually apply it because I know that it's all a little confusing. Let's do the first problem as a worked example. We'll kind of all do it together. Then the second one I'll have you guys do on your own.
Let's go ahead and analyze this carbon right here. I'm going to make that one my red one. What I'm wondering, first of all, is how many adjacent non-equivalent protons it has next to it. What we find is first of all, adjacent means it's within one space. So if I got to the left, there's nothing there. If I go to the right, I do have a carbon here. Does that carbon have any protons on it?
No. It turns out that in this case if you go to the left there's nothing, if you go to the right, there's nothing. That means for red, n is equal to zero. Now using the n plus one rule, that means that zero plus one equals one, which means that I'm going to get a singlet. I'm going to get just a single peak for the red hydrogens. Does that make sense so far? So I would expect it just to look like a single peak. Awesome.
So let's keep going. Now let's look at the blue hydrogens over here. Now, I drew the same thing. I said how many adjacent, non-equivalent protons does it have next to it. Well, if I go to the left, again nothing. I'm next to a carbon with nothing. But if I go to the right, do I have any hydrogens. Actually, yes. How many do I have? Three. I have three to the right.
That means that this would mean that n is equal to three. If I use the n plus one rule three plus one equals four. Four gives me a quartet I would expect that I would get a quartet from those protons. Those protons would be split into a quartet because I've got the three protons splitting plus one.
Now let's look at green. Green is over here If I go to the right, nothing. If I go to the left, how many protons do I have? Two. So that means that for green, n is equal to two. According to n plus one, it would be two plus one equals three, which would be what we call a triplet. That means that here I have a singlet, a quartet, and a triplet and that's what we mean by splitting, that every proton has its own unique shape based on the number of protons it's next to.
On top of that, if you wanted to predict the shapes of these singlets, quartets, triplets, it's easy because you could just use Pascal's triangle. You could say that a singlet – by the way, I'm not drawing these in any particular order. I'm not drawing chemical shift here. I'm just drawing shapes. The singlet would look like this. A quartet would look like this with one, two, two, one.
Actually, that's not what it says. That's wrong. Let's look at Pascal's triangle. Pascal's triangle actually says for a quartet, that should be one, three, three, one. I drew that all wrong. Let's do it again. So it should be one, three, three, one. Then my triplet, as we said earlier is one, two, one. Then the triplet should look like this.
Now. by the way, don't worry about the heights here. It doesn't have to be a specific height or whatever. I'm just trying to show the ratios between the different splits. Now you know what a singlet, a quartet, and a triplet would look like based on Pascal's triangle.
Now I'm going to have you guys do (b) on your own. Just so you know, (b) actually comes with one special instruction before you can solve it, which is that it turns out that heteroatoms have a special rule. Heteroatoms. Do you guys remember what a heteroatom is? It's important for the rest of this course. Heteroatoms are just non-carbon atoms. That would be nitrogen, sulfur, oxygen, phosphorous, etcetera
Heteroatoms, this is a big deal, do not split. You can't split through a heteroatom. Think of it like a wall. You can't split through it. That means that think of this as being a complete wall blocking off one side from the other. When I go to analyze how many different adjacent hydrogens are next to that, well if I go to the right, obviously there's nothing. But if I go to the left, there's also nothing because I hit a wall. That wall is the heteroatom. Same thing when I go to analyze this one. When I go to the right, I'm going to hit that wall.
Just think of that. That's a special rule that I just wanted to include that you can't split through a heteroatom. Other than that everything else is still the same. So go ahead and try to solve for the rest of (b) to try to figure out how many splits you'd get for a proton type one, proton type two and proton type three. Now the rest is up to you. Go for it. 

Concept: Example: Proton Splitting

3m
Video Transcript

All right, so proton type one. If I go to the left, nothing. If I go to the right, do I have any hydrogens? Yes, I do. I have two. That has nothing to do with the number two. It just happened to be two hydrogens there. That means that n is equal to two meaning that two plus one equals three, meaning that one proton type one should be a triplet. Hopefully all you guys got that.
Now red gets a little bit more complicated. I'm going to draw an arrow here to show what I'm talking about, but I have to move it a little bit away for more space. To the left, I actually have three protons. Three H. To the right, I have two H's. Correct? That means that I've got three on one side, two on the other. That means that n is actually equal to five. Total, right? Meaning that I have five plus one, which equals six. And is is going to be a sextet.
Since that's one we haven't drawn yet, just to show you, you could predict what a sextet would look like by Pascal's triangle. It would be a one, five, ten, ten, five, one split. I'll quickly try to do that. It's going to be tiny. It's going to be tiny, something like this. Something like that. We've got the tiny ones on the edges, then the five's and then the tens. So obviously, it might not be perfect, but that's pretty good. Notice that Pascal's triangle helped me to know that shape.
Then, finally, we've got proton type three. I'll draw another arrow. For proton type three, to the right I've got nothing. We just said that's a wall. To the left, I have two hydrogens. That means that n is going to be equal to two, which means it's going to be two plus one equals to three, which means it's going to be a triplet.
In this case, I've got triplet, sextet, triplet and do you guys know what this H is going to be on the alcohol? A singlet. Because I told you guys n is equal to zero all the time for heteroatoms, so it's always just going to be one peak by itself.
So anyway, this is the explanation of splitting that like I said is going to suffice for almost all of you out there. Now go ahead and watch the introduction to the next video just to confirm if you have to watch it or not, but just so you guys know, have confidence that if you practice this and get good at it, this is probably all you're going to need for your upcoming exam. Cool. Let's move on. 

Problem: Predict the splitting pattern (multiplicity) for the following molecule: 

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Problem: Predict the splitting pattern (multiplicity) for the following molecule:

3m

Problem: Which of the following compounds gives a 1H NMR spectrum consisting of only a singlet, a triplet, and a pentet? 

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Concept: Concept: Splitting with J-Values: Simple Tree Diagram

12m
Video Transcript

In this video we're going to discuss spin splitting with J values and with tree diagrams so essentially this is the complicated version of the spin splitting explanation, now that you understand the simple version you might be wondering do I really need to learn this more complicated version or not? And what I'll tell you is probably not, OK? What is typical is that professors will briefly mention J values and will briefly show a picture of a tree diagram as they're explaining spin splitting, what's less common is that a professor will tell you that you need to know specific J values or that you need to know how to draw a tree diagram if any of those two things come up then you should watch this video, OK? You can also just directly ask you Professor will I be asked to draw a tree diagram? And if the answer is no then you probably don't need to watch this video, OK? But in the event that you do, here we go, OK?

So coupling constants also known as J values, OK? Describe the amount of interaction that a proton will have another so it's kind of quantification of the interaction, OK? And here are some examples of common coupling constants that are on frequently reviewed so vicinal protons that would be 2 non-equivalent protons that are just next to each otherÕs so this would be like a typical splitting example that we would have seen in the last video that would have a split of anywhere from 6 to 8 hertz, OK? These interferences you know or these interactions are always described in a Hertz frequency, OK? Now Cis protons so that would mean protons that are split....that are separated by a Cis double bond usually interact anywhere from 7 to 12 and trans protons have a J value from anywhere between 13 to 18 so kind of like I ordered it here in order, now this is not a comprehensive list of all the J values if your professor says that they want you to know specific J values then you should definitely refer to his resources so that you can make sure that you know all the ones that they want you to know but these are the three most common that's for sure, OK? So remember that we discussed in the more simplistic version of spin splitting that you could use Pascal's Triangle to predict the shapes of splits that you get but it turns out that Pascal's triangle only works if you assume that all of your J values are exactly the same, OK? So basically the whole reason that we could get those very predictable splits is because we're assuming that everything is splitting exactly the same that all of your hertz all your J values are the same but once you introduce the idea of multiple J values, multiple coupling constants being involved with splitting the same proton pascals triangle no longer applies in fact Pascal's Triangle is actually going to give you the wrong answer because instead of everything splitting evenly you got different coupling constants layering on top of each other making weird shapes so in that case in order to predict what the split is actually going to look like you have to use a tool that we call a tree diagram, the tree diagram is our way of visualizing how the splits are going to work and how they're going to happen in order so that we can get the final shape of the split meaning that if we predicted that something was a quartet before the shape might be different now that we're using a tree diagram, OK? So what I want to show you first is drawing a simple tree diagram and I'm going to use a simple explanation of one that the N+1 rule would have worked on so for example in this molecule notice that it says that basically I'm trying to figure out how Ha my bolded proton is going to be split, OK? And what I notice is that well first of all Ha over here is that one going to split it? No we're not going to get split by the other Ha because that's not adjacent that's actually on the same carbon it's also not.... ItÕs its equivalent so I don't even have to look at that, OK? Now we also talked about how OH is a wall so we're not going to get split over here either so that means that this Ha is only going to be split on one side, correct? It's going to be split on the left side, how many protons will be split by? 3, notice that all 3 protons are the same exact type of proton they're all Homotopic, OK? And they all have the same exact J Coupling J value or coupling constant of 6 Hertz, OK? So that means that in this case are all of my J values the same? Yes, they are because I have 3 protons that are all splitting with a J value of 6, OK? So that means that Pascal's Triangle should actually work in this example I should not have to draw a tree diagram to figure out what it's going to look like, OK? If I use the N+1 rule here, N is equal to what number? 3+1 so that means that what type of split should I get if I have 3+1? It means it should be a quartet, OK? It should be a quartet according to N+1 and what shape does a quartet predict according to Pascal's triangle? That means it should be a ratio of 1:3:3:1 that should be familiar to you so far, OK? But now what I'm going to do is I'm actually going to draw the whole tree diagram for this so you can see how to actually correct so let's start off with Ha. Ha is singlet by itself, OK? This is Ha by itself this is before it gets split, OK? Now in this graph this is basically graph paper I could give it any unit I want but let's just go with I think I have enough space to make every single unit equal to 1, OK? So that means that in my first split, I'm going to split every I'm going to go down as many layers as I have splits or coupling constants to split with so in my first one I'm going to split 3 on one side 3 on the other making my first split that is represented by each Hb 1, OK? Notice that I'm going to have Hb1 Hb2 and HB3 and all 3 of these are going to have a generation to split, OK? So basically what I have so far is that I had a singlet and now it just turned into a doublet with a distance of 6 Hertz, OK? This is now a doublet if I were to draw exactly if I were to represent this as a peak it would look something like this it would be a peak here and a peak here a doublet but I'm not done if I were to end there though be what it looks like but I only split with the first proton I still need to split with the others so now let's go ahead and split with HB2. HB2 is going to split both of these into 6 so I'm going to get 6 over here and I'm going to get 6 over here, OK? What that's going to give me is that now this is the split that I get from HB2 and what I now get is I mean these are still a distance of 6 Hertz each I'm not going to write this every time I'm just going to write this again so you can see these are now distance of 6 Hertz each but the important part is that now what would this look like if I were to start drawing it right now? Well what I would have is a ratio of 1:2:1, now why do I put 2 in the middle? Well because notice that 3 of the splits actually merged into one line that means that the.... Basically the amplitude of the interference in that area is actually going to be double that of the ones on the periphery, does that 1:2:1 ratio look familiar? Yes that's the ratio of a triplets but we're not done yet because you still have that last proton to split with so I'm going to erase that for now and we're going to split with our last proton HB3 all these lines have to get split so I'm going to split....Ops I'm trying to use green here and I'm going to split this one, I'm going to split this one and I'm going to split this one, by the way guys just so you know I know you've never done this before the height that I'm using you notice how I was using 3 units each that's completely irrelevant I just decided to do that because that was how much space I was given the whole point is that you just try to keep even however much downward you're drawing just try to draw that with every generation of split, OK? So this is Hb3 our final split and what we notice is what are the ratios going to be now? Well think of this almost like Pascal's Triangle whatever the top was it's going to add up to the bottom split so that means that up here I had a 1:2:1 ratio that means that the splits here are going to be now a 1:3(1+2): 3:1 ratio, right? Because 1 and 2 again make 3 and I've got 1 on the side, does this ratio look familiar 1:3:3:1? Yea guys that is the ratio for our final answer which is a quartet which if I were to draw would look something like this...Looks familiar, right? This is a quartet so now a very legitimate question you should be asking me is Johnny if we already have Pascal's Triangle if we already have the N+1 rule why did you just go through this whole exercise of doing something that just gave me the same exact answer? Well because what I'm trying to show you is how you don't need to do this if all your J values are the same, if they're all the same please skip the hassle we can already do this with Pascal's Triangle, OK? So then why would I ever need to use a tree diagram? You use a tree diagram if you have different J values in the same split, OK? So let's go ahead and turn the page and see how that might be the case.

Concept: Concept: Splitting with J-Values: Complex Tree Diagram

9m
Video Transcript

So now moving on to drawing complex tree diagrams which really should be the only type of tree diagram you ever draw since I already proved to you that you don't need to draw a simple tree diagrams to get the right shape so as I said we're going to have to use multiple J values in the same problem in order to really justify using a tree diagram and when you have these multiple J values involved it's important to always split an order of the highest J value first and then go to your lower J values later so always go from the highest Hertz to the lowest Hertz, OK? Now before kicking this question off and really drawing the whole thing I want to analyze it according to N+1 and then compare our answer at the end to what N+1 would have predicted so it says here use a tree diagram to predict the splitting pattern of the bolded proton which in this case is Hc and first of all What Would N+1 one and Pascal's Triangle say about this split? What should it look like? Well if we go to the left how many protons are we getting split by? So we're getting split by the 0, right? Because there's nothing there if we go to the right how many protons are we getting split by? 2, Ha and Hb so that means according to N+1 and N equals 2 so then it should be 2+1=3 so that N+1 rule if I were just to stick with that should be a triplet, OK? And we know that triplets come with a 1:2:1 ratio so that's what Pascal's rule tells me and unfortunately if you stop there you would actually get this question completely wrong, why? Because look these J values for Ha and Hb are actually not the same they're different JHa does not equal JHb so I can't use Pascal's Triangle I can't use N+1, OK? I have to use a tree diagram so let's go ahead and do the tree diagram like we did before and which proton or which J value should I get to split with the first, should I split with the J value from Ha or the J value from Hb? Ha because it's bigger remember I said you always start with the bigger value first so I'm going to start off with the singlet that Hc would have given me, OK? So this is Hc if it wasn't being split it would just be a singlet but I'm going to start off with Ha the split from Ha is a value of 16, now I don't have enough I guess cubes whatever I don't have enough units here so that every single unit or box is going to be equal to 1 so instead let's use 2 let's say that every box is equal to a split to 2 Hertz, OK? So if we're trying to split by 16 Hertz at the beginning that means that I have to go 4 to the right and 4 to the left to make 8 boxes or equal to 16 so basically what I'm saying here is that one of these is equal to 2 Hertz hope you guys can Let me get away with that one, OK? So we're going to go 4 to the right for the left that's going to give me a split of that's Ha and that's going to give me a split of 16 Hertz, OK? And so far, what we get we are getting? We are getting a doublet, OK? So far so good looks like a double it looks like a 1:1 ratio so nothing too crazy, OK? This is exactly would expect to see with N+1, OK? Now the difference is that if I was using N+1 the split for Hb would have to be what number as well? 16, it would have to have the same coupling constant for N+1 in Pascal's Triangle to work but notice that my second coupling constant is a different value it's 10 so now let's go ahead and use the same strategy but now I have to go....Damn it I didn't use the right number I have to go 2.5 to the right and 2.5 to the left so I said damn it because I have to go 0.5, 2.5 to one side and 2.5 to the other so I can make 10 hertz on this side and I do the same thing 2.5 on one side 2.5 on the other so I can go 10 Hertz on the other now what's going on guys? What kind of shape am I getting, OK? That's actually it that's the final answer so notice first of all what are the ratios going to be here for these splits? It's going to actually be since nothing's overlapping it's going to be 1:1:1:1, OK? Now notice that if the J value for Hb had been the same as Ha I would have that overlap in the middle and I would get a triplet but since the second J value smaller I don't get the overlap and I get separate peaks instead, OK? So notice that what I get here if I were to draw it out actually looks like this it's just a bunch of single peaks but now how many do I get? I actually get 4 peaks instead of 3, I would have expected 3 but I'm getting 4 it turned out that this type of arrangement is actually called a doublet of doublets, OK? So basically you had doublet and then you split it into another doublet, OK? And any time you hear things like this Doublet of doublet there's even there's doublet of quartet there's triplet of triplet all this kind of stuff if you hear anything like that that has to do with J values being different, OK? If your J values are different from each other then you get weird shapes like doublet of doublets, OK? So now just once again to compare this N+1 would have told me that I'm going to get a triplet and a 1:2:1 ratio when really what I'm getting is a doublet of doublet with a 1:1:1:1 ratio so this is exactly the reason that we need to be able to draw tree diagrams because if your professor wants you to be able to use different J values N+1 just doesn't cut it and you need to actually draw the entire thing to know what the shape is, OK? Now just so you know it can get more complicated so imagine that instead of being split by 2 protons I throw in a third one so let's say that there was a JHd and it had another value of let's say 20 hertz or whatever let's say a smaller number let's say 8 hertz, OK? Then you just keep going and you keep splitting to do another layer until you split with all of your protons, OK? In the end of the day when you're drawing these things the most important part is that you can get your ratios and that you can draw it right if you don't remember the exact name of the weird arrangement that's usually not a big deal, OK? But if you draw it correctly then you should be fine, OK? So guys I hope that that helps to settle J Values No J values and I hope that you can see that there's actually really related it's just depends on how complex your professor wants to make your life, how complicated they want to make it the semester, OK? So that being said let's move on to the next topic.

Concept: Concept: Common Splitting Patterns

9m
Video Transcript

So guys, one more note on splitting. I'm sure that at this point you're getting pretty sick of this subject but it turns out that certain combinations of splits, when they're seen on the same proton NMR spectrum, are highly indicative of certain types of molecular structures. If we can learn these combinations of splits and learn to associate them with those structures, we can get way ahead with our knowledge level of analytical techniques. In fact, this is the kind of knowledge that can really catapult you to the top of your class because this is stuff that your classmates probably aren't going to be able to do right away.
Let's talk about four really important splitting patterns and what they mean when you see them. Here are the four that we're going to discuss. We're going to discuss what an ethyl group looks like, what an ethylene group looks like, isopropyl and quaternary.
Let's start off with ethyl which is probably the most common. It's probably the one that your professor mentioned in class. What an ethyl group typically looks like is that notice that an ethyl group is always CH2CH2CH3, so you've got a two next to a three. That means that if we're using n plus one, which we would because this isn't very complicated example with different J values, etcetera, what you're going to wind up getting is a triplet and a quartet.
This is what it would look like. You'd have a quartet somewhere and a triplet somewhere. The order of them doesn't matter. It doesn't matter what one's in front of the other. It just matters that you have both a triple and quartet on the same spectra. If you see a triple and a quartet in the same NMR spectra, then you have to start thinking to yourself there might be an ethyl group there. Why? Because we know that ethyl groups produce a pair of – a triplet with a quartet.
Now, in the same manner, an ethylene group would just be two next to a two. So that means that one would split into a triplet and the other one would also split into a triplet. They would split each other into the same thing. So if you see a triplet, triplet. That tells you that you might have a – dual triplets might tell you that you have an ethylene group present. So it's just – it's not always the case, but it's very likely.
Now, what about isopropyl. This one's actually the most distinctive. If you see this, you pretty much for sure have an isopropyl. That would be a combination of a doublet and a septet. Because notice that this hydrogen in the middle is being split by how many other hydrogens? Well, six. It's got three on the top plus three on the bottom. So three, plus three plus one, remember it's n plus one, equals seven. So we would expect to see a septet here.
Now let's look at the hydrogen at the top. The hydrogens at the top are only being split by one H, so then you'd get n plus one equals two. And you'd get a doublet. When you have isopropyl groups present, you're going to see a combination of doublets and septets. If you see a doublet and a septet in the same spectrum, you know for sure that you have an isopropyl group, which, by the way, can help a lot when it comes to structure determination, which is a topic for another conversation. But it's coming up.
Lastly, we have quaternary groups. And this would be an example of when you just have singlets kind of for no reason. If you just have singlets, a bunch of singlets, then that tells you that you must have hydrogens that aren't being split by anything.
Remember that we already talked about one type of molecule that can create singlets so that would be also or heteroatoms. So we know that heteroatoms can cause singlets as well. But in the absence of heteroatoms, if you don't have oxygen, you don't have nitrogen and you still have a bunch of singlets popping up everywhere then that tells you that you might have carbons that have no H's attached.
So, in this case, I'm calling it a quaternary group because it's got four things around it that aren't hydrogen. But I mean, there's other examples. It doesn't have to just be R groups. It could be a carbonyl and an R. It could just be meaning that you have four bonds to carbon that are not to H. That are something to other than H, which means that when you g ahead and try to split this thing, is it going to split? No, because it's next to a carbon without hydrogens, so it can't split. Got it?
Obviously, heteroatoms are our first thought when we see singlets, but if you can't figure out that it's a heteroatom, then you might want to look into the fact that this carbon might not have any hydrogens attached to it.
Guys, that's really it. That's just the common splitting patterns that we're going to use for structure determination. Now just as a really quick example, here's a sample NMR. Is there a common splitting pattern seen here that could help us to deduce the structure of the molecule before even looking at it. Now notice, I included the structure of the molecule so that's a huge hint. But, just by looking at the signals and the types of signals we have, could we already deduce some stuff about this molecule?
We'll do this as a worked example. Don't worry about starting, pausing the video or doing it yourself. Let's just talk about it. What do you have?
Well, we have a quartet. We have a singlet. And we have a triplet. Do any of these splits give us hint as to what the molecule could look like? Well, I see one big hint right away, which is I see that we have a singlet. What do singlets indicate? Well, singlets indicate either heteroatoms or carbons that don't have any hydrogens. In this case, do we have – let's say we were just given the molecular formula below, which would be C2H6O. Well, would we have an idea of where that singlet could be coming from? Yeah. We could think that that might be a heteroatom. We would think to ourselves maybe this is a heteroatom. Maybe it's an alcohol since I have an O present. That's just one way to think.
Then there's another hint. There's a quartet and a triplet on the same exact spectra. What does that tell us? That means that I need to suspect ethyl group. So already I have a suspicion that I might have maybe an alcohol and then I have a suspicion that I might have an ethyl group because I have a triplet and a quartet. What does an ethyl group plus an alcohol equal? My final structure.
Obviously, I'm being a little bit crazy with the way I'm applying this, meaning that I'm probably making some extra connections that you haven't learned how to make yet. But I'm just letting you know that just by using splitting patterns we have a huge heads up on what this structure already is. Obviously, you're never going to determine a structure just due to splitting patterns, but it's amazing how much extra help this provides when you understand it.
Awesome guys. I hope that made sense. Let's go ahead and move on to the next topic. 

1H NMR: Spin-Splitting Additional Practice Problems

Predict the spiltting pattern for the structure below in order of protons a, b, c:


a. doublet, doublet, singlet 
b. doublet, triplet, triplet 
c. triplet, triplet, triplet 
d. triplet, triplet, singlet 
e. doublet, doublet, doublet

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Distinguish whether or not hydrogens are diastereotopic. Then, you are requested to state the splitting pattern of a specific hydrogen (singlet, doublet, triplet, or doublet of doublet, etc.).

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Distinguish whether or not hydrogens are diastereotopic. Then, you are requested to state the splitting pattern of a specific hydrogen (singlet, doublet, triplet, or doublet of doublet, etc.).

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In the boxes provided, place that letter (A, B, C, etc.) that corresponds to the signals in the spectrum provided below.

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Splitting: For the structure below, indicate the splitting for each signal. Use these abbreviations for the splitting : s=singlet; d=doublet; t=triplet; q=quartet; etc. If there is complex splitting, indicate the total # of peaks and how the splitting is produced. e.g. d of t

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Splitting: For the structure below, indicate the splitting for each signal. Use these abbreviations for the splitting : s=singlet; d=doublet; t=triplet; q=quartet; etc. If there is complex splitting, indicate the total # of peaks and how the splitting is produced. e.g. d of t

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Splitting: For each structure below, indicate the splitting for each signal. Use these abbreviations for the splitting : s=singlet; d=doublet; t=triplet; q=quartet; etc. If there is complex splitting, indicate the total # of peaks and how the splitting is produced. e.g. d of t

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One of the following compounds will show a doublet as part of its  1H NMR spectrum. Which one? 

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