In this video, we're going to discuss spin splitting with j values and with tree diagrams. 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? What I'll tell you is probably not. 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. You can also just directly ask your professor, will I be asked to draw a tree diagram? If the answer is no, then you probably don't need to watch this video. But in the event that you do, here we go.

So coupling constants, also known as J values, describe the amount of interaction that a proton will have on another. It's kind of a quantification of that interaction. Here are some examples of common coupling constants that are frequently reviewed. Vicinal protons, that would be 2 non-equivalent protons that are just next to each other. 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. These interferences or these interactions are always described in hertz frequency. Now cis protons, 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 of anywhere from 13 to 18, so kind of 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 you can make sure that you know all the ones they want you to know. But these are the 3 most common.

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. 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 of 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, Pascal's 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've got different coupling constants layering on top of each other making weird shapes. 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.

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 plus one rule would have worked on. For example, in this molecule, notice that it says that basically I'm trying to figure out how my bolded proton is going to be split. Okay? And what I notice is that Well, first of all, over here, is that one going to split it? No. We're not going to get split by the other because that's not adjacent. That's actually on the same carbon. It's also not its equivalent. I don't even have to look at that. Now we also talked about how the oxygen is a wall, so we're not going to get split over here either. That means that this is only going to be split on one side, correct? It's going to be split on the left side. How many protons will it be split by? 3. Notice that all three protons are the same exact type of proton. They're all homotopic. And they all have the same exact J coupling J value or coupling constant of 6 hertz. 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. 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. If I use the n+1 rule here, n is equal to what number? 3+1. That means that what type of split should I get if I have 3+1? It means it should be a quartet. 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 to 3 to 3 to 1. That should be familiar to you so far. 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 it's actually correct. Let's start off with is a singlet by itself. This is by itself. This is before it gets split. 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. That means that in my first split, I'm going to go down as many layers as I have splits or coupling constants to split with. 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 Hb1. Notice that I'm going to have Hb1, Hb2, and Hb3. And all 3 of these are going to have a generation to split. 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. 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, that would 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. I'm going to get 6 over here and I'm going to get 6 over here. 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 a 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 to 2 to 1. Now why do I put a 2 in the middle? Well because notice that 2 of the splits actually merged into one line. That means that 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 triplet. But we're not done yet because we still have that last proton to split with. 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 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, the height that I'm using, you know also I was using 2 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 it even. However much downward you're drawing, just try to draw that with every generation of split. This is Hb3, our final split. 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. That means that up here I had a 1 to 1 ratio. That means that the splits here are going to be now a 1, 3, 3, 1 ratio because 1 and 2 add together make 3 and then I've got 1 on the side. Does this ratio look familiar? 1, 3, 3, 1. Yeah, 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 plus one rule, why do you just go through this whole exercise of doing something that just gave me the same exact answer? 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. Okay. 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. Let's go ahead and turn the page and see how that might be the case.