So now we're 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 simple tree diagrams to get the right shape. 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 in 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. 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. It says here, use a tree diagram to predict the splitting pattern of the bolded proton, which in this case is HC. First of all, what would n plus one rule and Pascal's Triangle say about the 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 0, right? Because there's nothing there. If we go to the right, how many protons are we getting split by? 2. And so that means according to n plus 1, n equals 2. So then it should be 2 +1 equals 3. So the n plus one rule, if I were just to stick with that, should be a triplet. We know that triplets come with a 121 ratio, so that's what Pascal's rule tells me. And unfortunately, if you stop there, you'd 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. I have to use a tree diagram. Let's go ahead and do the tree diagram like we did before. Which proton or which J value should I split with first? Should I split with the J value from HA or the J value from HB 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. 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 split from HA is a value of 16. Now I don't have enough cubes or whatever. I don't have enough units here so that every single unit or box is going to be equal to 1. Instead, let's use 2. Let's say that every box is equal to a split to 2 hertz. Okay? 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. Okay? So we're going to go 4 to the right, 4 to the left. That's going to give me a split of 16 hertz. So far, what are we getting? We are getting a doublet. So far so good. Looks like a doublet. It looks like a 1 to 1 ratio, so nothing too crazy. This is exactly what we would expect to see with n+1. 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 and 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 and a half to the right and 2 and a half to the left. 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 point 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? Okay. 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 is overlapping, it's going to be 1 to 1 to 1 to 1. 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 is smaller, I don't get that overlap and I get separate peaks instead. Notice 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 turns out that this type of arrangement is actually called a doublet of doublets. Basically, you had a doublet and then you split it into another doublet. Okay. And any time you hear things like this, doublet of doublets, there's even there's doublet of quartets, 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. If your J values are different from each other, then you get weird shapes like doublet of doublets. Now just once again to compare this, n plus 1 would have told me that I'm going to get a triplet and a 1 to 1 ratio when really what I'm getting is a doublet of doublets with a 1 to 1 to 1 to 1 ratio. 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 plus 1 just doesn't cut it. You need to actually draw the entire thing to know what the shape is. Now just so you know, it can get more complicated. Imagine that instead of being split by 2 protons, I throw in a third one. Let's say that there was a JHD and it had another value, let's say 20 hertz or whatever. Let's say a smaller number. Let's say 8 hertz. Then you would just keep going and you'd keep splitting to another layer until you've split with all of your protons. At 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. But if you draw it correctly, then you should be fine. So guys, I hope that that helps to settle J values versus no J values. And I hope that you can see that they're actually really related. It just depends on how complex your professor wants to make your life, how complicated they want to make it this semester. Okay? So that being said, let's move on to the next topic.
15. Analytical Techniques:IR, NMR, Mass Spect
1H NMR:Spin-Splitting Complex Tree Diagrams
15. Analytical Techniques:IR, NMR, Mass Spect
1H NMR:Spin-Splitting Complex Tree Diagrams - Online Tutor, Practice Problems & Exam Prep
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Splitting with J-Values:Complex Tree Diagram
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PRACTICE PROBLEMS AND ACTIVITIES (8)
- The spectrum of trans-hex-2-enoic acid follows. (a) Assign peaks to show which protons give rise to which pea...
- Draw a splitting diagram for the Hb proton if Jbc = 10 and Jba = 5.
- Draw a splitting tree, similar to Figures 13-32 and 13-33, for proton Hc in styrene. What is the chemical shi...
- Draw the expected signal for a hydrogen with the following coupling constants. (a) Hₐ : δ 5.34 (Jₐ꜀ = 12 , ...
- What coupling constant would you expect between Hₐ and H₆ in cyclopent-2-en-1-one?
- Draw a diagram like the one shown in Figure 14.12 to predict b. the relative intensities of the peaks in a qui...
- Draw a splitting diagram for Hb, where a. Jba = 12Hz and Jbc = 6Hz.
- Using a 60-MHz spectrometer, a chemist observes the following absorption: doublet, J = 7 Hz, at d 4.00 (a) W...