# Video: Identifying the Relative Intensities of Splitting an NMR Peak

Which of the following correctly describes the relative intensities of splitting an NMR peak? [A] Triplet - 1 : 3 : 1 [B] Pentet - 1 : 3 : 5 : 3 : 1 [C] Quartet - 1 : 4 : 4 : 1 [D] Sextet - 1 : 5 : 10 : 10 : 5 : 1

09:40

### Video Transcript

Which of the following correctly describes the relative intensities of splitting an NMR peak? A) Triplet: one to three to one. B) Pentet: one to three to five to three to one. C) Quartet: one to four to four to one. Or D) Sextet: one to five to 10 to 10 to five to one.

The first thing that we should try to remember is why NMR peaks split in the first place. NMR splitting is caused by spin-spin coupling. Remember that the spin of a nucleus could be up or down. And spin-spin coupling happens when this nucleus interacts with another nearby nucleus. For the purposes of this question, let’s discuss proton NMR coupling. Let’s begin by looking at how we form a doublet. A doublet may occur in a molecule such as this, where one of our protons has a single proton neighbor. Note that here the neighbor is Hb. Because Ha and the other Ha are in the same environment, they would be classed as equivalent.

Let’s imagine that this line is our peak for Ha, excluding any spin-spin coupling. When we introduce spin-spin coupling, we can see that this peak can be split by Hb. Hb causes the peak for Ha to be split into two other peaks. These correspond to the spin of Hb either aligning with Ha or against Ha. The gap between these two peaks is the coupling constant. So the result in spectrum will have this peak split into two, called a doublet. With each part of the doublet the same height as the other. We would call this a one-to-one ratio. So this is the principle of how spin-spin coupling leads to peak splitting. So let’s apply it to the first of our potential answers, the triplet.

We can actually use the same molecule as an example for a triplet peak. But this time, we’re looking at the peak corresponding to Hb. Because Hb has two neighbors which it can couple with, we’re going to get a triplet. If you add up how many neighboring nuclei Hb can couple with and then add one. That will give you the number of peaks that your overall NMR peak splits into. In this case, Hb has two suitable neighbors. And you add one to give you three. So there’ll be three lines in this peak, making a triplet. Let’s again begin with a single line corresponding to the peak in our spectrum. And now, we can subsequently add in the coupling.

As Hb couples with the first Ha, we get a similar doublet as before. However, in this case, Hb has another neighbor, which it can couple with, the second Ha. This means that we need to repeat this process again on each of these two lines. What we end up with is something like this. However, if Ha1 and Ha2 are equivalent and therefore have the same coupling constant. These two central lines will overlap and lie along each other. We can draw these lying end to end. And this gives us the rough intensities of our triplet peak. With the central line being twice as long as the outermost lines. This gives us a one to two to one ratio and automatically means that answer A is incorrect. If Ha1 and Ha2 are not equivalent and have different coupling constants, then these two central lines will not overlap exactly. This is when you would end up with something more like a doublet of doublets. But this is beyond the scope of this question.

Let’s move on to look at the quartet. In order to form a quartet, our nucleus needs three neighbors to couple to. Let’s assume they are all equivalent and draw a similar diagram. After applying the first spin-spin coupling, we have two peaks, both of similar intensities. After the second peak has been split, we have a triplet in a one to two to one ratio. If we draw this central line as it’s two constituent peaks, it makes this drawing slightly easier. Now, we have to split each of these four peaks once again. As you can see, we now have a single line on both of the outer edges. But in the center, we have three overlapping lines for each of the two peaks. This gives us a one to three to three to one ratio and results in a peak on our NMR spectrum, a bit like this. If we compare this to answer C, we can see that the answer is incorrect. The relative intensity ratio is one to three to three to one, not one to four to four to one.

Now we could move on and draw similar diagrams for the pentet, sometimes called a quintet, and the sextet. But it starts to get quite complicated and unwieldy. Luckily, we don’t have to. There’s a quicker way. Here, we turn to maths and specifically Pascal’s triangle. Pascal’s triangle begins with a one at the top and two ones on the second level. To form the subsequent levels, we just add together the two numbers, which are above. It will always have ones down the outer edges. So on the third level of the triangle, we’re going to have a one on either end and a number in the center. This number in the center is formed by adding together the two above it, in this case, one plus one. This gives us a two for the center of this level. And we can repeat this as we go down. Each level has one more number than the previous level. So the fourth level will have four numbers, ones on the outer edge and two central numbers. The central numbers are made from adding together the above two digits. This gives us one, three, three, one.

Now you might be asking, what on earth does this have to do with NMR splitting? But let’s compare Pascal’s triangle to the splitting patterns we’ve already investigated. Notice that the ratio of the relative peak intensities in a doublet matches the second level of Pascal’s triangle. The triplet matches the third level. And the quartet matches the fourth level. So by reading down Pascal’s triangle, we can predict what the relative intensities will be for lots of NMR peak splitting. From our triangle, the splitting in a pentet or a quintet will be the fifth level. The fifth level reads one to four to six to four to one. This does not match answer B. So B is incorrect.

Next, let’s look at the sixth level of Pascal’s triangle for our sextet. The numbers read one, five, 10, 10, five, one. And this does match answer D for a sextet. So without drawing lots of complicated diagrams, we can say that a sextet will have the relative intensities of one to five to 10 to 10 to five to one. The same is true as you carry on down Pascal’s triangle. And, of course, the one at the top corresponds to a singlet.

So the correct answer here is sextet, one to five to 10 to 10 to five to one.