Why does coupling intensity in H-NMR correspond to Pascal's triangle?

[tmpr]

All right, my 1st-year college chemistry class is just beginning NMR, and I really have no clue what's going on. But what caught my eye was how the relative intensities in hydrogen coupling is roughly predicted by Pascal's triangle. Is this because of probability?

For a quartet, the number of combinations for 3 up spins and 0 down spins would be $${3 \choose 0} = 1$$, for 2 up spins and 1 down spin would be $${3 \choose 1} = 3$$, and so on, right?


My textbook just states the fact and doesn't explain why.

 

[Ygggdrasil]

Yes, that's exactly the reason.

 

[quicknote]

First of all, it's great that you are asking questions and trying to understand the concepts in your chemistry class. NMR is a complex technique, and it can take time to fully grasp it. Let me explain why coupling intensity in H-NMR corresponds to Pascal's triangle.

To understand this, we need to first understand what coupling is in NMR. Coupling refers to the interaction between different nuclei in a molecule. In H-NMR, coupling occurs between the hydrogen atoms that are attached to the same carbon atom. This interaction results in the splitting of the NMR signal into multiple peaks, known as multiplets.

Now, let's look at Pascal's triangle. This is a mathematical concept that shows the number of combinations that can be made with a given set of elements. In the case of NMR, we can apply this concept to the number of spin states that can arise from the coupling of two or more nuclei.

For example, in a quartet, we have four peaks, and each peak represents a different spin state. These spin states are a result of the coupling between two nuclei. As you correctly pointed out, for a quartet, the number of spin states or combinations for 3 up spins and 0 down spins would be {3 \choose 0} = 1, and for 2 up spins and 1 down spin would be {3 \choose 1} = 3.

Now, if we look at the next level of Pascal's triangle, we see that the number of combinations for 1 up spin and 0 down spins would be {1 \choose 0} = 1, and for 0 up spins and 1 down spin would be {1 \choose 1} = 1. These two combinations give us the doublet peak in NMR.

Similarly, if we go to the next level, we get the triplet peak, and so on. This is because, in each level of Pascal's triangle, we are considering one less spin state than the previous level, which corresponds to the splitting of the NMR signal into one less peak.

Therefore, coupling intensity in H-NMR corresponds to Pascal's triangle because the number of spin states or combinations that can arise from coupling follows the same pattern as the number of combinations in Pascal's triangle. I hope this explanation helps you understand the connection between these two concepts. Keep asking questions and exploring the world of chemistry!