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

  1. Jul 26, 2009 #1
    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 [tex] {3 \choose 0} = 1 [/tex], for 2 up spins and 1 down spin would be [tex] {3 \choose 1} = 3 [/tex], and so on, right?

    My textbook just states the fact and doesn't explain why.
  2. jcsd
  3. Jul 27, 2009 #2


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    Yes, that's exactly the reason.
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