What are the coupling constants (J)?

Sep 25, 2015

The coupling constant $J$ is pretty much the peak-to-peak distance, usually reported in $\text{Hz}$. Matching it up with other nearly-identical coupling constants elsewhere in the spectrum usually tells you which protons are near which others.

For example: Peak Data:
$\text{Hz"" "" " "" " ppm"" ""Intensity}$

Proton A:
$\text{371.56"" ""4.149"" ""24}$
$\text{365.38"" ""4.080"" ""56}$
$\text{359.25"" ""4.012"" ""72}$
$\text{353.13"" ""3.943"" ""60}$
$\text{347.06"" ""3.876"" ""28}$

Proton B:
$\text{193.00"" ""2.155"" ""335}$

Proton C:
$\text{110.44"" ""1.234"" ""1000}$
$\text{108.19"" ""1.209"" ""27}$
$\text{104.31"" ""1.165"" ""939}$
$\text{102.06"" ""1.140"" ""25}$

What is shown here for proton A is that $\text{4.008 ppm}$ is the average chemical shift of the 1-4-6-4-1 pattern (according to Pascal's triangle) from the rightmost peak to the leftmost peak, and the entire signal there has multiple peaks. The peaks split like so:

1 $\to$ 1-1 $\to$ 1-2-1 $\to$ 1-3-3-1 $\to$ 1-4-6-4-1

It is not visible in this zoom, but the distance between each peak is roughly identical. This distance is the numerical equivalent of the coupling constant $J$ in $\text{Hz}$. For proton A:

$4.149 - 4.080 = \text{0.069 ppm}$
$4.080 - 4.012 = \text{0.068 ppm}$
$4.012 - 3.943 = \text{0.069 ppm}$
$3.943 - 3.876 = \text{0.067 ppm}$

Interestingly enough, if you look at protons C at the averaged $\text{1.200 ppm}$, you would also see that that doublet has the same $J$ value (ideally that doublet should have both peaks at identical intensities too, but the shimming was not perfect, so they are a bit off). For proton C:

$1.234 - 1.165 = \text{0.068 ppm}$

From the identical (or nearly-identical) coupling constant, you can determine which protons are "communicating" with each other and thus which protons they neighbor.

If you take this number and multiply it by the $\text{MHz}$ of your NMR, you get the coupling constant in $\text{Hz}$. So, if your NMR's magnetic field frequency is $\text{89.56 MHz}$ (like for this particular spectrum), then:

$\text{0.068 ppm" * "89.56 MHz}$

 = 0.068 ("Hz")/("MHz") * "89.56 MHz"

$= \textcolor{b l u e}{\text{6.09 Hz}}$

Indeed, for proton C, $110.44 - 104.31 = \text{6.13 Hz" ~~ "6.09 Hz}$.

Therefore, without seeing the structure of the analyzed molecule, you can still figure out that proton A and protons C are coupling/"communicating" with each other.