# 0.2 Nmr spin coupling  (Page 3/3)

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Frequency of peak c - frequency of peak d = 212.71 Hz – 205.65 Hz = 7.06 Hz

Frequency of peak b - frequency of peak c = 219.77 Hz – 212.71 Hz = 7.06 Hz

Frequency of peak a - frequency of peak b = 226.83 Hz – 219.77 Hz = 7.06 Hz

Average: 7.06 Hz

J (H-H) = 7.06 Hz

In this case the difference in frequency between each set of peaks is the same and therefore an average determination is not strictly necessary. In fact for 1 st order spectra they should be the same. However, in some cases the peak picking programs used will result in small variations, and thus it is necessary to take the trouble to calculate a true average.

To determine the coupling constant of the same multiplet using chemical shift data (δ), calculate the difference in ppm between each peak and average the values. Then multiply the chemical shift by the spectrometer field strength (in this case 60 MHz), in order to convert the value from ppm to Hz:

Chemical shift of peak c - chemical shift of peak d = 3.5452 ppm – 3.4275 ppm = 0.1177 ppm

Chemical shift of peak b - chemical shift of peak c = 3.6628 ppm – 3.5452 ppm = 0.1176 ppm

Chemical shift of peak a - chemical shift of peak b = 3.7805 ppm – 3.6628 ppm = 0.1177 ppm

Average: 0.1176 ppm

Average difference in ppm x frequency of the NMR spectrometer = 0.1176 ppm x 60 MHz = 7.056 Hz

J (H-H) = 7.06 Hz

Calculate the coupling constant for triplet in the spectrum for chloroethane ( [link] ) using the data from [link] .

Using frequency data:
Frequency of peak f - frequency of peak g = 74.82 Hz – 67.76 Hz = 7.06 Hz
Frequency of peak e - frequency of peak f = 81.88 Hz – 74.82 Hz = 7.06 Hz
Average = 7.06 Hz
J (H-H) = 7.06 Hz

Alternatively, using chemical shift data:
Chemical shift of peak f - chemical shift of peak g = 1.2470 ppm – 1.1293 ppm = 0.1177 ppm
Chemical shift of peak e - chemical shift of peak f = 1.3646 ppm – 1.2470 ppm = 0.1176 ppm
Average = 0.11765 ppm
0.11765 ppm x 60 MHz = 7.059 Hz
J (H-H) = 7.06 Hz

Notice the coupling constant for this multiplet is the same as that in the example. This is to be expected since the two multiplets are coupled with each other.

## Second-order coupling

When coupled nuclei have similar chemical shifts (more specifically, when Δν is similar in magnitude to J ), second-order coupling or strong coupling can occur. In its most basic form, second-order coupling results in “roofing” ( [link] ). The coupled multiplets point to or lean toward each other, and the effect becomes more noticeable as Δν decreases. The multiplets also become off-centered with second-order coupling. The midpoint between the peaks no longer corresponds exactly to the chemical shift.

In more drastic cases of strong coupling (when Δν ≈ J ), multiplets can merge to create deceptively simple patterns. Or, if more than two spins are involved, entirely new peaks can appear, making it difficult to interpret the spectrum manually. Second-order coupling can often be converted into first-order coupling by using a spectrometer with a higher field strength. This works by altering the Δν (which is dependent on the field strength), while J (which is independent of the field strength) stays the same.

## Bibliography

• E. D. Becker, High Resolution NMR: Theory and Chemical Applications , 3 rd ed., Academic Press, San Diego (2000).
• A. M. Castillo, L. Patiny, and J. Wist, J. Magn. Reson. , 2010, 209 , 123.
• H. Günther, NMR Spectroscopy , 2 nd ed., John Wiley&Sons Inc., New York (1994).
• P. J. Hore, Nuclear Magnetic Resonance , Oxford University Press Inc., New York (1995).
• N. E. Jacobsen, NMR Spectroscopy Explained: Simplifiedd Theory, Applications and Examples for Organic Chemistry and Structural Biology , John Wiley&Sons Inc., Hoboken, New Jersey (2007).
• J. Keeler, Understanding NMR Spectroscopy , John Wiley&Sons Inc., Hoboken, New Jersey (2005).

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