University of Ottawa NMR Facility Web Site

Please feel free to make suggestions for future posts by emailing Glenn Facey.



Tuesday, June 24, 2008

Spin-Spin Coupling Between Equivalent Nuclei

When many chemists are asked what is the 2JH-H coupling for compounds like methane, acetone, methylene chloride, dimethyl ether or DMSO, they will often return a look of confusion. "There is no coupling," they will say, "the proton spectrum is a singlet". Indeed the proton spectrum is a singlet for these compounds but 2JH-H is not equal to zero. The only reason that the coupling is not observed in the spectrum is because the chemical shifts of each proton are identical. The coupling can easily be measured by observing the spectrum of a partially deuterated isotopomer. The 2JH-H coupling constant is equal to 2JH-D multiplied by the ratio of the gyromagnetic ratios of 1H to 2H. This is illustrated in the figure below for methylene chloride.In fact, 2JH-H is -7.192 Hz not +7.192 Hz however, this cannot be determined simply by observing the spectrum. Both spectra were measured for dilute solutions with CDCl3 as solvent. The residual protons of CDCl3 were used as the chemical shift reference (7.26 ppm). The chemical shift difference between CH2Cl2 and CHDCl2 is due to an isotope effect.

5 comments:

stan said...

It should be noted that the indirect way of how you deduce the scalar coupling between the nuclides of an equivalent group, while correct for an estimate, can be affected by a J-coupling isotope effect (theoretically, those may be quite significant, I believe).
If one wants to be accurate and backed-up by experiments, there just does not seem to exist any way to determine the values of those unobservable couplings.
Or am I wrong?

Glenn Facey said...

Stan,

Thank you for your comment. I have never thought about how significant an isotope effect might be on a coupling constant. In order to evaluate such an isotope effect I suppose one must look at more than one isotopomer of a suitable compound and measure the couplings constants.

Glenn

Anonymous said...

Cynthia Jameson has a good article on isotope effects on spin-spin couplings and shieldings in the Encyclopedia of NMR. There is a primary effect (the one you have been discussing) and a secondary effect where isotopic substitution somewhere in a molecule affects the coupling observed between two other nuclei.

The effects are typically small, with the largest primary effect being about 10% (in the SnH3 anion, measured by Wasylishen and Burford, Can. J. Chem., 1987, 65, 2707)

Dave

Anonymous said...

Hey Glenn

Great blog! It has been my go to for NMR questions.

I am still struggling with concept of why magnetically equivalent protons don't produce a splitting pattern. Two isolated methylene protons, for instance; each has the potential to be spin up or down. So why, in the absence of chirality, do we not see a doublet of doublets?

I think that technically there is a dd, but the inner peaks of the dd overlap, and the outer peaks are so small they are unobserved, but I guess I don't really understand why that is.

Adam

Glenn Facey said...

Adam,

Thank you for the question. I am not really sure whether or not I am able to provide an adequate answer for you. When the chemical shift difference between two coupled spins is of the same order as the coupling constant between them, then a second order spectrum is obtained. When the spins are magnetically equivalent or even just isochronous, then the second order spectrum is a singlet. Here is a neat example:

http://u-of-o-nmr-facility.blogspot.ca/2008/07/second-order-1-nmr-spectra-of-isopropyl.html

All magnetically equivalent nuclei (for example all three protons of a methoxyl group) are isochronous and will therefore give a singlet.

I know that this is not much of an answer.

Glenn

Glenn