3
$\begingroup$

I was looking at the NMR spectrum of tetraethyltin [$\ce{Sn(CH2CH3)4}$], and found that the $\ce{Sn-H}$ coupling constant between $\ce{Sn}$ and $\ce{CH2}$ is larger than that between $\ce{Sn}$ and $\ce{CH3}$.

$\hspace{4.1cm}$SnEt4



$$\begin{array} {|r|r|}\hline {}^2J({}^{119}\ce{Sn-H}) & {}^3J({}^{119}\ce{Sn-H}) \\ \hline \pu{49.2 Hz} & \pu{68.6Hz} \\ \hline \end{array}$$

This trend is confirmed by literature values[1].

However, I am confused as to why this is happening. The magnitude of the coupling constant is determined by the Fermi contact mechanism, so surely the coupling constant over two bonds should be larger than the coupling constant over 3 bonds? The ethyl groups also should have free rotation, so any type of angle effects should be cancelled out on average.

So why is ${}^2J$ smaller than ${}^3J$ here?


Reference:

  1. G. Barbieri, F. Taddei, "117Sn, 119Sn–proton long-range coupling constants in the 1H nuclear magnetic resonance spectra of alkylhalogenostannanes", J. Chem. Soc., Perkin Trans. 2, 1972, 10, 1327
$\endgroup$
3
  • 1
    $\begingroup$ Presumably it’s not just Fermi contact interactions. Note that 2J and 3J of 13C-1H have the same pattern - there’s a paper on this which I will try to find... $\endgroup$
    – orthocresol
    May 16 at 10:42
  • $\begingroup$ @orthocresol Thank you. So, are there other mechanisms except Fermi contact for nuclear spin coupling? I was only aware of Fermi contact. And please mention the paper if you find it, that would be very helpful. $\endgroup$
    – S R Maiti
    May 16 at 14:04
  • 1
    $\begingroup$ This is quite a broad area and one that I'm not qualified to discuss myself, but here are a few relevant papers: doi.org/10.1021/jp0353284 (talks about C–H), doi.org/10.1021/jo200513q (talks about H–H where Fermi contact is the most important factor, but also summarises some of the other factors), doi.org/10.1002/chem.200306065 (discusses factors), doi.org/10.1021/ed084p156 (maybe the most accessible) $\endgroup$
    – orthocresol
    May 16 at 14:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.