In Clayden, Greeves, Warren Organic Chemistry Second Edition Page 815, it states:

In four- and three-membered rings, cis couplings are larger than trans ...

but I did not find an explanation for the statement. That being said, why is it true that cis coupling is larger than trans coupling in cyclopropanes and cyclobutanes? Previously, the textbook stated that coupling is a through-bond effect, not a through-space effect, which led me to reason that the trans coupling in cyclopropane should be larger than the cis coupling, as in the case of cyclohexane.

  • 1
    $\begingroup$ For one there is the normal (Karplus equation-like) coupling across bonds for $^3J_{\mathrm{H-C-H}}$. Yet, building a model of cyclopropane or cyclobutane shows that hydrogens «on one side» of these plate-like molecules are this much squeezed into proximity that likely there is additional coupling across space for those sharing the same hemisphere of the cycle which in sum increases the interaction of cisoide H nuclei over their trans interaction. Higher cycloalkanes are not this much pushed into one conformation, so the through-bond-only coupling is more significant for them. $\endgroup$
    – Buttonwood
    Aug 21 '19 at 21:05
  • $\begingroup$ @Buttonwood Thank you for the reply. So are you saying that coupling can also occur through space? $\endgroup$ Aug 21 '19 at 21:18
  • $\begingroup$ Coupling may occur both across bonds (as seen in multiplet splitting pattern of the typical monodimensional $^1H$-NMR spectrum), and coupling may occur across space; the later is the foundation for H,H-NOESY and H,H-ROESY two-dimensional NMR spectroscopy. The intensity of the corresponding coupling signals through space however decay rapidly over distance r by about 1/(r**6), which is why you need specialized techniques to register them. Today however they are commonly used in structure elucidation of organic compounds, typically acessed within less than 5 to 10 minutes per sample. $\endgroup$
    – Buttonwood
    Aug 21 '19 at 22:22

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.