Why in HNMR the coupling constant of trans protons (typically ~17Hz) is higher than the cis protons (typically ~10Hz)? Also, why the antiperiplanar protons in a cyclohexane derivative have higher coupling constant (typically ~ 10Hz) than the equatorial protons (typically ~2Hz) ?
This depends on the extent of overlap. In the trans conformer, the C-H bond is parallel, and so orbital overlap is greater compared to the cis case where the bonds are tilted away from one another.
This depends on the coupling interaction and the dihedral angle ( α ).
In general, with increasing the angle; J value increases.
In cyclohexane, Jaa=10-14 Hz, α = 1800; Jae= 2-6 Hz, α = 600; Jee= 2-5 Hz, α = 600.
For more information, see introduction to spectroscopy 5th edition by Pavia, page 349.
Generally J value depends upon how much overlapping of orbital take place and Further the dihedral angles present between the two atom or group. Both of these combined effects determine the J value.
why the antiperiplanar protons in a cyclohexane derivative have higher coupling constan Spacing between two NMR lines is the coupling constant. It depends upon the dihedral angles present between the two atom or group. By increasing dihedral angle "J value increases.Hence the antiperiplanar protons in a cyclohexane derivative have higher coupling constant than coperiplanar.
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