I observed a doublet with peaks at 5.6336 and 5.5886 ppm with the NMR instrument running at 300 MHz, and got a $J_\ce{PH}= \pu{13.5 Hz}$. Is this correct for this molecule (ethoxycarbonylmethyl(triphenyl)phosphonium bromide)? This value is larger than the hydrogen coupling constants for the C-C bonds. Is this larger coupling constant a consequence of the phosphorous?

Ethoxycarbonylmethyl(triphenyl)phosphonium Bromide


Yes, this is the 2JP-H coupling. You could verify this (if you had access to a NMR spectrometer) by running a 1H{31P} experiment; that is, a proton with 31P decoupling. This doublet would collapse back to a singlet. You would also notice some changes to the phenyl protons, as they all have some coupling to the 31P also. A 2J coupling of 13.5Hz is very typical for this type of 1H-31P coupling.

The magnitude of 1H-31P couplings varies enormously, depending on bond order, oxidation state of 31P and bonding arrangement.

  • 1JPH of PH3 = ~175Hz.
  • 1JPH of HP(O)(OMe)2 = ~1200Hz
  • 2JPH P(CHx)y are typically 1-20Hz

It is sometimes observed in 31P-1H (and 31P-13C) that high bond order couplings are larger than lower bond order couplings (ie 3JPH > 2JPH)


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