Journal of Organic Chemistry p. 3799 - 3804 (1986)
Update date:2022-08-17
Topics:
Cheng, Spencer
Hawley, M. Dale
Ph2C=N2 undergoes successive one-electron reductions in DMF-0.1 M (CH3)4NBF4 at subambient temperatures to give a relatively stable anion radical and an unstable dianion.In the absence of added proton donors, Ph2C=N22- undergoes rapid reaction to give an unobserved intermediate, believed to be Ph2CH-, that reacts with Ph2C=N2 to give Ph2CHN-N=CPh2(radical) (k = 5*104 M*s-1 at -37 deg C).Ph2C=N2(radical-anion) reacts under these conditions by abstraction of a hydrogen atom from a component of the solvent-electrolyte system (k = 0.4 s-1 at -23 deg C) to give Ph2C=NNH- as a longer lived intermediate.This species subsequently reacts with Ph2C=N2, giving first Ph2CHN-N=CPh2 and then the final product, Ph2C=NN=CPh2.The transformation of Ph2C=N2 into Ph2C=NN=CPh2 occurs by a chain process and is initiated by both Ph2C=NNH- and Ph2CHN-N=CPh2.Ph2C=NNH- and Ph2CHN-N=CPh2 are proposed to react with Ph2C=N2 by pathways that transfer hydride ion, either directly or indirectly, from the anion to the central carbon of Ph2C=N2.The final steps in the propagation cycle involve the loss of N2 from Ph2CHN2- and the subsequent coupling of Ph2CH- with Ph2C=N2 to regenerate Ph2CHN-N=CPh2.Ph2C=N2(radical-anion) and Ph2(C=N22- behave as ambient bases in the presence of Broensted acids which can effect their protonation and afford, depending upon whether the central carbon or the terminal nitrogen is protonated, Ph2CH2 or Ph2C=NNH2, respectively.The fraction of Ph2C=NNH2 formed increases with decreasing pKa of the proton donor.
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