Protonation of the Dinitrogen-Reduction Catalyst [HIPTN(3)N]Mo(III) Investigated by ENDOR Spectroscopy
Authors: Kinney, RA; McNaughton, RL; Chin, JM; Schrock, RR; Hoffman, BM
HERO ID: 1000995
Dinitrogen is reduced to ammonia by the molybdenum complex of L = [HIPTN(3)N](3-) [Mo; HIPT = 3,5-(2,4,6-iPr(3)C(6)H(2))(2)C(6)H(3)]. . . .
Dinitrogen is reduced to ammonia by the molybdenum complex of L = [HIPTN(3)N](3-) [Mo; HIPT = 3,5-(2,4,6-iPr(3)C(6)H(2))(2)C(6)H(3)]. The mechanism by which this occurs involves the stepwise addition of proton/electron pairs, but how the first pair converts MoN(2) to MoN═NH remains uncertain. The first proton of reduction might bind either at N(β) of N(2) or at one of the three amido nitrogen (N(am)) ligands. Treatment of MoCO with [2,4,6-Me(3)C(5)H(3)N]BAr'(4) [Ar' = 2,3-(CF(3))(2)C(6)H(3)] in the absence of reductant generates HMoCO(+), whose electron paramagnetic resonance spectrum has greatly reduced g anisotropy relative to MoCO. (2)H Mims pulsed electron nuclear double-resonance spectroscopy of (2)HMoCO(+) shows a signal that simulations show to have a hyperfine tensor with an isotropic coupling, a(iso)((2)H) = -0.22 MHz, and a roughly dipolar anisotropic interaction, T((2)H) = [-0.48, -0.93, 1.42] MHz. The simulations show that the deuteron is bound to N(am), near the Mo equatorial plane, not along the normal, and at a distance of 2.6 Å from Mo, which is nearly identical with the (N(am))(2)H(+)-Mo distance predicted by density functional theory computations.