Abstract

The reaction mechanism for the step-wise alcoholysis of BH4−, THF·BH3, Me2NH·BH3 and BH3 by ROH [ROH=CH3OH, CF3CH2OH (TFE) and (CF3)2CHOH (HFIP)] was studied computationally. The calculations were performed in gas phase at the DFT/M06/6-311++G(d,p) theory level taking into account non-specific solvent effects by SMD approach. The dihydrogen bonded complexes BH⋯HOR are the intermediates of this cascade borohydride alcoholysis, which set the proper orientation of the reactants molecules and direct their further activation. The consecutive introduction of RO groups instead of hydride ligands in [(RO)nBH(4−n)]− (n=0–3) decreases the dihydrogen bond strength due to the stabilization of borohydride HOMO orbital and the decrease of molecular electrostatic potential. Nevertheless the BH bond polarization and thermodynamic hydricity (hydride donor ability) increase with substitution, leading to the decrease of the reaction barrier. The HH bond formation can be considered as a result of concerted proton and hydride transfer in transition state. For BH4− alcoholysis the highest activation barrier was computed for the first reaction step: BH4−+HOR→H2+[(RO)BH3]− and the reaction is self-accelerating. The proton accepting and hydride donor abilities of neutral 4-coordinate borohydrides X·BH3 (X=THF, NHMe2) are lower than those of BH4−. Accordingly the activation barriers change in the order BH4−<THF·BH3<Me2NH·BH3 and at any reaction stage they remain higher than the barrier for the reaction of [(RO)BH3]− with the same alcohol. Together with low activation barrier for BX bond dissociation (X=O, N) this suggests the ligand exchange and the switch to the kinetically more favorable [(RO)BH3]− alcoholysis.

Full Text
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