Reaction of trans-Pd(styryl)Br(PMePh2)2 with styryl bromide affording 1,4-diphenylbutadiene.: An unexpected homocoupling process induced by P-C reductive elimination

Reaction of (Z)-styryl bromide (1) with PhB(OH)(2) in toluene at 80 C for 1 h in the presence of Pd(PPh3)(4) (1.5 mol %) and an aqueous solution of K2CO3 (3 equiv) affords (Z)-stilbene as the cross-coupling product in 99% yield. On the other hand, the same reaction of the (E)-isomer of 1 forms considerable amounts of homocoupling products (1,4-diphenylbutadiene (2, 22%) and biphenyl (27%)) in addition to the cross-coupling product ((E)-stilbene, 73%). The formation of 2 was examined by kinetic experiments using trans-Pd(CH=CHPh)Br(PMePh2)(2) (3) as a model of the presumed intermediate. Complex 3 reacts with 1 at 50 C for 5 h, giving 2 (91%) and trans-PdBr2(PMePh2)(2) (4, 92%), together with a small amount of Pd(eta(2)-PhCH=CHPMePh2)Br(PMePh2) (5, 8%). The reaction rate shows first-order dependence on the concentration of 3 and 5, respectively, but is independent of the concentration of 1. A novel homocoupling process induced by P-C reductive elimination from 3 giving 5 is proposed.