Reaction of Aryl Iodides with (PCP)Pd(II)—Alkyl and Aryl Complexes: Mechanistic Aspects of Carbon—Carbon Bond Formation

Abstract

AbstractThe reaction of the PCP‐type complex Pd(Me){2,6‐(iPr2PCH2)2C6H3}(3) with phenyl iodide results in the formation of Pd(I){2,6‐(iPr2PCH2)2C6H3} (5), methyl iodide, toluene, and biphenyl. Formation of Pd(Ph){2,6‐(iPr2PCH2)2C6H3}(4) is observed during the reaction by 31P NMR. Reaction of 4 with aryl iodides results in the formation of 5 and Ph–Ph, Ph–Ar, and Ar–Ar, products indicative of a radical reaction. Under pseudo‐first‐order conditions, the rates of the reactions follow the order p‐OMe > p‐Me > H > p‐NO2 > m‐Cl. The reaction is likely to involve electron transfer from 4 to the aryl iodide followed by fast decomposition of a postulated radical cation [Pd(Ph){2,6‐(iPr2PCH2)2C6H3}]+. (4+.) to give a phenyl radical and [Pd{2,6‐(iPr2PCH2)2C6H3}]+ (6+). Facile decomposition of the aryl iodide radical anion generates an aryl radical and I−. Recombination of aryl radicals gives rise to mixed biaryls, and 6+ combines with I− to give 5.