Heck possible initial reaction steps

The second Heck reaction involves a naphthyl iodide (Ar2 = 2-naphthyl) but the initial mechanism is much the same. However, the enol ether has two diastereotopic faces syn or anti to the aromatic substituent (Ar1) introduced in the first step. Palladium is very sensitive to steric effects and generally forms less hindered complexes where possible. Thus coordination of the palladium(II) intermediate occurs on the face of the enol ether anti to Ar1. This in turn controls all the subsequent steps, which must be syn, leading to the trans product. The requirement for syn p-hydride elimination also explains the regiochemical preference of the elimination. In this cyclic structure there is only one hydrogen (green) that is syn the one on the carbon bearing the naphthyl substituent is anti to the palladium and cannot be eliminated.. ... 

The mechanism for the Heck reaction is shown in Scheme 4.63. The first step involves the oxidative addition of an aryl or vinyl halide, R -X, to a palladium(O) species. This species normally contains an auxiliary donor, L, where L is often a phosphine. This may be preceded by a reduction of the metal if a palladium(II) salt is employed initially. Thereafter, two different pathways are possible depending on which group dissociates to provide a vacant coordination site for the incoming alkene. If a neutral ligand (such as a phosphine) detaches and the halide is retained, the active species immediately prior to the C-C coupling step is the neutral complex III-90. Conversely, if the anionic ligand (such as a halide) dissociates, the active species is the cationic complex III-91. 

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