Mechanism of Asymmetric Hydrogenation of α-(Acylamino)acrylic Esters Catalyzed by BINAP−Ruthenium(II) Diacetate

The mechanism of asymmetric hydrogenation of α-(acylamino)acrylic esters with Ru(CH3COO)2[(S)-binap] (BINAP = 2,2‘-bis(diphenylphosphino)-1,1‘-binaphthyl), giving the S saturated products in >90% ee, has been investigated by means of a kinetic study, deuterium labeling experiments, isotope effect measurements, and NMR and X-ray analysis of certain Ru complexes. The hydrogenation in methanol under a low H2 pressure proceeds via a monohydride-unsaturate mechanism that involves the initial RuH formation followed by a reaction with an olefinic substrate. The migratory insertion in the enamide−RuH chelate complex occurs reversibly and endergonically in an exo manner, giving a five-membered metallacycle intermediate. The cleavage of the Ru−C bond is achieved with either H2 (major) or CH3OH (minor). Both of the pathways result in overall cis hydrogenation products. The hydrogen at C(3) is mainly from an H2 molecule, and the C(2) hydrogen is from another H2 or protic CH3OH. The major S and minor R enantiomers are produced via the same mechanism involving diastereomeric intermediates. The turnover rate is limited by the step of hydrogenolysis of a half-hydrogenated metallacyclic intermediate. The participation of two different hydrogen donor molecules is in contrast to the pairwise dihydrogenation using a single H2 molecule in the Rh(I)-catalyzed reaction which occurs via a dihydride mechanism. In addition, the sense of asymmetric induction is opposite to that observed with (S)-BINAP−Rh(I) catalysts. The origin of this phenomenon is interpreted in terms of stereocomplementary models of the enamide/metal chelate complexes. A series of model stoichiometric reactions mimicking the catalytic steps has indicated that most NMR-observable Ru complexes are not directly involved in the catalytic hydrogenation but are reservoirs of real catalytic complexes or even side products that retard the reaction.