Strecker-Type Cyanation of Acetals Catalyzed by Tin(IV) Chloride─A Mechanistic Study

High Resolution Image Download MS PowerPoint Slide The mechanism of reaction between benzaldehyde dimethyl acetal (PhCH(OMe) 2 ) and trimethylsilylcyanide (TMSCN) in the presence of SnCl 4 ·5H 2 O was investigated by theoretical (DFT) and experimental (NMR) methods. Two competitive reaction channels were revealed: Strecker-type cyanation to PhCH(OMe)(CN) and acetal hydrolysis to benzaldehyde. The most favorable mechanism of cyanation includes coordination of TMSCN, TMS transfer to acetal, the C–O Me bond rupture to give [PhCH(OMe)] + and cis -[SnCl 4 (NC)(H 2 O)] −, and the ion recombination affording the final product. Hydrolysis includes the acetal protonation by a coordinated water molecule to form the [PhCH(OMe)] + ··· cis -[SnCl 4 (OH)(H 2 O)] − ion pair, ion recombination to hemiacetal, and elimination of the second MeOH molecule to give benzaldehyde. Further evolution of benzaldehyde as well as other intermediates (i.e., PhCH(OMe)(NC) and HCN) is also discussed. The tin(IV) catalyst plays a crucial role in both reaction channels. In the cyanation of acetal, the TMSCN coordination to the Sn(IV) center activates this species toward the Si–C bond cleavage while, in the acetal hydrolysis, the highly acidic nature of water coordinated to the Sn(IV) center facilitates the acetal protonation and the C–O bond cleavage as well as the proton transfer in the hemiacetal─the key steps in the formation of benzaldehyde.