Conformationally Locked Pyramidality Explains the Diastereoselectivity in the Methylation of trans-Fused Butyrolactones

Abstract

A stereoselectivity model inspired by the total synthesis of stemona alkaloids is developed to explain why enolate-derived 3,4-fused butyrolactones are methylated with a preference for syn alkylation. The model shows how conformational locking present in nonplanar enolate structures favors syn over anti methylation, due to less significant structural distortions in the syn pathway. The developed model was also successfully used to rationalize selectivities of previously documented methylation reactions.

Origin of High Diastereoselectivity in Reactions of Seven‐Membered‐Ring Enolates

“…Transition‐state calculations illustrate more subtle interactions that cause attack from the different faces to have such different energies. Attack from the favored peripheral β‐face would enable the transition state to avoid having the vicinal protons at C2 and C3 pass through an eclipsed transition state, as illustrated in the conversion of 26 to TS‐26β (Figure 1b, c) [24] . Conversely, attack from the disfavored α‐face through transition state TS‐26α would require the H−C2−C3−H torsional angle to pass through an eclipsed conformer on the way to the product (Figure 1b, c) [25] .…”

Origin of High Diastereoselectivity in Reactions of Seven‐Membered‐Ring Enolates

“…Attack from the favored peripheral β-face would enable the transition state to avoid having the vicinal protons at C2 and C3 pass through an eclipsed transition state, as illustrated in the conversion of 26 to TS-26β (Figure 1b, c). [24] Conversely, attack from the disfavored α-face through transition state TS-26α would require the HÀ C2À C3À H torsional angle to pass through an eclipsed conformer on the way to the product (Figure 1b, c). [25] This transition state TS-26α also suffers from a developing gauche interaction between the incipient bond to the electrophile and the C3À C4 bond of the seven-membered ring (C Me À C2À C3À C4 dihedral = 50°, Figure 1b, c).…”