Reaction pathway of aliphatic pinacol‐type rearrangement reexamined

Pinacol rearrangement is often written to proceed via 1,2‐Me migration to the tertiary cationic center, followed by deprotonation to give pinacolone. Computational study was carried out for model reactions to clarify why the migration of the OH group is not involved in the mechanistic scheme despite the fact that OH is a better migrating group than Me. It was found that the migratory aptitude of X in both XCMe₂‐CH₂Cl and XCMe₂‐CMe₂Cl is in the order, NH₂ > OMe > Ph > Me, indicating that a migrating group with n‐electrons has a larger aptitude than a π‐ or σ‐electron group. However, the reactivity differences became much smaller for XCMe₂‐CMe₂OMe, a model compound for aliphatic pinacol rearrangement. Calculations of MeOCMe₂‐CMe₂OMe revealed that three initial ionization steps, C? O heterolysis, concerted OMe migration and concerted Me migration, compete with each other. On the other hand, the ring‐opening step of the epoxide‐type intermediate formed via OMe migration was shown to have quite a large activation barrier. It was suggested that aliphatic pinacol rearrangement proceeds via the concerted Me migration route or the C? O heterolysis‐Me migration‐deprotonation route. Epoxide may form by the concerted MeO migration, but it would not be an important intermediate of pinacol rearrangement. Copyright © 2010 John Wiley & Sons, Ltd.