Abstract

Reaction of HO– with MeCOCHEt2 produces two enolate ions, MeCOEt2 and –CH2COCHEt2. The primary carbanion competitively eliminates C2H4 and C4H8, and forms C2HO–. The elimination of C2H4 is a stepwise reaction proceeding through a six-membered transition state; the first step (deprotonation) is rate-determining. The loss of C4H8 is a rearrangement reaction –CH2COCHEt2 [graphic omitted] –CH2COMe + EtCHCH2. The tertiary carbanion competitively eliminates H2, CH4, and C3H8. The losses of CH4 and C3H8 are stepwise processes occurring through six- and five-membered transition states, respectively. A double isotope fractionation experiment (2H, 13C) shows that both steps of the CH4 elimination are rate-determining.

Full Text
Published version (Free)

Talk to us

Join us for a 30 min session where you can share your feedback and ask us any queries you have

Schedule a call