Past exam of the chemistry B course of the University of Cambridge / 2022 / Paper 1 / B2 / b / Solution

Number the cyclopentenone carbonyl carbon $C_1$, its alkene carbons $C_2$ and $C_3$ with OR at $C_3$, and its saturated carbons $C_4$ and $C_5$. The ester alkene has alpha carbon $X$, bearing Me and CO2Et, and beta carbon $Y$, bearing the protected carbon substituent. lithium diisopropylamide generates the ordinary ketone enolate at the saturated alpha carbon $C_5$. The acetonide is retained.

The carbon end of this enolate attacks ester beta carbon Y. Its pi system shifts through X onto the ester oxygen, forming the first new bond $C_5-Y$ and an ester enolate at X. That enolate now attacks the original cyclopentenone beta carbon $C_3$ intramolecularly, giving the second new bond $X-C_3$. The original enone electrons move onto its ketone oxygen, and protonation of the final ketone enolate completes the product. The OR substituent is retained at the new bridgehead; it need not leave in either Michael addition.

The atom map provides a useful structural check. Between bridgeheads $C_3$ and $C_5$, the three paths are $C_3-C_2-C_1-C_5$, $C_3-C_4-C_5$, and $C_3-X-Y-C_5$. Their numbers of internal atoms are 2, 1 and 2, establishing the norbornane skeleton. Both original carbonyls survive and both initial alkenes are consumed. The two successive conjugate additions give this atom map without requiring a concerted cycloaddition.