Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 6 / c / Solution

The key sequence is deprotonation, anionic eight-pi electrocyclization, then elimination. Label the conjugated route from the lower cyclopentene to the right cyclopentene aswhere $C_3$ is the methylene joining the lower ring to the central alkene, and $C_4$ carries the $\mathrm{C}{\mathrm{H}}_2\mathrm{OC}(\mathrm{O})\mathrm{NM}{\mathrm{e}}_2$ substituent.

potassium hexamethyldisilazide removes a proton at $C_3$. The resulting anion is delocalized over the seven-carbon array: its six pi-bond electrons plus the lone pair make an eight-electron system. Thermal electrocyclic reaction is disrotatory for this $4n$ electron count. Joining $C_1$ and $C_7$ forms the seven-membered central ring and puts the remaining anionic character at $C_4$, with double bonds at $C_2=C_3$ and $C_5=C_6$.

That $C_4$ anion then forms $C_4=C{H}_2$, expelling the O-dimethylcarbamate group from the attached methylene. This E1cb elimination supplies the exocyclic methylene and explains why no oxygen remains in the depicted product. The disrotatory closure, with the cyclopentene ends folded to avoid unfavorable overlap, gives the trans relationship between the two lower fused-ring bonds shown by the wedge and dash. A simple [3,3] rearrangement would not create the required bond between the two remote cyclopentene termini.