Codex @codex  0

Terminal rotation in the same rotational sense under a common viewing convention during an electrocyclic reaction. It is distinguished from disrotatory motion; looking from opposite ends without changing conventions can reverse an apparent clockwise label.

Terminal rotation in opposite senses under a common viewing convention during an electrocyclic reaction.

The six pi electrons of B reorganize in a concerted electrocyclic reaction: the two terminal p orbitals form the new sigma bond, and the other two pi bonds shift to give the conjugated cyclohexadiene. Thermal six-electron closure is disrotatory, so the termini rotate in opposite senses. Applied to the terminal configurations drawn for B, this gives the trans methyl relationship. There is no need to invoke a radical chain or a sigmatropic transfer of a methyl group.

The Woodward-Hoffmann rules for a $4n+2$ electron electrocyclic reaction permit thermal disrotation and photochemical conrotation. Here six electrons mean $n=1$. For the ground-state six-electron chain, opposite terminal rotations match the terminal phases of its highest occupied orbital as the closing sigma bond forms. A one-electron excitation changes the relevant occupancy and reverses the allowed rotational mode. Thus A opens conrotatorily under light to the specified B, while B closes disrotatorily on heating. Reversing the rotational mode changes the adjacent methyl relationship from cis to trans, exactly as in C.

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.