Question

The correct statement about the intermediate formed in the following reaction is

• A.

  

  is formed through conrotatory opening of \(\mathrm{X}\)
• B.

  

  is formed through disrotatory opening of \(\mathrm{X}\)
• C.

  

  is formed through conrotatory opening of \(\mathrm{X}\)
• D.

  

  is formed through disrotatory opening of \(\mathrm{X}\)

Hint:

Aziridine ring-opening to an azomethine ylide is a 4\(\pi\)-electron electrocyclic process; under thermal conditions the Woodward-Hoffmann rules require conrotatory ring-opening, and the resulting ylide geometry follows from that rotation sense.

Answer 1

The Correct Option is A

The same answer can be reached by thinking about orbital symmetry directly, instead of just quoting the 4$\pi$/conrotatory rule from memory.

When the $\mathrm{C-C}$ sigma bond of the aziridine breaks to open the ring, the two electrons in that bond become the terminal p-orbitals of the new 3-atom pi system (the azomethine ylide), together with the $\mathrm{C=N}$ pi bond and the nitrogen lone pair contribution. Counting all the electrons delocalised over this 3-atom, 3-orbital system gives 4 electrons, so the ring-opening is governed by the same 4-electron (4$\pi$) electrocyclic selection rules as the butadiene-to-cyclobutene interconversion.

For a 4$\pi$ electrocyclic process, only one rotation mode keeps the terminal lobes overlapping with the right sign as the sigma bond stretches into the two new p-orbitals. Under thermal activation that mode is conrotatory; disrotatory opening only becomes symmetry-allowed once the system is promoted to an excited state by light. Since this problem runs on heat, not light, conrotatory is the mode in play, ruling out options (B) and (D).

Between the two remaining structures, the conrotatory pathway keeps the two ring substituents rotating the same way as the ring opens, walking the two $\mathrm{CO_2Me}$ groups out to opposite, extended ends of the newly linear 2-azaallyl cation, the geometry drawn in option (A). Option (C) shows a folded arrangement that this particular conrotatory opening does not produce; that folded shape would instead be the disrotatory outcome.

So orbital symmetry (thermal, 4 electrons, conrotatory) and the geometric consequence of that rotation both point to option (A) as the correct intermediate.