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We recently started learning about pericyclic reactions and this is a step from a mechanism I'm working on for an assignment. I believe that the arrows on the right are better than the arrows on the left because if I throw in some charges instead of arrows, the outcome I get on the right is somewhat more "stable?" for the reasons stated on the picture. Am I correct in my assumption? On the overall reaction, it's the 6 to 7 step that I'm interested in.

reaction mechanism

overall reaction

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  • $\begingroup$ Any reason you didn't draw out the radical version? Frankly, it doesn't make a difference because what experiment would you design to determine this? If you can't, then it doesn't matter. $\endgroup$
    – Zhe
    Commented Nov 26, 2016 at 22:03
  • $\begingroup$ Also, I don't believe the arrows you drew for the concerted mechanism lead to the partial charges you've drawn. $\endgroup$
    – Zhe
    Commented Nov 26, 2016 at 22:04
  • $\begingroup$ I meant that they lead to the same bonds being formed in either one direction or the other. $\endgroup$
    – Vlad
    Commented Nov 26, 2016 at 22:05

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The answer is rather bland and probably not what you want to hear. Electrocyclic reactions are called electrocyclic because they are cyclic. (Please don’t see cyclic circular reasoning in there.) They proceed in a cyclic, i.e. fully symmetric fashion. You simply cannot draw one arrow direction and claim it is the most logical or the only one possible. You could draw these arrows as you drew them, but you could also formulate a radical-based mechanism (each double bond splits into two radicals which combine with the two of the other double bond) at no loss.

Thus, there is no way to say whether left or right. The reaction happens; in a cyclic fashion.

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    $\begingroup$ Arrow-pushing mechanisms are crude oversimplifications at best. $\endgroup$ Commented Dec 2, 2016 at 11:15
  • $\begingroup$ @Jan Can the same be said about sigmatropic rearrangements e.g. [1,5] hydride shift? $\endgroup$
    – Anton
    Commented Apr 29, 2021 at 16:58
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    $\begingroup$ @Anton Quite obviously so. Which also neatly explains why they can be written both as hydride transfers and as proton transfers. $\endgroup$
    – Jan
    Commented May 1, 2021 at 21:05

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