4
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2‐methylidene‐2H‐pyran

According to me, in methylenepyran 8 π-electrons are present as there are 3 double bonds. Therefore, 2 × 3 = 6 π-electrons and 2 π-electrons as one of the two lone pairs of oxygen will participate in resonance.

But the correct answer is 10 π-electrons. How? Are both lone pairs of oxygen are being considered? If yes, then why?

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  • 2
    $\begingroup$ Your reasoning is fine, and 10 is nonsense. $\endgroup$ – Ivan Neretin Aug 23 at 6:27
  • $\begingroup$ I thought the same. $\endgroup$ – Garima Singh Aug 23 at 6:42
  • $\begingroup$ Can you quote the question statement along with options, or link the source, saying the other way. $\endgroup$ – rv7 Aug 25 at 6:12
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I am quoting from http://www.hutters-online.de/publikationen/dmchdo.html.

One requirement that is basic to this concept and is agreed upon by most authors is that there have to be two double bonds that are in conjugation with a third one, but not in a linear arrangement. The p electronic system thus forms a bifurcation. So the simplest cross-conjugated hydrocarbon would be 3-methylene-1,4-pentadiene (I) that can be viewed as an ethene disubstituted geminally with two vinyl groups:cross conjugation

Based on the above the following resonance structures are possible. These have 8 $\pi$ electrons in conjugation. You are correct.

resonance structures

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