# Are all lone pairs considered as π-electrons?

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?

• Your reasoning is fine, and 10 is nonsense. Aug 23, 2019 at 6:27
• I thought the same. Aug 23, 2019 at 6:42
• Can you quote the question statement along with options, or link the source, saying the other way. Aug 25, 2019 at 6:12

## 1 Answer

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:

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