-5
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Is the structure II a resonating structure of I? If yes, then how? Isn't the lone pair on N non conjugated so as not to participate in resonance?

I Structure I II Structure II

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    $\begingroup$ The bottom structure has a carbon with 5 bonds in the bottom ring. $\endgroup$ – Zhe Jul 21 '17 at 18:06
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    $\begingroup$ Actually there are carbon atoms with 5 bonds almost everywhere. $\endgroup$ – Alchimista Jul 21 '17 at 19:11
  • $\begingroup$ @Mithoron. Almost everywhere in the structures depicted in the figure posted by the OP. :) $\endgroup$ – Alchimista Jul 21 '17 at 20:35
  • $\begingroup$ Ah, yeah. I see poor skill with drawing program there. $\endgroup$ – Mithoron Jul 21 '17 at 20:43
  • $\begingroup$ @Zhe you know I am not hexaccordinate carbon for nothing. I like so much these structures with 5 bonds carbons :D $\endgroup$ – ParaH2 Jul 21 '17 at 20:43
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Note that both representations contain errors.

Under normal circumstances, the valence of uncharged carbon atoms in a molecule equal to four.

Hence, in the upper representation, the carbon bond to the nitrogen bears one valence to much. Keeping the rest of the structure drawn as close to the original, your first drawing should look like the following:

enter image description here

In the lower representation, both the carbons binding towards oxygen are depicted as if carbon were pentavalent, which under normal circumstances does not occur. In addition the central atom of nitrogen is drawn as if it were tetravalent. There are instances of tetravalent nitrogen atoms in organic compounds, however these are positively charged (like the tetrabutylammonium ion, or the iminium cation) on nitrogen. Indeed the second (assuming a motif of $\ce{C=N=C}$) would be something not frequently seen, but on paper could be represented for example as

enter image description here

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There is a carbon atom in the second structure with 5 bonds. It is not a real structure nor a resonance structure of 1

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    $\begingroup$ I count more than one instance of "pentavalent carbon". $\endgroup$ – Buttonwood Jul 21 '17 at 19:46
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    $\begingroup$ @Alchimista look at my username ^^ $\endgroup$ – ParaH2 Jul 21 '17 at 20:45
  • $\begingroup$ @Hexacoordinate-C. This is quite an outstanding nick! :) $\endgroup$ – Alchimista Jul 21 '17 at 20:46
  • $\begingroup$ @Alchimista yeah but it really exists ! :) $\endgroup$ – ParaH2 Jul 21 '17 at 20:47
  • $\begingroup$ I know but it is indeed outstanding. By the way it could have been a real question with mistake forced by some "new to the user" programme. It is good that Burtonwood made a clear picture $\endgroup$ – Alchimista Jul 21 '17 at 20:51

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