3
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N,N‐diethylbenzamide resonance

I know for a compound to show geometrical isomerism it should

  1. have site of restricted rotation;
  2. be connected to two different groups on either side of site of restricted rotation;
  3. the distance between the groups on the either side of site of restricted rotation should be different.

I think conditions 1 and 3 hold true here. I am confused if or not 2 is true.

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    $\begingroup$ (1) "Restricted rotation" is not a black or white thing; different things have different barriers to rotation. (2) What makes you think that one side of a benzene ring could be different from the other side? $\endgroup$ – orthocresol Aug 2 at 13:56
  • $\begingroup$ @orthocresol An empty p orbital , on the other side there is p orbital involved in π bonding $\endgroup$ – Jasmine Aug 2 at 14:00
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    $\begingroup$ Firstly, resonance forms don't exist individually; but that's not the main problem. More crucially, $$\ce{C+-C=C-C=C <-> C=C-C=C-C+}$$ $\endgroup$ – orthocresol Aug 2 at 14:13
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    $\begingroup$ i.stack.imgur.com/XR9J7.png - there is no way you can treat two sides of a benzene ring differently, ever. And there is no "double bond" there, by the way. It is only there in certain resonance forms, which contribute only a tiny bit to the overall resonance hybrid. Which brings us back to the point that restricted rotation is not something that can be answered with only a yes or no. $\endgroup$ – orthocresol Aug 2 at 14:34
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    $\begingroup$ Should really be looking at amide-iminol tautomerism for this compound $\endgroup$ – Beerhunter Aug 3 at 21:14

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