# Will N,N‐diethylbenzamide show geometrical isomerism?

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.

• (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? – orthocresol Aug 2 at 13:56
• @orthocresol An empty p orbital , on the other side there is p orbital involved in π bonding – Jasmine Aug 2 at 14:00
• Firstly, resonance forms don't exist individually; but that's not the main problem. More crucially, $$\ce{C+-C=C-C=C <-> C=C-C=C-C+}$$ – orthocresol Aug 2 at 14:13
• 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. – orthocresol Aug 2 at 14:34
• Should really be looking at amide-iminol tautomerism for this compound – Beerhunter Aug 3 at 21:14