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If a double bond which has two different substituents on each of it's two ends is showing resonance with one of the substituents would the double bond exhibit Geometrical isomerism?..if yes how so since the electron cloud is dispersed over them?..Moreover can the distances and relative configurations of the substituents even change because what exists would actually be a hybrid which would not even have a double bond?

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  • $\begingroup$ Resonating structures always have the same fixed sequence and relative position of nuclei, it is just the electron distribution around them that changes. That analysis must be done on the two isomers separately. But from beginning, as the idea and fact is that they are two different molecules! $\endgroup$ – Alchimista Jul 13 '19 at 9:53
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I am taking N-methyl-N-phenylacetamide as an example. The amide bond exhibits resonance. The lone pair on nitrogen is in conjugation with carbonyl group. In its resonance structure, it acquires $\ce{C=N}$ around which geometrical isomerism can be exhibited.

In N-methyl-N-phenylacetamide, $\ce{N-C}$ bond undergoes conformational rotation (as shown in pink, figure 1). This gives rise to another conformer. Both the conformers undergo resonance exhibiting different geometrical isomers.

That said, cyclic double bonds (here lactums) with 3-7 atoms in the ring are fixed in the cis or Z structure (figure 3). Even after resonance lactums with 3-7 atoms in the ring do not exhibit geometrical isomerism since the strain in the compound is very high.1

isomers and resonance structures

Reference

  1. http://www.siue.edu/~tpatric/Ch%2003%20Stereochem%20H%20T%20I.pdf
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    $\begingroup$ I've noticed this in other posts by you, so I thought I should point it out: The space comes after the punctuation mark. $\endgroup$ – Martin - マーチン Jul 13 '19 at 10:55
  • $\begingroup$ @Martin - マーチン , I will keep it in mind while posting my answers in future . $\endgroup$ – Chakravarthy Kalyan Jul 13 '19 at 14:19

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