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In my textbook, under the topic "Conditions for geometrical isomerism", the following has been given:

Groups responsible to show geometrical isomerism must be nearly in the same plane.

Is this statement really true? Because, we are well aware that disubstituted cycloalkanes which show geometrical isomerism are not planar. Eg: 1,2-Dimethylcyclopropane

Kindly explain with examples.

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    $\begingroup$ There is nothing to explain. People make blunders, and an example you've just found yourself. $\endgroup$
    – Ivan Neretin
    Aug 8, 2019 at 11:38
  • $\begingroup$ It is a poor definition but it is formally correct. What is important is rigidity. Two points always identify plane (s). There must be a plane *on which the groups cannot rotate, ie at the end a bond with rotation. $\endgroup$
    – Alchimista
    Aug 9, 2019 at 8:01
  • $\begingroup$ In my comment above "with" reads "without rotation", of course. $\endgroup$
    – Alchimista
    Aug 10, 2019 at 11:14

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To have geometrical isomers you need a structurally rigid bond. This bond can be double and this is a case where groups are in the same plane:

enter image description here

Or can be a cyclic compound where the carbon-carbon single bond is restricted(2 examples below):

1st example:

enter image description here

2nd example:

enter image description here

Your textbook definition is incomplete and I wouldn't use the word "must" there.

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    $\begingroup$ The captions of both structures in your first example is identical. Yet 2nd set of that example is 1,2-dimethylcyclopropane. It has both cis- and trans-isomers. Please change the captions accordingly. $\endgroup$
    – Mathew Mahindaratne
    Aug 8, 2019 at 21:00

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