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As far as I was taught, cis and trans isomers were categorised under stereoisomers and sub category of geometric isomers. Qualifying properties like, rigid and restricted double bond rotation (due to pi bond electron cloud), no terminal alkenes (those with double bond at the end of carbon chain).

But, here was a quote from a LibreTexts article:

If one carbon of the double bond has two identical groups such as 2 H's or 2 Cl's or 2 CH3 etc. there cannot be any geometric isomers.

enter image description here

What is the basis of above claim? Why cis and trans with identical groups can't be qualified as a geometric isomer? If not what is this? Am I missing something here?

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    $\begingroup$ "If one carbon of the double bond has two identical groups …" $\endgroup$ – Loong Jul 13 '17 at 4:27
  • $\begingroup$ @Loong I am still not getting there :/ so identical groups means, there's symmetry, it's isn't in geometry? so what is this then?... $\endgroup$ – bonCodigo Jul 13 '17 at 4:59
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Hmm, you seem to be having a problem visualizing the structures in the case of 2 identical groups on one carbon. But never fear, 3D molecules are here.

Check out chloroethene:

enter image description here

It has two identical groups (hydrogens) on the left side. If you tried to make a geometrical isomer for this, you'd get:

enter image description here

But hey! That's just the previous image flipped upside down! That's not an isomer, it's the same thing!

Indeed it's the same molecule. These molecules can't have a geometrical isomer because it possess an plane of symmetry:

enter image description here

So there you go. Go ahead and visit this link if you want an interactive 3D model that you can spin around.


Added based on comments from @Zhe, there's a plane of symmetry in following angle,

enter image description here

enter image description here

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    $\begingroup$ In the last figure, we should not refer to this is an "axis." It is a plane. $\endgroup$ – Zhe Jul 13 '17 at 13:30
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    $\begingroup$ Actually, reading this again, you should call it an axis, but it's just not an axis of symmetry for the single molecule. The two structures are related by an axis of rotation. $\endgroup$ – Zhe Jul 13 '17 at 14:58
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    $\begingroup$ This molecule does not have an axis of symmetry. It has an plane of symmetry. The lack of isomerism comes from the two substituents on the left side being identical. @bonCodigo $\endgroup$ – Zhe Jul 17 '17 at 3:31
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    $\begingroup$ @Zhe With reference to the molecule in the answer $\ce{C2H3Cl}$, I understand the identical substituents contributes to the lack of isomerism. And the axis of symmetry (for rotation) doesn't concern it. However I am concerned about your comment of axis of symmetry vs plane of symmetry (it's opening another door a pages of digging conceptual details...). Speaking of plane of symmetry, this molecule still doesn't have a plane of symmetry as the imaginary plane that bisects this molecule doesn't give mirror images of each other... (cut me in to half now : sagital plane, pun intended). Does it? $\endgroup$ – bonCodigo Jul 17 '17 at 4:00
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    $\begingroup$ @Zhe Actually now that I bisected the molecule horizontally (transverse...)... I see the light you mentioned. It has a plane of symmetry in that angle, but not from vertical though. $\endgroup$ – bonCodigo Jul 17 '17 at 4:03
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I take issue with the way a plane of symmetry is used in the currently accepted answer (see comments). So, here is my explanation of the question:

Have a look at some example molecules: $\ce{H2C=C(Cl)(CH3)}$, $\ce{H2C=C(Cl)(H)}$, $\ce{H2C=C(CH3)(CH3)}$. All of these do not show any geometrical isomerism. The first two have two identical groups (hydrogen) on the left side, while the last example has identical groups (hydrogen and methyl) on both sides.

Why does having identical groups on either side of the pi bond lead to no geometrical isomerism? (In fact, it is the only restrictive condition in pi-bond geometrical isomerism.)

We know that, to obtain the other geometrical isomer of a given compound, we just need to flip the groups on one side of the molecule. For example, here are the GIs of 2-butene:

enter image description here

You can also go in the reverse direction, and flip either side, with the same result. This rule of flipping-and-fixing will hold true for any pi-bonded molecule.

Now, let's apply this on 2-chloropropene:

enter image description here

Turns out, it's not a geometrical isomer! In fact, you can take the final molecule formed and rotate it 180deg (about an axis passing through a double bond) to obtain the original compound in the question. Recall that any molecules that become identical with each other on complete rotation are actually identical. Hence, in the above diagram, there's not two new isomers. It's still the same molecule, just shown rotated.

Note that if you had from the side where both groups are identical, you'd immediately seen both the molecules being identical:

enter image description here

Hence, this is how having identical groups on either side of the pi bond leads to no geometrical isomerism. Now, you should go back to the original example of 2-butene, and notice how rotating it through the pi bond, or any other axis does not lead to the same representation.

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