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enter image description here

My teacher explained it as restricted rotation between the double bonds. But I think there is more to it. After all, why is it considered to be a type of isomerism itself? Whether an alkene is cis or trans only depends on how you make the structure on paper. So what is the basic idea behind it?

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    $\begingroup$ What do you mean exactly by "it only depends on how you make the structure on paper"? There are no rotational nor mirror symmetries that can convert a cis isomer into a trans, so they are different compounds. Another way to look at it is that not all the parts of the molecule are in the same position relative to each other in the isomers, so there should naturally be different steric interactions. You can convert one into the other by rotation of the double bond, but that is highly restricted as you need to temporarily break the $\pi$ bond, resulting in a high bond rotation energy barrier. $\endgroup$ Nov 23 '13 at 12:17
  • $\begingroup$ I've tried to explain my point via paper 53. You see, I get your explanation but the problem arises when you draw it on paper. If you specify cis or trans, I will make the compound. But if you ask me that I make 2-butene and without seeing my structure you ask me is it cis or trans, then it would only depend on how I MAKE the compound. That's the problem I've experienced. $\endgroup$
    – Rohinb97
    Nov 24 '13 at 14:09
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    $\begingroup$ It is very hard to see which problem you have. If I ask you to draw a hat I cannot pretend you draw it in the colour I want. If I want a red hat I must ask for it. Else it cab be blu or whatever. It as an analogy E and Z are different ergo the E Z nomenclature to distinguish them. Moreover the title does not reflect your problem thus the nice answers cannot solve it . If your teacher ask you to draw but-2-ene is up to you. Or you draw both as differently from the hat analogy their are just two, it is manageable. $\endgroup$
    – Alchimista
    Feb 6 '18 at 16:32
  • $\begingroup$ @Alchimista The actual intention behind this question was the fact that sometimes my teacher "obviously" referred to some geometric isomer and I would be lost and be looking for properties of the compound to make myself understand why it was so obvious to him. $\endgroup$
    – Rohinb97
    Feb 7 '18 at 7:29
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Geometric isomerism occurs when two structures with the same connectivity are not interconvertible.

Cis-Trans isomerism is common and easy to recognize kind of geometric isomerism. The carbon-carbon double truly has limited rotation. The double bond in the alkene functional group consists of a $\sigma$-bond located within the plane of the molecule and a $\pi$-bond located perpendicular to the plane of the molecule. In the image below (created by David Pilz for Wikipedia), you can see the $\pi$-bond in a perpendicular plane.

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The $\pi$-bond is made from $p$-orbitals, so it cannot have the same symmetry as the $\sigma$-bond. You can see the formation of the $\pi$-bond from $p$-orbitals in this picture by Wikipedia user JoJan. In order for the $\pi$-bond to exist the $p$-orbitals must be aligned.

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In order for the molecule to rotate from $cis$ to $trans$, the bond axis would need to rotate so that the $p$-orbitals are no longer aligned and can no longer form the $\pi$-bond. This slide show from James Condon at Roane State(PDF) explains the same concept. Rotation around the double bond requires enough energy input to break the $\pi$-bond, at least $250 \ \text{kJ/mol}$, which is a lot for room temperature.

The clearest demonstration that geometric isomers are clearly different molecules is to look at some examples with drastically different properties. It is important to remember that the experimental observation that these compounds with very similar structures have very different properties predates the explanation that the double bond has limited rotation.

The two isomers of 1,2-dichloroethene have different physical properties due both the difference in polarity and the difference in shape:

$cis$-1,2-dichloroethane

enter image description here

  • Dipole moment = $1.9 \ \text{D}$
  • Boiling point = $60.2 \ ^\circ\text{C}$
  • Melting point = $-81.47 \ ^\circ\text{C}$
  • Density = $1.28 \ \text{g/cm}^3$

$trans$-1,2-dichloroethane

enter image description here

  • Dipole moment = $0.0 \ \text{D}$
  • Boiling point = $48.5 \ ^\circ\text{C}$
  • Melting point = $-49.44 \ ^\circ\text{C}$
  • Density = $1.26 \ \text{g/cm}^3$

However, my favorites are maleic acid and fumaric acid. Notice the extreme difference in water solubility as well as the difference in acidity. These molecules behave very different chemically.

Maleic Acid $cis$

enter image description here

  • Melting point = $135 \ ^\circ\text{C}$ (decomposes)
  • Density = $1.59 \ \text{g/cm}^3$
  • Solubility in water $788 \ \text{g/L}$
  • Acidity $\text{p}K_{a_1} = 1.9$, $\text{p}K_{a_2} = 6.07$

Fumaric Acid $trans$

enter image description here

  • Melting point = $287 \ ^\circ\text{C}$ (does not decompose)
  • Density = $1.635 \ \text{g/cm}^3$
  • Solubility in water $6.3 \ \text{g/L}$
  • Acidity $\text{p}K_{a_1} = 3.03$, $\text{p}K_{a_2} = 4.44$
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    $\begingroup$ Thanks for explaining geometrical isomers such a nice way @BenNorris but please look at the picture I've attached and the comment I posted below it. Although in real life you can differentiate between them but when you make 2 butene on paper and I ask you what it is, your answer would depend on how you've made your structure. That is the real problem. $\endgroup$
    – Rohinb97
    Nov 24 '13 at 14:14
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    $\begingroup$ Well, since cis and trans 2-butene are different compounds with different properties, you either need to be told which one to draw, be given enough information (for example some reactions are stereospecific and would generate one isomer and not the other based on the reactant), or it does not matter. If you are worried about when it matters and when it does not, ask your instructor about it. $\endgroup$
    – Ben Norris
    Nov 24 '13 at 19:51
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    $\begingroup$ @BenNorris In higher temperatures they should interconvert. So why they have different boiling point and not a common one? I mean suppose that from a given reaction we took only one isomer. Now we isolated this isomer and start heating it up. As we heat it up shouldn't interconversion be possible? So if we can find a temperature that can interconverts these two (reaching equilibrium) then we should expect a mixture with both of them. Then shouldn't this mixture have a common boiling point? $\endgroup$
    – Anton
    Jan 2 at 21:49
  • $\begingroup$ "Geometric isomerism occurs when two structures with the same connectivity are not interconvertible" Does this mean that a pair of enantiomers are also geometrical isomers? $\endgroup$ Mar 1 at 12:55
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    $\begingroup$ Correct me if I'm wrong; for some reason, the boiling points of all pairs of cis-trans isomers lie below the temperature required for interconversion. If this isn't the correct answer to @Anton's excellent question, could you explain it, Ben? $\endgroup$
    – harry
    May 5 at 15:45

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