Is the alkene with the representation $\ce{CH3CH=CHCH2CH3}$ a cis-trans isomer? If so, why? I mean, it is not like the other cis-trans alkenes, because all of them were symmetrical somehow?


First of all, let's talk about geometrical isomerism from the beginning, as I think that you don't understand it properly.

Geometrical isomerism is different spacial arrangements of atoms or groups across any bond under the following conditions:

  1. Restricted rotation about a bond in the molecule.
  2. Both, substituted atom or group on each C-atom, about which rotation is frozen or restricted should be different.
  3. In case of cumulenes, odd and even number of double bond and the distance between the atom or group should be taken into consideration.

Now coming to your question, $\ce{CH3-CH=CH-CH2-CH3}$ can show geometrical isomerism because it follows conditions 1 and 2 given above. It can therefore form cis and trans isomers.
Which of them will be the major product depends upon the stability of the compound.

As you can see, in the given conditions there was no mentioning that the compound has to be symmetrical to form cis-trans isomerism. You don't have to look for symmetry, but for the above conditions.

I mentioned earlier that geometrical isomerism is a different spacial arrangement of atoms or groups across any bond, se let me give some examples for that:

  1. Compounds containing $\ce{C=C}$, $\ce{C=N}$, $\ce{N=N}$
  2. Compounds containing cyclic structures, homocyclic compounds (benzene type), heterocyclic (eg. pyridine), fused rings (eg. naphthalene), etc.
  3. Compounds containing restricted rotation about single bonds due to steric hindrance. Attached groups are so bulky, such that on rotation repulsion increases thus, decreasing the stability.
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Yes, the formula could be for either:


enter image description here

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  • $\begingroup$ Sorry but I don't get it. I can't see the difference could you please clarify more ? $\endgroup$ – aazz20 Dec 11 '15 at 17:09
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    $\begingroup$ The trans-2-Pentene isn't drawn quite the same way as the cis version. Redraw it with the two H atoms and you should get it. see: en.wikipedia.org/wiki/Cis%E2%80%93trans_isomerism $\endgroup$ – MaxW Dec 11 '15 at 17:16
  • $\begingroup$ I'll add that the $\ce{-CH3}$ group can spin about the single $\ce{C-C}$ bond, but the $\ce{C=C}$ is "locked." So the hydrogen atoms are either on the same side (cis isomer), or opposite sides (trans isomer) of the $\ce{C=C}$ bond. $\endgroup$ – MaxW Dec 11 '15 at 18:36

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