Can someone explain how to calculate the number of geometrical isomers in allenes like $\ce{CH3-(CH=CH)4-CH3}$ or $\ce{CH3-(CH=CH)5-CH3}$

  • $\begingroup$ In the linear cases you drew, the answer is always two. They’re just sometimes called differently. (E/Z vs. R/S) $\endgroup$
    – Jan
    May 15 '15 at 19:23
  • $\begingroup$ But suppose one terminal group is in a different plane compared to the other,will still there be 2 geometrical isomers? I don't think so. $\endgroup$
    – user14857
    May 15 '15 at 19:25
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    $\begingroup$ Other than searching the net, what have you tried to figure this out? Have you drawn any compounds? $\endgroup$
    – jerepierre
    May 15 '15 at 19:28
  • $\begingroup$ Ah, sorry, I misunderstood ‘geometrical isomer’ But what is the reason to consider axial R and S differently from E and Z? Is ther eany? $\endgroup$
    – Jan
    May 15 '15 at 19:29
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    $\begingroup$ @SanchayanDutta Put that information in the question. Show some of the compounds you have drawn. You will get better answers, and your question won't be seen as homework. $\endgroup$
    – jerepierre
    May 15 '15 at 19:38

If we have one double bond in a hydrocarbon compound we have an olefin or alkene. Ethylene is the simplest example of this class of compounds. The carbons in the double bond and the 4 atoms attached to them lie in the same plane. One pair of cis-trans isomers is possible in compounds with a single double bond.

enter image description here

If we add a nother double bond directly on to the end of the double bond in ethylene such that an sp hybridized carbon is created in the process, then we have formed an allene. The 4 substituents at the end of the double bonds in allene lie in planes that are oriented 90° to one another. Look at the allene (bottom line) in the following drawing. One set of substituents (R3, R4) are located in the plane of the screen; the other set of substituents (R1, R2) are located in a plane perpendicular to the screen. Allenes cannot generate cis-trans isomers, but they can generate enantiomers.

enter image description here

Now let's add one more double bond to allene such that we now have 2 sp hybridized carbons (top line in the above drawing). We see that, like in ethylene, the 4 atoms connected to the double bond lie in the same plane as the double bond and one pair of cis-trans isomers is possible.

Any compound with 3 or more cumulative double bonds is a member of the cumulene family. Any cumulene with an odd number of double bonds is geometrically structured like ethylene (the 4 atoms connected to the double bond lie in the same plane as the double bond) and is capable of having one pair of cis-trans isomers. Cumulenes with an even number of double bonds are structured like allene (the 4 atoms connected to the double bonds lie in perpendicular planes) and cannot display cis-trans isomerization, but can have enantiomers.


As a general rule, an even number of cumulative double bonds with different groups on terminal carbon atoms will be optically active and odd number of cumulative double bonds with two different groups on terminal carbon will be optically inactive.

The reason behind this is : For a even number of cumulative double bond compound(generally: Allene) the plane of π bond of that Allene are perpendicular to each other. This geometry of this π bond causes the groups attached to terminal C atom to lie in perpendicular planes. Such Allene with different substitutents on end C are chiral .Such Allene do not show cis trans isomerism, but show optical activity. (Due to the presence of stereo centres not chiral centres.)

Penta12diene do not show optical activity ,since terminal carbon has two same groups. (H atoms)

  • 4
    $\begingroup$ Please consider using some punctuation. $\endgroup$
    – orthocresol
    Apr 27 '19 at 18:04

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