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How can we determine the geometrical isomers for this compound . It contains a ring and 3 double bonds . I am confused because i was told to consider 1 ring = 1 double bond. When i tried it i got it as $2^4$ but its not the right answer . Kindly explain how to find the geometrical isomers for the compound by taking the above compound as an example.

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    $\begingroup$ The correct answer is 2⁴ . $\endgroup$
    – Infinite
    Jan 31, 2022 at 14:31
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    $\begingroup$ You could consider 1,4-disubstituted cyclohexane = 1 double bond, if the two substituents are different. However, the 1.3-disubstituted cyclopentane is less symmetric, so you are off by one. $\endgroup$
    – Karsten
    Jan 31, 2022 at 19:15

1 Answer 1


The molecule contains three $\ce{C=C}$ double bonds which may be in either (E), or in (Z) configuration, a total of $2^3$ variations if taken alone (blue rectangles). The molecule contains two stereogenic centres, which may be either in (R), or in (S) configuration, a total of $2^2$ variations if taken alone (red rectangles):

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Since you want to run the permutations on double bonds and stereogenic centers as independent of each other, there are $2^5 = 32$ stereo isomers possible (five parameters, two levels each).


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