Determination of geometrical isomers for chains with rings and double bonds

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.

• The correct answer is 2⁴ . Jan 31, 2022 at 14:31
• 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.
– Karsten
Jan 31, 2022 at 19:15

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):
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).