# How many stereoisomers can exist for 4,5-bis(3-hydroxyoct-1-en-1-yl)cyclopentane-1,3-diol?

Q.10 For the given compound $$\ce{X}$$, the total number of optically active stereoisomers is ____.

This is from JEE Advanced 2018 second question paper.

I thought the answer to be at least 32 because of exactly 5 stereocentres present. One of the double bond cannot show geometrical isomerism possibly.

I did not find any plane of symmetry because of the different kind of double bonds.

• Two issues. (1) The image is not clear enough; it's hard to tell exactly which bonds are wavy versus wedge. (2) You're misreading the question. The wedge bonds are fixed. You can change configuration only for the wavy bonds. – Zhe May 22 '18 at 17:21
• I'm so sorry for the picture quality. I can clarify it, The long chain on the bottom of the cyclopentane system contain the 2 wavy bond S, all others are in dark bond. – Sarah jane May 22 '18 at 17:40
• Well, then there are 3 configurations to worry about, for a maximum of $2^{3}=8$. It can't be 32, but 7 is certainly reasonable. I'll let you figure out why it's 7 and not 8. – Zhe May 22 '18 at 17:43
• Okay, but where did you get the third stereogenic centre there? Those two in wavy and then? Does it show synanti about that sigma bond in cyclopentane. If it is 8 then it will be easy because there must be a plane of symmetry. – Sarah jane May 22 '18 at 17:54
• There's no third stereogenic center. But if the double bond isomerism that generates a separate isomer, that isomer can be chiral. If it is chiral, it is optically active... – Zhe May 22 '18 at 18:07