According to my book 2-bromo-2'-(1-chloroethyl)-6'-iodo-6-(prop-1-en-1-yl)-1,1'-biphenyl has eight possible spatial arrangements. How can this be so? I can think of only four – left two groups being planar with right two groups up-down and down-up, and vice versa. For this configuration of I and $\ce{CHClCH3}$ I can think of four, for I and $\ce{CHClCH3}$ interchanged I can think of four4 more, but that is not the same molecule and is just a configurational isomer.

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    There is a cis/trans thing in one substituent and a chiral center in another. – Ivan Neretin Aug 10 at 12:20
  • @IvanNeretin will that be so when the substituents are planar or up/down? I'm a bit confused about the planar part actually. – Hema Aug 10 at 12:29
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    Roughly speaking, the rings are perpendicular to each other. You may think that one is horizontal (I guess that's what you mean by planar) and another is up/down, or vice versa; it is all the same. – Ivan Neretin Aug 10 at 12:53

The rings are likely to rotate their plane so that they become perpendicular to each other due to steric hindrance. After rotation, due to absence of a plane of symmetry, the compound will be optically active.
There will be a dextro and a levo form of each of the four compounds you mentioned. So there will be 8 compounds in total.

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