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 four more, but that is not the same molecule and is just a configurational isomer.

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    $\begingroup$ There is a cis/trans thing in one substituent and a chiral center in another. $\endgroup$ – Ivan Neretin Aug 10 '18 at 12:20
  • $\begingroup$ @IvanNeretin will that be so when the substituents are planar or up/down? I'm a bit confused about the planar part actually. $\endgroup$ – Hema Aug 10 '18 at 12:29
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    $\begingroup$ 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. $\endgroup$ – Ivan Neretin Aug 10 '18 at 12:53
  • $\begingroup$ Double bond isomerism is a type of stereoisomerism as well, and certainly has to do with the spatial arrangement. $\endgroup$ – mykhal Sep 4 '18 at 14:10

There are three independent stereogenic centers or elements in this molecule,

  • chiral carbon atom of the chloroethyl group,
  • double bond of the propenyl group,
  • axis of the substituted biphenyl with bulky substituents at 2,2′,6,6′ positions.

Each of them has two configurations. It means that the total number of stereoisomers is 23 = 8. (No one of them is symmetrical and cannot be superimposed with its mirror image, no one can be superimposed with any other one.)

Fig. 1: Formulas I-VIII

  1. (1M)-2-bromo-2′-[(1R)-1-chloroethyl]-6′-iodo-6-[(1Z)-prop-1-en-1-yl]-1,1′-biphenyl
  2. (1M)-2-bromo-2′-[(1S)-1-chloroethyl]-6′-iodo-6-[(1Z)-prop-1-en-1-yl]-1,1′-biphenyl
  3. (1M)-2-bromo-2′-[(1R)-1-chloroethyl]-6′-iodo-6-[(1E)-prop-1-en-1-yl]-1,1′-biphenyl
  4. (1M)-2-bromo-2′-[(1S)-1-chloroethyl]-6′-iodo-6-[(1E)-prop-1-en-1-yl]-1,1′-biphenyl
  5. (1P)-2-bromo-2′-[(1R)-1-chloroethyl]-6′-iodo-6-[(1Z)-prop-1-en-1-yl]-1,1′-biphenyl
  6. (1P)-2-bromo-2′-[(1S)-1-chloroethyl]-6′-iodo-6-[(1Z)-prop-1-en-1-yl]-1,1′-biphenyl
  7. (1P)-2-bromo-2′-[(1R)-1-chloroethyl]-6′-iodo-6-[(1E)-prop-1-en-1-yl]-1,1′-biphenyl
  8. (1P)-2-bromo-2′-[(1S)-1-chloroethyl]-6′-iodo-6-[(1E)-prop-1-en-1-yl]-1,1′-biphenyl

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.
If your talking about the four molecules shown in the image, then a few of them are identical. Molecules (i) and (ii) [drawn above the line in the image] are identical because if you rotate the (i) molecule firstly in its plane by 180° and then perpendicular to its plane by 180° you get (ii) molecule. Similarly (iii) and (iv) are identical. enter image description here


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