Number of spatial arrangements of tetrasubstituted biphenyl

Question

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.

Answer 1

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.)

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