2-bromo-2'-(1-chloroethyl)-6'-iodo-6-(prop-1-en-1-yl)-1,1'-biphenyl

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.

  • 1
    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
  • 1
    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
  • Double bond isomerism is a type of stereoisomerism as well, and certainly has to do with the spatial arrangement. – mykhal Sep 4 at 14:10
up vote 5 down vote accepted
+50

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

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.