amansci

thanks

Thats where all the problem is.. What I have studied is stereoisomers classified into optical and geometrical isomers!

The actual compound was 2-(1-bromoethyl)-6-iodo-2,-nitro-6,-[(1E)-prop-1-en-1-yl]biphenyl. So 2 for chiral carbon (CHBrMe) , 2 for -CH=CH-CH3(cis-trans) and 2 for M-P configuration of biphenyl. Total I get 8 stereoisomers. Am I doing it the right way now?

Yeah , they do show chirality when there is hindrance due to groups at ortho positions and they dont have plane of symmetry, but I dont know how to count the number of stereoisomers for such system.

it was a question in my last exam. One of the benzene had -NO2 and -CH=CH-CH3 at ortho positions, another had -CHBrMe and -I. I was asked to write number of stereoisomers possible. I answered 4 ( 2 of cis-trans and 2 of that chiral centre.) am i missing something?

Do biphenyl system show geometrical isomerism?