amansci Posted December 4, 2011 Share Posted December 4, 2011 Do biphenyl system show geometrical isomerism? Link to comment Share on other sites More sharing options...
Fuzzwood Posted December 4, 2011 Share Posted December 4, 2011 Why do you think it will? Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted December 4, 2011 Share Posted December 4, 2011 Is this a homework question? If so, you will need to tel lus what your thoughts are and what has you puzzled with the question. Do you know what a geometric isomer is? Link to comment Share on other sites More sharing options...
amansci Posted December 4, 2011 Author Share Posted December 4, 2011 (edited) 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? Edited December 4, 2011 by amansci Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted December 4, 2011 Share Posted December 4, 2011 Stereoisomers are not the same as geometric isomers, so which one is it? If it is stereoisomers, then do you understand how biphenyl systems may possess chirality? Link to comment Share on other sites More sharing options...
amansci Posted December 5, 2011 Author Share Posted December 5, 2011 (edited) 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. Edited December 5, 2011 by amansci Link to comment Share on other sites More sharing options...
mississippichem Posted December 5, 2011 Share Posted December 5, 2011 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. Most chiral centers arise from having a tetrahedral center with four different groups attached to it. This one arises from an unsurpassable rotational barrier around an otherwise freely rotating bond. Im afraid the only way to work through this is by brute force with a pencil or model kit. There aren't THAT many isomers though so it is doable. Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted December 5, 2011 Share Posted December 5, 2011 (edited) Alright, well saying that there are two due to cis/trans isomerism is not correct. I'm not sure if you know how to properly assign configuration of biphenyls, so I'll give a quick overview just in case. Consider the following system: To assign the configuration of a chiral biphenyl system you need to do the following: Consider the positions of the substituents by looking down the single bond Draw a Fischer-type projection as follows: To assign priorities, near bonds (the NO2 and the CO2H in this case) come first. Assign them as per CIP priorities and then assign the far groups. You can assign (R) and (S) configuration that way. There are also (P) (plus) and (M) (minus) configurations, which are used specifically for biphenyls. In this, you view the compound as a helices and consider direction going from the first priority group to the third, as in the above diagram Are the groups you mentioned in any particular position, or were you just given the substituents and asked to figure it out? Assuming you know the position of the groups, you will have two possible stereoisomers, which arise from the chirality of the biphenyl system. The other thing to consider is that you have a substituent with a chiral centre, which means you now have a few possible combinations: (M) configuration of the biphenyl and (S) configuration at the stereocentre (M) configuration of the biphenyl and (R) configuration at the stereocentre (P) configuration of the biphenyl and (S) configuration at the stereocentre (P) configuration of the biphenyl and (R) configuration at the stereocentre So your answer of 4 is correct, though not for the reasons you've stated. Edited December 5, 2011 by hypervalent_iodine Link to comment Share on other sites More sharing options...
amansci Posted December 6, 2011 Author Share Posted December 6, 2011 Alright, well saying that there are two due to cis/trans isomerism is not correct. I'm not sure if you know how to properly assign configuration of biphenyls, so I'll give a quick overview just in case. Consider the following system: To assign the configuration of a chiral biphenyl system you need to do the following: Consider the positions of the substituents by looking down the single bond Draw a Fischer-type projection as follows: To assign priorities, near bonds (the NO2 and the CO2H in this case) come first. Assign them as per CIP priorities and then assign the far groups. You can assign (R) and (S) configuration that way. There are also (P) (plus) and (M) (minus) configurations, which are used specifically for biphenyls. In this, you view the compound as a helices and consider direction going from the first priority group to the third, as in the above diagram Are the groups you mentioned in any particular position, or were you just given the substituents and asked to figure it out? Assuming you know the position of the groups, you will have two possible stereoisomers, which arise from the chirality of the biphenyl system. The other thing to consider is that you have a substituent with a chiral centre, which means you now have a few possible combinations: (M) configuration of the biphenyl and (S) configuration at the stereocentre (M) configuration of the biphenyl and (R) configuration at the stereocentre (P) configuration of the biphenyl and (S) configuration at the stereocentre (P) configuration of the biphenyl and (R) configuration at the stereocentre So your answer of 4 is correct, though not for the reasons you've stated. 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? Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted December 6, 2011 Share Posted December 6, 2011 You said the question asked you for stereoisomers, didn't you? Geometric isomers - i.e. cis and trans isomers - do not fall under this definition and would therefore not be counted in your answer if this is the case. Link to comment Share on other sites More sharing options...
amansci Posted December 6, 2011 Author Share Posted December 6, 2011 Thats where all the problem is.. What I have studied is stereoisomers classified into optical and geometrical isomers! Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted December 6, 2011 Share Posted December 6, 2011 Well, that's embarrassing. I had my definitions a tad muddled. Geometric isomers fall under that definition, so you were right to include them. Alright, so now we have: (M) configuration of the biphenyl, (S) configuration at the stereocentre, cis double bond (M) configuration of the biphenyl, (S) configuration at the stereocentre, trans double bond (M) configuration of the biphenyl, (R) configuration at the stereocentre, cis double bond (M) configuration of the biphenyl, (R) configuration at the stereocentre, trans double bond (P) configuration of the biphenyl, (S) configuration at the stereocentre, cis double bond (P) configuration of the biphenyl, (S) configuration at the stereocentre, trans double bond (P) configuration of the biphenyl, (R) configuration at the stereocentre, cis double bond (P) configuration of the biphenyl, (R) configuration at the stereocentre, trans double bond Giving a total of 8. Link to comment Share on other sites More sharing options...
amansci Posted December 6, 2011 Author Share Posted December 6, 2011 thanks Link to comment Share on other sites More sharing options...
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