Do biphenyl system show geometrical isomerism?
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biphenyl system isomerism
#4 4 December 2011 - 04:48 PM
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?
This post has been edited by amansci: 4 December 2011 - 04:58 PM
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#6 5 December 2011 - 01:39 PM
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.
This post has been edited by amansci: 5 December 2011 - 01:40 PM
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#7 5 December 2011 - 02:13 PM
amansci, on 5 December 2011 - 01:39 PM, said:
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.
You've come a long way. Remember back when we defined what a velocity meant? Now we are talking about an antisymmetric tensor of second rank in four dimensions.
-Feynman Lectures on Physics II
-Feynman Lectures on Physics II
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#8 5 December 2011 - 02:31 PM
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:

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:
So your answer of 4 is correct, though not for the reasons you've stated.
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
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.
This post has been edited by hypervalent_iodine: 5 December 2011 - 02:32 PM
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#9 6 December 2011 - 02:29 PM
hypervalent_iodine, on 5 December 2011 - 02:31 PM, said:
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:

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:
So your answer of 4 is correct, though not for the reasons you've stated.
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
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?
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#12 6 December 2011 - 02:56 PM
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:
Giving a total of 8.
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.
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