Jump to content

biphenyl system


amansci

Recommended Posts

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 by amansci
Link to comment
Share on other sites

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 by amansci
Link to comment
Share on other sites

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

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:

 

biphenyl1-1.jpg

 

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:

biphenyl2-1-1.jpg

 

  • 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:

 

  1. (M) configuration of the biphenyl and (S) configuration at the stereocentre
  2. (M) configuration of the biphenyl and (R) configuration at the stereocentre
  3. (P) configuration of the biphenyl and (S) configuration at the stereocentre
  4. (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 by hypervalent_iodine
Link to comment
Share on other sites

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:

 

biphenyl1-1.jpg

 

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:

biphenyl2-1-1.jpg

 

  • 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:

 

  1. (M) configuration of the biphenyl and (S) configuration at the stereocentre
  2. (M) configuration of the biphenyl and (R) configuration at the stereocentre
  3. (P) configuration of the biphenyl and (S) configuration at the stereocentre
  4. (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

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:

 

  1. (M) configuration of the biphenyl, (S) configuration at the stereocentre, cis double bond
  2. (M) configuration of the biphenyl, (S) configuration at the stereocentre, trans double bond
  3. (M) configuration of the biphenyl, (R) configuration at the stereocentre, cis double bond
  4. (M) configuration of the biphenyl, (R) configuration at the stereocentre, trans double bond
  5. (P) configuration of the biphenyl, (S) configuration at the stereocentre, cis double bond
  6. (P) configuration of the biphenyl, (S) configuration at the stereocentre, trans double bond
  7. (P) configuration of the biphenyl, (R) configuration at the stereocentre, cis double bond
  8. (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

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.