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Why is it said that the presence of a carbon with 4 different groups is necessary for it to exist as an enantiomeric form?

Lets take 2-fluoropropane for example.

When I draw the structure of it in such a way so that the 2 CH3 groups are adjacent to each other (on paper ) and take its mirror image I see that I got a mirror image which is not superimposable on the original structure. My teacher told me I am only allowed to rotate it in the plane of the paper and only by 180° without lifting it out.

This seems to tell me that they are a pair of enantiomers( non superimposable mirror images).

On the other hand when I draw the structure such that the 2 CH3 groups are on opposite sides of the central carbon and then take a mirror image of it , rotate it by 180° in the plane of the paper I get the original structure. This tells me that they are not stereoisomers.

Also where is the mirror supposed to be placed - up down left right? Any convention (here I have placed the mirror on the right)?

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  • $\begingroup$ They are superimposible. There are no "opposite sides" of a tetrahedral carbon - all the groups are 109.5 degrees away from each other. For a stereogenic center on one atom there must be four different groups. Not all chiral molecules have stereogenic centers though. For example, hindered biphenyls and allenes- they have non-superimposible mirror images. $\endgroup$
    – RobChem
    Jan 2, 2015 at 15:00
  • $\begingroup$ I know that but even after drawing it so many times I am unable to show that. I have mentioned "on paper" whenever I mentioned opposite or adjacent. $\endgroup$
    – Karan Singh
    Jan 2, 2015 at 15:10
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    $\begingroup$ Post a picture of the pair of molecules that you can't superimpose. Then I may be able to see what you've done wrong. $\endgroup$
    – RobChem
    Jan 2, 2015 at 15:21

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When you say: 

 When I draw the structure of it in such a way so that the 2 CH3 groups are 
 adjacent to each other (on paper ) and take its mirror image I see that I got a 
 mirror image which is not superimposable on the original structure.

and then:

On the other hand when I draw the structure such that the 2 CH3 groups are on        
opposite sides of the central carbon and then take a mirror image of it , 
rotate it by 180° in the plane of the paper I get the original structure.

This tells me you are drawing structures where the bonds are at right angles to each other, like this:

     F                              CH3
     |                              |
 CH3-C-CH3        and           CH3-C-H
     |                              |
     H                              F

This will make the seeing of the structures much harder, you can't quite see the 'tetrahedral'. In fact both of the drawings you have indicated are the same, only the connectivity matters -- that is the central carbon is bonded to a fluorine, a hydrogen and two methyl groups.

This is a good place to use models. And if you don't want to buy tinker toys, tooth picks and colored marshmallows (or cheeses) will work just as well.

To resolve your conundrum, expand requirements such that if there exists any arrangement of the groups in your drawing that can form a superimposable mirror image, the molecule is not chiral.

Also consider using dashes and wedges, such as:

enter image description here

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