How do I identify #R# and #S# stereoisomers??? When is a carbon compound chiral?

1 Answer
Jul 31, 2017

You could have given us a bit more to go on.........

Explanation:

It is relatively straightforward to assign chirality to carbon compounds. Of course we have to have a depiction of the molecule.........and we ain't got one here. I can give you a few tips.

https://en.wikipedia.org/wiki/Chirality_(chemistry)

ANY carbon centre in a tetrahedral array that has four different substituents can generate a pair of optical isomers based on the geometry of its substituents. The one on the left (as we face it) is the #S# isomer, and the one on the right (as we face it) is the #R# isomer.

How? Well for each depicted stereoisomer, the substituent of least priority, the hydrogen, projects INTO the page. #Br# takes priority over #Cl# takes priority over #F#, and this orientation proceeds COUNTERCLOCKWISE around the chiral carbon, and hence a sinister geometry. The stereoisomer on the right hand side exhibits the opposite geometry, i.e. the #"R-stereoisomer"#.

Given such a stereoisomer, the interchange of ANY two substituents at one optical centre results in the enantiomer. Interchange again, (and it need not be the original two substituents) and you get the mirror-image of a mirror-image, i.e. the enantiomer of an enantiomer, that is the ORIGINAL isomer.

The big mistake that students tend to make with these problems #"IS NOT TO USE MOLECULAR MODELS"#, which are allowed examination materials in every test of organic and inorganic chemistry. It is hard to assess stereochemistry with a model; it is even harder to assess it without a model. And I assure if you go to the office of your organic prof., his or her desk will have a set of molecular models representing molecules in several stages of construction. The prof will have a fiddle with the models as an idea strikes them.

Of course you need to practise how to represent such models on the printed page UNAMBIGUOUSLY. Good luck.