Question

How to write R,S nomenclature or R,S configuration for a given projection? And how to determine whether the given two compounds are enantiomers or diastereomers using R,S configuration?

Answer 1

An easy way to find the R / S configuration of a molecule with more than one chiral center is with Fischer projection. A Fischer projection is a convenient two-dimensional drawing that represents a three-dimensional molecule. To make a Fischer projection, you view a chiral center so that two substituents are coming out of the plane at you, and two substituents are going back into the plane, as shown here. Then the chiral center becomes a cross on the Fischer projection. Every cross on a Fischer projection is a chiral center.

Creating a Fischer projection.
Fischer projections are convenient for comparing the stereochemistries of molecules that have many chiral centers. But these projections have their own sets of rules and conventions for how you can rotate and move them.

As shown here, the two main ways to rotate a Fischer projection are as follows:

Working with Fischer projections.
1. You can rotate a Fischer projection 180 degrees and retain the stereochemical configuration, but you cannot rotate a Fischer projection 90 degrees.
2. You can rotate any three substituents on a Fischer projection while holding one substituent fixed and retain the stereochemical configuration.

Here’s how to determine the configuration of a chiral center drawn in a Fischer projection. First, you prioritize each of the substituents using the Cahn-Ingold-Prelog prioritizing scheme.

According to the Cahn-Ingold-Prelog prioritizing scheme, the highest priority goes to the substituent whose first atom has the highest atomic number. (For example, Br would be a higher priority than Cl, because Br has a larger atomic number.)

Then, you put the fourth priority substituent on the top, and draw a curve from the first- to the second- to the third-priority substituent. If the curve goes clockwise, the configuration is R; if the curve goes counterclockwise, the configuration is S. To get the number-four priority substituent at the top of the Fischer projection, you have to use one of the two allowed moves diagramed in the second figure. (You can make a 180-degree rotation, or you can hold one substituent fixed and rotate the other three.) Two examples of the determination of the configuration from Fischer projections are shown here.

Determining R and S configurations from Fischer projections.
Drawing enantiomers: Recall that enantiomers are molecules that are nonsuperposable mirror images. Because these are mirror image molecules, and because stereoisomers can only have two absolute configurations, all the stereocenters of one enantiomer will be the mirror image of the other enantiomer. A mirror image stereocenter can be drawn by swapping two groups attached to the stereocenter, or by swapping the positions of the solid and broken wedges on the stereocenter. We can also draw the mirror image of the enantiomer using an imaginary mirror.
Example: Draw the enantiomer of (R)-2-chlorobutane. (R)-2-chlorobutane Solution: Let's switch the places of the methyl and ethyl groups. This gives us (S)-2- chlorobutane. (Check this by determining the stereocenter’s absolute configuration.) (R)-2-chlorobutane (S)-2-chlorobutane Swapping the wedges achieves the same result. Verify this by using models to compare the two representations of (S)-2-chlorobutane shown below. (R)-2-chlorobutane (S)-2-chlorobutane
Drawing Diastereomers Recall that stereoisomers differ in the position of the atoms in space, and that diastereomers are stereoisomers that are not enantiomers. We can draw the diastereomer(s) of a given structure by inverting one or more, but not all, of its stereocenters.
Example: Draw all the diastereomers of (2R,3R)-tartaric acid.
Solution: How many diastereomers can there be? Recall that for a molecule with n stereocenters, the molecule can have 2n stereoisomers. Thus, (2R,3R)-tartaric acid, which has two stereocenters, can have at most 22 = 4 stereoisomers. The (2S,3S) stereoisomer cannot be a diastereomer of (2R,3R) because both stereocenters have been inverted. This leaves two possibilities, each of which has one stereocenter the same as the given molecule: (2R,3S) and (2S,3R). The structures are drawn by inverting the stereocenter that is altered.In this case, the new diastereomers are meso compounds, and are identical. This happens when the two stereocenters have the same attachments.