Naming Enantiomers: The
R,S System of Nomenclature
1. Rank groups by atomic number of the
atom bonded to the chirality center.
Use the same system that was used
for the E and Z isomers of alkenes
© Prentice Hall 2001
Chapter 4
1
Naming Enantiomers: The R,S
System of Nomenclature
2. Orient the molecule
so that the group
(or atom) of lowest
priority is directed
away from you
Draw a curve from
group of highest
priority to the group
of second priority
© Prentice Hall 2001
Chapter 4
2
Naming Enantiomers: The R,S
System of Nomenclature
A similar set of rules exists for use with Fischer
Projections
1. If the group of lowest priority is on a vertical
line, just draw the curve from the group of
highest priority to the group of second highest
priority. If clockwise, configuration is R. If
counterclockwise, configuration is S.

(S)-2-bromobutane
© Prentice Hall 2001
Chapter 4
3
Naming Enantiomers: The R,S
System of Nomenclature
2. If the group of lowest priority is on a horizontal
line, follow step 1, but reverse the R & S.
Rule of thumb: If lowest priority is on a Vertical
bond, configuration is Very true. If lowest
priority is on a Horizontal bond, configuration
is Horribly wrong.
(R)-2-bromobutane
© Prentice Hall 2001
Chapter 4
4
Naming Enantiomers: The R,S
System of Nomenclature
3. In drawing the curve from the group of
highest priority to the group of second
highest priority, it is ok to go past the group
of lowest priority, but never past the group
of third priority.
© Prentice Hall 2001
Chapter 4
5
Naming Enantiomers: The
R,S System of Nomenclature


In working with a Fischer
Projection, a rotation of (R)-2-bromobutane (S)-2-bromobutane
90° or a horizontal flip
will switch the chirality
Rotation of a Fischer
projection by 180° in the
plane of the paper does (R)-2-bromobutane (S)-2-bromobutane
not switch the chirality
(R)-2-bromobutane (R)-2-bromobutane
© Prentice Hall 2001
Chapter 4
6
Drill on Problem 8
Indicate whether the following
structure has R or S configuration
2
4
CH(CH3)2
CH(CH3)2
C
C
CH3
Switch CH3
and
CH2CH3
CH3CH2
CH3CH2
3
groups
CH Br
1
© Prentice Hall 2001
CH3
CH2Br
2
R
Chapter 4
7
Drill on Problem 8
Indicate whether the following
structure has R or S configuration
4
H
Cl
CH 3 3
H3CH 2CH 2C
2
Cl
1
© Prentice Hall 2001
Chapter 4
R
8
Optical Rotation
  =
Τ
λ

lxc
T is the temp in
°C
 is the
wavelength
 is the measured rotation in
degrees
l is the path length in decimeters
c is the concentration in grams
per mL
© Prentice Hall 2001
[ ]T = specific rotation
Chapter 4
9
Optical Rotation and Absolute
Configuration
In general there is no relationship between the
R or S configuration of an enantiomer and the
direction it rotates polarized light
This must be determined by experiment
CH3
CH3
H
OH
H
-
COOH
OH
(S)-(+)-lactic acid
+
COO Na
(S)-(–)-sodium lactate
A mixture of equal amounts of two enantiomers
is called a racemic mixture
© Prentice Hall 2001
Chapter 4
10
Optical Purity
o bse rv e d sp e c ific ro t at io n
o p t ic al p u rit y =
sp e c ific ro t at io n o f t h e p ure e n an t io m e r
For example, if a sample of 2-bromobutane
has an observed specific rotation of +9.2°,
and the specific rotation of pure (S)-(+)
enantiomer is +23.1°, then
 9.2o
optical purity 
 0.40  40%
o
 23.1
© Prentice Hall 2001
Chapter 4
11
Compounds with More Than
One Chirality Center
© Prentice Hall 2001
Chapter 4
12
Meso Compounds
© Prentice Hall 2001
Chapter 4
13
Drill on Meso Compounds
Problem 27

Does the following compound have a
stereoisomer that is a meso compound?
Br
Br
No possibility of a plane of symmetry
No meso form
© Prentice Hall 2001
Chapter 4
14
Drill on Meso Compounds
Problem 27

Does the following compound have a
stereoisomer that is a meso compound?
Br
Br
Plane of symmetry
Meso form possible
© Prentice Hall 2001
Chapter 4
15