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Winthrop University
Department of Chemistry
Organic Chemistry Lab
CHEM 303
NOTE: For this exercise, the student only needs to turn in the answers to the questions included in
each part. No formal pre-lab or in-lab is necessary.
Enantiomeric Resolution Exercise
Required reading:
Bruice, 5th ed:
Sections 5.4 – 5.12, pp. 204 – 224.
Pavia, Lampman, Kriz, and Engel, 4th ed:
“Resolution of (±)--Phenylethylamine and Determination of Optical Purity” (Resolution of
Enantiomers section), pp. 331 – 332.
“Polarimetry”, pp. 818 – 827.
Lab Report
Address the questions included in each part, and show all calculations.
Introduction
Most common organic syntheses start from achiral materials. If, in a reaction, a chiral product is
generated from an achiral starting material or intermediate, a 50:50 mixture of enantiomers (a
racemic mixture) results. For example, addition of HBr to trans-2-butene (an achiral compound)
produces racemic 2-bromobutane:
CH3
CH3
+
H3C
HBr
H3C
Br
50% (R)-2-bromobutane
+
50% (S)-2-bromobutane
trans-2-butene (achiral)
Often, it is desirable to separate (resolve) a racemic mixture into its individual enantiomers. For
example, one enantiomer of a particular drug may have activity against a certain disorder, while the
other enantiomer may be less active or even harmful. A drug company may therefore need to make
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Winthrop University
Department of Chemistry
Organic Chemistry Lab
CHEM 303
the drug as a racemic mixture, then separate the individual enantiomers so only the active enantiomer
will be packaged and sold. This separation of a racemic mixture into its individual enantiomers is
known as resolution.
The resolution of a racemic mixture cannot be accomplished using standard physical means (e. g.
distillation, recrystallization, or chromatography), because enantiomers have identical physical
properties, except for the direction of rotation of plane polarized light. To distinguish between the
components of a racemic mixture, a chiral probe (something capable of distinguishing between
enantiomers) must be used. When a chiral probe is used to resolve a racemic mixture into its
enantiomers, it is called a resolving agent.
Traditionally, the method for resolving a racemic mixture is to react the racemate with an
enantiomerically pure compound (often a readily available natural product). When the enantiomers
that make up the racemate form bonds with the enantiomerically pure resolving agent, a mixture of
diastereomers is formed. Unlike enantiomers, diastereomers have different physical properties
(boiling point, solubility, etc.) and can be separated by typical physical methods. Once the
diastereomers are separated, they are subjected to conditions under which the reverse reaction occurs,
and the original enantiomers are recovered. The process can be described schematically as follows:
(R,S)
(Racemic)
+
(S’)
(Resolving
Agent)
(R,S’)
+
(S,S’)
Mixture of
Diastereomers
Separation
Recovery
(pure R)
Enantiomers are
“Resolved”
(R,S’)
(S,S’)
(- S’)
(pure S)
To illustrate the steps involved in an enantiomeric resolution, consider the resolution of a racemic
mixture of 2-aminobutane using (R,R)-(+)-tartaric acid (available in large quantities as a by product
of winemaking). 2-Aminobutane reacts with (R,R)-tartaric acid in hot water solution to form a
mixture of diasteromeric amine salts:
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Winthrop University
Department of Chemistry
Organic Chemistry Lab
CHEM 303
NH2
H3C

CO2H
H
CH3
+
H
OH
HO
+
NH3
H3C
H
H
+
H3N
OH
CH3 HO
H
+
CO2
H
H
CH3 HO
H3C
CO2H
CO2H
Racemic 2-aminobutane

CO2
(R,R)-(+)-tartaric acid
OH
H
CO2H
(R)-amine-(R,R)-tartrate
(S)-amine-(R,R)-tartrate
The diastereomeric salts have differing solubilities in water, so upon cooling the solution to room
temperature, the less soluble (S,R,R) salt crystallizes out and is isolated by filtration, leaving the more
soluble (R,R,R) salt in solution. The (R,R,R) salt is then isolated by evaporation of the solvent:
H
H3C
+
NH3
CO2
H

+
H3N
OH
CH3 HO
H
+
CO2
H
H
CH3 HO
H3C
CO2H
OH
H
CO2H
(R)-amine-(R,R)-tartrate
(S)-amine-(R,R)-tartrate
Crystallization/
Filtration

Evaporation of
mother liquor
(isolation of
less soluble
diastereomer)
+
H3N
H3C

H
CH3
CO2
H
OH
HO
H
CO2H
CO2
H
CH3
H
H3C
+
NH3
(R)-amine-(R,R)-tartrate
OH
HO
H
CO2H
(S)-amine-(R,R)-tartrate
The individual enantiomers are then recovered by treatment of the individual salts with aqueous
base:
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Winthrop University
Department of Chemistry
Organic Chemistry Lab
CHEM 303
+
H3N

CO2
H
H3C
CH3
H
OH
HO
H2N
KOH/H2O
H
H3C
CO2H
(S)-amine-(R,R)-tartrate
H
CH3
CH3
(S)-2-aminobutane

+
CO2
NH3
H3C
H
H
OH
HO
H
KOH/H2O
H
H3C
CO2H
NH2
CH3
(R)-2-aminobutane
(R)-amine-(R,R)-tartrate
The optical purity of each product can then be determined by polarimetry as described in Pavia,
Lampman, Kriz, and Engel, 4th ed. pp. 819 – 827 to evaluate the efficiency of the resolution.
Part 1
For the following problem, the following information will be helpful:
NH2
O
NH2
HO
H
CH3
CH3
H
OH
HO
O
(R)-(+)--Phenylethylamine
o
[]22
D = +40.3
mw = 121.18
+
H
CH3
(S)-(-)-Malic acid
[]D = -2.3o
mw = 134.09
(S)-(-)--Phenylethylamine
o
[]22
D = -40.3
mw = 121.18
+
O
NH3
H
O
NH3
HO
HO
O
O
HO
CH3
H
-
H
O
O
(S, S)--Phenylethylammonium malate
Solubility in water (25 °C) > 0.3 g/mL
HO
-
H
(R, S)--Phenylethylammonium malate
Solubility in water (25 °C) < 0.1 g/mL
(S)-(-)-Malic acid (also known as apple sugar) is used as a coating in many “extreme sour” candies
like MegaWarheads to give the initial “extreme sour” taste to the product. It is an abundant
chirally pure material, and therefore can be used as an inexpensive resolving agent for the
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Winthrop University
Department of Chemistry
Organic Chemistry Lab
CHEM 303
separation of enantiomeric amines. As the new chemist working at “Drugs Я Us”, you are given
the task of resolving racemic -phenylethylamine into its individual enantiomers.
1. Draw a flowchart (include structures) illustrating how you will resolve racemic
-phenylethylamine into (R)-(+)--phenylethylamine and (S)-(-)--phenylethylamine using
(S)-(-)-malic acid as the resolving agent.
2. After you carried out your resolution, you labeled the products you isolated “Sample A” and
“Sample B”. You subjected both samples to polarimetry ( = 589 nm (sodium D-line), 1-dm cell)
and obtained the following results:
Sample A: 1.00 g dissolved in methanol and diluted to 10.0 mL gave an observed rotation ()
at 22 °C of + 4.00°.
Sample B: 1.00 g dissolved in methanol and diluted to 10.0 mL gave an observed rotation ()
at 22 °C of – 2.42°.
For samples A and B, identify the predominant enantiomer and calculate the specific rotation
for each sample. Also, calculate the optical purity (enantiomeric excess) of each sample
as well as the % of each enantiomer present in each sample. Show your calculations.
Part 2
You were so successful at resolving racemic -phenylethylamine, your boss has given you another
task. She wants to know if you can suggest ways to resolve the following racemic compounds:
CH3
OH
(a)
N
N
(b)
O
CH3
H3C
nicotine
CH3
ibuprofen
Describe how you might resolve each of these racemic compounds. (Advice from the boss: a
flowchart (including structures) is a very good way to show the steps involved in this operation.)
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