312 112 Basic Organic Chemistry
Instructor: Chanokbhorn Phaosiri
e-mail: chapha@kku.ac.th
Office Hours: MW 9.00-10.00 AM
Office: SC1401
Topics: Alkynes, Aromatic Compounds, Stereochemistry
- Introduction
- Physical Properties
- Nomenclature - Preparation and Reactions
References :
1). McMurry, M. Organic Chemistry, 3rd ed. Brooks/Cole
Publising, Califonia, 1992.
2). Morrison, R. T.; Boyd, R. N. Organic Chemistry, 6th ed.
Prentice-Hall Inc., New Jersey, 1992.
3). Vollhardt, K. P. C.; Schore, N. E. Organic Chemistry,
3rd ed. W. H. Freeman Co., New York, 1998.
Online References :
1). http://academic.csuohio.edu/hehemannd/
(Dr. David G. Hehemann, Cleaveland State University)
2). http://www.tamug.tamu.edu/mars/chem227/aromatics.htm
(Dr. Melanie Lesko, Texas A&M University at Galveston)
3). http://www.uncwil.edu/chem/chm212martin/
(Dr. Martin, University of North Carolina at Wilmington)
4). http://www-personal.une.edu.au/~sglover/CHEM204_Ch7/
(Dr.Steve Glover, The University of New England)
Alkynes
O
H3CC C C C C
CH3
HC C C CH2CH3
OH
Capillin
3-Methyl-1-pentyn-3-ol
(Active against skin fungi)
(Hypnotic)
CH3 OH
C CH
H
H
H
HO
NH2
CH2 CH C CH
17-Ethynylestradiol
An amphetamine analog
(A birth control agent)
(Active in the central nervous system)
Structure of Alkynes
1. The functional group is a triple bond.
a. The triple bond is composed of two π bonds and
a σ bond.
i.
ii.
The π bonds are oriented perpendicular to each
other.
The two π bonds are electron-rich, they
undergo electrophilic addition like alkenes.
b. The general formula is: CnH2n-2
R
C
C
R
R = H, alkyl
Acetylene
(Ethyne)
H
C
C
H
1.20 Å
* The strength of the carbon-carbon triple bond is about 837 kJ/mol.
* It is the strongest and the shortest known carbon-carbon bond.
Nomenclature: The alkynes are named according to 2 systems
1. They are considered to be derived from acetylene by replacement
of one or both hydrogen atoms by alkyl groups.
H
C
C
C2H5
H3C
Ethylacetylene
H3C
C
C
CH3
Dimethylacetylene
C
C
CH(CH3)2
Isopropylmethylacetylene
2. For more complicated alkynes, the IUPAC names are used.
IUPAC names for Alkynes
1. The rule are exactly the same as for the naming of alkenes, except
that the ending -yne replaces -ene.
2. The parent structure is the longest continuous chain that contains
the triple bond.
3. Numbering provides the lowest number for the triple bond.
5
1
H3C
2
3
C
C
4
CH3
CH
4-Methyl-2-pentyne
CH3
CH2
4-Bromo-2-hexyne
CH3
H3C C CH2 C
CHCH2CH2
CH
4-Penten-1-yne
OH
CH
CH3
4,4-Dimethyl-2-pentyne
H3C C CH2 C
CH
CH3
4-Methyl-4-pentyn-2-ol
* Triple bonds have priority over double bonds.
* An -OH (hydroxyl group) has priority over the triple bond.
Physical Properties of Alkynes
1. The alkynes have physical properties that are essentially the same
as those of the alkanes and alkenes.
2. They are insoluble in water but quite soluble in the usual organic
solvents of low polarity.
3. They are less dense than water.
4. Their boiling points show the usual increase with the increasing
carbon number.
Preparation of Alkynes
Dehydrohalogenation of alkyl dihalides.
 Treatment of a dihalide with strong base, leads to
elimination of HX (X = Cl, Br).
 Just like alkene synthesis.
 The reaction can take place twice with a dihalide to
form an alkyne.
Cl
H
C
Cl
C
H
KOH
C
C
H3CHC CH2
Br2
H3CHC CH2
Br Br
KOH
H3CC CH
NaNH2
- KBr
- H2O
H3CHC CH + KBr
Br + H2O
Reaction of Alkynes
1. Addition Reactions
RC CR'
Y
YZ
YZ
RC CR'
Y
Z
RC CR'
Z
Y
Z
1.1 Addition of Hydrogen (Reduction to Alkenes)
Na, NH3(liq)
H
R'
Anti
C C
R
H
R
R'
RC CR'
H2
Lindlar catalyst
C C
H
H
Syn
 The trans alkene can be obtained by using Li or Na /
liquid NH3 as the reducing agent.
1. When lithium or sodium are dissolved in liquid
ammonia an intensely blue solution results.
2. These solutions are composed of the metal cations and
dissolved electrons which produce the blue color.
3. When an alkyne is added to this solution, an
electron adds to one of the π bonds to produce a
radical anion.
R
C
C
R
e
(Li)
R
C
C
R + Li+
4. The radical anion abstract a hydrogen from the solvent,
ammonia leading to a radical.
R
R
C
C
R
H
NH2
H
C
+ NH2-
C
R
5. The radical then reacts with a second dissolved electron to
form an anion which again abstracts a hydrogen to give the
final product.
R
H
C
R
e
C
C
R
R
R
H
C
R
C
R
H
C
H
NH2
H
C
H
C
R
* The cis alkene can be obtained by using Lindlar’s catalyst.
* Lindlar’s catalyst is a form of Palladium (Pd) that has been
deactivated by treatment with leadacetate and quinoline.
HH
+ H2
Surface of metal catalyst
* When a Platinum (Pt) catalyst is used, the alkyne generally
react with two molar equivalents of hydrogen to give an alkane.
1.2 Halogenation
X
X2
C C
X2
C C
(X = Cl, Br)
X
X
X
C C
X
X
Mechanism
C C
X2
(X = Cl, Br)
X
X
C
C C
C
X
X
1.3 Hydrohalogenation
C C
HX
(X = Cl, Br, I)
H X
HX
C C
C C
H X
H X
Mechanism
C C
H X
H
C C
H
X
Alkenyl Carbocation
C C
X
H X
H
C C
X
H X
C C
H
X
C C
H X
X
Carbocation
 The second carbocation, with the halogen atom attached,
is stabilized by the lone electrons on halogen atom.
 The charge on the alkenyl carbocation is centered in an
sp2 orbital, this is relatively high energy.
These two factors lead to the fact that the second
hydrogen attack occurs faster than the initial attack.
H C C H + HCl
HgCl2
H2C CHCl
Vinylchloride
(Chloroethylene)
H3CH2CC CCH2CH3 + HCl
NH4Cl
CH3COOH
Cl
C C
H3CH2C
CH2CH3
H
(Z)-3-Chloro-3-hexene
H3CH2CC C H + 2Br2
CCl4
CH3CH2CBr2CHBr2
1,1,2,2-Tetrabromobutane
1.4 Hydration
* Terminal alkynes
H
C
C
R
HgSO4
H2SO4
H
H
O
C
C
R
H
H2O
* Other alkynes
CH3
C
C
R
HgSO4
H2SO4
CH3
R
C
C
R
H2SO4
H2O
O
C
C
H
H2O
HgSO4
H
R
H
O
C
C
H
R
CH3
O
H
C
C
H
R
R
* Markovnikov regiochemistry is found for the hydration with the
OH group adding to the more highly substituted carbon and the H
attaching to the less highly substituted carbon.
* The intermediate in the reaction, a vinyl alcohol, then rearranges to
a ketone in a process called tautomerism.
H
O
C C
Enol tautomer
(less favored)
O
H
C C
Keto tautomer
(more favored)
Mechanism
H
H
R C C H
OH2
R C C
H
O
H
C C
Hg+SO42-
Hg+SO42-
R
-H+
H
O
H
C C H
R
H
H
O
H
C C
R
H
H3O+
O
H
C C
R
Hg+SO42-
1.5 Hydroboration/ Oxidation
CH3
C
C
H
1) BH3, THF
2) H2O2, OH
CH3
H
O
C
C
H
H
* Hydroboration of symmetrically substituted internal alkynes gives
vinylic boranes.
* The vinylic boranes are oxidized by basic hydrogen peroxide to
yield ketones (via enols)
* The reaction occurs in anti-Markovnikov fashion.
2. Reactions as Acids
2R
C C H + 2 Na
2 R C C Na
+
H2
Sodium acetylide
R C C H +
R C C Li
LiNH2
+
Lithium acetylide
R C C H
+
Ag
+
R C C Ag
Precipitate
+ HNO3
R C C H
+
+
Ag
HNH2
Table 1. Acidity of Simple Hydrocarbons
Type
Example
Ka
pKa
Alkyne
HC CH
10-25
25
Alkene
H2C CH2
10-44
44
Alkane
H3C CH3
10-60
60
Stronger acid
Weaker acid
Reactions of metal acetylides with primary alkyl halides
* Lithium or sodium acetylides can react with primary alkyl halides
* The alkyl group becomes attached to the triply bonded carbon,
and a new, larger alkynes has been generated.
RC C Li
HC C Li
R'X
CH3CH2Br
RC C R'
+ LiBr
HC C CH2CH3 + LiBr
* Acetykide ion alkylation is limited to the use of primary alkyl
bromides and iodides.
* Acetylides ions are sufficiently strong bases to cause
dehydrohalogenation instead of substitution when they react with
seconday and tertiary alkyl halides
H
+ CH3C CH + BrH
H
Br
+
H
CH3C CLi
H
C
C
CH3
3. Oxidative cleavage
An internal alkyne
R C C R'
KMnO4 or O3
O
R C OH +
O
R'
C OH
A terminal alkyne
R C C H
KMnO4 or O3
O
R C OH +
O
C O
Organic Synthesis
* Prepare octane from 1 pentyne
1). NaNH2, NH3
2). BrCH2CH2CH3, THF
H2/Pt in Ethanol
* Prepare cis-2-hexene from 1 pentyne
1). NaNH2, NH3
2). CH3Br, THF
H2/Pd in Ethanol
Aromatic Compounds
The original meanings of Aliphatic = Fatty, Aromatic = Fragrant
O
CH 3
H
Benzene
Benzaldehyde
From coal distillate
From cherries,
peaches, and almonds
Toluene
From Tolu balsam
“Benzene and compounds that resemble benzene in chemical behavior”
!! Benzene has been found to cause bone-marrow depression and consequent leukopenia.
CH3 O
O
H
OCH2 C N
S
H
N
H
O
H
HO
Estrone
CO2H
Penicillin V
(Female Steroidal Hormone)
H3CO
CH3
H3C
H3C
O
3
HO
CH3
Vitamin E
O
N CH3
H3CO
Morphine
(Analgesic)
OH
COOH
O
CH3
H3C
O
O
2-Acetyloxybenzoic acid
(Aspirin, Bayer Aspirin)
2-[4-(2-Methylrpopyl)phenyl]propanoic acid
(Ibuprofen)
OH
N-(4-Hydroxyphenyl)acetamide
(Acetaminophen, Tylenol)
HN
C
O
CH3
The Painkillers
Nomenclature
* Monosubstituted Benzenes (IUPAC)
 Monosubstituted aromatics are named using -benzene as the parent name.
 Alkyl substituted benzenes are named according to the chain length of the
alkyl group.
 If the alkylsubstituent has six carbons or fewer, it will be named as
alkylbenzene.
F
Fluorobenzene
NO2
Nitrobenzene
CH2CH3
Ethylbenzene
* Monosubstituted Benzenes (Some Common Names)
CH3
Toluene
CH(CH3)2
Cumene
OH
Phenol
COCH3
Acetophenone
NH2
Aniline
HC
Styrene
CH2
* If the alkyl substituent is larger than the ring (more than six
carbons), the compound is named as a phenyl-substituted alkane.
CH3
CHCH2CH2CH2CH2CH3
A phenyl group
CH2
2-Phenylheptane
CH2
Br
A benzyl group
Benzylbromide
* Disubstituted Benzenes
 Disubstituted benzenes can be named one of two ways.
 Each method describes the relationship of the two groups on
the six membered aromatic ring.
 Using the prefixes ortho-, meta-, or para-.
 Systematic numbering of the aromatic ring.
 The common names for monosubstituted benzenes can be
used as parent names for disubstituted aromatics.
X
ortho
ortho
meta
meta
3
2
Cl
2
Cl
ortho-Dichlorobenzene
1,2-Dichlorobenzene
para
O2N
1
1
NO2
1
I
4
3
I
2
meta-Dinitrobenzene
para-Diiodobenzene
1,3-Dinitrobenzene
1,4-Diiodobenzene
1 CH3
2
H3C
3
2
1
CH3
CH3
2
1
ortho-Iodoaniline
para-Cresol
(p-Cresol)
COOH
NH2
I
4 CH3
HO
meta-Xylene
(m-Xylene)
ortho-Xylene
(o-Xylene)
3
O2N
meta-Nitrobenzoic acid
CH3
Br
para-Bromotoluene
* Polysubstituted Benzenes
 Polysubstituted benzenes are named by numbering the
position of each substituent on the ring.
 The numbering is carried out to give the substituents
the lowest possible numbers.
 A substituted carbon is always C#1.
 For rings with common names, the carbon bearing the
substituent responsible for the common name is C #1.
Br
4
3
2
1
CH3
1
CH3
O2N 4 3 2
4-Bromo-1,2-dimethylbenzene
Br
4
3
NO2
4-Bromo-2-ethyl-1-nitrobenzene
1-Bromo-3-ethyl-4-nitrobenzene
(Wrong ???)
Cl
2-Chloro-1,4-dinitrobenzene
NO2
1
2 CH2CH3
1
NO2
Cl
5
4
2
3 Br
3-Bromo-5-chloronitrobenzene
O2N 6
5
CH3
1
4
O2N
2
3
NO2
Br 6
5
OH
1
4
2
3
Br
2,6-Dibromophenol
2,4,6-Trinitrotoluene
(TNT)
NH2
1
Br 6
5
2
3
4
Br
Br
2,4,6-Trbromoaniline
* Polynuclear Aromatic Hydrocarbons
Anthracene
OH
Napthalene


7
6


8
1
5
4 3

2

2-Napthol (-Naphthol)
SO3H


H2N
6-Amino-2-napthalenesulfonic acid
Stability of Benzene
 C6H6 indicates that several double bonds must be present and
should react the same way as alkenes.
KMnO4/ H2O
COOH
COOH
H3O+
OH
Cyclohexene
HCl
Ether
Cl
 Benzene shows none of the behavior characteristic of alkenes.
 Benzene does not undergo electrophilic addition reactions.
KMnO 4/ H 2O
H 3O +
No Oxidation
No Hydration
Benzene
HCl
Ether
No Addition
Table 1. Heats of Hydrogenation of Cyclic Alkenes
Reactant
Cyclohexene
1,3-Cyclohexadiene
Benzene
Product
Cyclohexane
Cyclohexane
Cyclohexane
H2/Metal Catalyst
H0hydrog (kJ/mol)
120
232
208
1.39 Å
1.10 Å
H 1200
C-C Bonds are 1.54 Å
C=C Bonds are 1.34 Å
H
1200
1200
H
H
H
H
1.5 Bonds on average
Representation of Benzene : The Resonance Approach
1.5 Bonds on average
H
H
H
H
H
H
H
H
H
H
H
H
Napthalene is Aromatic
Aromaticity and The Huckel 4n + 2 Rule
Criteria for simple aromatics
 Follow Huckel's rule, having 4n+2 electrons in the
delocalized cloud.
 Are able to be planar and are cyclic.
 Every atom in the circle is able to participate in
delocalizing the electrons by having a p orbital or an
unshared pair of electrons.
n=0, 2 electrons:
cyclopropenyl cation
n=1, 6 electrons:
n=2, 10 electrons:
Compounds that are "anti-aromatic"
Chemistry of Benzene
 Benzene undergoes substitution reactions with electrophile,
in contract with the corresponding reaction of alkenes
 Proceeding by addition of the electrophile and then by proton loss
to regenerate the aromatic ring.
 Many different substituents can be introduced onto the aromatic
ring :
Halogenation : X = -F, -Cl, -Br, -I
Nitration : -NO2 / Sulfonation : -SO3H
Alkylation : -R (alkyl group)
Acylation : -COR (acyl group)
Mechanism of Electrophilic Aromatic Substitution
H
H
H
H
H
E
H
H
E+X-
H
H
H
H
H
H
H
E
H+
H
H
H
?
1). Halogenation of Benzene : The need for a catalyst
Benzene is unreactive in the presence of halogens,
because halogens are not electrophilic enough to disrupt
the aromaticity.
Br
Br
FeBr3
A weak electrophile
Br
Br
FeBr3
Br Br
FeBr3
A strong electrophile
 Activation of Bromine by the Lewis Acid; FeCl3.
H
Br
Br
FeBr3
Br
H
FeBr3Br
 Electrophilic attack on Benzene by Activation Bromine
Br
H
Br
FeBr3
HBr
Br
 Bromobenzene Formation.
FeBr3
Chlorination
Cl2
FeCl3
HCl
Cl
H
Iododination
I2
H
CuCl2
HI
I
2). Nitration of Benzene : The attack by nitronium ion
HO N
O
H OSO3H
O
Nitric Acid
Sulfuric Acid
O
HO N
O
H
O
HO N
O
H
O N O
H
HOH
OSO3H
O N O
Nitronium Ion
NO2
OSO3H
H
NO2
3). Sulfonation of Benzene
2H2SO4
H
O
O
S
SO3 +
O
H 3O
+ HSO4
O
S O
O
H
SO3H
Sulfonation is reversible
H2O, cat. H2SO4, 100 oC
SO3H
H
HOSO3H
Benzenesulfonic acids have important use.
• Aromatic Detergent Synthesis
• Sulfa Drugs
N
R
SO3Na
SO2NH
H2N
N
Detergent
Sulfadiazine
(Antimalarial drug)
N
H2N
O
SO2NH
Sulfathoxazole
(Antibacterial agent)
CH3
4). Friedel-Craft Alkylation
* Haloalkane reacts with benzene in the presence of aluminium halide
RH2C X
RH2C
AlX3
AlX3X
XAlX3
H
H
AlX3
H
RCH2XAlX3
CH2R
X
CH2R
AlX3
CH2R
HX
* Limitations of Friedel-Craft Alkylation
X
X
An aryl halide
(ArX)
Y
Y = -NO2, -CN, -SO3H,
-COCH3,-COOH,
-COOCH3, -NR3
H3C
CH3
H3C
C
CH3
An vinyl halide
CH2Cl
AlCl3
CH3
C
CH2CH3
AlCl3
(65%)
Cl
(35%)
H
CH3CH2CHCH2
(Primary (1°) Carbocation)
H
CH3CH2CHCH2
(Secondary (2°) Carbocation)
5). Friedel-Craft Acylation
O
O
R
C
Cl
O
R
C
C
AlCl3
O
Cl
AlCl3
Acylium ion
R
RC
O
C
R
AlCl3
HCl
O
Cl
R
RC
O
C
Cl AlHCl3
ClAlCl3
To practice your brain…..
CH3
HNO3/H2SO4
?
CH3
Cl2/ FeCl3
?
OH
Bromination
?
Reactivity of Aromatic Rings
Substituents already presenton the ring have two effects.
 Reactivity : some make the ring more reactive than
benzene, and some make it less reactive.
 Orientations : The nature of the substituent
determines the position of the second substituent;
ortho-, meta-, or para-.
The powerful activating group will lead to
a trisubstituted product in nearly quantitative yield.
OH
OH
Br2 (xs)
Br
Br
( No catalyst needed)
H 2O
NH2
Br2 (xs)
Br
Br
NH2
( No catalyst needed)
H2O
(xs = excess)
Br
Br
CH3
NO2
(59%)
CH3
CH3
HNO3
H2SO4
(37%)
NO2
CH3
(4%)
NO2
F
Cl
Br
I
HNO3/H2SO4
NO2
2
3
Examples of Electrophilic Substitution Reactions of Benzenes
Aromatic-Aliphatic Compounds
 Halogenation
CH3
CH2Cl
Cl2
 or h
CH2CH3
CHClCH3
CH2CH2Cl
Cl2
 or h
Major Product
Minor Product
 Oxidation
CH3
HC CHCH
3
C CCH3
1) KMnO4, OH-, 
2) H3O+
COOH
* The sense of the directing power of substituents
can be changed.
NO2
NH2
Zn(Hg), HCl, or H2, Ni, or
Fe, HCl
O
Nitro
(Meta Director)
F 3C
C
OOH
Amino
(Ortho, Para Director)
O
R
CH2R
Zn(Hg), HCl, or
H2/Pd in Ethanol
CrO3, H2SO4, H2O
Alkanoyl
(Meta Director)
Alkyl
(Ortho, Para Director)
Stereochemistry
is the study of the three-dimensional structure of molecules.
 Stereoisomers
 Stereoselective and Stereospecific Reactions
 Enantiotopic and Diastereotopic Ligands and Faces
Stereoisomers
Isomers : Different compounds that have the same
molecular formula.
Stereoisomers : Isomers that have the same bonding sequence
but differ in the orientation of their atoms in space.
(http://www.paccd.cc.ca.us/instadmn/physcidv/psganapa/chem8a/STEREOCHEMISTRY.htm)
Conformational Isomers
 Different spatial orientations of the atoms of a molecule
that result from rotations or twisting about single bonds.
CH3CH2CH2CH3
H
H H
H
H
H
H
H
H
H3C
n-Butane
H H
180°
H
H
H
H
H
H
H
CH3
H H
HH H
H3C
H
H
HH H
CH3
H
H H
H
H
H
H
H
H H
180°
H
HH H
H
H
H
H
H
H
From 3D to 2D structures:
H H H H
H
H H H H
In/Out of A Plane
H
H
H
H
H
CH3
CH3
Newman Projection
H
H
CH3
H
H
H
H
CH3
Fisher Projection
CH3
H
CH3
H
Sawhorse Formula
How many conformers can n-butane have ?
Staggered (Anti)
conformation
Eclipsed
conformation
Staggered
(Gauche)
conformation
Eclipsed
conformation
A
B
C
D
Dr. William Reusch (http://www.cem.msu.edu/~reusch/VirtualText/sterisom.htm#isom4)
Ring Conformations
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
Boat Conformation
Axial
H
H
H
H
H
Cyclohexane
H
H
H
H
H
H
H
H
H
H
H
H H
H
Twist Boat Conformation
H
H
H
H
H
H
HH
H
H
Equatorial
Chair Conformation
Important aspects of conformational stereoisomerism
 Most conformational interconversions in simple molecules occur
rapidly at room temperature. Isolation is impossible.
 Staggered conformations about carbon-carbon single bonds are
more stable than the corresponding eclipsed conformations. The
higher energy of eclipsed bonds is known as eclipsing strain.
 In butane the gauche-conformer is less stable than the anticonformer by about 0.9 kcal/mol. This is due to steric hindrance.
Configurational Isomers : 1) Geometrical Isomers
H3C
CH2CH3
C
H
H3C
C
H
C
H
cis-2-pentene
H
C
CH2CH3
trans-2-pentene
Alkenes may exist as a pair of configurational
stereoisomers, often designated cis (on this side) and trans
(across).
 Each carbon of the double bond must have two
different substituent groups (one may be hydrogen).
Sometimes it is very hard to identify cis- or trans- alkenes
Cl
CH2CH3
C
H
H3C
C
C
Br
H
C
CH2CH3
 A competely unambiguous system, based on a set of
group priority rules, assigns a Z (German, zusammen for
together) or E (German, entgegen for opposite) to designate
the stereoisomers.
 Z is equivalent to cis and E is equivalent to trans.
Assign priorities to double bond substituents by
looking at the atoms attached directly to the double
bond carbons.
 The higher the atomic number of the immediate
substituent atom, the higher the priority.
For example, H– < C– < N– < O– < Cl–. (priority
increases left to right)
(Different isotopes of the same element are assigned a
priority according to their atomic mass.)
 If two substituents have the same immediate substituent
atom, move to the next atom (away from the double bond)
until a difference is found.
For example, CH3– < C2H5– < ClCH2– < BrCH2– < CH3O–
 Once the relative priorities of the two substituents on
each of the double bond carbons has been determined, a cis
orientation of the higher priority pair is designated Z, and a
trans orientation is termed E.
Cl
CH2CH3
C
H
H3C
C
C
Br
E isomer
H
C
CH2CH3
Z isomer
Configurational Isomers : 2) Optical Isomers
2.1) Enantiomers: non-superimposable (different) mirror
images; most of chemical and physical properties are
identical.
2.2) Diastereomers: are stereoisomers that are not mirror
images (all stereoisomers except enantiomers) and have
different chemical and physical properties.
Mirror
H
(-)-limonene
(lemon oil)
H
(+)-limonene
(oranges)
H
180° rotation
 Late 50’s, thalidomide was prescribed as an analgesic for
morning sickness and used extensively in Europe and Canada
despite strong warning that it not be used by pregnant women.
 By 1961, it was recognized as the cause for numerous birth
defects (~7-10,000 in 28 countries).
 Although two enantiomers have identical boiling points and
melting points, they rotate the plane of polarized light in opposite
directions. A polarimeter is used to measure the optical rotations of
enantiomers.
(http://www.cem.msu.edu/~reusch/VirtualText/sterism2.htm#isom12)
 Each enantiomer of a stereoisomeric pair is optically active
and has an equal but opposite-in-sign specific rotation.
 One enantiomer will rotate polarized light in a clockwise
direction, termed dextrorotatory (+), and its mirror-image
partner in a counter-clockwise manner, termed levorotatory (–).
 It is common practice to convert the observed rotation, α, to
a specific rotation, [α].
Specific rotation, [] = amount (degrees) that a substance
rotates plane polarized light expressed in a standard form.
Carvone from caraway: [α]D = +62.5º
Carvone from spearmint: [α]D = –62.5º
Lactic acid from muscle tissue: [α]D = +2.5º
Lactic acid from sour milk: [α]D = –2.5º
How can one identify enantiomerism?
All objects may be classified with respect to chirality
(from the Greek cheir = hand):
Chiral = Objects that are different from theirs mirror
image; i.e. golf clubs, scissors; enantiomers are chiral.
Achiral = Objects that are identical with theirs mirror
image; i.e. a pencil, a T-shirt.
Chiral molecule : (R)-lactic acid
Achiral molecule : water
Water (H2O)
Achiral molecules have either one or both of the following:
 Plane of symmetry
Center of symmetry
Achiral Objects and Molecules
Center of
Symmetry
Center of
Symmetry
Plane of Symmetry
Plane of Symmetry
CH3
H3C
COOH
HO
H
HO
H
COOH
Chiral molecules have chiral center (or stereo or stereogenic
center): an atom attached to 4 different atoms or groups.
(chiral carbon)
H
F
Br
CH3
H3CH2C
Cl
H3CO
H
Can you indicate stereocenters?
H
CH3
O
OH
Methylcyclohexane
Methylcyclohexane
Menthol
Designating the Configuration of Stereogenic Centers
The CIP system of nomenclature.
(R. S. Cahn, C. K. Ingold and V. Prelog)
 Each stereogenic center in a molecule is assigned a
prefix (R or S), according to whether its configuration is
right- or left-handed.
 The symbol R comes from the Latin rectus for right,
and L from the Latin sinister for left.
The assignment of the prefixes depends on the
application of two rules:
The Sequence Rule
The Viewing Rule
Right-Handed
Left-Handed
The Sequence Rule (The same as E-Z assignment)
 Assign sequence priorities to the four substituents by looking at
the atoms attached directly to the chiral stereogenic carbon atom.
 The higher the atomic number of the immediate substituent
atom, the higher the priority; H– < C– < N– < O– < Cl–.
 If two substituents have the same immediate substituent atom,
evaluate atoms progressively further away from the chiral center
until a difference is found.
i.e. CH3– < C2H5– < ClCH2– < BrCH2– < CH3O–.
 If double or triple bonded groups are encountered as
substituents, they are treated as an equivalent set of singlebonded atoms. i.e. C2H5– < CH2=CH– < HC≡C–
H
H
C C
R
IS TREATED AS
C C R
IS TREATED AS
O
C H
IS TREATED AS
H H
C C R
C C
C
C
C
O
C
O
C
C R
C
H
The Viewing Rule
 The chiral center must be viewed from the side
opposite the lowest priority group.
Numbering the substituent groups from 1 to 4, with 1
being the highest and 4 the lowest in priority sequence,
and put a viewers eye on the side opposite substituent #4.
 If the progression from 1 to 3 is clockwise, the
configuration at the stereocenter is R. Conversely the
counterclockwised progression is assigned as S.
1
HO H
H3C
CO2H
Assign Priorities
HO H
H3C
3
Lactic acid
4
CO2H
2
Twist the lowest
priority to the
back
HO H
H3C
CO2H
(R)-Lactic acid
Rotate Priorities
4
H
Br
H3C
CH2CH3
Priorities
1
2
3
2-Bromobutane
View &
Assign
1
(S)-2-Bromobutane
2
3
If you have troubles looking at the stereocenter, try Fischer Projections.
Press flat
W
X
W
C
Y
Z
Z
Z
C
X
W
X
Y
A Simple Mental Exercise for Fisher Projection
Y
Only two kinds of motions are allowed for Fischer projection.
1) Rotation on page 180˚ is allowed for Fischer projection.
COOH
H
OH
CH3
SAME AS
H
HO
CH3
COOH
1800
2) One group can be held steady while the other three rotate in either
a clockwise or a clockwise direction.
COOH
H
OH
CH3
COOH
SAME AS
CH3
HO
H
Assigning R, S configurations to Fischer projections.
 Assign priorities to the four substituents in the usual way.
 Perform one of the two allowed motions to place the group
of lowest (4th) priority at the top of the Fischer projection
if it is necessary.
 Determine the direction of rotation in going from priority
1 to 2 to 3, and assign R (clockwise) or
S (counterclockwise).
HO H
H3C
Looking up
CO2H
OH
H3C
Lactic acid
H
COOH Rotate
clockwise
4
H
HO
COOH
1
2
CH3
H
Br
H3C
CH2CH3
2-Bromobutane
3
4
H
H3C
Br
CH2CH3
3
2
1
Compounds Having Two or More Stereogenic Centers
H
OH
1
HN
Stereocenter 1
3
CH3
Stereocenter 2
1
(1R), (2S)-(-)-Ephedrine
2
3
2
2
1
4
3
H
(-)-Ephedrine
1
4
CH3
2
4
2
3
4
1
Can you assign absolute configurations of these compounds?
H3C
HO H
H NH
CH3
(+)-Ephedrine
HO H
CH3
HN H
CH3
(+)-Pseudoephedrine
H3C
H OH
H NH
CH3
(-)-Pseudoephedrine
(+)-Pseudoephedrine and (-)-Pseudoephedrine are
diastereomers of (+)-Ephedrine.
Diastereomers
 Stereoisomers that are not mirror images of each other.
 Diastereomers have similar chemical properties.
 Diastereomers have different physical properties:
melting points, boiling points, solubities in solvent, etc.
 Diastereomers can be separated by fractional distillation,
or crystallization.
H
Ph
H3CHN
R
S
H
Mirror
OH
H
CH3
HO
H
S
R
CH3
Ph
NHCH3
(-)-Ephedrine
(+)-Ephedrine
Diastereomers
Diastereomers
H
HO
H3CHN
S
S
Ph
H
CH3
(+)-Pseudoephedrine
H
Ph
H
R
R
CH3
OH
NHCH3
(-)-Pseudoephedrine
Relationships Between Stereoisomers
Stereoisomers
Enantiomeric with
Diastereomeric with
(1R), (2S)-(-)-Ephedrine (1S), (2R)-(+)-Ephedrine (1S), (2S)-(+)Pseudoephedrine
and (1R), (2R)-(-)Pseudoephedrine
Cahn-Ingold-Prelog R/S notation = Specifies absolute
configuration of a chiral center;
there is no correspondence between R and + or S and –
COOH
H R OH
HO R H
COOH
(2R, 3R)-(+)Tartaric Acid
mirror
COOH
HO S H
H S OH
COOH
(2S, 3S)-(+)Tartaric Acid
COOH
HO S H
HO R H
COOH
(2S, 3R)-(+)Tartaric Acid
mirror
COOH
H R OH
H S OH
COOH
(2R, 3S)-(+)Tartaric Acid
meso-Tartaric Acid
(+)-Tartaric Acid:
[α]D = +12º
m.p. 170 ºC
(–)-Tartaric Acid:
[α]D = –12º
m.p. 170 ºC
meso-Tartaric Acid: [α]D = 0º
m.p. 140 ºC
Meso Isomer:
an achiral molecule with 2 or more chiral centers and
an internal plane of symmetry; the molecule is achiral.
COOH
S
HO
H
R
HO
H
COOH
180o
(2S, 3R)-(+)-Tartaric Acid
COOH
R
H
OH
H S OH
COOH
(2R, 3S)-(+)-Tartaric Acid
Identical
Number of stereoisomers = a molecule with n
stereogenic centers (and for which a meso isomer isn’t
possible) will have 2n stereoisomers.
Racemix Mixtures (Racemate)
 A 50:50 mixture of the two enantiomers.
 Denote by the symbol (±) or prefix d, l.
 Must show zero optical rotation (optically inactive).
enantiomeric excess (ee)
= (specific rotation of the mixture/ specific rotation of
the pure enantiomer in excess) × 100
= (moles of major enantiomer - moles of other
enantiomer / Total moles of both enantiomers) × 100
* Let's suppose that we have a reaction mixture which we
measure for rotation, and the observed specific rotation of
the reaction mixture is measured to be +8.52°. The specific
rotation of the S-configured enantiomer was said to be
-15.00° . Determine the enantiomeric excess.
enantiomeric excess = (+8.52 / +15.00) × (100) = 56.8 %
in excess of R-isomer
* This means that 56.8% of the entire mixture is in excess of
the R-configured isomer
Resolution of Racemic Mixtures
There are two basic ways that one can separate the
enantiomers in a racemic mixture:
 Biological Resolution : Uses a microbe which metabolizes
one specific enantiomer leaving the other alone.
 Chemical Resolution : The racemate is converted to two
diastereoisomers. Once separated the diasteriosomeric
forms are converted back to enantiomers in separate
containers.