organic chemistry dr v.o. nyamoricheminnerweb.ukzn.ac.za/files/chem120 organic 2010... · 7/11/2010...
TRANSCRIPT
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Organic Chemistry Dr V.O. Nyamori
Textbook by i) Bruice “Organic Chemistry” 5th Edition
ii) Hart et al. “Organic Chemistry: A Short
Course” 12th Edition
• The study of carbon‐containing compounds and their properties.
• The vast majority of organic compounds contain chains or rings
of carbon atoms.
• They form the basis of, or are important constituents of many
products (plastics drugs petrochemicals food explosives
Course 12 Edition
products (plastics, drugs, petrochemicals, food, explosives,
paints, to name but a few) and, with very few exceptions, they
form the basis of all earthly life processes.
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Soap/detergent ‐surfactant C17H35COO‐
Sugar ‐glucose C6H12O6
Medicine ‐ ascorbic acid HC6H7O6
e.g.
Carbon: group 14, atomic no. 6Recall: Electronic configuration for carbon?
Periodic Table
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Structure of Carbon Compounds
Three hybridization states
C C C C C C
1 54 Å 1 20 Å1 33 Å
each satisfies the octet rule for each carbon!!
1.54 Å 1.20 Å1.33 Å
Hart et al. “Organic Chemistry: A Short Course”, 12th Edition,
Chapter 1.14 ‐ 1.18
Geometries of carbon compounds
sp3 Tetrahedral 108.5°
sp2 Trigonal planar 120°
sp Linear 180°
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Hydrogens are in a tetrahedral
arrangement around the sp3
hybridized carbon atom
Methane: CH4
hybridized carbon atom.
Hydrogens bond to the carbon sp3
orbitals with 1s orbitals.
sp3 Hybridizationcarbon
Energy
sp3
2p
2s
1s
Hybridization
nergy
2p
2s H b idi ti
sp2 Hybridizationcarbon
En
sp22s
1s
Hybridization
2p
Ethene: C2H4
sp Hybridizationcarbon
sp
p
2s
1s
Hybridization
Energy
Ethyne: C2H2
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HYDROCARBONS
• compounds composed of only carbon and hydrogen
• chain of carbon atoms bonded to enough hydrogen atoms
to satisfy the octet rule for each carbonto satisfy the octet rule for each carbon
• chain is bent because of the 109.5° C–C–C tetrahedral angle
CCH
H
HH
CH
CHHC
CH2
CC
H3C
e.g.
C
CCC
CC
H
H
H
H
HH
H CH3
Line notation
HYDROCARBONS
• Hydrocarbons with all single carbon‐carbon bonds(no double or triple bonds)
Alkanes
• They contain the maximum number of hydrogen atoms
(no double or triple bonds)
• Alkanes are SATURATED
Alkenes, alkynes and aromatic compounds
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• UNSATURATED hydrocarbons
‐ they ARE NOT ALKANES
• contain carbon‐carbon multiple bonds
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Past exam Question
C‐1: _____ C‐3: _____
C‐2: _____ C‐4: _____
Example
HBr
H
1. Indicate the hybridization for carbons 1 – 10 and theirrespective geometry. Include bond angles in your answer.
CC
C
CC
CC
C
O
CN
F H
O
ClH
H
H
H
12
3
46
78
910
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Hydrocarbons
• Four basic types:‐ Alkanes
CnH2n+2
‐ Alkenes
Alkynes
C2H6 Ethane
CnH2n
C2H4 Ethene 120°
‐ Alkynes
‐ Aromatic hydrocarbons
C2H2 Ethyne
CnHn
CnH2n
HH
H
H H
H
C6H6 Benzeneor
Organic Nomenclature
• Three parts to a compound name:
1 2 3
1. Prefix
2. Base/parent
3. Suffix
Chapter 2: “Organic Chemistry” 5th Edition , Bruice P. Y.
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Organic Nomenclature ‐ IUPAC Rules
Suffix: Tells what type of compound it is.
Prefix : Tells what substituent(s) are attached, if any.
prefix base suffix
Base/parent: Tells how many carbons are in the longest continuous chain.
prefix base suffix
What substituent? How many
carbons?
What family?
Alkanes
• Only van der Waals force: London force.
• Boiling point increases with length of chain.
• Nomenclature suffix “‐ane”
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To Name a Compound…1. Determine what type of
compound it is.
2. Find the longest chain in the molecule.
3. Number the chain from the end nearest the first substituent encountered.
4. List the substituents as a prefix in alphabetical orderprefix in alphabetical order along with the number(s) of the carbon(s) to which they are attached.
CH2CH
CH2H2C
CH2
H3C
H3C
CH3Name??
ExampleCH3
CH2CHCH3
If there is more than one type of substituent in the molecule,
CH CH
CHCH3
CH2 CH3
CH3
CH3
list them alphabetically i.e. name of substituent, not prefix for
frequency e.g. di, tri, tetra, etc...are not considered.
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Cycloalkanes
• Carbon can also form ringed structures.
• Five‐ and six‐membered rings are most stable.
– Can take on conformation in which angles are
very close to tetrahedral angle.
– Smaller rings are quite strained.
l f l
cyclohexane cyclopentane cyclopropane
General formulaCnH2n
CYCLIC ALKANES
How do we name….CH3
CH2CH3
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Alkenes
VSEPR Theory
• Contain at least one carbon–carbon double bond
120°
• Unsaturated
– Have fewer than maximum number of hydrogens
– The C atoms on double bond are sp2 hybridized
Structure of Alkenes
• Unlike alkanes, alkenes cannot rotate
freely about the double bondfreely about the double bond.
– Side‐to‐side overlap makes this
impossible without breaking ‐bond.
C C
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Structure of Alkenes
This creates geometric isomers
diff i h i ldifference in the spatial arrangement of groups about the double bond
Cis‐ isomer “Z”‐isomer
Z‐2‐Pentene
Trans‐ isomer “E”‐isomer
E‐2‐Pentene
NAMING ALKENES
1. Find the longest unbranched carbon chain containing the
d bl b ddouble bond.
2. Number the carbon atoms in the main chain.
• Name chain according to number of carbon atoms.
add ‐ene as a suffix
• Start from the end of the chain that is closest to the
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Start from the end of the chain that is closest to the
double bond.
location of the double bond is numbered with the
lowest‐numbered carbon in the double bond.
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Example
H
Name this alkene
C
H
CHCH3 C
H
CH
CH2
CH2
CH
CH3
CH3
H2C
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Name this alkene
Example
C CHCH3 C
H
CH CH2 CH3
CH3
1. The longest unbranched chain containing the double
bond which is the functional group (suffix ‐ ene)
Name this alkeneH
CH2
CH
H2C
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bond which is the functional group (suffix ‐ ene)
2. The chain numbering starts closest to the double bond.
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3. There are two substituent groups on this alkene:
C CHCH3 C
H
CH CH2 CH3
CH3
4. Compose the name…..
• Add the substituent groups alphabetically to the alkene name
H
CH2
CH
H2C
name of the alkene is?
• Specify the position of each group on the main chain
Properties of AlkenesExample: C4H8
2‐Methyl‐1‐propenebp. ‐7 ⁰C
1‐Butenebp ‐6 ⁰C
Cis‐2‐Butenebp +4 ⁰C
Trans‐2‐Butenebp +1 ⁰C
Structure also affects physical properties of alkenes
Can we have more than one double bond?
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Alkenes exhibit cis‐trans isomerism.
GEOMETRICAL ISOMERS
Trans‐ isomer “E‐” Cis‐isomer “Z‐”
• Identical substituents on
opposite sides of the
double bond
C
H
CCH3
CCH3
CCH3
• Identical substituents on
same side of the double
bond
CH3 HC
HC
Htrans‐“E‐”
cis‐“Z‐”
Stick diagram
Geometric isomers of Alkenes• Cis‐alkenes have similar higher priority elements or group (or
Z‐isomer i.e. higher priority elements but not necessarily
similar) in the chain on the same side of the molecule.
• Trans‐alkenes have similar higher elements or group (or E‐Trans alkenes have similar higher elements or group (or E
isomer i.e. higher priority elements but not necessarily similar)
in the chain on opposite sides of the molecule.
• Priorities are assigned by the atomic numbers of the atoms
bonded to the carbon in the double bond.
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Examples1. Name the following alkenes and determine whether there
are geometric isomers i.e. either Trans‐ (E‐) or Cis‐ (Z‐)isomers.
H
a) Br
Hb)
F
c)
Certain groups of atoms give a molecule a....FUNCTION
Acidic, basic, alcohol, etc…
FUNCTIONAL GROUPS
Each functional group is specified by a suffix or prefix
The GROUPS are called functional groups.
depicted on the nomenclature of the organic molecule
Functional groups are given an order of priority to
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decide on which is the suffix.
Please refer to your textbooks on priority preference by:
i) Bruice “Organic Chemistry” 5th Edition
ii) Hart et al. “Organic Chemistry: A Short Course” 12th Edition
HOMEWORK!
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Group / Family
FormulaStructural Formula
Prefix Suffix Example
Alkane RH alkyl- -ane
Ethane
Alkene R2C=CR2 alkenyl- -ene
Alkyne RC≡CR' alkynyl- -yne
Ethene
Benzene derivative
RC6H5
RPhphenyl- -benzene
2-phenylpropaneisopropylbenzene
Ethyne
Group / Family
FormulaStructural Formula
Prefix Suffix Example
Haloalkane RX halo-alkyl
halide
Fluoroalkane RF fluoroalkyl
Chloroethane
Ethyl chloride
FluoromethaneFluoroalkane RF fluoro-
fluoride
Chloroalkane RCl chloro-alkyl
chloride
Bromoalkane RBr bromo-alkyl
bromide
Methyl fluoride
Chloromethane
Methyl chloride
Bromomethane
M th l b idbromide
Iodoalkane RI iodo-alkyl
iodide
Methyl bromide
Iodomethane
Methyl iodide
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Group / Family
Group FormulaStructural Formula
Prefix Suffix Example
Primary amine
RNH2 amino- -amine
Methylamine
Amines
Secondary amine
R2NH amino- -amine
Tertiary amine
R3N amino- -amine
Dimethylamine
Trimethylamine
Quaternaryammonium
ionR4N+X-
ammonio--ammonium
H3C
N
H CH3
CH3
Cl
Trimethyl-ammonium
chloride
Group / Family
FormulaStructural Formula
Prefix Suffix Example
Alcohol ROH hydroxy- -ol
M th l
Ketone RCOR' keto-, oxo- -one
Methanol
ButanoneMethyl-
ethyl ketone
Aldehyde RCHO aldo- -al
EthanalAcetaldehyde
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Group / Family
FormulaStructural Formula
Prefix Suffix Example
Carboxylic acid
RCOOH carboxy- -oic acidEthanoic acid
Acetic acid
Acyl halide RCOX haloformyl- -oyl halide
Eth ROR' lkalkyl alkyl
Ethanoyl chlorideAcetyl chloride
Ether ROR' alkoxy-y y
ether
Ester RCOOR'alkyl
alkanoate
EthoxyethaneDiethyl ether
Ethyl butanoateEthyl butyrate
Primary (1°) alcohols and amines
General structure
Alcohol H
Example
HAlcoholR1 C
H
OH
1°
C
H
H
OHCH3CH2
HH
Amine
1°
N
H
H
CH3CH2R1 N
H
H
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Secondary (2°) alcohols and amines
General structure
Alcohol H
Example
HAlcoholR1 C
R2
OH
2° Alcohol
C
H
CH3
OHCH3CH2
HHAmine
2° Amine
N
H
CH3
CH3CH2R1 N
H
R2
Tertiary (3°) alcohols and amines
General structure
Alcohol1
R3
Example
CH3
R1 C
R2
OH
3° alcohols
C
CH3
OHCH3CH2
CH3R3
Amine
3° Amine
N
CH3
CH3
CH3CH2R1 N
R3
R2
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Quaternary amines
General structure
R2 CH
Example
AmineR4 N
R2
R1
R3+ N
CH3
CH3
CH3CH3CH2
+
Exercise
1. Draw the structures of the following alcohols and amines
and classify them as either 1°, 2°, 3° or quaternary
a) Pentan‐1‐ol
b) Dimethylamine
c) 3‐Ethyl‐hexan‐3‐ol
d) Diethyl‐methyl‐amined) Diethyl methyl amine
e) Butan‐2‐ol
f) Triethylmethly ammonium ion
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a) Pentan‐1‐ol
Solutions
b) Dimethylamine c) 3‐Ethyl‐hexan‐3‐ol
d) Diethyl‐methyl‐amine e) Butan‐2‐ol f) Triethylmethly‐ammonium ion
Naming Hydrocarbons with Functional Groups
Name the other substituent groups, using the
prefixes for alkyl groups and the prefixes for any
other functional groups
Specify the position of each group on the main chain.
Add the substituent groups alphabetically to the name
f th lk ( lk lk ) l ith th
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of the alkane (or alkene or alkyne) along with the
frequency of each group.
Example…..
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Example: Name this organic molecule:
CH3 CHCH
OH
CH3
NO2
3 3
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Example: Name this organic molecule:
CH3 CHCH
OH
CH3
NO2
1. This molecule contains a ……………. and …………. group
3 3
Only the group has priority.
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Only the …………… group has priority.
So this is an ................
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Example: Name this organic molecule:
CH CHCH
OH
CH
NO2
longest carbon chain containing the hydroxyl
CH3 CHCH CH3
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g g y y
group has ………… carbons.
Therefore its a “….……….. ”
An alcohol suffix is ‐OL
CH3 CHCH
OH
CH3
NO2
3. Number the carbons, starting NEAREST the
functional group.
The hydroxyl group is on position …
…so this is a ………… or .…………
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4. This molecule has one substituent
A nitro group on position …
………………………… …………………….or
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Example:
Name this organic
molecule ?? CH CH CH CH C
CH2
CH2C
CH3
CH3
O
H2N
molecule ?? CH3 CH CH CH2
CH3
C
CH2
CH3
OH
1. Identify functional groups
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2. Determine longest carbon chain that contains the highest priority functional group.
Example: Name this organic molecule:
CH
CH2C
CH3
CH3H2N
CH3 CH CH CH2
CH3
C
CH2
CH2
CH3
OH
O
The longest carbon chain has 8 carbons
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The longest carbon chain has 8 carbons.
BUT………………...
THIS CHAIN DOES NOT CONTAIN THE ‐CO2H group
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Example: Name this organic molecule
CH
CH2C
CH3
CH3H2N
CH3 CH CH CH2
CH3
C
CH2
CH2
CH3
OH
O
The longest carbon CONTAINING the ‐CO2H
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g 2
chain has …. carbons.
so this molecule is based on a ..………… ..
CH3 CH CH CH2 C
CH2
CH2C
CH3
CH3
O
H2N
Highest priority functional group is a carboxyl group
3 2
CH3 CH2
CH3
OH
Suffix OIC ACID
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Suffix ‐OIC ACIDHence… Heptanoic acid
Now number chain…….
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CH3 CH CH CH2 C
CH2
CH2C
CH3
CH3
O
H2N
Heptanoic acid
3. Number the carbons starting with the functional group
3 2
CH3 CH2
CH3
OH
The carboxyl group is on position 1The carboxyl group is on position 1,
do not include in the name because
the carboxyl group is always a terminal group.
CH3 CH CH CH2 C
CH2
CH2C
CH3
CH3
O123
4
5 6 7H2N
Heptanoic acid
h
3 2
CH3 CH2
CH3
OH
4. This molecule has four substituents
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Compose the name
ALPHABETICAL LIST
INTERPETING AN IUPAC NAME…...
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WHAT IS THE STRUCTURAL FORMULA OF
BUTANONE ?
CH3 CH2C
O
CH3
Carbonyl CANNOT Be at the END!! WHY?????
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C CH2CH2
O
CH3H
BUTANAL
Examples
Give the correct IUPAC name for the following compounds.
Bra) b)) )
H
OH
c) d)
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O
H CH3
Examples
Give the correct IUPAC name for the following compounds.
f) g)H
O
CH3
Cl
)
h) i) O
O
OHh) i) O
O
Examples
Draw structural formulae for the following compounds:
a) 2,3,5‐trimethylhexane
b) (Z)‐3‐chloro‐hept‐2‐ene
c) 3‐ethylnonanol
d) 2,3‐dimethylpentanoic acid
e) methylhexanoatee) methylhexanoate
f) 3‐iodohexanal
g) pentan‐2‐one
h) 3‐aminopentane T1
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Solutions
2,3,5‐trimethylhexane (Z)‐3‐chloro‐hept‐2‐enea) b)
3‐ethylnonanol 2,3‐dimethylpentanoic acidc) d)
Solutions
methylhexanoate 3‐iodohexanale) f)
O OI
pentan‐2‐oneg) h)
O
OI
H
3‐aminopentane)
OH2N
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ISOMERS
(a) Structural isomers
Molecules with the same chemical formula but different
Two types: (a) Structural isomers (b) Stereoisomers
CONSTITUTIONAL ISOMERS
Constitutional isomers have different properties:
e.g. butane (C4H10) has 2 structural isomers
Molecules with the same chemical formula but different
bonds between the atoms
Now called...
These are…..?
CH3 CH2 CH2 CH3 CH3 CH CH3
CH3
n‐Butane: C4H10 2‐methylpropane: C4H10
bp = ‐12 °C mp = ‐159 °Cbp = 0 °C mp = ‐138 °C
Solution:
ExampleHow many constitutional isomers are formed from C5H12? Draw their structures.
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CONSTITUTIONAL (STRUCTURAL) ISOMERS
The general formula for ALKANES is…... 22 nnHCn = 1, 2 and 3 1 ISOMER
The number of ISOMERS increases with n…..
,
n = 4 2 ISOMERS
n = 5 3 ISOMERS
n = 6 5 ISOMERS
n = 7 9 ISOMERSn = 40
n = 8 18 ISOMERS
n = 9 35 ISOMERS
n = 10 75 ISOMERS
n = 20 366,319 ISOMERS
n = 40 62,491,178,805,831 ISOMERS
HOMEWORK: DRAW THE ISOMERS OF C40H82 !!!
Constitutional isomers for multibonds
C6H10
Example: Alkene
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Constitutional isomers for multibonds
Example: Alkyne
C6H6
H3C C C C C CH3
C C C C CH2
CH3
H
2,4‐Hexadiyne
Hexa‐2,4‐diyne
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C C C C CH2H
1,3‐Hexadiyne
Hexa‐1,3‐diyne
OPTICAL ISOMERISM
Optical isomerism arises when molecules have a
structure such that the mirror image is notstructure such that the mirror image is not
superimposable on the original molecule.
occurs whenever there are four different groups
bound to the same tetrahedral carbon atom.
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Some terminology…….
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OPTICAL ACTIVITY IN ORGANIC COMPOUNDS
Stereogenic centre has four different groups attached
to a tetrahedral carbon atom
ZX
W
Y
C
Stereogenic centre
Chiral carbon atom
*
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The carbon involved is called a chiral carbon or
stereogenic carbon and the molecule is known as a chiral
molecule.Example: 2‐butanol
Y
2‐Butanol
Dash shows bond going Bonds aligned to
Perspective formula
C
CH CH
OH
H3C
Dash shows bond going backwards from the viewer
*
gthe asymmetric
center in the plane
CH2CH3H
3
Solid wedge represents a bond extending out towards
the viewerChiral centre
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CH
2‐Butanol
Fischer projection
Bonds aligned to Bond going
CH2CH3HO
CH3
*
gthe asymmetric
center in the plane
Bond going backwards from the
viewer
Bond extending d h
H
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Chiral centre
out towards the viewer
2‐butanol has two optical isomers.
A pair of isomers called enantiomers ‐
ENANTIOMERISM in ORGANIC CHEMISTRY
non‐superimposable mirror images of each other.
View in 3‐DView in 3‐D“mirror”
OH
C
OH
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C
CH2CH3H
H3CC
HCH3CH3CH2
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Identify the chiral carbon (stereogenic centre) and draw the
structural formula for each of the following molecules:
(a) 1‐chloroethanol (b) 2,3,5‐Trimethyl‐hexane
( ) h l l h (d) di hl l
Question
Solution
(c) Methylcyclohexane (d) 1,3‐dimethlycyclopentane
C CH3HO
H
(a) 1‐chloroethanol
*
*
(b) 2,3,5‐Trimethyl‐hexane
3
Cl
* *No chiral carbon
(d) 1,3‐dimethlycyclopentane(c) Methylcyclohexane
NAMING OPTICAL ISOMERS
Stereogenic centre creates twomolecular optical isomers
Cl ClTwo
configurations
How do we name these isomers??
CH
OHH3C
CH
HO CH3
configurations
Enantiomers
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Solution: Use R‐S nomenclature system for designating
the configuration
We assign priorities as in the E, Z system……...
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1. Assign relative priorities to each of the four groups on
the stereogenic carbon to describe the configuration.
CAHN‐PRELOG‐INGOLD R,S‐NOMENCLATURE
The priorities are given by rules:
• Higher atomic numbers are given higher priorities.
• If necessary, the second atom in each substituent is
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y,
used to determine the priorities.
2. Draw the molecule with the lowest‐priority group
pointing directly into the page….
CAHN‐PRELOG‐INGOLD R,S‐NOMENCLATURE
and the other three groups pointing out of the page in
an arrangement like a steering wheel.
Example: 2‐ Butanol…..CHCCH
OH
CH
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CH2CCH3 CH3
H
Draw molecule as a wedge and dashed line diagram
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O O
2‐ butanol
CH2CCH3
OH
CH3
H
p
Hart et al. “Organic
Chemistry: A Short Course” 12th Edition. Chapter 5
H3CCH2CH3
C
OH
HH3CH2C
C
OH
CH3
H
H
73
DO NOT FORGET THE OTHER ISOMER……….
THESE ARE THE TWO…. ENANTIOMERS...or ....OPTICAL ISOMERS
1. Assign priorities:
O > CMe = CEt > H
C,H,H > H,H,H
2‐ butanol
C
CH2CH3
OH
H, , , ,
CEt > CMe
CH3
3
74
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CHCH2CH3
C
OH
H
1
24
O > CEt > CMe > H
1. Assign priorities to each group:
2‐ Butanol
CH33
CH CHC
OH1
2
2. Redraw the molecule with the lowest priority group facing in.
Et Me
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H3C CH2CH33
Now what?????
H C CH CH
OH
1
2C
2‐ Butanol
3. Look at the direction in which the priorities decrease.
If they decrease in a clockwise direction, the
H3C CH2CH33
(R)‐2‐Butanol
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stereogenic centre is called “R” or rectus
which is Latin for “right.”Or….
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CH3CH3CH2
OH
1
2 3C
2‐ Butanol
3CH3CH2
If the priorities decrease in a counter‐clockwise
di ti th t i t i ll d “S”
(S)‐2‐Butanol
direction the stereogenic centre is called “S” or
sinister, which is Latin for “left.”
Example:
NH2 NH2
What are the configurations of the following chiral molecules?
CH3
CH2CH3
CH
CH3CCH2CH3
H
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H3C CH2CH3
C
NH2CH3
CH2CH3
CH2N
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Example: Give the configuration of the stereogenic
centre in each of the following molecules:
CH2OHCl
CH
ClC OH
O
CHCH2OHHO C
O(A) (B)
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THEY LOOK DIFFERENT BUT ARE THEY???
2‐chloro‐3‐hydroxy‐propanoic acid
A)
CH
CH2OH
ClC OH
O
1. Assign priorities to each group:
O
ClC
CH2OH
C OH
2. Redraw the molecule with the
lowest priority group facing in.
3. Look which way the priorities decrease
8080
C OH
O
The priorities decrease anti‐clockwise, so this centre is “.…”
decrease.
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1. Assign priorities to each group
CHCH2OH
ClB)
2. Redraw the molecule with CH2OHHO C
Othe lowest priority group
facing in.
C
Cl 3. Direction of the priorities
decrease?C
CH2OHHO C
O
The priorities decrease anti‐clockwise, so this
centre is “…”
Example: Give the configuration of the stereogenic
centre in each of the following molecules:
ClCH2OH B)A)
2‐chloro‐3‐hydroxy‐propanoic acid
CHCH2OHHO C
O
CH
ClC OH
O
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THEY LOOK DIFFERENT BUT ARE THEY???
2‐chloro‐3‐hydroxy‐propanoic acid
WHAT IF?????
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If we flip the COOH and H ???????
CH2OHA) CH2OHA’)
CH
ClC OH
O
C H
Cl
C
O
HO
CH2OH CH2OH
ClC
C OH
O
CC
O
HO Cl
What is the structural formula of (R)‐2‐chloro‐2‐butanol?
DRAW MOLECULAR STRUCTURE
EXAM QUESTION
Solution:
84
ASSIGN PRIORITIES
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(R)‐2‐chloro‐2‐butanolAssign priorities
Draw steering wheel with lowest priority group pointing in…….
MAKE SURE PRIORITY GOES
85
DRAW MOLECULE…..
MAKE SURE PRIORITY GOES CLOCKWISE FOR R
C
DRAW MOLECULE TO SEE ALL GROUPS…..C
Remember lowest priority group h “d h d” b d
Make sure priority goes clockwise for “R”
C
has a “dashed” bond
C
(R)‐2‐chloro‐2‐butanol
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[2R,3R] [2S,3S] [2R,3S] [2S,3R]
a b c dCH CH
CH3 CH3
CHCH3 CH
Br Cl
CH3* *
R
R
C
C
CH3
CH3
H
Cl
Br
H
S
S
C
C
CH3
CH3
Br
H
H
Cl
R
S
C
C
CH3
H
H
Br
ClR
SC
C
CH3
Br
Cl
H
H
enantiomeric pairs enantiomeric pairs
• Enantiomeric pair differ only in optical activity
RCH3
SCH3
RCH3
SCH3a b c d
Diastereomers b & ca & c
b & da & d
R
S
C
C
CH3
H
H
Br
ClR
SC
C
CH3
Br
Cl
H
H
R
R
C
C
CH3
H
Cl
Br
H
S
S
C
C
CH3
Br
H
H
Cl
Diastereomers ‐ are distinct chemical compounds, differing not only in optical activity but also in mp, bp., , solubility etc…
molecule with n stereogenic centres may exist inmaximum of 2n stereisomeric forms, with maximum of2n/2 enantiomeric pairs
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Meso compoundCHHOC CH COH
OH OH
OO* *
RCO2H
SCO2H
RCO2H
SCO2H
R
R
C
C
CO2H
OH
H
H
HO
S
S
C
C
CO2H
H
OH
HO
H
R
S
C
C
CO2H
OH
OH
H
HR
SC
C
CO2H
H
H
HO
HO
170 °C 170 °C 140 °C
+12° ‐12° 0°
meso compound ‐ an achiral (optically inactive)
diastereomer of compound with stereogenic centres
arises because 4 different groups making each of C‐2 & C‐3
stereogenic are same 4 different groups…(!)
Meso compound
RC
CO2H
OHHS
C
CO2H
HHOR
C
CO2H
OHHS
C
CO2H
HHO
RC
CO2H
HHOS
C
CO2H
OHHS
C
CO2H
OHHR
C
CO2H
HHO-------------------- -------------------- -------------------- --------------------
Enantiomers, Chiral Identical, achiral, Meso form
possess plane of symmetry bisecting central C‐C bond
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Shows arrangements in space
Newman projections
e.g. Ethane C2H6
H H HH
60° Staggered H H
HH
HH
H
H
“dash‐wedge”
H
HH
HH
H
“sawhorse”
0°
Newman
Staggeredconformation
Eclipsed
H
H
H
H
H
H H
HHH H
“dash‐wedge”
H
HHH
H H“sawhorse”
0
Newman
Eclipsed conformation
HH
HHH
H
a
e
e aa
e "flip"16
5
3
a
e
e
a e
aa
e
e
16
5
43
2
Cyclohexane conformations
ae
a
e
a
e
42
aa
e
H
HH
Chair conformations
H H
HH
H
HH
H
H
HBoat conformation
Cyclohexane
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Examples1,2‐Dimethylcyclopentane
CH3 CH3 H
H
H CH3 H H
H
H
H H
H
H H
3
CH3H
H H
H
H
H H
Different bond pattern
Structural (constitutional)
Summary of isomerism
Hart et al. “Organic Chemistry: A Short Course” 12th Edition, page 52‐54
pattern
isomers
( )isomer
Stereoisomer
Same bond pattern
Interconvertible by single bond rotation
Not interconvertibleby bond rotation
Conformers (rotamers)
Configurational isomers
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Exercise
1. Draw the structures of
a) (Z)‐3‐methyl‐2‐pentene c) (E,Z)‐2,4‐Heptadiene
b) (S)‐2‐bromopropan‐1‐ol d) (2S,3R)‐3‐Bromobutan‐2‐ol
2. Using the Newman projection draw the structure a
staggered conformation of butane.
3. Using the Fischer projection draw a structure of
(S)‐2‐methylpentanoic acid.
4. Draw the structure for the cis and trans isomers of
1‐bromo‐2‐chlorocyclopropane
Solutions
1. a) (Z)‐3‐methyl‐2‐pentene b) (S)‐2‐bromopropanol
c) (E,Z)‐2,4‐Heptadiene d) (2S,3R)‐3‐Bromobutan‐2‐ol
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Staggered conformation of butane
Solutions
2.
Newman projectionp j
3. (S)‐2‐methylpentanoic acid
Fischer projection
(4) 1‐bromo‐2‐chlorocyclopropane