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10/4/2011
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CHEM 120 CHEMICAL REACTIVITY
Prescribed book

ORGANIC CHEMISTRY
T.L. Brown, H.E. LeMay Jr, B.E. Bursten, C.J. Murphy,
S. Langford and D. Sagatys, Chemistry: The Central
Science: A Broad Perspective, 2nd Edition, Pearson,
Australia, 2010.
Dr. Vincent O. Nyamori
Recommended books
19 Lectures
5 Tutorials
2 Quiz
1 Test
Organic Chemistry
•
Surfactant
C17H35COO‐
H. Hart, L.E. Craine, D.J. Hart and C.M. Hadad, Organic Chemistry - A
Short Course, 12th Edition, Houghton-Mifflin (students not intending to
continue with Chemistry).

P.Y. Bruice, Organic Chemistry, 6th Edition,
Pearson/Prentice Hall (Chemistry major students).

Any other relevant Organic textbook - Library
worksheets
http://cheminnerweb.ukzn.ac.za/Firstyear/chemonetwenty.aspx
•

Carbon
1s22s22p2
The study of carbon‐containing
compounds and their properties.
The vast majority of organic
compounds contain chains or rings of
carbon atoms.
Glucose
C6H12O6
Ascorbic acid
HC6H7O6
3
Periodic Table
4
1
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Structure of Carbon Compounds
Geometries of carbon compounds
sp2
sp3
Three hybridization states Trigonal planar
Tetrahedral
C
C
C
C
sp2
sp3
1.33 Å
1.54 Å
each satisfies the octet rule for each carbon!
C
C
sp
1.20 Å
C
C
 “Organic Chemistry: A Short Course” 12th
Edition, by Hart et al. Chapter 1.14 ‐ 1.18
 “Chemistry: The Central Science: A Broad Perspective”, 2nd
Edition, by LeMay et al. Chapter 21.1 & Chapter 23.1 pg 852
sp
Linear
5
Methane: CH4
carbon
 Hydrogens are in a tetrahedral arrangement around the sp3
Energy
hybridized carbon atom.
2s
carbon
carbon
sp2
Ethene: C2H4
sp Hybridization
2p
Hybridization
Energy
2p
Energy
Hybridization
1s
sp3 Hybridization
1s
sp2 Hybridization
2p
 Hydrogens bond to the carbon sp3
orbitals with 1s orbitals.
2s
6
sp3
2s
Hybridization
sp
1s
7
Ethyne: C28H2
2
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HYDROCARBONS
HYDROCARBONS
• compounds composed of only carbon and hydrogen
Alkanes
• chain of carbon atoms bonded to enough hydrogen atoms to satisfy the octet rule for each carbon
• chain is bent because of the 109.5° C–C–C tetrahedral angle
e.g.
CH2
H
H
180°
C
C
H
H
C
C
sp3
H
109.5°C
H
C
H 120°
H
C
sp2
C
HC
H3 C
CH3
H
How many bonds are • Hydrocarbons with all single carbon‐carbon bonds
(no double or triple bonds)
• Alkanes are SATURATED
• They contain the maximum number of hydrogen atoms
Alkenes, alkynes and aromatic compounds CH
C
sp
“Chemistry: The Central Science: A Broad Perspective” ‐ Chapter 21
• UNSATURATED hydrocarbons
• contain carbon‐carbon multiple bonds
H
Π = 3 bonds
θ = 16 bonds
“Chemistry: The Central Science: A Broad Perspective”
‐ Chapter 23 & 24
‐ they ARE NOT ALKANES
Line notation
9
10
Example Past exam Question
1. Indicate the hybridization for carbons 1 – 10 and their
respective geometry. Include bond angles in your answer.
Br
H
H
C
1
2
H
3
3
O
Cl
C3
C
C
C
sp
C‐1: _____ C‐3: _____
sp2
sp
sp
C‐2: _____ C‐4: _____
H
C
O
Solution
H
4
F
C
Hybridization
11
9
7
C
6
8
C
10
N
H
Geometry Bond angles
Carbon 1, 2 and 6:
sp3
Tetrahedral Carbon 3, 4 and 9: sp2
Trigonal planar Carbon 7, 8 and 10:
sp
Linear
109.5°
120°
180°
12
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Organic Nomenclature
Hydrocarbons
• Four basic types:
• Three parts to a compound name:
‐ Alkanes
CnH2n+2
1
C2H6 Ethane
‐ Alkenes
3
2
1. Prefix
CnH2n
C2H4 Ethene
2. Base
120°
‐ Alkynes
CnHn
C2H2 Ethyne
3. Suffix
‐ Aromatic hydrocarbons
CnH2n‐2 C6H6 Benzene
or
H
H
H
H
H
 Chapter 2: “Organic Chemistry”, 5th Edition , Bruice P. Y.
H
Organic Nomenclature
 Chapter 21: pg. 807‐9: “Chemistry: The Central Science: A Broad Perspective”, 2nd Edition, LeMay et al.
Organic Nomenclature
Suffix: Base/parent: Tells what type of compound it is.
Tells how many carbons are in the longest continuous chain.
suffix
base
What family?
15
suffix
How many carbons?
What family?
16
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Organic Nomenclature ‐ IUPAC Rules
Alkanes
Prefix: Tells what substituent(s) are attached, if any.
prefix
What substituent?
base
suffix
How many carbons?
What family?
Properties
17
To Name a Compound…
3. Number the chain from the end nearest the first substituent encountered.
H2C
Name??
CH
H3C
CH2
CH3
3‐Methylheptane 2‐Ethylhexane
CH3
CH
CH2
CH
CH
CH3
CH
CH3
4. List the substituents as a prefix
in alphabetical order along with the number(s) of the carbon(s) to which they are attached.
CH2
E1
CH3
2. Find the longest chain in the molecule. ‐ Base
CH2
Only van der Waals force: London force.
Boiling point increases with length of chain.
Combust to give mainly CO2 and H2O
Nomenclature suffix “‐ane”
Example
1. Determine what type of compound it is. ‐ Suffix
H3 C
•
•
•
•
19
CH2
CH3
CH3
 If there is more than one type of
substituent in the molecule, list them
alphabetically i.e. name of substituent,
not prefix for frequency e.g. di, tri,
tetra, etc...are not considered.
3‐Ethyl‐2,4,5‐trimethylheptane
5
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CYCLIC ALKANES
Cycloalkanes
• Carbon can also form ringed structures.
How do we name….
• Five‐ and six‐membered rings are most stable.
methyl
CH3
– Can take on conformation in which angles are
very close to tetrahedral angle.
General formula
CnH2n
4
1‐Ethyl‐3‐methylcyclohexane
cyclohexane
cyclopentane
H
H
C
H
H
H
H
C
C
H
H
C
C
C
C
C
C
H
1
Alkenes
VSEPR Theory
O
H
C
6
2
22
Carbon can form multiple bonds with itself or other
atoms, e.g. N or O.
H
5
3
cyclopropane
Unsaturated Hydrocarbons

ethyl
CH2CH3
Ethylmethylcyclohexane
– Smaller rings are quite strained.
H
C
H3 C
CH 3
C
H
H 3C
C
H3 C
120°
N
S
CH3
CH 3

Multiple bonds affect, physical and chemical
properties (reactivity of organic molecules).

Hydrocarbons that contain one or more double or
triple bonds are called unsaturated hydrocarbons.
• Contain at least one carbon–carbon double bond
• Unsaturated
– Have fewer than maximum number of hydrogens
– The C atoms on double bond are sp2 hybridized
 Chapter 23: “Chemistry: The Central Science: A Broad Perspective”
2nd Edition, by LeMay et al.
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Structure of Alkenes
Structure of Alkenes
 This creates geometric isomers
• Unlike alkanes, alkenes cannot rotate freely about the double bond.
– Side‐to‐side overlap makes this impossible without breaking ‐bond.
C
 difference in the spatial arrangement of groups about the double bond
C
Z‐2‐Pentene
 Cis‐ isomer “Z”‐isomer
 Trans‐ isomer “E”‐isomer
E‐2‐Pentene
25
26
Example
NAMING ALKENES
1. Find the longest unbranched carbon chain containing the double bond. • Name chain according to number of carbon atoms.
Name this alkene
 add ‐ene as a suffix
2. Number the carbon atoms in the main chain. • Start from the end of the chain that is closest to the
double bond.
 location of the double bond is numbered with the
CH3
7
1. The longest unbranched chain containing the double bond is seven carbons long, so this is a heptene.
2. The chain numbering starts closest to the double bond.
 Carbon #2 is the lowest‐numbered carbon in the
double bond, so this is a 2‐heptene or hept‐2‐ene
lowest‐numbered carbon in the double bond.
27
H
1
2
CH3
CH3 C 3 4
C CH CH CH2
5
6
H
CH
H2C CH2
28
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Properties of Alkenes
3. There are two substituent groups on this alkene:
1
CH3
H2C
Example: C4H8
5‐methyl
H
2
C 3 4
C CH
H
CH
CH3
5
CH CH2
6
CH2
7
CH3
4‐cyclopropyl
Compose the name…..
• Add the substituent groups alphabetically to the alkene name
2‐Methyl‐1‐propene
bp. ‐7 ⁰C
1‐Butene
bp ‐6 ⁰C
Cis‐2‐Butene
bp +4 ⁰C
Trans‐2‐Butene
bp +1 ⁰C
• Specify the position of each group on the main chain
 Structure also affects physical properties of alkenes
 the alkene is:
4‐cyclopropyl‐5‐methyl‐2‐hept ene
Can we have more than one double bond?
Chapter 22.2: “Chemistry: The Central Science: A Broad Perspective” 2nd Edition
29
Geometric isomers of Alkenes
GEOMETRICAL ISOMERS
Alkenes exhibit cis‐trans isomerism.
Trans‐ isomer “E‐”
Cis‐isomer “Z‐”
• Identical substituents on
opposite sides of the double
bond
H trans‐ CH3
C
CH3
“E‐”
• Identical substituents on
same side of the double
bond
cis‐
C
CH3
CH3
C
H
H
“Z‐”
1,3‐butadiene
C
• Cis‐alkenes have similar higher priority elements or group in
the chain on the same side of the molecule (or Z‐isomer i.e.
have higher priority elements but not necessarily the same on
the same side of the molecule)
• Trans‐alkenes have similar higher priority elements or group in
the chain on opposite sides of the molecule (or E‐isomer i.e.
have higher priority elements but not necessarily the same on
opposite sides of the molecule).
H
e.g.
Stick diagram
 Priorities are assigned by the atomic numbers of the atoms
bonded to the carbon in the double bond.
C5H10
31
1‐Pentene
(Z)‐2‐Pentene
(E)‐2‐Pentene
32
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Examples
FUNCTIONAL GROUPS
1. Name the following alkenes and determine whether there
are geometric isomers are either Trans‐ (E‐) or Cis‐ (Z‐)
isomers if applicable.
1
2
1
a) b)
2
1
1
 Each functional group is specified by a suffix or prefix
2
H
(E)‐ 3‐Methylhex‐3‐ene
(Z)‐ 3‐Bromopent‐2‐ene
Equal priority
c)
 No geometric isomers formed.
Group /
Formula
Family
Alkane
RH
Structural
Formula
Prefix
Suffix
alkyl-
-ane
33
Example
Ethane
Alkene
R2C=CR2
alkenyl-
 depicted on the nomenclature of the organic molecule
 Functional groups are given an order of priority to decide on which is the suffix.
HOMEWORK!
3‐Fluoro‐2‐methylpent‐2‐ene
F
-ene
Hint: Please refer to your textbooks !!
• “Chemistry: The Central Science: A Broad Perspective”, 2nd Edition, by LeMay et
al. Chapter 24.1, 25.1, 26.1 & 26.3
• “Organic Chemistry” 5th Edition, by Bruice
34
• Hart
et al. “Organic Chemistry: A Short Course” 12th Edition
Group /
Family
Formula
Haloalkane
Structural
Formula
Prefix
Suffix
RX
halo-
alkyl
halide
Chloroethane
Ethyl chloride
Fluoroalkane
RF
fluoro-
alkyl
fluoride
Fluoromethane
Methyl fluoride
Chloroalkane
RCl
chloro-
alkyl
chloride
Chloromethane
Methyl chloride
Bromoalkane
RBr
bromo-
alkyl
bromide
Iodoalkane
RI
iodo-
alkyl
iodide
Ethene
Alkyne
RC≡CR'
alkynyl-
-yne
Ethyne
Benzene
derivative
RC6H5
RPh
phenyl-
FUNCTION
Acidic, basic, alcohol, etc…
The GROUPS are called functional groups.
2
H
Br
 Certain groups of atoms give a molecule a....
-benzene
2-phenylpropane
isopropylbenzene
Example
Bromomethane
Methyl bromide
Iodomethane
Methyl iodide
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Group /
Family
Group
Formula
Primary
amine
(1º)
RNH2
Secondary
amine
(2º)
R2NH
Tertiary
amine
(3º)
R3N
Amines
Structural
Formula
Prefix
Suffix
amino-
-amine
Example
Group /
Family
Formula
Methylamine
Alcohol
ROH
Group /
Family
Formula
Carboxylic
acid
RCOOH
Acyl halide
Ester
ROR'
RCOOR'
hydroxy-
-ol
Example
Dimethylamine
amino-
Ketone
RCOR'
keto-, oxo-
-one
Aldehyde
RCHO
aldo-
-al
2-Butanone
Methylethyl ketone
-amine
Trimethylamine
CH3
N
R4N+X-
-ammonium
ammonio-
Structural
Formula
Prefix
Suffix
carboxy-
-oic acid
H
Cl
CH3
Trimethylammonium
chloride
Example
haloformyl-
Ethanal
Acetaldehyde
Primary (1°) alcohols and amines
alkoxy-
Alcohol
H
1
R
alkyl
alkanoate
C
H
OH
H
-oyl halide
alkyl alkyl
ether
Example
General structure
Ethanoyl chloride
Acetyl chloride
Ether
Suffix
-amine
Ethanoic acid
Acetic acid
RCOX
Prefix
Methanol
amino-
H 3C
Quaternary
ammonium
ion
Structural
Formula
1° alcohols
CH3CH2
Propan‐1‐ol
1‐Propanol
C
H
Ethoxy ethane
Diethyl ether
Ethyl butanoate
Ethyl butyrate
Amine
1
R
N
H
1° amine
OH
H
H
CH3CH2
N
Ethylamine
H
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Secondary (2°) alcohols and amines
H
1
R
OH
R2
2° Alcohol
CH3CH2
1
R
Alcohol
C OH
R
1
R
C
H
CH3CH2
N
R2
2° Amine
CH3
OH
R2
3° alcohols
Butan‐2‐ol CH3
2‐Butanol
H
Amine
3
H
C
Amine
CH3
+
R4 N
1
R
N
1
R
N
CH3
Ethyldimethylamine
42
Exercise Example
CH3CH2
CH3
CH3CH2
3° Amine
1. Draw the structures of the following alcohols and amines and classify them as either 1°, 2°, 3° or quaternary CH3
R3
C OH
CH3
2‐Methylbutan‐2‐ol
2‐Methyl‐2‐butanol
R2
Ethylmethylamine
R2
Amine
CH3CH2
R3
N
Quaternary amines
General structure
Example
General structure
Example
General structure
Alcohol
Tertiary (3°) alcohols and amines
N
+
a) Pentan‐1‐ol
CH3
b) Dimethylamine
CH3
Ethyltrimethylammonium ion
c)
3‐Ethylhexan‐3‐ol
d) Diethylmethylamine
e)
Butan‐2‐ol
f) Triethylmethyl ammonium ion
43
44
11
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Solutions
a) Pentan‐1‐ol
H
HO
C
H
H
H
C
C
H
H
H
C
H
b) Dimethylamine
H
C
H
CH2 C
CH3
2° amine
e) Butan‐2‐ol
CH3
CH2CH3
H
CH2CH3
C
2° alcohol
OH
CH2
3° alcohol
CH2
CH3
f) Triethylmethyl
ammonium ion
CH2CH3
H3C
CH2CH3
OH
3° amine
CH3
N
CH3CH2
Name this organic molecule:
CH3
OH
NO2
CH
CH
 Specify the position of each group on the main chain.
 Add the substituent groups alphabetically to the name of the alkane (or alkene or alkyne) along with the frequency of each group.
CH2CH3
Example…..
Quaternary amine
A45
Example:
Naming Hydrocarbons with Functional Groups
 Name the other substituent groups, using the
prefixes for alkyl groups and the prefixes for any
other functional groups
CH2
N
CH3
H
d) Diethylmethylamine
N
CH3
H
1° alcohol
H3C
c) 3‐Ethylhexan‐3‐ol
46
Example:
Name this organic molecule:
OH
CH3
CH3 CH
NO2
CH
CH3
1. This molecule contains a hydroxyl and a nitro group
Only the hydroxyl group has priority.
So this is an ALCOHOL…..
47
48
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Example:
Name this organic molecule:
OH
CH3 CH
OH
NO2
CH
CH3 CH
1
2
CH3
1. This molecule contains a hydroxyl group and a nitro group
this is an ALCOHOL.

 The hydroxyl group is on position 2
4. This molecule has one substituent
A nitro group on position 3
Therefore its a “butanol”.
49
3‐nitro‐2‐butanol
CH3
1.
CH CH2
H2N
CH3
CH2
C
CH2
CH3
CH2
O
CH
CH3
C
OH
CH
CH3
CH O
CH
CH2
C
OH
CH3
CH3
Identify functional groups
carboxyl group
Highest priority
CH3
CH2
CH
3‐nitrobutan‐2‐ol
Name this organic molecule:
CH3
CH3
or
50
Example:
Name this organic molecule
C
Butanol
CH3
4
3. Number the carbons, starting NEAREST the .
functional group
 An alcohol suffix is ‐OL
H2N
CH
3
…so this is a 2‐butanol or butan‐2‐ol
longest carbon chain containing the hydroxyl group has four carbons.
Example:
NO2
1. Molecule contains a carboxyl group and an amino group.
The carboxyl group has highest priority, hence amino group
a carboxylic acid
Now determine longest carbon chain
51
52
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Example:
Name this organic molecule:
Example:
Name this organic molecule
CH3
H2N
C
CH3
CH2
CH3
CH2
CH3
CH
CH C
CH2
C
CH3
OH
CH
O
CH CH
C
OH
CH2
CH3
CH3
CH3
The longest carbon chain has 8 carbons.
The longest carbon CONTAINING the ‐CO2H
chain has 7 carbons.
BUT………………...
so this molecule is based on a heptane.
THIS CHAIN DOES NOT CONTAIN THE ‐CO2H group
53
54
Heptanoic acid
CH3
H2N
C
CH2
CH3
CH
CH3
CH CH3
H2N
5
C
6
CH2
7
CH3
4
CH2
CH3
CH2
O
CH
CH2
C
CH3
OH
CH3
Highest priority functional group is a carboxyl group CH
CH3
3
CH 2
CH
CH2
1
O
C
OH
CH3
3. Number the carbons starting with the functional group
The carboxyl group is on position 1,
Suffix ‐OIC ACID
Hence… Heptanoic acid
Now number chain…….
55
CH3
CH2
O
CH
CH2
CH3
H2N
do not include in the name because
the carboxyl group is always a terminal group.
56
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Heptanoic acid
CH3 5‐methyl
5
H2N
5‐amino
C
6
CH2
WHAT IS THE STRUCTURAL FORMULA OF
7
CH3
4
CH2
CH3
3
CH
CH 2
CH
CH3
3‐isopropyl
1
KETONE GROUP
OH
CH2
CH3
BUTANONE ?
O
C
CH3
4. This molecule has four substituents
i.e. at carbons 2, 3 and two at 5
C
H
INTERPETING AN IUPAC NAME…...
O
Br
d)
Alcohol
(2°)
OH
(Z)‐3‐methylhex‐2‐ene
CH2
3,3‐dimethylbutan‐2‐ol
H
CH3
Ketone
CH3
O
h)
Carboxylic Cl
acid
3‐methylhexan‐2‐one
O
i)
Ester
OH
O
2‐chloropentanoic acid
59
g)
2‐ethylpentanal
6‐bromo‐4‐ethyl‐2,3,7‐trimethyloctane
H
CH2
Aldehyde
f)
Alkene
c)
BUTANAL
Give the correct IUPAC name for the following compounds.
Bromoalkane
Alkylbromide
2,3,7‐trimethyloctane
WHY?????
Examples
Give the correct IUPAC name for the following compounds.
b)
CH3
58
Examples
Alkane
CH2
O
5‐amino‐ 2‐ethyl‐ 3‐isopropyl‐ 5‐methyl heptanoic acid
a)
C
CANNOT BE AT END!!!!
Compose the name
ALPHABETICAL LIST
57
O
4 CARBONS
2‐ethyl
O
ethylbutanoate
60
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Examples
Solutions
Draw structural formulae for the following compounds:
a) 2,3,5‐trimethylhexane
a) 2,3,5‐trimethylhexane
b) (Z)‐3‐chlorohept‐2‐ene
Cl
b) (Z)‐3‐chlorohept‐2‐ene
c) 3‐ethylnonanol
H
d) 2,3‐dimethylpentanoic acid
d) 2,3‐dimethylpentanoic acid
c) 3‐ethylnonanol
e) methyl hexanoate
O
f) 3‐iodohexanal
OH
g) pentan‐2‐one
h) 3‐aminopentane
61
OH
62
Solutions
e) methyl hexanoate
ISOMERS
Two types: (a) Structural isomers (b) Stereoisomers
f) 3‐iodohexanal
O
I
O
(a) Structural isomers
O
 Molecules with the same chemical formula but different bonds between the atoms
Now called... CONSTITUTIONAL ISOMERS
H
 Constitutional isomers have different properties:
g) pentan‐2‐one
e.g. butane (C4H10) has 2 structural isomers
h) 3‐aminopentane
These are…..?
O
n‐Butane: C4H10
H2 N
CH3 CH2 CH2 CH3
63
bp = 0 °C mp = ‐138 °C
2‐methylpropane: C4H10
CH3 CH
CH3
CH3 bp = ‐12 °C mp = ‐159 °C
64
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Example
CONSTITUTIONAL (STRUCTURAL) ISOMERS
How many constitutional isomers are formed from C5H12? Draw their structures.
Solution:
THREE CONSTITUTIONAL ISOMERS
1) CH3 CH2 CH2 CH2 CH3
2)
CH3 CH CH2 CH3
CH3
2‐methylbutane
n‐pentane
3)
The general formula for ALKANES is…...
n = 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 ISOMERS
n = 40 62,491,178,805,831 ISOMERS
CH3
C n H 2n 2
n = 8 18 ISOMERS
n = 9 35 ISOMERS
CH3 C CH3
HOMEWORK: DRAW THE ISOMERS OF C40H82 !!!
CH3 2,2‐dimethylpropane
n = 10 75 ISOMERS
66
n = 20 366,319 ISOMERS
65
Constitutional isomers for multibonds
Constitutional isomers for multibonds
Example: Alkene
Example: Alkyne
C6H10
C6H6
(E,E)‐2,4‐hexadiene
H3 C
(E)‐1,3‐hexadiene
C
C
C
C
CH3
2,4‐hexadiyne
hexa‐2,4‐diyne
(Z,Z)‐2,4‐hexadiene
CH3
H
(Z)‐1,3‐hexadiene
C
C
C
C
CH2
1,3‐hexadiyne
hexa‐1,3‐diyne
(Z,E)‐2,4‐hexadiene
1,5‐hexadiene
67
68
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OPTICAL ACTIVITY IN ORGANIC COMPOUNDS
OPTICAL ISOMERISM
 Optical isomerism arises when molecules have a
structure such that the mirror image is not
superimposable on the original molecule.
 Stereogenic centre has four different groups attached
to a tetrahedral carbon atom
W
Stereogenic centre
 occurs whenever there are four different groups bound to the same tetrahedral carbon atom.
 The carbon involved is called a chiral carbon or
stereogenic carbon and the molecule is known as a chiral
molecule.
Example: 2‐butanol
70
Perspective formula
Fischer projection
Wedge‐dash notation
2‐Butanol
2‐Butanol
Solid wedge represents a bond extending out towards the viewer
Simple line for Bonds aligned to the asymmetric center in the plane
C*
H3C
H
X
Y
69
OH
C*
Z
Some terminology…….
Dash shows bond going backwards from the viewer
Chiral carbon atom
Bond going backwards from the viewer
HO
Bond extending out towards the viewer
CH2CH3
CH3
*
Bonds aligned to the asymmetric center in the plane
CH2CH3
H
Chiral centre
Chiral centre
71
72
18
10/4/2011
Question
ENANTIOMERISM in ORGANIC CHEMISTRY
Identify the chiral carbon (stereogenic centre) and draw the
structural formula for each of the following molecules:
• 2‐butanol has two optical isomers.
• A pair of isomers called enantiomers ‐
(non‐superimposable mirror images of each other)
(a) 1‐chloroethanol
H
OH
HO
C
C
H
CH3CH2
CH2CH3
(d) 1,3‐dimethlycyclopentane
CH3
H
E
Chapter 22.3: “Chemistry: The Central Science: A Broad Perspective” 2nd
(b) 2,3,5‐trimethylhexane
*
View in 3‐D
OH
H3C
(b) 2,3,5‐trimethylhexane
Solution
“mirror”
View in 3‐D
(a) 1‐chloroethanol
(c) Methylcyclohexane
C* CH3
Cl
(c) methylcyclohexane
(d) 1,3‐dimethlycyclopentane
No chiral carbon
*
*
74
Edition
NAMING OPTICAL ISOMERS
CAHN‐PRELOG‐INGOLD R,S‐NOMENCLATURE
 Stereogenic centre creates two molecular optical isomers
1. Assign relative priorities to each of the four groups on the stereogenic carbon to describe the configuration.
Cl
C*
H
Enantiomers
*C
OH
H3C
Two configurations
Cl
HO
H
CH3
How do we name these isomers??
Solution: Use R‐S nomenclature system for designating
the configuration
We assign priorities as in the E, Z system……...
75
The priorities are given by rules:
• Higher atomic numbers are given higher priorities.
• If necessary, the second atom in each substituent is used to determine the priorities.
Chapter 22.2: “Chemistry: The Central Science: A Broad Perspective” 2nd Edition
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10/4/2011
CAHN‐PRELOG‐INGOLD R,S‐NOMENCLATURE
2‐ butanol
OH
2. Draw the molecule with the lowest‐priority group pointing directly into the page….
CH3
CH2 CH3
H
OH
and the other three groups pointing out of the page in an C
Chapter 22; pg. 829‐839. “Chemistry: The Central Science: th Edition.
A Broad Perspective” 2
p
OH
arrangement like a steering wheel or Mercedes‐Benz symbol
C
Example: 2‐Butanol…..
CH3
C
CH2 CH3
H
Draw molecule as a wedge and dashed line diagram
2‐ butanol
H
OH
CH3
H
CH3
THESE ARE THE TWO…. ENANTIOMERS
...or ....OPTICAL ISOMERS
78
2‐Butanol
1 OH
1. Assign priorities to each group:
O > CMe = CEt > H
CH2CH3
H3CH2C
DO NOT FORGET THE OTHER ISOMER……….
1. Assign priorities:
C
CH2CH3
H3C
OH
C
C
H
C
4 H
C,H,H > H,H,H
3
CEt > CMe
CH3
2
CH2CH3
O > CEt > CMe > H
1 OH
C
H3C
3
2
2. Redraw the molecule with the lowest priority group facing in.
CH2CH3
Now what?????
79
80
20
10/4/2011
1
C
H3C
1
2‐Butanol
OH
2
C
2
CH3
CH3CH2
CH2CH3
3
2‐Butanol
OH
3
(R)‐2‐Butanol
(S)‐2‐Butanol
3. Look at the direction in which the priorities decrease.
 If they decrease in a clockwise direction, the stereogenic centre is called “R” or rectus
 If the priorities decrease in a counter‐clockwise
direction the stereogenic centre is called “S” or sinister, which is Latin for “left.”
which is Latin for “right.”
Or….
81
82
What are the configurations of the following chiral molecules?
Example:
NH2
N > CEt > CMe > H
C
H
CH3
CH2CH3
(R)‐2‐butanamine
1
2
H2N
CH2CH3
centre in each of the following molecules:
NH2
H3C
C OH
CH2CH3
CH2OH
HO C
O
(A)
2
C
H
Cl
(S)‐2‐butanamine
3
C
H
CH2CH3
H
Cl
CH2OH
C
H3C
(B)
O
2‐chloro‐3‐hydroxy‐propanoic acid
C
C
A83
Example: Give the configuration of the stereogenic
THEY LOOK DIFFERENT BUT ARE THEY???
3
CH3
1
NH2
84
21
10/4/2011
1. Assign priorities to each group:
A)
CH2OH
4
3
O,O,O > O,H,H
Cl > Ccarb > Calc > H 85
C OH
Cl
1
4
C
H
1
3
C
C OH
1
The priorities decrease anti‐clockwise, so this centre is “S”
(S)‐2‐chloro‐3‐hydroxy‐propanoic acid
Example: Give the configuration of the stereogenic centre in each of the following molecules:
O
3
CH2OH
A)
Cl
Cl
B)
C
H
3. Direction of the priorities decrease?
CH2OH
3. Look which way the priorities decrease.
O
Cl
C
2
Cl
2. Redraw the molecule with the lowest priority group facing in.
O
HO C
2
Cl > Ccarb > Calc > H CH2OH
HO C
2
1. Assign priorities to each group
2. Redraw the molecule with the lowest priority group facing in.
CH2OH
86
B)
Cl > Ccarb > Calc > H O
1
Ccarbox > Calc
1. Assign priorities to each group:
2
Cl
C OH
O
Go to next atoms on tied carbons
Cl
CH2OH
C
H
Cl > Calc = Ccarb > H C
H
3
A)
H
C OH
O
C
CH2OH
HO C
O
2‐chloro‐3‐hydroxy‐propanoic acid
 The priorities decrease anti‐
clockwise, so this centre is “S”
THEY LOOK DIFFERENT BUT ARE THEY???
WE JUST FOUND OUT THAT THEY ARE IDENTICAL!!!!!
(S)‐2‐chloro‐3‐hydroxy‐propanoic acid
88
WHAT IF?????
22
10/4/2011
If we flip the COOH and H ???????
A)
A’)
CH2OH
C
H
HO
C OH
O
Cl
3
CH2OH
C
3
HO
C
2
O
“S” configuration
2
2
H3C
C
Cl
O
“R” configuration 89
CH3
Cl 1
Draw steering wheel with lowest priority group pointing in…….
OH
1
Cl
91
MAKE SURE PRIORITY GOES CLOCKWISE FOR “R” configuration
2
C
CH2CH3
3
DRAW MOLECULE…..
H3C
C
3
CH2 CH3
1
ASSIGN PRIORITIES
90
OH
Cl
3
CH2
OH
Cl
1
OH
4
1
Assign priorities
(R)‐2‐chloro‐2‐butanol
4
DRAW MOLECULAR STRUCTURE
Solution:
CH2OH
C
2
H
What is the structural formula of (R)‐2‐chloro‐2‐butanol?
Hint: Make use of perspective formula or dash‐wedge notation.
Cl
O
C OH
1
C
C
CH2OH
Cl
EXAM QUESTION
2
DRAW MOLECULE TO SEE ALL GROUPS…..
C
CH2CH3
3
 Remember lowest priority group has a “dashed” bond
Make sure priority goes clockwise for “R”
OH
Perspective formula
or
Dash‐wedge notation
C
H3C
Cl
CH2CH3
(R)‐2‐chloro‐2‐butanol
92
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10/4/2011
* *
Diastereomers
CH3 CH CH CH3
Br Cl
[2R,3R]
a
Br
R
b
CH3
C H
H C Cl
R
S
c
CH3
Br
H C Br
Cl
CH3
S
Cl
C H
a & d
[2R,3S] [2S,3R] [2S,3S]
R
S
CH3
CH3
d CH3
C H
H C Br
C H
H C Cl
S
R
CH3
CH3
enantiomeric pairs enantiomeric pairs • Enantiomeric pair differ only in optical activity
Chapter 22.2, page 838‐9: “Chemistry: The Central Science: A Broad Perspective”
Meso compound
O
93
O
* *
a CH3
Br
R
C H
H C Cl
R
CH3
R
H C OH
HO C H
R
CO2H
Enantiomeric 170 °C
+12°
pairs S
HO C H
H C OH
S
CO2H
170 °C ‐12°
R
CO2H
S
CO2H
H C OH
HO C
H
H C OH
HO C
H
S
c
H C Br
Br
S
Cl
S
C H
CH3
Cl
R
S
CH3
d CH
3
C H
H C Br
C H
H C Cl
CH3
S
R
CH3
Meso compound
HOC CH CH COH
CO2H
b CH3
b & d
 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 in
maximum of 2n stereisomeric forms, with maximum of
2n/2 enantiomeric pairs
94
OH OH
CO2H
b & c
a & c
CO2H
140 °C
0°
R
CO2H
Meso compounds
 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…(!)
95
R
CO2H
S
CO2H
R
CO2H
S
CO2H
H C OH
HO C H
H C OH
HO C H
-------------------- -------------------- -------------------- -------------------HO C H
H C OH
H C OH
HO C H
R
S
CO2H
CO2H
Enantiomers, Chiral
S
CO2H
R
CO2H
Identical, achiral, Meso form
 possess plane of symmetry bisecting central C‐C bond
96
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Newman projections
Cyclohexane conformations
 Shows arrangements in space e.g. Ethane C2H6
 Staggered
H
conformation
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
“dash‐wedge”
Examples
H
CH3
H
H
H
H
H
3
4
H
H
Cis‐1,2‐Dimethylcyclopentane
a 2
e
e
a
a
a
H
H
H
H
H
H
H
H
Cyclohexane
H
H
Boat conformation
Summary of isomerism
98
Hart et al. “Organic Chemistry: A Short Course” 12th Edition, page 52‐54
Different bond pattern H
Structural (constitutional) isomer
isomers
H
H
1
e
e
H
H
H
e
6
3
4
e
e
a
a
5
Chair conformations
H
H
H
H
Newman97
H
e
"flip"
e
H
H
H
a
5
H
“sawhorse”
CH3
a
0°
H
H
2
e
H
1,2‐Dimethylcyclopentane
CH3
e
6
HH
H
H
e
1
a
H
Newman
H
H
H
H
H
“dash‐wedge” “sawhorse”
 Eclipsed conformation
60°
a
a
a
Same bond pattern
CH3
Trans‐1,2‐Dimethylcyclopentane
Stereoisomer
Interconvertible by single bond rotation
99
Conformers (rotamers)
Not interconvertible
by bond rotation
Configurational isomers
100
25
10/4/2011
Exercise
Solutions
1. Draw the structures of
1. (a) (Z)‐3‐methyl‐2‐pentene
(b) (S)‐2‐bromopropanol
H
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 of a
staggered conformation of butane.
CH3
CH3
H2C
(c)
(E,Z)‐2,4‐heptadiene
H
CH3
H3C
Solutions
C
H
C
HO
H
Br
102
(4) 1‐bromo‐2‐chlorocyclopropane
Staggered conformation of butane
CH3 60°
H
H
H
Cis‐1‐bromo‐2‐
chlorocyclopropane
Newman projection
H
H
Br
Cl
CH3
H
H
*
COOH
H
H
(S)‐2‐methylpentanoic acid
H 3C
OH
(d) (2S,3R)‐3‐bromobutan‐2‐ol 4. Draw the structure for the cis‐ and trans‐isomers of
101
1‐bromo‐2‐chlorocyclopropane
3.
H
C
Br
CH3
3. Using the Fischer projection draw a structure of
(S)‐2‐methylpentanoic acid.
2.
C
H
H3C
Trans‐1‐bromo‐2‐
chlorocyclopropane
H
Br
H
H
Cl
H
Fischer projection C H 2C H 2 C H 3
http://cheminnerweb.ukzn.ac.za/Firstyear/chemonetwenty.aspx
103
104
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Five important organic reactions

Substitution reaction:
Arrow Notation
Substitution of one atom or group of
 Uses
 shows redistribution of valence electrons a molecule
atoms by another atom or group of atoms.

Addition reaction:
 defining resonance
Addition of one molecule or atom to another
 chemical reaction direction
to give a new molecule . normally on double or triple bond.

 defining a reaction mechanism.
Elimination reaction: Elimination of two atoms or group of atoms
from a molecule (reverse of the addition reaction).

Oxidation and reduction reactions: Involve the loss or gain of

Two common types of electron redistribution:

From a bond to an adjacent atom, and

From an atom to an adjacent bond.
electron density by a carbon respectively.

Rearrangement reaction: Conversion of one structure to an
isomeric structure.
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Terminology
Bond breaking
 Nucleophile (Nu):- electron rich species and/or negatively
charged species
A
B
A
+
B
A
B
A+ +
B-
Bond forming
A
+
B
A
Homolytic bond forming
B
A+ + B-
O
-OH, Br-, -NH2,
Heterolytic bond breaking
Homolytic bond breaking
examples
O
C
 Electrophile (E+):- electron poor species and/or positively
charged species
A
Heterolytic bond forming
B
examples
O
H+, M+, H3O+
C
C
C
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
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10/4/2011
Reactions of Alkanes
PowerPoint to accompany
Chapter 21
Alkanes:
Scaffolds for
Organic
Chemistry

Because of
unreactive.

Forcing conditions are required for a reaction


structural
role,
alkanes
are
relatively
They make excellent nonpolar solvents.

Combustion: give mainly H2O and CO2

Free-radical reactions: chain reactions
In order to be able to predict the product of a reaction, it is
important to be able to distinguish between carbon (and
hydrogen) types in an alkane.

Primary, secondary, tertiary and quaternary carbon atoms.

Primary, secondary and tertiary hydrogens.
Combustion
Classification
 hydrogen and carbon atoms in alkanes
its


Combustion of alkane gives CO2 and H2O.
Combustion reactions are exothermic.
e.g. ethane
2C2H6(g) + 7O2(g)
4CO2(g) + 6H2O(l)
∆H = -2855 kJ
Question
Write a balance equation for the combustion of butane
Solution
Figure 21.25
C4H10(g) + 6½ O2(g)
4 CO2(g) + 5 H2O(l)
2C4H10(g) + 13 O2(g)
8 CO2(g) + 10 H2O(l)
Homework Try combustion of pentane C5H12 ??
28
10/4/2011
Combustion - Napalm


The commonly quoted composition is 21% benzene,
33% gasoline, and 46% polystyrene. This mixture is
difficult to ignite.
Free-Radical Reactions

Consider the following halogenation reactions:
Reaction 1
Movie – Vietnam.
dark
CH4 + Cl2
no reaction
Reaction 2
CH4 + Cl2
•
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

The free-radical halogenation is also an example
of a radical chain reaction.

Such reactions are characterised by three steps
 Initiation
Light
λ
CH3Cl + CH2Cl2 + CHCl3 + CCl4
This type of reaction is called a substitution reaction.
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
 Propagation steps - a series of new reactions
which generate other radicals
e.g.
 Propagation
 Termination

radical
Initiation produces the first free-radicals by a
homolytic bond cleavage.
Cl2
Cl
Cl
λ
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
radical
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
29
10/4/2011
 Termination steps: stop the formation of free-radicals
neutral molecular species formed
e.g.
Free-Radical Reactions

Given the opportunity, free-radical reactions are regiospecific.

More so for bromination than chlorination.
radical pair
3°> 2° > 1°
most stable
least stable
1°> 2° > 3°
radical pair
most stable
least stable
R
C
R = akly group - EDG
X
C
e.g X = halide - EWG
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
PowerPoint to accompany
Chapter 23
Electrophilic addition and Substitution
reactions

The presence of carbon-carbon double or triple bonds in a
compound markedly increases its chemical reactivity.

Most characteristic reactions of alkene and alkynes are addition
reactions:
Alkenes and
Alkynes


Halogenation

Hydrohalogenation

Hydration
The electron-rich nature of a multiple bond leads to enhanced
reactivity with electrophiles (electron-loving species).
30
10/4/2011
Mechanism for HX Addition rxn

Two-steps mechanism
1.
First step is slow, rate-determining step.
2.
Second step is fast.
Alkenes
reactions
slow
H
Br
carbocation
Fast
Figure 23.20
carbocation
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Addition of HX

In the first step, -bond breaks and new C—H bond
and cation form.
Regioselectivity

A occurs when a non-symmetric alkene is used.

The preference based on intermediate carbocation stability:
 Primary


In the second step, a new bond forms between negative
Brown,
LeMay, Bursten,
Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
positive
carbon.
bromide ion and
<
(Less Stable)
Figure 23.22
Secondary
<
Tertiary
(Most Stable)
The regioselectivity is stated as Markovnikov’s Rule:
“In the addition of HX to an alkene, the hydrogen adds to the
carbon atom of the double bond bearing the greater number of
hydrogen atoms bonded directly to it.”
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
31
10/4/2011
Addition of HX
1. Which of the following is the most stable carbocation and why?
CH3
H
1
C
H
Exercise
H
+
CH 2
2
C
Br
Soln:
CH3
+
CH2
(A )
(D)
• It
is a tertiary
carbocation.
(B )
CH3
H 3C
+
C
CH3
• Inductive
effect
electron rich groups
(3˚)
from
CH3
+
(C )
Major
product
Minor
product
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Soln:
C
C
Cl
(A)
minor
product
CH3 H
H 3C
CH3
C
H 3C
HCl
2-methylbut-2-ene
 The hydration of alkenes and alkynes requires the addition of a
strong acid catalyst e.g. H2SO4
2-chloro-3-methylbutane
C
H
2. What will be the major product from the reaction by reflux in dark
of 2-methylbut-2-ene with hydrochloride in a carbon tetrachloride
solution? Name and draw the structure of the product.
Addition of H2O
CH3 CH3
H
(D )
hydronium ion
(electrophile)
CH 3 CH 3
CCl4
Cl
C
C
CH 3 H
H
(B)
major
product
 The first step is rate determining.
 Leads to the formation of a carbocation.
2-chloro-2-methylbutane
32
10/4/2011
Addition of H2O (Reversible)

Second step is fast

Carbocation reaction with excess H2O (nucleophile)

leads to the formation of an oxonium ion.
 The fact that the reaction is always described by equilibria
means that the overall reaction is reversible.
 That is, in the presence of H2SO4, alcohols can be dehydrated to
yield alkenes.
oxonium ion
 Loss of H+ from the oxonium ion leads to formation of an alcohol
and regenerates the hydronium ion, hence it is catalytic reaction.
butan-2-ol
2-butanol
hydronium ion
regenerated
Formation of Alkenes
Formation of Alkenes
 Previously, we have discussed the formation of
alcohols from alkenes.
 The reverse process is also possible.
 This dehydration reaction is also known as an
elimination reaction.
I- > H2O > Br- > Cl- >> F- > CH3COO- > OH Requires the elimination of a good leaving group.
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33
10/4/2011
Exercise
Halogenation
1. Draw and name the major product in the following reactions
OH
(a)
H+ / H2O
OH
(b)
H
 One of the easiest and most
dramatic reactions is the
addition of Br2 to an alkene.
pentan-2-ol
2-pentanol
H 2SO4
cyclohexene
Heat
 Colour change
 Useful test for alkenes
Figure 23.23
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Reactions of Alkynes
 Many similar reactions as alkenes.
 Reactions involve the replacement of -bonds with -bonds.
Example
H3 C
H3C
C
C
Br2
CH 3
CH2Cl2
Br
Br
C
C
Br
CH3
(E)-2,3-dibromobut-2-ene
(1)
CH 3
H3C
Br
(2)
CH2Cl2
Br
H2C
Cl
hex-2-yne
C
H2C
C
CH3
CH2
Cl2
C
F
Br
C
C
Br
Br 2
Br
F
1,1,2,2-tetrabromo1,2-difluoroethane
Br
C
Br
CH3
Cl
(E)-2,3-dichlorohex-2-ene
F
CH2Cl2
CH2
C
H3C
CH2Cl2
Br2
CH3 Br
2,2,3,3-tetrabromobutane
C
2. How would you form 1,1,2,2-tetrabromo-1,2-difluoroethane using
bromine as one of your reagent?
Solutions
Br
C
Problems
1. What will be the product for a reaction between hex-2-yne and one
mole of chlorine in dichloromethane under anhydrous condition? Draw
the structure of the product and provide it IUPAC name.
C
Br 2
CH2Cl 2
F
(E)-1,2-dibromo-1,2-difluoroethene
F
C
C
F
1,2-difluoroethyne
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Catalytic Hydrogenation

Reaction requires molecular hydrogen (H2) and a metal
catalyst
Problems
Provide the major organic product(s) in the reactions below
(1)
Br
(2)
Br 2
CH2Cl 2
HO
Br 2
(3)
H 2O
PowerPoint to accompany
Br
1,2-dibromo-3-ethylpentane
3-ethylpent-1-ene
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
3-ethylpentane
3-ethylpent-1-ene
 Ni, Pd, Pt, Ru typical metals used
 Because a gas, liquid and solid metal are used, this
reaction is known as a heterogeneous catalysis
H2
Pd/C
3-ethylpent-1-ene

Br
1-bromo-3-ethylpentan-2-ol
Conventions in Organic Chemistry
Chapter 24
Alcohols, Ethers
and Haloalkanes
Figure 24.1
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35
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Alcohol Properties

The difference in electronegativity between oxygen and
hydrogen allows alcohols to hydrogen bond. Compared to
alkanes they:
Alcohol Solubility

There are differences in solubility:

Like dissolves like
Table 24.2
 Are more soluble in water
 Have higher boiling and melting points
Figure 24.4
Ether Properties
Ether Nomenclature
 Ethers are more polar than alkanes but are unable
to hydrogen bond.
 They are chemically inert.
 Both these properties mean that ethers are
excellent solvents!
Figure 24.10
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36
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Reactions of Alcohols
Reactions of Alcohols
 Alcohols react with sodium metal to form
alkoxides and H2(g).
Figure 24.12
 Alkoxides are excellent nucleophiles and
strong bases.
Figure 24.13
Homework
Fill in the reagents and
name type of reaction !!
2 CH3CH2O
H
+
2 Na
2 CH3CH 2O-Na+ + H2(g)
Sodium ethoxide
Basicity of Alcohols
 An alcohol can act as a weak base, generating an oxonium ion
under acidic conditions, which then loses water to form
carbocation intermediate.
CH3CH2 O H
+
H+
H
O+
H
CH3CH2
CH3CH2+
+
oxonium ion
H2O
carbocation
2 CH3CH 2O
H
+
2 Na
2
CH3CH 2O-Na+
+ H2(g)
 The stability of the carbocation is an enormous driving force for
the nature of the reactivity of alcohols.
Sodium ethoxide
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Alcohols to Haloalkanes

Haloalkanes can be prepared from alcohols.

Haloalkanes are important for introducing alkyl groups in
chemical reactions.

Two reactions are common

Dehydration of Alcohols

Dehydration of alcohols is an acid catalysed process. The ratedetermining step is the elimination of water to form the corresponding
carbocation. Removal of a small group from a larger molecule is called
elimination reaction.

Water is classed as a good leaving group because of its stability and
relatively unreactive nature. Hydroxide ion, on the other hand, does not
eliminate easily and so is classed as a poor leaving group.
Depend on the type of alcohol

oxonium ion
Elimination reaction
Nucleophilic Substitution (Haloalkanes)
 Nucleophilic substitution reactions of haloalkanes proceed by attack of the
nucleophile at the carbon bearing the halogen.
 Because of the difference in electronegativity between carbon and any of
the halides, this bond is polarised such that the carbon becomes mildly
electrophilic.
 This is a SN2 or biomolecular (2), nucleophilic (N), substitution (S)
reaction.
Order of leaving group:
I- > H2O > Br- > Cl- >> F- >
CH3COO- > OH- > NH2-
Example
1. Show the reaction mechanism between sodium hydroxide
and 1-bromoethane.
Soln
H
C
H
Br
CH3
+
NaOH
mechanism
SN2
Mechanism
Figure 24.15
1-bromoethane.
Transition state
Transition state
ethanol
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University of KwaZulu-Natal, Westville Campus, Durban
2. Use reaction mechanisms to explain the why the inversion of the
carbon reaction centre is observed when (S)-2-bromobutane is
reacted with NaOH.
Examinations : November 2010
CHEMICAL REACTIVITY - CHEM120
Duration: 3 hours Total Marks for Examination: 100
Internal Examiners: V.O. Nyamori, H.G. Kruger, M.D. Bala, B.S. Martincigh,
P.G. Ndungu and G.D. Dawson
Soln
HO-
CH 2CH 3
C
H 3C
H
Br
(S)-2-bromobutan
CH2CH3
HO
C
Br
CH3
Transition state
_
QUESTION 6
Primary organic halides react with nucleophiles through a SN2
mechanism. In the mechanism inversion of the carbon reaction centre is
observed.
Provide the mechanism of the SN2 mechanism below.
(4)
CH2CH3
HO
C
H
CH3
H
(R)-butan-2-ol
C
H2C
H3C
Br
H
(a)
starting material
 Inversion:- the (S)-configuration now becomes (R)-configuration.
HO
SN2
(b)
transition state
(c)
product(s)
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Four Factors affecting the SN2 reaction rate
 Substrate
 The nucleophile attacks from the back of the substrate and hence to
maximise the rate of the SN2 reaction, the back of the substrate must
be as unhindered as possible.
 Order of reactivity: Primary > secondary > tertiary substrates
(Most reactive)
(not reactive)
 Nucleophile
 Solvent.

Solvents may or may not surround a nucleophile, thus hindering or
not hindering its approach to the carbon atom.

Polar aprotic solvents because polar protic solvents will be solvated
by the solvent hydrogen bonding to the nucleophile and thus
hindering it from attacking the carbon with the leaving group.
 Leaving group

The more stable the leaving group is, the more likely that it will take
the two electrons of its carbon-leaving group bond with it when the
nucleophile attacks the carbon.

Therefore, the weaker the leaving group is as a conjugate base,
and thus the stronger its corresponding acid, the better the leaving
group. Examples of good leaving groups are therefore the halides
(except fluoride) and tosylate, whereas HO- and H2N- are not
 Steric hindrance affects the nucleophile's strength e.g. methoxide
anion is better nucleophile compared to tert-butoxide.
 Nucleophile strength is also affected by charge and electronegativity:
nucleophilicity increases with increasing negative charge and
decreasing electronegativity, e.g. nucleophile OH- > H2O and I- > Br(in polar protic solvents).
39
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University of KwaZulu-Natal, Westville Campus, Durban
Examinations : November 2010
CHEM120 - CHEMICAL REACTIVITY
Section B
Example
QUESTION 6
Primary organic halides react with nucleophiles through a SN2 mechanism.
In the mechanism inversion of the carbon reaction centre is observed.
(4)
Provide the mechanism of the SN2 reaction below.
H
H 3C
C
H 2C
Br
H
(a)
starting material
HO
SN2
(b)
transition state
(c)
product(s)
Soln
Nucleophilic Substitution (Haloalkanes)
SN1 reaction mechanism
R1
Table 24.4
Nu
+
R3
C
R1
R2
C
R3
LG

This allows two different avenues for
the nucleophilic attack, one on either
side of the planar molecule.
R3

Nu or
R2
carbocation
R1
C
R1
Nu
and
Nu
C
R3
If the reaction takes place at a
R2
R2
stereocenter yielding a racemic
mix of enantiomers.
Racemic mixture: Both (R) and (S)-configuration
enantiomeric mixtures
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Scope of the reaction
Example

The SN1 mechanism tends to dominate when the central carbon
atom is surrounded by bulky groups. SN2 reaction hindered by steric
factor.

Bulky substituents on the central carbon favour carbocation
formation because of the relief of steric strain that occurs.

The resultant carbocation is also stabilized by both inductive
stabilization and hyperconjugation from attached alkyl groups.

The SN1 mechanism dominates in reactions at tertiary alkyl centers
and is further observed at secondary alkyl centers in the presence
of weak nucleophiles.

The carbocation intermediate formed in the reaction's rate limiting
step is an sp2 hybridized carbon with trigonal planar molecular
geometry.
Step 1
Separation of the leaving group
Carbocation
Step 2
Nucleophilic attack
oxonium ion
Step 3
Deprotonation
Product : 2-methylpropan-2-ol
Haloalkanes to Alkenes: -Elimination

The process of elimination of a good leaving group.

Haloalkanes undergo elimination under basic conditions. This
process is called -elimination reaction. It involves the
dehydrohalogenation of the haloalkane.

For the reaction to proceed, a hydrogen atom on the carbon atom
adjacent to the carbon bearing the halide must exist and be
removed.

The -elimination reaction is a E2 or biomolecular (2) elimination
(E) reaction.

When there are two sets of -hydrogens present in the
haloalkane, there is only one product formed and it is determined
by the Zaitsev’s Rule:

The major alkene formed by dehydrohalogenation is the more
substituted one.
major
minor
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HX
C2H5O-Na+
+
R
R
R
C
C
H
tetrasubstituted
HX
C2H5O-Na+
+
R
Substitution vs Elimination
R
X
R
R
C
C
H
X
 Both Substitution reactions (SN1 & SN2) and elimination are important,
however, each of these reactions does not work in isolation.
H
trisubstituted
HX
disubstituted
disubstituted
disubstituted
Increasing
degree of
substitution
R
C2H5O-Na+
HX +
C2H5O-Na+
+
R
monosubstituted
HX
+
H
H
C
C
H
X
C2H5O-Na+
H
ethene
Figure 24.16
R
H
H
H
C
C
H
X
 E.g. the addition of strong base to an alkyl halide yields an alkene:
R
H
C
C
H
H
X
H
C
C
H
X
H
R
 This elimination reaction competes with a substitution reaction which,
in this case, forms ethers:
H
X = Cl, Br, I
Substitution vs Elimination

Mechanism shows a trigonal planar intermediate formed by
concerted bond breaking and bond making processes

There are usually stereochemical consequences
Elimination vs Substitution

Thus, there is a competition between elimination and
substitution reactions and the result is dependant on:

Size of the alkoxide

The accessibility of the carbon bearing the halide

The accessibility of the -hydrogen
Figure 24.17
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Elimination vs Substitution

PowerPoint to accompany
We looked at the
hydration of alkenes
Chapter 25
 The reverse reaction
is also possible!
Aldehydes and
Ketones
 The presence of other
nucleophiles add another
dimension to the reaction.
Figure 24.19
Aldehydes
Carbonyl Compounds
 Contain C=O double bond.
O

Characterised by the functional group CHO.

Contains a carbonyl group

C
Both carbon and oxygen are sp2 hybridised.
 Include many classes of compounds:
aldehydes, ketones, carboxylic acids, esters, amides.
O
O
C
C
R
H
Aldehyde
R
Ketone
R'
R
O
O
C
C
O
H
Carboxylic acid
R
O
O
Ester
R'
R
C
Amide
N
R'
Figure 25.1
R''
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Aldehyde Nomenclature
Ketones
 Name the longest chain containing the carbonyl group
and replace ane with al.
 Name the longest chain containing the carbonyl
group and replace ane with one.
 Must, by definition, occur at one of the ends of a
molecule. Remember that aldehyde group takes priority
over both methyl and hydroxyl groups.
 Carbonyl group is bonded to alkyl groups on either
side.
OH
O
H
H
H
H
H
O
progesterone
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Ketone Nomenclature

Longest chain containing the carbonyl group.
Replace ane with one.

Constitutional isomers occur for C5 or greater.
H
O
O
camphor
testosterone
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Preparation of Aldehydes and Ketones
 Oxidation of alcohols using a range of reagents
• Most are chromium(VI) compounds (CrO3).
 Again, different types of alcohols yield different products
• Primary alcohols can lead to aldehydes and carboxylic acids.
• Secondary alcohols lead to ketones.
• Tertiary alcohols do not oxidise.
Figure 25.2
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44
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Preparation of Aldehydes and Ketones
Preparation of Aldehydes and Ketones
 Using pyridinium chlorochromate (PCC)
Preparation of Aldehydes and Ketones
ozonolysis
 Aldehydes and ketones can also be prepared in good yields by the
 Secondary alcohols can be oxidised by PCC, Jones
Reagent (CrO3) or K2Cr2O7 to yield the corresponding
ketone.
reaction of alkenes with ozone, followed by a mild reduction using
dimethyl sulfide. This reaction is called ozonolysis
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Reduction Reactions
Reactions Aldehydes and Ketones


In general
Aldehydes and ketones can be reduced to give alcohols

Catalytic hydrogenation is one way
Reduction reaction
Examples
Question
Reduction Reactions
Provide the structure of the reaction products from the following
equations.
O

Hydrides:
OH
(A)
H2
H
Pd/C or Pt
H
NaBH4
OH
O
LiAlH4
H2
(B)
Pd/C or Pt
OH
O
H2
(C)
Pd/C or Pt
O
(D)
OH
H2
Pd/C or Pt
H
H
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Figure 25.5
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Examples
Question
Tautomerism in Aldehydes and Ketones
Provide the structure of the reaction products from the following
equations.
OH
O
(A)
LiAlH 4
H
H
 Aldehydes and ketones bearing -hydrogen are able to undergo
tautomerism to form enols (i.e. compounds bearing an alkene and an
alcohol). Enols are constitutional isomers of their respective keto form
(i.e. aldehyde or ketone).
 Tautomerism is the process by which two isomers are interconverted by
a formal movement of an atom or group bearing in mind valency rules.
Hence, an enol and its keto form are tautomers of the same structure.
OH
O
LiAlH 4
(B)
OH
O
LiAlH 4
(C)
O
LiAlH 4
OH
(D)
H
H
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia


Aldehydes and ketones are in equilibrium with their enol forms and this
equilibrium is acid catalysed. The process involves two separate proton
transfers:
The extent of tautomerisation is dependent on the aldehyde or ketone:
Halogenation of Aldehydes and Ketones

Aldehydes and ketones undergo a substitution reaction at the αcarbon in the presence of halogen in good yield.

This reaction is regiospecific in that only the α-hydrogen is
substituted under these conditions. The rate determining step in
this reaction is the conversion of the ketone to its enol isomer,
e.g.
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Carboxylic Acids
PowerPoint to accompany

Have hydroxyl group bonded to carbonyl group.

Carboxylic acids are weak acids.
Chapter 26
Carboxylic Acids
and their
Derivatives
Carboxylic Acids
Figure 26.3
Carboxylic Acids - Acidity
Figure 26.4
Table 26.2
Figure 26.5
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Carboxylic Acids - Acidity
Question
Which of the following organic acids is expected to be the
strongest and why? Provide an explanation
(A)
H
H
(B)
C
C
O
H
ethanoic acid
acetic acid
Soln
H
F
F
(C)
C
C
O
H3C
H
C
C
O
pKa = 4.8
H
2,2-dimethylpropanoic acid
2,2,2-trifluoroacetic acid
pivalic acid
O
F
C
carboxylate anion
CH3
CH3
F
2,2,2-trifluoroethanoic acid
(B)
 Stronger acid than alcohols because of resonance stabilisation.
O
O
O
F
C
O-
+
H+
F
 Inductive effect from the substituents on the α-carbon.
Figure 26.7
 F atom is the most electronegative, hence, stabilizes the
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carboxylate anion Brown,
generated
Carboxylic Acids - Reactions with Bases
Examples
Question
Provide the structure of the reaction products from the following
equations.
O
O
(A)
NaOH
HO
O
NaOH
+
Na-O
O
(B)
OH
O-Na+
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Esters
Esters and Esterification
 Products of reaction between carboxylic acids and alcohols.
O
R
C
Examples
O
O
H
+
HO
R'
R
C
O
+
R'
H
O
H
Table 26.3
 Found in many fruits, perfumes and many other everyday
products:
Figure 26.9
Ester Hydrolysis

A
Carboxylic Acid Derivatives
Saponification = hydrolysis of ester using aqueous base.
Figure 26.12
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50
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Acid Anhydrides
Acid Chlorides


Formed by treatment of a carboxylic acid
with thionyl chloride.
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Formed by the reaction of an acid
chloride and carboxylic acid or by
dehydration.
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Questions
(a) Provide the structures for the reaction products from the following
equations.
O
O
(i)
+
HO
O
OH
O
(ii)
Chapter 28
O
F Dilute NaOH
O
OH
H2O
+
HO
F
(b) Provide appropriate starting compounds or reagents used to form the
products in the following equations.
O
O
CH3
C
HO
?
PowerPoint to accompany
SOCl3
CH3
O
CH2
C
Cl
CH2
?
H3C
O
C
OH
O
CH3
C
H 3C
O
CH2
NitrogenContaining
Compounds of
Biological
Relevance
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Biological Chemistry
Biological Chemistry

Research
at
the
interface
of
biology
and

chemistry.

The principles of chemistry are applied to
understand the molecular basis of biological
processes.
We will investigate:





Amines and amides
Amino acids
Peptides, proteins and enzymes
Vitamins
DNA
Figure 28.1
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Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Amines
 General formula:
 Can be aliphatic, heterocyclic or
aromatic:
• Next slide shows a number of simple
amines.
• Classified as 1o, 2o, 3o
• Quaternary (4o) ammonium salts are also
likely.
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Figure 28.3
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Amine Nomenclature
Figure 28.4
Amine Nomenclature

Simple amines are named using the same guidelines for alcohols.

In many instances, the amino group is named as a substituent.
Reactivity of Amines
 Amines react with acids to form ammonium salts.
 Amines react with alkyl halides to form ammonium salts.
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Formation of Amides
Amides

Amides are derivatives of carboxylic acids.

They are classified as 1o, 2o and 3o
H3C
O
C
H
C
H3C
N
N
CH3
C
H3C
H
H
acetamide
N-methylacetamide
1°
Amides are formed by the reaction of amines with “activated”
carboxylic acids.
O
O

N
CH3
CH3
O
N,N-dimethylacetamide
2°
3°
C
H3C
Cl
+
O
H
NCH2CH3
H
C
H 3C
N
CH2CH 3
+
HCl
H
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
CHEM 120 CHEMICAL REACTIVITY
ORGANIC CHEMISTRY
Dr. Vincent O. Nyamori
19 Lectures
5 Tutorials
2 Quiz
1 Test
worksheets
http://cheminnerweb.ukzn.ac.za/Firstyear/chemonetwenty.aspx
THANK YOU
and
All the BEST
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