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An Introduction to Organic Chemistry
An Introduction to Organic Chemistry 8
81 An Introduction to Organic Chemistry
Organic chemistry is the study of compounds
containing carbon with the exception of simple
compounds e.g. carbonates (CO ), carbon
dioxide (CO ) and carbon monoxide (CO).
There are over 6 million known organic
compounds. Nomenclature is therefore very
Here are some basic guidelines that should help
in the naming of the simple compounds you will
come across during this course. You will get
practice at this in your tutorials.
1) Find the longest carbon chain in the molecule.
This will give you the base of the name:
No of C atoms Name
82 An Introduction to Organic Chemistry
2) Determine the principle functional group and
functional group formula becomes
alkane C-C -ane
alkene C=C -ene
alkyne C≡C -yne
alcohol -OH -anol
aldehyde -CH=O -anal
ketone C=O -anone
carboxylic acid -COOH -anoic acid
Position is indicated, where necessary, by
numbering the carbons in the main chain.
Position need not be indicated for alkanes, as
they have no functional group, and aldehydes
and acids, as they are terminal functional
groups. Positioning numbers are flanked by
dash signs. Multiple positions for a given
functional group are separated by commas
and indicated by the prefixes di, tri, tetra,
penta, hexa, hepta, octa , nona and deca.
3) Ancilliary functional groups are given in
alphabetical order, with their position at the
beginning of the name.
functional group formula prefix
methyl -CH methyl
ethyl -C H ethyl
83 An Introduction to Organic Chemistry
propyl -C H propyl
butyl -C H butyl
pentyl -C H pentyl
hexyl -C H hexyl
heptyl -C H heptyl
octyl -C H octyl
nonyl -C H nonyl
decyl -C H decyl
fluorine -F fluoro
chlorine -Cl chloro
bromine -Br bromo
iodine -I iodo
amine -NH amino
hydroxyl -OH hydroxy
cyanide -CN cyano
benzyl -CH C H benzyl
2 6 5
phenyl -C H phenyl
Empirical and Molecular Formulae
Quantitative elemental analysis tells us what
elements make up a compound and in what
The percentage of each element present in a
compound is determined by total combustion. C,
H, S and N burn to give CO , H O, SO and NO .
2 2 2 2
The quantities of these gases may readily be
measured and this leads to information that can
be used to calculate the % composition and
hence empirical and molecular formulae.
84 An Introduction to Organic Chemistry
How is this done? First some definitions:
One mole of a substance is 6.02 x 10 particles
of that substance. This huge value is termed
Avogadro’s number. One mole of any substance
has a mass equal to the relative molecular mass
(RMM) of that substance in grams.
Relative molecular mass is the sum of the
relative atomic masses (RAMs) of the constituent
elements in the compound.
e.g. for ethanol C H OH
-1 -1 -1
RMM = (2 x 12.010 g mol ) + (6 x 1.006 g mol ) + (15.999 g mol )
= 46.057 g mol
0.152 g of an organic compound X containing
only C, H and O produces:
0.223 g of CO
0.091 g of H O
upon total combustion. Calculate the empirical
formula of the compound X.
Consider the CO
-1 -1 -1
CO RMM = 12.010 g mol + 2 x 15.999 g mol = 44.008 g mol
0.223 g of CO = 0.223 g / 44.008 g mol = 5.07 x 10 mol
5.07 x 10 mol of CO were produced from 5.07 x 10 mol of C
The mass of C = 5.07 x 10 mol x 12.010 g mol = 0.061 g
% C in X = 100% x 0.061 g / 0.152 g = 40.1%
85 An Introduction to Organic Chemistry
Consider the H O
-1 -1 -1
H O RMM = 2 x 1.006 g mol + 15.999 g mol = 18.011 g mol
0.091 g of H O = 0.091 g / 18.011 g mol = 5.05 x 10 mol
5.05 x 10 mol of H O were produced from 1.01 x 10 mol of H
The mass of C = 1.01 x 10 mol x 1.006 g mol = 0.010 g
% C in X = 100% x 0.010 g / 0.152 g = 6.7%
Consider the O
% O in X = 100% - 40.1% - 6.7% = 53.2%
Mass % are used to calculate mole % which
yield the empirical formula or simplest ratio of the
C H O
relative mass % 40.1 6.6 53.2
divide by RAM 12.010 1.006 15.999
relative mole % 3.3 6.6 3.3
divide by smallest 1 2 1
This gives the ratio 1:2:1 and the empirical
formula CH O. The molecular formula could be
any multiple of the empirical formula e.g.
C H O , or C H O since these would all have
2 4 2 3 6 3
the same percentage mass ratios.
5.05 x 10 mol of C means 5.05 x 10 mol of X in 0.152 g
-3 -1 -1
RMM of X = 0.152 g / 5.05 x 10 g mol = 30.10 g mol
The molecular formula is also CH O and X is
actually methanal or formaldehyde.
86 An Introduction to Organic Chemistry
Structural and Isomerism Structural
Different arrangements of atoms for a given
molecular formula are often possible. Such
compounds are called isomers.
Example one: C H
CH CH CH CH CH CHCH CH
3 2 2 3 3 3 3
H H H H H H
C C H
H C C
H C H
H H H H H
87 An Introduction to Organic Chemistry
Example two: C H O
CH CHOH CH OCH
3 2 3 3
H H H H H
H O H
Both exemplify structural isomerism. C H is a
molecular formula as it shows constituent atoms.
CH HC CH CH is a structural formula as it
3 2 2 3
shows constituent atoms AND connectivities.
It is also possible to arrange the atoms in
molecules with the same structural formulae
such that they have different spatial orientation.
This is known as stereoisomerism.
There are two distinct types of stereoisomer:
geometric and optical.
88 An Introduction to Organic Chemistry
It is possible for single C-C bonds to rotate
freely, however, double C=C bonds cannot.
Thus if the two carbon atoms of a C=C bond
carry different groups, it becomes possible to
orientate these groups in two ways to create
H COOH HOOC H
H COOH H COOH
Z ( Zusammen - together) E ( Entgegen - opposite)
Geometric isomers have different physical and
89 An Introduction to Organic Chemistry
A carbon atom attached to four different groups
(substituents) is termed a chiral centre.
Two different non-superimposable mirror images
These mirror images are called enantiomers.
Enantiomers have identical physical properties,
except for the direction in which they rotate the
plane of plane polarised light.
Rotation to right termed dextro or d
Rotation to left termed laevo or l
They have identical chemical properties except
towards optically active reagents.
90 An Introduction to Organic Chemistry
If a compound contains a chiral centre but does
not rotate the plane of plane polarised light then
it must be an equal mixture of d- and
l-enantiomers. Such a mixture is termed a
racemic mixture or a racemate.
Stereochemistry is crucially important to the
pharmaceutical industry. The drug thalidomide,
prescribed to pregnant women as a powerful
sedative from 1956 exists as two enantiomers.
One was the powerful sedative. The other
caused human transmutation…
91 An Introduction to Organic Chemistry
The simplest organic molecule is methane CH .
The ground electronic state of carbon suggests it
should form 2 bonds as there are two unpaired
How and why does carbon form 4 bonds?
Promotion of one of the two 2s electrons
increases energy but the formation of four bonds
causes a four-fold decrease.
How can the four bonds formed be
92 An Introduction to Organic Chemistry
This mixing of an s orbital and three p orbitals to
produce four hybrid orbitals is called sp
hybridisation. ALL tetrahedral carbon and
nitrogen atoms in organic chemistry are sp
ALL trigonal carbons such as those found in
double bonds are sp hybridised. The unused p
orbital on each carbon overlaps to form the π
part of the double bond, e.g. ethene.
93 An Introduction to Organic Chemistry
ALL linear carbons such as those found in triple
bonds are sp hybridised. The unused p orbitals
on each carbon overlap to form the π parts of the
triple bond, e.g. ethyne (acetylene).
94 An Introduction to Organic Chemistry
Hydrocarbons are a family of compounds
containing only hydrogen and carbon. There are
two main classes:
aliphatic and aromatic
Within the aliphatic class there are both
saturated and unsaturated hydrocarbons.
A homologous series of saturated compounds
with general molecular formula C H (where n
is an integer).
methane CH , ethane C H , propane C H , butane C H , pentane
4 2 6 3 8 4 10
C H , hexane C H , heptane C H , octane C H , nonane C H ,
5 12 6 14 7 16 8 18 9 20
decane C H , etc.
Homologous series: a series of compounds in
which each successive compound differs from
the previous one by a CH unit.
At n = 4 it becomes possible to arrange the
carbon skeleton differently i.e. it becomes
possible for structural isomers to exist. The
result is termed branching of the C-C backbone.
95 An Introduction to Organic Chemistry
How does this affect a physical property such as
the boiling point?
n-pentane 2-methylbutane 2,2-dimethylpropane
36 °C 28 °C 9 °C
As branching increases, the strength of the van
der Waals interactions between molecules
decreases, resulting in the lowering of boiling
Natural gas - principally methane CH
Petroleum oil - mixture up to ca. n = 40
Separation of crude oil is achieved by fractional
distillation. This forms the basis of the petroleum
and petrochemical industries.
Relatively, alkane chemistry is very limited.
Their main use is as fuels for combustion or
Methane domestic gas supply
Propane LPG (liquid propane gas)
Butane camping stove gas
96 An Introduction to Organic Chemistry
Complete combustion yields carbon dioxide and
water. Incomplete combustion is dangerous as it
produces carbon monoxide. Always provide a
good supply of air to any process in which a
hydrocarbon is being burned
2C H (g) + 7O (g) Æ 4CO (g) + 6H O(l)
2 6 2 2 2
2C H (g) + 5O (g) Æ 4CO(g) + 6H O(l)
2 6 2 2
Chemically this is more accurately described as
oxidation. Both reactions produce energy in the
form of heat and are said to be exothermic.
In the presence of a halogen and ultraviolet light
a series of reactions take place:
Cl Æ 2Cl· I
Cl· + CH Æ CH · + HCl P
CH · + Cl Æ CH Cl + Cl· P
3 2 3
Cl· + CH Cl Æ CH Cl· + HCl P
CH Cl· + Cl Æ CH Cl + Cl· P
2 2 2 2
Cl· + CH Cl Æ CHCl · + HCl P
2 2 2
CHCl · + Cl Æ CHCl + Cl· P
2 2 3
Cl· + CHCl Æ CCl · + HCl P
CCl · + Cl Æ CCl + Cl· P
3 2 4
Cl· + Cl· Æ Cl T
CH · + CH · Æ C H T
3 3 2 6
97 An Introduction to Organic Chemistry
The first step is homolytic fission of the halogen
to produce halide radicals. This is termed the
intitiation step, I. Radicals are extremely
reactive species with single unpaired electrons
denoted ·. Radicals react with non-radical and
radical species in propagation and termination
steps, P and T respectively. The result is an
extensive, indiscriminate mixture of halogenated
hydrocarbons that is very expensive to separate.
Reactivity: F Cl Br I
2 2 2 2
Provides an albeit prohibitively expensive route
to useful compounds since polarity (and thus
chemical reactivity) has been introduced.
This is enormously important industrially. The
bonds in longer chain alkanes are cleaved using
heat in a process called pyrolysis. The process
produces smaller more useful hydrocarbon
98 An Introduction to Organic Chemistry
A homologous series of unsaturated compounds
with general molecular formula C H (n is an integer
greater than 1) that contain a double bond.
ethene C H , propene C H , butene C H , pentene C H , hexene
2 4 3 6 4 8 5 10
C H , heptene C H , octene C H , nonene C H , decene C H ,
6 12 7 14 8 16 9 18 10 20
Alkenes are defined by the presence of a C=C
double bond. In the laboratory, they can be
prepared via the dehydration of alcohols by
CH CH OH Æ CH =CH + H O
3 2 2 2 2
One of the most important principles in organic
chemistry is the understanding of how reactions
happen at a molecular level. This is termed the
Mechanisms are represented by "pushing
electrons" between and/or around molecules.
The arrows MUST be accurately drawn to show
both the origin and destination of the electrons.
99 An Introduction to Organic Chemistry
So what is the mechanism for the dehydration of
H H H
H H H H
A mechanism can NEVER be proven, only
supported by experimental evidence.
The mechanism shown above may be applied to
the dehydration of any alcohol with acid to yield
Alkenes are much more reactive than alkanes.
Most of their chemistry involves addition to the
C=C double bond.
Addition of H to C=C in the presence of a
suitable catalyst e.g. Pd activated charcoal or
Raney Ni (treat nickel aluminium alloy with hot NaOH)
H C=CH + H Æ H C-CH
2 2 2 3 3