Unit 11-1
Unit 11
Fundamentals of Organic Chemistry
Introduction
（1）What is Organic Chemistry ?
Organic Chemistry is the chemistry of compounds which contain the element carbon.
Most of these compounds contain hydrogen and many also contain oxygen, nitrogen or other
elements. There are a few compounds containing carbon, however, which are not normally classified as
organic compounds. Carbon monoxide, carbon dioxide and the metal carbonates are examples.
Although organic substances such as sugars, alcohol and vinegar had been known for thousands of
years, it was not until the eighteen century that organic compounds were first isolated. One of the first
scientists to study organic compounds was the self-taught chemist, Carl Wilhelm Scheele (1742-1786).
He obtained and purified a number of organic acids and other organic compounds from plant and animal
sources. He isolated 2-hydroxypropanoic acid (lactic acid) from milk and showed that this acid was the
cause of turning milk sour.
During the eighteen century chemists believed that organic compounds could only be synthesized by
means of a ‘life force’ in living cells. This was called the vitalistic theory of organic chemistry.
However, in 1828, the German chemist Friedrich Wohler (1800-1882) prepared urea (carbamide) by
heating an aqueous solution of ammonium cyanate :
NH4CNO

CO(NH2)2
This was the first synthesis of an organic compound. It heralded the decline of the vitalistic theory.
The term ‘biochemistry’ is now used for the chemistry of living things and life processes.
All living things contain organic compounds. Furthermore, many of the modern products and
materials upon which we depend are organic.
Some organic compounds :
Naturally occurring
carbohydrates, proteins, fats and oils,
vitamins
Synthetic
plastics, many medicines and drugs,
insecticides, many dyes
A measure of the importance of organic chemistry nowadays can be gauged from the almost
exponential growth in the number of known organic compounds over the last century :
Year
1880
1910
1940
1960
1970
1980
1990
Number
12000
150000
500000
1000000
2000000
5500000
7000000
A knowledge of organic chemistry enables chemist to develop and manufacture drugs, agricultural
chemicals, anesthetics and other chemicals whose effects on life processes are important to humans.
Unit 11-2
（2）The unique nature of carbon
Why is that carbon can form such a vast number of naturally occurring and synthetic compounds ?
The answer lies in its unique ability to catenate : to bond with itself and form stable long-chain and
ring structures. Carbon also has the ability to form single, double and triple bonds not only with itself
but also with other elements such as oxygen and nitrogen.
There are three important properties of carbon that enable it to form so many stable carbon
compounds :
1. Catenation
The ability of carbon to form strong bonds to itself means that it can form chains and rings of
varying size. This property is called catenation. The stability of the single, double and triple
carbon-carbon bonds can be seen by comparing the bond enthalpies in table :
Bond
C-C
Bond enthalpy 346
kJmol-1
C=C
610
CC
835
Si-Si
226
Si=Si
S-S
318
272
(estimated)
C-H
413
Note the strength of the C-C bond compared with that of the Si-Si and S-S bonds. Note too the
high strength of the C-H bond : all but a handful of the vast number of carbon compounds also contain
hydrogen. In the presence of air, carbon compounds are not stable relative to their oxidation products,
carbon dioxide and water. For example, methane :
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
H = -890 kJmol-1
Although methane is energetically unstable relative to its combustion products, it does not react with
air until heated to quite high temperature. This is because the reaction between methane and oxygen has
a high activation energy which must be supplied before the reaction will proceed. Thus, most carbon
compounds are energetically unstable in the presence of air, but kinetically stable.
2.
Carbon can form four covalent bonds
Carbon is also unique in its ability to hybridize.
hybridization to form :
Its four bonding electrons can undergo
four single bonds --
one double bond and two single bonds --
one triple bond and one single bond --
3.
Carbon has a fully shared octet of electrons in its compounds
Carbon atoms have no lone pairs or empty orbitals in their outer shells, so that they are unable to
form dative bonds. This properties is responsible for the kinetic stability of its compounds.
Unit 11-3
Section 11.1
Bonding and Structure
（1）Orbital hybridisation of carbon
The concept of hybridisation theory involves the following points :
1. Hybridisation is the mixing of pure orbitals to form new hybrid orbitals which are equivalent and have
definite orientations in space.
2. Only orbitals which lie close together in energy can be used in the construction of hybrid orbitals.
3. The number of hybrid orbitals always equals the number of component atomic orbitals.
sp3 hybridisation
It is formed by one 2s orbital and three 2p orbitals of a carbon atom to give four equivalent hybrid
sp3 orbitals pointing towards the vertices of a regular tetrahedron :
1.
sp2 hybridisation
It is formed by one 2s orbital and two 2p orbitals of a carbon atom to give three equivalent sp2
hybrid orbitals which are pointing towards the vertices of an equilateral triangle :
2.
3.
sp hybridisation
It is formed by one 2s orbital and one 2p orbital of a carbon atom to give two equivalent sp hybrid
orbitals of linear structure :
Unit 11-4
（2）Structures and shapes of hydrocarbons
1.
Saturated hydrocarbons
Hydrocarbons whose molecules contain only single bonds are known as saturated hydrocarbons.
In saturated hydrocarbons the outermost shell electrons of each carbon atoms have hybridised to form
four equivalent sp3 orbitals which are arranged tetrahedrally about the nucleus. This geometry gives
idealized bond angles of 109.5o.
Energy
___ ___ ___ ___
sp3 sp3
sp3 sp3
hybridized state
___ ___ ___ ___
2s
2p 2p 2p excited state
___ ___ ___ ___
2s
2p 2p 2p ground state
When an sp3 carbon atom forms bonds, it does so by overlapping each of its four sp3 orbitals (each
with one electron) with orbitals from other four atoms (each orbital in turn containing one electron).
Therefore it is capable of forming four single bonds. The molecular orbital of a single bond is
symmetrically around the axis passing through the nuclei of two bonded atoms. It is known as a 
bond.
Structures and shapes of saturated hydrocarbons :
Methane CH4
In methane, each sp3 orbital of carbon overlaps with a 1s orbital of hydrogen. Each of the resultant
sp3-1s molecular orbital is symmetrically around the axis passing through the nuclei of the carbon and
hydrogen atoms. The C-H bonds in methane are  bonds.
Structural formula
H
H C H
H
condensed structural formula
3-dimensional structure (shape)
H
CH4
H
C H
H
Unit 11-5
Ethane C2H6
Ethane contains two sp3 carbon atoms. These two carbon atoms form a C-C  bond by the overlap
of one sp3 orbital from each carbon. Each carbon atom has three remaining sp3 orbitals, and each of
these overlap with a 1s orbital of a hydrogen atom to form a C-H  bond.
Structural formula
H
H
H
C
C
H
H
condensed structural formula
3-dimensional structure (shape)
CH3CH3
H
Propane C3H8 and Butane C4H10
Propane and butane are examples of alkanes that are sometimes called unbranched chain alkanes.
Because of the tetrahedral arrangement of the bond pairs around carbon atoms, their chains are zigzagged
in shape.
Structural formula
H
condensed structural formula 3-dimensional structure (shape)
H
H
H
C
C
C
H
H
H
H
CH3CH2CH3
H H H H
H C C C C H
CH3CH2CH2CH3
H H H H
Cyclohexane C6H12
Cyclohexane is an example of cyclic alkane. If the cyclohexane ring were flat, a serious bond
strain would be resulted and all the hydrogen atoms on the ring would be eclipsed. There are many
shapes that a cyclohexane ring can assume, the most stable one is the ‘chair’ form with minimum bond
strain and all hydrogen atoms staggered.
 Draw the 3-dimensional structure of 2-methylpropane which is a branched chain alkane.
Unit 11-6
2.
Unsaturated hydrocarbons
Hydrocarbons whose molecules contain double bonds or triple bonds are known as unsaturated
hydrocarbons.
Carbon-carbon double bond : the sp2 hybridization
Formation of the C=C bonds
The atomic orbitals of carbon can also hybridise to give three identical sp2 orbitals arranged
symmetrically in a plane at an angle of 120o to each other, with the remaining electron in a p orbital at
right angles above and below this plane.
Energy
___ ___ ___ ___
sp2 sp2
sp2 2p hybridized state
___ ___ ___ ___
2s
2p 2p 2p excited state
___ ___ ___ ___
2s
2p 2p 2p ground state
Structure and shape of the ethene C2H4 molecule :
In ethene, one of the three sp2 orbitals overlaps with an sp2 orbital of a second carbon atom to form a
 bond. The other sp2 orbitals of the two carbon atoms each overlaps with the 1s orbital of a hydrogen
atom to form four  bonds. The 2p orbitals of the carbon atoms overlap above and below the plane to
form a  bond. A  bond is formed by lateral overlap of the p orbitals and has a plane of symmetry.
A C=C double bond consists of a  bond and a  bond. A  bond is stronger than a  bond as it is
formed between the lines of centres of the bonding atoms and greater orbital overlap is possible.
Shape of the ethene molecule :
 Draw the 3-dimensional structure of propene and indicate the  and  bonds in the molecule.
Unit 11-7
Carbon-carbon triple bond : the sp hybridization
Formation of the CC bonds
The hybrid orbitals formed from the 2s and one of the 2p orbitals of carbon atom consists of two
identical collinear sp orbitals with lobes directed in opposite directions on either side of the nucleus.
The two remaining unpaired electrons are in 2p orbitals at right angles to each other.
Energy
___ ___ ___ ___
sp
sp 2p 2p hybridized state
___ ___ ___ ___
2s
2p 2p 2p excited state
___ ___ ___ ___
2s
2p 2p 2p ground state
sp
sp
Structure and shape of the ethyne C2H2 molecule :
In the ethyne C2H2 molecule, one of the two sp orbitals overlap with an sp orbital of a second carbon
atom to form a  bond, while the two remaining sp orbitals(one on each carbon atom) form  bonds with
hydrogen atoms. The unhybridised 2p orbitals overlap laterally and two  bonds are formed in planes at
right angles above, below and on either side of the linear molecule.
A CC triple bond consists of a  bond and two  bonds. These  bonds draw the carbon atoms
closer together, making the CC bond both shorter and stronger than the carbon-carbon double and single
bonds.

The following two compounds have the same molecular formula C5H8 but different structures :
CH3
CH3
H
C C C
CH3
H
C CH CH C
H
H
H
Give the hybridisation states of all carbon atoms and draw the 3-dimensional structures for both
compounds.
Unit 11-8
3.
Aromatic hydrocarbons
Organic compounds are divided into two main classes : the aliphatic and the aromatic.
The molecules of both saturated and unsaturated aliphatic compounds are open-chain (acyclic) or
simple rings (cyclic) structures.
The term aromatic refers to a major class of unsaturated cyclic organic compounds which includes
benzene and its derivatives. Benzene is an unsaturated, but highly stable, cyclic molecule containing six
carbon atoms. Aromatic hydrocarbons are hydrocarbons which have benzene ring in their structures.
They are sometimes called arenes.
Structure and shape of the benzene C6H6 molecule :
Benzene C6H6 is a cyclic compound with six carbon atoms joined in a ring. Each carbon atom is
2
sp hybridised, and the ring is planar. Each carbon atom has one hydrogen atom bonded to it and an
unhybridised 2p orbital perpendicular to the plane of the  bonds of the ring. Each of these six 2p
orbitals can contribute one electron for  bonding. This type of p-orbital system leads to complete
delocalisation of all six  electrons giving rise to a highly stable structure of the benzene molecule.
It is known that all carbon-carbon bond length in benzene is the same, 0.140 nm. All six bonds are
longer than C=C double bonds, but shorter than C-C single bonds. From the bond lengths plus a body of
other evidence, chemists have concluded that benzene is a symmetrically molecule with six ring  bonds
(C-C bonds) and six  bonds of C-H bonds. Instead of alternating double and single bonds, the six 
electrons are completely delocalised in a cloud of electronic charge above and below the ring.
 bonds skeleton (C-C : sp2-sp2 ; C-H : sp2-1s)
 bonds (6  electrons delocalised)
Table : Correlation of bond length and bond order for benzene
Benzene is just one example of aromatic compounds that contain aromatic  clouds. If an
unsaturated ring system contains 4n + 2  electrons, complete delocalisation of  electrons and
subsequently stabilised structure is obtained.
Example : Pyridine (C5H5N) and naphthalene (C10H8) are also classified as aromatic compounds :
Unit 11-9
Delocalisation energy of benzene
An indication of the enhanced stabilization conferred by the delocalisation of the  electrons is given
by the enthalpy change of hydrogenation of benzene to form cyclohexane.
+
3 H2

H = -208 kJ mol-1
If there were no delocalisation of the  electrons and there were three alternative  bonds in the ring,
the enthalpy change of hydrogenation would be expected to be three times that of cyclohexane :
H = -120 kJ mol-1
The difference between the theoretical, 3 x (-120) = -360 kJ mol-1, and the experimental enthalpy
change of hydrogenation is the delocalisation energy and is a measure of this increased stability.
Enthalpy
kJ mol-1

The enthalpy changes of combustion of benzene and cyclohexene determined experimentally are
- 3267.4 kJ mol-1 and - 4128.1 kJ mol-1. Together with the data of combustion for carbon and
hydrogen,
C(s) + O2(g) → CO2(g)
ΔHOc = - 393.5 kJ mol-1
1
H2(g) +
O2(g) → H2O(l)
ΔHOc = - 285.8 kJ mol-1
2
calculate the enthalpy changes of formation of benzene and cyclohexene.
From the above calculations, the enthalpy change of formation of benzene is found to be less
positive than that of cyclohexene. This shows that benzene is energetically more stable than
cyclohexene despite that it has 'three double bonds' in the molecule.
Unit 11-10
Section 11.2
Functional groups and homologous series
（1）Functional groups
A functional group is an atom, a group of atoms, or a bonding arrangement, which is responsible
for specific properties of an organic compound or class of compounds.
The structure of a typical carbon compound can be considered as two parts : a saturated
carbon-hydrogen ‘skeleton’, which is comparatively unreactive, and a reactive part consisting of one or
more functional groups. Two examples are shown below :
H
H
C
H
H
C
H
H
H
C
C
H
O
H
H
H
H
H
H
C
C
C
C
H
H
H
H
C
N
H
Table 1 lists three simplest functional groups, which contain only carbon-carbon bonds. In the
formulae of these groups, the open-ended bond-lines represent attachment to other carbon atoms or
hydrogen atoms. The name of an organic compound usually contains clues that indicate which
functional groups are present.
Functional group
Double bond
Class names
Alkenes
Name-clues
-ene
C C
Example
ethene
H
C C
H
Triple bond
C C
Alkynes
Benzene
aromatics
arenes
-yne
H
H
Ethyne
HC C H
-benz-
methylbenzene
phen-
phenol
Unit 11-11
Table 2 lists functional groups which contain bonds linking carbon atoms to halogen, oxygen and
nitrogen atoms. The functional groups in this table are attached only to carbon atoms, as indicated by
the open-ended bond-lines.
Functional group
Carbon-halogen halogeno
-X
-F
bonds
-Cl
-Br
-I
Carbon-oxygen hydroxyl
bonds
ether
-OH
C-O-C
carbonyl
-CO-
C O
-CHO
Class names
Name-clues
halogeno-compo fluorounds
chlorobromoiodo-
Examples
chloromethane CH3-Cl
hydroxy
compounds
(alcohols)
-ol
ethanol
CH3CH2-OH
hydroxyethanoic acid
ethers
-oxy
carbonyl
compounds :
ketones
-one
aldehydes
-al
hydroxy-
1-iodobutane
CH3CH2CH2CH2-I
ethoxyethane
CH3CH2-O-CH2CH3
propanone
ethanal
carboxyl
-COOH
carboxylic acids -oic acid
ester
-COOR
esters
alkyl
acyl halides
-oyl halide
ethanoyl chloride
amides
-amide
ethanamide
acyl halide
amide
-COX
-CONH2
propanoic acid
CH3CH2COOH
-oate
ethyl ethanoate
CH3COOCH2CH3
acid anhydride
acid anhydrides -oic anhydride ethanoic anhydride
(-CO)2O
Carbon-nitrogen amino
bonds
-NH2
primary amines
ethylamine
-NHR
-amine
amino-
-NR2
secondary
amines
tertiary amines
cyano
-CN
nitriles
-nitrile
nitro
-NO2
nitro compounds nitro-
diethylamine
trimethylamine
ethanenitrile
nitroethane
Unit 11-12
（2）Homologous series and physical properties
Homologous series is a series of related organic compounds in which the formula of each member
differs from the preceding member by -CH2-. Members of a homologous series have the same
functional group and the same general formula. They show similarities in chemical properties and
trends in physical properties. The similarity in chemical properties is explained by the same functional
group in each compound. The trend in physical properties is explained by the change in intermolecular
forces with the increasing length of carbon chains.
Example : A homologous series of alcohols CnH2n+1OH
Name
Molecular formula Structural formula
methanol
CH3OH
CH3OH
ethanol
C2H5OH
CH3CH2OH
propan-1-ol
C3H7OH
CH3CH2CH2OH
butan-1-ol
C4H9OH
CH3CH2CH2CH2OH
pentan-1-ol
C5H11OH
CH3CH2CH2CH2CH2OH
Boiling point ℃
65
78
97
118
138
The physical properties of organic compounds depend on both the structure of their functional
groups and the length of their carbon chains.
Melting points and Boiling points
Most organic compounds are covalent molecules and exist as gases, liquids or relatively low melting
solids at room temperature. This is because the organic molecules are held together by intermolecular
forces only. As van der Waals’ forces increase with increasing molecular size and molecular mass, both
melting point and boiling point increase with increasing length of carbon chains in a homologous series.
For instance, the boiling points of the unbranched chain alkanes are shown below :
Alkane CnH2n+2
Methane CH4 Ethane C2H6 Propane C3H8 Butane C4H10 Pentane C5H12
-89
-42
0
36
Boiling point / ℃ -164
Alcohols and carboxylic acids, both of which contain the -OH group, have abnormally high boiling
points. This is due to intermolecular hydrogen bonding.
Example
methoxymethane ethoxyethane
Butan-1-ol
butanoic acid
CH3OCH3
CH3CH2OCH2CH3 C4H9OH
C3H7COOH
35
118
164
Boiling point / ℃ - 25
Solubility
Non-polar organic compounds such as the hydrocarbons are generally immiscible with water.
However, they do dissolve in non-polar solvents such as trichloromethane and methylbenzene.
The more polar organic compounds such as the alcohols and carboxylic acids tend to be more
soluble in polar solvents such as water. This is due to the formation of hydrogen bonds between the
solute and solvent molecules. However, the solubility decreases rapidly as the length of carbon chains
increases.
Density
As the length of carbon chains and thus the relative molecular masses of organic compounds
increase along a homologous series, the densities of the compounds increase. The branched chain
isomers usually have lower densities than the unbranched chain isomers. This can be explained by the
decrease in contact surface area between branched chain isomers and thus weaker van der Waals forces
are expected.
Alcohol
Butan-1-ol
2-methylpropan-1-ol
2-methylpropan-2-ol
-3
-3
0.802 gcm
0.789 gcm-3
Density at 20℃ 0.810 gcm
Unit 11-13
Section 11.3
Systematic Nomenclature
An organic compound may have a trivial name, e.g. wood spirit; a radicofunctional name, e.g.
methyl alcohol; or a substitutive name, e.g. methanol. The systematic nomenclature described here is
substitutive nomenclature, since internationally this is the most favoured system.
The systematic name of an organic compound comprises three parts :
(a) the beginning - one or more prefixes indicating substituent groups;
(b) the middle part - the ‘stem’ or ‘root’ derived from the parent hydrocarbon;
(c) the ending - one or more suffixes indicating unsaturation and the ‘principal group’.
Organic nomenclature begins with the principal group, for if one is present it determines the identity
of the hydrocarbon (the stem) and the numbering of the molecule.
Table 1
Prefixes and Suffixes for some principal groups
Group
Suffix
Prefix
Formula ＊
-ammonium
----Cations
Carboxylic
acids
- COOH
benzoic acid
carboxy
- (C)OOH
Sulphonic acid - SO3H
Esters
-carboxylic acid
- COOR
-oic acid
-sulphonic acid
Sulpho
alkyl -carboxylate
alkoxycarbonyl
Acyl halides
- (C)OOR
alkyl
- COX
-carbonyl halide
- (C)OX
- CONH2
-oyl halide
-carboxamide
aminocarbonyl
Nitriles
Aldehydes
- (C)ONH2
-amide
- CN
-carbonitrile
- (C) N
-nitrile
- CHO
-carbaldehyde
Ketones
-al
- COR
-one
COOR
COX
CO NH2
benzamide
cyano
CN
benzonitrile
methanoyl
- (C)HO
-----
-oate
halocarbonyl
Amides
Aromatic compound
-----
CHO
benzaldehyde
-----
oxo
Alcohols
- OH
-ol
hydroxy
OH
phenol
Amines
- NH2
-amine
amino
NH2
phenylamine
＊Note : Carbon atoms in parentheses are numbered as part of the parent hydrocarbon.
Unit 11-14
Table 2
Groups which may be named only as prefixes
Formula
-F
-Cl
-Br
-I
-OR
-NO2
-N=NRC6H5Table 3
Prefix
fluoro
chloro
bromo
iodo
alkoxy
nitro
azo
alkyl
phenyl
Hydrocarbon chains
Carbon atoms
1
2
3
4
5
6
7
8
9
10
Stem name
methethpropbutpenthexheptoctnondec-
Endings
For substituent groups
For saturated compounds
For unsaturated compounds :
one double bond
two double bonds
three double bonds
one triple bond
two triple bonds
-yl
-ane
-ene
-adiene
-atriene
-yne
-adiyne
Carbon atoms
11
12
13
14
15
16
17
18
19
20
Stem name
undecdodectridectetradecpentadechexadecheptadecoctadecnonadecicos-
Example
buta-1,3-diene
hexa-1,3,5-triene
penta-1,4-diyne
1.
The principal group, and the final suffix
Some groups can be named as suffixes or prefixes, while others can be named only as prefixes (see
Table 1 and Table 2). If a group is present which can be named as a suffix, this is the principal group
and it is named as the final suffix. If more than one such group is present, the one highest in Table 1 is
the principal group.
Example :
HOCH2CN
If no such group is present, the final suffix becomes -ane for saturated compounds, or -ene or -yne
for unsaturated compounds.
Examples :
CH3CH2NO2
CH2=CHCl
Unit 11-15
2.
The parent hydrocarbon and the stem
The parent hydrocarbon may be a chain or a ring system and its name forms the stem. Stem names
for parent chains are given in Table 3. In order of priority the parent hydrocarbon should :
(a) contain the principal group if one is present, or the most principal groups if there is more than one;
(b) contain the maximum number of multiple bonds if any are present; and
(c) be as long as possible.
Examples :
CH2
OH
C
CH2
OH
CH3
CH2 CH3
CH2 CH2 OH
3.
Numbering atoms in the parent hydrocarbon
Lowest possible numbers are allocated in the following strict order of priority to :
(a) the principal group if one is present
Examples : CH2=CH-CH2-OH
CN
COOH
CHO
(b) double and triple bonds taken together if one or more is present
Examples :
CH3-CH=CH-CCH
CH3-CC-CH=CH2
(c) double bonds before triple bonds
Example : CH=CH-CH2-CCH
(d) prefixes taken together
Examples :
CH CH CH2 Cl
CH3
3
Br
Br
NO
(e) prefixes taken in alphabetical order
Examples : Br-CH2-CH2-CH2-Cl
2
Unit 11-16
4. Adding the suffixes
(a) The name of the principal group, if one is present, will be the final suffix.
Example :
CH3CHO
(b)
If unsaturation is present, the order is double bond(s), followed by triple bond(s), followed by
principal group.
Example :
CH2=CH-CC-CH2NH2
(c) Each suffix (except -ane) will normally preceded by one or more numbers indicating the
position in the parent hydrocarbon.
Examples :
CH2=CH-CH2-CH2-OH
CH3CH(OH)CH(OH)CH2CH3
5. Adding the prefixes
(a) The names of all other substituents are added as prefixes in alphabetical (not numberical) order.
Example :
CH3CHBrCH2Cl
(b) The prefixes ‘di’, ‘tri’, ‘tetra’ etc. are ignored when considering alphabetical order.
Example :
CH3CH2C(CH3)2CH2CI3
6. Use of numbers, commas and hyphens
(a) A number immediately precedes the name of the group or multiple bond to which it refers, and
is separated from it by a hyphen.
Example :
CH3CH2CHClCH3
(b) When there is more than one type of substituent, numbers in the middle of the name are both
preceded and followed by a hyphen.
Example :
CH3CHBrCH2CHClCH2OH
(c) When there is more than one of a particular substituent, the numbers indicating the position of
each are separated by commas.
Example :
CH3CCl3
(d) Some or all numbers may be omitted.
Examples : The principal group includes a terminal carbon atom in the parent chain.
CH3CH2CH2CH2CHO
HOOCCH2CH2CH2CH2COOH
Examples : A single substituent named either as a principal group or a prefix is at position 1 on a benzene
or a cyclohexane ring.
CH3
OH
NO
2
NO
2
Examples : No ambiguity occurs.
CH3CH2COCH3
CH3CH2CH(CH3)CH3
Unit 11-17
7. When to omit or retain an ‘e’
(a) Omit ‘e’ immediately before a suffix if the suffix begins with a vowel (this includes y).
Examples :
CH3CH2CH2CH2COCH3
CH3CH2CH=CH-CCH
(b) Leave ‘e’ in place if the suffix begins with a consonant.
Examples :
CH3CH2COCH2COCH3
CH3CH2CH2CN
8.
Use of brackets
They are generally used for clarity.
Example :
CH3 CH2 CH2 CH2 CH CH2 CH2 CH2 CH2 CH3
CH2
HC CH3
CH3
 Write the systematic names of the following organic compounds :
Structural formula
Name
1.
CH2(OH)CH2CH(OH)CHC=CH2
2.
CH3 CH2 CH2 CH CH3
O
CH3
3.
CH3 C
4.
O
Br
Br CH2 C O CH3
CH2 CH COOH
O
5.
CH3
HO CH2 C CH CH C
C COOH
CH2 CH2 CH2 CH3
6.
COOH
CH3
Cl
7.
COO CH3
8.
CH2 OH
9.
CH3CH2NH2
Unit 11-18
Section 11.4
Isomerism
（1）Structural isomerism
If the compounds with the same molecular formula have their atoms attached in different orders,
they have different structures and are said to be structural isomers of each other.
1. Structural isomers containing the same functional group
Chain isomerism
The structural isomers differ in the arrangement of the carbon atoms. In general, a branched chain
isomer has a lower boiling point than a unbranched chain isomer; the more numerous the branches, the
lower the boiling point. This is because with branching the shape of the molecule tends to approach that
of a sphere; and as this happens the surface area decreases, with the result that the van der Waals’ forces
become weaker and are overcome at a lower temperature.
Example 1 : Structural isomers for C4H10
butane b.p. 0℃
Example 2 : Structural isomers for C5H12
pentane b.p. 36℃
methylpropane b.p. -12℃
methylbutane b.p. 28℃
dimethylpropane b.p. 9.5℃
Position isomerism
Structural isomers have the same carbon skeleton and belong to the same homologous series, but
differ in the position of the functional group.
Example 1 : Structural isomers for chloropropane, C3H7Cl
1-chloropropane
2-chloropropane
Example 2 : Structural isomers for propanol, C3H7OH
propan-1-ol
propan-2-ol
Example 3 : Structural isomers for disubstituted benzene, C6H4X2
Example 4 : Structural isomers for trisubstituted benzene, e.g. C6H3Br3
Unit 11-19
2. Structural isomers containing different functional groups
Functional group isomerism
The structural isomers have the same molecular formula but belong to different homologous series,
i.e. they differ in the nature of the functional group.
Example 1 : Alcohols and ethers
Two structural formulae can be written for the molecular formula C2H6O . Methoxymethane is an
ether and exists as a gas that has been used as an aerosol propellant and a refrigerant. Ethanol is an
alcohol and exist as a liquid that is used as a solvent and in alcoholic beverages.
methoxymethane b.p. - 25℃
ethanol
b.p. 78℃
Example 2 : Aldehydes and ketones, e.g. C3H6O
propanal (an aldehyde)
propanone (a ketone)
Example 3 : Carboxylic acids and esters, e.g. C3H6O2
propanoic acid
methyl ethanoate (an ester)
ethyl methanoate (an ester)
Example 4 : Acyclic and cyclic hydrocarbons
All acyclic alkanes have the general formula CnH2n+2 , where n = the number of carbon atoms
in the molecule. The presence of a ring or a double bond reduces the number of hydrogen
in the formula by two for each double bond or ring; that is, a compound with the general
formula CnH2n contains either one double bond or one ring. For example, structural isomers
for C4H8 are :
A compound with the general formula CnH2n-2 might have one triple bond, two rings, two
double bonds, or one ring plus one double bond. For example, three of many possible
structural isomers for C8H14 are :
Write the structural formulae for the following organic compounds :
Molecular formula
C3H8O
Alcohols
Ethers
C4H8O
Aldehydes
Ketones
C4H8O2
Carboxylic acids
Esters
Structural formula
Unit 11-20
（2）Stereoisomerism
In stereoisomerism the isomers have the same molecular formula and the same structural formula,
but differ in the spatial arrangement of their groups. There are two kinds of stereoisomerism :
geometrical isomerism and enantiomerism.
1.
Geometrical isomerism
Geometrical isomers are compounds with the same molecular and structural formula, but differ in
the spatial arrangement of their groups. They are not mirror images and usually arise from the rigidity
of C=C bond in organic compounds.
Rigidity of C=C bond leading to cis/trans isomers
In a molecule of ethene, the two carbon atoms are linked by a  bond and a  bond.
The shape of the  orbital ensures that :
(i) all six atoms in the molecule lie in one plane,
(ii) the carbon atoms cannot be rotated relative to one another about the bond axis without breaking
the  bond. This would require energy (about 250 kJmol-1) in the form of heat or radiation.
Because of this restriction, an alkene substituted on different sides of the double bond do not
interconvert, but may have two possible structures differing in the spatial arrangement of the substituted
atoms or groups.
Example 1 : Geometrical isomers for 1,2-dichloroethene
C2H2Cl2
molecular formula structural formula
cis-1,2-dichloroethene trans-1,2-dichloroethene
(geometrical isomers)
In the cis form, the two chlorine atoms lie on the same side of the C=C axis; in the trans form they
lie on opposite sides. They differ only in the spatial arrangement of the atoms around C=C bond.
Example 2 : Structural and geometrical isomers for butene, C4H8
Structural isomers
CH2=CH-CH2-CH3
CH3-CH=CH-CH3
Geometrical isomers
None
Name
but-1-ene
cis-but-2-ene
m.p. ℃ b.p. ℃
-185.4
-6.3
-138.9
3.7
trans-but-2-ene
-105.6
0.9
Trans-alkenes are usually higher melting than their cis isomers because of higher symmetry, but their
boiling points differ very little.
Unit 11-21
Condition for geometrical isomerism
Geometrical isomerism is possible in any molecule containing a C=C double bond with two different
substituents on each carbon.
Structural formula
Geometrical isomers
Cab=Cab
cis form
trans form
Cab=Cac
cis form
trans form
Cab=Ccd
Caa=Cbd
Write all the structural formulae of hexene, C6H12.
For each structural formula, indicate whether
there are cis/trans isomers.
A typical example of geometrical isomerism : cis- and trans-butenedioic acids
The two isomers contain the same functional groups, but the different spatial arrangement of the
groups give rise to difference in chemical properties.
The most important chemical difference is the relative ease of anhydride formation. The cis-isomer
readily forms a cyclic anhydride on heating to 170℃. The reaction is reversed on adding water.
The principal differences in the physical properties are summarized in table :
Property
cis-butenedioic acid
trans-butenedioic acid
130
286(dec.)
m.p.
℃
-3
density
gcm
1.590
1.635
-3
solubility in water g 100cm
79
0.7
-3
-2
Ka1
mol dm
1.2 x 10
9.5 x 10-4
Ka2
mol dm-3 5.9 x 10-7
4.2 x 10-5
The cis-isomer has lower m.p., lower density and higher solubility than that of the trans-isomer.
These properties can be explained by the existence of intramolecular hydrogen bonding in the cis-isomer,
which reduces the chance of intermolecular hydrogen bonding, and thus weaken interactions between
molecules of the cis-isomer.
The cis-isomer is a stronger acid than the trans-isomer for the first dissociation owing to stabilization
of the resulting anion by intramolecular hydrogen bonding. This order is reversed for the second
dissociation. The lower acid strength of the cis-isomer for the second ionization is partly a result of the
stability of the monoanion, but it also reflects the difficulty of removing the second proton from a
carboxyl group which is close to the site of the anionic charge.
Unit 11-22
2.
Enantiomerism
Enantiomers are compounds with the same molecular and structural formula, but differ in the
spatial arrangement of their groups. They are non-superimposable mirror images and usually arise
from the presence of chiral carbon atom in organic compounds.
Chiral carbon atom is a saturated carbon atom with four different attached atoms or groups.
Example 1 : Enantiomers for butan-2-ol
Butan-2-ol has two enantiomers, whose molecules are non-superimposable mirror images. This is
because the four groups attached to C-2 (H, OH, CH3, CH3CH2) are all different, making the molecule
asymmetric (without plane and axis of symmetry). C-2 is called an asymmetric carbon atom or a chiral
carbon atom.
mirror
Structural formula of butan-2-ol
A pair of enantiomers
Example 2 : Enantiomers for 2-chlorobutanal
mirror
Structural formula of 2-chlorobutanal
A pair of enantiomers
Two enantiomers are identical in all their normal physical or chemical properties, and differ only in
their behaviour toward agencies which are themselves chiral. Historically, the existence of enantiomers
was first discovered through their differing behaviour towards plane-polarized light, which has chirality.
For this reason, enantiomerism is usually called optical isomerism.
Optical activity
Plane-polarized light is produced by passing a beam of ordinary light through certain crystals, e.g.
calcite or ‘Polaroid’. As light advances through space, they create an electromagnetic field which may
be resolved into a series of sine-waves oscillating in all planes perpendicular to the direction of motion.
Polarizing crystals, when correctly oriented, transmit only the components of these waves lying in a
particular plane, and the emergent beam is the plane-polarized light. When a plane-polarized beam pass
through a chiral medium, its plane of polarization becomes rotated about the axis of the beam. Any
material which produces this rotation is said to be optically active. Optically active compounds may
rotate plane-polarized light clockwise (+) or anticlockwise (-) as seen by an observer looking towards the
source. Enantiomers are optically active and rotate plane-polarized light in opposite direction.
Unit 11-23
Polarimeter and measurement of optical rotation
Optical rotation is measured by a polarimeter. Light from a monochromatic source is polarized by
a polarizing lens or prism, then passes through the sample cell to a rotatable analyzing lens or prism
coupled to a 360o scale. The analyzer and polarizer are first set at right angles, so that no light is
transmitted. The sample is then introduced, in liquid form or in solution, and the analyzer is rotated until
the light is once again extinguished. The angle through which it has to be turned corresponds to the
angle of rotation, , produced by the sample.
For a given substance, the value of  varies directly with the number of molecules through which the
light passes, i.e. with the path length of the cell and the concentration of the solution. Optical rotation is
usually expressed as specific rotation, [] :
[] t =
Example :
The specific rotations of the enantiomers of 1-phenylethanol measured at 27℃ with
wavelength 589.3 nm are :
Racemic mixture
A 1 : 1 mixture of two enantiomers is called a racemic mixture. It is optically inactive, because
the opposite optical rotations of the two chiral forms cancel each other out. Racemic mixture differ in
crystal structure from either pure enantiomer.
Example : A 1 : 1 mixture of (+)-alanine and (-)-alanine is optically inactive.
The structural formula of glucose (C6H12O6) is given below.
there in a molecule of glucose ?
How many chiral carbon atoms are
Is glucose optically active ?
OH OH OH OH OH
CH2CH CH CH CH C
O
H
There are four chiral carbon atoms in a glucose molecule. Glucose is optically active and rotates
plane-polarized light clockwise. The trade name of glucose is ‘dextrose’ where ‘dextro’ means
clockwise (+) and ‘ose’ means sugar.
Note : The term dextro or d- was formerly used for (+), and laevo or l- for (-).
Unit 11-24
Section 11.5
Preparation, separation and purification of organic compounds
（1）Preparation
Heating under reflux
Since many organic reactions occur slowly at room temperature, it
is quite common to carry out the reaction by heating under reflux. The
higher temperature speeds up the reaction.
A water condenser is attached vertically to the reaction flask. The
flask contains the reaction mixture. This may be dissolved in an
organic solvent. The mixture is heated to its boiling point and
maintained at this temperature for the duration of the reaction.
The vapour from the mixture condenses in the condenser and
continuously drips back into the flask.
Anti-bumping granules
A few small pieces of porcelain or pumice stones can be added to
the reaction mixture to make it boil more evenly. These anti-bumping
granules provide surface area for the vapour bubbles to form and thus
avoid overheating and bumping of reaction mixture.
Water condenser
Entry of water at the lower end of the condenser is essential so that the jacket is filled progressively
up to the top. If water enters at the upper end of the jacket may drain faster than it is filled, thus
becoming unevenly cooled.
Safety precautions when preparing organic compounds
1. Reactions should be performed in a fume cupboard.
2. Use electric heater instead of naked flames.
3. Safety clothing and goggles should be worn.
4. It is also important to know the location of fire-fighting equipment and how to use it.
Unit 11-25
（2）Separation and Purification
On completion of a reaction the reaction flask will often contain not only the desired product but also
other products, by-products of side reactions, unreacted reactants and the solvent. Various techniques
are used for the separation and purification of the required product.
Simple distillation
If the required product is the only volatile substance in the reaction flask, it can be separated from
the other substances by distillation. The vapour is heated to its boiling point. Its vapour distils over,
condenses and is collected. For liquids which boils at temperature lower than about 140℃ at
atmosphere pressure, a water condenser is used. If the boiling point of the liquid is higher than 140℃
an air condenser is used. An air condenser is a straight glass tube without a jacket. Since many
organic liquids are flammable, an electric heating mantle is commonly used.
Vacuum distillation
Some organic liquids thermally decompose at temperatures below their boiling points at atmospheric
pressure. In such case it is necessary to distil the liquid at reduced pressure in order to lower the boiling
point. To do this a vacuum pump is connected to a side arm in the collection flask. This method is
called vacuum distillation.
Fractional distillation
If the required product is one of component of a mixture of two or more volatile liquids then
fractional distillation can be used to separate the components.
Unit 11-26
Solvent extraction
If the desired product is the only component in a
reaction mixture which is soluble in a particular solvent
then it can be isolated by solvent extraction. The
technique is particularly useful for preparing an organic
product from an aqueous solution containing inorganic
impurities. The solution is shaken with an organic solvent
which is immiscible with water but in which the product is
soluble. Ethoxyethane(ether) is often used for this
purpose. After extraction, ether can be distilled off in a
fume cupboard by using an electric heater and water bath.
The ether vapour should be led to the sink.
Drying agent
When an organic compound is extracted from an aqueous
solution using an organic solvent the solvent often becomes
wet. Before it can be distilled off to leave the pure product
the solvent must be dried. Anhydrous magnesium sulphate,
anhydrous calcium chloride and sodium metal are the common
drying agents. Anhydrous CaCl2 can not be used with
alcohols, phenols and amines because it reacts with them.
Crystallisation
The crude solid is dissolved in the minimum amount of
hot solvent in order to give an almost saturated solution. The
solution is then filtered to remove insoluble impurities and then
allowed to cool. The solid crystallises out.
The crystals are filtered under reduced pressure using a
Buchner funnel and flask. They are either dried or
recrystallised to obtained a more pure product.
Mixed melting point determination
A pure organic solid melts over a degree or two whereas
an impure solid melt over a range of five or more degree
Celsius. Most pure solids have characteristic sharp melting
points. The presence of even a small amount of a second
component can lower the melting point. This fact can be
used to determine the identity of unknown organic
compounds by the method of mixed melting points. The
unknown compound is matched with a known compound with
the same melting point. Approximately equal amounts of
each are powdered and mixed together thoroughly. The
melting point is determined. If it is not sharp and is lower
than that of the two separate samples, then the samples are not
identical. If the melting point is sharp and is not lowered
then the two samples are identical.
Unit 11-27
Calculating the percentage yield of product
In organic chemistry, the actual yield of purified product is often well below the theoretical yield.
% yield =
There are four main reasons for a lowering of the percentage yield :
1. the reactants may not be completely pure, e.g. some moisture may be present;
2. incomplete reaction : the reactants may not react completely;
3. side reactions which give rise to by-products;
4. loss of product during purification, e.g. small quantities of solid left on the filter paper or funnel
during recrystallization.
Often, product yields can be improved by carrying out the synthesis using an excess of one(or more)
reactant.
Example : When 15 g of butan-1-ol and 10 g of ethanoic acid were refluxed together in the presence of
concentrated sulphuric acid, 17.8 g of butyl ethanoate were formed. Calculate the percentage
yield. (Mr of butan-1-ol =74, ethanoic acid = 60, butyl ethanoate = 116)
 A set of ‘quick fit apparatus’ is shown in figure.
Give the names for its components.
1. air /steam inlet tube
2. thermometer
5. adapter with T connection 6.
9. stillhead
10. drying tube
3.
4.
7. sintered glass funnel 8.
11.
12. pear shaped flask
with angled neck
15. air condenser
16.
13.
14. screwcap adapter
Unit 11-28
Section 11.6
Reaction mechansim
（1）Reaction types
Part of the attraction of organic chemistry lies in the variety of chemical changes that occur among
organic compounds. Equally fascinating is the fact that most of these reactions can be rationalized and
unified around a few basic principles. An organic reaction could be generalized in the equation :
substrate
+ reactive species
products
(a functional group)
conditions
(a new functional group)
Most of the organic reactions can be classified as substitution, addition or elimination.
Substitution
In a substitution reaction, an atom or group of atoms of an organic compound is replaced by
another species.
C A
B
C B
A
Addition
In an addition reaction, atoms or groups add to the adjacent carbons of a multiple bond.
C C
A B
C C
A B
Elimination
An elimination reaction involves the removal of a pair of atoms or groups from adjacent carbon atoms.
This necessarily results in the formation of a multiple bond.
C C
A B
C C
A B
（2）Breaking covalent bonds and reaction intermediates
Since organic compounds are predominantly covalent compounds, their reactions inevitably involve
the breaking and formation of covalent bonds.
Homolytic fission
In homolytic fission the two shared electrons in the bond are split equally between the two atoms.
Examples :
The resulting species are known as free radicals. They are reaction intermediates and many only
exist for a split second.
A free radical is defined as an atom or group of atoms which possesses one or
more unpaired electrons.
Heterolytic fission
In heterolytic fission the two shared electrons in the bond are split unequally between the two atoms.
One of the atoms keeps both electrons. As a result it acquires a negative charge. The other atom is
deficient in one electron and thus has a positive charge. A species which contains a carbon atom with a
negative charge is known as a carbanion. One with a positive charge on a carbon atom is called a
carbocation (or carbonium ion).
Examples :
Unit 11-29
（3）Types of reactive species which attack organic compounds
Free radicals
Free radicals are formed whenever a covalent bond is broken homolytically. For example, chlorine
free radicals are formed when the element chlorine is exposed to ultraviolet light or heated to temperature
above 400℃ :
A free radical is a highly reactive species and will immediately attack a substrate for completing its
octet.
Electrophiles
An electrophile is defined as a species which is electron seeking, i.e. it seeks out an electron-rich
centre in the substrate for its attack. Examples are :
Electrophile
An electrophile usually contains an atom possessing a positive charge or a partial positive charge.
must be able to form a strong bond with a carbon atom, otherwise there cannot be a stable reaction
product.
It
Nucleophiles
A nucleophile is defined as a species which seeks out a part of substrate that is positively charged for
its attack. Examples are :
Nucleophile
A nucleophile usually contains an atom possessing a negative charge or a partial negative charge.
must be able to form a strong bond with a carbon atom.
It
（4）Reaction mechanisms
A chemical equation describes what happens, whereas a reaction mechanism describes how it happens.
Consider, for example, the possible ways, or reaction mechanisms, by which hydrogen bromide can add
to ethene :
Do the hydrogen bromide add simultaneously ? Does the hydrogen bond first, followed by the
bromine ? Or is it possible that the bromine bonds first, followed by the hydrogen ? Do the hydrogen
and bromine add as neutral or charged species ? Are any short-lived intermediate species formed during
the steps of the reaction ? A reaction mechanism answers these questions in describing the step-by-step
process of the reaction. It is postulated and then supported by experiment.
Unit 11-30
（5）Factors affecting reactivity of organic compounds
A number of factors influence the breaking and formation of bonds and thus the reactivity of organic
compounds.
Inductive effect
Inductive effect applies only to single covalent bonds between unlike atoms. Such bonds are
polarized due to the different electronegativities of the two atoms.
Example :
This shift in electron density from one atom to another and the resultant polarization of the bond is
known as the inductive effect.
Most atoms and groups are more electronegative than carbon and thus withdraw electrons from
carbon. This withdrawal of electrons is called the negative inductive effect (-I effect) or
electron-withdrawing effect.
Examples :
However, some atoms and groups are less electronegative than the carbon atom and thus donate
electrons to the carbon atom. This is known as the positive inductive effect (+I effect) or
electron-releasing effect. Alkyl groups are known to have a positive inductive effect. The effect
increases with the number of alkyl groups substituted on the carbon atom :
Mesomeric effect (Resonance effect)
Mesomeric effect occurs in molecules with multiple bonds and operate through  orbitals. This
effect results in the redistribution of  electrons and leads to the stabilization of molecules and some
radicals and carbon ions.
Example : Hydrolysis of 3-bromopropene using aqueous alcoholic sodium hydroxide
Mechanism :
1.
2.
The intermediate carbocation is stabilized by delocalisation of  electrons and thus activation energy
for the first step (the r.d.s.) is lowered.
Steric effect
The most important example of steric effect is steric hindrance. Steric hindrance can occur when
large groups on a molecule ‘get in the way’ and thus hinder the reaction. An example is provided by the
2,6-dimethylbenzoic acid :
Because of the presence of the adjacent methyl groups, the carboxylic group is not free to rotate about
its bond between the carbon atom and the benzene ring. It is thus fixed in a position at right angles to
the benzene ring. As a result of these steric factors, the 2,6-disubstituted benzoic acids are resistant to
normal methods of esterification.
Unit 11-31
Section 11.7
Organic acids and organic bases
（1）Relation of structural characteristics of organic compounds and their acid
properties
1.
Acidic properties of alcohols
The chemistry of the alcohols is characterized by the reactions of their functional group, i.e. the
hydroxyl group. Alkoxy-hydrogen fission releases a proton and thus alcohol functions as an acid.
R-O-H
+ H2 O
The degree of polarity of the -OH group is dependent upon the electron-releasing or withdrawing
powers of the group to which it is attached. Electron-releasing groups inhibit the withdrawal of
electrons away from the hydrogen atom of the hydroxyl group and impair its facility to release a proton.
Since all alkyl groups are electron-releasing, it is therefore the acidic strength of alcohols is in the order :
Acidic strength :
Name
PKa
primary alcohols
water
14
methanol
15.5
 secondary alcohols
ethanol
16

tertiary alcohols
2-methylpropan-2-ol
18
phenol
10
Alcohols, being much weaker acids than water, do not react aqueous solutions of sodium hydroxide,
sodium carbonate or sodium hydrogencarbonate. They are neutral to litmus paper.
Comparison of the acidic properties between alcohols and phenol
The greater acidity of phenol is explained by the formation of phenoxide ion which is stabilized by
delocalisation of electrons (resonance).
OH
+ H2 O
Phenol is soluble in aqueous sodium hydroxide, with which it forms a solution of sodium phenoxide.
However, its inability to liberate carbon dioxide from hydrogencarbonate enables phenol to be
distinguished from carboxylic acids.
Substitution of electron-withdrawing groups in the benzene ring of phenol, especially in the 2- and
4- positions, enables the ring in turn to withdraw more electrons from the oxygen atom, thus stabilizing
the phenoxide ion further and promoting the ionization process.
pKa
phenol
10.0
2-chlorophenol 2-nitrophenol 2,4,6-trinitrophenol 4-methylphenol
8.5
7.2
0.42
10.3
Unit 11-32
2.
Acidity of carboxylic acids
Carboxylic acids are weak acids, dissociating only slightly in aqueous solution :
RCOOH(aq)
+
H2O(l)

RCOO-(aq)
These equilibria lie heavily to left hand side.
1.3% of acid molecules dissociate into ions.
+
H3O+(aq)
For example, in 0.10 M ethanoic acid, only about
Comparison of the acidity of carboxylic acids with alcohols
Acid
ethanoic acid
benzoic acid
phenol
-3
-5
-5
Ka
mol dm
1.7 x 10
6.3 x 10
1.0 x 10-10
Like phenols, carboxylic acids react with metals and alkalis, giving salts.
with alkalis. For examples :
a.
COOH
+ NaOH(aq)
ethanol
1.0 x 10-16
Alcohols do not react
(s)
b.
OH
+
NaOH(aq)
(aq)
However, carboxylic acids are much stronger acids than phenols, and this is shown by their ability to
liberate carbon dioxide from a hydrogencarbonate (a weak base). Indeed, this reaction can be used to
distinguish carboxylic acids from phenols. For examples :
a.
COOH
+ NaHCO3(aq)
(s)
b.
OH
+
NaHCO3(aq)
(aq)
Carboxylic acids, phenols and alcohols have an acidic nature. They are all oxoacids, that is, they
lose a proton via the fission of an -OH bond. This is more likely to happen if the O-H bond breaks
easily and the negative charge on the oxoanion is delocalised, thereby increasing the stability of the
oxoanion and the acid strength.
In carboxylic acids and phenols, the acyl (RCO-) and phenyl (C6H5-) groups respectively have the
ability to pull electron cloud towards itself, thus weakens the O-H bond and encourages the loss of a
proton. Of more importance is the stability of the oxoanions. Electron delocalisation occurs in each
carboxylate and phenoxide ion, thereby spreading out its negative charge. The greater oxoanion stability
is achieved in the case of carboxylate ion where there are two oxygen atoms to share the negative charge.
R-OH +
O H
H 2O
+
O
H2 O

H3O+
+

H3O+
+
R C
R-O-
O
O
R C
O H
+
H2 O

H3O+
+
O
increasing stability of the oxoanion
Unit 11-33
Influence of substituents, viz. alkyl and chloro groups, on acidity
Acid
methanoic acid
ethanoic acid
propanoic acid
2,2-dimethylpropanoic acid
chloroethanoic acid
dichloroethanoic acid
trichloroethanoic acid
benzoic acid
4-nitrobenzoic acid
4-methylbenzoic acid
Formula
HCOOH
CH3COOH
CH3CH2COOH
(CH3)3CCOOH
ClCH2COOH
Cl2CHCOOH
Cl3CCOOH
C6H5COOH
NO2 C6H4COOH
CH3 C6H4COOH
Ka at 25℃
1.6 x 10-4
1.7 x 10-5
1.3 x 10-5
1.0 x 10-5
1.4 x 10-3
5.0 x 10-2
2.3 x 10-1
6.3 x 10-5
4.0 x 10-4
4.6 x 10-5
mol dm-3
pKa
3.8
4.8
4.9
5.0
2.85
1.3
0.7
4.2
3.4
4.3
The strength of a carboxylic acid, Y-COOH, depends on the nature of the Y group.
If the Y group has an electron-releasing effect, the O-H bond will be strengthened and the carboxylate
ion will be encouraged to accept a proton. Thus, the acid dissociates to a lesser extent and will be
weaker. For example, the acid strength decreases as the alkyl chain is lengthened and with number of
alkyl groups :
CH3 O
H
CH3 O
O
decreasing
CH3 C C
H C C
CH3 C C
acid strength
OH
OH
CH3 OH
H
H
pKa
4.8
4.9
5.0
Conversely, increasing the electron-withdrawing power of the Y group both assists proton loss and
stabilises the resulting carboxylate ion. Hence the corresponding acid increases in acid strength. For
example, the strengths of the monohalogenated ethanoic acids increases with the electronegativity of the
halogen atom :
H-CH2COOH
pKa
I-CH2COOH
4.8
3.2
Br-CH2COOH
Cl-CH2COOH
2.9
increasing
acid strength
2.85
Similarly, the electron-withdrawing power of the Y group increase as more chlorine atoms are
attached.
 Arrange the following acids in order of increasing acid strength :
a.
1. ClCH2CH2COOH 2. ClCH2COOH 3. ClCH2CH2CH2COOH
b.
1. 2-iodopropanoic acid
4. 2-bromopropanoic acid
c.
1. 2,4-dichlorobenzoic acid
4. 4-chlorobenzoic acid
2. 2-methylpropanoic acid
5. 2-chloropropanoic acid
2. benzoic acid
3. propanoic acid
6. 3-iodopropanoic acid
3. 4-methylbenzoic acid
Test : To distinguish alcohol, phenol and carboxylic acid
Test
R-OH
Add NaOH(aq) No reaction
Add NaHCO3(aq) No reaction
C6H5OH
Soluble, forming phenoxide ion.
No reaction
RCOOH
Soluble, forming carboxylate ion.
CO2 gas released.
Unit 11-34
（2）Relation of structural characteristics of organic compounds and their base
properties
1.
Base properties of amines
Like ammonia, amines are weak bases. They react with strong acids to form salts.
dil. H2SO4(aq)
R-NH2
R-NH3+(aq)
or dil. HCl(aq)
Example :
CH3CH2NH2(aq)
+ HCl(aq) →
2.
Comparison of the basic strength of ammonia, primary aliphatic amines and phenylamine
When tested with universal indicator, aqueous solutions of primary aliphatic amines and ammonia
are found to be weak bases. Base dissociation constants, Kb, for primary amines represent the following
equilibrium :
RNH2(aq)
+ H2O(l)  RNH3+(aq) + OH-(aq)
Remember the higher Kb and the lower pKb, the stronger the base.
Amine
ammonia
methylamine
ethylamine
phenylmethylamine
phenylamine
Formula
NH3
CH3NH2
CH3CH2NH2
C6H5CH2NH2
C6H5NH2
Kb at 25℃ mol dm-3
1.8 x 10-5
4.4 x 10-4
5.4 x 10-4
2.2 x 10-5
4.2 x 10-10
pKb
4.8
3.4
3.3
4.7
9.4
pH range
10 - 12
10 - 12
10 - 12
10 - 12
about 7
The values in table are typical of a general pattern in basicity, namely,
increasing basic strength
Primary aliphatic amines and phenylmethylamine are bases of strength stronger than ammonia.
In primary aliphatic amines, alkyl groups are able to push electrons away from themselves, and this
has two effects. Firstly, an alkyl group can assist protonation by increasing electron density on nitrogen.
Secondly, an alkyl group will help stabilize the resulting alkylammonium ion by dispersing its positive
charge. Both effects would aid proton-acceptance, making the base stronger.
R
N H
H
R
H
N H
H
In phenylamine, the lone pair of electrons on nitrogen is held in a p orbital. This can overlap
sideways with benzene’s  electron ring. Due to this delocalisation, the lone pair is less available for
dative covalent bonding with a proton. Thus, phenylamine is a poor proton-acceptor, making it a very
weak base.
Note that phenylmethylamine, though aromatic, is a slightly stronger base than ammonia. This is
because its -NH2 group is in an aliphatic side chain and not bonded directly to the benzene ring.