CHM167/chapter8/HaslizaYusof
ORGANIC CHEMISTRY
8.1 Introduction



the organic compounds are devided into several groups known as homologues series
the homologous series is a series of organic compounds that have similar structural features but differ
from adjacent member by a (-CH2) group
each member of a homologous series has the same chemical properties and can be represented by a
general formula
Homologous Series
General Formula
Example
Alkanes
CnH2n+2
n = 1, 2, 3,….
CH4, n = 1
C C
Alkenes
CnH2n
n = 2, 3, 4,….
C2H4, n = 2
C C
Alkynes
CnH2n-2
n = 2, 3, 4,…
C2H2, n = 2
C C
Alcohol
CnH2n+1OH
n = 1, 2, 3,….
CH3OH, n = 1
C OH
Carboxylic acid
CnH2n+1COOH
n = 1, 2, 3,…
CH3COOH, n = 1
C COOH
Aldehyde
RCHO
R = alkyl group
CH3CHO, n = 2
RCOR
R = alkyl group
CH3COCH3, n = 3
RNH2
R = alkyl group
CH3NH2
Ketone
Primary amines
hydrocarbon: compound that contain only carbon and hydrogen
have 4 categories: alkanes, alkenes, alkynes and aromatic compounds
1
O
R C H
8.2 Hydrocarbons


Functional Group
O
R C R
H
R N
H
CHM167/chapter8/HaslizaYusof
8.3 Alkanes

alkanes are hydrocarbon containing only single bonds (CC)

also called saturated hydrocarbons because they are saturated with hydrogen

general formula: CnH2n+2, n = 1, 2, 3, ….

the simplest alkanes: methane, where n = 1, C1H2(1)+2 = CH4

the alkanes are named from the Greek numbers according to the number of carbon atom present

the suffix –ane is added to the end of each name to indicate that the molecule identified is an alkane
8.3.1 Nomenclature

used the IUPAC (International Union of Pure and Applied Chemistry)

in this system, the longest continuous chain of carbon atoms called the parent chain- determines the
base name of the compound

the prefixes for the base names depends on the number of carbon atoms in the parent chain

base names for alkanes always have the ending –ane

groups of carbon atoms branching off the base chain  substituents
Table: Common Alkyl Groups
Condensed Structural Formula
Name
CH3
methyl
C2H5
CH2CH3 or
ethyl
F
fluoro
Cl
chloro
Br
bromo
I
iodo
2
CHM167/chapter8/HaslizaYusof

guide to naming alkanes:
STEP 1
Write the alkane name of the longest continuous
chain of carbon atoms
Number the carbon atom starting from the end nearest
a substituent
Give the location and name of each alkyl group
(alphabetical order) as a prefix to the name of the
parent chain (base chain)
STEP 2
STEP 3
Example:
CH3
CH3
STEP 1
CH CH2 CH2 CH3
Write the alkane name of the longest continuous chain of carbon atoms. In this
alkane, the longest chain has five carbon atoms, which is pentane
CH3
CH3
STEP 2
CH
CH2 CH2 CH3
pentane
Number the carbon atoms starting from the end nearest a substituent
CH3
CH3
CH CH2 CH2 CH3
2
1
STEP 3
3
4
5
pentane
Give the location and name of each substituent as a prefix to the alkane name.
Place a hyphen between the number and the substituent name
CH3
CH3
1
CH
2
CH2 CH2 CH3
3
4
5
2-methylpentane
List the substituents in alphabetical order
CH3 Cl
CH3
CH CH2 CH2 CH3
1
2
3
4
3 – chloro – 2 – methylpentane
5
Use a prefix (di - , tri - , tetra -) to indicate a group that appears more than once. Use commas to separate
two or more numbers
CH3 CH3
CH3
1
CH CH2 CH2 CH3
2
3
4
2,3 – dimethylpentane
5
When there are two or more substituents, the parent chain is numbered in the direction that gives the
lowest set of numbers
Br
CH3
5
CH3
CH CH2 C CH3
4
3
2
Br
1
2,4 – dibromo – 2 – methyl pentane
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CHM167/chapter8/HaslizaYusof
8.3.2 Drawing Structural Formulas for Alkanes

guide to drawing alkane formula
STEP 1
Draw the parent (main) chain of carbon atoms
STEP 2
Number the chain and place the substituents on the
carbons indicated by the numbers
STEP 3
Add the correct number of hydrogen atoms to give
four bonds to each carbon atom
Example:
Write the condensed structural formula for 2,3 – dimethylbutane
STEP 1
Draw the main (parent) chain of carbon atoms. For butane, we draw a chain of
four carbon atoms and number it
C C C C
1
STEP 2
2
3
4
Number the chain and place the substituents on the carbons indicated by the
numbers. The first part of the name indicates two methyl groups (CH3-),
one on carbon 2 and one on carbon 3
CH3
C C C C
1
2
3
4
CH3
STEP 3
Add the correct number of hydrogen atoms to give four bonds to each carbon
atom
CH3
CH3 CH CH CH3
1
2
3
4
CH3
2,3 - dimethylbutane
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CHM167/chapter8/HaslizaYusof
Exercise
1. Give the IUPAC name for each of the following
a)
F
b)
CH3
CH3 C CH3
CH3 CH CH2 CH3
CH3
c)
d)
CH3 Cl
CH3
CH2
Br
CH CH CH3
CH3
CH2
e) CH3(CH2)5CH3
2. Draw a condensed structural formula for each of the following alkanes
a) 2 – methylbutane
b) 3,3 – dichloropentane
c) 2,3,5 – trimethylhexane
d) 3 – iodopentane
e) 2-methylpropane
f) 2-bromo-2,3-dichloropentane
5
Cl
CH CH2 CH CH3
CHM167/chapter8/HaslizaYusof
8.3.3 Isomers

same formula, different structure

the structural formulae of methane, ethane & propane can be written only in one way

that means, 3 hydrocarbons do not have isomers
a) alkanes with four carbon atoms
o
the hydrocarbon with molecular formula C4H10 has two isomers: butane & 2-methylpropena
H H H H
H C C C C H
CH3
H H H H
Butane
CH3 CH CH3
2 – methylpropane
b) alkanes with five carbon atoms
o
o
the molecular formula: C5H12
has 3 isomers: pentane, 2 – methylbutane & 2,2 – dimethylpropane
8.3.4 Reactions
Reagent
Bromine, Br2
Effect on alkanes
No effect in the dark. Bromine slowly decolourised in
sunlight. Halogenation reaction
H H
H C C H
+
Br2
H H
Chlorine, Cl2
uv
light
H Br
H C C H
+
HBr
+
HCl
H H
No effect in the dark. Reaction occurs in sunlight.
Halogenation reaction
H H
H C C H
+
Cl2
uv
light
H Cl
H C C H
H H
H H
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CHM167/chapter8/HaslizaYusof
8.3.5 Cycloalkanes

Molecular formula has two less H atoms than straight chain, CnH2n
H H
C
H C C H
H
H
H H
H C C H
H C C H
H H
cyclopropane
H H
H
C
H C
C H
H
H C
C H
C
H
H
H H
cyclobutane
cyclohexane
8.4 Alkenes

containing at least one double bond between carbon atoms ( C = C )

the simplest alkene is ethane, C2H4, also called ethylene

the suffix –ene is added to the end of each name to indicate that the molecule identified is an alkene
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CHM167/chapter8/HaslizaYusof
8.4.1 Nomenclature

guide to naming alkenes
STEP 1
Name the longest carbon chain with a double bond (C=C)
STEP 2
Number the carbon chain starting from the end nearest a
double bond
STEP 3
Give the location and name of each substituent (alphabetical
order) as a prefix to the name
Example:
CH3
CH3 CH CH CH CH3
STEP 1
Name the longest carbon chain that contains the double bond. There are five
carbon atoms in the longest chain. Replacing the corresponding alkane
ending with ene gives pentene
CH3
pentene
CH3 CH CH CH CH3
STEP 2
Number the longest chain from the end nearest the double bond. The number of
the first carbon in the double bond is used to give the location of the double
bond
CH3
CH3
5
STEP 3
2 – pentene
CH CH CH CH3
4
2
3
1
Give the location and name of each substituent (alphabetical order) as a prefix to
the alkene name. The methyl group is located on carbon 4
CH3
CH3
5
4 – methyl – 2 – pentene
CH CH CH CH3
4
2
3
1
Exercise
Give the IUPAC name for each of the following
1.
CH3
CH3
C CH2
2.
CH3
CH3
C C CH3
CH3
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CHM167/chapter8/HaslizaYusof
8.4.2 Isomers

cis - and trans –
1. 2-butene
a) cis – 2 – butene
b) trans – 2- butane
2. 2,3-dichloro-2-butene
a) cis-2,3-dichloro-2-butene
b) trans-2,3-dichloro-2-butene
8.4.3 Reactions
1. halogenation
a) Br2

the decolorisation of bromine solution:
H H
H C C H
H H
+
Br2
H C C H
Br Br
Ethene
1,2 - dibromoethane
2. hydrogenation
H H
H C C H
H H
+
H2
H C C H
H H
8.4.4 Test to Confirm the Presence of the C=C
1. Bromine in CCl4
2. KMnO4 in acid
8.5 Alkyl Halide


containing the halogen elements (group 17); F, Cl, Br, I
example:
H H
H H
H H H H
H C C H
H C C H
H C C C C H
H Cl
H F
Br H H H
9
CHM167/chapter8/HaslizaYusof
8.6 Alcohol


containing the hydroxide group, - OH
general formula: CnH2n+1OH
H
H
H C OH
H
R C OH
H
R C OH
H
primary alcohol, 1o
R
secondary alcohol, 2o
tertiary alcohol, 3o
8.6.1 Nomenclature
CH3
Example:
CH3
C CH2 OH
H
Name the longest carbon chain containing the – OH group. Replace the e in the
alkane name with ol
STEP 1
CH3
CH3
C CH2 OH
propanol
H
Number the longest chain starting at the end closest to the – OH group
STEP 2
CH3
CH3
3
C CH2 OH
2
1
1 - propanol
H
Name and number other substituents relative to the – OH group
STEP 3
CH3
CH3
3
C CH2 OH
2
1
2 – methyl – 1- propanol
H
CH3
CH3
CH CH CH3
OH
8.6.2 Reactions
1. Oxidation
H
H C H
H
+
O2
H
H
2. Reduction
O
H C H
H C OH
H
+
H2
H C OH
H
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CHM167/chapter8/HaslizaYusof
3. Halogenation
CH3
CH CH3
+
OH
2 - propanol
HI
CH3
+
CH CH3
H2O
I
2 - iodopropane
Hydrogen
iodide
8.7 Aldehyde
O

have carbonyl group (

add the ending – al

general formula: R C H
C
) bonded to a least one H atom
O
8.7.1 Nomenclature
STEP 1
Name the longest carbon chain containing the carbonyl group by replacing the e in
the alkane name with al.
No number is needed for the aldehyde group because it always appears at the end
of the chain
STEP 2
Name and number any substituents on the carbon chain by counting the carbonyl
carbon as carbon 1
CH3
CH3 O
CH3
3
CH3
CH C H
2
5
1
2 – methylpropanal
8.7.2 Reactions
1. Oxidation
H
O
H C OH
+
O2
H C H
H
2. Reduction
O
H C H
H
+
H2
H C OH
H
11
O
CH CH2 CH2 C H
4
3
2
1
CHM167/chapter8/HaslizaYusof
8.8 Ketone
O


have a carbonyl group (
add the ending – one
C

general formula:
) on the interior of the chain
O
R C R
8.8.1 Nomenclature
STEP 1
Name the longest carbon chain containing the carbonyl group by replacing the e in
the alkane with one
STEP 2
Number the parent chain starting from the end, nearest the carbonyl group. Place
the number of the carbonyl carbon in front of the ketone name (propanone and
butanone do not require numbers)
STEP 3
Name and number any substituents on the carbon chain
O
CH3
O
O
C CH3
CH3
propanone
CH2
CH3
C CH3
CH2
1
2
butanone
1
3
CH3
1
4
CH C CH2 CH3
2
8.8.2 Reaction
1. Oxidation
OH
CH3 CH CH3
3
4
5
2 – bromo – 3 - pentanone
3 - methylbutanone
O
+
4
Br O
C CH CH3
2
3
3 - pentanone
O CH3
CH3
C CH2 CH3
O2
CH3
12
C CH3
5
CHM167/chapter8/HaslizaYusof
8.9 Carboxylic Acid
O

containing a carboxyl group

add the ending – ioc

general formula:
C OH
O
R C OH
8.9.1 Nomenclature
STEP 1
Name the longest carbon chain containing the carboxyl group and replace the e
of the alkane name by oic acid
STEP 2
Number the carbon chain beginning with the carboxyl carbon as carbon 1
STEP 3
Give the location and names of substituents on the parent chain
O
CH3 O
H C OH
CH3
Methanoic acid
8.9.2
CH C OH
2 – methylpropanoic acid
Reaction
1. Oxidation
H
CH3 C OH
O
+
O2
CH3
H
13
C OH
CHM167/chapter8/HaslizaYusof
TUTORIAL 8
1. Draw the structural formula for each of the following organic compounds
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
k)
2-methylpropane
2-bromo-2,3-dichloropentane
3-methyl-2-hexene
cis-2-butene
2,2,3-trimethylpentane
1-bromo-1-chlorobutane
2-methyl-2-butene
propanal
2,4-heptadiene
2-propanol
2-methyl-2-butanol
2. Draw and name all alkane isomer of the compound with molecular formula of C5H12
3. Name of the following compound
a) CH3(CH2)5CH3
b) CH3CH(OH)CH2CH3
c) CH3CH2CH2CHO
4. Write the structures of the cis and trans isomers for the 2 – butene.
5. Give one test to confirm the presence of the C=C bond in hydrogen.
6. Draw structures containing four carbon atoms for each of the following classes of compound:
a)
b)
c)
d)
e)
Alkane
Alkyl halide
Alcohol
Alkene
Carboxylic acid
7. Write the condensed structures and the IUPAC names for all five possible alcohols with general
formula C5H12O.
8. Identify the functional group of each of the following compounds
a)
CH3CH2COCH2CH2CH3
b)
CH3CH2CH2CHO
c)
CH3CH2COOH
d)
CH3CH2CH CH2
e) CH3CH2CH2OH
f) CH3CH2CH3
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CHM167/chapter8/HaslizaYusof
9. Predict the product of each of the following reactions
a) CH3CH2CH=CH2 + Br2  ___________
b) CH3CH2CH3 + Cl2  __________ + HCl
c) CH3CHO + H2  ___________
d) CH3COCH2CH3 + H2  _____________
e) CH3CH2CH2CH2OH + O2  _______________ + H2O
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