LECTURE NOTE SK025 (ORGANIC) KMKt

Example 2:

2

Cl Cl
31

45

1,3-dichlorocyclopentane
NOT

1,4-dichlorocyclopentane

31

STEP 4: When three or more substituents are

present, begin at the carbon with

substituent that leads to the lowest

set of locants.

Example 1: CH2CH Locants:
chloro 1 4
2 33 ethyl 3 2
3 methyl 4 1
2
1 4

Cl 4 1 CH

5 63
6
5

1-chloro-3-ethyl-4-methylcyclohexane

4- chloro-2-ethyl-1-methylcyclohexane 32

Example 2:

H3 CH2CH3
C

1

52

43

CH

1-ethyl-1,3-dimethylcy3clopentane
NOT

3-ethyl-1,3-dimethylcyclopentane

33

STEP 5:

If……

i) a single ring system is attached to a
single chain with a greater number of
carbon atoms,

@

ii) more than one ring systems are
attached to a single chain,

☞ Then it is appropriate to name the compound
as cycloalkylalkane.

34

Example:

CH2CH2CH2CH2CH2CH3

cyclopentylhexane

1,3-dicyclohexylpropane

35

Physical Properties of
Alkanes

36

Learning Outcomes

d) Explain the following physical properties:
i) boiling point of:
- alkanes based on molecular weight/size;
- isomeric alkanes
- alkanes and cycloalkanes

ii) solubility in water and organic solvents

PHYSICAL PROPERTIES BOILING POINT alkanes based on
molecular weight, Mr

isomeric alkanes

alkanes & cycloalkanes

SOLUBILITY in water

in organic 38
solvents

Boiling point of alkanes increase smoothly
with increasing number of C atoms
(molecular weight / molar mass)

Alkane Molecular Structural formula Boiling
formula point (oC)
Methane
Ethane CH4 CH4 - 162
Propane C2H6 CH3CH3 - 89
Butane C3H8 CH3CH2CH3 - 42
Pentane C4H10 CH3CH2CH2CH3 - 0.5
Hexane C5H12 CH3(CH2)3CH3 36
Heptane C6H14 CH3(CH2)4CH3 69
Octane C7H16 CH3(CH2)5CH3 98
Decane C8H18 CH3(CH2)6CH3 126
C10H22 CH3(CH2)8CH3 174 39

Effect of increasing number of C atoms on
boiling point…

• The boiling points of alkanes show a regular
increase with increasing molecular weight
because the number of carbon increase.

• the molecule has bigger surface area in contact.
• the stronger Van der Waals attractive forces.
• more energy required to overcome the attractive

forces.
• the higher the boiling point.

40

Effect of branching on boiling point for
isomeric alkanes ...

Alkane

Boiling 35 oC 28 oC 9.5 oC
point

41

Effect of branching on boiling point ...

• more branches, molecule become more compact.

• surface area in contact are reduced.

• this causes the branched alkanes to have a
weaker Van der Waals attractive forces.

• less energy required to overcome the attractive
forces.

• the lower the boiling point.

No. of branches ↑, Surface area ↓, Van der 42 42
Waals forces ↓, boiling point ↓

Cyclic alkanes have higher boiling point
than the corresponding straight chain.

Alkane CH2CH2CH2CH2CH2CH3 cyclohexane

hexane 81 oC

Boiling 69 oC
point

43

Why cyclic alkanes have higher boiling point ??
• cyclic alkanes has larger surface area in contact

than the corresponding straight chain alkane.
• the stronger Van der Waals attractive forces.
• more energy required to overcome the attractive

forces.
• the higher the boiling point.

44

Solubility in Water and Organic Solvents..

• Alkanes and cycloalkanes are almost totally
insoluble in water because they are:
(i) non-polar molecules
(ii) unable to form hydrogen bond with H2O.

• Alkanes and cycloalkanes are generally dissolve in
non-polar solvents.

• Examples of non-polar solvent: benzene, chloroform,
carbon tetrachloride, etc.

45

45

Chemical Properties
of Alkanes

46

Learning Outcomes

e) Write a balance chemical equation for the
combustion of alkane in:

i) excess oxygen ii) limited oxygen
f) Explain the halogenation reaction of alkanes

*include bromination and chlorination

g) Explain the monosubstitution of alkane containing
equivalent and non-equivalent type of hydrogen
atoms
h) Illustrate the free radical monosubstitution
mechanism of alkanes

CHEMICAL PROPERTIES OF ALKANES in excess O2

COMBUSTION in limited O2

HALOGENATION reagent :
Br2 @ Cl2
Type of reaction :
free radical substitution condition :

under uv light @
hv

product: 48
haloalkane

Combustion – in excess oxygen

• Reactant : alkane and excess O2
• Product: CO2 and H2O

Example:

49

49

Combustion – in limited oxygen

• Reactant : alkane and limited O2
• Product: CO and H2O

Example:

Combustion – in very limited oxygen

• Reactant : alkane and limited O2
• Product: C (soot) and H2O

Example:

50

50

Halogenation

Type of reaction : Free radical substitution
Reagent : X2 ( X= Br2 @ Cl2)
Condition : under uv light @ hv
Product : haloalkane

General equation:

R–H + X2 R–X + HX

alkane haloalkane

51

Example: Halogenation

i. CH4 + Cl2 CH3Cl + HCl

ii. CH3CH3 + Br2 CH3CH2Br + HBr

Example (i) and (ii) have one haloalkane
product only since the reactants (alkane)
contain identical hydrogen atoms.

52

52

Example: Halogenation
iii CH3CH2CH3 + Cl2
CH3CH2CH2Cl +

(minor product)

CH3CH(Cl)CH3 + HCl

(major product)

Example (iii) has two haloalkane products
because the reactant (alkane) contains non-
identical hydrogen atoms.

CH3CH(Cl)CH3 is the major product due to the
stability of free radical.

53

53

ALKANES

iv CH3CH2CH2Cl (45%)
. (55%)
CH3CH2CH3 + Cl2 1–chloropropane (minor) (99%)

propane + 54

fast and unselective! CH3CHCH3 54
Cl
CH3CH2CH3 + Br2
2–chloropropane (major)
propane
CH3CHCH3
Slow and selective
Br

2–bromopropane

Reaction Mechanism
of Alkanes

55

Example 1 : CH3Cl + HCl
CH4 + Cl2

Steps are … Most important !
1. Initiation
2. Propagation
3. Termination

56

Step 1 – Initiation

-uv light or heat provides the energy needed for
homolytic bond cleavage.
-Free radicals are formed.
-Reaction begins !

Cl Cl Cl • + • Cl

57

Step 2 - Propagation
One radical generates formation of another.

(i) Cl • + HCl + •CH3

(ii) CH3 • + Cl Cl CH3 Cl + •Cl

58

Step 3 - Termination
Recombination of two free radicals:

(i) CH3 • + Cl • CH3 Cl

(ii) CH3 • + • CH3 CH3 -CH3

(iii) Cl • + • Cl Cl-Cl

59

Example 2 :
Write a complete mechanism for the halogenation
reaction of 2-methylpropane with bromine.
Answer :

Step 1 - Initiation

60

Step 2 - Propagation
(i)

(ii)

61

Step 3 - Termination

(i)
(ii)

(iii)

62

Example 2 :

Bromination reaction of certain alkanes can be used for
laboratory preparations, for example in the preparation of
bromocyclopentane from cyclopentane.
Give the mechanism for the reaction.

Answer : Br • + • Br

Step 1 : Initiation

Br Br

63

Step 2 : Propagation ● + H—Br
H + ● Br
Br
● + Br—Br
+ ● Br

64

Step 3 : Termination Br2
Br ●• + ●• Br

● +●

● + ●• Br Br

65

TYPES OF
REACTION

5.2 ALKENES

66

Preparation IUPAC Physical
Nomenclature Properties
1.Dehydration of alcohols
– include Mechanism 5.2 : Chemical Test
2.Dehydrohalogenation ALKENES
of haloalkanes

Chemical Reaction 1.Baeyer’s test

A. Addition reaction of alkenes 2.Bromine in CH2Cl2
1. Hydrogenation 3.Bromine water

2. Halogenation – In inert solvent(CH2Cl2)
3. Halogenation – In water (halohydrin formation)

4. (a) hydrogen halides (Markovnikov‘s Rule)- Mechanisme

(b) Addition of HBr to alkenes in the presence of peroxide

(Anti-Markovnikov)

5. Acidified water– Mechanisme

B. Oxidation of alkenes

67

Learning Outcomes

(a) Give the name of alkenes according to the IUPAC
nomenclature

(b) Give the structural formulae of the following alkenes
i) straight chain and branched alkenes (parent
chain ≤ C10)
ii) Cyclic alkenes (C3 – C6)
iii) Simple dienes (C4 – C6)

(c) Explain boiling point of isomeric alkenes.

*limit to cis-trans isomers only

68

Alkenes

- General formula CnH2n , n ≥ 2.
- Functional group : C=C double bond
- C═C ☞ 1 σ bond and 1 π bond

- Restricted rotation of carbon-carbon double
bond causes cis-trans isomerism

Cycloalkenes

- General formula CnH2n-2
- Isomeric to alkynes CnH2n-2

69

IUPAC
Nomenclature

70

STEP 1:

Determine the parent name by selecting the
longest chain that contains the double bond and
change the ending ‘-ane’ in alkane to‘-ene’.

STEP 2:

When the chain contains more than three carbon
atoms, numbering is needed to indicate the
location of the double bond.

71

876 54 3 21

3-octene 72
5-octene

STEP 3:

Indicate the position of the substituent by the
number of the carbon atoms to which they are
attached.

4 32 1

2-methyl-2-butene 73
3-methyl-2-butene

STEP 4:

If the alkene contains more than one
double bond, change the ending ‘-ene’ to:
☞ diene – if there are two double bonds.
☞ triene – if there are three double bonds.

74

1,3-butadiene
1,3-dibutene

1,3,5-heptatriene
1,3,5-triheptene

75

STEP 5:

In Cycloalkenes

Number the carbon atoms with a double
bond as 1 and 2, in the direction that gives
the substituent encountered first with a
small number.

76

2
31

45

1-methylcyclopentene

2-methylcyclopentene

77

1 2
1
2
6

3

54 5 63
4
CH3 CH3

3,5-dimethylcyclohexene

4,6-dimethylcyclohexene

78

STEP 6:

When alkenyl groups are perform as a
substituent it is known as:

RCH=CHR -H RCH=CH-
alkene alkenyl

CH2=CH- CH2=CHCH2-
vinyl group allyl group

79

STEP 7:

Prefixes cis- and trans- are used if the alkene
shows geometrical isomerism.

cis-2-pentene trans-2-pentene

80