Reaction of Alkene

5.2 Alkenes Hydrocarbons

IUPAC Nomenclature IUPAC name
Alkanes IUPAC names end with –ene suffix. ethene
propene
Structural formula 1-butene
CH2=CH2
CH3CH=CH2 1-pentene
CH3CH2CH=CH2 1-hexene
CH3CH2CH2CH=CH2 1-heptene
CH3CH2CH2CH2CH=CH2 1-octene
CH3CH2CH2CH2CH2CH=CH2 1-nonene
CH3CH2CH2CH2CH2CH2CH=CH2 1-decene
CH3CH2CH2CH2CH2CH2CH2CH=CH2
CH3CH2CH2CH2CH2CH2CH2CH2CH=CH2

46

47

48

5.2 Alkenes Hydrocarbons

IUPAC Nomenclature

• When the alkene has more than one C=C
• - ene is changed to

diene ( two C=C bonds)
triene ( three C=C bonds)

4 321

1,3-butadiene

1 2 3 45 67

1,3,5-heptatriene

49

5.2 Alkenes Hydrocarbons

IUPAC Nomenclature

Substituents with C=C bonds (alkenyl groups); common names are used.

methylene
vinyl

allyl

1 46
C H2 CH CH2 C H C H2 CH C H CH3

H2C CH 4-vinyl-1,6-octadiene

50

5.2 Alkenes Hydrocarbons

IUPAC Nomenclature

In cycloalkenes, the carbon atoms of the ring double bond are numbered 1 and 2 in
the direction that gives the substituent encountered first the smaller number.

3-methylcyclopentene

5-ethyl-3-methylcyclohexene

NOT

4-ethyl-6-methylcyclohexene

51

5.2 Alkenes Hydrocarbons

IUPAC Nomenclature

• For stereoisomeric alkenes, prefix cis and trans are used to distinguish the
two isomers.

cis-2-pentene trans-2-pentene

cis-1,2-dichloroetene trans-1,2-dichloroetene

52

5.2 Alkenes Hydrocarbons

Physical Properties of Alkenes

• Alkenes of C1 - C4 are gases

C5 - C14 liquid

> C15 solid

• Boiling of alkenes increase smoothly with molecular weight.

Molecular sizes / contact surface area increases

Strength of van der Waals forces increase

• Cis have higher boiling points compared to trans-isomers.

Reason : Cis isomer has net dipole moment, it is polar molecule that
leads to stronger van der Waals forces.

Cis isomer Trans isomer 53

5.2 Alkenes Hydrocarbons

Stability of trans- and cis- Alkene

Cis alkene is less stable than trans isomers because the electron repulsion between
alkyl groups on the same side of the double bond (stearic effect).

Greater steric repulsion Less steric repulsion

Cis isomer Trans isomer

54

5.2 Alkenes Hydrocarbons

Preparation of Alkenes

Two methods to prepare alkenes through elimination reaction;
i) Dehydration of alcohols
ii) Dehydrohalogenation of haloalkanes

55

5.2 Alkenes Hydrocarbons

Preparation of Alkenes
1st method: Dehydration of alcohols (eliminate H2O)

elimination of H2O ( H and OH are removed from adjacent carbon atoms)
alcohol reacts with hot, concentrated H2SO4 or H3PO4
(dehydrating agents)

+ H2O

example The relative ease of alcohols undergo dehydration:
3° ROH > 2° ROH > 1° ROH

+ H2O

Propene 56

5.2 Alkenes Hydrocarbons

Preparation of Alkenes

2nd method: Dehydrohalogenation of Haloalkanes (eliminate HX)

elimination of hydrogen halide from alkyl halide or haloalkane (RX).
using hot, alcoholic KOH or NaOH

+ HBr

example

+ HBr

Propene 57

5.2 Alkenes Hydrocarbons

Preparation of Alkenes

• The relative ease of alcohols undergo dehydration:
3° ROH > 2° ROH > 1° ROH

• This behaviour related to the relative stabilities of carbocations that form in the
reaction.
Carbocations stability; 3° > 2° > 1°

• Certain alcohols undergo dehydration to produce more than one alkenes

• from the mixture of alkenes produced, the more stable alkene is the major product
(predicted by Saytzeff’s rule)

58

5.2 Alkenes Hydrocarbons

Preparation of Alkenes

What Is Saytzeff’s Rule ?

‘In elimination reaction, the alkene with more highly substituted
carbon-carbon double bond (greater number of alkyl groups
attached to the C=C) is the major & more stable product.’

>>>>

di (geminal)

(trans) (cis)

>>

59

5.2 Alkenes Hydrocarbons
Preparation of Alkenes

Example 1

Example 2

Which one is the major product??

Saytzeff’s rule

‘In elimination reaction, the alkene with more highly substituted carbon-carbon

double bond (greater number of alkyl groups attached to the C=C) is the major &

more stable product. 60

5.2 Alkenes Hydrocarbons

Reaction Mechanism

General rule for writing mechanisms:
draw all the bonds involved
show one step at a time
use curve arrow to show movement of electron

61

5.2 Alkenes Hydrocarbons

Preparation of Alkenes

Example 1:

Chemical Equation for Dehydration of 2-butanol

62

63

64

65

5.2 Alkenes Hydrocarbons

Exercise 1:

Chemical Equation for Dehydration of 2-methyl-1-butanol

+

+ H2O

66

67

1 carbocation

3 carbocation 68
(more stable)

69

5.2 Alkenes Hydrocarbons

Exercise:

1. Write the equation and give the IUPAC name for the alkene formed in each of the
following reaction
a) i. CH3CH(OH)CH2CH3
ii. (CH3)3COH
iii. CH2(OH)CH(CH)3CH2CH3

b) i. CH3CH(Cl)CH2CH3
ii. (CH3)3CCl
iii. CH2(Br)CH(CH3)CH2CH3

2. State Saytzeff rule. Predict the more stable alkene

i. 2-methyl-2-pentene or 2,3-dimethyl-2-butene
ii. trans-2-hexene or 2-methyl-2-pentene
iii. 1-hexene or 3-hexene

70

5.2 Alkenes Hydrocarbons

Exercise:

3. Outline a synthesis of propene from each of the following:
i. Propyl chloride ii. Isopropyl chloride
iii. Propyl alcohol iv. Isopropyl alcohol

4. Outline a synthesis of cyclopentene from:
i. Bromocyclopentene ii. Cyclopentanol

5. Complete the following equation;

i.

ii.

71

5.2 Alkenes Hydrocarbons

Exercise:

6. Dehydration of 2-methylcyclopentanol gives a mixture of three alkenes; A, B and C.
State the major product. Explain the rule used to determine the major product.
Propose mechanisms to account for the formation of the major product.

Answer:

72

73

5.2 Alkenes Hydrocarbons

Chemical Properties of Alkenes

• Alkenes are more reactive than alkanes:
– alkenes are unsaturated, contain C=C; double bond made up of bond &
bond
– bond is a weaker bond ( electrons less strongly attracted to the carbon’s
nucleus; electrons are delocalised), thus easier to be broken.
– Thus C=C is a site of high electron density (nucleophilic site), becomes the
site of electrophilic attack.
– Reactions of alkenes are mainly electrophilic addition reaction following
Markovnikov’s rule.

+ AB AB

74

5.2 Alkenes Hydrocarbons
Chemical Properties of Alkenes

Markovnikov’s rule

‘In electrophilic addition of alkenes, the electrophile adds to the
carbon atom of the carbon-carbon double bond with greater number
of hydrogen atoms.’

75

5.2 Alkenes Hydrocarbons

Chemical Properties of Alkenes

a) Hydrogenation (+H2)

• Addition of H2(g) in the presence of catalyst (Pt/Pd/Ni)
• 1 mole of alkene with one C=C requires 1 mole of H2(g)

+ H2(g) Pt

Example: HH

+ H2(g) Pd

+ H2 (g) Ni

76

5.2 Alkenes Hydrocarbons

Chemical Properties of Alkenes

b) Halogenation (+ X2) X2 = Cl2 or Br2

i) Halogenation in inert solvent (eg CH2Cl2)

• React rapidly at at room temperature or even in the dark to form
vicinal dihalides.

+ X2 CH2Cl2

XX

Example:

+ Br2 CH2Cl2

Reddish

brown Br Br
colourless

Bromine test is used to test for the presence of C = C 77
Observation: reddish brown colour of bromine disappears

5.2 Alkenes Hydrocarbons

Chemical Properties of Alkenes

c) Halogenation (+ X2 (aq))

ii) Halogenation in aqueous solution

• Addition halogen in water.
• Halohydrin is the major product

+ X2 H2O +

X OH XX
halohydrin (minor)

Example: (major)

+ Br2 H2O CH3
+ Br

(major) Br 78
(minor)

5.2 Alkenes Hydrocarbons

Chemical Properties of Alkenes H OH

d) Hydration (+ H2O /H+) - (acidified water)

• reaction to prepare secondary & tertiary alcohols
• product determined by Markovnikov’s rule

+ H2O H2SO4

Example:

79

5.2 Alkenes Hydrocarbons

Chemical Properties of Alkenes

e) Hydrohalogenation (+ HX)

• Reactivity : HI > HBr > HCl
• Product formed is determined by Markovnikov’s rule

‘In an electrophilic addition of HX to an alkene, H+ adds to the carbon
of the C=C bond with the greater number of hydrogen atoms.’

+ HX

HX

Example:

80

5.2 Alkenes Hydrocarbons

Chemical Properties of Alkenes

e) Hydrohalogenation (+ HBr)

• In the presence of peroxides, ROOR (eg: H2O2), the addition of HBr to
alkenes will occur in an anti-Markovnikov manner, that is the hydrogen atom
of HBr adds to the C atom of C=C with fewer hydrogen atoms.

Example:

+ HBr H2O2

Br H

**** Only for the addition of HBr in the presence of peroxides, ROOR 81
(eg: HOOH, CH3OOCH3,)

5.2 Alkenes Hydrocarbons
Chemical Properties of Alkenes
Oxidation of alkenes

• Reaction of alkenes with oxidizing agents like KMnO4, O3
• Determining the position of C = C

• Oxidation & cleavage of C=C to determine of the position of the C=C double bond

• 3 types of oxidation:
a) KMnO4, OH-, cold(Baeyer’s test)

b) Ozonolysis using (i)O3 (ii)Zn, H2O or (i) O3 (ii) (CH3)2S
c) KMnO4, H+, hot solution

82

5.2 Alkenes Hydrocarbons

Chemical Properties of Alkenes + MnO2
i) KMnO4, OH-, cold (Baeyer’s test)
Brown
KMnO4(dilute) precipitate
Cold, OH-
(purple) diol
• 1, 2-diol (glycol) is formed.

Baeyer’s test is used to test for the presence of C = C
Observation: purple colour of KMnO4 is decolourise

83

5.2 Alkenes Hydrocarbons

Chemical Properties of Alkenes

ii) Ozonolysis (Ozone, O3)

• Oxidative cleavage of alkenes; carbonyls (ketone or aldehyde are formed.

1. O3 C =O + O=C

2. Zn, H2O

1. 1. O3 CH3 CH3
C =O + O=C CH3
2. Zn, H2O H3C

2. CH3 CH3 1. O3 CH3 CH3

HC C CH2CH3 2. Zn, H2O H C=O + O= C CH2CH3
84

5.2 Alkenes Hydrocarbons
Chemical Properties of Alkenes

ii) Reaction of alkenes with hot KMnO4

• Oxidative cleavage of alkenes; ketone or carboxylic acid or CO2 are formed

KMnO4 , H+(conc.) C= O + O=C
hot

1. KMnO4 , H+(conc.) CH3 CH3
O=C CH3
hot H3C C =O +

2. KMnO4 , H+(conc.) CH3 CH3
O=C CH3
hot HO C=O +

KMnO4 / H+(conc.) CH3

3. hot CO2 + H2O + O=C CH385

5.2 Alkenes Hydrocarbons
Reaction Mechanism

General rule for writing mechanisms:
draw all the bonds involved
show one step at a time
use curve arrow to show movement of electron

86

5.2 Alkenes Hydrocarbons

Hydrohalogenation Reaction

Reaction
Equation

Reaction Step 1: Protonation to form carbocation
Mechanism

Step 2: Reaction of carbocation with nucleophile

87

5.2 Alkenes Hydrocarbons

Hydration Reaction

Reaction
Equation

Reaction Step 1: Protonation to form carbocation
Mechanism

Rearrangement

88

5.2 Alkenes Hydrocarbons
Hydration Reaction

Reaction Step 2: Reaction of carbocation with nucleophile
Mechanism

Step 3: Deprotonation-loss of proton, H+ to form alcohol

89

5.2 Alkenes Hydrocarbons

Chemical Test

Unsaturation Test for Alkenes:
• Baeyer’s test
• Bromine tests

90

5.2 Alkenes Hydrocarbons

Unsaturation Test for Alkenes
i) Baeyer’s test (KMnO4, OH-, cold)

HH KMnO4, OH-
cold
H C C CH3 + MnO2

(purple) Brown
precipitate
diol

Observation : The purple colour of KMnO4 solution decolourised and
brown precipitate is formed.

Exercise:

KMnO4, OH- 91
cold

5.2 Alkenes Hydrocarbons

Chemical Properties of Alkenes
ii) Bromine Test (Bromine in CH2Cl2)

Br2
CH2Cl2

Observation : The reddish brown colour of bromine solution disappear.
Exercise:

Br2
CH2Cl2

92

5.2 Alkenes Hydrocarbons

Chemical Properties of Alkenes
iii) Bromine Water Test (Br2 in H2O)

HH Br2 +
H C C CH3 H2O

(major)

Observation : The yellowish brown colour of bromine solution disappear.

Exercise:

Br2
H2O

Unsaturation Tests are tests to distinguish alkenes from alkanes and benzene93