LECTURE NOTE SK025 (ORGANIC) KMKt

cis-1,3-dimethylcyclopentane

trans-1,3-dimethylcyclopentane

81

Boiling point of
Alkenes

82

Boiling Point of Isomeric Alkenes:

• Cis and trans alkenes have different physical properties.
• For example, cis-2-butene has a higher boiling point

(4 0C) than trans-2-butene (1 0C).

• This differences arises because the C-C single bond
between an alkyl group and one of the double bond
carbons of an alkene is slightly polar.

• In cis isomer, the two C-C bond dipoles reinforce each
other, yielding a small net molecular dipoles.

Example trans-2-butene
: cis-2-butene CH3 H
C═C
HH
H CH
C═C
No net dip3ole
CH3 CH Lower boiling point (0.9 0C)

A small net3dipole
Higher boiling point (3.7 0C)

cis-molecule has higher boiling point because the
molecule is polar, therefore the intermolecular
forces (van der Waals forces) are relatively stronger.

In a trans isomer, the two bond dipoles
cancel each other.

* Boiling point:
cis-isomer > trans- isomer

Table show the boiling point for a few
stucture of alkene

Alkene Structure Boiling point
(oC)
Ethene
Propene CH2=CH2 -104.0
1-Butene CH3-CH=CH2 -47.0
Isobutene CH2=CH-CH2-CH3 -6.5
Trans-2-Butene CH2=CH(CH3)2 -7.0
Cis-2-Butene CH3-CH=CH-CH3 0.9
1-Pentene CH3-CH=CH-CH3 3.7
Trans-2-Pentene CH2=CH-CH2-CH2-CH3 30.0
Cis-2-Pentene CH3-CH=CH-CH2-CH3 36.0
1-Hexene CH3-CH=CH-CH2-CH3 37.0
CH2=CH-CH2-CH2-CH2-CH3 63.5
1-Heptene
CH2=CH-CH2-CH2-CH2-CH2-CH3 115.0

Preparation of
Alkenes

86

Learning Outcomes

(d) Describe the preparation of alkenes through :
i) Dehydration of alcohols;
ii) Dehydrohalogenation of haloalkanes.

(e) Illustrate the mechanism of (d) I
(f) State the Saytzeff’s rule.
(g) Explain major product formed using Saytzeff’s

rule

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TYPE OF Elimination
REACTION
PREPARATION OF ALKENES
NAME OF 1. Dehydration of alcohols
REACTIONS ☞ loss of H2O from alcohol

2. Dehydrohalogenation of
haloalkanes

☞ loss of HX from haloalkane

NAME OF RULE THAT Saytzeff’s rule
BEING FOLLOWED
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1. Dehydration of alcohols

reagent conc. H2SO4 or conc. H3PO4
condition heat

general conc. H2SO4
equation

☞ The reaction produces a mixture of product.

☞ The major product obtained can be predicted by
using Saytzeff’s rule.

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Saytzeff’s Rule

The major product in elimination reaction is the
most stable alkenes which has the most highly

substituted double bond.
(greater number of alkyl groups attached to the C=C)

STABILITY INCREASES

90

Example 1: Determine the major and minor product

2 alkyl groups 1 alkyl group
(major product) (minor product)

Example 2: Determine the major and minor product

3 alkyl groups 2 alkyl groups
(major product) (minor product)

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Examples : conc. H2SO4

(i)

(ii)

major product

minor product
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(iii)

CH3 H conc. H2SO4 CH3 CH3
CH3 C C CH3 Δ
CH3 C C CH3 H2C C CHCH3
CH3 OH
CH3 CH3
(major (minor
product) product)

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REARRANGEMENT

1,2-methanide shift 1,2-hydride shift

the migration of -CH3 the migration of -H
(methanide ion) from the (hydride ion) from the
C atom that adjacent to C atom that adjacent
the carbocation. to the carbocation.

***Rearrangement step occur to form
more stable carbocation

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Example :

CH3+ 1,2-methanide shift CH3

+
CH3 C CH CH3 CH3 C CH CH3

CH3 CH3
2o carbocation 3o carbocation

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Example :

H 1,2-hydride shift H
++
CH3 C CH CH3 CH3 C CH CH3

CH3 CH3
2o carbocation 3o carbocation

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Example 1:

Dehydration of 2–butanol gives a mixture of two alkenes.
Propose mechanisms to account the formation of major
alkene (major product).

major product

2-butanol

minor product

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Step 1: Protonation of –OH group

HH HH
H3C C C •O+• H
H3C C C ••O•• H + H •O+• H
H CH3 H
H CH3 H protonated alcohol

28

Step 2: Formation of carbocation

H H •O+• H HH + •• •O• H
H3C C C H3C C C+

H CH3H H CH3 H

carbocation

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Step 3: Formation of alkene

☞ C atom that adjacent to the carbocation loses
proton, H+

HH HH H •O+• H
H3C C C+ H
H3C C = C CH3 +
H CH3 major product

+

H ••O•• H

100

Example 2:

Write the mechanism for the formation of major product for
dehydration of 3,3-dimethyl-2-butanol.

Step 1: Protonation of –OH group

CH3 +H •O+• H CH3
CH3–C—CH–CH3
CH3–C—CH–CH3
CH3•••O•H H CH3••O+ H

H

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Step 2: Formation of carbocation

CH3 CH3 H

CH3–C—CH–CH3 CH3–C—C+ H–CH3 + •• •O•
CH3••O+ H
H CH3 H

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Step 3: Rearrangement

CH3 CH3

CH3–C—C+ H–CH3 1,2-methanide shift CH3–C+ —CH–CH3

CH3 CH3

2o carbocation 3o carbocation

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Step 4: Formation of alkene

☞ C atom that adjacent to the carbocation loses
proton, H+

CH3 H CH3
CH3–C+ —C–CH3 + H ••O•• H
CH3–C═C–CH3 +
CH3
CH3

H •O+• H
H 104

Example 3:

Show the mechanism for the formation of alkene Y
from alcohol X.

OH CH2–CH3
+ H2O
CH–CH3 conc. H2SO4
Δ

XY

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Step 1: Protonation of –OH group

H H
••+O H
•• O•• •O+• H
CH–CH3
CH–CH3 + H
+
H
•• •O• H
Step 2: Formation of carbocation H

H + •• •O• H
+•• O H H 106
CH–CH3
CH–CH3
+

Step 3: Rearrangement

H C+ H–CH3 1,2-hydride shift H
+ CH–CH3

2o carbocation 3o carbocation

Step 4: Formation of alkene

H ••O•• H + HH H
+ CH–CH3 CH–CH3

+H •O+• H
H 107

2. Dehydrohalogenation of alkyl halide

reagents Alcoholic OH- @ OR-
e.g: NaOH, KOH,KOCH3, CH3CH2ONa

condition reflux

general
equation

☞ The reaction produces a mixture of product.
☞ The major product obtained can be predicted by

using Saytzeff’s rule.

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Example 1: + HBr
Example 2 :

major product

minor product + HBr

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Chemical Properties
of Alkenes

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CHEMICAL PROPERTIES OF ALKENES 1) ELECTROPHILIC catalytic hydrogenation
ADDITION
halogenation
☞ in inert solvent
☞ in water
hydrohalogenation

Acidified water

ozonolysis

2) OXIDATION

hot, acidified KMnO4

cold, dilute alkaline KMnO4

(Baeyer’s Test) 111

Learning Outcomes

(h) Explain the addition reaction of alkenes with:

i. hydrogen in the presence of catalyst;

ii. halogen (Cl2 or Br2) in inert solvent (CH2Cl2);
iii. halogen (Cl2 or Br2) in water;
iv. hydrogen halides (HCl or HBr);

v. acidified water
(i) State Markovnikov’s rule

(j) Predict the products formed according to the
Markovnikov’s rule

(k) Illustrate the mechanism of electrophilic addition of

(h) iv and (h) v.

(l) Predict the product of the reaction between alkene and

hydrogen bromide in the presence of hydrogen peroxide/

acid peroxide according to anti-Markovnikov’s rule. 112

Reagent Electrophilic Addition
Product Catalytic Hydrogenation
Example:
H2 , Pt / Pd / Ni / Pd
alkane

H2, Pt

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Reagent Halogenation in Inert Solvent
Product
Example: X2, CH2Cl2 (X2 = Cl2, Br2 )
vicinal dihalide
(i)

(ii) CH2Cl2

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Halogenation in Inert Solvent

When the equation below are tested with Br2 in CH2Cl2,

Name of chemical test : Bromine test
Observations : Reddish brown colour of bromine

decolourised

115

Reagent Halogenation in Water
Product
X2, H2O (X2 = Cl2, Br2 )
halohydrin

Example 1:

***Follow Markovnikov’s Rule : 116
-OH carbon with fewer no. of H atoms

halogen carbon with greater no. of H atoms

Markovnikov’s Rule

In the electrophilic addition of alkenes, the
electrophile adds to the C atom of the C=C with

greater number of H atoms

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Halogenation in Water

Example 2:

118

Halogenation in Water

When the two equations below are tested with Br2 in H2O,

Name of chemical test : Bromine test 119

Observations : Reddish brown colour of bromine
decolourised

Reagent Hydrohalogenation
Product
Reactivity HX (X = Cl, Br)
haloalkane
HCl < HBr

Example 1:

120

Hydrohalogenation

Example 2 :

***Follow Markovnikov’s Rule : 121
halogen carbon with fewer no. of H atoms

H carbon with greater no. of H atoms

Reaction Mechanism : Hydrohalogenation

Example 1:
Write the mechanism for the hydrohalogenation reaction
of propene.

Answer:
Step 1 : Formation of carbocation and halide ion

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Reaction Mechanism : Hydrohalogenation

Step 2 : Halide ion reacts with the carbocation by
donating an electron pair

50

Reaction Mechanism : Hydrohalogenation

Example 2:

Write the mechanism for the major product of the
following reaction:

CH3–CH–CH═CH2 + H—Cl Cl
CH3 CH3–C–CH2–CH3

CH3

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Reaction Mechanism : Hydrohalogenation

Answer:

Step 1 : Formation of carbocation and halide ion

H H
CH3–C–CH═CH2
+ H—Cl slow + + Cl-
CH3
CH3–C–CH—CH3

CH3

Step 2 : Rearrangement

H 1,2-hydride shift H

+ + 125

CH3–C–CH—CH3 CH3–C–CH—CH3

CH3 CH3

2o carbocation 3o carbocation
(more stable)

Reaction Mechanism : Hydrohalogenation

Step 3 : Halide ion reacts with the carbocation by
donating an electron pair

H fast Cl
CH3–C–CH2–CH3
CH3–C+–CH—CH3 + Cl-
CH3
CH3

126

Hydrohalogenation
Anti-Markovnikov’s Rule

In the presence of peroxides, ROOR (eg: H2O2) the
addition occurs in an anti-Markovnikov rule

Example : + H—Br H2O2 Br
CH3–CH2–CH2
CH3–CH═CH2

Anti-Markovnikov’s Rule : 127
H atom added to C atom (Less H)
MAJOR product

***only occurred in the addition of HBr to an alkene

Reagent Acidified water
Product
H2O, dilute H2SO4 / H3PO4
alcohol

Example 1:

HH HH

H C C H + H2O dilute H2SO4 H C C H

H OH

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Acidified water

Example 2:

(major) (minor)

***Follow Markovnikov’s Rule :
-OH carbon with fewer no. of H atoms
H carbon with greater no. of H atoms

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Reaction Mechanism : Acidified water

Example 1:
Write the mechanism for the hydration reaction of propene.

Answer:
Step 1 : Formation of carbocation

130