Alkenes
Introduction
• Alkenes are unsaturated hydrocarbons that
contain one or more carbon-carbon double
bonds C=C, in their structures
• Alkenes have the general formula CnH2n
Nomenclature of alkenes and
cycloalkenes
• The IUPAC names of the alkenes end with –
ene and the location of the double bond takes
precedence over that of the side chains in
numbering the parent chain.
• The parent chain is the longest chain
• For open chain alkenes, the parent chain is
numbered from the end that gives the lowest
number to the first carbon of the double bond
to be reached.
CH3
CH3
1
C
H
C
2
3
H2
C
H2
C
4
5
CH3
6
3-methylhex-2-ene
CH3
CH3
H2
C
C
H
C
H
3-methylpent-1-ene
CH2
• For cyclohexenes, position 1 is always given to
one of the two carbons at the double bond.
• The ring atoms are numbered from the carbon
1 through the double bond in whichever
direction to reach a substituent first.
CH3
3
4
2
3-methylcyclohexene
5
1
6
Lets try
H 3C
C
CH2
C H 2C H 3
CH3
H 3C
C
CH3
Br
C
H
H2
C
CH3
The existence of cis-trans isomerism
• Cis-trans isomerism have the same molecular
formula and the same order of attachment of
atoms but differ in arrangement of their
atoms in space
• An alkene shows cis-trans isomerism if each
carbon atom of the double bond has two
different groups attached to it
Isomer
Structure
Cis
H 3C
Trans
H
CH3
C
C
C
H
CH3
H 3C
H
C
H
Name
Melting point (˚C)
Cis but-2-ene
-139
Trans but-2-ene
-106
Boiling point (˚C)
4
1
• Trans isomer is more stable than cis-isomer because
the two larger methyl groups are on the opposite side of
the double bond. Hence, there is less strain on the
molecule
• Cis-isomers are polar molecules because the bond
moments do not cancel out. The molecule thus has a net
dipole moment
• Trans-isomers are usually non polar because the bond
moment cancel out each other. The molecule has no net
dipole moment
• The cis-isomer will generally have a higher boiling
point because of its higher polarity which leads to
stronger dispersion forces between the molecules
• The cis-isomer will generally have a lower melting
point because of its lower symmetry. It is more difficult
to fit the cis-isomer into the crystalline structure
H 3C
CH3
C
H
H
CH3
C
C
H
Cis but-2-ene
Has a net dipole moment
C
H 3C
H
Trans but-2-ene
Has no net dipole moment
Exercise
• Name each compound below and show the
configuration about each double bond using
the cis-trans system
H
C H 2C H 3
C
C H 3C H 2
H 3C H 2C H 2C
C H 2C H 3
C
C
H
H
C
H
General preparation of alkenes
1. Dehydration of alcohol
A lcohol
dehydration
A lkene
May be achieved by heating with excess concentrated
H2SO4 or H3PO4 or Al2O3 with heat to 225 oC.
H
H
alum ina
H
C
C
H
H
H
C
+
C
heat
H
H
H
H
OH
H
C on cen trated H 2 S O 4
H
C
C
C
OH
1 70 o C
H
H
H
P ropan-1-ol
H
H 2O
2. Haloalkanes with KOH/ethanol
• Haloalkanes undergo elimination reaction
when refluxed with purely KOH/ethanol.
R-X + KOH alkene + K-X + H2O ; X = Br, I, F, Cl
H
H
H
H
C
C
C
H
H
Br
H
B rom o propane
+
KOH
re fu lx
H
H
H
H
C
C
C
H
P rop ene
H
+ K B r + H 2O
H
H
CH3 H
C
C
C
H
Cl
H
H
+
2-chloro-2-methylpropane
KOH
refu lx
H
H
CH3 H
C
C
H
C
H
+ K C l + H 2O
Physical properties of alkenes
1. Alkenes have lower melting and boiling
temperatures than alkanes
2. Weak dispersion forces between molecules.
The forces are so weak that the lower alkenes
(ethene, propene, butene) exist as gas at
room temperature and pressure
3. The melting and boiling point increase with
increasing number of carbon
Reactivity of alkenes
• The cloud of electron which forms the p bond
in alkene lies above and below the plane of
the molecule. In this position, the p electrons
are susceptible to attack by electrophilies
• Addition reaction are the most common
reactions of alkenes
• The π bond of the carbon-carbon double bond
breaks in an order to release electrons to form
two new σ bonds with the reactant molecule
and a saturated molecule is formed
C
C
+
X
Y
C
C
X
Y
Addition reactions of alkenes
1. Addition of Hydrogen (Hydrogenation)
• The addition of hydrogen to an alkenes can
be carried out by passing hydrogen and the
alkenes, both in the gaseous state, over a
metal catalyst
N i / Pt
C
C
+
H
H
C
C
H
H
150 o C 5 atm
• Example hydrogenation
H
H
H
H
C
C
H
H
N i / Pt
+
C
C
H
H
H
H
150 o C 5 atm
H
H
ethane
ethene
H
H
H
H
C
C
H
H
N i / Pt
H 3C
C
C
CH3 +
H
H 3C
H
o
150 C 5 atm
ethene
ethane
CH3
Exercise
• Write down the structure of the products, if
any, of the following reactions
C H 3C H
CHCH3
Ni
+
H2
1 5 0 o C , 5 atm
Ni
+ H2
1 5 0 o C , 5 atm
2. Addition of halogen (Halogenation)
• Both chlorine, Cl and bromine, Br, add rapidly
to the double bond at room temperature
without the need of any catalyst
C
C
X2 = Cl2 or Br2
+
X2
room tem perature
C
C
X
X
• For example
H
H
H
C
C
H
+ C l2
E thene
H
H
H
C
C
Cl
Cl
1,2-dichloroethane
Br
+ B r2
ro o m tem p
Br
cy c lo h e x en e
H
1 ,2 -d ib ro m o cy c lo h ex a n e
• The mechanism of addition of bromine to ethene
is as follows:
The mechanism is described as electrophilic addition.
Because the bromine molecule acts as an electrophile
when it is attracted to the electron-rich carbon-carbon
double bond
When a bromine molecule approaches an ethene
molecule, the p electron cloud of ethene interacts
with the approaching bromine molecule, causing a
polarisation of the Br-Br bond.
H
H
H
C
C
H
B r 
Br

The double bond of ethene induces a
dipole on Br2
Electron move from the double bond towards the
partially positively charge (Brδ+) and at the same
time electrons in the Br-Br bond are repelled to
the partially negatively charged bromine atom
(Brδ-). This results in the formation of two ions, a
carbocation and a bromide ion, Br-. A carbocation
is a species that contains a carbon atom with only
three bonds to it and bearing a positive charge.
H
H
H
H
C
C
slow
H
C
C
H
H
H
+ B r-
B r 
Br
Br

electrophilic attack
carbocation
brom ide ion
The carbocation is very unstable and quickly
combines with the bromide ion, Br- to form 1,2dibromoethane.
H
H
H
H
C
C
Br
Br
fast
H
C
Br
C
H
B r-
H
H
1,2 -dibro m o ethan e
Exercise
• Name and draw structural formulae for the
products of the following halogenation
process
C H 3C H
CHCH3
+
+ B r2
B r2
3. Reaction with hydrogen halides
• Hydrogen halide readily add to the carboncarbon double bond in the cold to give
haloalkanes
ethene
H
H
H
C
C
H aloalkane
H
+
H
X
H
H
C
C
H
X
cold
H
H
• The rate of addition decreases in the order
H-I > H-Br > H-Cl because the bond energy of
the hydrogen halides increases from H-I to
H-Cl
• Addition halide to unsymmetrical alkenes
– When propene reacts with a hydrogen halide such
as hydrogen bromide, HBr, there are two possible
products
H
H
H
H
C
C
H
Br
co ld
H 3C
C
C
H
H
Br
H 3C
co ld
H
1-b ro m o p ro p an e (m ino r)
H 3C
H
H
C
C
Br
H
H
2 -bro m o p ro p an e (m ajo r)
– Markovnikoff’s rule
• In the addition of hydrogen halide to a carbon-carbon
double bond in an unsymmetrical alkene, the hydrogen
atom of the hydrogen halide adds to the carbon atom
of the double bond with the greatest number of
hydrogen atoms.
U nsym m etrical double bond
H 3C
H
H
C
C
H
C arbon atom w ith the greatest num ber of hydrogen atom s
– Example of Markovnikoff’s rule
H 3C
H2
C
C
H
CH2
+ HBr
H 3C
H2
C
H
C
CH3
not H 3 C
H2
C
H2
C
CH2
Br
Br
CH3
CH3
CH3
+ HCl
Cl
n ot
Cl
– Stability of carbocation
• When the double bond in propene attacks the
hydrogen in the acid, two different carbocations can be
formed. The initial step in this reaction involves the
formation of the carbocation. Two different carbocation
can be formed.
H 3C
H
H
C
C
H
+
H
Br
H 3C
H
H
C
C
+
H
B r-
H
p rim ary carb o catio n (less stab le)
H 3C
H
H
C
C
H
+
B r-
H
S eco nd ary carb o catio n (m o re stab le)
– The secondary carbocation are more stable than
the primary. This is because in secondary there
are two alkyl groups pushing electrons onto the
positively charged carbon atom.
– The secondary carbocation combine with Br- to
form 2-bromopropane.
H 3C
H
H
C
C
H
H
+
B r-
H 3C
H
H
C
C
Br
H
2 -b ro m opro p an e
H
– The order of stability carbocations is as follows:
R
R
C
H
R
>
R
Tertiary
C arbocation
R
C
>
R
H
Secondary
C arbocation
increasing stability
H
C
>
H
Prim ary
C arbocation
H
# T he R group is alkyl group
C
H
M ethyl
C arbocation
Example
• Arrange the following alkenes in order of
increasing reactivity on addition of hydrogen
bromide.
CH2
H 2C
ethene
CH2
(C H 3 ) 2 C
2-m ethylpropene
C H 3C H CH CH3
but-2-ene
– The relative stabilities are directly related to the
stabilities of the intermediate carbocations
formed.
– 2-methylpropene is the most reactive because it
forms the tertiary carbocation, the most stable
carbocation
(CH 3 ) 2 CC H 3
– But-2-ene forms the secondary carbocation
CH 3 CH CH 2 CH 3
– Ethene forms the primary carbocation, thus it is
the least reactive CH 3 CH 2
– The order of increasing reactivity is
ethene <but-2-ene < 2-methylpropene
Exercise
• When but-1-ene,
H 3C
H2
C
C
H
CH2
reacts with hydrogen bromide, what are the
structures of the two possible intermediate
carbocations formed?
• Which one of these two ions is more stable?
Explain your answer.
• What is the major product?
Exercise
• Name and draw the structural formula for the
major products
C H 3C H
H 3C
CHCH3
C
CH2
+ HI
+ HBr
CH3
CH2
+ HCl
4. Reaction with bromine water
• An alkene reacts readily at room temperature
with an aqueous solution of bromine to
produce a mixture of products, bromoalcohol
and dibromoalkane. The bromine water
would be decolourised.
C
C
+ 2 B r2 + H 2 O
C
C
Br
OH
B rom oalcohol
+
C
C
Br
Br
D ibrom oalkane
+ HBr
+ H 2O
C
C
+ B r2
C
Br
C
+ B r-
C
C
Br
OH
B rom oalcohol
C
C
Br
Br
D ibrom oalk ane
• The addition reaction occurs in the way that
yields more stable carbocation. Follow the
Markovnikoff’s rule.
Exercise
• Write and overall equation for the reaction of
propene with bromine water
C H 3C H
CH2
+ 2B r 2
+ H 2O
room conditions
5. Reaction with concentrated suphuric (VI) acid
• An alkene undergoes addition reaction with
concentrated sulphuric (VI) acid in the cold to
give alkyl hydrogensulphate.
cold
C
C
+
H 2SO 4
C
C
O
O
S
H
O
A lkyl hydrogensulphate
O -H
• When added to the water and warmed, the
alkyl hydrogensulphate is converted to an
alcohol. This is a hydrolysis reaction. Breaking
up a compound by reacting it with water.
C
C
O
O
w arm
+ H 2O
S
C
C
H
OH
+
H 2S O 4
H
O
A lkyl hydrogensulphate
O -H
A lcohol
• The addition of concentrated sulohuric acid to
unsymmetrical alkenes follows Markovnikoff’s
rule
6. Reaction with stem (hydration)
• When steam and ethene passed over a
catalyst, ethanol is produced. A temperature
of 330 oC and a pressure of 60 atm are used
in the presence of phosphoric acid catalyst.
H 3P O 4
H 2C
CH2
+ H 2O
H 3C
330 o C , 60 atm
• The hydration of alkenes follow
Markovnikoff’s rule
H2
C
OH
• For example, the hydration of propene
H 3C
H
H
C
C
H
+ H 2O
H 3P O 4
H 3C
H
H
C
C
OH
H
H
330 o C , 60 atm
H
H
C
C
H
OH
not
H 3C
H
Exercise
• Name and draw structural formula for the
major products of these addition reactions of
alkenes
C H 3C H
H 3C
CHCH3
C
CH2
+ H 2O
H 3PO 4
330 o C , 60 atm
+ H 2O
H 3P O 4
330 o C , 60 atm
CH3
7. Reaction with potassium manganate (VII)
(Oxidation)
• An alkene react with dilute potassium
manganate in an acidic or alkaline solution to
give a diol
C
C
+ [O ]
dilute M nO 4 - / H +
+ H 2O
room tem p
C
C
OH
OH
8. Reaction with oxygen
• Like all hydrocarbon, alkenes burns in air to
produce carbon dioxide and water
Alkene + O2  CO2 + H2O
Thank you..