Post on 23-Dec-2015
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 CH
CH2C
H2C CH3
CH3
1 2 3 4 5 6
3-methylhex-2-ene
CH3
H2C C
HCH
CH2
CH3
3-methylpent-1-ene
• 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
1
23
4
5
6
3-methylcyclohexene
Lets tryH3C C
CH3
CH2
H3C C
CH3
CH
H2C CH3
Br
CH2CH3
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 Cis Trans
Structure
Name Cis but-2-ene Trans but-2-ene
Melting point (˚C) -139 -106
Boiling point (˚C) 4 1
C C
CH3
H
H3C
H
C C
CH3
H
H
H3C
• 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
C C
CH3
H
H3C
H
C C
CH3
H
H
H3C
Cis but-2-ene Trans but-2-ene
Has a net dipole moment Has no net dipole moment
Exercise
• Name each compound below and show the configuration about each double bond using the cis-trans system
CH3CH2
C C
CH2CH3H
H
C C
CH2CH3H3CH2CH2C
HH
General preparation of alkenes
1. Dehydration of alcohol
Alcoholdehydration
Alkene
May be achieved by heating with excess concentrated H2SO4 or H3PO4 or Al2O3 with heat to 225 oC.
C Calumina
heatC C
H
H
H
H
+ H2O
H
H
H
H
H
OH
C
H
H
H
C
H
H
C
H
H
OH
Propan-1-ol
Concentrated H2SO4
170oC
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
C
H
H
H
C
H
H
C
H
Br
H + KOHrefulx
C
H
H
H
C
H
C
H
H + KBr + H2O
Bromopropane Propene
C
H
H
H
C
CH3
Cl
C
H
H
H + KOHrefulx
C
H
H
H
C
CH3
C
H
H + KCl + H2O
2-chloro-2-methylpropane
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
YX
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
C C + H H C C
HH
Ni / Pt
150 oC 5 atm
• Example hydrogenation
C C
H
H
H
H
+ H H C C
H H
H
HH
H
Ni / Pt
150 oC 5 atm
ethene ethane
C C + H H C C
H H
CH3
HH
H3C
Ni / Pt
150 oC 5 atm
ethene ethane
H3C
H H
CH3
Exercise
• Write down the structure of the products, if any, of the following reactions
CH3CH CHCH3 + H2Ni
150oC, 5 atm
+ H2
Ni
150oC, 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 + C C
XX
room temperatureX2
X2 = Cl2 or Br2
• For example
C C
H
H
H
H + Cl2 C C
H
H
H
H
Cl Cl
Ethene 1,2-dichloroethane
+ Br2
Br
Br
cyclohexene 1,2-dibromocyclohexane
room temp
• 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.
C C
H
H
H
H
Br
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.
C C
H
H
H
H
Br
Br
slowC C
H
H
H
H
Br
+ Br-
electrophilic attack carbocation bromide ion
The carbocation is very unstable and quickly combines with the bromide ion, Br- to form 1,2-dibromoethane.
C C
H
H
H
H
BrBr-
fastC C
H
H
H
H
Br Br
1,2-dibromoethane
Exercise
• Name and draw structural formulae for the products of the following halogenation process
CH3CH CHCH3 + Br2
+ Br2
3. Reaction with hydrogen halides• Hydrogen halide readily add to the carbon-
carbon double bond in the cold to give haloalkanes
• 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
ethene Haloalkane
C C
H
H
H
H C C
H
H
H
H
H X
+ H X cold
• Addition halide to unsymmetrical alkenes– When propene reacts with a hydrogen halide such
as hydrogen bromide, HBr, there are two possible products
C C
H
H3C H
H
H Brcold
C C
H
H3C H
H
H Br
1-bromopropane (minor)
C C
H
H3C H
H
Br H
2-bromopropane (major)
cold
– 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.
C C
H
H3C H
H
Carbon atom with the greatest number of hydrogen atoms
Unsymmetrical double bond
– Example of Markovnikoff’s rule
H3CH2C C
HCH2 + HBr H3C
H2C
HC CH3
Br
not H3CH2C
H2C CH2
Br
CH3
+ HCl
CH3
Cl
CH3
Cl
not
– 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.
C C
H
H3C H
H
+ H Br C C
H
H3C H
H
H
+ Br-
C C
H
H3C H
H
H
primary carbocation (less stable)
Secondary carbocation (more stable)
+ Br-
– 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.
C C
H
H3C H
H
H
+ Br-C C
H
H3C H
H
HBr
2-bromopropane
– The order of stability carbocations is as follows:
R C
R
R
> R C
R
H
> R C
H
H> H C
H
H
TertiaryCarbocation
SecondaryCarbocation
PrimaryCarbocation
MethylCarbocation
# The R group is alkyl group
increasing stability
Example
• Arrange the following alkenes in order of increasing reactivity on addition of hydrogen bromide.
– The relative stabilities are directly related to the stabilities of the intermediate carbocations formed.
CH3CH CHCH3H2C CH2
ethene(CH3)2C CH2
2-methylpropene but-2-ene
– 2-methylpropene is the most reactive because it forms the tertiary carbocation, the most stable carbocation
– But-2-ene forms the secondary carbocation
– Ethene forms the primary carbocation, thus it is the least reactive
– The order of increasing reactivity is ethene <but-2-ene < 2-methylpropene
(CH3)2CCH3
CH3CHCH2CH3
CH3CH2
Exercise
• When but-1-ene,
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?
H3CH2C C
HCH2
Exercise
• Name and draw the structural formula for the major products
CH3CH CHCH3 + HI
H3C C
CH3
CH2 + HBr
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 Br2 + H2O C C
Br OH
+ C C
Br Br
+ HBr
Bromoalcohol Dibromoalkane
• The addition reaction occurs in the way that yields more stable carbocation. Follow the Markovnikoff’s rule.
C C C C
Br OH
C C
Br Br
Bromoalcohol
Dibromoalkane
+ Br2 C C
Br
+ H2O
+ Br-
Exercise
• Write and overall equation for the reaction of propene with bromine water
CH3CH CH2 + 2Br2 + H2Oroom 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.
C C + H2SO4 C C
H
O
S
O
O O-H
Alkyl hydrogensulphate
cold
• 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.
• The addition of concentrated sulohuric acid to unsymmetrical alkenes follows Markovnikoff’s rule
C C
H OH
+ H2O + H2SO4
C C
H
O
S
O
O O-H
Alkyl hydrogensulphate
warm
Alcohol
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.
• The hydration of alkenes follow Markovnikoff’s rule
H2C CH2 + H2OH3PO4
330 oC, 60 atmH3C
H2C OH
• For example, the hydration of propene
H3C C H3PO4
330 oC, 60 atm
C + H2O
H H
H H3C C C
H H
H
HOH
not
H3C C C
H H
H
H OH
Exercise
• Name and draw structural formula for the major products of these addition reactions of alkenes
CH3CH CHCH3 + H2OH3PO4
330 oC, 60 atm
H3C C
CH3
CH2+ H2O H3PO4
330 oC, 60 atm
7. Reaction with potassium manganate (VII) (Oxidation)
• An alkene react with dilute potassium manganate in an acidic or alkaline solution to give a diol
+ H2OC C + [O]dilute MnO4
- / H+
room temp C C
OH OH
8. Reaction with oxygen• Like all hydrocarbon, alkenes burns in air to
produce carbon dioxide and waterAlkene + O2 CO2 + H2O
Thank you..