1435-1436
2014-2015
Alkenes
Learning Objectives
Chapter two discusses the following topics and the student by the end of this
chapter will:
 Know the structure, hybridization and bonding of alkenes
 Know the common and IUPAC naming of alkenes
 Know the geometry of the double bond i.e. cis/trans isomerization
 Know the physical properties of alkenes
 Know the different methods used for preparation of alkenes (elimination
reactions ; dehydrogenation, dehydration and alkenes stability (Zaitsev’s rule)
play an important role in understanding these reactions
 Know the addition reactions of alkenes and the effect of Markovnikov’s rule
in determining the regioselectivity of this reaction.
Alkenes
tructure Of Alkenes
They are unsaturated hydrocarbons – made up of C and H atoms
and contain one or more C=C double bond somewhere in their
structures.
Their general formula is CnH2n - for non-cyclic alkenes
Their general formula is CnH2n-2 - for cyclic alkenes
3
Alkenes
sp2 Hybridization Of Orbitals In Alkenes
The electronic configuration of a carbon atom is 1s22s22p2
Thus
2s2
2s1
2p2
promotion
2p3
3 x sp2
hybridization
Trigonal planar
2p
Alkenes
Orbital Overlap In Ethene
 In ethylene (ethene), each carbon atom use an sp2 orbital
to form a single
C-C bond. Because of the two sp2 orbitals overlap by endto- end the resulting bond is called σ bond. The pi (π) bond
between the two carbon atoms is formed by side- by-side
overlap of the two unhybridized p- orbitals (2p–2p ) for
maximum overlap and hence the strongest bond, the 2p
orbitals are in line and perpendicular to the molecular plane.
 This gives rise to the planar arrangement around C=C
bonds. Also s orbitals of hydrogen atoms overlap with the
sp2 orbitals in carbon atoms to form two
C-H bonds with each carbon atom.
 The resulting shape of ethene molecule is planar with
bond angles of 120º and C=C bond length is 1.34 Å
5
Alkenes
Orbital Overlap In Ethene
sp2 hybridized carbon atoms
two 2p orbitals overlap to form a pi
bond between the two carbon atoms
6
two sp2 orbitals overlap to form a sigma
bond between the two carbon atoms
s orbitals in hydrogen atoms overlap with the
sp2 orbitals in carbon atoms to form C-H
bonds
the resulting shape is planar with bond
angles of 120º and C=C (1.34 Å)
Alkenes
Nomencalture Of Alkenes And Cycloalkenes
1. Alkene common names:
CH3
H2C
CH2
Common: Ethylene
CH3-CH
CH2
H3C
Propylene
C
CH2
Isobutene
Substituent groups containing double bonds are:
 H2C=CH– Vinyl group
 H2C=CH–CH2– Allyl group
Br
Cl
Common: Allyl bromide
7
Vinyl chlorride
Alkenes
2. IUPAC Nomenclature Of Alkenes
 Find the longest continuous Carbon chain containing the double bond this
determines the root name then add the suffix -ene.
 Number the C- chain from the end that is nearer to the double bond. Indicate
the location of the double bond by using the number of the first atom of the
double bond just before the suffix ene or as a prefix.
1
H2C
2
3
4
CH CH2CH3
But-1-ene or 1-Butene
(not 3-Butene)
1
2
CH3CH
3 4
5
6
CHCH2CH2CH3
Hex-2-ene or 2-Hexene
(not 4-Hexene)
 Indicate the positions of the substituents using numbers of carbon atoms to
which they are bonded and write their names in alphabetical order (N.B.
discard the suffixes tert-, di, tri,---when alphabetize the substituents) and if
more than one substituent of the same type are present use the prefixes di- or
tri or tetra or penta,--- to indicate their numbers.
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Alkenes
8
Cl
7
1
4
6
2
5
H3C
1
6
4
C
2
CH
3
Br
1
3
CH3-CH2-CH2-CH=CH-C-CH3
6
7
5 4 3 2 1
Br
OCH3 = (CH3CHCHCH2OCH3)
2
CH3
1-Methoxy-but-2-ene
(not 4-Methoxy-but-2-ene)
2,2-Dibromo-3-heptene
(not 6,6-Dibromo-4-heptene)
1
2
1
3
5
4
CH3
4
2-Methyl-but-2-ene
or 2-Methyl-2-butene
(not 3-Methyl-2-butene)
1CH
3
3-Chloro-2-hexene
(not 2-Chloro-1-methyl-1-pentene)
6-Methyl-2-octene
4
5
3
2
3
CH3
6
7
9
8
2,3,7-Trimethyl-non-3-ene
(not 2-Isopropyl-6-methyl--2-octene)
3
2
5-Methylcyclopenta-1,3-diene
An ''a'' is added due to inclusion of di
put two consonants consecutive
3
2
1
5
4
CN
4-Cyano-2-ethyl-1-pentene
(not 2-Ethyl-4-cyano-1-pentene)
Alkenes
 In cycloalkenes the double bond carbons are assigned ring locations #1 and #2.
Which of the two is #1 may be determined by the nearest substituent rule.
1
CH3
1
H3C
6
2
5
3
4
CH3
3,5-Dimethyl-cyclohexene
(not 4,6-Dimethylcyclohexen)
(not 1,5-Dimethyl-2-cyclohexen)
1-Methyl cyclopentene
(not 2-Methylcyclopeneten)
 If the substituents on both sides of the = bond are at the same distance, the
numbering should start from the side that gives the substituents with lower alphabet
the lower number.
2
1
3
4
7
6
10
5
3-tert-Butyl-7-isopropyl-cycloheptene
(not 3-Isopropyl-7-tert-butylcycloheptene)
Alkenes
 When the longer chain cannot include the C=C, a substituent name
is used.
1
1
CH
CH2
3
6
5
Vinyl-cyclohexane
2
CH
CH2
4
3-Vinyl-cyclohexene
Alkenes
Geometrical Isomerism In Alkenes
 G. I. found in some, but not all, alkenes
 It occurs in alkenes having two different groups / atoms attached to each carbon
atom of the = bond
A
B
C
A=C or B=D
No Cis or transe
(G. I. X)
G. I. x
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D
A≠C B≠D,
A =B or C=D Cis / A =D or C= B transe
G. I. 
G. I. X
G. I. 
G. I. 
Alkenes
Geometrical Isomerism In Alkenes
 It occurs due to the Restricted Rotation of C=C bonds so the groups on either
end of the bond are ‘fixed’ in one position in space; to flip between the two groups
a bond must be broken.
X
Geometrical isomers can not convert to each at room temperature.
13
Alkenes
Types Of Geometric Isomerism
A) Cis / trans isomerism in alkenes
 Exhibited by alkenes having two H’s and two other similar groups or
atoms attached to each carbon atom of the = bond (or generally the
alkene have only two types of atom or groups i.e. ABC=CAB)
 Cis prefix used when hydrogen atoms on both carbon atoms are on
the SAME side of C=C bond
 Trans prefix used when non-hydrogen groups / atoms are on the
opposite sides of C=C bond
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Cis-But-2-ene
Trans-But-2-ene
Alkenes
H
H
Cl
H
Cl
Cl
H
Cl
cis-1,2-Dichloro-ethene
trans-1,2-Dichloro-ethene
Cis
Groups / atoms are on the
Same Side of the double
bond
Trans
Groups / atoms are on
Opposite Sides across the
double bond
=
H
Cis-Oct-4-ene
15
H
Trans-Oct-4-ene
Alkenes
B) Z/ E isomerism in alkenes
If the groups attached to the C=C are different, we distinguish
the two isomers by adding the prefix Z (from German word
Zusammen) if the higher-priority groups are together in the
same side or E (from German word Entgegen) if the higherpriority groups are opposite sides depending on the atomic
number of the atoms attached to each end of the C=C.
Atoms with higher atomic numbers receive higher priority
I> Br > Cl > F > O > N > C > H
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Alkenes
I
Cl
Br
CH3
I
CH3
Br
Cl
Z-2-bromo-1-chloro-1-iodopropene
E-2-bromo-1-chloro-1-iodopropene
O
CH3
HC
CH3
I
CH2
Cl
H
HO
Z
Z
17
Alkenes
Exercise
Q1-Which of the following compounds can exhibit cis / trans isomerism
a) 2-Methylpropene
b) 1-Butene
c) 2-Methyl-2-pentene
d) 2-Butene
e) 3-Methyl-2-hexene
Q2- Name the following compounds according to IUPAC system
a)
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b)
c)
Alkenes
1
3
4
2
3
5
2
Br
4
1
3-Bromo propene
2-Ethyl-4-methyl pentene
CH3
5
4
Cl
CH3
6
1
2
3
5
4
3
CH3
4-Chloro-3,6-dimethylcyclohexene
Br
C
H
3-Chloro-2,5-dimethylcyclohexene
1
H
C
3
Br
C
1
2
Cl
CH3
6
C
H
3
H
C
CC
2 3
H
H
H
Cis-1,2-Dibromoethene
Geometrical isomerism
2
C
H
C
H
2
3
45
E-3-Methyl-2-pentene
4
Cl
C
Cl
1,1-Dichloroethene
not geometrical isomerism
6
1
3
5
Trans-trans-2,4-heptadiene
Cis-cis-2,5-heptadiene
7
Trans-1,3,5-heptatriene
Home work
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Alkenes
Physical Properties of Alkenes




20
Alkenes are nonpolar compounds thus:
Insoluble in water
Soluble in nonpolar solvents ( hexane, benzene,…)
The boiling point of alkenes increase as the number of carbons
increase.
Alkenes
Preparation Of Aalkenes
1- Dehydration of alcohols ( removal of OH group and a proton from
two adjacent carbon atoms ) using mineral acids such as H2SO4 or
H3PO4
+
CH 3CH 2OH
H / heat
H2C
Ethanol
CH2
Ethene
OH
+
H / heat
+ H2O
H
cyclohexanol
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+ H2O
cyclohexene
Alkenes
Zaitsev’sRule
 If there are different protons can be eliminated with the hydroxyl
group or with halogen atom, in this case more than one alkene can be
formed, the major product will be the alkene with the most alkyl
substituents attached to the double bonded carbon.
H2C
CH3 + H2O
H3C
CH3
H / Heat
1- Butene Minor
OH
H3C
CH3 + H2O
2- Butene Major
Zaitsev rule: an elimination occurs to give the most stable, more highly
substituted alkene
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Alkenes
2- Dehydrohalogenation of alkyl halides using a base
or NaOH
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Alkenes
3. Dehalogenation of vicinal dihalides
Zn/AcOH
Br
Br
For example: Dehalogenation of 1,2-Dibromobutane leads to the formation of 1Butene. In the presence of catalyst.
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Alkenes
Reactions Of Alkenes
Oxidation Reactions
KMnO4
Reactions of Alkenes
Ozonolysis
Addition(Electrophilic)
reaction:
- Hydrogenation
- Halogenation
- Hydrohalogenation
- Halohydrin formation
-Hydration
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Alkenes
An electrophile, an electron-poor species,(from the Greek
words meaning electron loving). It is a species (any
molecule, ion or atom) that accept a pair of electrons to form
a new covalent bond.
, H , B r , C l , I , e tc ., A lC l 3 , B F 3 , F e C l 3 , F e B r 3 , e tc .
C
A nucleophile, an electron-rich species, ,(from the Greek
words meaning nucleus loving). It is a species (any molecule,
ion or atom) that donate an electron pair to form a new
covalent bond.
C
, OH , Br , Cl , I , etc., H2N, HS, etc., H2O ,
CH3OH , RNH2 , R2NH , R3N ,
, etc.
26
Alkenes
Electrophilic Addition Reaction
1- Additions To The Carbon-Carbon Double Bond
1.1 Addition Of Hydrogen: Hydrogenation
A
A
+
A
H2
Pt or Ni or Pd
A
A
A
A
H
H
An alkane
An alkene
H2C
H3C
CH2
A
+
Pt
H2
CH2
+
H2
Pt
CH3
H3C
CH3
CH3
CH3
+
CH3
H3C
H2
Pt
CH3
Cis-1,2-Dimethyl cyclohexane
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Alkenes
1.2.Addition of Halogens( Halogenation)
A
A
A
+
A
H3C
X2
A
A
CH3
A
A
X
+
Cl 2
CCl 4
(X= Cl or Br)
X
Cl
H3C
CH3
Cl
Br
+
Br2
CCl 4
Br
CH3
Br
CH3
+
CH3
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Br2
CCl 4
Br
CH3
Trans-1,2- Dibromo-1,2-Dimethyl cyclohexane
Alkenes
1.3. Addition of Hydrogen Halides
 Only one product is possible from the addition of these strong acids to symmetrical alkenes
such as ethene, 2-butene and cyclohexene.
A
A
A
+
A
A
A
A
HX
(x= Cl or Br or I)
A
H
X
Cl
+
HCl
CH3
H3C
H
+
H
HI
I
 However, if the double bond carbon atoms are not structurally equivalent, i.e.
unsymmetrical alkenes as in molecules of 1- propene, 1-butene, 2-methyl-2-butene and 1methylcyclohexene, the reagent may add in two different ways to give two isomeric
products. This is shown for 1-propene in the following equation.
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Alkenes
Br
CH3CHCH3
HBr
CH3CHCH3
2o Carbocation
maijor
Br
CH3CH=CH2
Br
CH3CH3CH3
CH3CH2CH2Br
minor
1o Carbocation
Stability of carbocation
H
C
3
C
H
C
3H
3
C
C
H
C
H
3
C
H
3
o
3
30
o
2
C
H
C
H
2
2 C
H
3
o
1
Alkenes
 However when the addition reactions to such unsymmetrical alkenes are
carried out, it was found that 2-bromopropane is nearly the exclusive
product. Thus it said the reaction proceeded according to Markovnikov’s
rule
Markovnikov’s rule stats that : In addition of unsymmetrical reagent to
unsymmetrical alkenes the positive ion adds to the carbon of the alkene
that bears the greater number of hydrogen atoms and the negative ion
adds to the other carbon of the alkene.
CH3
CH3
CH3 +
H3C
Cl
HCl
CH3
H3C
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Alkenes
1.4. Addition of HOX halogen in aqueous solution ( -OH, X+):
Halohydrin formation
 Only one product is possible from the addition of HOX acids (formed from
mixture of H2O and X2) to symmetrical alkenes such as ethene and cyclohexene.
Symmetrical akenes
A
A
A
+
A
H2 O / X2
A
A
A
A (x= Cl or Br )
OH X
Cl
+
H2O / Cl2
CH3
H3C
OH
32
Alkenes
 However, addition reactions to unsymmetrical alkenes will result in the formation of
Markovonikov’s product preferentially.
Unsymmetrical akenes
+
H2O / Cl2
OH
Cl
+
CH2Br
H2O / Br2
OH
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Alkenes
1.5. Addition of H2O: Hydration
 Only one product is possible from the addition of H2O in presence of acids as
catalysts to symmetrical alkenes such as ethene and cyclohexene.
Symmetrical akenes
A
A
A
+
A
H2 O
H
A
A
A
A
H
OH
OH
+
H2 O
H
CH3
H3C
H
 However, addition reactions to unsymmetrical alkenes will result in the formation of
Markovonikov’s product preferentially.
Unsymmetrical akenes
H
H
34
CH3
+
H2O
OH
CH3
Alkenes
Oxidation Reaction:
1- Ozonolysis (Oxidative cleavage):
This reaction involves rupture of the C=C to give aldehydes or ketones according to the
structure of the original alkene.
A
A
A
+
A
A
A
O3
A
A
O
O
O
( A= H or R)
i) O3
O +
ii) Zn /H2O
H
i) O3
ii) Zn /H2O
i) O3
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Zn /H2O
- H2O2
O +
O
ii) Zn /H2O
O
O
O
A
A
O +
A
O
A
Alkenes
2- Oxidation with KMnO4 (Oxidative addition):
OH
KMnO4 / OH
OH
Cis- doil
36
Alkenes
Thank You for your kind attention !
Questions?
Comments
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