Download Unsaturated Hydrocarbons Alkenes

Dr. Shatha I Alaqeel
108 Chem
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.
2
Structure Of Alkenes (Olefines)
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
cyclic alkenes
- for non-
Their general formula is CnH2n-2
alkenes
- for cyclic
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)
Nomenclature Of Alkenes and
Cycloalkenes
1. Alkene common names:
CH3
H2C
CH2
CH3-CH
Common: Ethylene
CH2
Propylene
Substituent groups containing double bonds are:
H2C=CH– Vinyl group
H2C=CH–CH2– Allyl group
H3C
C
CH2
Isobutene
Br
Common: Allyl bromide
Cl
Vinyl chlorride
The substituent is named in a similar way to the parent alkene.
It is named based on the number of carbon atoms in the branch
plus the suffix -yl. i.e. alkenyl. Vinyl group (=Ethenyl group)
7
IUPAC RULES
 Select the longest continuous Carbon chain containing the
double bond , the ending of the name is changed from alkane
to alkene .
 The C- chain is numbered starting from the end closet 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
2 3 4
1 2
3 4 5 6
CH3CH
CHCH2CH2CH3
H2C
CH CH2CH3
Hex-2-ene or 2-Hexene
But-1-ene or 1-Butene
(not 4-Hexene)
(not 3-Butene)
 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
8
tri or tetra or penta,--- to indicate their numbers.
8
Cl
7
1
4
6
2
5
H3 C
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)
9
 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.
CH3
1
1
2
H3C
1-Methyl cyclopentene
(not 2-Methylcyclopeneten)
6
2
5
3
4
CH3
3,5-Dimethyl-cyclohexene
(not 4,6-Dimethylcyclohexen)
(not 1,5-Dimethyl-2-cyclohexen)
CH3
5
4
5
4
Cl
3
2
3
1
1
2
CH3
4-Chloro-3,6-dimethylcyclohexene
NOT
CH3
1 6
6
3-Chloro-2,5-dimethylcyclohexene
3
2
5-Methylcyclopenta-1,3-diene
An ''a'' is added due to inclusion of di
put two consonants consecutive
10
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
7
4
6
5
3-tert-Butyl-7-isopropyl-cycloheptene
(not 3-Isopropyl-7-tert-butylcycloheptene)
 When the longer chain cannot include the C=C bond, a substituent name is
used
1
2
1
CH
CH2
3
6
5
Vinyl-cyclohexane
CH
CH2
4
3-Vinyl-cyclohexene
11
Geometrical Isomerism (G.I) In Alkenes
G. I. found in some, but not all, alkenes
It occurs in alkenes having two different groups / atoms attached to each carbon
A
B
atom of the = bond
C
G. I. x
G. I. X
A = C or B = D
No Cis or trans
(G. I. X)
D
G. I. 
G. I. 
A≠C B≠D
A = B or C = D Cis or Z
A =D or C = B trans or E
G. I. 
12
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
Geometric (cis-trans) isomerism
* The restricted rotation about C=C and the planer geometry
give rise to a type of isomerism.
* The prefix cis- is used when the two similar atoms or groups
are on the same side of the double bond.
* The prefix trans- is used when they are on opposite sides
of the double bond.
Cis-But-2-ene
Trans-But-2-ene
=
Cis-Oct-4-ene
14
H
H
Trans-Oct-4-ene
E-Z NOTATION FOR
GEOMETRIC ISOMERISM
If the groups attached to the C=C are different, we
distinguish the two isomers
*If
the two first-priority atoms are together on the
same side of the double bond, you have Z isomer.
*If the two first-priority atoms are on opposite sides
of the double bond, you have E isomer.
*The priorities of the substituents are determined by
the atomic number with atoms of higher atomic
number having higher priority.
I> Br > Cl > F > O > N > C > H
15
1
H3C
CH3
C
2
H
C
3
CH2CH3
4
5
E-3-Methyl-2-pentene
16
Alkenes
*
Exercise
Q1-Which of the following compounds can exhibit cis / trans isomerism
a)
b)
c)
d)
e)
2-Methylpropene
1-Butene
2-Methyl-2-pentene
2-Butene
3-Methyl-2-hexene
Q2- Name the following compounds according to IUPAC system
a)
17
b)
c)
*
Physical Properties of Alkenes
physical states
C1-C4 are gases
C5-C17 are liquids
more than 18 carbon atoms are solids.
*
Solubility
Alkenes are non polar compounds. Insoluble in
water. Soluble in non polar organic solvents (
hexane, benzene,…).
They are less dense than water.
*
Boiling point
The alkenes has a boiling point which is a small
number of degrees lower than the corresponding
alkanes. The boiling point of alkenes increase as
the number of carbons increase.
18
Preparation Of Alkenes
Elimination
1- Dehydration of alcohols: ( removal of OH group and a proton from
two adjacent carbon atoms ) using mineral acids such as H2SO4 or
H3PO4 (H+).
+
CH 3CH 2OH
H / heat
H2C
Ethanol
CH2
+ H2O
Ethene
OH
+
H / heat
+ H2O
H
cyclohexanol
cyclohexene
19
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
20
2- Dehydrohalogenation of alkyl halides using a base:
(removal halogen atom with hydrogen atom ) Heating an alkyl
halide with a solution of KOH, in alcohol, yields an alkenes.
21
3- Dehalogenation of vicinal dihalides
alcohol
Example:
22
Reactions Of Alkenes
Substitution reaction
Oxidation Reactions
KMnO4
Reactions of Alkenes
Ozonolysis
Epoxidation
Addition(Electrophilic) reaction:
- Hydrogenation
- Halogenation
- Hydrohalogenation
- Halohydrin formation
-Hydration
23
1-Substitution reactions
Halogenation at High Temperature
Involve the saturated alkyl chain at High temperature or UV light
Addition reaction:
low temperature
Absence of light
CH2=CH-CH3
Substitution reaction:
High temperature
or Ultraviolet light
24
Electrophilic Addition Reaction
2- Additions To The Carbon-Carbon Double Bond
2.1 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
25
2.2.Addition of Halogens( Halogenation)
A
A
+
A
X2
CCl4
A
A
A
A
A
X
(X= Cl or Br)
X
Cl
H3C
CH3
+
Cl 2
CCl 4
H3C
CH3
Cl
Br
+
Br2
CCl 4
Br
26
2.3. Addition of Hydrogen Halides (Hydrohalogenation)
 when HX is added to symmetrical alkenes, there is only one possible product from
this addition by strong acids 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 1-methylcyclohexene, the reagent may add in two different ways to give two
isomeric products. This is shown for 1-propene in the following equation.
27
CH3CHCH3
HBr
Br
CH3CHCH3
2o Carbocation
Br
CH3CH=CH2
CH3CH2CH3
maijor
Br
CH3CH2CH2Br
minor
1o Carbocation
Stability of carbocation
H3C
CH3
C
CH
CH3
CH3
3o
28
H3C
2o
CH2CH2
1o
CH3
Markovnikov’s rule
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
29
Anti Markovnikov’s Rule
In the addition of an acid to an alkene the hydrogen will
go to the vinyl carbon that already has the lowest
number of hydrogens.
30
2.4- Addition of Sulfuric acid
heat
(H-OSO3H)
Addition of Sulfuric acid to alkenes also follows Markovnikov’s rule,
as the example
H2O
heat
Propene
Isopropyl hydrogen sulfate
2-Propanol
31
2.5. Addition of HOX ( -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
 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
32
OH
2.6. 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
H
H2 O
A
A
A
A
H
OH
OH
+
H
H2 O
CH3
H3C
H
 However, addition reactions to unsymmetrical alkenes will result in the formation
of Markovonikov’s product preferentially.
Unsymmetrical akenes
H
H
CH3
+
H2O 33
OH
CH3
3-Oxidation of alkenes
3.1- Ozonolysis ( Oxidation of alkenes by ozone O3 )
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
Zn /H2O
- H2O2
O
A
A
O +
O
A
A
O
( A= H or R)
i) O3
O +
ii) Zn /H2O
O
H
i) O3
ii) Zn /H2O
i) O3
O +
O
O
ii) Zn /H2O
O
34
3.2- Oxidation of alkenes with Permanganate (Baeyer test)
OH
KMnO4 / OH
OH
3.3- Oxidation of alkenes with peroxy acid
RCO 3H
O
Epoxide
H3O
+
OH
anti hydroxylationOH
35
Polymerization (Free radical)
A)
B)
36
Thank You for your kind
attention !
Questions?
37