Alkenes
1. Introduction
2. Recall: Structure and Bonding
3. Degree of Unsaturation
4. Nomenclature of Alkenes
5. Cis/trans Isomerism of Alkenes
6. Cahn-Ingold-Prelog E/Z System
7. Physical Properties of Alkenes
8. Reactions of Alkene
a. Electrophilic Addition Reactions
i. Markovnikov’s Rule
ii. Anti-Markovnikov’s Rule (Free-radical Reaction)
b. Carbocations as Intermediates in the addition of HX to alkenes
c. Rearrangements of carbocations
9. Addition of Halogens
10. Halohydrin Formation
11. Hydroboration-Oxidation
12. Oxymercuration-Reduction
13. Hydrogenation of Alkenes
14. Hydroxylation of Alkenes
15. Oxidative Cleavage of Alkenes – Ozonolysis
16. Addition of carbenes to Alkenes: Preparation of cyclopropane
17. Allylic bromination of Alkenes
18. Polymerization of alkenes
19. Preparation of Alkenes
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1. Introduction
Unsaturated hydrocarbons have one or more carbon-carbon double or triple bonds and
contain fewer hydrogen atoms than alkanes. There are three classes of unsaturated
hydrocarbons, namely alkenes, alkynes and arenes (benzenes).
Unsaturated Hydrocarbons
alkenes
alkynes
CnH2n
CnH2n-2
arenes (benzenes)
Alkenes have at least one carbon-carbon double bond and alkynes one triple carboncarbon triple bond. Arenes, usually called aromatics, will be given a closer later.
Quite a number of alkenes are useful in nature. For example, ethylene is a plant hormone
that induces the ripening of fruits; α-pinene is a major component of turpentine; β-carotene
is the orange pigment responsible for the color of carrots and serves as a source of vitamin
A.
carotene
α -pinene
Some pheromones – compounds produced by an organism for the purpose of
communicating with other organisms of the same species, are alkenes. A pheromone
2
may act as a sex attractant or set an alarm or mark a trail of food. For example, have
you ever considered why ants follow each other in a line?
1. Recall: Structure and Bonding
Unknown 21/8/07 14:09
Formatted: Bullets and Numbering
sp2 orbital
π bond
sp2 orbital
+
p orbital
p orbital
σ bond
sp2 +sp2 = σ bond
p + p = π bond
Bond angle H-C-H = 120 º
H-C-C = 120 º
Bond lengths C=C = 1.34 Å
C-H = 1.10 Å
The presence of double bonds in alkenes confers on them the restricted rotation of the
carbon-carbon double bond. Thus conformational isomers are not easily recognizable in
alkenes. Consequently, alkenes show cis/trans isomerism in which each carbon atom of
the double bond has two different groups attached to it. For example, 2-butene shows
cis/trans isomerism. If the two methyl groups are on the same side of the double bond, the
molecule is said to exhibit cis isomerism and if the methyl are on opposite sides of the
double bond the molecule is said to exhibit trans isomerism.
H3C
CH3
H3C
H
H
cis-2- butene
H
H
CH3
trans-2-butene
2.1 Degree of Unsaturation
The degree of unsaturation is the number of multiple bonds and/or rings in a molecule.
This number can be calculated depending on the number and kinds of atoms in
molecules.
3
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Formatted: Bullets and Numbering
a. Alkenes that contain only carbon and hydrogen
Degree of unsaturation (DU) = #(C+1) - #H/2
For example, cyclohexene, C6H10, DU = (6+1) – 10/2 = 2
The degree of unsaturation of cyclohexene is two which means that cyclohexene may have
one triple bond, two double bonds, or a ring and a double bond. Cyclohexene has a ring
and a double bond.
b. The presence of an oxygen atom does not affect the degree of unsaturation
of a molecule. In other words, ignore the presence of the oxygen atom.
e.g., C6H11OH has a DU = (6+1) – 12/2 =1 (a ring or a double bond)
OH
OH
c. If a halogen is present, use the following formula
DU = #(C+1) - #(X+H)/2, where X is the halogen.
e.g., C4H6Br2 has a DU = (4+1) – (2+6)/2 = 1 (a double bond or
ring)
d. In the presence of nitrogen atoms, the formula is modified as follows
DU = (C+1) + N/2 – (X+H)/2
e.g., C5H9N has a DU = (5+1) +1/2 – (0+9)/2 = 2 (a ring + double or
2 double bonds or 1 triple bond)
Problem 2.
Write out the structures for the following compounds: C4H6Br2’ C5H9N
Problem 2
Calculate the degree of unsaturation for each of the following compounds
a. C3H4Cl4
b. C5H8N2
c. C8H14
d. C5 H6
e. C20H32
f. C3H4
2.2 Nomenclature of Alkenes
Alkenes are named following a series of rules similar to those enumerated in alkanes with
the suffix –ene instead of –ane. The basic rules are:
1. Name the longest parent chain containing the double bond using the suffix –ene.
H
CH2CH2CH2CH3
C
C
H3C
CH2CH2CH3
The parent name is heptene (longest chain containing the double bond), not octene
since it does not contain the double bond.
2. Number the carbon atoms in the chain beginning at the end nearest the double
bond. If the double bond is equidistant from the two ends, begin numbering at the
end nearer the first branch.
4
1
H3C
2
C
3
C
H
4 5 6
CH2CH2CH3
H3C
H
H3C
1
H
2
CH
3
C
4
C
5 6
CH2CH3
H
3. Write out the name in full naming the substituents according to their positions in
the chain and listing them in alphabetical order. Indicate the position of the double
bond by giving the number of the first alkene carbon atom.
4. If more than one double bond is present, indicate the position of each and use the
suffixes –diene, -triene, -tetraene, etc.
1
H3C
H
2
C
3
C
4 5 6
CH2CH2CH3
H3C
H
H3C
1
2-Hexene
2
CH
H
3
C
4
C
H
5 6
CH2CH3
2-methyl-3-Hexene
CH3 H
H2C
1
CH2
C
C
4
2
3
2-methyl-1,3-Butadiene
Problem
Name the following compound.
H
CH2CH2CH2CH3
C
H3C
C
CH2CH2CH3
Cycloalkenes are named in a similar way. We number the cycloalkenes such that the
double bond is between C1 and C2 and the first branch point has as low a value as
possible.
1
1
2
2
1-methylcyclohexene
not
2-methylcyclohexene
4
3
1,4-cyclohexadiene
5
There are some low-molecular weight alkenes that do not conform to the IUPAC rules of
nomenclature but they are of common usage. For example, ethylene (ethane) is accepted
by the IUPAC because it has been used for so long. Other examples include groups like:
vinyl group
H2C
CH
H2C
CH
CH2
H2C
CH
Cl
H2C
CH
CH2
allyl group
e.g.,
vinyl chloride
OH
allyl alcohol
The table below indicates common names of some alkenes
Compound
CH2=CH2
CH3CH=CH2
(CH3)2C=CH2
CH2=C(CH3)CH=CH2
CH3CH=CHCH=CH2
CH2=CHCH2=CH-CHCH=C-CH2CH2=
CH3CH=
Systematic Name
ethane
propane
2-methylpropene
2-methyl-1,3-butadiene
1,3-pentadiene
ethenyl
Propenyl
Propynyl
methylene
Ethylidene
Common Name
ethylene
propylene
isobutylene
isoprene
piperylene
vinyl
allyl
propargyl
Both common and systematic names are recognized by IUPAC.
Problem
Give the IUPAC names for the following compounds.
1. CH2=CHCH(CH3)C(CH3)3
2. CH3CH2CH=CH(CH3)CH2CH3
3. CH3CH=CHCH(CH3)CH=CHCH(CH3)2
Problem
Draw the structures corresponding to these IUPAC names.
a. 2-methyl-1,5-hexadiene
b. 3-ethyl-2,2-dimethyl-3-heptene
c. 2,33-trimethyl-1,4,6-octatriene
d. 3,4-diisopropyl-2,5-dimethyl-3-hexene
e. 4-tert-butyl-2-methylheptane
Problem
Provide systematic names for these cycloalkenes.
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a.
b.
c.
2.3 Sequence Rules: E/Z System of Nomenclature
Consider the following compound. Try to name this compound as cis or trans.
H3C
C2H5
H
CH3
It can be labeled trans because the two identical groups are on opposite sides of the double
bond. It can also be labeled cis because larger groups are on the same side of the double
bond. Consequently, the naming of this compound is ambiguous as far as the cis/trans
nomenclature is concerned. A set of rules were proposed to assign priorities to the groups
attached to the carbon-carbon double bond of the alkenes. They postulated that if groups
of high priority are on the same side of the double bond, the molecule is said to be in the Z
(“zussamen” in German) conformation or geometry and if they are on opposite sides, the
molecule is said to be in E (“entgegen” in German) conformation.
high
low
Z
high
high
low
low
low
E
high
How do we prioritize the groups of substituents that are attached to the double bond? The
following rules are proposed:
1. Atoms of higher atomic number have higher priority. Consider the atoms or groups
of atoms attached to the following structure.
H
CH3
The carbon atom attached to the double bond has a higher priority than the H atom.
The following examples illustrate this point.
1 H3C
2
H
CH2CH3 1
2
H
1 H3C
2
(Z)-2-pentene
H
Cl 1
CH3 2
(E)-2-chloro-2-butene
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For (Z)-2-pentene, the groups of higher priority are on opposite sides of the double
bond, i.e., - CH3 and – CH2CH3 groups.
2. An isotope of higher atomic mass receives higher priority. For example, D
(deuterium or 2H), an isotope of hydrogen has a higher priority than the 1H atom.
3. If atoms attached to the alkene carbon are identical, the next atom in each group is
considered and so on.
CH(CH3)2
CH2CH2CH2CH3
Consider the structure above. Which of the groups have a higher priority? Which atom
makes the difference?
4. For the purposes of assigning priority, triple bonds have higher priority than double
bonds and double bonds have high priority than single bonds.
N > C
C > C
O > C
C
C
C
C
C
N
The E,Z nomenclature of compounds with more than one double bond is illustrated by
the following molecule.
H
CH3
H
CH
H3C
CH2CH3
H
H
(2Z,5E)-4-methyl-2,5-octadiene
Study the name of this compound very well. Can you figure out how it was named?
Easy. Right? Check out the priorities on the double bonds. Can you see them? Answer
the question before you continue. If not, you are not ready for the next one. All right,
pal, name the compound below.
CH3
CH3
H
CH
H
CH2CH3
H
H
Problem
Which member in each set is higher in priority?
1. – H or – Br
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2. – Cl or – Br
3. – NH2 or – OH
4. – CH2OH or – CH3
5. – CH2OH or – CHO
Rank the sets of substituents in order of Cahn-Ingold-Prelog priorities.
1. – CH3, - CH2CH3, - CH=CH2, - CH2OH
2. – COOH, -CH2OH, -C=N (nitrile), - CH2NH2
3. –CH2CH3, - C=CH (triple bond), - C=N (nitrile), -CH2OCH3
Assign E or Z configurations for these alkenes.
H3C
CH2OH
H
CN
a.
b.
H3CH2C
Cl
H3C
CH2NH2
Give the structures of
i.
(2Z,4Z)-2,4-octadiene
ii.
(2Z,5Z)-2,5-octadiene
2.4 Physical Properties of Alkenes
Physical properties of alkenes are similar to those of alkenes except for dipole
moments and melting points. The low-molecular weight alkenes are flammable and
nonpolar.
Let us compare some physical properties of 1-hexene and n-hexane.
Table showing some alkenes and physical properties
Property
Compound
Boiling point
Melting point
Density
Solubility in water
Dipole moment
63.4 ºC
- 139.8 ºC
0.673 g/mL
negligible
0.46 D
68.7 ºC
- 95.3 ºC
0.660 mg/mL
negligible
0.085 D
How we account for the higher dipole moments in alkenes? Electron density lies closer to
2
3
the nucleus in sp orbitals than it does in sp orbitals. Alkyls groups attached to double
bonds are polarized towards the double bond.
H3C
polarization of electrons
toward the trigonal carbon atom
results in bond dipole
2.5 Relative Stabilities of Alkene Isomers
Cis-alkenes are less stable than their trans counterparts because of steric (spatial)
interference between two bulky substituents are on the same side of the double bond. Cisand trans-2 butene is used to demonstrate the relative stability of alkenes.
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H3 C
H3 C
CH3
H
H
CH3
trans-2-butene
H
H
cis-2-butene
In the cis isomer, the bulky methyl groups on the same side of the double bond tend to
interfere with each other making the molecule unstable relative to the trans isomer where
there less interference between the hydrogen atom and the methyl group.
Comparatively, the alkene with the greatest number of alkyl groups on the double bond is
usually the most stable, i.e.,
R
R
R
H
>
R
R
R
H
>
R
R
R
R
H
>
>
R
R
H
H
H
H
>
H
H
H
H
An alkyl group stabilizes an electron-deficient carbocation (see later for explanation of
this term) in two ways:
1. through the inductive effect
2. through the partial overlap of filled orbitals with empty ones.
The inductive effect is a donation of electron density through sigma bonds of the
molecule. The positively charged carbon atom withdraws some electron density from the
alkyl groups bonded to it.
H3C
CH3
C
CH3
An alkyl substituent with filled orbitals can overlap with π bond and a properly oriented CH σ bond of a neighboring substituent. This type of overlap between a p-orbital and a
sigma bond is called hyperconjugation.
overlap with filled sp3 orbital
vacant p orbital
C
2.6 Reactions of Alkenes
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CH3
2.6a Addition Reactions of Alkenes
Alkenes act as nucleophiles, i.e., electron-rich species which donate a pair of electrons to
an electron-deficient carbon atom called an electrophile. This reaction can be compared to
an acid-base reaction in which the alkene acts as a Lewis base and the electrophile as the
Lewis acid.
A reaction mechanism describes in detail how a reaction occurs. It attempts to elucidate
which bonds are broken and formed and at what rate these processes occur. The typical
mechanism for the addition reaction is as follows: When an electrophile, E+, approaches
the weakly held pi electrons of the double bond of the alkene, the electron pair is donated
to the electrophile forming a dative covalent sigma bond. A carbocation is formed, which
is subsequently attacked by the nucleophile to form a neutral addition product.
E
E
alkene
C
E+
electrophile
C
C
C
eqn 1.0
Nu
Nu
carbocation intermediate
There are two steps in this mechanism: Step 1 consists of the formation of the carbocation
intermediate and in step 2, the carbocation is attacked by an electron-rich species. We will
illustrate the steps in this mechanism by the reaction of 2-butene with a hydrogen halide.
Step 1.
H
The π electrons of the
alkene form a bond with
the proton (electrophile)
from HX to form a
cabocation and a halide ion.
+
H
X
slow
+C
+
C
X
-
carbocation
eqn 2.1
Step 2.
The halide ion (nucleophile)
reacts with the carbocation by
donating an electron pair to
form the product (alkyl halide).
H
X
-
+
+C
C
H
fast
C
C
X
eqn 2.2
The energetics of the reaction can be shown by the following diagram (Figure xx).
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Define and indicate transition states on the diagram and show breaking and forming of
bonds
This reaction depends on the presence of a strong electrophile to react with the π bond
generating a carbocation. Since there is a rate-determining step, this kind of reaction is
called an electrophilic addition reaction to an alkene. Here are some examples.
+
HCl
ether
Cl
propene
2-chloropropane
eqn 2.3
Br
HBr/ether
cyclohexene
bromocyclohexane
eqn 2.4
HI is usually generated in situ by the reaction of KI with H3PO4.
I
KI/H3PO4
eqn 2.5
2.6b Orientation of Electrophilic Addition: Markovnikov’s Rule
In electrophilic addition reactions to unsymmetrical alkenes, it not obvious where the
electrophile or nucleophile attaches itself to the alkene. For example, 2-methylpropene
may react with HCl to yield 1-chloro-2-methylpropane in addition to 2-chloro-2methylpropane as indicated above, but it did not. We say that the reaction is regiospecific
when only one of the two possible products of an addition is obtained.
HCl
ether
Cl
+
2-chloro-2-methylpropane
Cl
1-chloro-2-methylpropane
(not observed)
eqn 2.6
Markovnikov observed many of such addition reactions and proposed a rule which can be
summarized today as follows: in an electrophilic addition reaction to an unsymmetrical
alkene, the electrophile adds in such a way as to generate the most stable intermediate. In
other words, the hydrogen atom becomes attached to the carbon atom with fewer alkyl
groups. In biblical terms, it may be stated as “to he who hath more hydrogen atoms, more
hydrogen atoms shall be given.” No offence intended.
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H3C
CH2CH3
H3C
Cl /ether
H
H3C
H
CH2CH3
H
H3C
Cl
H
CH3
I
CH3
KI/H3PO4
H
When both ends of the double bond have the same degree of substitution, a mixture of
products is formed.
Br
H
H
+
HBr/ether
H
H
Br
Let us look at the mechanism for the following reaction which can be applied to others of
a similar nature.
Administrator 2/10/07 10:15
H
Cl
Cl
Cl
carbocation
Step 1: The alkene (nucleophile) attacks the halide to create a carbocation and a halide ion
Step 2: The halide ion (chloride ion) attacks the highly reactive carbocation to yield the
final product.
Notice that the more highly substituted tertiary carbocation is formed. The primary
carbocation is less stable and it is less likely to be formed.
Give the carbocations for the examples above. Indicate which of them are more stable.
Problems
Predict the products of the following reactions.
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Comment [1]: Step 1: The alkene (nucleophile)
attacks the halide to create a carbocation and a
halide ion
Step 2 : The halide ion (chloride ion) attacks the
highly reactive carbocation to yield the final
product.
+ HCl
?
+ HBr
HBr/ether
b.
a.
?
KI/H3PO4
d.
c.
?
?
What alkenes would you start with to prepare
i. bromocyclohexane
ii. CH3CH2CH(Br)CH2CH3
iii. 1-iodo-1-ethylcyclohexane
2.6c Free-radical Addition of hydrogen halides to Alkenes: Anti-Markovnikov Addition
In the presence of peroxides, hydrogen halides add to alkenes to form anti-Markovnikov
products.
HBr
peroxide
eqn 2.
Br
1-bromo-2-methylpropane
Peroxides give rise to free radicals that confer a different mechanism to the addition
reaction of alkenes. In the presence of peroxides, alkenes undergo a chain-reaction
mechanism which consists of three steps: initiation, propagation and termination steps.
Initiation
In the presence of heat or ultraviolet light, the peroxide (commonly dibenzoyl peroxide)
dissociates to yield free radicals.
heat
R
O
O
R
O
2 R
eqn 2.0
free radical
R
O
+
H
R
Br
OH
+ Br
eqn 2.
Propagation
The bromine radical thus generated reacts with the alkene to yield an alkyl radical which
reacts with excess HBr to give a neutral product and a bromine radical and the reaction
continues until one of the radicals is not sufficient enough for the reaction to continue.
Br
+
eqn 2.
Br
+
H
+ Br
Br
H
Br
Br
14
eqn 2.
addition product
Termination
There are a number of ways the reaction can be terminated but essentially in this reaction,
two radicals react to form a product.
Br +
Br
Br2
Br
Br
+
Br
Br
eqn 2.
Can you guess other reactions that would be part of the termination step?
For example, 2-methyl-2-pentene reacts with hydrogen bromide in the presence of
peroxides to yield 2-bromo-3-methylpentane instead of the 2-bromo-2-methylpentane.
+ Br
+
Br
secondary radical
(less stable)
Br
tertiary radical
HBr
H
Br
2-bromo-3-methylpentane
Notice that the less stable secondary free radical would have formed the Markovnikov
product. The reversal of regiochemistry in the presence of peroxides is called the peroxide
effect.
2.6d Carbocations
Carbocations are reactive intermediates that have a positively charged carbon atom. They
are often referred to as carbonium ions. They are classified as follows:
R
C
H
R
R
H
R
R
C
C
R
H
secondary 2°
tertiary 3°
1°
This classification depends on the number of alkyl groups attached to the positively
charged carbon atom. In a primary carbocation, the positively charged carbon atom is
primary
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