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
1
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
 -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
sp2 orbital
 bond
sp2 orbital
+
p orbital
 bond
p orbital
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
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
4 5 6
CH2CH2CH3
H3C
H
H3C
1
3
C
H
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
4 5 6
CH2CH2CH3
H3C
H
H3C
1
3
C
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.
6
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
7
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
8
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
the nucleus in sp2 orbitals than it does in sp3 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.
9
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
10
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
C
E+
electrophile
C
C
C
eqn 1.0
Nu
alkene
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.
The π electrons of the
alkene form a bond with
the proton (electrophile)
from HX to form a
cabocation and a halide ion.
H
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).
11
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.
12
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
HBr/ether
H
+
H
H
Br
Let us look at the mechanism for the following reaction which can be applied to others of
a similar nature.
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.
13
+
HCl
HBr/ether
?
+ HBr
?
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.
eqn 2.
HBr
peroxide
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
primary
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
15
attached to an alkyl group. Two alkyl groups are attached to the positively charged carbon
atom in secondary carbocation and three alkyl groups are attached to the positively
charged carbon atom in tertiary carbocations.
We have mentioned above that 3°carbocations are more stable than 2° carbocations which
in turn are more stable than 1° carbocations. Free radicals follow the same pattern of
stability.
2.6e Rearrangement
Consider the addition of HCl to 3-methyl-1-butene. The major product is not 2-chloro-3methylbutane but the major product is 2-chloro-2-methylbutane.
H
H
HCl
ether
eqn 2.
+
Cl
H
Cl
2-chloro-3-methylbutane
(40%)
2-chloro-2-methylbutane
(60%)
What has happened? There has been a rearrangement. A hydrogen atom has moved with
its electron pair from the carbon adjacent to the positively charged carbon atom. This
transfer is called a hydride (H-) shift.
________________________________________________________________________
(i) Formation of a secondary (2°) carbocation
H
H
H
+
Cl
Cl
2°
eqn 2.
(ii) Hydride shift (shift of a hydrogen atom with its electron pair to an adjacent deficient
carbon atom) to form a more stable tertiary (3°) carbocation.
H
rearrangement
H
3°
eqn 2.
(iii) The more stable tertiary carbocation accepts the chloride ion to yield the final product.
16
Cl
Cl
H
2-chloro-2-methylbutane
3° H
eqn 2.
________________________________________________________________________
Rearrangements are favored because the more stable tertiary carbocation is favored than
the secondary carbocation.
Not only hydride shifts are possible but alkyl shifts do occur to. In this case, an alkyl
group with its electron pair shifts from an adjacent carbon atom to the cation (eqn 2. ).
Alkyl shifts are called Wagner-Meerwein rearrangements.
_______________________________________________________________________
H
Cl
+ Cl
2°
3,3-dimethyl-1-butene
Methyl shift
Cl
Cl
3°
2-chloro-2,3-dimethylbutane
eqn 2.
2.7 Addition of Halogens to Alkenes
In the presence if a variety of solvents, halogens react with alkenes by electrophilic
addition to yield vicinal dihalides
X
eqn 2.
+
X2
CCl4
X
The halogens are usually chlorine or bromine and the reaction takes place rapidly at room
temperature. The addition of bromine to an alkene serves as a simple laboratory test for
unsaturation. A solution of bromine in carbon tetrachloride (CCl4) is red. On addition of a
few drops of an alkene to this solution, the reddish brown color of bromine disappears,
i.e., the solution becomes colorless.
17
Addititon of bromine to cycloalkenes illustrate anti addition stereochemistry: the two
bromine atoms add on opposite faces of the double bond. In syn addition, the two bromine
atoms add on the same face of the double bond.
Br
eqn 2.
Br2/CCl4
Br
cyclopentene
trans-1,2-dibromocyclopentane
Cl
Cl2/CHCl3
eqn 2.
Cl
cyclohexene
trans-1,2-dichlorocyclohexane
To explain the stereo chemistry, it is proposed that the reaction intermediate is not a
carbocation but a cyclic bromonium ion, in which the posistive charge resides on the
bromine atom, not the carbon atom.
C
C
Br
bromonium ion
For example, ethylene reacts with bromine as follows:
1. Reaction of ethylene reacts with bromine to form the bromonium ion and a
bromide ion.
H
H
C
H
C
H
H
+
Br
+
Br
H
H
H
Br
bromonium ion
2. Nucleophilic attack of the bromide ion on the bromonium ion
18
Br
eqn 2.
H
H
H
Br
H
H
Br
Br
eqn 2.
H
Br
bromonium ion
H
H
Alkyl groups on the carbon-carbon double bond release electrons, stabilize the transition
state for the formation of the bromonium ion and increase the reaction rate. Thus the
reaction of alkenes with bromine follow the order:
(CH3)2C=C(CH3)3 > (CH3)2C=CH2 > CH3CH=CH2 > CH2=CH2
Notice that because carbocations are not formed, no rearrangement products are expected
The addition of a halogen to an alkene is stereospecific. A stereospecific reaction is a
reaction in which a stereoisomeric form of a reactant reacts in such a way that it leads to a
specific stereoisomeric form of a product. A stereoselective reaction is one in which one
stereoisomer is formed or destroyed in preference to others (eqn 2.).
Br
+ Br2
CCl4
Br
trans-product
eqn 2.
2.8 Formation of Halohydrin
In aqueous solution, chlorine, bromine and iodine react with alkenes to yield vicinal
halohydrins (1,2-haloalcohols).
X2/H2O
HO
X
eqn 2.
In this reaction HO-X+, which is formed in aqueous solution, is added to the double bond.
______________________________________________________________________
Cl2
H3C
C
CH2
H3C
CH
CH2Cl
H2O
eqn 2.
________________________________________________________________________
H
OH
The mechanism is similar to that of addition halogens to alkene. This involves the
formation of a bromonium ion and the subsequent attack of the ion by the water molecule
and loss of a proton.
19
Br
+ Br2
Br
O
H
H2O
H
Br
Br
+
OH
HBr
eqn 2.
Water usually attaches itself to the more highly substituted carbon atom because this
carbon atom has a greater degree of carbocation character than the less substituted
counterpart.
Problem
Complete the following reaction.
1.
2.
+
+
Cl2
Br2
H2O
?
H2O
?
Problem
Arrange the compounds 2-methyl-1-butene, 2-methyl-2-butene, 3-methyl-1-butene in
order of decreasing reactivity toward bromine.
What product would you expect from the reaction of Cl2 with 1,2-dimethylcyclohexene?
Show the stereochemistry of the product.
2.9 Hydration of Alkenes
2.9a Acid-catalyzed Hydration of Alkenes
20
Water can add to alkenes to yield alcohols in the presence of concentrated mineral acid as
a catalyst. The reaction follows Markovnikov’s rule and the mechanism of the reaction is
as follows.
(i) The alkene accepts a proton from the hydronium ion forming a carbocation
O
H
+
H
H
(ii) The neutral water molecule attacks the carbocation and the intermediate loses a proton
yielding an alcohol.
H
H
H2O
OH
O
2.9b Oxymercuration-Demercuration
This is a method of converting an alkene to an alcohol. The alkene is treated with a
mercury salt (mercury (II) acetate) followed by reduction (demercuration) with sodium
borohydride.
1. Hg(OAc)2, H2O
C
2. NaBH4, OH-
C
eqn
H
OH
AcO- = CH3COO- (the conjugate base of acetic acid, CH3COOH). The mercury (II) acetate
dissociates as follows:
O
H3C
C
O
O
Hg
O
C
O
O
CH3
H3C
C
O
+ H3C
C
O
Hg+
eqn
The first step in this reaction is the formation of a cyclic, mercurinium ion (compare with
the bromonium ion). Carbocations are unlikely intermediates in this reaction since
rearrangement products are not obtained.
Oxymercuration is regiospecific: HgOAc becomes attached to the less substituted carbon
atom of the alkene and the OH group of water becomes attached to the more substituted
carbon atom. Markovnikov’s rule is respected.
21
R
H
C
R
Hg(OAc)2
C
C
H
Hg+
R
C
Hg
C
CH2
CH2
H
H
H
H
R
Hg
=
CH2
H2O
OAc
OAc
OAc
H2O
addition
R
H
C
R
HgOAc
CH2
deprotonation
H
C
CH2
OH
O
H
HgOAc
H
organomercury compound
(isolable)
H2O
eqn
R
R
H
C
HgOAc
NaBH4, OH-
H
C
OH
CH2
R
OH
CH3
NaBD4, OD-
H
C
OH
D
CH2
eqn
In eqn the use of sodium borodeuteride indicates the position of the entering deuterium.
Examples:
1. Hg(OAc)2, H2O
2. NaBH4, OHOH
22
eqn
OH
1. Hg(OAc)2, H2O
2. NaBH4, OHAc
(O
Hg
-
H
,O
) 2,
H4
aB
N
O
H2
HO
HgOAc
1. Hg(OAc)2, H2O
2. NaBH4, OH-
OH
Problem
1. Predict the product of the last example above if acidified water was used as the
hydrating agent.
2. Ethers could be prepared using the oxymercuration-demercuration method.
Suggest a possible mechanism for the preparation of the following ether from an
appropriate alkene.
OCH3
3. Starting with an appropriate alkene, show all steps in the synthesis of each of the
following alcohols by oxymercuration-demercuration.
a. tert-butyl alcohol
b. isopropyl alcohol
c. 2-methyl-2-butanol
4. An oxymercuration-demercuration reaction is anti-stereospecific. Justify.
2.9c Hydroboration-Oxidation
Hydroboration is the addition of boron hydride, BH3, commonly known as borane, to an
alkene to form an organoborane. The trigonal boranes have an empty p orbital on the
boron atom and act as Lewis acids.
23
F
H
B
B
H
F
F
H
borane
boron trifluoride
Figure … Borane and boron trifluoride act sa Lewis acids.
BF3 is commercially available but free BH3 is not available as it spontaneously dimerizes
to diborane (B2H6), a foul-smelling, toxic gas. Borane forms a complex with ethers and
this complex is used as a source of borane.
H
B
H
H3B
O
O
H
Borane-THF complex eqn
Lewis acid
Lewis base
Figure … Formation of borane-ether complex
When the borane-ether complex reacts with an alkene, borane adds with the double of the
alkene to yield an initial alkylborane (a monoalkyl organoborane)
C
C
H3B
+
O
H
BH2
C
C
+ O
alkylborane
eqn
Figure.. Hydroboration: The elements of BH3 have been added across the double bond
Eventually, a trialkylborane is formed which is oxidized to an alcohol when treated with
hydrogen peroxide in base.
H
BH2
C
C
+
C
C
H
BH
C
C
C
C
+
C
C
C
C
dialkylborane
C
3
OH
+
C
H2O2, OH-
B(OH)3
H
B
C
C
trialkylborane (BR3)
24
eqn
The addition of borane to an alkene is a syn-addition: the B-H bond adds on the same face,
with the boron attaching itself to the less substituted carbon atom of the alkene. An antiMarkovnikov product is formed.
H3C
C
CH2
H3C
1. BH3/THF
CH
2. H2O2/OH-
CH2
H3C
H3C
CH3
BH3/THF
OH
CH3
H2O2/OH-
H
H
H
B
H
OH
Hydroboration is a one-step, concerted addition of boranes to alkenes. In this mechanism,
bond breaking and bond forming occur simultaneously.
H3C
H
H
H3C
H
H
H
H3C
H
H
H
H
H
H H
B
B H
H
H
B
Four-center transition state
H
Formation of πcomplex: donation of π
electron to the empty
2p-orbital of the boron
eqn
Boron adds to the
less substituted
carbon atom
Oxidation of Trialkylboranes
R
R
R
R
+
B
R
R
O
O
H
B
R
R
B
O
O
O
+
H
OH-
R
Borate ester
unstable intermediate
eqn
1. The boron atom accepts an electron pair from the peroxide ion to form an unstable
intermediate
2. An alkyl group migrates from the boron atom to the adjacent oxygen atom as a
hydroxide departs from the unstable intermediate to form an alkyl borate ester.
3. Eventually all the alkyl groups migrate to form the trialkylborate which undergoes
hydrolysis to yield an alcohol.
25
OR
+ 3 OH
B
RO
H2O
33 ROH + BO3
OR
trialkylborate
eqn
Problem
Draw the structural formulas for the trialkylborane and alkene that give the following
alcohols under the reaction conditions shown.
a.
an alkene
BH3
b.
an alkene
BH3
trialkylborane
H2O2/NaOH
trialkylborane
H2O2/NaOH
OH
OH
c. Complete the following reaction.
H3O+
1. BH3/THF
2. H2O2, OH-
d. Provide a mechanism for the formation of the following trialkylborane.
CH3
CH3
+
CH
H3C
B
CH3
H3C
C
C
CH3
BH2
CH
CH3
CH3
2.10 Hydrogenation of Alkenes
Alkenes react with hydrogen in the presence of an appropriate catalyst to yield the
corresponding alkanes. The addition of hydrogen to an unsaturated hydrocarbon is called
hydrogenation. This reaction can be achieved by two different catalytic processes:
heterogeneous and homogeneous catalysis. In heterogeneous catalysis, a solid insoluble
catalyst, such as platinum or palladium on charcoal is used. In homogeneous catalysis, a
soluble catalyst, such as Wilkinson’s catalyst, RhCl(Ph3P)3, is used.
26
H2
C
C
H
H
C
C
H
H
(heterogeneous catalysis)
Pd/C
H2
RhCl(Ph3P)3
Wilkinson's catalyst
(homogeneous catalysis)
Figure Reaction of H2 with alkenes in the presence of heterogeneous and homogeneous
catalyst.
P(c6H5)3
RhCl(Ph3P)3 =
(C6H5)3P
Rh
Cl
(Ph = C6H5)
P(C6H5)3
Chlorotris(triphenylphosphine)rhodium(I) = Wilkinson's catalyst
The hydrogenation reaction is stereoselective as both hydrogen atoms add on the same
face of alkenes – a syn addition occurs.
CH3
CH3
H
H2
H
Pd/C
CH3
CH3
cis-2,3-dimethylcyclohexane
(the only product)
Figure Syn-addition of hydrogen to an alkene. The trans-product is not formed.
The Wilkinson’s catalyst
1. catalyzes the hydrogenation of alkenes and does not affect other functional groups,
e.g.,
H2 (1 atm)
RhCl(Ph3P)3
2. Terminal double bonds are hydrogenated much more rapidly by the bulky rhodium
catalyst than are internal double bonds, e.g.,
27
H2 (1 atm)
OH
OH
RhCl(Ph3P)3
Diimide (HN=NH), a nonmetallic reducing agent, which very unstable under normal
conditions, can be used for hydrogenation as well.
HN
NH
H
H
Problem
Complete the following reactions, paying particular attention to regiochemistry and
stereochemistry, where appropriate
H2 (excess)
H2
a.
b.
?
Pd/C
?
RhCl(Ph3P)3, benzene
H2
?
RhCl(Ph3P)3, benzene
H2 (excess)
c.
?
Pt/C
2.11 Hydroxylation
Hydroxylation of alkenes involves the addition of an –OH group to each of the alkene
carbon atoms. Alkaline potassium permanganate (KMnO4) or osmium tetroxide, OsO4,
add to alkenes to form five-membered intermediates which eventually decompose to yield
vicinal 1,2-diols (glycols). Hydroxylation occurs with syn stereochemistry and cis
products are common.
O-K+
O
MN
O
O
C
C
Os
O-K+
O
O
O
O
Mn
O
O
C
C
O
O
C
C
O
Os
O
O
C
C
A cyclic osmate ester
A cyclic manganate ester
28
Figure… Addition of potassium permanganate and osmium tetroxide to alkenes yielding
five-membered intermediates.
The manganate ester decomposes in alkaline solution to give 1,2-diols, while the osmate is
treated with aqueous sodium sulfite (Na2SO3) to yield, yielding the same product.
O-K+
O
HO
Mn
O
OH
H2O
O
C
OHC
C
+
MnO2
C
1,2-Glycol
O
O
HO
Os
O
O
H2O
OH
C
+ H2OsO4
C
Na2SO3
C
C
1,2-Glycol
Examples:
HO
HO
KMnO4
CH3OH/H2O
NaOH, 20°C
2. H2O, Na2SO3
HO
H
H3C
Problem
How would you prepare the following compounds? Show the starting alkene and the
reagents you would use.
Me
HO
OH
a.
HO
c.
b.
OH
OH
OH
OH
1. OsO4
OH
OH
Et
2.12 Oxidation of Alkenes with Peroxyacids
Check out carboxylic acids with general formula
O
R
C
Peroxycarboxylic acids have the general formula
OH
O
R
C
O
29
O
H
H
H
which is very similar to the hydrogen peroxide structure (HOOH) and therefore, an
oxidizing agent. Peroxycarboxylic acids, therefore, react with alkenes to give threemembered cyclic ethers, ususlly called epoxides or oxiranes. These small ring compounds
are quite reactive and sometimes very explosive. The rings can be destroyed under acidic
and basis conditions.
The common peroxy acids used for this synthesis are m-chloroperbenzoic acid and
trifluorperacetic acid.
O
O
C
O
O
H
C
F3C
O
O
Trifluoroperacetic acid
H
Cl
m-Chloroperbenzoic acid
The mechanism of the reaction involves a one-step transfer of the oxygen atom in the
peroxy acid to the double bond of the alkene.
C
H
+
C
O
O
+
CF3CO2H
C
C
O
C
Epoxide (Oxirane)
O
F3C
Figure… Reaction of a peroxyacid with an alkene to form an epoxide (a cyclic ether).
Epoxidation is a stereoselective reaction. For example, (Z)-3-hexene is oxidized to cis-3diethyloxirane and (E)-3-hexene gives trans-3-diethyloxirane.
H
O
H
m-Cl-PBA
H
H
dioxane
cis-3-diethyloxirane
H
O
m-Cl-PBA
dioxane
H
H
+
H
H
H
O
trans-3-diethyloxirane
(racemic mixture)
On reaction with a nucleophile or an acid, the epoxide opens up to give a trans-product.
For example, cis-3-diethyloxirane will easily open in the presence of strong acid to give a
1,2-diol
30
HO
O
H
H
H
H3O+
H
or OH-
OH
The more substituents on the double bond of the alkene, the more easily it reacts with the
peroxyacid. For such unsymmetrical alkenes, the oxirane ring can be opened in acid or
base to yield different products.
+
acid (H3O )
OH
O
OH
R
base (Nu:)
R
Nu
OH
In acid, the nucleophile
generally becomes attached
to the more substituted carbon
In base, the nucleophile
generally becomes attached
to the less substituted carbon
Figure … An unsymmetrical epoxide usually opens with different regiochemistries.
For example, 1,1-dimethyloxirane reacts with nitride ion and hydrochloric acid to yield the
following compounds.
O
N3
NaN3/H2O
SN2
OH (41%)
H3C
H3C
HCl
OH +
Cl
(55%)
Cl
OH
(45%)
The nitride ion attacks the less sterically hindered carbon atom but in the acid medium,
protonation of the oxirane takes place first. The ring can then open up to form two
carbocations. The more stable carbocation (3º) yields the major product.
Problem
Suggest the mechanism for the attack of (a) N3- and (b) HCl to an unsymmetrical oxirane.
Make sure that you indicate the intermediates in each case.
Problem
Predict the major products in the following reactions. Write structures showing the
pertinent stereochemistry, where appropriate.
31
H3C
CH3
m-Cl-PBA (1 molar equiv.)
dioxane
a.
?
H3C
m-Cl-PBA
CH2Cl2
H
b.
?
H
m-Cl-PBA (1 molar equiv.)
c.
dioxane
?
Sharpless Epoxidation
Epoxidation by Sharpless’ method uses a pinch of titanium isopropoxide, tert-butyl
peroxide, and one enantiomer of a tartaric ester to react with an allylic alcohol
(RCH=CHCH2OH). This reaction is very useful as the two enantiomers of the ester
tartaric lead to two products of oxiranes with different stereochemistries.
O
H
O
EtOOC
CH2OH
H
Allylic alcohol
+
HO
H
OH
H Ti[OCH(CH3)2]4
(CH3)3COOH
COOEt
CH2Cl2
O
O
H
O
H
CH2OH
Diethyl tartrate
(S,S) isomer
Figure… Asymmetric oxidation of alkenes using Sharpless’s method.
2.13 Oxidative Cleavage of Alkenes: Ozonolysis
The C-C double bond can be converted into other functional groups by oxidative cleavage
using ozone and acidic or neutral solutions of potassium permanganate. Ozone adds
rapidly to an alkene forming a cyclic intermediate called a molozonide (or primary
ozonide). Once formed molozonides rapidly rearrange to form more stable ozonides which
are cleaved by either Zn/CH3CO2H or CH3SCH3 (reductive conditions) and by H2O2
(oxidative conditions) to yield carbonyl compounds. If the ozonides are reduced,
32
aldehydes or ketones are formed depending on the nature of the alkene. Oxidation of the
ozonides yields mostly carboxylic acids or ketones.
Ozone is a 1,3-dipole which functions as a ultraviolet (UV) let absorber in the atmosphere.
Thus this gas screens us from the dangerous UV radiation which causes skin cancer in
humans. Its resonance forms are:
O
O
O
O
O
O
O
O
O
Figure … Resonance structures of Ozone
The 1,3-dipolar addition of ozone to an alkene gives a molozonide, a five-membered ring
containing three oxygen atoms in a series. The molozonide simply rearranges to give a
relatively more stable ozonide with two oxygen atoms in series.
O
O
O
O
O3, - 78°C
rearrangement
O
ozonide
molozonide
O
O
O
O
Figure .. Formation of the ozonide through the molozonide.
As earlier said, the ozonide is cleaved under reactive conditions to yield either aldehydes
or ketones depending on the nature of the alkene.
R
R
reductive
workup
R
H
O
O
O
R
ketone
H
aldehyde
R
R
O
O +
R
oxidative
workup
R
O +
HO
carboxylic acid
O
R
ketone
Figure … Reduction and Oxidation of Ozonides to give either aldehydes, ketones or
aldehydes.
33
For example,
CH3
CH3
H2C
1. O3, -78°C, 16 h
C(CH3)3
O
2. Zn, H2O
C(CH3)3
H
H
1. O3, -30°C
H2C
2. (CH3)2S, 0 C
°
O
1. O3, -30 °C
2. (CH3)2S, 0 C
°
O
O
H
OH
1. O3, CH2Cl2, -78°C
H2C
2. H2O2, CH3COOH
O
H
COOH
1. O3, CH3OH, -78° C
2. H2O2, HCOOH
COOH
H
The net result of oznolysis is that the carbon-carbon double bond is cleaved, and oxygen
atom becomes doubly bonded to each of the original alkene carbon atoms.
Problem
Indicate the product(s) of the following reaction.
1. O3, -30 °C
?
°
2. (CH3)2S, 0 C
1. O3, CH3OH, -78° C
?
2. H2O2, HCOOH
The alkene can also be cleaved by using a neutral or an acidic solution of potassium
permanganate (KMnO4) giving carbonyl compounds. If hydrogen atoms are present on the
double bond, carboxylic acids are produced.
34
H
COOH
+
KMnO4, H3O
COOH
H
2.14 Oxidative Cleavage of 1,2-Diols
1,2-Diols can be cleaved by reaction with periodic acid, HIO4 to yield carbonyl
compounds.
O
OH
O
HIO4
O
I
OH
O
H
O
H
OH
H
O
Cyclic periodate
intermediate
2.15 Addition of Carbenes to Alkenes: Synthesis of Cyclopropanes
A carbine is a neutral molecule in which a carbon atom is surrounded by six valence
electrons. It can be represented as follows:
C
R
R
R
3
empty sp orbital
R
R
R
an electron
in each sp3 orbital
B
A
Figure… The structure of carbene showing (a) the electron pair in an sp 3 orbital and (b) an
electron in each sp3 orbital.
Structure A will act as a Lewis acid and B as very reactive diiradical. A carbenes add to an
alkene to a cyclopropane.
R
+
R
C
R
R
Figure Synthesis of Cyclopropane from carbenes
A substituted carbene is generated by treatment of chloroform CHCl3 with a strong base
such as potassium tert-butoxide.
35
t-BuOK + CHCl3
:CCl2
:CCl3
The carbene thus produced reacts with alkenes to give chlorinated cyclopropanes as shown
in the examples below.
Cl
+ CHCl3
Cl
t-BuOK
Cl
+ CHCl3
t-BuOK
Cl
Nonhalogenated cyclopropanes are prepared by Simmoms-Smith’s reaction.
Diiodomethane (methylene iodide), CH2I2, is treated with zinc-copper alloy to form
(iodomethyl)zinc iodide, ICH2ZnI. In the presence of an alkene, the CH2 of the
(iodomethyl)zinc iodide, attaches itself to the double bond and yields the cyclopropane.
+ CH2I2
Zn(Cu)
ether
+ ZnI2
Problem
What product will you expect from each of the following reactions?
CHCl3
t-BuOK
a.
?
b.
CH2I2
Zn(Cu)
?
2.16 Dienes
Dienes are alkenes with two double bonds. They can be classified into three categories:
1. Cumulated dienes (cumulenes or allenes) are dienes in which two double bonds are
adjacent to each other, e.g., 1,2-pentadiene
2. Isolated dienes are dienes in the double bonds are separated by more than one
single bond, as in 1,4-pentadiene.
3. Conjugated dienes are dienes in which the two double bonds are separated by one
single bond as in 1,3-pentadiene.
1,2-pentadiene
(Cumulated
or allenes)
1,4-pentadiene
(Isolated)
36
1,3-pentadiene
(Conjugated)
In conjugated dienes, the π-electron system is highly delocalized over the whole system
and this adds to the stability of the molecule, i.e., delocalization increases the stability of
the molecule.
Figure … Delocalization of π electrons in a conjugated system.
Addition of Bromine to Conjugated Dienes
There are usually two products: 1,2- and 1,4- addition product.
+
Br
Br
Br
Br
1,4-product
1,2-product
Br-
Br-
Br
BrH2C
Br
Br
(63%)
(37%)
Figure… Addition products of Bromine to Dienes
The 1,4 –product could be a cis or trans alkene with the trans product being the major
product. The reaction is temperature-dependent. At low temperature (~ 15 ºC) the 1,2product is predominant and at high temperature (~ 40 ºC) the 1,4-product predominates.
2.16a Diels-Alder Reaction
This reaction involves the reaction of an alkene to a diene to form a cyclic structure
(adduct). For example,
A
A
B
X
Y
B
+
Diene
Y
X
Dienophile
Diels-Alder adduct
The alkene that adds to the diene is called the dienophile and the reaction is termed a
cycloaddition reaction. The Diels-Alder reaction is highly stereospecific. The reaction is a
syn addition and the configuration of the dienophile is retained in the product. This
reaction can easily take place if there are electron-withdrawing groups on the dienophile.
37
O
O
+
H
H
O
+
H
O
O
O
O
H
O
O
O
H
OCH3
+
OCH3
H
OCH3
H
H
O
OCH3
O
cis-dienophile
cis-adduct
2.17 Polymerization of Alkenes
Polymerization is a process in which molecules of a low molecular weight compound
(monomer) react with themselves over and over again to form large molecules called
polymers with recurring structural units.
n H2C
initiator
CH2
CH2
CH2
n
This is an addition polymer as no atoms of the monomer unit have been lost.
Polymerization of alkenes takes place, mostly, by a free-radical mechanism (see the
peroxide effect).
Initiation
heat
R
O
O
R
O
2 R
eqn 2.0
Propagation
RO + H2C
CH
RO
CH2
R
CH
R
38
RO
CH2
CH + H2C
CH
R
R
repeating
units
RO
CH2
CH
R
CH2
C
RO
CH2
R
CH
R
Termination
i.
Radicals combine
ii.
Chain accepts a H. radical
iii.
Chain loses a H. radical
The table below indicates some polymers produced by free-radical polymerization.
Table xx Common Polymers produced by Free-radical Polymerization
Monomer
H2C
CH2
Polymer
polyethylene
H2C
CHCH3
polypropylene
H2C
C(CH3)2
polyisobutene
H2C
CHCl
polyvinyl chloride
(PVC)
H2C
CHCN
Polyacrylonitrile
(orlon, acrilan)
polystyrene
H2C
CH
O
H2C
CH
H2C
C
H2C
CH3
CCl2
F2C
CF2
OCCH3
O
OCCH3
Uses
sheets and films, blow-molded
bottles, injection-molded toys ans
housewares, wire and cable
coverings, shipping containers
carpets, car parts, packaging,
toys and housewares
adhesives (used on plastic
bandages)
plastic pipes and fittings, floor
tiles, records, coatings, films and
sheets
clothing particularly sweaters
Packaging and containers
(Styrofoam), insulation,
appliance parts, toys
polyvinyl acetate
adhesives, latex paints
polymethyl
methacrylate
(plexiglas, lucite)
clear, transparent and tough
objects
polyvinylidene
food packaging
chloride (Saran)
polytetrafluoroehylene Coating for cooking utensils,
(Teflon)
electric insulators, lenses for
high-intensity discharge lamps
39
n
2.18 Allylic Bromination of Alkenes
Compounds with allylic hydrogen atoms react with N-bromosuccinimde (NBS) in CCl4
under free-radical conditions (heat or UV light) to produce compounds where the double
bond remains intact.
H
Br
O
N
+
Br
O
heat
NBS
+
N
+
CCl4
O
NBS
H
O
h
Br
The mechanism proposed for the reaction is as follows:
H
Br
+
H
H
H
C
CH
CH2
C
CH
CH2
CH2
allylic free radical
H
H
O
O
N
Br + HBr
Br2
+
N
O
O
H
H
CH2
CH
H
C
+
Br
CH2
Br
CH
C
Br + Br
H
H
Figure.. Proposed Mechanism for the Allylic Bromination of Alkenes.
This reactions occurs very easily because of the stability of the allylic free radical.
40
CH
C
H