Reaction of alkenes
1. Addition Reactions
• The characteristic reaction of alkenes is addition
• Because alkenes are electron rich, they react with
electrophiles.
Electrophilic Addition
• Step 1: Pi electrons attack the electrophile.
• Step 2: Nucleophile attacks the carbocation.
carbocation
1.a. Addition of HX (Hydrohalogenation
Hydrohalogenation)
C C
+
H
X
C C
H
X
haloalkane
Mechanism of Addition of HX
• Step 1: Protonation of the double bond.
• Step 2: Nucleophilic attack of the halide on the carbocation.
Regioselectivity—Markovnikov’s
Markovnikov’s Rules
• Markovnikov’s rule states that in the addition of HX to an
unsymmetrical alkene, the H atom adds to the less
substituted carbon atom—that
that is, the carbon that has the
greater number of H atoms to begin with.
• Markovnikov’s Rule (extended):
(extended) In an electrophilic addition to
the alkene, the electrophile adds in such a way that it
generates the most stable intermediate.
intermediate
• In the addition of HX to an unsymmetrical alkene, the H atom
is added to the less substituted carbon to form the more
stable, more substituted carbocation.
carbocation
Free-Radical Addition of HBr
• In the presence of peroxides,
peroxides HBr adds to an
alkene to form the “anti-Markovnikov”
Markovnikov” product.
• Peroxides produce free radicals.
radicals
• Only HBr has the right bond energy.
• The HCl bond is too strong, so it will add according
to Markovnikov’s rule, even in the presence of
peroxide.
• The HI bond tends to break heterolytically to form
ions, it too will add according to Markovnikov’s
rule.
Free-Radical Initiation
• The peroxide bond breaks homolytically to form the first radical:
• Hydrogen is abstracted from HBr.
Propagation Steps
• Bromine adds to the double bond forming the
most stable radical possible:
CH3
CH3
C
CH
CH3
+
Br
CH3
CH3
C
CH
CH3
Br
tertiary radical (more stable)
• Hydrogen is abstracted from HBr:
H
Br
C
H
C
H
+
CH3
H Br
H
Br
C
H
H
C H
CH3
+
Br
CH3
CH3
C
CH
CH3
+
Br
CH3
CH3
C
CH
Br
CH3
tertiary radical (more stable)
CH3
CH3
C
Br
CH
CH3
secondary radical (less stable)
not formed
• The intermediate tertiary radical forms faster because it
is more stable.
1.b. Hydration
• The Markovnikov addition of water to the double bond forms
an alcohol.
• Rearrangements can ocur.
• This is the reverse of the dehydration of alcohol.
• Uses dilute solutions of H2SO4 or H3PO4 to drive equilibrium
toward hydration.
Mechanism for Hydration
• Step 1: Protonation of the double bond forms a carbocation.
• Step 2: Nucleophilic attack by water.
water
• Step 3: Deprotonation by nucleophilic attack of water.
water
Orientation of Hydration
CH3
CH3 C
CH CH3
+
H
H O H
The protonation follows
Markovnikov’s rule: The hydrogen
is added to the less substituted
carbon in order to form the most
stable carbocation.
CH3
CH3 C
CH CH3
H
3o, more stable
CH3
CH3 C
H
CH CH3
2o less stable
not formed
Rearrangements
CH3
CH3
C CH
CH3
CH2
50% H2SO4
Rearrangement:
CH3
CH3 C CH CH3
CH3
2o, less stable
CH3
CH3 C
CH CH3
CH3
3o, more
e stable
CH3
CH3
C CH CH3
OH CH3
CH3
CH3 C
CH CH3
OH CH3
2,3-dimethyl-2-butanol
(major product)
• Rearrangements can occur when there are carbocation
intermediates.
• A methyl shift after protonation will produce the more stable
tertiary carbocation.
Solved Problem 1
Show how you would accomplish the following synthetic conversions.
conversion
(a) Convert 1-methylcyclohexene
methylcyclohexene to 1-bromo-1-methylcyclohexane.
1
Br
Solution
This synthesis requires the addition of HBr to an alkene with
Markovnikov orientation. Ionic addition of HBr gives the
correct product.
Solved Problem 2
Convert 1-methylcyclohexanol
methylcyclohexanol to 1-bromo-2-methylcyclohexane.
Solution
This synthesis requires the conversion of an alcohol to an alkyl bromide
with the bromine atom at the neighboring carbon atom. This is the antiMarkovnikov product, which could be formed by the radical-catalyzed
addition of HBr to 1-methylcyclohexene
methylcyclohexene.
1-Methylcyclohexene
Methylcyclohexene is easily synthesized by the dehydration of 11
methylcyclohexanol. The most substituted alkene is the desired product.
Solution (Continued)
The two-step
step synthesis is summarized as follows:
1. c. Addition of Halogens
• Cl2, Br2, and sometimes I2 add to a double bond to
form a vicinal dibromide.
Mechanism of Halogen Addition to Alkenes
• The intermediate is a three-membered
membered ring called the
halonium ion.
Halogenation of cycloalkenes
1.d. Formation of Halohydrin
(Addition of halogen and water)
Mechanism of Halohydrin Formation
Alkenes in Organic Synthesis
Suppose we wish to synthesize 1,2-dibromocyclohexane from
cyclohexanol.
Working backwards from the product to determine the starting
material from which it is made is called retrosynthetic analysis.
Working backwards:
1) 1,2-Dibromocyclohexane, a vicinal dibromide, can be prepared by
the addition of Br2 to cyclohexene.
Working forwards:
2) Cyclohexanol can undergo acid-catalyzed
acid
dehydration to form
cyclohexene.
Cyclohexene is called a synthetic intermadiate, or simply an
intermadiate, because it is the product of one step and the starting
material of another. We now have a two-step sequence to convert
cyclohexanol to 1,2-dibromocyclohexane
dibromocyclohexane, and the synthesis is
complete. Take note of the central role of the alkene in this
synthesis.
A two-step synthesis:
1.e. Addition of hydrogen (Reduction)
C C
+
H
H
Catalyst
C C
H H
• The addition of H2 occurs only in the presence of a metal catalyst,
and thus it is called catalytic hydrogenation.
hydrogenation
• The catalyst consists of a metal—usually
metal
Pd, Pt, or Ni, adsorbed
onto a finely divided inert solid, such as charcoal.
• H2 adds in a syn fashion.
• The hydrogen and the alkene are adsorbed on the metal surface.
• Once adsorbed, the hydrogens insert across the same face of
the double bond and the reduced product is released from the
metal.
• The reaction is a syn addition since both hydrogens will add to
the same side of the double bond.
2. Epoxidation
• Epoxidation is the addition of a single oxygen atom to an
alkene to form an epoxide.
• Epoxidation is typically carried out with a peroxyacid.
• The usual reagent is peroxybenzoic acid.
Nomenclature of Epoxides:
• Epoxides can be named in three different ways—As
ways
epoxyalkanes, oxiranes, or alkene oxides.
oxides
• To name an epoxide as an epoxyalkane,
epoxyalkane first name the
alkane chain or ring to which the O atom is attached, and
use the prefix “epoxy” to name the epoxide as a substituent.
Use two numbers to designate the location of the atoms to
which the O’s are bonded.
• Epoxides bonded to a chain of carbon atoms can also be
named as derivatives of oxirane,
oxirane the simplest epoxide having
two carbons and one oxygen atom in a ring.
• The oxirane ring is numbered to put the O atom at position
one, and the first substituent at position two.
• No number is used for a substituent in a monosubstituted
oxirane.
• Epoxides are also named as alkene oxides,
oxides since they are
often prepared by adding an O atom to an alkene. To name
an epoxide in this way:
• Mentally replace the epoxide oxygen with a double bond.
• Name the alkene.
• Add the word oxide.
Mechanism of epoxydation with peracid:
R
R
H
H
O
R
O
H
Both of two bond
form simultaneously
O
H
R
R
O
H
R
O
H
O
Epoxide formation from halohydrin cyclization
3. Oxidation reactions
3. a. Hydroxylation
• Alkene is converted to a syn-1,2-diol
diol
• Two reagents:
– Osmium tetroxide, OsO4, followed by hydrogen peroxide or NaHSO3
– Cold, dilute solution of KMnO4 in base.
Mechanism
• The oxygens both approach from the same side of the alkene
• Sometimes NAHSO3 (sodium bisulfite)
bisulfite is shown as a co-reactant. Its
purpose is to break down the resulting cyclic Os compound into the
diol and osmium salt.
Mechanism
+
hydrolysis
oxidation
KMnO4
(purple)
reduction
(Brown)
Permanganate test for C=C double bond
Baeyer Test
POSITIVE TEST
KMnO4
purple
Show the presence
of double bond
MnO2
(+)
KMnO4
(-)
+
brown
OH OH
+
NEGATIVE TEST
No double bond
MnO2
3. b. Oxidative cleavage by KMnO4
R
H
R
•
•
•
•
•
H
C
C
C
C
R1
R2
H
H
KMnO4
H3O
+
KMnO4
H3O
+
O
R C OH
O
R C OH
+
+
O
R1 C
CO2
Permanganate is a strong oxidizing agent.
Glycol initially formed is further oxidized.
Disubstituted carbons become ketones.
ketones
Monosubstituted carbons become carboxylic acids.
Terminal ═CH2 becomes CO2.
R2
3. b. Oxidative cleavage by KMnO4
Some Example
3. b. Oxidative cleavage by KMnO4
Some Example
3.c. Ozonolysis
Common reducing agents are zinc (Zn) or dimethyl sulfide (CH3)2S
The key intermadiates in ozonolysis
• Ozonolysis of dienes or other polyenes results in oxidative cleavage of all C=C bon
3.c. Ozonolysis
Some Example
3.c. Ozonolysis
Some Example
Solved Problem 3
Ozonolysis–reduction of an unknown alkene gives an equimolar mixture of
cyclohexanecarbaldehyde and 2-butanone.
butanone. Determine the structure of the original
alkene.
Solved Problem 3
Solution
We can reconstruct the alkene by removing the two oxygen atoms of the carbony
groups (C=O) and connecting the remaining carbon atoms with a double bond.
One uncertainty remains, however: The original alkene might be either of two
possible geometric isomers.
Polymerization
• An alkene (monomer) can add to another molecule like itself to form a
chain (polymer).
• Three methods:
– Cationic, a carbocation intermediate
– Free radical
– Anionic, a carbanion intermediate (rare)
Cationic Polymerization
Some Common Addition Polymers
Commercial use of ethylene
Commercial use of propylene
O
propylene oxide Ag catalyst
H / H2O
+
OH
OH
propylene glycol
polymerize
O2
n
polypropylene
H2O
catalyst
OH
H
iso-propyla
alcohol
(IPA)
O2
catalyst
O
acetone