Organic Chemistry, 5th Edition
L. G. Wade, Jr.
Chapter 8
Reactions of Alkenes
Jo Blackburn
Richland College, Dallas, TX
Dallas County Community College District
2003, Prentice Hall
Reactivity of C=C
• Electrons in pi bond are loosely held.
• Electrophiles are attracted to the pi
electrons.
• Carbocation intermediate forms.
• Nucleophile adds to the carbocation.
• Net result is addition to the double bond.
=>
Chapter 8
2
Electrophilic Addition
• Step 1: Pi electrons attack the electrophile.
E
C
C
+
+
E
C
C +
• Step 2: Nucleophile attacks the carbocation.
E
C C+
+
_
Nuc:
Chapter 8
E Nuc
C C
3
=>
Types of Additions
=>
Chapter 8
4
Addition of HX (1)
Protonation of double bond yields the most
stable carbocation. Positive charge goes to
the carbon that was not protonated.
CH3
CH3 C CH CH3
+
H
CH3
CH3 C CH CH3
H Br
+ Br
X
_
CH3
CH3 C CH CH3
+
H
Chapter 8
=>
5
Addition of HX (2)
CH3
CH3 C CH CH3
H Br
CH3
CH3 C CH CH3
+
H
CH3
+ Br
_
CH3
CH3 C CH CH3
+
H
_
Br
CH3 C CH CH3
=>
Br H
Chapter 8
6
Regiospecificity
• Markovnikov’s Rule: The proton of an
acid adds to the carbon in the double
bond that already has the most H’s. “Rich
get richer.”
• More general Markovnikov’s Rule: In an
electrophilic addition to an alkene, the
electrophile adds in such a way as to
form the most stable intermediate.
• HCl, HBr, and HI add to alkenes to form
Markovnikov products.
=>
Chapter 8
7
Free-Radical
Addition of HBr
• In the presence of peroxides, HBr adds
to an alkene to form the “antiMarkovnikov” product.
• Only HBr has the right bond energy.
• HCl bond is too strong.
• HI bond tends to break heterolytically to
form ions.
=>
Chapter 8
8
Free Radical Initiation
• Peroxide O-O bond breaks easily to
form free radicals.
R O O R
heat
R O
+
O R
• Hydrogen is abstracted from HBr.
R O
R O H
+ H Br
+ Br
=>
Chapter 8
Electrophile
9
Propagation Steps
• Bromine adds to the double bond.
Br
+
C C
C C
Br
• Hydrogen is abstracted from HBr.
C C
+
H Br
C C
Br
+
Br
Br H
Chapter 8
Electrophile =>
10
Anti-Markovnikov ??
CH3
CH3 C CH CH3
CH3
CH3 C CH CH3
Br
+
Br
X
CH3
CH3 C CH CH3
Br
• Tertiary radical is more stable, so that
intermediate forms faster.
=>
Chapter 8
11
Hydration of Alkenes
+
H
C C
+ H2O
H OH
C C
alkene
alcohol
• Reverse of dehydration of alcohol
• Use very dilute solutions of H2SO4 or
H3PO4 to drive equilibrium toward
hydration.
=>
Chapter 8
12
Mechanism for Hydration
H
H
C C
+
C C
+
H O H
+
H2O
H
+
H O H
H
+
C C
+
H2O
C C
H
+
H O H
C C
+
H
H O
+
H2O
C C
Chapter 8
+
H3O
+
=>
13
Orientation for Hydration
• Markovnikov product is formed.
H
CH3
CH3 C CH CH3 + H O+ H
CH3
CH3
CH3 C CH CH3
H
CH3
CH3 C CH CH3 + H2O
+
H
H2O
CH3 C CH CH3
+O H
H
H
H2O
=>
O H
Chapter 8
14
Indirect Hydration
• Oxymercuration-Demercuration
Markovnikov product formed
Anti addition of H-OH
No rearrangements
• Hydroboration
Anti-Markovnikov product formed
Syn addition of H-OH
=>
Chapter 8
15
Oxymercuration (1)
• Reagent is mercury(II) acetate which
dissociates slightly to form +Hg(OAc).
• +Hg(OAc) is the electrophile that attacks
the pi bond.
O
CH3
O
O
C O Hg O C CH3
CH3
_
C O
O
+
Hg O C CH3
=>
Chapter 8
16
Oxymercuration (2)
The intermediate is a cyclic mercurinium
ion, a three-membered ring with a
positive charge.
OAc
+
C C
+
Hg(OAc)
Hg
C C
=>
Chapter 8
17
Oxymercuration (3)
• Water approaches the mercurinium ion from
the side opposite the ring (anti addition).
• Water adds to the more substituted carbon to
form the Markovnikov product.
OAc
+
Hg
C C
OAc
Hg
Hg
C C
+
H2O
OAc
H2O
C C
H O
O
H
H
Chapter 8
=>
18
Demercuration
Sodium borohydride, a reducing agent,
replaces the mercury with hydrogen.
OAc
Hg
4
C C
H
_
4
+ NaBH4 + 4 OH
C C
O
O
H
H
+
NaB(OH) 4
_
+ 4 Hg + 4 OAc
=>
Chapter 8
19
Predict the Product
Predict the product when the given alkene
reacts with aqueous mercuric acetate,
followed by reduction with sodium
borohydride.
CH3
D
(1) Hg(OAc) 2, H2O
(2) NaBH4
anti addition
Chapter 8
OH
CH3
D
H
=>
20
Alkoxymercuration Demercuration
If the nucleophile is an alcohol, ROH,
instead of water, HOH, the product is an
ether.
C C
(1) Hg(OAc) 2,
CH 3OH
Hg(OAc)
H3C
H
(2) NaBH4
C C
C C
O
O
H3C
=>
Chapter 8
21
Hydroboration
• Borane, BH3, adds a hydrogen to the
most substituted carbon in the double
bond.
• The alkylborane is then oxidized to the
alcohol which is the anti-Mark product.
C C
(1) BH3
(2) H2O2, OH
-
C C
C C
H BH2
H OH
=>
Chapter 8
22
Borane Reagent
• Borane exists as a dimer, B2H6, in equilibrium
with its monomer.
• Borane is a toxic, flammable, explosive gas.
• Safe when complexed with tetrahydrofuran.
H
2
O
+
2
B2H6
+
-
O B H
=>
H
THF . BH3
THF
Chapter 8
23
Mechanism
• The electron-deficient borane adds to
the least-substituted carbon.
• The other carbon acquires a positive charge.
• H adds to adjacent C on same side (syn).
=>
Chapter 8
24
Actually, Trialkyl
CH3
H
H3C C C H
H3C
H
C C
3
H3C
H
+ BH3
H
B
H
H C
H3C C
H
C
H
CH3
H
H C CH3
CH3
Borane prefers least-substituted carbon due to
steric hindrance as well as charge distribution.
=> 25
Chapter 8
Oxidation to Alcohol
• Oxidation of the alkyl borane with basic
hydrogen peroxide produces the alcohol.
• Orientation is anti-Markovnikov.
CH3 H
CH3
C C H
B
H
CH3 H
H2O2, NaOH
H2O
CH3
C C H
OH
H
=>
Chapter 8
26
Predict the Product
Predict the product when the given alkene
reacts with borane in THF, followed by
oxidation with basic hydrogen peroxide.
CH3
(1) BH3, THF
-
D
(2) H2O2, OH
syn addition
Chapter 8
H
CH3
OH
D
=>
27
Hydrogenation
•
•
•
•
Alkene + H2  Alkane
Catalyst required, usually Pt, Pd, or Ni.
Finely divided metal, heterogeneous
Syn addition
Chapter 8
=>
28
Addition of Carbenes
• Insertion of -CH2 group into a double
bond produces a cyclopropane ring.
• Three methods:
Diazomethane
Simmons-Smith: methylene iodide and
Zn(Cu)
Alpha elimination, haloform
=>
Chapter 8
29
Diazomethane
N N CH2
N N CH2
diazomethane
N
N
heat or uv light
CH2
H
N2
+
C
H
carbene
C
C
H
C
H
C
C
H
C
H
Extremely toxic and explosive.
Chapter 8
=>
30
Simmons-Smith
Best method for preparing cyclopropanes.
CH2I2
+
Zn(Cu)
ICH2ZnI
a carbenoid
CH2I2
=>
Zn, CuCl
Chapter 8
31
Alpha Elimination
• Haloform reacts with base.
• H and X taken from same carbon
+-
CHCl3 + KOH
Cl
K CCl3 + H2O
C
C Cl
Cl
Cl
Cl
+
-
Cl
CHCl3
Cl
KOH, H2O
Cl
Chapter 8
=>
32
Stereospecificity
Cis-trans isomerism maintained around
carbons that were in the double bond.
H
C C
H3C
H
CH3
CHBr3
NaOH, H2O
H
C C
H3C
Br
H
CH3
Br
=>
Chapter 8
33
Addition of Halogens
• Cl2, Br2, and sometimes I2 add to a
double bond to form a vicinal dibromide.
• Anti addition, so reaction is
stereospecific.
Br
C C
+
Br2
C C
=>
Br
Chapter 8
34
Mechanism for
Halogenation
• Pi electrons attack the bromine
molecule.
• A bromide ion splits off.
• Intermediate is a cyclic bromonium ion.
C C
+ Br
Br
Chapter 8
Br
C C
=>
+ Br
35
Mechanism (2)
Halide ion approaches from side opposite
the three-membered ring.
Br
Br
C C
C C
Br
Br
=>
Chapter 8
36
Examples of
Stereospecificity
=>
Chapter 8
37
Test for Unsaturation
• Add Br2 in CCl4 (dark, red-brown color) to
an alkene in the presence of light.
• The color quickly disappears as the
bromine adds to the double bond.
• “Decolorizing bromine” is the chemical test
for the presence of a double bond.
=>
Chapter 8
38
Formation of Halohydrin
• If a halogen is added in the presence of
water, a halohydrin is formed.
• Water is the nucleophile, instead of
halide.
• Product is Markovnikov and anti.
Br
C C
H2O
Br
Br
C C
C C
O
O
H
H
H
+
+ H3O
=>
H2O
Chapter 8
39
Regiospecificity
The most highly substituted carbon has
the most positive charge, so nucleophile
attacks there.
=>
Chapter 8
40
Predict the Product
Predict the product when the given alkene
reacts with chlorine in water.
CH3
Cl2, H2O
D
OH
CH3
D
Cl
=>
Chapter 8
41
Epoxidation
• Alkene reacts with a peroxyacid to form
an epoxide (also called oxirane).
• Usual reagent is peroxybenzoic acid.
O
C C
+ R C O O H
O
O
C C
+ R C O H
=>
Chapter 8
42
Mechanism
One-step concerted reaction. Several
bonds break and form simultaneously.
O
C
O
C
C
O
C
R
O
H O
C
+
H
C
R
O
=>
Chapter 8
43
Epoxide
Stereochemistry
Since there is no opportunity for rotation
around the double-bonded carbons, cis
or trans stereochemistry is maintained.
O
H
CH3
C C
H
Ph
C O O H
CH3
H
CH3
Chapter 8
O
C C
=>
H
CH3
44
Opening the
Epoxide Ring
• Acid catalyzed.
• Water attacks the protonated epoxide.
• Trans diol is formed.
OH
C C
O
H
+
H3O
O
O
C C
C C
OH
H
C C
=>
O
H2O
H
Chapter 8
H
H2O
45
One-Step Reaction
• To synthesize the glycol without
isolating the epoxide, use aqueous
peroxyacetic acid or peroxyformic acid.
• The reaction is stereospecific.
O
OH
CH3COOH
H
H
Chapter 8
OH
=>
46
Syn Hydroxylation
of Alkenes
• Alkene is converted to a cis-1,2-diol,
• Two reagents:
Osmium tetroxide (expensive!), followed by
hydrogen peroxide or
Cold, dilute aqueous potassium
permanganate, followed by hydrolysis with
base
=>
Chapter 8
47
Mechanism with OsO4
Concerted syn addition of two oxygens to
form a cyclic ester.
O
C
O
O
C
Os
C
O
Os
C
O
O
O
H2O2
C OH
C OH
Chapter 8
O
+ OsO4
=>
48
Stereospecificity
If a chiral carbon is formed, only one
stereoisomer will be produced (or a pair
of enantiomers).
H
CH2CH3
(1) OsO4
C
(2) H2O2
C
H
CH2CH3
H
C
OH
H
C
OH
=>
CH2CH3
CH2CH3
meso-3,4-hexanediol
cis-3-hexene
Chapter 8
49
Oxidative Cleavage
• Both the pi and sigma bonds break.
• C=C becomes C=O.
• Two methods:
Warm or concentrated or acidic KMnO4.
Ozonolysis
• Used to determine the position of a
double bond in an unknown.
=>
Chapter 8
50
Cleavage with MnO4• Permanganate is a strong oxidizing
agent.
• Glycol initially formed is further oxidized.
• Disubstituted carbons become ketones.
• Monosubstituted carbons become
carboxylic acids.
• Terminal =CH2 becomes CO2.
=>
Chapter 8
51
Example
H
CH3
C C
CH3
CH3
H CH3
KMnO4
(warm, conc.)
H3C C C CH3
OH OH
H
H3C C
CH3
+
O
O
H3C C
Chapter 8
C CH3
O
OH
=>
52
Ozonolysis
• Reaction with ozone forms an ozonide.
• Ozonides are not isolated, but are
treated with a mild reducing agent like
Zn or dimethyl sulfide.
• Milder oxidation than permanganate.
• Products formed are ketones or
aldehydes.
=>
Chapter 8
53
Ozonolysis Example
H
CH3
C C
CH3
O
H
O3
C
C
CH3
H3C
CH3
O O
CH3
Ozonide
(CH3)2S
H
C O
H3C
O C
O
CH3
+
CH3
Chapter 8
CH3
S
CH3
=>
DMSO
54
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)
=>
Chapter 8
55
Cationic Polymerization
Electrophile, like H+ or BF3, adds to the
least substituted carbon of an alkene,
forming the most stable carbocation.
H
H
C C
CH3
H
H
H
+
H C C
H
H O
CH3
H
H
H
C C
CH3
H
H CH3 H CH3
H C C C C
H
H H H
Chapter 8
=>
H
56
Radical Polymerization
In the presence of a free radical initiator,
like peroxide, free radical polymerization
occurs.
H
H
C C
Ph
H
H
+
RO C C
H
H
Ph
H
RO
H
C C
Ph
H
H Ph H Ph
RO C C C C
H H H
Chapter 8
=>
H
57
Anionic Polymerization
For an alkene to gain electrons, strong
electron-withdrawing groups such as nitro,
cyano, or carbonyl must be attached to the
carbons in the double bond.
O
O
-
OH
H
H
C C
COCH3
CN
H
COCH3
HO C C
H
O
+
CN
H
H
C C
COCH3
CN
O OCH3 O
H C
H COCH3
HO C C C C
Chapter 8
H CN H CN
=>
58
End of Chapter 8
Chapter 8
59