Chapter 8
Reactions of Alkenes
Jo Blackburn
Richland College,Dallas,TX
Dallas County Community College District
2003,Prentice Hall
Organic Chemistry,5th Edition
L,G,Wade,Jr.
Chapter 8 2
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 3
Electrophilic Addition
Step 1,Pi electrons attack the electrophile.
C C + E + C
E
C +
C
E
C + + N u c,
_
C
E
C
N u c
=>
Step 2,Nucleophile attacks the carbocation.
Chapter 8 4
Types of Additions
=>
Chapter 8 5
Addition of HX (1)
Protonation of double bond yields the most
stable carbocation,Positive charge goes to
the carbon that was not protonated.
X
=>
+ B r
_
+
+
C H 3 C
C H 3
C H C H 3
H
C H 3 C
C H 3
C H C H 3
H
H B r
C H 3 C
C H 3
C H C H 3
Chapter 8 6
Addition of HX (2)
C H 3 C
C H 3
C H C H 3
H B r
C H 3 C
C H 3
C H C H 3
H
+
+ B r
_
C H 3 C
C H 3
C H C H 3
H
+
Br
_
C H 3 C
C H 3
C H C H 3
HB r
=>
Chapter 8 7
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 8
Free-Radical
Addition of HBr
In the presence of peroxides,HBr adds
to an alkene to form the,anti-
Markovnikov” product.
Only HBr has the right bond energy.
HCl bond is too strong.
HI bond tends to break heterolytically to
form ions,
=>
Chapter 8 9
Free Radical Initiation
Peroxide O-O bond breaks easily to
form free radicals.
+R O H B r R O H + B r
O OR R +R O O R
heat
Hydrogen is abstracted from HBr.
Electrophile
=>
Chapter 8 10
Propagation Steps
Bromine adds to the double bond,
+
C
B r
C H B r+ C
B r
C
H
B r
Electrophile =>
C
B r
CC CB r +
Hydrogen is abstracted from HBr.
Chapter 8 11
Anti-Markovnikov
Tertiary radical is more stable,so that
intermediate forms faster,=>
C H 3 C
C H 3
C H C H 3 B r+
C H 3 C
C H 3
C H C H 3
B r
C H 3 C
C H 3
C H C H 3
B r
X
Chapter 8 12
Hydration of Alkenes
Reverse of dehydration of alcohol
Use very dilute solutions of H2SO4 or
H3PO4 to drive equilibrium toward
hydration,
=>
C C + H 2 O
H
+
C
H
C
O H
a l k e n e
a l c o h o l
Chapter 8 13
Mechanism for Hydration
+C
H
C
+
H 2 O C
H
C
O H
H
+
+ H 2 OC
H
C
O H
H
+
C
H
C
O
H
H 3 O+
+=>
C C OH H
H
+
+
+ H
2
O
C
H
C
+
Chapter 8 14
Orientation for Hydration
Markovnikov product is formed.
+
C H
3
C
C H
3
C H C H
3 OH H
H
+
+
H
2
O+
H
C H
3
C H
C H
3
CC H
3
H 2 O
C H 3 C
C H 3
C H C H 3
HO
H
H
+
H 2 O
C H 3 C
C H 3
C H C H 3
HO
H =>
Chapter 8 15
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 16
Oxymercuration (1)
Reagent is mercury(II) acetate which
dissociates slightly to form +Hg(OAc).
+Hg(OAc) is the electrophile that attacks
the pi bond.
C H 3 C
O
O H g O C
O
C H 3 C H 3 C
O
O
_
H g O C
O
C H 3
+
=>
Chapter 8 17
Oxymercuration (2)
The intermediate is a cyclic mercurinium
ion,a three-membered ring with a
positive charge.
C C
+
H g( O A c ) C C
H g
+
O A c
=>
Chapter 8 18
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.
C C
H g
+
O A c
H 2 O
C
O
+
C
H g
H
H
O A c
H 2 O
C
O
C
H g
H
O A c
=>
Chapter 8 19
Demercuration
Sodium borohydride,a reducing agent,
replaces the mercury with hydrogen.
C
O
C
H g
H
O A c
4 4 C
O
C
H
H
+ N a B H 4 + 4 O H
_
+ N a B ( O H )4
+ 4 H g + 4 O A c
_
=>
Chapter 8 20
Predict the Product
Predict the product when the given alkene
reacts with aqueous mercuric acetate,
followed by reduction with sodium
borohydride.
C H 3
D
( 1 ) H g ( O A c )2,H 2 O
( 2 ) N a B H 4
=>
O H
C H 3
D
H
anti addition
Chapter 8 21
Alkoxymercuration -
Demercuration
If the nucleophile is an alcohol,ROH,
instead of water,HOH,the product is an
ether.
C C
( 1 ) H g ( O A c )2,
C H 3 OH
C
O
C
H g( O A c )
H 3 C
( 2 ) N a B H 4
C
O
C
H 3 C
H
=>
Chapter 8 22
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 ) B H 3
C
H
C
B H 2
( 2 ) H 2 O 2,O H -
C
H
C
O H
=>
Chapter 8 23
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.
THF THF,BH3
O B 2 H 6 O + B -
H
H
H
+2 2=>
Chapter 8 24
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 25
Actually,Trialkyl
Borane prefers least-substituted carbon due to
steric hindrance as well as charge distribution,
=>
C C
H 3 C
H 3 C
H
H
+ B H 3
B
CC H
C H 3
H 3 C
H
H
C
CH
H
H
C H 3
C H 3
C
C
H
H
H 3 C
C H 3
H
3
Chapter 8 26
Oxidation to Alcohol
Oxidation of the alkyl borane with basic
hydrogen peroxide produces the alcohol.
Orientation is anti-Markovnikov.
C H 3 C
C H 3
H
C
H
H
B
H 2 O 2,N a O H
H 2 O
C H 3 C
C H 3
H
C
H
H
O H
=>
Chapter 8 27
Predict the Product
Predict the product when the given alkene
reacts with borane in THF,followed by
oxidation with basic hydrogen peroxide.
C H 3
D
( 1 )
( 2 )
B H 3,T H F
H 2 O 2,O H -
=>syn addition
H
C H 3
D
O H
Chapter 8 28
Hydrogenation
Alkene + H2? Alkane
Catalyst required,usually Pt,Pd,or Ni.
Finely divided metal,heterogeneous
Syn addition
=>
Chapter 8 29
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 30
Diazomethane
Extremely toxic and explosive,=>
N N C H 2 N N C H 2
d i a z o m e t h a n e
N N C H 2
h e a t o r u v l i g h t
N 2 +
c a r be n e
C
H
H
C
H
H
C
C
C
C
C
H
H
Chapter 8 31
Simmons-Smith
Best method for preparing cyclopropanes.
C H 2 I 2 + Z n ( C u ) I C H 2 Z n I
a c a r be n o i d
C H 2 I 2
Z n,C u C l
=>
Chapter 8 32
Alpha Elimination
Haloform reacts with base.
H and X taken from same carbon
C H C l 3 + K O H K + - C C l 3 + H 2 O
CC l
C l
C l C l -+C
C l
C l
C l
C l
C H C l 3
K O H,H 2 O
=>
Chapter 8 33
Stereospecificity
Cis-trans isomerism maintained around
carbons that were in the double bond.
C C
H
C H 3
H
H 3 C N aO H,H 2 O
C H B r 3
C C
H
C H 3
H
H 3 C
B rB r
=>
Chapter 8 34
Addition of Halogens
Cl2,Br2,and sometimes I2 add to a
double bond to form a vicinal dibromide.
Anti addition,so reaction is
stereospecific.
CC + B r
2 C C
B r
B r
=>
Chapter 8 35
Mechanism for
Halogenation
Pi electrons attack the bromine
molecule.
A bromide ion splits off.
Intermediate is a cyclic bromonium ion.
CC + B r B r CC
B r
+ B r=>
Chapter 8 36
Mechanism (2)
Halide ion approaches from side opposite
the three-membered ring.
CC
B r
B r
CC
B r
B r
=>
Chapter 8 37
Examples of
Stereospecificity
=>
Chapter 8 38
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 39
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.
CC
B r
H 2 O
CC
B r
O
H H
H 2 O
CC
B r
O
H
+ H 3 O
+
=>
Chapter 8 40
Regiospecificity
The most highly substituted carbon has
the most positive charge,so nucleophile
attacks there.
=>
Chapter 8 41
Predict the Product
Predict the product when the given alkene
reacts with chlorine in water.
C H 3
D
C l 2,H 2 O
=>
O H
C H 3
D
C l
Chapter 8 42
Epoxidation
Alkene reacts with a peroxyacid to form
an epoxide (also called oxirane).
Usual reagent is peroxybenzoic acid.
CC + R C
O
O O H CC
O
R C
O
O H+
=>
Chapter 8 43
Mechanism
One-step concerted reaction,Several
bonds break and form simultaneously.
O
C
O
R
H
C
C
OO
H
O
C
O
RC
C
+
=>
Chapter 8 44
Epoxide
Stereochemistry
Since there is no opportunity for rotation
around the double-bonded carbons,cis
or trans stereochemistry is maintained.
CC
C H 3 C H 3
H H P h C
O
O O H
CC
C H 3 C H 3
H H
O
=>
Chapter 8 45
Opening the
Epoxide Ring
Acid catalyzed.
Water attacks the protonated epoxide.
Trans diol is formed.
CC
O
H 3 O +
CC
O
H
H 2 O
CC
O
O H
H H
H 2 O
CC
O
O H
H
=>
Chapter 8 46
One-Step Reaction
To synthesize the glycol without
isolating the epoxide,use aqueous
peroxyacetic acid or peroxyformic acid.
The reaction is stereospecific.
C H 3 C O O H
O
O H
H
O H
H
=>
Chapter 8 47
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 48
Mechanism with OsO4
Concerted syn addition of two oxygens to
form a cyclic ester.
C
C
O s
O O
OO
C
C
O O
OO
O s
C
C
O H
O H
+ O s O 4H 2O 2 =>
Chapter 8 49
Stereospecificity
If a chiral carbon is formed,only one
stereoisomer will be produced (or a pair
of enantiomers).
C
C
C H 2 C H 3
H C H 2 C H 3
C
C
C H 2 C H 3
C H 2 C H 3
O H
O H
H
HH 2 O 2
H
( 2 )
( 1 ) O s O 4
c is - 3 - h e x e n e m e s o - 3,4 - h e x a n e di o l
=>
Chapter 8 50
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 51
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 52
Example
CC
C H 3 C H 3
H C H 3 K M n O 4
( w a r m,c o n c,)
C C
C H 3
C H 3
O HO H
H 3 C
H
C
O
H 3 C
H
C
C H 3
C H 3
O
C
O
H 3 C
O H
+
=>
Chapter 8 53
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 54
Ozonolysis Example
CC
C H 3 C H 3
H C H 3 O 3
C
H 3 C
H
O O
C
C H 3
C H 3
O
Ozonide
+
( C H 3 ) 2 S
C
H 3 C
H
O C
C H 3
C H 3
O C H 3 S
O
C H 3
DMSO
=>
Chapter 8 55
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 56
Cationic Polymerization
Electrophile,like H+ or BF3,adds to the
least substituted carbon of an alkene,
forming the most stable carbocation.C C
C H 3
H
H
H
O
H
H
H
C
H
H
H
C
C H 3
H
+ C C
C H 3
H
H
H
C
H
H
H
C
C H 3
H
C
H
H
C
C H 3
H
=>
Chapter 8 57
Radical Polymerization
In the presence of a free radical initiator,
like peroxide,free radical polymerization
occurs.
C C
P h
H
H
H
R O
C
H
R O
H
C
P h
H
+ C C
P h
H
H
H
C
H
R O
H
C
P h
H
C
H
H
C
P h
H
=>
Chapter 8 58
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.
C C
C O C H 3
C N
H
H
O
O H
-
C
H
H O
H
C
C O C H 3
C N
O
+ C C
C O C H 3
C N
H
H
O
C
H
H O
H
C
C
C N
C
H
C
C O C H 3
C NH
OO O C H
3
=>
Chapter 8 59
End of Chapter 8