Reactions of Alkenes I
Reading: Wade chapter 8, sections 8-1- 8-8
Study Problems: 8-47, 8-48, 8-55, 8-66, 8-67, 8-70
Key Concepts and Skills:
•
Predict the products of additions to alkenes, including regiochemistry and
stereochemistry.
•
Propose logical mechanisms to explain the observed products of alkene addition
reactions, including regiochemistry and stereochemistry
Lecture Topics:
I.
Additions to pi bonds
The pi bond of Alkenes requires 63 kcal/mol to break; C-C sigma bond is 83 kcal/mol,
thus the reactivity of alkenes is dominated by the addition of reagents that cleave the pi
bond to form more stable sigma bonds, such as C-H bonds (100 kcal/mol). Additions to
alkenes are thus driven by thermodynamics- the tendency to form stronger bonds.
A. Electrophilic additions
Strong electrophiles shift the electron density of the double bond. Since the electron
density of a pi bond is above and below the internuclear axis, the electrons in pi bonds are
not as tightly held as those in sigma bonds. Thus the approach of a strong electrophile (or
Lewis acid, bearing a partial or formal positive charge, to the pi bond distorts the electron
cloud toward the electrophile, leaving one of the carbons of the double bond with a
partial or formal positive charge (carbocation):
E+
E+
1. Addition of HBr to alkenes
Addition of HBr (HCl, HI) to alkenes is an exothermic process which occurs
regiospecifically. According to the Markovnikov rule, the proton ends up on the carbon
with more hydrogen attached. Alternatively, this may be restated by saying that
electrophilic addition of H-X to a double bond occurs in such a way that the more stable
carbocation intermediate is produced. Halogen adds to the more highly substituted end of
the double bond
I
HI
CH3
HCl
Cl
H
H3C
H3C
H3C
HBr
H3C
H
H3C
CH3
Br
CH3
H
CH3
Br
H
H3C
H
not produced
H+
Br-
H3C
H
H3C
H3C
CH3
H3C
H3C
H
H
H
H
H3C
Br
CH3
CH3
Br-
2. Free radical additions to double bonds: Anti-Markovnikov regiochemistry
This process was discovered as the “peroxide effect” and only works for HBr addition
across alkenes. Peroxides contain a weak O-O bond which homolytically splits upon
heating to generate an oxygen radical, initiating a radical chain reaction. Again, reaction
of bromine radical (Br•, the electrophile) occurs in such a way that the more substituted
radical (more stable radical) is produced; reaction of this radical with H-Br generates
anti-markovnikov product and Br•, which continues the radical chain reaction.
H3C
H
H
H
H3C
HBr
H
Br
H
ROOR
heat
Anti-Markovnikov product
H
H
Initiation:
O
heat
O
O
O
2
O•
O
O
O
H
+ Br•
Br
OH
O•
Propagation:
H
H 3C
H
H
H
H
H3C
+ Br•
Br
Br
H
H
2° radical
X
H 3C
H3C
H
H
Br
1° radical
H
Br
H
H
H
+
Br•
H
Thus, Br• adds to the less substituted end of the double
bond to give the more stable (more substituted) radical.
Radical stabilites:
3°>2°>1°
Note again that this peroxide-promoted radical reaction is much faster than ordinary
electrophilic addition of H-X only for HBr addition to double bonds.
II.
Hydration of Alkenes
A. Acid-catalyzed hydration of alkenes is the microscopic reverse of alcohol dehydration.
The equilibrium is driven in the direction of the alcohol product in the presence of excess
water. Markovnikov regiochemistry is observed; each step is an equilibrium
Principle of Microscopic reversibility  the forward and reverse reaction must follow
the same pathway in microscopic detail.
H
H
O
H
H 3C
H
H
O
H3C
O
H
H
H
H
O
H
H
H
H 3C
H
H3C
CH3
H
H3C
CH3
H3C
CH3
3° carbocation
X
H3C
H 3C
H
H
H3C
H 3C
H
H
CH3
H
O
CH3
Markovnikov Product
2° carbocation
Acid-catalyzed hydration of alkenes is usually not a practical process, since presence of
cationic intermediate indicates that rearrangements and polymerization may occur:
Rearrangement:
H 3C
H 3C
H
H
H
O
H
H 3C
CH2
H3C
H3C
H 3C
H
CH3
methyl
shift
H 3C
H 3C
H
CH3
CH3
H2O
H3C
H 3C
Cationic polymerization
HO
H
CH3
CH3
H+
polymers
B. Oxymercuration/Demercuration of Alkenes: Markovnikov Hydration
Treating alkenes with mercuric acetate in water or an alcohol solvent is a mild method of
obtaining Markovnikov hydration or alkoxylation products. No rearrangements or
polymerization is observed in this process, since there is no carbocation intermediate. The
reaction occurs through the intermediacy of a mercurinium ion, formed from the reaction
of the alkene with +Hg(OAc). Because of the presence of a partial positive charge on the
more substituted carbon of the mercurinium ion intermediate, the nucleophile (water or
an alcohol) attacks the more substituted carbon to give the anti-hydroxy-organomercurial
intermediate. The C-Hg bond is reduced with NaBH4 in NaOH to give the alcohol/ether
product, in which Markovnikov regiochemistry is observed. Note the anti relationship
between the OH (OR) and Hg(OAc) in the intermediate.
+Hg(OAc)
Hg(OAc)2
+ -OAc
OAc
H3C
H
Hg
+Hg(OAc)
H
Hg(OAc)
H 3C
H
H 3C
H3C
H 3C
-H+
δ+
H 3C
H
O
H
H
H
H2O
NaBH4, HO-
Mercurinium ion
H 3C
H
H 3C
H
H
O
H
Markovnikov Product
OAc
OAc
+Hg(OAc)
NaBH4, HO-
CH3 Hg
Hg
CH3
H
δ+
H 3C
D
D
H 3C
D
H3CO
H3CO
CH3O
H
D
Anti relationship
between OCH3 and H
Anti relationship
between OCH3 and Hg
Yields in the oxymercuration/alkoxymercuration/demercuration process are high, the
reaction occurs under mild conditions, and no rearrangements or polymerizations are
observed
C. Hydroboration: Anti-Markovniknov hydration of an alkene
BH3 is a reagent in which the boron atom is electrophilic (it contains an empty p-orbital
on B, since BH3 lacks an octet). The reaction occurs in a single step through a fourcentered transition state in which the electrophilic boron atom attacks the less hindered
carbon atom of the alkene. In so doing, the electron density of the double bond is shifted
toward the B atom, leaving the more highly substituted carbon atom with a partial
positive charge. This regiochemistry of boron addition to alkenes is also favored by
sterics. All three B-H bonds on BH3 are reactive, so with 1/3 of an equivalent of BH3 to
alkene, a trialkyl borane results. The second step of the reaction is treatment of the
organoborane with basic peroxide, which oxidizes the C-B bond, producing ultimately
boric acid and the alcohol product.
H
H
empty p orbital
H
H
B
H
H
H 3C
δ+
H 3C
B
H
Higher energy T.S.
H
H
H
H3C
H
H3C
B
H
BH3•THF
H3C
H3C
H
H
H
H
B
H 3C
δ+
H
H
H
H 3C
H
H
O
alkene pi bond acts as a nucleophile in attacking Lewi acid (electrophilic)
BH3
H3C
CH3
H
H
H
B
H
2 more times
H
H
H
H3C
H 3C
H
H
H
H
H2O2
HO-
B
H3C
H 3C
H 3C
H 3C
H
H
H
A trialkyl borane
CH3
CH3
H
H
H
Anti-Markovnikov
product
Stereochemically, borane addition to alkenes results in a syn addition of H and OH across
the double bond:
BH3•THF
H
D
CH3
D
CH3
D
B
H
H
H
H3C
B
H
H
H2O2
HO-
D
H
H3C
O
H
Syn addition of H,OH
across alkene
Hydroboration is a Stereospecific Reaction
H3C
1. BH3•THF H C
3
H
H
OH
H3C
S
+
H
2. H2O2
HO-
H3C
R
CH3
CH3
R
H
S
OH
E
H3C
1. BH3•THF
H
H3C
2. H2O2
HO-
CH3
OH
R
R
H3C
H3C
+
S
CH3
H
H
S
H
OH
Z
Hydroboration is also stereospecific: stereoisomeric reactants lead to stereoisomeric
products
III.
Catalytic hydrogenation
Addition of H2 across alkenes requires a transition metal catalyst. The syn addition of
hydrogen gas occurs at the metal surface in a stepwise process. Hydrogen is added to
H
D
D
D
H2, cat. Pd or Pt
D
H
the same side of the alkene that is adsorbed to the metal surface.
Heterogenous catalysis takes place when the reactants (alkene, H2 gas) are in a different
phase from the reactants
Homogenous catalysis takes place when the reactants and catalyst are in the same phase.
Wilkinson’s catalyst is an example of a homogenous catalyst.
A chiral catalyst may be employed to enantioselectively reduce double bonds. This
process is very important in chemical industry. A chiral catalyst consists of a transition
metal complexed to a chiral ligand, typically a chiral bis-phosphine.
ClRh(PPh3)2
OH
H3C
H
(Wilkinson's
catalyst)
OH
R
H
CH3
OH
S
racemic
Cl2Ru(P*)
H3C
P* = chiral
bis-phosphine
OH
H
OH
R
96% e.e.
Additional Problems for practice:
1.) Write a mechanism which accounts for the formation of the observed
product(s) in the following reactions:
a.
+
+
HCl
Cl
Cl
b.
Br
+
HBr
2.) Show how you could prepare the following compounds from an alkene:
H3CO
a.
c.
CH3
b.
OH
d
CH3CH2C(CH3)2
CH3
Br
Br
OH
e.
CH3CH2CHCHCH2CH3
CH3
3.) Draw mechanisms for and show the products of the following reactions.
Remember to indicate the regio- and stereochemistry of the products
CH3
H2, Pt
a.
CH3
b.
1. Hg(OAc)2
CH3CH2OH
2. NaBH4,
HO-
1. BH3•THF
H
CH3CH3
2. H2O2, HO-
c.
HBr
HBr
ROOR
heat
e.
1. BH3•THF
2. H2O2, HO-
f.
1. BH3•THF
2. H2O2, HO-