Reactions of Alkenes II
Reading: Wade chapter 8, sections 8-9- 8-16
Study Problems: 8-47, 8-48, 8-55, 8-66, 8-67, 8-70
Key Concepts and Skills:
•
Propose logical mechanisms to explain the observed products of alkene addition
reactions, including regiochemistry and stereochemistry
•
Use clues provided by the products of reactions such as ozonolysis to determine
the structure of an unknown alkene
•
When more than one method is available for a chemical transformation, choose
the better method and explain its advantages
Lecture Topics:
I.
Additions to pi bonds: Carbenes
Carbenes react with pi bonds to form cyclopropanes. There are three methods for
generating carbenes:
1. heating of diazo compounds: Diazomethane produces methylene upon
heating:
heat
N
N
N
CH2
N
H
+ N2
C
CH2
H
methylene
H
H
H 3C
CH2CH3
H3CH2C
C
H
H3C
+
H3CH2C
H
H3C
CH2CH2
H
CH2CH3
H3CH2C
H
H
enantiomers
2. Simmons-Smith Reaction: insertion of zinc into a dihaloalkane
H
C
"I–CH2–Zn–I"
Zn + CH2I2
H
methylene
CH2I2
Zn
H
H
H
H
3. α-elimination reaction:
Br
Br
C
Br
HOBr
H
Br
C
C
Br
Br
Br
dibromomethylene
Br
CH3
H
C
Br
CH3
Br
Br
H
+
CH3
Br
Br
H
Stereochemistry of cyclopropanation: cis groups in the alkene remain cis in the
cyclopropane; trans groups in the alkene remain trans in the cyclopropane
II. Alkene Halogenation
Addition of X-X across the pi bond
Pi bond acts as a nucleophile; a bromonium ion intermediate gets attacked by
nucleophilic bromine; the resulting dibromide has trans stereochemistry:
Backside displacement
H
BrH
H
Br
Br––Br
Br
H
H
H
Br
trans-1,2-dibromo
cyclohexane
Bromonium ion intermediate
The reaction is stereospecific: different stereoisomers of starting materical produce
different steroisomers of products:
CH3
H
H 3C
H3C
Br2
H
Br
Br
H
H
meso
H
H
H 3C
H 3C
Br2
CH3
H
S
S
Br
CH3
Br
Br
+
H
R
H
H3C
H
CH3
CH3
R
Br
Halohydrin synthesis:
• Involves the interception of the bromonium ion by water (or alcohols), which are
stronger nucleophiles than Br-. Again, anti (trans) stereochemistry is observed in the
product; unsymmetrical alkenes give Markovnikov orientation in the products:
H
H
H
H
Br–Br
Br
Br
Br+
H2O
H
H
+
OH
H
OH
H2O
trans bromohydrin
enantiomers
H
δ++
H
H 3C
+ Cl2
H
+ H2O
H
H
Cl+
H
H 3C
H2O
δ+
H
Cl
H
H
HO
+
Cl
chloronium ion
• The nucleophile attacks the more highly substituted (more electrophilic)
carbon atom
H
H 3C
H
CH3
H
H
OH
II.
Oxidation of Alkenes  Addition of oxygen
Three flavors:
1. epoxidation
2. dihydroxylation
3. Oxidative cleavage
1. Epoxidation
Epoxide (oxirane) formation takes place when an alkene is treated with an organic
peracid. In addition to the epoxide product, a carboxylic acid by-product is formed.
Epoxidation is a one-step process (concerted bond formation and bond breaking), and the
reaction is therefore stereospecific: groups that are cis in the alkene are cis in the epoxide,
and groups which are trans in the alkene are trans in the epoxide.
O
meso
H3C
CH3
O
R
O
O
H3C
O
CH3
+
H
H
H
H
H
cis-2-butene
O
R
O
O
O
H
O
CH3
H3C
H
O
O
O
H3C
H
Reagent of choice:
R
H
O
R
H
CH3
CH2Cl2
H3C
H
H
Cl
H
H
MCPBA
H
CH3
H3C
H
O
O
H
H
+
CH3
H3C
O
CH3
H
enantiomeric trans-epoxides
Acid-catalyzed opening of epoxides: synthesis of trans-1,2-diols
Anti-diols are obtained by backside attack of water on activated epoxides formed by
protonation of the epoxide oxygen. To favor formation of the trans 1,2-diol, peracetic
acid is used in aqueous acid; to favor formation of just the epoxide, MCPBA is used in a
non-polar solvent.
H
CH3COOOH
H3O+
H
O H
H
H
H
H
H
O
H
O H
O
H
H
trans diol available from reaction
of alkenes with peracetic acid in water
H
Anti orientation
of hydroxyl groups
HO
OH
H
(±)-trans-1,2-cyclopentanediol
epoxides available from reaction of alkenes with MCPBA in CH2Cl2
H
MCPBA
O
+enantiomer
CCl4
H
2. Direct Syn Hydroxylation of Alkenes
Two methods exist for syn hydroxylation:
i. Osmium tetroxide is used in catalytic quantities in the presence of
an oxidant to oxidize alkenes. An intermediate osmate ester is
formed; H2O2 hydrolyzes the osmate ester and oxidizes osmium
back to OsO4. Syn 1,2 diol products are obtained.
cis-glycol
OsO4
H
H
H2O2
H
O
O
HO
+ OsO4
OH
syn diol product
Os
O
H
O
H2O2
H
H
O
O
Os
O
O
Osmate ester
OsO4 is expensive and toxic, but it gives excellent yields of the cis
diol from alkenes.
ii. Permanganate hydroxylation: treatment of alkenes with dilute, cold
aqueous KMnO4 gives moderate yields of the cis-1,2-glycol
product. KMnO4 is cheaper and less toxic than OsO4
KMnO4
HOCH3
H
HOCH3
H
O
O
O
HO
Mn
Mn
-O
O
-O
CH3
H
+ MnO2
OH
50%
cis-glycol
O
O
Manganate ester
HO- serves to hydrolyze the intermediate Manganate ester
3. Oxidative Cleavage
a. Treatment of alkenes with a warm solution of concentrated KMnO4 leads to an
oxidative cleavage of alkenes.
An alkene carbon with two alkyl groups is oxidized to a ketone (R2C=O)
An alkene carbon with one alkyl group is oxidized to a carboxylic acid (RC(=O)OH)
An alkene carbon with no alkyl groups and two hydrogen atoms is oxidized to CO2 and
H2O.
O
-O
warm, conc.
Mn
O
O
O
+
O
KMnO4
H
Aldehydes are oxidized to acids in the presence of KMnO4
KMnO4
O
HO
KMnO4
OH
OH
O
O
+ CO2 + H2O
O
b. Ozonolysis – Treatment of alkenes with ozone (O3) leads to oxidative cleavage, giving
aldehydes and ketones. An intermediate ozonide is produced; dimethyl sulfide (DMS) is
added in a second step to reduce the ozonide to aldehydes and ketones. Dimethyl sulfide
prevents aldehydes from being further oxidized to carboxylic acids by ozone. To predict
the products of an ozone cleavage, simply erase the double bond and put two carbonyls
(C=O) in its place.
R
R'
R
H
O
O-
R
R
O
+ O3
R'
R O R'
H
R
O
O
H
O
O
ozonide
molozonide
O
H3C
O
R R'
R
Example:
1. O3
H
H O
O
2.DMS
O
+
S
CH3
O
O
+
H
H3C
S
CH3
+ H2C=O
Ozonolysis provides confirmation
of the location of olefins within a molecule
CH3
1. O3
O
H
2.DMS
ketone and aldehyde
generated
O
1. O3
H
2.DMS
dialdehyde generated
O
O
H
Additional Problems for practice:
1. Show the structures of alkenes that give the following products upon
reaction with warm, concentrated KMnO4
(a) CH3CH2COOH + CO2
O
(c) (CH3)2C=O +
(b) (CH3)2C=O + CH3CH2CH2COOH
2.) Show how you could prepare the following compounds from an alkene:
O
COOH
(c)
COOH
(a)
H
O
OH
(d)
(b)
OH
OH
(e)
CH3
OH
H
3.) Draw reasonable mechanisms for the following reactions:
Br
OCH2CH3
Br2, CH3CH2OH
+ enantiomer
(a)
H
1. Br2, H2O
(b)
H
2. NaOH, H2O
Br2, H2O
(c)
O
OH
O
Br
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-