ozone. ozone electric discharge or cosmic rays.. : : : equivalent resonance structures + + - -
TRANSCRIPT
OZONEOZONE
OO
O OO
O
OZONEOZONE
O2 O3
electric discharge
orcosmic rays
.... ....
.......... :
::
EQUIVALENT RESONANCE STRUCTURES
+ +-
-
OZONOLYSISOZONOLYSIS
O3
C
OO
O
C
R2 R4
R1 R3
C C
R1
R2 R4
R3
a molozonide(1,2,3-trioxolane)
spontaneousrearrangement
O
CO
C
O
R1
R2
R3
R4
an ozonide(1,2,4-trioxolane)
O
CO
C
O
R1
R2
R3
R4
Zn
CH3COOH+C O
R1
R2
O C
R4
R3
FORMATION OF AN OZONIDE
OzonolysisOzonolysis
HYDROLYSIS OF THE OZONIDE (WORKUP)
unstable
aldehydes or ketones
O
O
O
..
..
..:
:
+
-CH2Cl2
0o CO
O
O
CO
O
OCO C
CO
O
OO
O
:....
..
..
....
.. ..
.. ......
..
..
: :
::
:::
..
::
+ +
- -
FORMATION OF AN OZONIDE - MECHANISMFORMATION OF AN OZONIDE - MECHANISM
ozone
molozonide
ozonide
carbonyloxide
ketone oraldehyde
UNSTABLE
Addition isconcerted
O3O O
OCH3
Ph CH3
CH3
CH3
CH3
Ph
CH3
OVERALL RESULT OF THE OVERALL RESULT OF THE REACTION WITH OZONEREACTION WITH OZONE
O O
O
The double bond isreplaced with theozonide ring.
molozonide ozonide
forms initially forms after rearrangement
OZONIDE AND MOLOZONIDE STRUCTURESOZONIDE AND MOLOZONIDE STRUCTURES
SOME EVIDENCE FOR THE MECHANISM
-
+
:....
..C
O
O
-
+
:....
..
CO
O
:..
..
:..
..
CO
O
O
OC
..
..
hexane
EVIDENCE FOR THE MECHANISMEVIDENCE FOR THE MECHANISM
When ozonolysis is performed in an aliphatic hydrocarbon solvent(hexane) dimers of the carbonyl oxide intermediate sometimes form.
carbonyl oxide dimer
insoluble
This dimerization proves the existence of the carbonyl oxide intermediate.
CH3
CH3
CH3
CH3
O O
OCH3
CH3 CH3
CH3
O O
OPh
Ph CH3
CH3
O3
C OPh
Ph
CH2Cl2
If a foreign ketone is placed in the solution (e.g., benzophenone)it becomes incorporated into a portion of the ozonides formed.
benzophenone
EXPECTED
Can you explain this with a mechanism ?HINT: The benzophenone reacts with the carbonyl oxide intermediate.
YET MORE MECHANISTIC EVIDENCEYET MORE MECHANISTIC EVIDENCE
+
TRY THIS ON YOUR OWN
traps thecarbonyl oxide
When an unsymmetrical alkene undergoes ozonolysis, it is not uncommon for three different ozonides to form.
EXPECTED
cis & trans
Can you draw mechanismsthat explain the formation all three ozonides ?
O3O O
OCH3
Ph CH3
CH3
O O
OCH3
Ph Ph
CH3
O O
OCH3
CH3 CH3
CH3
CH3
CH3
Ph
CH3
MORE EVIDENCEMORE EVIDENCE
TRY THIS ON YOUR OWN
The alkene mustbreak in two andrecombine.
WORKUP PROCEDURES FORWORKUP PROCEDURES FOR OZONOLYSISOZONOLYSIS
WORKUP PROCEDURES FOR OZONOLYSISWORKUP PROCEDURES FOR OZONOLYSIS
Two types of work-up (decomposition of the ozonide) are possible :
Oxidative Workup Add aqueous acid (H3O+)
Reductive Workup
Two methods :A) Zn, acetic acid or Zn, H2O
B) Pd/H2 followed by aqueous acid
H2SO4 + H2O
After the ozonide is formed it is hydrolyzed (work-up).
CH3COOH
OXIDATIVE WORKUP
Just add H3O+ ( = H20 and acid )
Aldehydes are oxidized to carboxylic acids.
Formaldehyde is oxidized to carbon dioxide, which is lost as a gas.
OXIDATIVE WORKUP (acid)OXIDATIVE WORKUP (acid)
These oxidations occur because H2O2 is a hydrolysis product. The general hydrolysis mechanism follows.
O O
OO O
H3O+
You do not have to known this mechanism.
OXIDATIVE WORKUPOXIDATIVE WORKUP
+
+
+ CH3
CH3
O
CH3
CH3O
H
O O
H H
O OCH3
CH3OH
H H
H
PhO
OH
Ph
H
O OCH3
CH3OH
H
O O
OH
Ph CH3
CH3
HO
H
H
O
H
H
O O
OH
Ph CH3
CH3
HO
H
O O
OH
Ph CH3
CH3
H
+
+
+ H+O3
O O
OH
Ph CH3
CH3
CH3
CH3
Ph
H
O O
OH
Ph CH3
CH3
H
+ H+
+ H+
continued …...
(acid + water)
O O
H H
H
PhO +
O
PhO
H
Aldehydes are easily oxidized by the H2O2 that is produced.
oxidizingagent
Ketones are not oxidized by hydrogen peroxide.
HYDROGEN PEROXIDE IS A PRODUCT HYDROGEN PEROXIDE IS A PRODUCT OF THE OXIDATIVE WORKUPOF THE OXIDATIVE WORKUP
If formaldehyde is produced, it is oxidized to CO2 and H2O.
O O
H H
CH
HO + C
O
OO
H
H
CO
OO
H
H
H+
CO2
H2O+
OXIDATIVE METHODS DESTROY FORMALDEHYDE
Carbonic acid is unstable anddecomposes to CO2 and H2O.
twomoles
gas
REDUCTIVE WORKUP
METHOD A Add Zn and H2O (or acetic acid)
METHOD B Reduce the ozonide with Pd / H2 , and then add acid ( H3O+ ).
With either method, aldehydes survive intact and are not oxidized.
REDUCTIVE WORKUPREDUCTIVE WORKUP
There are two methods of reductive workup.
The zinc “scavenges” the peroxide before it can act.
O O
OCH3
CH3 CH3
CH3
O O
OCH3
CH3 CH3
CH3
H H
+ ZnO2Zn H2O+ +
REDUCTIVE WORKUP - METHOD AREDUCTIVE WORKUP - METHOD A
With the REDUCTIVE WORKUPS, no H2O2 is produced.
Notice that the O-O bond is broken so that no H2O2 is formed during the subsequent hydrolysis.
O O
OCH3
CH3 CH3
CH3
O O
OCH3
CH3 CH3
CH3
H H
H2
Pd
REDUCTIVE WORKUP - METHOD BREDUCTIVE WORKUP - METHOD B
With the REDUCTIVE WORKUPS, no H2O2 is produced.
The hydrogenation step cleaves the O-O bond.
Since the O-O bond is broken, no H2O2 is formedduring the hydrolysis.
CH2
O
C
O
HH
O
+ CO2
EXAMPLESEXAMPLES
O3 Zn / H2O
1) O3
2) H3O+
ORO3 1) Pd/H2
2) H3O+
REDUCTIVEWORKUP
OXIDATIVEWORKUP
aldehydesurvives
formaldehyde CO2
USING OZONOLYSIS FORUSING OZONOLYSIS FOR STRUCTURE PROOFSTRUCTURE PROOF
AT ONE TIME OZONOLYSIS WAS WIDELY USED AT ONE TIME OZONOLYSIS WAS WIDELY USED FOR STRUCTURE PROOF BY DEGRADATIONFOR STRUCTURE PROOF BY DEGRADATION
O
O O
CH3
O
CH3
+ +O3
CH2Cl2
H3O+
Unknowncompound
Broken apart ( or degraded ) tosimpler pieces that are easier toidentify.
The original structure can bededuced by reassembling thepieces.
“At one time” =before spectroscopy.
Simpler KnownCompounds
C7H12
C7H14
1) O3 / CH2Cl2
2) H3O+
Pd / H2
CH3 C
O
CH2CH2CH2CH2 C
O
OH
C
O
CH3
C
O
OH
CH3
H
PROBLEM TO SOLVEPROBLEM TO SOLVE 6-ketoheptanoic acid
or 6-oxoheptanoic acid
answer
C8H12 C
O
HOC
O
OH
C
O
HOC
O
OHCH3 CH3
+H2 Pd
O3 H3O+
C8H16
WHAT WAS THE ORIGINAL STRUCTURE ?WHAT WAS THE ORIGINAL STRUCTURE ?
oxidativeworkup
C
O
HOC
O
OHCH3 CH3
O
C C
O
HO OH
CH3 CH3
oxidizedduringwork-up
OXIDATION OF ACETYLENESOXIDATION OF ACETYLENES
ACETYLENESACETYLENES
C C CH2 PhPh
C
O
O HCH2C
O
OH
KMnO4 or1) O3, CH2Cl22) H3O+
Oxidation of acetylenes, whether by KMnO4 or ozone, normally yields carboxylic acids.
OZONE AND SMOG
SO2CO2
NO2
NOR-CH=CH2
COOLER AIR
WARMER AIR
O3
temperature inversiontraps pollutants
Temperature Inversion: Air above land is cooler than air above.
FORMATION OF SMOG - FORMATION OF SMOG - OZONE IS A COMPONENTOZONE IS A COMPONENT
H2O
incompletely burnedhydrocarbons
reacts with unburnedhydrocarbons
Spruce, Cedar, Fir or Pine Forest
TerpenesO3
NATURAL SOURCES
reacts withterpenes
temperature inversiontraps bioemissions
OXIDATION OF SIDE CHAINS OXIDATION OF SIDE CHAINS ON AROMATIC RINGSON AROMATIC RINGS
CH2 CH2 CH3
O
O H
BENZENE RINGSBENZENE RINGSUnder normal conditions of ozonolysis, or treatment by KMnO4, benzene rings are not oxidized. They arequite resistant to reaction due to the presence ofaromatic ring resonance.
KMnO4 / 50o C /2 hours
When using KMnO4 on a benzene ring that has a side chain, the side chain is oxidized before the ring is affected.
The exception is rings with -OH, -OCH3, -NH2 and similar O/N groups, which oxidize quite readily.
Ozone, will not attack the saturated side chain.
BENZENE RINGSBENZENE RINGS
CH2 CH2 CH3
O
O H
O3 / CH2Cl2 /0o
O3 / CH2Cl2 /20o
long time
However, under morevigorous conditions thebenzene ring can beozonized.
C CH2 CH2 CH3
O
OHCO25 +
SELECTIVITYSELECTIVITY
SELECTIVITYSELECTIVITY
CH CH CHO
CH CH COOH
COOH
COOHHOOC
C
O
OHCOOH
COOHHOOC
H2O2
O3 / CH2Cl2 /0o
oxidizesaldehyde
cleaves double bond
H3O+
oxidizes aldehydes(oxidative work-up)
more vigorousozonolysis :
cleaves benzene ringcleaves double bondoxidizes aldehydes
1)
2)
O3 / 20o/ CH2Cl22 hours
cinnamaldehyde
cinnamic acid
benzoic acid
oxalic acid
MORE SELECTIVITYMORE SELECTIVITY
CH CH CHO
CH CH CHO
OHOH
CHO
CHOOHC
COOH
COOHHOOC
1) OsO4
2) NaHSO3
H2O
1) O3 / CH2Cl2 / 0o
2) Zn / CH3COOH
KMnO4
H2SO4
30 min
aldehydes survive(reductive work-up)
aldehyde survives(OsO4 is mild) aldehydes are
oxidized by KMnO4
CH CH
O
CHORCO3H
cinnamaldehyde
epoxidation
benzaldehyde