Chem 206 D. A. Evans Olefin Addition Reactions: Part–2 ! Problems of the Day: R. Noyori Bull. Chem. Soc. Japan 47, 2617, (1974) 28 : 72 97 : 3 R 2 AlH N Me O OH N Me N Me H H OH Rationalize the stereochemical outcome of the indicated reaction. LiAlH 4 M–H H Cl 1 A B RCO 3 H H Cl 2 O H Cl 3 RCO 3 H O favored disfavored Problem 579. The following publication (J. Org. Chem. 1991, 56, 5553) reported the surprisingly selective olefin epoxidation illustrated below. In this reaction, olefin B in 1 was found to be much less reactive than olefin A. Using your knowlege of stereoelectronic effects, provide an explanation for the reduced reactivity of olefin B in diene 1. Problem 313. Overman and co-workers recently reported the indicated selective epoxidation in conjunction with a synthesis of briarellins A and E, a new family of diterpenes (JACS 2003, 125, 6650). It should be noted that the Al(t-BuO) 3 /(t)-BuOOH reagent system is both highly diastereoselective and site selective. It is also relevant to the mechanism of the reaction that the ring-trisubstituted olefin lacking an allylic oxygen substituent would normally be more prone to epoxidation with a peracid than the acyclic trisubstituted olefin. Me O H H R R Me H O H HO Me O H H Me TIPSO H Me H H HO t-BuOOH 4 Å sieves toluene, -20˚C Al(t-BuO) 3 " " Part. Provide a general mechanism illustrating how the Al(t-BuO) 3 /(t)-BuOOH reagent epoxidizes olefins. Three-dimensional drawings are recommended. Part B. Provide a general mechanism illustrating how the above epoxidation proceeds and provide the stereochemistry (") of the product epoxide along with a stereochemical analysis of the noted face selectivity. D. A. Evans Friday, October 7, 2005 http://www.courses.fas.harvard.edu/~chem206/ ! Reading Assignment for week A. Carey & Sundberg: Part B; Chapter 4 "Electrophilic Additions to C–C Multilple Bonds" Chemistry 206 Advanced Organic Chemistry Lecture Number 9 Olefin Addition Reactions–2 ! Sharpless Epoxidation continued..... ! Hydrogenation ! Olefin Bromination Hoveyda, Evans, & Fu (1993). Substrate-directable chemical reactions. Chem. Rev. 93: 1307-70 (pdf) J. M. Brown, Angew. Chem. Int. Edit. 26, 190-203 (1987) (Handout) Investigation of the early Steps in Electrophilic Bromination through the Study of the Reaction of Sterically Encumbered Olefins R. S. Brown, Accts. Chem. Res. 1997, 30, 131 (handout) H. Yamamoto et.al, Angew. Chem. Int. Ed. 2005, 44, 4389-4391 (pdf)
Transcript
Chem 206D. A. Evans Olefin Addition Reactions: Part–2
! Problems of the Day:
R. NoyoriBull. Chem. Soc. Japan 47, 2617, (1974) 28 : 72
97 : 3R2AlH
NMe
OOH
NMe
NMe
H
H OH
Rationalize the stereochemical outcome of the indicated reaction.
LiAlH4
M–H
HCl
1
AB
RCO3HHCl
2O
HCl
3
RCO3H
Ofavored disfavored
Problem 579. The following publication (J. Org. Chem. 1991, 56, 5553) reported the surprisingly selective olefin epoxidation illustrated below. In this reaction, olefin B in 1 was found to be much less reactive than olefin A. Using your knowlege of stereoelectronic effects, provide an explanation for the reduced reactivity of olefin B in diene 1.
Problem 313. Overman and co-workers recently reported the indicated selective epoxidation in conjunction with a synthesis of briarellins A and E, a new family of diterpenes (JACS 2003, 125, 6650). It should be noted that the Al(t-BuO)3/(t)-BuOOH reagent system is both highly diastereoselective and site selective. It is also relevant to the mechanism of the reaction that the ring-trisubstituted olefin lacking an allylic oxygen substituent would normally be more prone to epoxidation with a peracid than the acyclic trisubstituted olefin.
Me
O
H HR
RMe H
O
H
HO
Me
O
H H
Me
TIPSO
H
Me HH
HO
t-BuOOH
4 Å sievestoluene, -20˚C
Al(t-BuO)3
"
"
Part. Provide a general mechanism illustrating how the Al(t-BuO)3/(t)-BuOOH reagent epoxidizes olefins. Three-dimensional drawings are recommended.Part B. Provide a general mechanism illustrating how the above epoxidation proceeds and provide the stereochemistry (") of the product epoxide along with a stereochemical analysis of the noted face selectivity.
D. A. EvansFriday,October 7, 2005
http://www.courses.fas.harvard.edu/~chem206/
! Reading Assignment for week
A. Carey & Sundberg: Part B; Chapter 4"Electrophilic Additions to C–C Multilple Bonds"
Chemistry 206
Advanced Organic Chemistry
Lecture Number 9
Olefin Addition Reactions–2
! Sharpless Epoxidation continued.....
! Hydrogenation
! Olefin Bromination
Hoveyda, Evans, & Fu (1993). Substrate-directable chemical reactions. Chem. Rev. 93: 1307-70 (pdf)
J. M. Brown, Angew. Chem. Int. Edit. 26, 190-203 (1987) (Handout)
Investigation of the early Steps in Electrophilic Bromination through the Study of the Reaction of Sterically Encumbered Olefins
R. S. Brown, Accts. Chem. Res. 1997, 30, 131 (handout)
Conformational Analysis and Reactivity: Curtin-Hammett PrincipleK. A. Beaver, D. A. Evans Chem 206
J. I. Seeman, Chem Rev. 1983, 83, 83-134.
J. I. Seeman, J. Chem. Ed. 1986, 63, 42-48.Leading References:
How does the conformation of a molecule effect its reactivity?
Consider the following example:
Do the two different conformers react at the same rate, or different rates? What factors determine the product distribution?
(1)kB
kA
minormajor
k2k1PA A B PB
Consider two interconverting conformers, A and B, each of which can undergo a reaction resulting in two different products, PA and PB.
The situation:
See also Eliel, pp. 647-655
13 Me–I13 Me–I
NMe
NMe
!G°
!GAB‡
!G1‡
!G2‡
Rxn. Coord.
En
erg
y
PA A B
PB
k1, k2 >> kA, kB: If the rates of reaction are faster than the rate of interconversion, A and B cannot equilibrate during the course of the reaction, and the product distribution (PB/PA) will reflect the initial composition.
=[PB]
[PA]
[B]o[A]o
In this case, the product distribution depends solely on the initial ratio of the two conformers.
Me
O
NMe
ON
Me
Me
major product
minor product
less stable more stable
Me–Br
N
O
Me
Me
Me
H
N
O
Me
Me
H
Me
Padwa, JACS 1997 4565
Me–Br
!G = -3.0 kcal/mol(ab initio calculations)
Case 1: "Kinetic Quench"
We'll consider two limiting cases:
(1) The rate of reaction is faster than the rate of conformational interconversion
(2) The rate of reaction is slower than the rate of conformational interconversion
-78°C
While enolate conformers can be equilibrated at higher temperatures, the products of alkylation at -78° C always reflect the initial ratio of enloate isomers.
If the rates of conformationall interconversion and reaction are comparable, the reactants are not in equilibrium during the course of the reaction and complex mathmatical solutions are necessary. See Seeman, Chem. Rev. 1983, 83 - 144 for analytical solutions.
steric hindrance
HH
K. A. Beaver, D. A. Evans Chem 206Curtin - Hammett: Limiting Cases
k1, k2 << kA, kB: If the rates of reaction are much slower than the rate of
interconversion, (!GAB‡ is small relative to !G1
‡ and !G2‡), then the ratio of
A to B is constant throughout the course of the reaction.
!G°
!!G‡
!G1‡ !G2
‡
Rxn. Coord.
En
erg
y
PA
A
B
PB
kB
kA
minormajor
k2k1PA A B PB
d[PA]dt
= k1[A] andd[PB]
dt= k2[B] we can write:
d[PA]
d[PB]=
k1[A]d[PA]d[PB]
k1[A]=
k2[B]
d[PA]d[PB]k1
=k2
Since A and B are in equilibrium, we can substitute Keq =
Keq k1=
k2Keq
k2[B]
[PB]
[PA]
Using the rate equations
or
Integrating, we get
[B]
[A]
To relate this quantity to !G values, recall that !Go = -RT ln Keq or Keq =
e-!G°/RT, k1 = e-!G1
‡/RT, and k2 = e-!G2
‡/RT. Substituting this into the above
equation:
k1
=k2
Keqe-!G
2/RT
e-!G1/RT
(e-!G°/RT) e-!G2/RTe-!G°/RTe!G
1/RT=
e-(!G2 + !G°-!G
1)/RT=
Where !!G‡ = !G2
‡+!G°-!G1‡
[PB]
[PA]
[PB]
[PA]or e-!!G/RT=
[PB]
[PA]
The Derivation:
Curtin - Hammett Principle: The product composition is not solely
dependent on relative proportions of the conformational isomers in
the substrate; it is controlled by the difference in standard Gibbs
energies of the respective transition states.
(2)
(3)
(4)
When A and B are in rapid equilibrium, we must consider the rates of reaction of the conformers as well as the equilibrium constant when analyzing the product ratio.
(1)
Case 2: Curtin-Hammett Conditons
Within these limits, we can envision three scenarios:
• If the major conformer is also the faster reacting conformer, the product from the major conformer should prevail, and will not reflect the equilibrium distribution.
• If both conformers react at the same rate, the product distribution will be the same as the ratio of conformers at equilibrium.
• If the minor conformer is the faster reacting conformer, the product ratio will depend on all three variables in eq (2), and the observed product distribution will not reflect the equilibrium distribution.
This derivation implies that you could potentially isolate a product which is derived from a conformer that you can't even observe in the ground state!
Combining terms:
!GAB‡
=
slow slow
fast
Some Curtin-Hammett ExamplesK. A. Beaver, D. A. Evans Chem 206
k1 k2
Keq = 10.5
Ratio: 5 : 95
++
more stableless stable
fasterslower
minor product major product
Me3C
H
MeN
NMe
H
Me3C
O –
N
Me
H
Me3C
Me3C
H
Me
NO –
Oxidation of piperidines:
major product minor product
++
13 Me–I13 Me–I slowerfaster
less stable more stable
NMe
NMe
NMe13Me
NMe 13Me
Tropane alkylation is a well-known example.
When the equilibrium constant is known, the Curtin-Hammett derivation can be used to calculate the relative rates of reaction of the two conformers. Substituting the above data into [PB]/[PA] = k2K/k1, the ratio k2/k1 ~ 2.
Note that in this case, the more stable conformer is also the faster reacting conformer!
The less stable conformer reacts much faster than the more stable conformer, resulting in an unexpected major product!
JOC 1974 319
Tet. 1972 573Tet. 1977 915
i-Pr2N O
Me
Hi-Pr2N O
Me
Lis-BuLi
(-)-Sparteine
i-Pr2N O
Me
Li•sparteine i-Pr2N O
Me
Li•sparteine
slowerfaster
i-Pr2N O
Me
Cl
i-Pr2N O
Me
82 - 87% ee
This is a case of Dynamic Kinetic Resolution: Two enantiomeric alkyl lithium complexes are equilibrating during the course of a reaction with an electrophile.
Beak, Acc. Chem. Res, 1996, 552
Enantioselective Lithiation:
N
N
(-)-Sparteine
Enantioselectivities are the same, regardless of whether or not the starting material is chiral, even at low temperatures. Further, reaction in the absence of (-)-sparteine results in racemic product.
Note that the two alkyllithium complexes MUST be in equilibrium, as the enantioselectivity is the same over the course of the reaction. If they were not equilibrating, the enantioselectivity should be higher at lower conversions.
Because sparteine is chiral, these two complexes are diastereomeric and have different properties.
H2O2
Mechanism of Asymmetric HydrogenationK. A. Beaver, D. A. Evans Chem 206
Rh
O
P
H
P
H
Ph
HN
Me
Rh
O
H
P
P
H
Ph
NH
Me
The asymmetric hydrogenation of prochiral olefins catalyzed by Rhodium is an important catalytic process.
Ph
NHAcMeO2C
Ph
NHAcMeO2C[L2Rh]+
> 95% ee Rh
O
P
PPh
NH
MeEnantioselectivities are generally very high when the ligand is a chelating
diphosphine. (ee's are given for S,S-CHIRAPHOS)
Rh
O
P
PPh
HN
Me
Rh
S
SP
PS,S Ph
NHAcMeO2C
coordinationcoordination
hydrogen addition
hydrogen addition
migration
reductive elimination
reductive elimination
migration
Ph
NHAcMeO2C
Ph
NHAcMeO2C
-L2RhS2-L2RhS2
S
> 95% ee
Observations:
• Complex 2 is the only diasteromer observed for the catalyst-substrate complex (1HNMR, X-Ray crystallography) in the absence of hydrogen
• The enantioselectivity is strongly dependant on the pressure of H2, and degrades rapidly at higher hydrogen pressures
MeO2CCO2Me
CO2MeMeO2C
• The observed enantiomer is exclusively derived from the minor complex 2
These observations may be explained using the Curtin - Hammett Principle
Halpern, Science, 217, 1982, 401
Rh
S
H
P
P
O
Me
NH
CO2Me
CH2Ph
Rh
S
H
P
P
O
Me
HN
MeO2C
PhH2C
S
When a chiral ligand is used, there are two diastereomeric complexes which may be formed:
Rh
O
P
PPh
HN
Me
CO2Me
Rh
O
P
PPh
NH
Me
MeO2C
Ph
NHAcMeO2C
Ph
NHAcMeO2C
S
H2 H2
observed product
major complex1 2
* *
faster slower
(NMR, X-Ray)
+ S + S
minor complex
majorminor
fast slow
R
R
R
Sharpless Epoxidation (V+5)D. A. Evans Chem 206
Aldrichimica Acta, 12, 63 (1979)
RDS
The Sharpless Epoxidation
+
O
OV
ORO
O
O
O
V
RO
O
HO
ROR
V
OR
O
O
RO
OHO
O
OR
V
ROO
OR
HO
!
!
!
!!
–ROHROOH
slowChem 3D Transition State
43
2
1
O–C2–C3–C4 = 41°The Sharpless estimate: ~50°
O
O V
OR
O-OtBu
V
O
O
OO
RO
tBu
O
tBu
VRO
OO
O
HO OHO
t-BuOHt-BuOOH
! The literature precedent: Sheng, Zajecek, J. Org. Chem. 1970, 35, 1839
4 : 1VO(acac)2
80 oC
1 : 1Catalyst
OH OHO
OHO" "
TBHP
Mo(CO)6
80 °C
Stereoselection 98:2 (90 % yield)
! Next step: Sharpless, Michaelson JACS 1973, 95, 6136
Regioselection 20:1
80 °C
VO(acac)2O
OHMe Me
Me
OH
Me
Me
Me
OH OH
O
"
"
Mo(CO)6
TBHP
TBHP
a,b The relative rate data apply only to a given column.
Values in parenthesis refer to the ratio of syn:anti epoxide.
krela,b (diastereoselectivityc )
10.0 (98 : 2)11.0 (98 : 2)
--0.07 (40 : 60)
>200 (98 : 2)4.5 (98 : 2)
1.001.00
0.42 (60 : 40)
0.046 (37 : 63)
0.55 (92 : 8)
1.00
SubstrateVO(acac)2peracid
Relative Rates (Diastereoselectivities) for the Epoxidation of Cyclohexene Derivatives JACS 1973, 95, 6136
OH
OH
OAc
Mo(CO)6
Epoxidation of Acyclic AlcoholsD. A. Evans Chem 206
! Allylic Alcohols:
~ 120 °
40-50 °
! Estimate
71 : 29
95 : 5
t-BuOOH / VO(acac)2
m-CPBA
RatioReagent
t-BuOOH / Mo(CO)6 84 : 16
erythrothreo
Reagent+
Me
Me OH OHMe
MeO O
Me
Me OH
" "
! RCO3H Transition States: ! ~ 120 °
TSminorTSmajor C
Me
H
HC
Me
H
OH
C
Me
CH
Me
HH
OH
! V(+) Transition States: ! ~ 45 °
TSminorTSmajor C
HO
H
HCMe
H MeC
HO
CH
MeH
H
Me
K. B. Sharpless & CoworkersTetrahedron Lett. 1979, 20, 4733.
K. Oshima & CoworkersTetrahedron Lett. 1980, 21, 1657, 4843.
