lecture 23
TRANSCRIPT
Chem 106E. Kwan Lecture 23: Enolates and C–C Bond Formation
October 29, 2010.
Enolates and C–C Bond Formation
Scope of Lecture
Eugene E. Kwan
Helpful References
enolates andC–C bond formation
Key Questions
chiral imidealkylation
enolatetautomerismintermolecular
Michael reactions
enolateacylation
HWEolefination
chiral amidealkylation
-amino acidsynthesis
the Claisencondensation
HOMe
Bn
1. How can chiral imide or amide enolates be used to prepare optically active alkanes?
NO
O OMe
MeMe
?
2. Why do these intermolecular Michael reactions give high diastereoselectivity?
O
OLi
Me
O
Me+–78 °C
THF O
O
Me
Me O
85%, 95:5 syn:anti
O
OLiMe
O
Me+–78 °C
THF/HMPA O
O
Me
Me O
73%, 13:87 syn:anti
Z enolate
E enolate
3. What are the thermodynamic driving forces behind the Claisen condensation?
RO
O
Me
+ RO
O
MeRO
O
Me
OMe RO H+
I thank Professor David A. Evans (Harvard) for the use ofsome material from Chem 206 for this lecture.
PhNCH3
OMe
OH
R
R'
1. "Pseudoephedrine as a Practical Chiral Auxiliary for the Synthesis of Highly Enantiomerically Enriched Carboxylic Acids, Alcohols, Aldehydes, and Ketones." Myers, A.G.; et al. J. Am. Chem. Soc. 1997, 119, 6496-6511.
2. "Intermolecular Michael Reactions: A Computational Investigation." Kwan, E.E.; Evans, D.A. Org. Lett. 2010 doi: 10.1021/ol102017v
3. "Claisen Ester Condensation Equilibria - Model Calculations" Garst, J.F. J. Chem. Educ. 1979, 56, 721-722.
4. "Olefin Synthesis with Organic Phosphonate Carbanions." Boutagy, J.; Thomas, R. Chem. Rev. 1974, 74, 87-99.
Chem 106E. Kwan, D.A. Evans Lecture 23: Enolates and C–C Bond FormationDiastereoselective Enolate AlkylationQ: How does one alkylate a carboxylic acid derivative diastereoselectively?
N
OR*
R*
R
Overall, the target transformation is:
HO
OR
?HO
OR
MeIMe
Of course, this means that the auxiliary has to be added andthen removed. However, one advantage is that the productis less sensitive to epimerization.
One solution to this problem is to use a chiral amide auxiliary:
HO
OR
N
OR*
R*
R
Me
OR
MeHO
For example, an -alkyl aldehyde might epimerize, but an-alkyl amide will not because of 1,3-allylic strain. Recallthe stereoelectronic requirement for deprotonation:
OR
ON
Me
MeH
R
MeA(1,3)
minimized
ground state:stable but unreactive
severeA(1,3)
interactiontransition state:
reactive but strained
ON
Me
MeMe
H
R
R
H
Me
base conformationnecessary fordeprotonation
If R=H, then this is perfectly viable. However, if R is an amide,then A(1,3) strain ensures that the acidic C-H bond remainsin the -plane of the amide, preventing deprotonation:
OH
HH
Me
easily accesses a conformationwhich can be deprotonated
As we will see, these auxiliaries can be cleaved or otherwiseconverted to a variety of carboxylic acid derivatives includingaldehydes.
Another advantage of amides is that they form Z-enolates quiteselectively. An early advance was made by Evans in thealkylation of prolinol amide enolates (TL 1980 21 4233):
or othercarboxylic acid
derivatives
O
NMe
HO 2 equiv LDABnBr
OLi
NMe
OLi
O
NMe
HO
Bn
aq. HClO
HOMe
Bn75%, 76% de
The amide hydrolysis is facilitated by intramolecular N to Oacyl transfer, and then by intramolecular H-bonding:
O
NMe
HO
Bn
O
OMe
BnNH
H
The diaselectivity of this process depends on: (1) thegeometrical purity of the starting enolate; (2) the selectivity ofthe alkylation itself; and (3) any losses incurred duringcleavage of the auxiliary. Here, the problem seems to be thesecond factor.
Chem 106E. Kwan, D.A. Evans Lecture 23: Enolates and C–C Bond FormationDiastereoselective Enolate AlkylationIf oxazolidinone imides are used instead, much more selectivealkylations can be realized (Evans JACS 1982 104 1737)
NO
O O
CH3 NO
O O
CH3Bn
MeMe
LDABnBr
MeMe
92%,>99:1 dr
It has been suggested, but not proven, that lithium is chelated inboth the ground and transition states. Various oxazolidinonesof different enantiomers are commercially available. However,less activated electrophiles like ethyl iodide are problematic:
NO
O OCH3 NO
O OCH3
BnCH3
NaHMDSEtI
Me
53%>99:1 dr
Ph Ph
(Sodium enolates are more reactive.) Presumably, theincreased acidity of imides over amides is responsible for thedecrease in reactivity (note that basicity and nucleophilicity areclosely related, but not the same--one is thermodynamic andthe other is kinetic; see Jaramillo, J Phys Org Chem 2007 201050).Myers has developed a solution to this problem involvingpseudoephedrine amides (JACS 1997 119 6496):
PhNCH3
OMe
OH
Me LDA, LiCln-BuI
PhNCH3
OMe
OH
Me
n-Bu
80%,99:1 dr
Advantages: (1) compounds are crystalline and can bepurified by recrystallization; (2) both enantiomers of psuedo-ephederine are available; (3) reactions can be done at 0 °C;(4) products can be converted to a variety of useful carboxylicacid derivatives; (5) iterative use gives 1,3-n-substitutedcarbon chains (see below).
conversion to carboxylic acid derivatives
PhNCH3
OMe
OH
R
R'
H2SO4,reflux (harsh)
HO
OR
R'
A variety of non-epimerizing conditions are available:
or (n-Bu)4NOHt-BuOH/H2O
(milder)
carboxylic acid
LAB
- LAB: lithium amidoborohydride, formed by LDA + NH3·BH3- the pseudoepherdrine amide is analogous to a Weinreb amide (see next lecture on tetrahedral intermediates)
HOR
R'alcohol
LiAlH(OEt)3HR
R'aldehyde
O
MeLi
ketoneMe
OR
R'
iterative synthesis of 1,3-n-substituted carbon chains
X*
OMe
Myers Synlett 1997 5 457
X*
O
Me
BnBnBr LABHO
Me
Bn PPh3, I2imidazole
AI
Me
Bn
A
X*
O
Me Me
Bn etc.
The auxiliary is powerful enough to override any inherent"matched" or "mismatched" effects.
Chem 106E. Kwan Lecture 23: Enolates and C–C Bond FormationPseudoephedrine GlycinamideThis methodology can be used to synthesize -amino acids(Myers JACS 1997 119 656; JOC 1999 64 3322):
PhNCH3
OMe
OH NH2
pseudoephedrineglycinamide
LiHMDS
LiCl, THF–78 °C
PhNCH3
OMe
OH
N
PhNCH3
OLiMe
OLi
NH2
PhNCH3
OMe
OLi NHLi
0 °C
Br
Br
The lithium amide is the kineticproduct; this does not lead to thedesired C-alkylated product.
However, equilibration can give the right product:
PhNCH3
OMe
OLi NHLi
PhNCH3
OMe
OH
NH280%, 93% de (crude)69%, 99% de (recrystallized)
These products can be hydrolyzed without loss of optical purity:
PhNCH3
OMe
OH
NH21. 0.5 M NaOH, reflux2. Boc2O, dioxane
HO
ONHBoc
91%, 99% de99% de
Q: What are the advantage to this method?(1) Both enantiomers of the auxiliary are cheap.
(2) The enamides required for asymmetric hydrogenation
R NHCOR'
CO2R'' cat.*H2
RCO2R''
NHCOR'
are not always available:
(3) No protecting groups are required and this sequence works well for a variety of electrophiles.
Q: What are the disadvantages of this method?
(1) There is no access to -amino acids bearing quaternary stereocenters. These can be obtained using asymmetric Strecker methods (Jacobsen Nature 2009 461 968):
R
N
Ph
Ph
H
NH
NH
S
CF3
CF3
O
NMe
Ph
Ph
tBu
KCN, AcOH, H2O, PhMe, 0 °C, 4-8 h
The hydrocyanate can be hydrolyzed in a similar way:
1. H2SO4, HCl, 2-3 d2. NaOH, NaHCO3
3. Boc2O, dioxane, 16 g4. recrystallize
87-90% ee
0.5 mol%
R CN
HNCHPh2
R CN
HNCHPh2
R CO2H
NHBoc
98-99% ee
The cheap enantiomer of tert-leucine (S) gives theexpensive enantiomer (R):
tBu CO2H
NHBoc
CO2H
NHBoc
CO2H
NHBoc
tBuNH2
(2) The procedure requires dry pseudoephedrine glycinamide (hygroscopic) and glycine methyl ester, which is polymerizable. However, these issues have been worked out.
Chem 106E. Kwan Lecture 23: Enolates and C–C Bond FormationPseudoephedrine GlycinamideThis example serves to illustrate how enolate-based reactionscan be optimized. Here is the optimized sequence comparedwith the unoptimized one. Direct alkylation of pseudoephedrineglycinamide hydrate is targeted, as it is an air-stable,free-flowing, and crystalline solid (Myers, JOC 1999 64 3322):
MeO
ONH2·HCl
free-baseand distill
MeO
ONH2
polymerizable
PhNCH3
OMe
OH NH2·H2O
PhNHCH3
Me
OH
LiOMe, LiCl;H2O
LiOtBu, THF;H2O, crystallize
pseudoephedrineglycinamide hydrate
(stable)
The old procedure dries the hydrate azeotropically (84 wt%MeCN/16% water; bp=76 °C; 80% PhMe/20% water; bp=84 °C)
PhNCH3
OMe
OH NH2·H2O
azeotropic dryingin MeCN; PhMe Ph
NCH3
OMe
OH NH2
1.95 equiv LiHMDS,6 equiv LiCl, THF, 0 °CRX
PhNCH3
OMe
OH
NH2
R
The improved procedure uses an extra equivalent of base toremove the water, rather than dehydrating the glycinamide in aprevious step:
PhNCH3
OMe
OH NH2·H2O
3.2 equiv LiHMDS,3.2 equiv LiClRX
PhNCH3
OMe
OH NH2·H2O
Access to Quaternary StereocentersAlthough there is no reported way to use this approach tomake -amino acids, it is possible to do the Myers alkylationwith tetrasubstituted enolates (JACS 2008 130 13231).The enolate geometry has been established by trapping:
PhNCH3
OMe
OHMe
Me
1. LDA, LiCl, 0 °C
2. DMPU, (iPr)2SiCl2 –40 °C
NCH3
O
MeMe
SiOPh
Me
iPr
iPr
Presumably, deprotonation occurs through a conformation which incurs considerable 1,3-allylic strain. It is unclear ifaggregates are involved. However, the other epimer givesthe opposite enolate geometry:
19:1 Z isomer
PhNCH3
OMe
OHMe
Me
1. LDA, LiCl, 0 °C
2. DMPU, (iPr)2SiCl2 –40 °C
NCH3
OMe
SiOPh
Me
iPr
iPr
1:15 E isomerMe
The behavior of these enolates is different on alkylation, too:
Chem 106E. Kwan Lecture 23: Enolates and C–C Bond FormationAccess to Quaternary Stereocenters
PhNCH3
OMe
OHMe
Me
LDA, LiCl; BnBr PhN
OMe
OHMe
MeBn
95%, 10:1 dr1 h at –40 °C19:1 Z isomer fast, selective
PhNCH3
OMe
OHMe
Me
LDA, LiCl; BnBr PhN
OMe
OHMe
BnMe
89%, 1:5 dr4.5 h at –40 °C
1:15 E isomer slowerless selective
The selectivity for the top alkylation decreases over the courseof the reaction from 19:1 at 65% conversion to 10:1 at fullconversion, suggesting that some equilibration to the E enolatemay be occurring. With an excess of enolate relative to alkylbromide, higher selectivities are observed. However, thereason for the differences in selectivity is unclear.
Another way to generate the enolates is via conjugate addition.Again, enolates are alkylated from a common diastereoface:
PhNCH3
OMe
OH Me
1. MeLi, LiCl2. RLi3. R'X
PhN
OMe
OH R'R
Me
The Michael ReactionIn terms of "charge affinity," the Michael reaction is a vinylogousform of alkylation. Compared to the aldol reaction, it forms 1,5-rather than 1,3- dioxygenation patterns:
O+ R X
ORalkylation
O
(+)(–) (+)
O+
Oaldol
O
(+)(–)
(+)
O
RH
OH
R
MichaelO
(+)(–)
O
(+)(+)(–)O
+O O
OM+
O O OM
ketoneenolate
ketoneenolatetype of enolate preserved
What are the relative thermodynamics of these reactions?One must consider the reactions before quenching:
OM
+O
RH
O OM
R
Overall, an aldol reaction replaces a weak C=C bond (theenolate's) with a stronger C–C bond (the red bond). The samething happens in a Michael reaction:
On bond strength alone, this replacement should be worthabout –20 kcal/mol. However, one must also consider aciditydifferences. The ketone to ketone Michael reaction shownproduces an enolate of the same type, so there is no difference.What about the aldol reaction?
OM
+O
RH
O OM
R
acetone pKa:26.5
isopropanol pKa:29.3
Because the starting material is a bit more acidic than theproduct, there is a bit of a penalty: by this estimate, 4 kcal/molat room temperature. However, these penalties do notrepresent an insurmountable barrier:
Chem 106E. Kwan, D.A. Evans Lecture 23: Enolates and C–C Bond Formation
OM+
OR
O O OM
OR
-ketoesterenolate
esterenolatenew enolate is less stable
OR
O
CO2R
The Michael ReactionA perfectly reasonable Michael reaction is between a -keto-ester enolate and an ,-unsaturated ester acceptor:
ethyl acetoacetatepKa=14.2
t-butyl acetatepKa=30.3
In this case, the revised estimate is –20 + 22.5 = +2.5 kcal/molat room temperature. Of course, these crude analyses do notaccount for further proton transfers, solvation, etc. In fact, inthe above reaction, a downstream proton transfer converts theester enolate back to a -ketoester enolate:
OM+
OR
O
OM O
OR
OR
OCO2R
O OM
ORCO2R
energy
This means that the overall thermodynamic barrier is unaffected,but the kinetic barrier is raised. This explains why, ester enolateMichael reactions can occur at –78 °C, but -ketoester enolateMichael reactions require elevated temperatures and harsherbases.
Mechanism of the Michael ReactionWhat is the mechanism of the Michael reaction? It depends onwhat the enolate actually looks like. We can imagine a numberof possible tautomers:
all O-bound, open TS
OLi
OLi
OLi O
OLi
O
O
OLi O
OLi
OLiO
LiO
O OLi
O- to C- bound, closed 6-membered TS
all O-bound, closed8-membered TS
C- to O- bound, closed 6-membered TS
OLi
1-(O)syn
O
1-(C)
LiO
Li
1-(O)anti
3
OLi
"true" 3 usesp orbital on Li
O
Li
ion pair - Li isattracted to negative
charge on carbon
vs.
This gives rise to several possible mechanisms:
Q: Could these six-membered transition states explain the highlystereoselective nature of these reactions?
Some computational studies have suggested that 3 enolatesare the energy minimum for lithioimines (Collum JACS 2006 1286939). This is a serial solvation study, in which an increasingnumber of Me2O ligands is coordinated to lithium:
Chem 106E. Kwan, D.A. Evans Lecture 23: Enolates and C–C Bond FormationMechanism of the Michael Reaction
G
° (k
cal/m
ol)
Me
N Li
Me
NLi
Me
NS·Li
Me
N Li·S
Me
N Li·S2
+8.5
+17.0
+4.0
+0.3+0.8
0.0
+8.3+S
+S
i-Pr
+8.2i-Pr i-Pr
i-Pr
Me
NS2·Li i-Pr
i-Pr
Me
NS3·Li i-Pr
Me
NLi·S3i-Pr
B3LYP/6-31g(d)
+S
However, a similar serial solvation study on acetone does notshow the same trend (top right).
How can we be sure that we have checked every possiblestructure? A systematic search in the gas phase suggests thatthe 3 structure is the energy minimum for lithioacetaldehyde(next page, MP2/6-31G*, Anslyn OL 2006 8 3461).
G°
(kca
l/mol
)
Me
O Li
Me
O
Li
Me
OLi·S
Me
OLi·S2 Me
OLi·S3Me
O Li·SMe
O Li·S2
+15.4
+19.9
+4.7+3.6
+1.0+0.6
0.0
B3LYP/6-31g(d)
+S
+S
+S
O
Li
H
C-O-Li bond angle
180°OLi
H
120°O
Li
H60°
C-C-O-Li dihedral angle
O
Li
H
O
Li
H
150°30°
Chem 106E. Kwan, D.A. Evans Lecture 23: Enolates and C–C Bond FormationMechanism of the Michael ReactionLithioacetaldehyde (MP2/6-31G* enthalpy, gas phase):
Lithioacetone (M05-2X/6-31G* energy, 3 dimethyl ethers):
Conclusion: Lithium enolates are exclusively O-bound in etheral solutions.
gas phase: 3-bound enolate
60 80 100 120 140 160 1800
20
40
60
80
100
120
140
160
180
Li-O-C angle (degrees)
Li-O
-C-C
dih
edra
l ang
le (d
egre
es)
1.4
2.8
4.2
5.6
10
15
20
35
enthalpy(kcal/mol)
(a)
Li
60 80 100 120 140 160 1800
20
40
60
80
100
120
140
160
180
Li-O-C angle (degrees)
Li-O
-C-C
dih
edra
l ang
le (d
egre
es)
1.4
2.8
4.2
5.6
10
15
20
35
electronic energy(kcal/mol)
global min.:-0.4 kcal/mol
(b)
Chem 106E. Kwan, D.A. Evans Lecture 23: Enolates and C–C Bond FormationStereochemical Course of Michael Reactions
Me
O
Me
LiSS
O O
Me
+0.0
O
Me
Li SS
+3.3
Me
O
LiOMe
S
S-13.6
-19.9
free
ener
gy (k
cal/m
ol)
O LiO
Me
S
S
+4.0
E enolate
Z enolate
Me
+3.1
+3.7
O
Me
LiO
S
S
OLi
OMeH
Me
S
S
OOMe Li
Me
S
S
Si
Re
Re
Si
startingmaterials
s-cisTSs
s-transTSs
Assuming an open transition state, here are the numbers forthis prototypical Michael reaction (M05-2X/6-31G*, threedimethyl ether ligands on lithium, denoted as S):
Me
OLi+
Me
O
Me
O OLi
Me
The four transition states represent the enone reacting in ans-cis or s-trans fashion with either diastereoface of theenolate. For this relatively unhindered system, there islittle preference for any particular geometry. Note that s-cistransition states lead to Z enolates and s-trans transition stateslead to E enolates. Kwan and Evans OL 2010 ASAP
How do the computations perform for reactions which haveactual selectivities which can be compared against?
O
OLi
Me
O
Me+–78 °C
THF O
O
Me
Me O
85%, 95:5 syn:anti
O
OLiMe
O
Me+–78 °C
THF/HMPA O
O
Me
Me O
73%, 13:87 syn:anti
Z enolate
E enolate
Here is the scheme for the Z enolate (the E enolate givessimilar results):
LiS
OOS
MeO
Me
O
OR
MeMe
O
R
Li
S
S
160.0
+6.3
+0.6
+9.0
+13.8
starting materials(Z enolate)
s-transtransition
states
Re
Si
Si
Re
s-cistransition
states
favored TS
-11.5
-12.0
-22.8
-24.0
1,2-anti
1,2-syn
E enolateproducts
Z enolateproducts
1,2-anti
1,2-syn
Chem 106E. Kwan, D.A. Evans Lecture 23: Enolates and C–C Bond FormationStereochemical Course of Michael Reactions
Here are the predictions vs. the real results:
O
OLi
Me
O
Me+–78 °C
THF O
O
Me
Me O
85%, 95:5 syn:anti
O
OLiMe
O
Me+–78 °C
THF/HMPA O
O
Me
Me O
73%, 13:87 syn:anti
Z enolate
E enolate
predicted: +4.9 kcal/molexperimental: +1.1 kcal/mol
predicted: -4.5 kcal/molexperimental: -0.7 kcal/mol
Why are the selectivities too high?
(1) The calculations are just wrong.
(2) The calculations are OK, but there is leakage through open transition states.
(3) The calculations reflect the instrinsic selectivity of geometrically pure enolates, but the enolates in the flask are not geometrically pure:
OMe
OLi
MeO
MeOLi
MeMeO
O
MeLDA/THF: 91:9LDA/HMPA/THF: 16:84
Where do the computed selectivities come from? The s-cispreference reflects the unfavorability of having a trans doublebond in an eight-membered ring.
Here is the favored transition state for the Z enolate:
Here is the disfavored transition state for the E enolate. Thes-cis requirement gives rise to a substantial steric interaction:
Chem 106E. Kwan, D.A. Evans Lecture 23: Enolates and C–C Bond FormationThe Claisen Condensation"Claisen Ester Condensation Equilibria - Model Calculations"Garst, J.F. J. Chem. Educ. 1979, 56, 721-722.Not to be confused with the Claisen rearrangement, the Claisencondensation refers to the acylation of an enolate with acarbonyl group. What are the thermodynamics of thisprocess? Is strong base needed?
Fundamentally, one must consider a number of competingequilibria, shown below for the generic self-condensation of anester. Here is the overall reaction:
RO
O
Me
+ RO
O
MeRO
O
Me
OMe RO H+
In terms of bond strengths, and ultimately, free energies, theproducts are less stable than the reactants (estimate: +10kcal/mol based on a fragement-based method). This may bedue to the loss of resonance energy in one ester or theincreased polarity of the products. Whatever the reason,it is not surprising that for straight chain ethyl esters, theremoval of ethanol drives the reaction to completion, raisingyields from 40% to 80% (Hauser Org. Reac. 1942, 1, 266).
Note that unlike the aldol reaction, which can be done in eitheracid or base, the Claisen reaction must be done in base. Theoverall reaction is really the sum of three components. Thefirst is the deprotonation of the ester by a base B with a metalcounterion M:
RO
O
Me
+ M B RO
OM
Me
+ B H
Of course, this produces an enolate and the conjugate acid ofthe base. Many Claisen condensations are done in hydroxylicsolvents like methanol with bases like sodium methoxide.
If we assume the pKa's of isopropanol and tert-butyl propiponatein water (16.5 and 24.5, respectively) are representative, thenthis equilibrium is disfavored by (24.5-16.5)x1.4 = 11 kcal/mol.If the reaction is performed in an aprotic solvent, we can assumethat pKa's in DMSO are representative (29.3 vs. 24.5), and theequilibrium is favored by 4.8 kcal/mol. With a stronger base liketritylsodium, the equilibrium is favored in both solvents (pKa =31.5 in water or 30.6 in DMSO). To summarize:
RO
O
Me
+ M B RO
OM
Me
+ B H
base = alkoxide in water (+11 kcal/mol)base = alkoxide in DMSO (–5 kcal/mol)
base = tritylium in water (–15 kcal/mol)base = tritylium in DMSO (–6 kcal/mol)
The second step involves the actual Claisen step:
RO
OM
Me
+ RO
O
MeRO
O
Me
OMe + RO M
This produces the -ketoester product as well as an alkoxide.Finally, a proton transfer is possible:
RO
O
Me
OMe + RO M RO
O
Me
OMMe RO H+
The overall effect of these two steps combines the intrinsicthermodynamics of the Claisen condensation itself (+10 kcal/mol) with the thermodynamics of forming a more stable enolate.
Q: The pKa of ethyl acetoacetate is 11 in water and 14.2 inDMSO. What are the expected free energy changes forsteps two and three in water and DMSO?
O
Me
Step 1: Enolization
Chem 106E. Kwan, D.A. Evans Lecture 23: Enolates and C–C Bond FormationThe Claisen CondensationQ: The pKa of ethyl acetoacetate is 11 in water and 14.2 inDMSO. What are the expected free energy changes forsteps two and three in water and DMSO?
RO
OM
Me
+ RO
O
MeRO
O
Me
OMMe RO H+
(1) The "bond energy component" is +10 kcal/mol.
(2) The "acidity component" is (11-24.5)x1.4 = -18.9 kcal/mol in water and (14.2-30.3)x1.4 = -22.5 kcal/mol in DMSO.
(3) Thus, we predict G = -3.5 kcal/mol in water and -12.5 kcal /mol in DMSO for steps two and three.
(4) The overall reaction with alkoxide must also take into account the free energy of deprotonation. The overall standard free energy changes are +7.5 kcal/mol (water) and -11.5 kcal/mol in DMSO.
(5) If trityl anion is used, the the energy change becomes -18.5 kcal/mol (water) and -18.5 kcal/mol (DMSO).
RO
OM
Me
+ RO
O
MeRO
O
Me
OMe + RO M
RO
O
Me
OMe + RO M RO
O
Me
OMMe RO H+
Step 2: Claisen Condensation
Step 3: Proton Transfer
pKa(ester) = 24.5 in water 30.3 in DMSO
pKa(ketoester) = 11 in water 14.2 in DMSO
The sum of these steps is:
Therefore, the reaction is affected by three factors:
(1) the initial favorabilty of ester deprotonation;(2) the intrinsic cost of forming of a -ketoester; and(3) the eventual favorability of proton transfer
So the reaction is favored by having a base which is strongenough to deprotonate the ester. Even then, one needs astoichiometric amount of base (the reaction is not really"catalytic") because the product is more acidic than the startingmaterial. Thus, one uses stronger basse like potassium hydride(Brown Synthesis 1975 326) or forcing conditions (continualremoval of ethanol).
RO
O
Me
+ M B RO
O
+ B HO
MeIn this case, one loses the benefit of the third step. For alkoxidein water, the new estimate is 10+11=+21 kcal/mol, which isunsurmountable. For example, this reaction is done with KH inTHF (pKa ~ 36), and does not work with alkoxide in water:
Me
Me Me2
O
O2 O
O O
MeMe Me
This reaction also takes advantage of the changes in pKa ongoing from water to THF. An alternative strategy uses Lewisacids to lower the pKa of the starting ester (Tanabe JACS 2005 127 2854). Crossed Claisen reactions are possible:
2 h, rt>95%
R OH
O NaHCl3CCOCl
R O
O O
CCl3
RO
OMe
TiCl4Bu3N R
OMe
OTiLn
CH2Cl2MeO
O O
MeMe Me
0
One can also imagine a scenario in which the product has noenolizable protons:
Chem 106E. Kwan, D.A. Evans Lecture 23: Enolates and C–C Bond FormationAlternatives for -Dicarbonyl SynthesisAs rosy as this thermodynamic analysis is, the truth is that theperformance of the crossed Claisen reaction is not terrific, partlybecause esters are not very electrophilic. Unfortunately, if acylchlorides or carbonates are used, O-acylation is seen. In thissection, three alternatives to this are presented.
RO
OLi O
Cl R+
RO
O
O
R+
RO
O O
R
O-acylationproduct
(undesired)
C-acylationproduct
(undesired)
The Mander Reagent
Chelation and Carbon DioxideIn Rathke's solution to this problem (JOC 1985 50 4877), the pKa of the ester is lowered by Lewis acid coordination, andthen deprotonated by an amine. This is called "soft enolization"and will be discussed in detail in Lecture 24. The electrophileis carbon dioxide:
RO
OMe
MgCl2NEt3 R
OMe
OMgLn
O C O+
RO
O
OH
O
RO
O
O
OMg
The magnesium also serves to protect the final product fromdecarboxylation. The product can then be alkylated to producea stable -ketoester.
-Ketoacids are unstable, so it is desirable to access theproduct directly. Mander has introduced (TL 1983 24 5425)methyl cyanoformate, which combines with lithium enolates togive exclusively C-acylated products:
RO
OLi O
NC OMe+
RO
O OLi
OMeCNRO
O O
OMe
The intermediate must be stable, since only one equivalent ofbase is required. Therefore, the tetrahedral intermediate mustbe stable. If it weren't stable, then either the product malonateor H-CN would quench any unreacted enolate.
Clearly, the tetrahedral intermediate is stabilized by chelation,but so is the intermediate in a regular Claisen:
RO
O OLi
OMeCN RO
O OLi
OMeR
morestablethan
A big part of the difference must be the leaving group abilities ofcyanide vs. methoxide (DMSO pKa's: 12.9 vs. 27.9). Based onthis, one might conclude that cyanide is the better leaving groupthan methoxide. These are thermodynamic arguments. Whatdo the kinetic data say? Stirling has reported (JCS Chem.Commun. 1975 940) the rates of E1cb elimination:
PhO2SX
log (k/kOPh)PhO2S + X
R=+PPh3 +4.4
R=OMe -3.9R=CN < -7
This does not agree withthe thermodynamic data:cyanide leaves slower.
Based on these rate data, one would conclude that thetetrahedral intermediate partitions to alkyl cyanide, sincemethoxide is apparently a better leaving group. However, therequirements of E1cb reactions, which go through carbanions,may be quite different from these reactions, which go throughoxyanions.
The Roskamp Reaction
R H
O
N
O
OtBuH
+
N
SnCl2
N2
O
OtBu
O
R
ClSn
H
O
OtBu
O
R
This clever reaction avoids the problems of condensationentirely (Roskamp JOC 1989 54 3258):
It proceeds via a formal 1,2-hydrogen shift. Yields are good toexcellent and the reaction works on sensitive substrates.However, it will not produce mono- or di-alkyl substitution at theactive methylene of the -ketoester.
Chem 106E. Kwan Lecture 23: Enolates and C–C Bond Formation
The Horner-Wadsworth-Emmons Reaction"Olefin Synthesis with Organic Phosphonate Carbanions."Boutagy, J.; Thomas, R. Chem. Rev. 1974, 74, 87-99.
Myers, A.G. et al. "Horner-Wadsworth-Emmons Olefination."Chem 215 Handout. http://www.chem.harvard.edu/groups/myers/page8/page8.html (accessed October 2010).
One of the most important reactions involving doubly stabilizedcarbanions is the HWE reaction:
ROP
OMe
O O
OMe
+O
H RRO2C
R
E olefin
How does the reaction work? Studies show that the additionof the phosphonate anion is rate-limiting. A mesomericallyelectron withdrawing group is required to avoid the formation ofstable -hydroxyphosphonates:
+ POMe
O
OMeHObase
POMe
O
OMeEWG
The stereochemical rationalization is as follows:
This is a very reliable reaction. The product is a phosphate,which is water soluble and can be easily removed by extraction.In contrast, the Wittig reaction produces triphenylphosphineoxide, which must be removed by chromatography.
+
R H
O
rate-determining
step
P(O)(OR)2
H
EWG
MO
RH
P(O)(OR)2
H
EWG
MO
HR
O P(OR)2
OM
M
R EWG
Z olefin
O P(OR)2
OM
R EWG
E olefin
(1) Rate-determining and selectivity-determining addition of the stabilized enolate to the aldehyde is occurring.
(2) Interconversion of the intermediates can occur through epimerization (this is impossible for a substituted HWE reagent), but not through retro-carbonyl addition.
(3) The overall selectivity depends on the selectivity of the initial addition and the lifetime of the intermediates.
selectivity-determining
step
equilibration viaepimerization is
possible
O P(OR)2
OM
R EWG
O P(OR)2
OM
R EWG
REWG
REWG
Q: How are stabilized phosphonates prepared?
Q: How can reagents be chosen to optimize E/Z selectivity?
Q: What are good reaction conditions?
Chem 106E. Kwan, D.A. Evans Lecture 23: Enolates and C–C Bond FormationThe Horner-Wadsworth-Emmons ReactionMichaelis-Arbusov ReactionThere are number of possible preparations, but this is the mostcommon. It starts from an -bromo carbonyl compound:
O
OBr
P(OMe)3
O
OPMeO
MeOO
CH3
Br
MeOP
OMe
O O
OMe
However, yields are typically moderate (60-70%), often becausea substantial amount of the methyl dimethyl phosphonate forms.What is the source of this side reaction?
O
OBr
P(OMe)3Me
POMe
O
OMe
This occurs because methyl bromide is generated during thereaction. Alkylation of trimethyl phosphite with acyl -bromideoccurs at a rate that is competitive with alkylation with methylbromide:
POMe
OMe
MeOH3C Br
Me
PMeOMeO
OCH3
Br
MeP
OMe
O
OMe
This can be reduced, but not completely eliminated, with the useof triethyl phosphite instead.
E vs. Z SelectivityThese olefinations are very versatile:
Stabilized Wittig: E olefinsUnstabilized Wittig: Z olefins (E possible with Schlosser mod.)
Stabilized HWE: E olefins (phosphonate OR with R=alkyl) Z olefins (Still-Gennari: R=CH2CF3 or Ando: R=OAr)
Anything that will increase the reversibility of the reaction willfavor the E isomer:
(1) higher reaction temperatures(2) DME vs. THF(3) potassium vs. sodium or lithium
The Still-Gennari Modification (TL 1983 24 4405)
ROP OCH2CF3
O O
OCH2CF3
+O
H RRO2C
RKHMDS, 18-cr-6THF, –78 °C
The Ando Modification (JOC 1998 63 8411)This is a more recent improvement that gives higher selectivitiesand uses phenol-based starting materials:
ROP OPh
O O
OPh+
O
H RRO2C
RNaH, THFTHF, –78 to
–10 °C
The use of 2-tert-butylphenoxy phosphonates gives evenhigher Z:E selectivities (Touchard Eur JOC 2005 9 1790).Epimerizable AldhydesThe Masamune-Roush conditions avoid the epimerizationproblems caused by strong base and are suitable for making Eor Z olefins (TL 1984 25 2183):
RO2CR
LiCl, DIPEAMeCN, rtRO
P ORO O
OR+
O
H R
Patterson has also reported the use of barium hydroxide forepimerizable aldehydes (Synlett 1993 774). A more recentmodification comes from the Myers group (OL 2005 7 4281):
RO2CR
n-BuLi, HFIPDME, –14 °CRO
P ORO O
OR+
O
H R
Lithium hexafluoroisopropoxide is a very weak base and giveshigher E selectivities than other conditions without epimerization.