Logical disconnections
16
SYNTHESIS OF TARGET MOLECULES BY TWO FUNCTIONAL GROUPS DISCONNECTIONS: When a target molecule (TM) contains two functional groups, the best disconnection is one that incorporates both groups and provides two synthons. Such disconnections are called two group disconnections. Definition of two groups disconnection : When we use one functional to help disconnect another group in a bifunctional target molecule, then the disconnections are known as two group disconnections. In bifunctional target molecules two group disconnections are more efficient than one group disconnections. For instance, we can consider the following target molecule. For target molecules consisting of two fragments joined by a heteroatom, we should disconnect the bond next to the heteroatom. Therefore, in the following TM, we can disconnect on either side of the ether oxygen atom to get synthons. However, the disconnection b is the best choice. Cause: Disconnection a does not correspond to a reliable reaction because it might be very difficult to control chemo- selective alkylation of a primary hydroxyl group in the presence of the secondary one. In the forward synthesis by path b, nucleophilic attack on the less hindered terminal carbon atom of the epoxide, the reagent of disconnection b, can give the desired molecule selectively. SYNTHESIS OF β-HYDROXY CARBONYLS AND α,β-UNSATURATED CARBONYLS : BY ALDOL REACTION: The aldol reaction of aldehydes and ketones is one of the most important methods for the synthesis of β-hydroxy carbonyls and α,β-unsaturated carbonyls. For example, O H H 3 C 2 aq. NaOH O H H 3 C OH (Aldol) H 3 O + O H H 3 C (Crotonaldehyde) With Acetaldehyde: O CH 3 H 3 C 2 aq. NaOH O CH 3 H 3 C OH (Diacetone alcohol) H 3 O + O CH 3 H 3 C With Acetone: H 3 C CH 3 (4-Methyl-pent-3-en-2-one) O Ph OH TM a b Disconnection b Disconnection a O Ph OH + Br HO Ph OH O Ph OH + OH Ph O O Ph OH O OH Ph O NaH O Ph O H 2 O workup
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Transcript of Logical disconnections
SYNTHESIS OF TARGET MOLECULES BY TWO FUNCTIONAL GROUPS DISCONNECTIONS:
When a target molecule (TM) contains two functional groups, the best disconnection is one that incorporates both
groups and provides two synthons. Such disconnections are called two group disconnections.
Definition of two groups disconnection: When we use one functional to help disconnect another group in a bifunctional
target molecule, then the disconnections are known as two group disconnections. In bifunctional target molecules two
group disconnections are more efficient than one group disconnections. For instance, we can consider the following
target molecule. For target molecules consisting of two fragments joined by a heteroatom, we should disconnect the
bond next to the heteroatom. Therefore, in the following TM, we can disconnect on either side of the ether oxygen
atom to get synthons. However, the disconnection b is the best choice.
Cause: Disconnection a does not correspond to a reliable reaction because it might be very difficult to control chemo-
selective alkylation of a primary hydroxyl group in the presence of the secondary one.
In the forward synthesis by path b, nucleophilic attack on the less hindered terminal carbon atom of the epoxide, the
reagent of disconnection b, can give the desired molecule selectively.
SYNTHESIS OF β-HYDROXY CARBONYLS AND α,β-UNSATURATED CARBONYLS:
BY ALDOL REACTION:
The aldol reaction of aldehydes and ketones is one of the most important methods for the synthesis of β-hydroxy
carbonyls and α,β-unsaturated carbonyls. For example,
O
HH3C2
aq. NaOHO
HH3C
OH
(Aldol)
H3O+ O
HH3C
(Crotonaldehyde)
With Acetaldehyde:
O
CH3H3C2
aq. NaOHO
CH3
H3COH
(Diacetone alcohol)
H3O+ O
CH3H3C
With Acetone:
H3C
CH3
(4-Methyl-pent-3-en-2-one)
OPh
OH
TM
a b Disconnection bDisconnection a
OPh
OH
+
Br
HOPh
OH
O
Ph
OH
+
OH
Ph
O
OPh
OH
OOHPh
O
NaH OPh
OH2O
workup
Mechanism:
Base catalyzed aldol reaction:
Acid catalyzed aldol reaction:
Mechanism of the dehydration of β-hydroxy carbonyl product:
Acid catalyzed dehydration:
Base catalyzed dehydration:
Mixed aldol reactions:
On the other hand, if only one of the two aldehydes has an α-hydrogen, then two aldol products are formed.
When one of the reactant involved in mixed aldol reaction is aromatic aldehyde such as benzaldehyde, furfural etc, the
reaction is known as Claisen-Schmidt condensation. In this case aromatic aldehyde cannot act as a nucleophilic
component due to the absence of α-protons for enolization.
O
CH2Raq. NaOH
HOH-
O
CH2R
O
CH2R
(enolate anion)
O
CH3R O
R
O
CH3R
O
R
OH
CH3R
(X)
H2O
OHOOH
OHH
OOH OOH - OH O
E1cb
OOH OH3O+ OOH2 (-H2O) O
E1H
OH2
H2O
O
CH3R
OH
CH2R(enol)
OH
CH3R OH
R
OH
CH3R
O
R
OH
CH3R
H+ OH
CH2R
H
- H+
R CHO R' CHO+OH
R' R'
OH
CHO
R R
OH
CHO
+R R'
OH
CHO
R' R
OH
CHO
++
(Self -condensation products) (Crossed condensation products)
PhCHO
OH+CH3CHO + PhCHO
OH H3C CHO
OH
α-hydrogen no α-hydrogen
iii)
OHO
CH3Ph
Ph
OH O
Ph+
O
HPh EtOH, heat
Ph
O
Phi)
ii)
Ph
Ph O
O
+
Ph
Ph
OOH
EtOH, heatO
Ph
Ph
Ph
Ph
O
CH3
OH+
O
HPh EtOH, heatPh
O
Intramolecular aldol reactions:
Synthesis of
Retrosynthesis:
Synthesis:
Work backward and identify the starting materials of the following products by aldol condensation and also point out
which syntheses are particularly feasible.
Answer:
i)
Synthesis of the given molecule with these substrates will not be successful. Because when the carbanion derived from
ethyl acetate mixed with acetone, proton transfer occurs more rapidly than the condensation occurs and forms the
O
O
NaOEt/ EtOH O
O
ElectrophilicNucleophile O
OHH3C H+
(-H2O)
O
CH3
O
O
NaOEt/ EtOH O
OO
OHH3CNaOEt/ EtOH
O
O
OH
O
(Formed)
Strained andnot formed
1,4-dicarbonyl compound
notH
O
KOH/MeOH
heat
O O
CHO
more reactivealdehyde group
H
O
O
ii)i)Me
Me CO2Et O
iii)
O OH
PhPh
Retrosynthesis:
Me
Me CO2EtO
Me
Me+ CH3CO2Et
CHOO2N
CHOO2N FGI CHOO2N
OH Retro-Aldol CHOO2N
OH
+
CHOO2N CH3CHO
CHOO2N
LDA/THF
- 78 oC H
O1.
2. Aq. NH4Cl
(work up)
O
H
CHOO2N
OHKHSO4
dehydration
CHOO2N
carbanion of acetone. The carbanion of acetone reacts with ethyl acetate to give Claisen condensation product after
final work up with dilute acid.
So the given molecule can be synthesized by aldol reaction. But other methods such as Reformasky reaction,Wittig
reaction or Knoevenagal reaction may be used for this purpose.
ii)
Synthesis of the given molecule by mixed aldol condensation using one molecule of acetone and two molecules of
benzaldehyde is feasible. This is because only acetone has α-protons and benzaldehyde does not have any α-protons
and therefore, the carbanion of acetone reacts with highly reactive benzaldehyde to give the desired product. The self-
condensation of acetone is not feasible because of its unfavourable equilibrium constant.
iii)
Synthesis of
Retrosynthesis:
Synthesis:
CH3CO2EtNaOEt
CH2CO2Et
O
O+ CH3CO2Et
O O
CH3
NaOEtO O
CH3
H3O+O O
CH3
Retrosynthesis:
O OH
+O O
H
Retrosynthesis:
O+ 2 PhCHO
O
PhPh
Ph
O
PhPh
NaOEtO
PhCHO
ONaOEt
PhCHO
LDA/THF
- 78 oC
O1.
2. Aq. NH4Cl
(work up)
O OH
O
HO
Synthesis:
Ph CO2H
O
FGI Retro-Aldol+
Ph CO2H
O
Ph CO2H
OOH
Ph
OH
CO2H
O
Ph
O
H CO2Et
O
2. Aq. NH4Cl
(work up)
Ph CO2H
OPh
O
H
CO2Et
O EtO
CO2Et
O1.
Ph CO2Et
OOH1. Dil. KOH
2. H3O+/heat
Synthesis of
How will you bring about the following transformation?
Answer:
Synthesis of
Path a is the best choice, because:
The starting materials of path a are cyclohexanone and benzyl, which are simple starting materials and readily available.
But 2-benzoylcyclohexanone, one of the starting materials of path b, is complicated and it has to be synthesized from
CHO
from
CHO
CHO
CHO
Synthesis:
CHO
CHOAq. NaOH
CH
CHO
O
CHO
OH2O
CHO
OHE1cbOHCHO
OH- OH
CHO
Reconnection
OsO4
NMO
OH
OH
NaIO4
Retrosynthesis:
O
NaOEt/EtOHO
1. O3
2. Me2S; CH2Cl2O
O
Aldol deactionOH
H+
Dehydration
O
PhPh OH
OO
PhPh OH
OO
Retrosynthesis:
Path a Path b
a
b
OPh
Ph
O
OH+ Ph
O
OH
+Ph
O
OPh
Ph
O
O
Ph
O
OPh
NO2
cyclohexanone and ethyl benzoate by Claisen condensation. Again, since 2-benzoylcyclohexanone is an unsymmetrical
1,3-diketone, nucleophilic addition of enolate of nitroalkane may occur at both carbonyl to provide mixture of products.
BY REFORMATSKY REACTION:
The reaction of α-halo ester with an aldehyde or ketone having α-hydrogen or without having α-hydrogen to form β-
hydroxy ester in the presence of zinc metal is referred to as the Reformatsky reaction.
Mechanism:
When a nitrile is allowed to react with the zinc enolatederived from ethyl bromoacetate the corresponding β-keto ester
is obtained. This is known as Blaise reaction.
Suggest suitable method for the synthesis of the following compounds using the given reaction.
Answer:
CO2Et Using Reformatsky reaction
+ Br CO2Et
ZnEther-Benzene
Reflux
dil H+
CO2Et
O
CO2Et
OZnBrCO2Et
OH
Ac2O, heat
O+ Br CO2Et
ZnEther-Benzene
RefluxCO2Et
OZnBr Dil. H+
CO2Et
OH
β-Hydroxy ester
O
EtO
Dil. H+
Br
Zn O
EtO
ZnBr
THF
O
EtOZn
Br
O
Zn O
Zn
EtO
OEt
BrBr
O
O
Dimer of zinc enolate
O
OZn
O
EtO
BrO
O
ZnOBr
OOEtO O
EtO
ZnBr O
O OH
EtO
β-Hydroxy ester
CN+ Br CO2Et
1. Zn(0), THF, reflux
2. Aq. HCl,
CO2Et
O
PhPh OH
OO
Synthesis:
OOPh
Ph
O
OLDA/THF
-78 oC
1.
2. H2Oworkup
Transformation of
Answer:
How would synthesize the following molecule using Reformatsky reaction?
BY KNOVENAGEL REACTION
Condensation of aldehydes or ketones with a compound having active methylene group in the presence of a weak base
to form an α,β-unsaturated compound is known as Knoevenagel condensation. Weak bases like amines or buffer
systems containing an amine and a weak acid catalyze these reactions.
Mechanism:
Describe the synthesis of the following compound with proper retrosynthetic analysis.
O CH2CO2Et
O
BrCH2CO2Et
Zn, THF-Benzeneheat
HO CO2EtKHSO4
Heat
Then dil HCl
CO2Et
Pd/ heat
CO2Et
CO2H
O
+CO2Et
CN cat. pyrrolidine CO2Et
CN
+
CO2Et
CN
N
H O NH - H+
+ H+
O N
- H+
+ H+HO N N
(Iminium ion)
+N
H
CO2Et
CN+
NH H
N
+
CO2Et
CN NEtO2C
CN
+ NH H
NEtO2C
CN
+N
HH H
NEtO2C
CN
H+ N
H H
EtO2C CN
O
+N
HRegenerated
catalyst
CO2Et
MeO1.
2.HO
CO2Et
Ph
3. HO CO2Et4. O
CO2Et
O
OMe
OMe
Answer:
Synthesis of α,β-Unsaturated Ketones via Mannich Bases:
The Mannich reaction is the condensation of an enolizable carbonyl compound with an iminium ion derived from the
reaction between an aldehyde (usually formaldehyde) and a secondary amine in presence of an acid to give, after
basification, an amino methyl derivative.
Mechanism:
Convert acetophenone to PhCOCH=CH2 via a Mannich base.
Answer:
CO2HCO2EtEtO2C
O
CO2Et
CO2Et+
Retrosynthesis:
CO2HCO2EtEtO2CForward synthesis:
O
CO2Et
CO2Et+
Py, Piperidine
heat
H3O+
heat
O
Ph
O
Ph NEt2
Me-IO
Ph N(Me)Et2I
Ag2O
heat
O
Ph
i) HCHO, Et2NH/H+
heat
ii) OH-
O
HH+ R2NH2
+Cl-OH
HH+ R2NH
H2O
HH NR2
HO
HH NHR2
- H2OH2C NR2H2C NR2
(iminium ion)
O
Ph
+ H+ OH
Ph
H2C NR2 OH
Ph NR2
O
Ph NHR2
Cl
OH-O
Ph NR2
- H+
OH
Ph + H+
- H+ O
Ph NR2
- H+
+ H+
(+ HCl) (- HCl)
Synthesis of α,β-Unsaturated Ketones via Wittig reaction:
The Wittig reaction, which combines a ketone or an aldehyde with a Wittig reagent, is used to create alkenes. If the
Wittig reagent itself contains an ester group, then the resulting product will be an α,β-unsaturated ester. Such Wittig
reagents are known as “stabilized ylides” since the ester electron-withdrawing group offers resonance stabilization to
the negative charge.
While the Wittig reaction typically gives the (Z)-alkene as the major product, stabilized ylides usually give the (E)
stereoisomer. For example,
The related Horner–Wadsworth–Emmons (HWE) reagent, a phosphonate-stabilized ylide that is prepared from P(OEt)3
rather than PPh3, may be used for this purpose. This reagent has the advantage of being more reactive (reacts with both
aldehydes and ketones) and offering better stereoselectivity.
Two possible routes can be proposed for the synthesis of the (E)-isomer of ethyl cinnamate using Wittig reaction.
Route II is more efficient because
i) The Wittig reagent of the route II is a stabilized yilde, and therefore, its reaction with benzaldehyde is stereoselective
and gives the desired (E)-alkene as the major product.
ii) The alternate route “I” would result in a more complicated starting material and would give the (Z)-alkene as the
major product.
Other examples:
CHO
CO2H
iii) EtO2C Br
PPh3, THFPPh3EtO2C
Br
1. PhLi
CHO2.
CO2Et hydrolysis CO2H
(Only trans product is formed)
OPh
Ph
Ph
Ph CHO
H
Ph
CO2Et
Route IIRoute IPh3P CHCO2Et + PhCHOPh3P CHPh+
O
CHOEtO
H
Ph
CO2Et
PhCHOO
OEtBr
PPh3O
OEtPh3P
Br
n-BuLi
THF
O
OEtPh3P
Synthesis:
R H
O+
O
OEtPh3P
O
OEtR
(E)-Alkene
O
OEtPh3P
O
OEtPh3P
O
OEtPh3P
Stabilized Wittig reagent
O
+
O
OEt(EtO)2P
OCO2Et
SYNTHESIS OF 1,3-DICARBONYLS:
Two alternative two groups disconnections can be possible for 1,3-dicarbonyls (Consonant systems): one lead to two
synthons of natural polarity and the other lead to two synthons of unnatural polarity. The disconnection leading to
natural synthons is the best choice because of the simplicity in synthesis and the synthetic equivalents of natural
synthons are readily accessible.
Claisen condensation is most commonly used for the synthesis of 1,3-dicarbonyl compounds.
CLAISEN CONDENSATION:
The ester having α-hydrogen atoms undergo base catalyzed self condensation reaction to form a 1,3-dicarbonyl
compound alternatively known as β-ketoester. This self condensation of ester is known as claisen condensation.
Mechanism:
i) CH3CH2OHCl2, MeOH, H+
heat
1. Ph3P/ether
2. n-BuLi
THF -10oC
OEt
EtO
Ph3PO
Ph
PhOEt
EtO
Ph
Ph
H+, H2O
Ph
Ph
CHO
Cl
OEt
OEt
O
OEt
O
OEt
O
OEt
O
OEt
O
OEt
OEt
O
+ EtO +EtOH
OEt
O O
+OEt
O O
EtOH+OEt
O O
OEt
O O
H+
OEt
O O
pKa = 25 pKa = 15.5
pKa = 10.7
highly acidic proton
less acidic protonsmore acidic proton
EtO
Na+
- EtO-
O OO O+
O O+
Naturalsynthon
unnaturalsynthon
unnaturalsynthon
Naturalsynthon
O
X
O
EtO2CO
Or
Li
SS
NO2Or
O
X OrNO2
Describe the synthesis of the following compound with proper retrosynthetic analysis.
Show two retrosynthetic pathways differing in the position of disconnection for the following compound. Which
pathway will lead to efficient synthesis and why?
The path a would lead to the efficient synthesis because: i) path a involves only a single substrate, ethyl
phenylacetate, which is inexpensive and readily available; ii) path a leads to a single step synthesis; iii) dibenzyl
ketone, one of the starting materials of path b would probably have to be synthesized in a multistep synthesis.
Analyze the following molecule and determine what starting material would be required for its synthesis by a Claisen
condensation. Then decide which, if any, of the possible Claisen condensation would be a reasonable route to the
desired product.
O
Ph Ph
O
O
Ph Ph
ORetrosynthesis:
O
MePh+
EtO Ph
O O
Ph Ph
OO
MePh+
EtO Ph
O
Synthesis:1. NaOEt (1 eq.),
xylene, heat
2. H3O+
Ph Ph
CO2Et
O
Synthesis:1. NaOEt (1 eq.),
xylene, heat
2. H3O+ Ph Ph
CO2Et
O
PhO
OEt+
Ph
CO2Et
Retrosynthesis:
Ph Ph
CO2Et
O
Path a
a
PhO
OEt+
Ph
CO2EtPath b b
Ph PhO
+O
OEtEtO
BrPh2 +O
OEtH
Via Grignardsynthesis
O Ofrom pinacol
Retrosynthesis:
O O O
+EtO
O O OHHO
1. NaOEt (1 eq.),xylene, heat
2. H3O+
Synthesis:
OHHO conc. H2SO4
O
O OO+
EtO
O
Pinacol-pinacolone
rearrangementPinacol
i) I2/Aq. NaOH
ii) H3O+
iii) EtOH/H+
OEt
O
O
Me
O
The retrosynthetic analysis shows that the target molecule can be synthesized either by the intermolecular Claisen
condensation (by path a) or intramolecular Claisen condensation.
Of the two possible routes, path a would not be a reasonable route to the synthesis of the desired product.
Path b would be reasonable.
Synthesize the following compound form the indicated starting material.
Q. Synthesize the following target molecule using readily available starting materials and reagents. Show all possible
retrosynthetic routes.
O
Me
O
ab
Path a Path bO
+ CH3CO2EtCO2Et O
Me
Retosynthesis:
Me
CO2EtO
NaOEt NaOEt
CH2
CO2EtO
Me
CO2EtO
O
CO2EtCH2
O
OEtO
O
Me
OEtO
O
Me
O
O
O
- EtO
- EtO
+ EtO
+ EtO
O
OO
Me
O
H3O+
O
Me
O
O
O
+
More stable(Major)
Less stable(Minor)
ONaOEt + EtO
O
Me
OH3O
+O
Me
OO O
OEtO
Me
O
O
Ph
O
Ph
CO2H
CO2HDil HNO3
heat
EtOH
H+ CO2Et
CO2Eti) NaOEt/EtOH
ONa
ii) PhCH2Br
CO2Et
O
PhEtO2C
i) dil NaOH
ii) H3O+
iii) Heat
O
OEt
O
SYNTHESIS OF 1,5-DICARBONYLS:
1,5-Dicarbonyls (Consonant systems) can be disconnected at ipso- or α-position with respect to one of the two carbonyl
groups. Of the four possible routes, the route that uses ethyl acetoacetate as the nucleophilic component and αβ-
unsaturated carbonyl compound as the electrophilic component is the best one.
1,5-Dicarbonyl compounds can be synthesized by direct attack of the enolate anion derived from ethyl acetoacetate to
the α,β-unsaturated carbonyl compounds which function as the electrophile.
Michael Addition:
Conjugate addition (1,4 addition) of a stabilized carbanion nucleophile derived from active methylene compound with
activated carbon-carbon multiple bonds (Michael acceptor) is known as Michael addition reaction.
Mechanism
Suggest two alternative possible pairs of starting materials for the synthesis of compound A by Michael reaction. Which
pair will you prefer for actual synthesis and why?
O CO2Et
CO2Et+
i) NaOEt, EtOH
ii) H3O+
12
345
(1,5-dicarbonyl compound)
Me
OEtO2C CO2Et
H2CCO2Et
CO2Et
+ EtO
- EtOHHC
CO2Et
CO2Et
Me
O Me
OEtO2C CO2Et
Me
OHEtO2C CO2Et
Me
OEtO2C CO2Et
H OEt(cat.)+ EtO
Regeneratedcatalyst
O O α-disconnectionipso-
disconnection
O O+
O O+
OR
O O+
O O+
OR
O
XBrMg
LiBr
OO
OOSS
O
CO2EtO
Easily accessible
O
XBrMg
OO
In the following 1,5-dicarbonyl, a disconnection can be made at either alpha carbon between the two carbonyls to get
two pair of Michael acceptors and Michael donors. Retrosynthetic analyses for the synthesis of the following molecules
are shown below
The question is which of the two pairs would be best choice in practice?
The choice of pair depends on the following factors:
a) The softer nucleophile tends to give conjugate addition and the harder nucleophile tends to give carbonyl addition. -
CH(CO2Et)2 is a softer nucleophile than -CH2COCH3 and hence
-CH(CO2Et)2 has a tendency to give conjugate addition
product while -CH2COCH3 gives carbonyl addition products.
b) CH2(CO2Et)2 gives corresponding enolate anion in the presence of weaker base but formation of enolate anion from
CH3COCH3 requires a strong base and drastic reaction conditions.
Consequently, starting materials of path b will be the best choice for the synthesis.
Q. Analyze the following molecules and determine what starting materials would be required for their synthesis.
ROBINSON ANNULATION:
Robinson annulation reaction is the process of formation of a new six membered ring involving intramolecular aldol
condensation of the initial Michael addition product and subsequent dehydration. For example,
PhCOOEt
O Ph
COOEt
Path aPath b
a
b
Ph
O
PhH2C
COOEt
COOEt+
Ph
O COOEt
Ph COOEt+
Synthesis:
Ph
O
Ph H
O+
i) Aq. NaOH
ii) H3O+ Ph
O
Ph
HCCOOEt
COOEt
PhCOOEt
O Ph
COOEtAldol reaction
Michaelreaction
Aq. NH4Cl
PhCOOEt
O Ph
COOEt
O
CO2H1.
O
CO2H2. 3.
CO2EtPh
O
Ph
4.
OMe
OO5 EtO2C
OPh
CN
If the four-carbon methyl vinyl ketone unit within the cyclohexenone ring is present in a target molecule, it is an
indication that the TM could be prepared by Robinson annulation reactions. For example;
Problems :-
1. Synthesize the following molecules from the indicated starting materials
2. Synthesize the following molecules using Robinson annulation.
Complete the following reactions scheme.
Answer:
Outline steps for the synthesis of following compound from the indicated starting materials. Also show the
retrosynthetic analysis.
O
MeCO2Et
O
CO2Et
fromi)ii)
O
O
from acyclic molecules
O
CHO
Odil KOH
(B)
H+
-H2O
(C)
O
O H
Michaeladdition O
CHO
O
OH
Aldolreaction
O
HO
O
(A)
O
O
O
H
OCH2
O
O
CO2Et
Cyclohexenonemoiety
O
CO2Et
OO
CO2EtO
+
Retrosynthesis:
Synthesis:
O
CO2EtO
+EtONa/EtOH
Heat
O
CO2Et
O
Michael reactionAldol reaction
(-H2O)
O
CO2Et
Synthesize the following molecule using Robinson annulation. Show retrosynthetic analysis also.
Answer:
OMe Me
O
CO2Et
O
+
1. NaOEt, EtOH
2.O
CO2Et
O-H2OO
Retrosynthesis:
Forward synthesis:
OMe Me
OMichael reaction
FGA
OMe MeOH
OMe Me
O
FGA
OMe Me
O
CO2Et
CO2Et
OHO Me Me
Heat OMe Me
CO2HH+/H2O (-CO2)
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