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.
Br OH
a OH b O Disconnection a b Disconnection O Ph + + TM OH OH O Ph Ph
OH O HO Ph Ph 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.
O O OH H O 2 O NaH Ph O OH O Ph Ph workup
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,
With Acetaldehyde:
O OH O + O aq. NaOH H3O 2 H3C H H3C H H3C H (Aldol) (Crotonaldehyde) With Acetone:
O OH O + CH O aq. NaOH H C H3O 3 2 3 H C CH CH H C CH 3 3 H3C 3 3 3 (Diacetone alcohol) (4-Methyl-pent-3-en-2-one)
Mechanism :
Base catalyzed aldol reaction : O
O O O R CH3 O O O OH aq. NaOH H2O R CH2 R CH 2 R CH2 R CH3 R CH3 H R R - (enolate anion) OH (X)
Acid catalyzed aldol reaction :
OH O + OH R CH H OH 3 OH OH -H+ O OH R CH 3 R CH2 R CH R CH R CH 2 R 3 3 H (enol) R
Mechanism of the dehydration of β-hydroxy carbonyl product :
Acid catalyzed dehydration:
+ OH O O OH O H3O 2 (-H2O) H2O O
E1 H OH2 Base catalyzed dehydration:
OH O OH OH O OH O - OH O E cb 1 H OH Mixed aldol reactions:
CHO CHO CHO CHO OH + R R + + + R CHO R' CHO R' R' R' R R R' OH OH OH OH
(Self-condensation products) (Crossed condensation products)
On the other hand, if only one of the two aldehydes has an α-hydrogen, then two aldol products are formed.
OH Ph H C CHO 3 CH3CHO + PhCHO + CHO OH OH
α-hydrogen no α-hydrogen 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 O OH Ph Ph Ph Ph i) + Ph CH EtOH, heat 3 Ph H OH O O
Ph Ph Ph O Ph OH ii) + O O EtOH, heat Ph O Ph Ph Ph O O OH iii) + Ph CH3 Ph H EtOH, heat O
Intramolecular aldol reactions :
H C CH3 O NaOEt/ EtOH O 3 OH H+
(-H2O) O O Nucleophile O O Electrophilic
OH O O H3C NaOEt/ EtOH NaOEt/ EtOH O OH
O O O O 1,4-dicarbonyl compound O Strained and not formed (Formed)
O O O H KOH/MeOH H not heat O O CHO more reactive aldehyde group Synthesis of O2N CHO
Retrosynthesis:
OH OH Retro-Aldol O2N CHO O2N CHO + O2N CHO FGI
O2N CHO CH3CHO
Synthesis:
O2N CHO OH O N CHO 1. O N CHO KHSO4 2 LDA/THF 2 O O dehydration o H - 78 C H 2. Aq. NH4Cl (work up)
Work backward and identify the starting materials of the following products by aldol condensation and also point out which syntheses are particularly feasible.
O Me CO2Et O OH i) ii) iii) Ph Ph Me
Answer: Retrosynthesis: Me i) Me CO2Et O + CH3CO2Et Me Me 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
carbanion of acetone. The carbanion of acetone reacts with ethyl acetate to give Claisen condensation product after final work up with dilute acid.
O
NaOEt O CH CO Et CH CO Et + CH3CO2Et 3 2 2 2
O O H O+ O O O O 3 NaOEt CH 3 CH3 CH3 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) Retrosynthesis: O O + 2 PhCHO Ph Ph 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.
O O O NaOEt NaOEt PhCHO Ph Ph Ph PhCHO iii) Retrosynthesis:
O OH O O + H
Synthesis: O
1. H O OH O LDA/THF O
o - 78 C 2. Aq. NH4Cl (work up)
Synthesis of O
Ph CO2H
Retrosynthesis:
Retro-Aldol OH O O FGI OH O + CO H Ph CO H Ph CO2H Ph 2 2
O O Synthesis: Ph H CO2Et O 1. 1. Dil. KOH O Ph H OH O O EtO O Ph CO H + 2 CO Et Ph CO2Et 2. H3O /heat CO2Et 2 2. Aq. NH4Cl (work up)
Synthesis of CHO
from
Retrosynthesis:
CHO CHO Reconnection CHO
Synthesis: CHO CHO NaIO4 Aq. NaOH OsO4 OH CHO CH O NMO OH
CHO CHO CHO CHO OH H2O - OH E cb O 1 OH OH
How will you bring about the following transformation? O
Answer: O O O NaOEt/EtOH + 1. O3 H
2. Me2S; CH2Cl2 Aldol deaction Dehydration O OH
Synthesis of O O
Ph Ph OH
Retrosynthesis:
O O O Path b O O Ph Path a O Ph b + + Ph Ph OH a Ph Ph OH OH
O O O NO Ph 2 Ph Ph Ph O O
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
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.
Synthesis: O Ph O Ph O 1. O O LDA/THF O Ph -78 oC 2. H O 2 Ph OH workup
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 .
Zn OZnBr + OH O Ether-Benzene Dil. H + CO Et CO2Et Br CO2Et 2 Reflux β-Hydroxy ester Mechanism :
EtO O
O Zn O Zn O ZnBr O Br Br Zn Br EtO Zn O EtO THF EtO Br O OEt
Dimer of zinc enolate O
O Br O Br O Zn OEt Zn O OH + Dil. H O O O O Zn EtO O EtO Br EtO β-Hydroxy ester O
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 .
O 1. Zn(0), THF, reflux CN CO2Et + Br CO2Et 2. Aq. HCl,
Suggest suitable method for the synthesis of the following compounds using the given reaction.
CO Et 2 Using Reformatsky reaction
Answer:
Zn O OZnBr + OH Ether-Benzene dil H CO Et + CO2Et 2 Br CO2Et Reflux
Ac2O, heat
CO2Et
Transformation of
O CH2CO2Et
Answer: CO2Et CO Et O 2 HO CO2Et BrCH CO Et KHSO4 Pd/ heat 2 2 Heat Zn, THF-Benzene heat Then dil HCl How would synthesize the following molecule using Reformatsky reaction? OMe O CO Et O 2 3. 4. 2. HO CO2Et 1. CO2Et HO MeO OMe Ph CO2Et 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.
O CN cat. pyrrolidine CO2Et + CO2Et CN
Mechanism :
CN CN + + N CO Et N CO2Et 2 H H H
O H O N O N HO N N + + N H -H + H + + + H -H+ (Iminium ion)
CN CN N EtO C + 2 N CN EtO2C N + N + N H H H H H CO2Et
EtO2C CN CN
EtO C N + 2 + N N H H H H Regenerated catalyst
Describe the synthesis of the following compound with proper retrosynthetic analysis.
CO2H
Answer: Retrosynthesis: CO Et CO H EtO2C 2 2 O
CO2Et + CO2Et
Forward synthesis: CO Et CO2H EtO2C 2 O + Py, Piperidine H3O CO2Et heat + heat CO2Et
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 :
HO O OH + R NH +Cl- 2 2 + R2NH H NHR2 H H H H H
-H2O H2O H C NR H2C NR2 2 2 H NR2 (iminium ion) H
O + H+ OH H C NR + -H+ OH 2 2 OH -H O Ph -H+ Ph + Ph + H+ + H Ph NR2 Ph NR2
(+ HCl) (- HCl)
O O OH-
Ph NR2 Ph NHR2 Cl
Convert acetophenone to PhCOCH=CH 2 via a Mannich base.
Answer:
+ i) HCHO, Et2NH/H O O O Ag O O heat Me-I 2 Ph N(Me)Et Ph Ph Ph NEt2 2 heat ii) OH- I
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. Stabilized Wittig reagent O O O Ph3P Ph3P Ph3P OEt OEt OEt
While the Wittig reaction typically gives the (Z)-alkene as the major product, stabilized ylides usually give the (E) stereoisomer. For example,
O O O + Ph3P R OEt R H OEt (E)-Alkene
The related Horner–Wadsworth–Emmons (HWE) reagent, a phosphonate-stabilized ylide that is prepared from P(OEt) 3 rather than PPh 3, 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.
O CO Et O O 2
(EtO)2P + OEt Two possible routes can be proposed for the synthesis of the (E)-isomer of ethyl cinnamate using Wittig reaction.
Ph O Route I Route II Ph P CHCO Et + PhCHO + Ph3P CHPh 3 2 EtO CHO H CO2Et 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. Synthesis: O O Ph O PPh3 n-BuLi PhCHO Ph P Ph3P Br 3 OEt THF OEt OEt H CO2Et Br Other examples:
CHO
CO2H
PPh ,THF hydrolysis 3 1. PhLi CO2Et CO2H iii) EtO C EtO2C PPh3 2 Br CHO Br 2. (Only trans product is formed)
Ph Ph CHO O Ph Ph
1. Ph P/ Ph 3 Ph + OEt ether O Cl2, MeOH, H Ph3P Ph i) CH CH OH OEt 3 2 Ph OEt heat Cl OEt 2. n-BuLi EtO EtO THF -10oC
+ H ,H2O
Ph CHO
Ph
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. O O O O O O + +
unnatural unnatural Natural Natural synthon synthon synthon synthon
NO 2 O NO2 O O O Or Or Or S S X EtO C X 2 Li
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 :
O O O + EtO EtOH + O OEt OEt OEt OEt pKa = 25 pKa = 15.5 O OEt O
more acidic proton OEt less acidic protons - EtO-
+ O O O O O O Na O O + EtOH EtO + OEt OEt OEt OEt
pKa = 10.7 H+ highly acidic proton O O OEt
Describe the synthesis of the following compound with proper retrosynthetic analysis.
O O Ph Ph
Synthesis: Retrosynthesis: 1. NaOEt (1 eq.), xylene, heat O O O O O O O O + + + Ph Ph Ph Me EtO Ph Ph Me EtO Ph 2. H3O Ph Ph
Show two retrosynthetic pathways differing in the position of disconnection for the following compound. Which pathway will lead to efficient synthesis and why?
CO2Et
Ph Ph O
Retrosynthesis: Path b b CO Et Path a O 2 OEt CO2Et + Ph Ph Ph + Ph Ph EtO OEt O a O Ph O Via Grignard synthesis O
2 Ph Br + H OEt 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. Synthesis: 1. NaOEt (1 eq.), xylene, heat CO2Et OEt CO2Et Ph + + Ph Ph Ph 2. H3O O O
O O from pinacol
Retrosynthesis: O O O O O HO OH + EtO
Synthesis:
HO OH i) I2/Aq. NaOH conc. H2SO4 OEt ii) H O+ Pinacol- O 3 O pinacolone iii) EtOH/H+ Pinacol rearrangement
O O 1. NaOEt (1 eq.), O O xylene, heat + EtO + 2. H3O 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 O
Me
The retrosynthetic analysis shows that the target molecule can be synthesized either by the intermolecular Claisen condensation (by path a) or intramolecular Claisen condensation. Retosynthesis: O O O O Path b CO2Et Path a a + CH3CO2Et Me Me b Of the two possible routes, path a would not be a reasonable route to the synthesis of the desired product.
CO Et CO2Et CO2Et CO2Et 2 NaOEt NaOEt CH2 EtO O O O O O O Me Me CH2 Me
- EtO
O O + EtO O O O O EtO O Me + EtO - EtO Me O O O
+ H3O O O O + Me O
More stable Less stable (Major) (Minor) Path b would be reasonable. O O O O O O O O + O + EtO H3O NaOEt OEt Me Me Me
Synthesize the following compound form the indicated starting material.
O
Ph
O ONa EtO C Dil HNO3 EtOH i) NaOEt/EtOH 2 CO2H CO2Et CO2Et Ph + heat CO2H H CO2Et ii) PhCH2Br i) dil NaOH + ii) H3O iii) Heat
O
Ph
Q. Synthesize the following target molecule using readily available starting materials and reagents. Show all possible retrosynthetic routes. O O
OEt
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.
O S S O O O O Br X Li BrMg
O O O O + + ipso- α-disconnection disconnection O O OR OR O O O O + +
O O O O O CO2Et BrMg X
Easily accessible 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.
2 i) NaOEt, EtOH 1 Me O CO2Et 3 + 5 4 O CO Et ii) H O+ EtO C 2 3 2 CO2Et (1,5-dicarbonyl compound)
Mechanism Me Me + EtO (cat.) O H OEt Me CO2Et CO2Et H2C HC O OH + EtO CO2Et CO2Et - EtOH EtO2C EtO C CO2Et 2 CO2Et Regenerated catalyst Me
O
EtO2C CO2Et
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?
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
O O Ph b COOEt Path b Path a O COOEt COOEt + H2C Ph + Ph Ph Ph COOEt a COOEt Ph COOEt 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(CO 2Et) 2 is a softer nucleophile than CH 2COCH 3 and hence CH(CO 2Et) 2 has a tendency to give conjugate addition - product while CH 2COCH 3 gives carbonyl addition products. b) CH 2(CO 2Et) 2 gives corresponding enolate anion in the presence of weaker base but formation of enolate anion from
CH 3COCH 3 requires a strong base and drastic reaction conditions.
Consequently, starting materials of path b will be the best choice for the synthesis. Synthesis: COOEt HC O O i) Aq. NaOH O COOEt O Ph + + COOEt Ph Ph H ii) H3O Ph Ph Michael Ph COOEt Aldol reaction reaction Aq. NH4Cl
O Ph COOEt Ph COOEt Q. Analyze the following molecules and determine what starting materials would be required for their synthesis.
O O O CO H CO2Et 2 Ph 2. 3. 1. Ph CO2H
OMe CN
5 EtO2C 4. O O Ph O
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,
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;
Retrosynthesis: Cyclohexenone moiety
O O O O O CO2Et +
CO Et 2 CO2Et
Synthesis: O O O O EtONa/EtOH O (-H2O) CO2Et +
Heat CO Et CO2Et Aldol reaction 2 Michael reaction
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: O H CHO O O O dil KOH O O OH CH2 H Michael CHO O addition O O (A) Aldol reaction
HO H+ O O -H2O O O (C) (B)
Outline steps for the synthesis of following compound from the indicated starting materials. Also show the retrosynthetic analysis. O O CO2Et Me CO2Et ii) from acyclic molecules i) from O O
Synthesize the following molecule using Robinson annulation. Show retrosynthetic analysis also.
O Me Me Answer: Retrosynthesis: CO Et FGA 2 FGA CO2Et O O O Me Me O O O O Me MeOH Me Me Me Me +
O
Forward synthesis:
CO2H CO2Et 1. NaOEt, EtOH CO2Et + H /H2O (-CO2) 2. Heat O O -H2O O O O HO Me Me Me Me O Michael reaction