Synthon or Disconnection or RetrosynthesisApproach in Organic Synthesis
Presented by: Ms. Sarika MohiteB.Pharm , M.Pharm IIIrd Semester
Guided by:Dr . Amit G. Nerkar,
Associate Professor in Medicinal Chemistry,Sinhgad Technical Education Society’s
Smt. Kashibai Navale College of Pharmacy, Kondhwa(Bk),Pune-48, Maharashtra, India
� An analytical approach in organic synthesis in which the target molecule is broken into fragments through a series of logical disconnection to get the best possible &likely starting materials ( Synthon)
� An analytical operation that breaks a bond & converts a molecule into possible starting materials
� It is exactly the reverse of chemical synthesis .Therefore also called Reterosynthesis
�Terminologies
Disconnection – An operation involving breaking of bonds between atoms
Synthon – An idealized fragment usually an ion or radical obtainedby disconnection
Reagent – Actual comp (chemicals) used in practice for a synthesise.g.- Synthon-Me⁺
Reagent-Me2SO4
FGI or FGE – Functional group interconversion or functional group equivalence usually written on double arrow
This means substitution of functional group by another one equivalent to it.e.g.- -COOH -CN
-NH2 -NO2-Cl -OH
�Basic rule �Disconnection of a bond should be such that stable fragment ion are obtained e.g. two mode of disconnection A&B
FGI
C CO2N RC
-O2N C
+R
C+
O2N C-
R
A
B
A will be preferred as carbocations are stabilized by electron donor gr. R,OR ,etc, While carbinions are stabilized by electron withdrawing gr. NO2,CN,COOR etc.
� Number of fragments generated through a disconnection should be as minimum as possible
R R'
O O
R
C+
O
CH 2-
R'
O
R
O
OC 2H 5
R'
O
halo
R'
O
O H
R'
O
O
R'
O
1
2
3
45
+
R R'
O O
R
C+
O
CH2-
R'
O
R
O
R'
O
12
� Always a C-hetero atom (O,S,N) bond is broken ,with the electron pair being transferred to the heteroatom (as heteroatom are more electro negative than C they can accommodate the electron pair )
C N C+
N-
C Cl NH
.. ..
..
� Some time a disconnection doesn’t generate sufficient stabilised fragments butsuch fragment can be obtained using FGI or introducing additional electronwithdrawing & removing them .
R NH2 R
CH2+
NH3
R CH2+
R
Cl
NH2
CH2NO2
R X
(not sufficiently stabilised)
( introduction of electrowithdrawing groups
a
b
-
-
R-O-H
RNH2
R-CH2
( vely charged alkyl)
R-CH2-X(alkyl halide)
R-CH2-OH(alcohol)
R-CH2OC2H5(ether)
R-C=O( vely charged acyl) R-CO-X(acylhalid)
R-CO-OH(acid)
R-CO-OC2H5(ester)
R-CO-O-CO-R(anhydride)
O2N-CH-R O2N-CH2-R
R-O
R-NH
+
+
-
-
-
� The –ve & + ve fragments generated by disconnection are replaced by recognizable &Meaningful chemical entities.
- vely charged fragments are considered equivalent to their protonated species .
CH3CH3
O
Cl CH3
O
CH3
BaseCH2
-
O
CH3
O
OH
O
CH3
e.g OH
O
CH3
CH2
-
O
CH3CH2
+OH+
CH3
O
CH3
Cl
OH
C-
CH3 C+
CH3
O
OH
OH
OH OH
O
+
CH3CH3
O
Cl
�Examine relationship between gr. i.e, which gr. is the proper directing gr.( disconnect it last ) to get the target molecule here thus order is important.e.g.
�The most electron withdrawing gr. to be disconnected first ( i.e. to be added last in the analysis ) e.g.
Analysis
Synthesis
�If FGI is needed do it at an appropriate stage to get the desired effect on orientation
Here CCl3 is meta director , but it FGI ,CH3 is P- directional. Therefore, do FGI prior to C-Cl disconnection.
Analysis
synthesis
�Avoid sequences that may lead to unwanted reaction at other sites of the molecule
Therefore b to be adopted as nitration of benzaldehyde may lead to side reaction i.e.,Oxidation CHO COOH
CH3
b
Toluene
�For compound consisting of two parte by heteroatom ,disconnect next to the heteroatom e.g.- Chlorbeniside
Analysis
Synthesis
�Multiple step synthesis –avoid chemo selectivity problem
�In this structure with two ether & an amine? functional gr. it requires several disconnection to take it back to simple comp. The question is which do we do first?
�Here there are four reasonable disconnection one at each of the ether gr. ( a, b) or on either side of the amine.( c, d)
�Both a & b posses problem of chemo selectivity as it would behard to alkylate the phenol in the presence of basic nitrogen atom. In between c & d , c appears to be the better choice because the next disconnection after d will have to be an alkylation of O in the presence an NH2 gr.
�To avoid chemo selectivity problem like this , we want to try & introduce reactive gr. late in the synthesis.
�Protection allows us to over come simple problem of chemo selectivity .� It easy to reduce the keto - ester 1 to alcohol 2 with nucleophilic reagent such as NaBH4 that attack only the more electrophilic ketone.
1
2
3
�To making alcohol 3 by reducing the less electrophilic ester is not so easy but protection of ketone as an acetal 4 a functional gr. that does notreact with nucleophiles allows reduction of the ester with more nucleophilic LiAlH4.
1
4
53
Protecting groupOH
OH
�Qualities needed in the protecting group
�It must be easy to put in.
�It must be resistant to reagents that would attack the unprotected functional group.
�It must be easily removed
Protecting group
Ethers and amides as protecting group
�Protection of alcohol and amides look simple.
�Methyl ether & simple amides are easy to make &are very resistant to a wide variety of reagent.
�These protecting gr. used when the molecule is robust enough to take the deprotection condition.
� If aniline is brominated the 2,4,6-tribromo derivative is formed.
� The yield is quantitative but we are more likely to want mono- bromination protection is needed against over reaction .
�The amide is easily made ,bromination goes only in the para position & the hydrolysis Does not destroy the benzene ring.
Protecting group
� Achilles heel strategy
�Achilles heel for an ether is commonly the DHP gr. that make the ether into an acetal . Dihydropyran (DHP ) 1 , is protonated an carbon 2 to give the cation 3 that captures the alcohol to give the mixed 4 acetal.
�After the reaction the hydrolysis needs only the weak aq. acid used for acetal.�The secret is that the weak acetal bond (b in 5) is cleaved rather than the strong ether bond ( a in 5)
1 2 3 4
5
Protecting group
Chemoselectivity
If a molecule has two reactive group & we want to react one of them & not the other we need chemoselectivity.� If 2 group have unequal reactivity , the more reactive can be made to react alone.
The amide 2 is paracetamol the popular analgesic. Amine are more Nucleophilic than phenol so reaction with acetic anhydride gives the amide we want without any of ester 3. The aminophenol 1 can be made by methods.
OH
NH CH3
O
NH2
OH OH
N+
O-
O
OH
C - N nitration
FGI C - N
amide
NH2
OH OH
NH CH3
O
OH
OCH3
O
+
?
The synthesis is straight forward. Nitration of phenol needs only dilute nitric acid & the reduction is best carried out catalytically.
OH CH3
NO2 NH2
OH
OH
NH CH3
O
HNO3
dilute
H2/ Pd/C
Ac2O
�If one functional group can react twice the product of the first reaction will compete with reagent .The reaction will stop cleanly after one reaction only if the Starting material is more reactive than the product.�Reaction of alkyl halide with NaSH or Na2S can not usually be made to stop after one alkylation as the anione of the first product is at least as nucleophilic as HS- or S2- .this is obivous in reaction with Na2S.Less obviously with NaSH the first reaction gives 1 gives the thiol 2 but this is in equilibrium with RS- & a second displacement 3 give the sulfide 4.
Br R SH- R SH H
+S+ R S
-R Br
S R
R
1 23
4
�Problem from gidelines 1& 2 may be solved by protecting group
�If we want to react the less reactive of two functional group we protect the more reactive .� If we want a reagent to react once when it could react twice we protect reagent. a protection group is something added to a functional group that reduces or eliminate unwanted reactivity .�e.g amino acid chemistry . The amine is more nucleophilic than carboxylic acid , so if we want to use the carboxylic group as a nucleophile , we must protect the amino group to Benzyl chloroformate 1 is often used in this way . It cleanly acylate the amino group to give the carbamate 3 if compare 1& 2 the carbonyl group be coming less electrophilic .
Ph
O
Cl
O
+ NH2
COOH
RPh
O
O
NH
COOH
R
12
OR
O
NH
COOH
R
3
One –Group C-X disconnection
� We disconnect a bond joining the heteroatom (X) to the rest of the molecule : a C-O ,C-N,C-S disconnection .
� The corresponding reaction are mostly ionic involving nucleophilic displacement by SN1 ,SN2 or carbonyl substitution with amine, alcohol, and thiols on carbon electrophiles.
� The normal polarity of disconnection 1 will be a cationic carbon synthon 2 & anionic heteroatom synthon 3 represented by acyl or alkyl halides 4 as electrophiles and amine , alcohol or thiols 5 as nucleophiles.
R X R+
+ X- RHal + HX
12 3 4 5
synthons reagents
Synthesis of Ethers
�The question of which bond to disconnect can be much more significant in thesynthesis of ether .with many ether , like the gardenia perfume compound 1 , it doesn’t matter much . The starting material will be an alcohol 3 or 4 and alkyl halide 2& or 5
�The reaction will be carried out by treating the alcohol with a base strong to form the anion – sodium hydride is favourite as the hydride ion ( H- ) is extremely hard & act only as a base , never as a nucleophile . Either chloride is available , both react in SN2 reaction . We prefer route ‘a’ as benzyl chloride 2 is more reactive & can not undergo elimination whlile 5 just might .
Ph O
CH3
CH3
a b
Ph Cl OH CH3
CH3
+
Ph OHCl CH3
CH3
+
a
b1
2 3
45
OH CH3
CH3
O-
CH3
CH3
-Ph Cl
OH CH3
CH3Ph O
CH3
CH3
3
6
2
1
One –Group C-X disconnection
Synthesis of sulfides
� Unsymmetrical sulfide 1 need the same disconnection we have just used for ether. The anion 3 of a thiol 4 will combine with an alkyl halide 2 to make a new C-S bond .
� The reaction is much easier with sulfur . Thiols are more nucleophilic towards saturated carbon than are alkoxides and the risk of elimination is much less.
� The acaricide (kill mice & ticks) chlorobenside 5 is disconnected to give an acidic thiophenol 6 & reactive alkyl halide 7. the synthesis merely combine these two in ethanol with NaOEt as base.
S R2
R1
C - S R1
halogen + S-
R2
SH R2
1 2 3 4
Cl
S
Cl
SH
+
Cl
5 6 7
One –Group C-X disconnection
1,1 dix disconnection
One gr. C-C disconnection I -Alcohol
XOH
R
OH
X
R
-
-
+
+
O
O
1,2 dis
C- X
C - C
1
2
1
2
5
6
76
R P
OH
O
(OR)2
1,1 dix
C- X R
O
+ P-
O
(OR)2-
1 2
R1
R2
O H
R1
O
+ R 2
-
C - C
34
� For compound with two heteroatom's joined to the same carbon , we used a 1,1 diXdisconnection 1 removing one heteroatom to reveal a carbonyl compound , here an aldehyde & a heteroatom nucleophile 2 replacing the heteroatom by R2 , disconnect in the same way to reveal the same aldehyde & same nucleophilic carbon reagent 4
probably R2MgBr.
� For compound with 1,2 relationship we used an epoxide 6 at the alcohol oxidation level in combination with a heteroatom nucleophile disconnecting the corresponding C-Cbond 7 ,we use the same epoxide & carbon nucleophile such as RLi or RMgBr.
X
R
O
XCH2
+
R
O
+
1,2-diX
X- C
-
89
R1
R2
O
Br R1CH2
-R2
O
C - C
+10
11
X CH3
O
X-
+CH2
CH3
O
C - X
12 13
1 - 3 diX
R CH3
O
CH2
CH3
O
R-
+C - C
1413
The same 1,2 dix relationship at the carbonyl level was 8 disconnected to give carbon electrophile 9 ,probably an α –bromoketone & heteroatom nucleophile.we generally Preferred nucleophilic heteroatoms but we can use nucleophile or electrophilecarbon atoms whichever better . Here we can should much rather use the
nucleophilic carbon synthon 11 as it is an enolate.
The 1,3 diX relationship 12 was quickly recognized as conjugate addition to the enone .The corresponding C-C disconnection 14 uses the same enone 13 but the nucleophiliccarbon species should be a copper derivative ; RCu, R2CuLi or RMgBr with Cu(I)Br.
One gr. C-C disconnection I -Alcohol
CH3
CH3
OH
CH3
CH3 CH3
CH3
MgBr +CH3 CH3
O
+CH3 Li
CH3
CH3
CH3
OH
C - C
C - C
15
1617
R
O
R
OH
CHO OHBr
FGIFGIC - C C - C
1,1 C-C disconnections- The synthesis of alcohol
Aldehyde & ketone The simplest route to aldehyde & ketone using the same strategy is oxidation of an Alcohol. So the analysis involve FGI back to the alcohol & then a C-C disconnection of one of the bond next to the OH group.
One gr. C-C disconnection I -Alcohol
Disconnection 3 shows that any alcohol may be disconnected at a bond to theOH group .Isomeric alcohol 15 &17 can both be made from acetone using perhaps a Grignard reagent 16 in the first case & available BuLi in the second.
R COOH R CN R Br + CN-
R MgBr + CO2R Br
12
3 4 1
C - CFGI
C - C
FGI
Carboxylic acid
Ph CH3
OH
PhMgX + O
CH3CH2
CH32 C - O
12 3
1,2 C - C
� Same disconnection 3 can be used for carboxylic acids with CO2 as the electrophile for a Grignard reagent 2 .Switching polarity by FGI to the nitrile 4 ,the same disconnection now uses cyanide ion as the nucleophile but the same alkyl halide 1 was used to make the Grignard reagent.
1,2 C-C disconnection the synthesis of alcohol
One gr. C-C disconnection I -Alcohol
Alcohol 1 is used in perfumery & can be disconnected at the next butone bondto the alcohol group with the idea of using the epoxide 2 made from the but-1-ene 3
CH3 CH3
CH3
OH
?ab
cd
e
Only one of the five bonds 1a is good choice & for two reasons .To achieve the greatest simplification in our disconnection so that get back simple starting material . This make the synthesis as short as possible.
So disconnect bonds that are-� Towards the middle of the molecule. This breaks the molecule into two reasonably equal parts & is much better than simply lopping one atom of the end. � At a branch point in the molecule : this is more likely to give simple straight chain material. Here we get the aldehyde 2 & grignard reagent 3 coming from the straight chain halide 4. both 2 & 4 are commercially available.
CH3 CH3
OH
CH3
CH3
H
O
+ MgBr CH3
Br
CH3CH3
FGI
C - C
1a
2
3
4
General strategy – Choosing a disconnection
N
OH
CH3
OCH3
N
O
CH3
+OCH3
MgBr
C - C
1,1
56 7
�The series of drug based on bicyclic structure 5 has an excellent disconnection between the two ring.
Symmetry� The symmetrical tertiary alcohol 1 can be made from two molecule of the Grignard reagent 2 & one of ethyl acetate. Then back to the alcohol 3 by FGI & a connection at the branch point give starting material .
CH3 CH3
CH3
CH3
CH3
OH
CH3 MgBr
CH3
CH3 OH
CH3
O
+
CH3
CH3MgBr
C - C
C - C
FGI
+MeCOOEt
1,1-
1,2-
12
3
4
5
General strategy – Choosing a disconnection
Summary of guideline for the good disconnections�Make the synthesis as short as possible �Use only disconnection corresponding to the known reliable reaction .�Disconnect structural C-X bonds first & try to use two gr. disconnections�Disconnect C-C bonds using FGI in the molecule.a) aim for the greatest simplification , if possible
disconnect near the middle of the mol.Disconnect at a branch point Disconnect ring from chain
b) Use symmetry (if any)�Use FGI to make disconnection easier �Disconnection back to available stating materials or ones that can easily be made.
General strategy – Choosing a disconnection
R X
O
R OCH3
O
+ X-
1,1 dix
C - X
1
R1
R2
O
R1 OCH3
O
+ R2
-
C - C
2
One group C-C disconnection II- carbonyl compound
Disconnection deals with carbonyl compound chiefly aldehyde & ketone by tworelated disconnection . Started by comparing the acylation of heteroatom's by acidderivatives such as ester .
1,1-dix disconnection
1,2 – dix disconnection XR
O
X +CH2
+R
O
-1,2 dix
R1
R2
O
Br R1 +CH2
-R2
O
4
6
C - C
3
5
Synthesis of aldehyde & ketone by acylation at carbon
The disconnection 2 is not useful because as MeO is the best leaving groupfrom the tetrahedral intermidiate 7, the ketone 2 is formed during the reactionThe ketone is more electrophilic than the ester so it reacts again & the product is tertiary alcohol 8.
R1 OMe
O
R1
OMe
R2O-
R1 R2
O
R1
R3
R2OHR2MgBr
or RLi
RLi
R3MgBr
7
2 8
One group C-C disconnection II- carbonyl compound
Carbonyl compound by alkylation of enols
�Disconnection 1 again uses the natural polarity of the carbonyl group but atthe next bond 1. So we use some enolate derivative 2 in an alkylation reaction.�The problem is that ketone is it self electrophilic & the self condensation by the aldol reaction is generally preferred to alkylation.
�First of all to convert the ketone 3 completely into some enolate derivative so that there is no ketone left for self condensation .�Lithium enolate 4 & anians 6 of 1,3 dicarbonyl compound 5 act as the enolate anion of acetone.
R1
R2
O
1,2 C - C R1 Br +CH2
O-
R2
1 2
CH3R2
O
R1 Br
baseTM 1
synthesis (no good)
3
CH3 CH3
O
CH3 CH3
OLi
i-Pr2NLi
CH3
CH2
COOEtCH3 OEt
OO-
3
4
5 6
EtO-
One group C-C disconnection II- carbonyl compound
Carbonyl compound by conjugate addition
�Conjugate addition of a heteroatom to the enone 2b give the 1,3 relationship in 1 & the same process with a carbon nucleophile gives 3.
� we can use either organo-lithiums or Grignard reagent as the carbon nucleophilesbut we need copper (I) to ensure conjugate addition without Cu (I) both
nucleophiles are inclined to add directly to the carbonyl group.� Disconnection of the ketone with conjugate addition in mind could remove the
vinyl group or the methyl group. There are two reasons why we prefer ‘a’ .� The addition is likely to occur from the opposite face of the molecule to he COOEt
group & that is why we want the vinyl group .
X R
CH2
1
X +CH2 R
O
2
1,3 dix
C - X
-
R1
R2
O R1-
+CH2 R2
O
32
Corresponding c-c disconnection
One group C-C disconnection II- carbonyl compound
� Conjugate addition to 3 might occurs at the β position but it could equally well occurs at the very exposed δ position .
� The starting material is also available hagemann’s ester
O
CH3
CH2
EtOOC
b
a
EtOOC
CH2OOMgBrCH2
EtOOCC - C
a
C - C
b
δβ
One group C-C disconnection II- carbonyl compound
� How to react one specific part of a single functional group & no other .This is Regioselectivity.
� We have seen that anions of phenol 2 are alkylate at oxygen to give ether 3. while enolate anions 5 are alkylated at carbon to form a new C-C bond 6.
OH O-
Base CH3
OCH3
o- alkylation
c-alkylation
CH3
O
COOEt
CH3
O-
COOEt
Base CH3 CH3
O
COOEt
CH3
1 2 3
4 5
6
Regioselectivity
Carbon nucleophiles in conjugate addition
� The very basic & aggressive nucleophilic organo-lithiums tend to do direct additionto all α ,β unsaturated carbonyl compound
� If we react 1 with grignard reagent with Cu (I) catalysis we get 2 as product thelithium enolate 3 giving us the opportunity to add an electrophile to make 4
If the electrophile is proton , the product still 2.
R1 R2
R O
R1R2
O
R1 R2
R OLi
R1 R2
R OLi
E
123
4
RMgBr
cat CuBr
R2 CuLi
E+
Regioselectivity
Regioselective alkylation of ketone
� Lithium enolates & 1,3 dicarbonyl compound both will help us to solve the Regioselectivity problem in the alkylation of unsymmetrical ketone.
� Suppose we want make 1 at first sight it appears that we must alkylate an unsymmetrical ketone on the more substituted side but if we remove the benzyl group & add our activating COOEt group to give 2 it is clear that we can make this by another alkylation & the activating group ill promote both.
CH3
CH3
Ph
O
CH3
CH3
O
COOEt
CH3
O
COOEt
+
Br Ph
+
PrBr
1
23
Regioselectivity
CH3
O
COOEt
3
NaOEt
PrBrCH3
CH3
O
COOEt
NaOEtBnBr
CH3
CH3
Ph
OCOOEt
NaOH
H2OH+
heat
CH3
CH3
Ph
O
2
1
� Benzyl is the more reactive bromide so it makes sense to add it last since making the quaternary carbon will be difficult.
Regioselectivity
Regioselectivity in nucleophilic addition to enones
� The problem of getting direct (1,2) or conjugate( 1,4 or michael ) addition to α ,β unsaturated compounds such as enones 1 can be solved without finding abstrusstrategies by choice of reagent.
R1 R2
Nu O
R1R2
O
R1R2
OHNu
12 3
Nu-
1,4-or Michael or
conjugate addition 1,2- or direct addition
Nu-
Regioselectivity
� Disconnection is often best found by reverse reaction mechanism . you may draw the arrow either clockwise or anticlockwise but one start from the alkene. It makes sense to draw this arrow first. � The disconnection is 1 for the general case & 2 for specific case, revealing a diene 3 & dienophile 4. These reagent 3 & 4 need. Only to be heated together in a sealed tube to give 2 .
1
2 3 4
� This is two group disconnection because it can be carried out only when two features are present in the target molecule .
� The cyclohexene ring & electron withdrawing group outside the ring & on the opposite side to the alkene. The relationship between these features must be recognized.
Two gr. C-C disconnection I –Diels Alder Reaction
Stereospecificity� The reaction occurs in one step so there is no chance for either diene or thedienophile to rotate & the stereochemistry of each must be faithfully reproduced in the product .The two Hs in 1 are cis because they were cis in the starting anhydride .The two Hs in 3 are trans in the diester 2.
12 3
Two gr. C-C disconnection I –Diels Alder Reaction
FGI on Diels – Alder Product The cyclic ether comes from the diol that can be made by reduction of various Diels –Alder adducts such as the anhydride.
Two gr. C-C disconnection I –Diels Alder Reaction
O
H
H
CH3
H
H
CH3OH
OHO
H
H
CH3
O
O
O
O
O
+CH2
CH2
CH3
ether
C - O FGI
D A
Two gr. C-C disconnection I –Diels Alder Reaction
CH2
CH2
CH3
O
O
O
heat
O
H
H
CH3
O
OH
H
CH3OH
OH
O
H
H
CH3
LiAlH4
TsCl
NaOH
Synthesis
� Target molecule of two main type 1 Hydroxyketone & 1,3 or β – diketones 4 bothhave a 1-3 relationship between the two functionalised carbons both can be disconnected at one of the C-C bonds between functional group to reveal the enolate 2 of one carbonyl compound reacting with either an aldehyde 3 oracid derivative 5 such as an ester.
1 23
4 2 3
Most stable carbonyl comp
Most nucleophilic enols or enolates
Most electrophilic carbonyl comp
Most stable enolsor enolates
Two gr. C-C disconnection II: 1,3 –Difunctionalised compound
5
β - Hydroxy carbonyl compound : The Aldol reaction
� With the compound 1 only one of the two C-C bonds in worth disconnecting , the one next to the hydroxyl carbon .
�A simple example without any selectivity is ketone 6 which disconnects to the enolate 7 & the ketone 8. It is easy to see that 7 is enolate of 8 so this is a self condensation .We simply need to reduce a small amount of enolate 7 in the presence of much unenolised ketone 8 & the reaction will occur .
6 7 8
Two gr. C-C disconnection II: 1,3 –Difunctionalised compound
R1 R2
O O
R1CH2
-
OCH2
+
R2
O
1,5 -diCO
C - C
CH2
R2
O
d2enolate
a3
Two gr. C-C disconnection III – 1,5 difunctionalised compound conjugate addition & Rabinson annelation
1
2
3
� The odd number relationship means are we still use Synthon of natural polarity .
� The 1,5 –diketone 1 disconnect to a d2 synthon, an enolate & an a3 synthon 2 it represented by the reagent 3. The conjugation in the enone makes the terminalcarbon atom electroplilic.
H OEt
O O
a
bH CH2
-
O
CH2
OEt
O
+EtO CH2
-
O
CH2
H
O
+1,5 -diCO
C - C
1,5 -diCO
C - C
Specific enol equivalence good at michael addition 1,3 –Dicorbonyl compound
If we want to make 1 we have a choice between adding an enolate equivalent ofaldehyde 5 to an unsaturated ester 4 or an enolate equivalent of ester 3 to an unsaturated aldehyde 2. We prefer the first 1a as unsaturated ester 4 ismore likely to do conjugate addition .an enamine would be good choice for 5
1 2 345
Two gr. C-C disconnection III – 1,5 difunctionalised compound conjugate addition & Rabinson annelation
O
OCH3
CH3
O
O
OCH3
12
3
46
1,5 diCO
CH3
CH2
O
+
C-
O
O
CH3
analysis
12
3
4
3
5
Robinson annelation
�Combining aldol & michael reaction in one sequence is very powerful, particularly if one of the reaction is cyclisation .The Robinson annelation makes new ring in the compound like 1 that were needed to synthesize steroids.�Disconnection of the reversal tri -ketone 2 having 1,3 & 1,5 dicarbonyl relationship �1,3 disconnection would not remove any carbon atom but the 1,5 at the branch point gives a symmetrical β -diketone that should be good at conjugate addition.
Two gr. C-C disconnection III – 1,5 difunctionalised compound conjugate addition & Rabinson annelation
synthesis
O
O
CH3 CH3R
CH2
CH3
O
O
OCH3
12
3
45
R2NH
O
O
CH3
OHH2 O
O
OCH3
Two gr. C-C disconnection III – 1,5 difunctionalised compound conjugate addition & Rabinson annelation
NH
CH3CH3
COOR COOR
Ar
2 C - N
enaminesCH3 CH3O O
COORCOOR
Ar
COOR
Ar
CH3 O
COOR
O
CH3
+
1 2 34
Heterocycles made from 1,5 dicarbonyl comp
� A family of calcium channel antagonist based on the general structure 1 is widely used to combat high blood pressure.
�Disconnecting the structure C-N bonds we discover symmetrical 1,5 diketone 2 So disconnection of either appropriate bond give the same starting material &
enone 3 an acetoacetate ester 4
Two gr. C-C disconnection III – 1,5 difunctionalised compound conjugate addition & Rabinson annelation
R1
R2
O
O
R1C
-
O
+ X R2
O
1,2-diCO
R1
R2
O
HO
1,2-diCO R1C
-
O
+ H R2
O
CH3
O
CH3
CH3OH
FGI
Hydration CH
CH3
CH3
OH
1,2-diCOCH C
-
+ CH3 CH3
O
Two gr. disconnection IV -1,2 difunctionalised compound
� In simple case of 1,2 diketone ,or an a- hydroxy –ketone 4. there is one C-C bondbetween the functionalized carbon. so, while we can use an acid derivative 3 or
� An aldehyde 5 for one half of the mol, we are forced to use a synthon ofunnatural polarity, the acyl anion 2 for other half.
1 2
3
4 2
3
21
� The hydroxy – ketone that could come from the acetylenic alcohol by hydration & hence from acetone with the anion of acetylene acting as the acyl anion equivalent.
R1
R2
OH
OH
FGI
electrophilic addition R1
R2
witting
R1- CHO + Ph3P+
R2FGI
Br R2
Method from alkenes.
R1
R2
OH
H2N
R1
O
R2 CH3
CH3C - N
1,2 dix FGI
electrophilic addition
Many possible starting material
Epoxide give rise to many 1,2 difunctionalised comp such as 6 with control over stereochemistry . Reaction of the epoxide give the anti stereochemistry in 6 in contrast To the syn stereochemistry in 1.
Two gr. disconnection IV -1,2 difunctionalised compound
OH
OH
OH
NH
CH3
CH3
C - N
reductive amination
OH
CHO
OH
OH
OH
OH
OH
CH3
Functionalisation1,2 diCO1
2
3
α Functionalisation of carbonyl compound
Metaproterenol 1 is an adrenaline analogue used as bronchodilater .the might be inserted by reductive amination on the aldehyde 2 & this might aΑ –functionalisation of the available ketone .
Two gr. disconnection IV -1,2 difunctionalised compound
R1
R2
O
O
R1 CH2
-
O
+
CH2
+R2
O
R1 CH2
O-
X
R2
O
enolate
1,4 diCO
1 2
3
4
5
Two gr. disconnection V: 1,4 difunctionalised comp.
�The problem of unnatural polarity also arise in making C-C disconnection for the synthesis of 1,4 difunctionalised compound. �If we start with 1,4 diketone 1, disconnection in the middle of the molecule gives a synthon with natural polarity 2 represented in real life by an enolate 4 & one of unnatural polarity, the synthon 3 represented by same reagent of the kind such as α - haloketone 5.
O
COOEtCH
-
O
+CH2
+COOEt
O- Br
H
H COOEt
1,4 diCO
12
3
4
5
Reaction of enol (ate)s with reagents for a2 synthons
A simple example would be the keto-ester 1 disconnect the bond at the branch point & that suggest the synthon 2&3 . The reagent for 3 can be bromoester 5 but we shall need to choose our enolate equivalent carefully .If should not too basic as the marked proton in 5 between Br & COO2Et are rather
acidic.
Two gr. disconnection V: 1,4 difunctionalised comp.
NH
OO
Ph
Ph
HOOC COOH
Ph
NC COOH
Ph
COOH+CN
-
2 C - NFGI
Conjugate addition of acyl anion equivalents
The anticonvulsant phensuximide 1 being an imide , comes from adicarboxylic acid 2 with 1,4 relationship between the two carbonyl gr. changing one to cyanide we get back to cinnamic acid as the available Starting material.
1 2
Two gr. disconnection V: 1,4 difunctionalised comp.
R1
R2
O
O
R1
C+
O
CH2
-R2
O
+
R2
O-
12
3
4
R2
OSiMe3
5
Direct addition of homoenolates�The same disconnection but of the opposite polarity requires some
acylating agent for synthon this no problemas we have various derivative at our disposal but the nucleophilic synthon 3 or homoenolate, is
another matter. �There is no stabilisation of the anione as drawn but if were
to cyclise to oxyanion 4, it would be rather more stable & there is evidence trapping with silicon to give 5 .
Two gr. disconnection V: 1,4 difunctionalised comp.
synthesis of 1,2 & 1,4diCO comp by oxidative cleavage.
If we wanted to add bromketones 4 to enolate 3 to make the 1,4 dicarbonylcompound 5 .We could not use a lithium enolate because it would be too basicno such difficulties exist in the reaction of enolate with allylic halides such as 2 Any enol equivalent will do as there are no acidic hydrogen & allylic halides are good electrophilic for the SN2 reaction.
CH3
CH3
CH2CH3
OCH3
CH3
CH2 CH3
CH3
O-
CH3
CH3
OCH3
CH3
OCH3
O
1
2
3
4
5
Reconnection ; joining the target molecule back up to something to reveal the precursor ,so consider the synthesis of the cis- enone 1 a structure found in insect pheromones, perfumes , flavourings . A witting reaction would ake the cis –alkenefrom phosphonium salt 2 but the ketoaldehyde 3 would need protection , perhaps as the acetal 4.
Reconnection
R
CH3
OPh3P
R
+
+ OHC CH3
O
O
CH3
OH
CH3OHC
1 23
4
witting
FGI
The problem is how to protect the ketone rather than the aldehyde & the answer is Protect it when the aldehyde is not there .Reconnection to the alkene achieves this & the ketone can be made by reaction of some enolate with allyl bromide.
O O
CH3OHC
4 O O
CH3
CH2
reconnect
O
CH3
CH2
CH2
Br+ CH3 CH2
-
O
FGI
C - C
Reconnection
O
O
O
O OHC
O
COOEt
COOEt O
COOEt
COOEt
Ph
reconnect
aldol
O
COOEt
COOEt
CH3
CH2
-CH3
O
Br COOEt
Br COOEt
+
12 3
45
6
The extraordinary polycyclic tetraketone 1staurone was made from 2 The 1,2 diCO relationship in 2 is an ideal candidate for reconnection in this style .
The aldol disconnection 3 reveals methyl ketone with 1,4 diCO relationship That could be made by double alkylation of some enolate of acetone with ethyl bromoacetate 6. the synthesis used benzyl acetoacetate for the double alkylation So that the benzyl ester 8 could be specifically cleaved by hydrogenation to give 4 . Condensation with unenolisable benzaldehyde is unambiguous & ozone does the rest.
Reconnection
CH3
CO2Bn
O
2(NaH)
Br COOEt2( )Me
O
COOEt
COOEtBnO2C
Me
O
COOEt
COOEt
H2Pd/C
O
COOEt
COOEt
Ph
PhCHO
base
OHC
O
COOEt
COOEt
2
78
4
3
O3Me2S
Reconnection
Two gr. C-C disconnection VI- 1,6 dicarbonyl compound
R1
R2
O
OR1
CH2
+
O
CH2
-R2
O+
?
12
34
5
6
1
2 3
�1,6 dicarbonyl compound disconnect in the middle we might be relived to see an a3 synthon 2 easily recognised as an enone in real life , but the d3 synthon 3 with unatural polarity, can cause problem ,so use reagent for 3 that does conjugate addition .
�Disconnecting else where is no help as the true difficulty is that the two carbonyl group are too far apart for this approach.
�Strategy of reconnection is needed the main strategy for the synthesis of 1,6 diCO compound
O
CH3
CH3
CH3
CH3
O
CH3
CH3
CH3
CH3
O
CH3
CH3
CH3
OHC1
2
3
4
56aldol
67
8
R1
R2
O
O
12
34
5
6
R1
R2
weaken ?
R1
R24 5
1
�Reconnect intramolecularly the marked atom C-1 & C-6 form a ring 4 & the bond between these atom must be made weaker than any other bond in the molecule Ironically we can do this by making it a double bond 5
�Bicyclic ketone made from the simple enone that had to be made. Aldol disconnection reveals the keto- aldehyde.
Two gr. C-C disconnection VI- 1,6 dicarbonyl compound
CH3
CH3
CH3
O
11
MeLi
H+H+
CH3
CH3
CH3
CH3
9
O3
Me2S
CH3
O
CH3
CH3
CH3
OHC1
2
3
4
56
CH3
O
CH3
CH3
CH3
KOHMeOH
synthesis
CH3
O
CH3
CH3
CH3
OHC1
2
3
4
56
CH3
CH3
CH3
CH3
CH3
CH3
CH3
CH3
OH
CH3
CH3
CH3
O
89 10
11
� This is 1,6 dicarbonyl compound so reconnection to the cyclohexene 9 is needed FGI & removal of the methyl group reveals a simple cyclohexanone 11 .
Two gr. C-C disconnection VI- 1,6 dicarbonyl compound
CH2
CH2
+CH2
O
R R
O
R
O
HOOC
HOOCO3
H2O2
12
34
The Diels –Alder Route to 1,6 dicarbonyl compound
� Normally we have to make the cyclohexene , need for oxidative cleavage & one ofthe best route to such is Diels – Alder reaction .Generalized example would beozonolysis of alkene .The product has a 1,6 relationship between two carboxylic acids.
� Since Diels –Alder adduct have a carbonyl group outside the ring ,the cleavage product also have 1,5 & 1-4 diCO relationship & would be a matter for personal judgment which of those should be disconnected instead if you choose that alternative strategy.
Two gr. C-C disconnection VI- 1,6 dicarbonyl compound
MeOOC
MeOOC OMe
OMe
H
H
OMe
OMe
H
H
O
O
O
H
H
OOO+
CH2
CH2
1
56
7
8
reconnection
1,6 diCO
FGI
D-A
� diester required for synthesis of the antibiotic pentalenolactone reconnecting the ester gives the cyclohexene . We must change the two ether group into carbonyl group & one starting material is , diels alder adduct of butadiene & maleic anhydride .
Two gr. C-C disconnection VI- 1,6 dicarbonyl compound
O
O
O
H
H 7
OMe
OMe
H
H
6
MeOOC
MeOOC OMe
OMe
H
H
5
LiAlH4
NaH,Mel
O3MeOH,H2O2
CH2N2
OO
O
O
COOHCH3
CH3
COOHHOOC
CH3
HO
CH3
OH
COOH
COOHHOOC
CH3
O
CH3
COOH
2(C - O )
ester
1,1 dix
9 10 11
� The synthesis followed this pattern with the ether 6 being made immediately after the reduction of 7 & the ester made with diazomethane CH2N2 after oxidative cleavage.
� The bicyclic double lactone used as precursor for all four heterocyclic ring in synthesis of Vit. B12 .disconnection of both lactones reveals a ketone .
Two gr. C-C disconnection VI- 1,6 dicarbonyl compound
COOHHOOC
CH3
O
CH3
COOH
11
COOHCH3
CH3 O
CH2
CH2
1
+
CH3
CH3 CH2
COOHreconnect
1,6 diCO
1
2
3 4
5
6
1
2
3
4
5
6
diels
alder
1213
� The ketone 11 in fact has 1-4,1-5 & 1-6 relationship & if we redraw in 11a to see 1,6 relationship clearly Being careful to get the sterochemistry right,we can reconnect to the cyclohexene 12 & hence , byDiels- Aldrdisconection , find the reactive dienophil disconnection 13 .The methyl & czrboxylic group are cis in 12 & must be cis in 13
Two gr. C-C disconnection VI- 1,6 dicarbonyl compound
Cl
OEt
OEt
MeHN
OEt
OEt
Cl
OEt
OEt
N OEt
OEt
EtO
CH3
OEt
1
2
3
MeNH2
Introduction to ring synthesis: Saturated Heterocycles
� Synthesis of butanone by reaction of the primary alkyl chloride with MeNH2 was likely to give a poor yield . The problem is that the product 2 is also a nucleophile &will react at similar rate with alkyl chloride as does MeNH2. The reaction is intermolecular & so bimolecular.
Cyclisation reaction
MeHN Cl
N+
CH3H
N
CH3
base
Cl
MeHN..
� The very similar reaction of 4 gives exclusively the pyrrolidine 5.The reaction 6 is now intramolecular a unimolecular cyclisation in fact & is greatly preferred to any bimolecular processes.
Introduction to ring synthesis: Saturated Heterocycles
CH3 Br
CH3
ClCH3
O
CH3
CH3
ClOH
CH3
O
CH3CH3
CH3
CH3
CH3
CH3
CH2
NaOHMg.OEt
1 2
3
4
Three membered ring
The simple formation of epoxide 3 by the action of peroxyacids such asmCPBA on alkenes 4.They can equally well be made by cyclisation of chloro- alcohols 2 as in the Cornforth addition of a Grignard reagent to an α-chloroketone & cyclisation in base.
Introduction to ring synthesis: Saturated Heterocycles
Ph
N
Ph
NH
X
Ph
O
O
+PhCHO
C - N C - N
Reductive amination
aldol
1
2
3
4
Four membered ring
Upjohn’s analgesic & antidepressant tazadoline 1 contain a foue member cyclic amine , an azetidine , simple disconnection of C-N bonds give 2 & then enone 3,the aldol product from cyclohexanone 4 & benzaldehyde.
Introduction to ring synthesis: Saturated Heterocycles
NH
O NH2 O
X
NH3
+COOEt
Cl-
NH2 O
OEt
NH
O
base
1 2
3
4
1
Five membered ring
Lactam come from acid derivative . The compounds such as amino ester arenot stable as the free amine but are usually isolated as salt such as hydrochloride .When treated with base, give the free amine which promptly cyclise to the lactam.
Introduction to ring synthesis: Saturated Heterocycles
S 2 C - S X X
OH OH Cl Cl
SPCl3 Na2S
The unsaturated rings does have a double bond but it is not next to the heteroatom. It is an allylic rather than a vinylic sulfide so two disconnection at the alcohol oxidation level suggest the doubly allylic starting material
Introduction to ring synthesis: Saturated Heterocycles
NH
O
CH3
CH3
CH3
CH3CH3 CH3
CH3CH3
O
CH3
CH3
O
CH3
CH3
O
CH3
CH3
O
2( 1,3 diX)
2( C - N)
2 aldol
O
CH3
CH3
O
CH3
CH3 CH3 CH3 CH3CH3CH3
O
NH
O
CH3
CH3
CH3
CH3
O
CH3
CH3NH3
CaCl2
Six membered ring
Intramolecular reaction comes in the synthesis of tetramethyl piperidine ,removal of nitrogen with conjugate addition of ammonia to the dienone opens the possibility of a double disconnection to reveal three molecule of acetone.
Introduction to ring synthesis: Saturated Heterocycles
O
NH
O
RR O
NH2
COOR R OH
NH2
+COOR
Cl
O
R
+ NH3
The morpholine derivative 1 has an obvious amide disconnection to 2 & less obvious 1,2 diX disconnection at the ether to 3. This is obviously an epoxide 4 adduct with ammonia.
Introduction to ring synthesis: Saturated Heterocycles
Seven membered ring
N
Pr
COOMe
Cl Cl
N
CH3
COOMe
Pr
Cl CH3
F
NH
MeOOC
Pr1
2 3
4
+
aldolSnAr
�The compound 1 with only one nitrogen in the ring are more interesting synthetically & are needed for an anti-HIV drug .�Initial C=C disconnection is followed by C-N disconnection between the ring &nitrogen 2. This is possible because nucleophilic aromatic substitution work wellon aryl fluorides with ortho or para electron – withdrawing gr. such as the aldehyde 3.
Introduction to ring synthesis: Saturated Heterocycles
N
O
Pr
Cl CH3
F
3
NH
O-OC
Pr
Cl
N
CH3
COOMe
Pr
2
HClNaOH
Mel,K2CO3
MeONa
(MeO)2CO
N
Pr
COOMe
Cl
6
1
The compound 4 are unstable & cyclise rapidly to the lactam . So lactam 5 used as starting material opening the lactam with NaOH gave the anion 6 of 4 that added to 3 to give 2 & hence that aldol product 1 in base.
Introduction to ring synthesis: Saturated Heterocycles
NH
O
NH2
COOH
CH3
CONH2
C - Namide
a C - N
b
reductive amination
N
O
CF3
COOH
OHCHO
COOH
COOH
H
+F3C
NH2
1
2
3
The related benzazapinones 1 can be made by formation o C-N bonds in Two different ways. Amide formation from compound like 2 is not surprising but reductive amination between the amide nitrogen & the aldehyde in 3 is atestament to the efficient of cyclisation even when a seven- membered ring is theproduct.
The intermediate 1 as an intermediate in the synthesis of a drug for the treatment of osteoporosis ,chose the double disconnection because we had way making singal enantiomers of the diacid 2
Introduction to ring synthesis: Saturated Heterocycles
CH3 CH3
O
CH3
CH3
OHCH3
OH
CH3
CH3
CH3
CH3O
CH3
Rearrangement in synthesis
R ROH O
?
The typical pinacol formed from acetone is important because it rearranges in acid to give a tertiary alkyl ketone known as ‘ pinacolone’ . The A key step is methyl migration as one of the OH group is lost.
The crowded alkenes can be made by dehydration of alcohol & hence from the ketone &RLi or RMgX as ketone has a t- alkyl substituent it is candidate for the pinacol approach.
Pinacol rearrangement
O
a b
OH
OH
OH
OH
O
a
b
PinacolFGI
�The best way to do the disconnection is to reverse rearrangement & there aretwo way to do this a & b .�The diol can be made by pinacol dimerisation of cyclopentanone while diol would be the product of dihydroxylation of the alkene.
Rearrangement in synthesis
O OH
OH
O
BuOH
Bu
SoCl2
Pyridine
BuLi
H+H+Mg/Hg
THF
The pinacol dimerisation use to make the diol & the pinacol rearrangementto make the spirocyclic ketone.
Rearrangement in synthesis
O O
Cl
COOR
O
H OR
O-
Cl
RO halogenationRO
-
The Favorskii rearrangment
Halogenation of cyclohexanon gives the α –chloroketone treatment of such compound with nucleophilic alkoxide gives ring contracted ester .The enolate of cyclise to give an unstable cyclopropanone that reacts immediately with alkoxide to cleave one of the weak C-C bonds in the three membered ring.
Rearrangement in synthesis
R1 R2
O
R1 R2
NO2
R1 R2
NH2
base,(o)
or H2O+TiCl333
or H2O + H2SO4
H2,Pd/C
12
3
Aliphatic Nitro compound in synthesis
H2N
CH3
CH3
NH2
O2N
CH3
CH3
NO2
O2N
Cl+
CH3 NO2
CH3
FGI C - C
1 2 3
4
Reduction of nitro compound
Few aliphatic nitro compound are wanted as target molecule in their own right but the nitro group is important in synthesis it can be converted into two functional group in great demand : amine 3 , by reduction ,and ketone 1 by various form of hydrolysis
The sequences of alkylation followed by reduction gives an amine and the special advantage of this strategy is that it can lead to t- alkyl amines .The appetite Suppressant 1 can be disconnected next to the tertiary center after the amine arechanged to nitro compound 2 . 2-Nitro propane 4 is available.
CH3 NO2
CH3
O2N
Cl O2N
CH3
CH3
NO2
H2N
CH3
CH3
NH2
1
NaOEt
RaNi
H2
The synthesis uses alkylation by a benzylic halide & the reduction of both nitro group is done catalytically with Raney nickel in the same step
Aliphatic Nitro compound in synthesis
CH2 CH2
OH
CH
OH
O
+C
-CH
FGIFGI
C - C1
2
3
4
5
Use of Acetylenes(Alkynes)
CH CHCH
OH
CH2
OH
CH2
NaNH2,NH3
O
Lindar
H2,Pd/BaSO4
poison
KHSO4
Dienes can be made by witting reaction and also by the addition of vinyl lithium or Grignard reagent to ketone followed by dehydration of the allylic alcohol product .Derivative s of acetylenes can do the same job .The first disconnection is the same but reagent for synthon 5 replace the vinyl metal derivative.
Synthesis of dienes
O
NH
O
F3C
Cl OH
NH2
F3C
ClO
NH2
F3C
Cl
+ CH
CHCH
X
CH
OH
C - N
C - OC - C
C - CFGI
12
3
4
4
56
Alkynes contain Anti –AIDS
Efavirenz
Reverse transcriptase inhibitor efavirenz 1 is one of a new generation of Anti-AIDS drug . Disconnection of two structural C-O bonds reveal 2that is clearly the adduct of an acetylene 4 & ketone 3 .but question is , how to do we make 4? We have not yet met three membered ring but cyclisation of carbon Nucleophiles onto CH2 with a leaving group work well.
Use of Acetylenes(Alkynes)
CH3 CH3
O
CH2
+
CH3
O
CH2
CH3
O
12
3
C - X
1,3 diX
X
OH
R
CH2
+
OH
RCH3
O X
O
R
CH2
+
O
R
Br
O
R
C - X C - X
1,2 diX 1,2 diX
4 56 7 8
9
R1
R2
OH
X
R1
C+
R2
OHR
1
R2
O
R1
X
O
R1
C+
O
R1
Cl
O
C - XC - X
1,1 diX1,1 diX
1011 12 13
14
15
Reversal of polarity, Cyclisation
We needed three types of synthon depending on the di-X relationship in the target molecule. For the 1-3-diX relationship we used just one synthon 2, for 1,2- diX we used related synthons 5 & 8 ,& for the 1,1 –dix two more 11, 14 .The synthons for 1,3-diX & 1,1-diX relationship could be turned into reagent 3,12 & 15 simply by using the Natural electrophilic behaviour of the carbonyl group . The synthons 5,8 for 1,2 –diX relationship could not be turned into reagent so easily: reagent 6 does not Resemble synthon 5 while synthon 8 look very unstable & such intermediatescan not be made
Reversal of polarity-synthesis of epoxide & halo carbonyl compound
Br
Br
O
Br
CH3
O
Br
+ MeCOClC-Br
Friedel- Crafts
bromination
Br Br
CH3
O
Br
Br
O
MeCOCl
AlCl3
Br2
HOAc
1 2 3
3 2 1
Halogenation of ketone
The halogenation of ketones must be carried out in acid solution to avoid polyhalogenation . So the synthesis of reagent 1 , used to make derivatives of carboxylic acid ,is simple providing that we notice the directing effect of two group on the benzen ring in 2 and disconnection with Friedel -Craft .
The synthesis is very straightforward : no bromination occurs on the ring as would be expected in the absence of lewis acid .Enols react with bromine withoutthe need of any atalysis.
Reversal of polarity, Cyclisation
Cyclisation reaction
CH3
OH + NR2
OHCH3
O
NR2
1 2
3
H+
OH
NR
OH
R
O
NRR
O
NH
4 56
The rate of cyclisation to form 3, 5, 6 & 7 membered rings are greater than the rates of corresponding biomecular reaction . This is kinetics but the smaller loss of entropy Is also a factor . We should not expect a good yield in an acid – catalysed ether formation from two alcohol . If the reaction worked at all ,we should get dimer ofeach alcohol as well as the mixed ether 1.
But if reaction were a cyclisation of diol then thing would br very different .The rate of the cyclisation will be much greater so even this unpromising reaction shouldgo well .and no regioselectivity problem would arises .If the side chain on nitrogen were different 4 we should still get the same product 5 regardless of which OH group were protonated & which acted as the nucleophile . The parent compound 6is morpholine & this unit is present in many drugs such as the analgesic phenadoxone.
Reversal of polarity, Cyclisation
Carbon Heteroatom Disconnection
Many heterocyclic rings are made by the formation of a carbon-heteroatom bond & it is important when planning this to get the oxidation level of the carbon electrophile rightIf we disconnected either C-N bond of the pyrrole we get back to ketone & an amine
NH
R
NH2
R
O
C - N
enamine
NH
N
RC - N
NH2
NH R
O
In disconnection of pyrroles ,disconnection of both C-N bond gives a very Reasonable intermediate ,the 1,4-diketon ,then it treatment with ammonia it give pyrroles.On the other hand ,if the furan is needed ,no heteroatom needs to be added & treatment with acid cyclise the ketone to the furan.
NH
R2R
1 NH3 + R1
R2
O OOR1
H+
enamine
C - N
Aromatic heterocycles
Pyrroles- e.g. ClopiracDisconnection of two C-N bonds reveals the diketone available as ‘acetonylacetone’& the simple aromatic amine .The synthesis is to mix the two together . This synthesis makes N-substituted pyrroles available
If the 1,4 –dicarbonyl compound is unsymmetrical ,then it disconnect at branch pointwith the idea of using a d’ reagent for BuCHO in conjugate addition to the enone.
Aromatic heterocycles
Thiazoles
When there are two different heteroatom in a five –membered ring , question ofregioselectivity often arise .The unsymmetrical thiazole might be disconnected at the imine to give the unstable primary enamine & then at the thioester to give an acylating agent & the undoubtedly very unstable , but it must have SH & NH2 on the same side of the alkene for cyclisation to be possible.
But the more heteroatom's the more alternative , we could disconnect the enamine first & the C-S bond second . This suggest a reasonable α- halo- ketone & an unstable -looking imine .Fortunately this is just a tautomer of the thioamides. Thoughthioketones are unastable , thio- amides are stable to extra conjugation.
Aromatic heterocycles
This is strategy followed is most thiazole synthesis ,The regioselectivity issue is which way the reagent combine . There are two possibilities: the sulfur could attack either the ketone or the saturated carbon atom as can the nitrogen but sulfur is excellent at SN2 reactions while nitrogen is better at addition to carbonyl so 1 & not 9is product. No intermediates are isolated: once either the C-S or C-N bond is formed ,cyclisation & aromatization are fast .This means that aromatic heterocycles are easier to make than the non -aromatic ones.
9 1
e.g. Fentiazac doing both disconnection at once we gat available thiobenzamide & the α- haloketone. This can be made from the parent ketone, available by a fridel –craft reaction using cyclic anhydride.
Aromatic heterocycles
NNH
R2
R1
NH2-NH2 + R1
O
R2
O R1
O
CH3
X
O
R2
+2 C - N
imineenamine
1,3 -diCO
1
23 4 5
Pyrazoles
�The disconnection of pyrazoles 1 is straightforward & leads to hydrazine 2 in combination with 1,3- dicarbonyl compound 3 . Simply disconnected to an enol 4 &
acylating agent 5.
� Six membered ring- Pyridines
�Disconnection of both C-N bones of pyridine gives an ene-dione but alkene has to be cis for cyclisation to be possible & conjugated cis-enones are rather unstable �It is usually easier to remove the double bond to reveal the saturated 1,5 diketone
that can be made by conjugate addition of an enolate to enone
NR1 R2
2 C - N
R1 O R2O R1 O R2O
R1 CH2
-
O
CH2
+
R2
O
+
CH2R2
O
FGI
1,5 diCO
imine enamine
Aromatic heterocycles
Treatment of the diketone with ammonia gives the dihydropyridine that is very easily oxidized by a variety of oxidant to the pyridine itself. A hydrogen from C-4 is very easily removed as the product is aromatic .
If you don’t want to be bothered with the oxidation, you can use hydroxylamine instatedof ammonia .The intermediate is now unstable & eliminates water very easily . One of the two marked Hs at the C-4 is lost is lost as a proton with cleavage of the weak N-O bond to give the pyridine & water.
R1 O R2O
NH2OH
N
HH
R1
R2
OH
NR1 R2
- H2O
R1 O R2O
NH3
NH
HH
R1
R2
O2(air) or Ce(IV)
quinone or
bromine
NR1 R2
Aromatic heterocycles
The bicyclic pyridine gives the diketone by disconnection & FGA. Disconnection at the branch point suggests some enolate equivalent of cyclohexanone & the enone.Vinyl ketone are unstable & we often prefer to use the Mannich base instead.
N Ph O
Ph
O
CH-
O
CH2
CH2 Ph
+1,5 diCO
2 C - N
imine enamine
Aromatic heterocycles
Pyrimidine
The compound Aphox .That kills greenfly without harming ladybirds ,is a pyrimidine. Disconnection of the ester side chain reveals a pyrimidine that we should rather draw as a pyrimidone .Disconnection of two C-N bonds gives simple starting materials , available dimethyl guanidine & acetoacetate derivative.
NN
CH3
CH3
NMe2
O NMe2
O
C - O
ester NN
CH3
CH3
NMe2
OH
NHN
CH3
CH3
NMe2
O
NH NH2
NMe2
+CH3
O
COOH
CH3
2 C - Namide enamine
Aromatic heterocycles
NaOEt
Mel
CH3
O
COOEt
CH3
CH3COOEt
O
NHN
CH3
CH3 O
NMe2
Cl NMe2
O
NN
CH3
CH3
NMe2
O NMe2
O
The synthesis used ethyl acetoacetate which was methylated & cyclised with guanidine to gives Aphox directly.
Aromatic heterocycles
Benzene -fused Heterocycles –Indoles
�The most imp. Heterocyces fused to benzene ring are the indole 1 . The obvious enamine disconnection gives 2 which would certainly cyclise to the indole but how are we to make 2 ? As result of this difficulties ,many special reaction have be invented to make indoles & the most imp is the Fischer indole synthesis .
� A phenylhydrazone 3 of a ketone or aldehyde is treated with acid or Lewis acid & the product is an indole.
NH
R
R
NH2
O
?C - N
enamine
12
NHN
CH3 R
NH
R
HOAc
or ZnCl2
3 1
Aromatic heterocycles
CH3 CH3
O
PhNHNH2
NHN
CH3 R
HOAc
or ZnCl2NH
NH
CH2 R
NH
R
NH
HR
NHNH2
43
5
67
Aromatic heterocycles
�The phenylhydrazone 3 , formed from the ketone 4 & PhNHNH2, tautomeriseso an enamine that can undergo a sigmatropic rearrangement, with cleavage of the weak N-N bond 3 , to give an unstable intermediate 6 that aromatises to 7.
�Cyclisation of the NH2 group onto the imine & loss of ammonia gives the indole.
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Fentanyl
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Ciprofloxacin
Ibuprofen
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