retrosynthesis

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


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


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.


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.


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


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


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.


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


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-


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


� 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


� 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


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.


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


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


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.


synthesis

O

O

CH3 CH3R

CH2

CH3

O

O

OCH3

12

3

45

R2NH

O

O

CH3

OHH2 O

O

OCH3


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


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.


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 .


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.


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


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 .


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 ketoaldehyde.


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 .


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.


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 .


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 .


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


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.


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.


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.


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.


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


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.


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.


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.


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.


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


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.


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.


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.


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.


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.


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.


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- haloketone & an unstable -looking imine .Fortunately this is just a tautomer of the thioamides. Thoughthioketones are unastable , thioamides are stable to extra conjugation.


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.


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


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


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


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


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.


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


CH3 CH3

O

PhNHNH2

NHN

CH3 R

HOAc

or ZnCl2NH

NH

CH2 R

NH

R

NH

HR

NHNH2

43

5

67


�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.

Terfenadine

Losartan

Captopril

Fentanyl

Rosiglitazon

Ciprofloxacin

Ibuprofen

Thank you for your patience and attention!!!!!!!!!!!

retrosynthesis

Documents

n c c r o

chetero atom o

alkylation of o

r ch2nh2

h3c c

acyl r

c clhn

nh3rnh2 r b r x