Retrosynthesis or the discconection approach

Synthon or Disconnection or Retrosynthesis
Approach in Organic Synthesis
Presented by: Ms. Sarika Mohite
B.Pharm , M.Pharm IIIrd Semester
Guided by:
Dr . Amit G. Nerkar,
M.Pharm, Ph.D., CCP., FAFSER,
Founderand Director

An analytical approach in organicsynthesis in which the target molecule is broken
into fragments through a series of logical disconnection to get the best possible &
likelystarting materials ( Synthon)
An analytical operation that breaksa bond & convertsa molecule into possible
starting materials
 It is exactly the reverseof chemical synthesis .Therefore alsocalled
Reterosynthesis
Terminologies
Disconnection – An operation involving breaking of bonds between atoms
Synthon – An idealized fragment usuallyan ion orradical obtained
bydisconnection
Reagent – Actual comp (chemicals) used in practice fora synthesis
e.g.- Synthon-Me⁺
Reagent-Me2SO4

FGI or FGE – Functional group interconversion or functional group
equivalence usually written on doublearrow
This means substitution of functional group byanother one equivalent to it.
e.g.- -CN
-COOH
-NH2
-Cl
-NO2
-OH
Basicrule
Disconnection of a bond should besuch that stable fragment ion are
obtained e.g. two mode of disconnection A&B
FGI
C C
2
O N R
-
O2N C
+
C R
+
C
O2N
-
C R
A
B
A will be preferred as carbocationsare stabilized byelectrondonorgr. R,OR ,etc,
While carbinionsare stabilized byelectronwithdrawing gr. NO2,CN,COOR etc.

Numberof fragmentsgenerated through adisconnectionshould beas minimum as
possible
R R'
O O
R
C
+
O
2
H C
-
R'
O
R
O
OC 2H 5
O
R'
halo
R'
O
O H
R'
O
O
R'
O
1
2
3
4
5
+
R R'
O O
R
C+
O
H2C
-
R'
O
R
O
R'
O
1
2

Always a C-hetero atom (O,S,N) bond is broken ,with the electron pair being
transferred to the heteroatom (as heteroatom are moreelectro negative than C
theycan accommodatethe electron pair )
C C
+ -
C Cl HN
N
..
N
..
..

Some time a disconnection doesn’t generate sufficient stabilised fragments but
such fragment can be obtained using FGI or introducing additional electron
withdrawing & removing them .
R NH2 R
CH2
+
NH3
R CH2
+
R
Cl
R X
CH2NO2 ( introduction of electrowithdrawing groups
a
b
-
NH2 (not sufficiently stabilised)
-

R-O-H
RNH2
( vely charged alkyl)
R-CH -OH(alcohol)
2
R-CH2OC2H5(ether)
( R-CO-X(acylhalid)
R-CO-OH(acid)
R-CO-OC2H5(ester)
R-CO-O-CO-R(anhydride)
O2N-CH-R 2
O N-CH2-R
R-O
R-NH
+
R-CH2
R-CH2-X(alkyl halide)
R-C=+O
vely charged acyl)
-
-
-
The –ve & + ve fragmentsgenerated bydisconnectionarereplaced by recognizable&
Meaningfulchemical entities.
- velycharged fragmentsareconsidered equivalent to their protonated species .

H3C CH3
O
Cl CH3
O
H3C
Base
CH2
-
O
H3C
O
OH
O
H3C
e.g
OH
O
H3C
CH2
-
O
H C
3
H2C
+
OH
+
CH3
O
H3C
Cl
OH
-
H C
3
+
C
CH3
O
HO
OH
HO OH
O
C +
H3C
CH3
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 mostelectron withdrawing gr. to bedisconnected first ( i.e. to beadded last in
theanalysis ) e.g.
Analysis
Synthesis

If FGI is needed do itatan appropriatestage toget thedesired effectonorientation
Analysis
Here CCl3 is meta director , but it FGI ,CH3 is P- directional. Therefore, do FGI
priorto C-Cl disconnection.
synthesis

Avoid sequences that may lead tounwanted reaction atother 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

Forcompound consisting of two parte by heteroatom ,disconnect nextto the
heteroatom e.g.- Chlorbeniside
Analysis
Synthesis

Multiple stepsynthesis –
avoid chemo selectivity
problem
In this structure with twoether
& an amine? functional gr. it
requires several disconnection to
take it back to simplecomp.
Thequestion is which dowedo
first?
Here there are four reasonable
disconnection one ateach of the
ether gr. ( a, b) oron eitherside
of the amine.( c, d)
Both a & b posses problem of
chemo selectivity as it would be
hard toalkylate the phenol in the
presenceof basic nitrogen atom. In between c & d , cappears to be the better choice
becausethe nextdisconnection after d will have to bean alkylation of O in the
presence an NH2 gr.
Toavoid chemo selectivity problem like this , we want to try & introduce reactive gr.
late in the synthesis.

Protection allows us to overcomesimple
problem of chemoselectivity .
It easy to reduce the keto - ester 1 toalcohol 2 1
with nucleophilic reagent such as NaBH4 that
attack only the more electrophilic ketone.
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. thatdoes not
react with nucleophiles allows
reductionof the ester with more
nucleophilic LiAlH4.
1
4
5
3
Protecting group
H
OH

Qualities needed in the protecting group
It must beeasy to put in.
It must be resistant to reagents that would attack the unprotected
functional group.
It must beeasily removed
Protecting group

Protecting group
Ethers and amides as protecting group
Protection of alcohol and amides look simple.
Methyl ether & simple amides areeasy to make &
arevery resistant to awidevariety of reagent.
These protecting gr. used when the molecule is robus
enough to take the deprotectioncondition.
 If aniline is brominated the 2,4,6-tribromoderivative is formed.
Theyield is quantitative butweare more likely towant mono- bromination
protection is needed against over reaction .
Theamide is easily made ,brominationgoes only in the para position & the hydrolysis
Does not destroy the benzenering.

1 2 3 4
5
Protecting group
 Achilles heel strategy
Achilles heel foran ether is commonly the DHP gr. that make the ether into
an acetal . Dihydropyran (DHP ) 1 , is protonated an carbon 2 togive the cation 3 tha
capturesthe alcohol to give the mixed 4 acetal.
After the reaction the hydrolysis needs only the weak aq. acid used for acetal.
Thesecret is that the weak acetal bond (b in 5) is cleaved rather than the strong
ether bond ( a in 5)

Chemoselectivity
If a molecule has two reactive group & wewant to reactone of them & not the otherwe
need chemoselectivity.
 If 2 group have unequal reactivity , the more reactive can be made to reactalone.
HO
NH CH3
O
NH2
HO HO
N
O
-
Theamide 2 is paracetamol the popularanalgesic. Amine are more
Nucleophilic than phenol so reaction with aceticanhydride gives the amide we
want without anyof ester 3. Theaminophenol 1 can be made byO
methods.
+
HO
nitration C - N
C - N FGI
amide
NH2
HO HO
NH CH3
O
OH
O
O
+
H3C
?

The synthesis is straight forward. Nitration of phenol needs only dilute nitric
acid & the reduction is best carried out catalytically.
HO H3C
NO2 NH2
HO
HO
NH CH3
O
dilute
HNO3
H2/ Pd/C
Ac2O


If one functional groupcan react twice the productof the first reaction will
competewith reagent .The reaction will stop cleanlyafterone reaction only if the
Starting material is morereactive than the product.

Reaction of alkyl halide with NaSH or Na2S can not usually be made to stop afterone
alkylation as the anione of the first product is at leastas nucleophilicas
HS- or S2- .this is obivous in reactionwith Na2S.
Lessobviously 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 HS
-
+
H S
R SH + R S
-
R Br
1 2
3
R
S R
4

Problem fromgidelines 1& 2 may besolved by protecting group

If we want to react the less reactive of two functional group we protect t
h
e
more
reactive .

If wewanta reagent to react oncewhen it could reacttwice weprotect 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 , s
o
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 .
O
Cl
Ph O
+ H N
2
COOH
R
Ph
O
O
NH
1
2
R
COOH
O
OR
NH
R
COOH
3

One –Group C-X disconnection
 We disconnecta bond joining the heteroatom (X) to the restof 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 beacationiccarbon synthon 2 & anionic
heteroatom synthon 3 represented byacyl oralkyl halides 4 as electrophiles and
amine , alcoholor thiols 5 as nucleophiles.
R X + X
- RHal + HX
1
+
R
2 3 4 5
synthons reagents

Synthesisof Ethers

The question of which bond to disconnect can be much more significant in the
synthesis of ether .with many ether , like the gardenia perfume compound 1 , it doesn’t
matter much . The starting material will bean 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
theanion – 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
undergoelimination whlile 5 just might .
Ph O
CH3
CH3
a b
Ph Cl HO CH3
CH3
+
Ph OH
Cl CH3
CH3
+
a
b
1
2 3
4
5
HO CH3
CH3
CH3
CH3
- O
-
Ph Cl
HO CH3
CH3
Ph O
CH3
CH3
3
6
2
1

Synthesisof 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 makea new C-S bond .

The reaction is much easier with sulfur . Thiols are more nucleophilic towards
saturated carbon than are alkoxidesand the risk of elimination is much less.
The acaricide (kill mice & ticks) chlorobenside 5 is disconnected to give an acidic
thiophenol 6 & reactivealkyl halide 7. the synthesis merelycombine these two
in ethanol with NaOEtas base.
R
1
C - S
1
+ S R
- 2
HS R
2
2
S R
1
R halogen
2 3 4
Cl
S
Cl
SH
+
Cl
5 6 7

One gr. C-C disconnection I -Alcohol
X
OH
R
OH
X
R
-
-
+
+
O
1,2 dis
C- X
C - C
1
2
1
2
5
6
7
6
R
OH
O
P(OR)
2
1,1 dix
C- X R
O
+ -
P(OR)2
O
-
1 2
1
R R
2
O H
R
1
O
+ R 2
-
C - C
3
4
1,1 dix disconnection
For compound with two heteroatom's joined to the same carbon , we used a 1,1 diX
disconnection 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 levelin
combination with a heteroatom nucleophile disconnecting the corresponding C-C
bond 7 ,we use the same epoxide & carbon nucleophile suchas RLi or RMgBr.
O

X
R
O
X
+
R
O
-
1,2 diX
X- C
- H2C
+
8
9
1
R
R
2
O
Br R1
H2C
-
R2
O
C - C
+
10
11
X CH3
X
-
+
H2C
CH3
O O
C - X
12 13
1 - 3 diX
R CH3
O
H C
2
CH3
O
R
-
+
C - C
14
13
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 nucleophileor electrophile carbon
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 nucleophilic
carbon speciesshould be a copper derivative ; RCu, R2CuLi or RMgBr with Cu(I)Br.

H3C
H C
3
OH
CH3
CH3
H3C
CH3
MgBr +
H C
3 CH3
O
+ H3C Li
H3C
CH3
H3C
OH
C - C
C - C
15
16
17
R
O
R
OH
OH
Br
FGI
FGI
C - C CHO C - C
Aldehyde & ketone
The simplest route to aldehyde & ketone using the samestrategy is oxidation of an
Alcohol. So the analysis involve FGI back to the alcohol & then a C-C disconnection of
oneof the bond next to the OH group.
1,1 C-C disconnections- The synthesis of alcohol
Disconnection 3 shows that anyalcohol may bedisconnectedat a bond to the
OH group .Isomeric alcohol 15 &17 can both be made from acetone using
perhapsa Grignard reagent 16 in the first case & available BuLi in the second.

-
+ CO2
R Br
1
R MgBr
2
4
R Br + CN
1
C - C
R CN
FGI
R COOH
3
C - C
FGI
Ph CH3
OH
+ O
CH3
H2C
CH3
2 C - O
1
2 3
1,2 C - C
PhMgX
Carboxylicacid
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 samealkyl 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 bedisconnected at the next butone bond
to the alcoholgroupwith the idea of using the epoxide 2 made from the but-1-ene 3

CH3
CH3
HO
?
a
b
c
d
H3C e
Onlyone 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 shortas possible.
Sodisconnect bonds thatare-

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. Hereweget thealdehyde 2 & grignard reagent 3 coming from the
H3C CH3
OH
straight chain
CH3
H C
3
halide 4. both 2 & 4 arecommercially available.
H
O
+ MgBr CH3
CH3
H3C
FGI
C - C
1a
2
3
4
Br
General strategy – Choosing a disconnection

N
OH
H3C
N
O
H3C
+
OCH3
MgBr
C - C
1,1
5
6 7

The series of drug based on bicyclic structure 5 has an excellent disconnection
between the two ring.
OCH3
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
branchpoint give starting material .
H3C CH3
CH3
H3C
CH3
OH
CH3
H C
3
CH3 OH
H3C
O
CH3
+
H3C MgBr
C - C
C - C
MgBr FGI
+
MeCOOEt
1,1-
1,2-
1
2
3
4
5
General strategy – Choosing a disconnection

General strategy – Choosing adisconnection
Summaryof guideline for the good disconnections
Make the synthesis as shortas possible
Useonlydisconnectioncorresponding to the known reliablereaction .
Disconnectstructural C-X bonds first & try to use twogr. disconnections
Disconnect C-C bonds using FGI in the molecule.
a) aim for the greatestsimplification , if possible
disconnect near the middleof the mol.
Disconnectata branch point
Disconnect ring from chain
b) Usesymmetry (if any)
Use FGI to make disconnection easier
Disconnection back to availablestating materials orones that can easily be made.

R X
O
R OCH3
O
+ X
-
1,1 dix
C - X
1
R
1
R
2
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 two
related disconnection . Started bycomparing the acylationof heteroatom's byacid
derivatives such asester .
1,1-dix disconnection
1,2 – dix disconnection X
R
O
H2C
+
+
R
O
X-
1,2 dix
R1
R2
O
Br R1 +
H2C
-
R2
O
4
6
C - C
3
5

Synthesisof aldehyde & ketone byacylationat carbon
The disconnection 2 is not useful because as MeO is the best leaving group
from the tetrahedral intermidiate 7, the ketone 2 is formed during the reaction
The ketone is moreelectrophilic than the ester so it reactsagain & the product
is tertiaryalcohol 8.
R1
O
OMe
O R2
R1
-
R1 R2
O
R3
2
HO R
R1
R2MgBr
or RLi
OMe
RLi
R3MgBr
7
2 8

Carbonyl compound byalkylation of enols
Disconnection 1 again uses the natural polarity of the carbonyl group but at
the next bond 1. So we usesome enolatederivative 2 in an alkylation reaction.
The problem is that ketone is it self electrophilic & the self condensation
bythe aldol reaction is generallypreferred toalkylation.
First of all to convertthe ketone 3 completely into some enolatederivative so
that there is no ketone left for self condensation .
Lithium enolate 4 & anians 6 of 1,3 dicarbonyl compound 5 actas the enolateanion
of acetone.
R1
R2
O
1,2 C - C R1 Br
+
H C
2
O
-
R2
1 2
H3C R2
O
1
R Br
base
TM 1
synthesis (no good)
3
H3C CH3 H3C
O
i-Pr2NLi
H3C
COOEt
H3C OEt
O
O
-
O
3
CH3
4
Li
CH2
5 6
EtO-

we can use either organo-lithiums or Grignard reagent as the carbon nucleophiles
but we need copper (I) to ensure conjugate addition without Cu (I) both
nucleophilesare inclined toadd directly to the carbonyl group.
 Disconnection of the ketone with conjugate addition in mind could remove t
h
e
vinyl groupor the methyl group. Thereare two reasonswhywe prefer ‘a’ .
 The addition is likely to occur from the opposite face of the molecule to he COOEt
group & that is whywewant thevinyl group .
X R
1
+H2C R
Carbonyl compound by conjugateaddition

Conjugate addition of a heteroatom to the enone 2b give the 1,3 relationship in1 &
the same processwith acarbon nucleophilegives 3.
O
CH2
2
1,3 dix
C - X
X
-
R2
R1 O R1
-
+ H2C R2
O
3
2
Corresponding c-cdisconnection

 Conjugate addition to 3 might occurs at the β position but it could equally w
ell
occurs at theveryexposed δ position .
 Thestarting material is alsoavailable hagemann’s ester
O
CH3
H C
2
EtOOC
b
a
EtOOC
O
O
MgBr
H2C
EtOOC
C - C
a
C - C
b H2C
δ
β

 How to react one specific part of a single functional group & no other .This i
s
Regioselectivity.
 We have seen that anions of phenol 2 are alkylate at oxygen to
give ether 3. whileenolateanions 5 are alkylated at carbon to form a new
C-C bond 6.
OH O
-
Base CH3
OCH3
o- alkylation
c-alkylation
H3C
O
COOEt
H C
3
O
-
COOEt
Base CH3 H3C
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 addition to
all α ,β unsaturated carbonyl compound
 If we react 1 with grignard reagent with Cu (I) catalysis weget 2 as product the
lithium enolate 3 giving us the opportunity to add an electrophile to make 4
If theelectrophile is proton , the productstill 2.
R1 R2
R O
R1
O
R2 R1 R2
R OLi
R1 R2
R OLi
1
2
3
4
RMgBr
cat CuBr
R2 CuLi
+
E
E
Regioselectivity

Regioselectivealkylation of ketone
 Lithium enolates & 1,3 dicarbonyl compound both will help us to solve the
Regioselectivity problem in thealkylation 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 ouractivating COOEt group to give 2 it is clearthatwe can
make this byanotheralkylation & the activating group ill promote both.
H3C
CH3
Ph
O
H3C
CH3
O
COOEt
H3C
O
COOEt
+
Br Ph
+
PrBr
1
2
3
Regioselectivity

H3C
O
COOEt
3
NaOEt
PrBr
H C
3
CH3
O
COOEt
NaOEt
BnBr
H3C
CH3
Ph
O
COOEt
NaOH
H2O
H
+
heat
H3C
CH3
Ph
O
2
1
 Benzyl is the more reactive bromide so it makes sense to add it last since
making the quaternarycarbon will bedifficult.
Regioselectivity

Regioselectivity in nucleophilicaddition toenones
The problem of getting direct (1,2) orconjugate( 1,4 or michael ) addition to
α ,β unsaturated compounds suchas enones 1 can be solved without finding abstrus
strategies bychoiceof reagent.
R1 R2
Nu O
R1 R2
O
R1
R2
OH
Nu
1
2 3
Nu-
1,4-or Michael or
conjugate addition 1,2- or direct addition
Nu-
Regioselectivity

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 cyclohexenering & electron withdrawing
group outside the ring & on the opposite side to the alkene. The relationship between
these features must be recognized.
Twogr. C-C disconnection I –DielsAlder Reaction

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
senseto 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
Twogr. C-C disconnection I –DielsAlder Reaction
Stereospecificity
The reaction occurs in one step so there is no chance for either diene or the
dienophile to rotate & the stereochemistry of each must be faithfully
reproduced in the product .The two Hs in 1 arecis because they werecis in the
starting anhydride .The two Hs in 3 are trans in the diester 2.

FGI on Diels – Alder Product
Thecyclic ethercomes from the diol that can be made by reduction of various
Diels –Alder adducts suchas the anhydride.
Twogr. C-C disconnection I –Diels Alder Reaction
O
H
H
H
H
CH3
O
H
H
CH3
O
O
O
O
+
H2C
H2C
CH3
HO
ether
HO
CH3 C - O FGI
D A
O

Twogr. C-C disconnection I –Diels Alder Reaction
H2C
H2C
CH3
O
O
O
heat
O
H
H
CH3
O
O
H
H
CH3
HO
HO
H
CH3
O
H
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
theenolate 2 of one carbonyl compound reacting with either an aldehyde 3 or
acid derivative 5 such as an ester.
1 2
3
2
4 3
Most nucleophilicenolsor
enolates
Mostelectrophilic
carbonyl comp
Moststable
carbonyl comp
Most stableenols
or enolates
Twogr. C-C disconnection II: 1,3 –Difunctionalised compound
5

6 7 8
Twogr. C-C disconnection II: 1,3 –Difunctionalised compound
β - Hydroxy carbonyl compound : The Aldol reaction
 With the compound 1 onlyoneof the two C-C bonds in worth disconnecting , t
h
e
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 amountof enolate 7 in the presence
of much unenolised ketone 8 & the reaction will occur.

R1 R2
O O
R1
CH2
-
O
H2C
+
R2
O
1,5 -diCO
C - C
H2C
O
2
d enolate
a
3
Twogr. C-C disconnection III – 1,5 difunctionalised compound conjugate addition &
Rabinsonannelation
1
2
3
R2
Theodd numberrelationship means arewestill use Synthon of natural polarity .
The 1,5 –diketone 1 disconnect to a d2 synthon, an enolate & an a3 synthon 2 i
t
represented by the reagent 3. The conjugation in the enone makes the terminal
carbonatom electroplilic.

H OEt
O O
a
b
H
-
O
H C
2
OEt
O
+
CH2 EtO
-
O
H2C
H
O
+
CH2
1,5 -diCO
C - C
1,5 -diCO
C - C
Specific enol equivalence good at michael addition 1,3 –Dicorbonyl compound
If wewant to make 1 we have achoice between adding an enolateequivalent of
aldehyde 5 to an unsaturated ester 4 oran enolateequivalent of ester 3 to
an unsaturated aldehyde 2. We prefer the first 1a as unsaturated ester 4 is
more likely to do conjugateaddition .an enamine would begood choice for 5
1 2 3
4
5
Rabinsonannelation

O
O
CH3
H3C
O
O
O
CH3
3
4
6
1,5 diCO
1
H C
3
CH2
O
C
-
+
O
H3C
analysis
1
2
3
O
4
2
3
5
Rabinsonannelation
Robinsonannelation
Combining aldol & michael reaction in one sequence is very powerful, particularly
if oneof the reaction is cyclisation .The Robinsonannelation makes new ring in
thecompound like 1 that were needed to synthesize steroids.
Disconnectionof 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 begood at conjugateaddition.

synthesis
O
O
H3C
H C
3
R
CH2
H3C
O
O
O
CH3
1
2
3
4
5
R2NH
O
O
CH3
OH
H2 O
O
O
CH3
Rabinsonannelation

N
H
CH3
H C
3
COOR COOR
Ar
2 C - N
enamines
H3C CH3
O O
COOR
COOR
Ar
COOR
Ar
H C
3
COOR
O
CH3
+
O
1 2 3
4
Rabinsonannelation
Heterocycles made from 1,5 dicarbonyl comp
 A family of calcium channel antagonist based on thegeneral structure 1 is
widelyused tocombat high blood pressure.
Disconnecting the structure C-N bondswe discover symmetrical 1,5 diketone 2
So disconnectionof eitherappropriate bond give the samestarting material &
enone 3 an acetoacetateester 4

R1
R2
O
O
R1
C
-
O
+ X R2
O
1,2-diCO
R1
R2
O
1,2-diCO R1
C
-
O
+ H R2
O
H C
3
O
CH3
CH3
HO
FGI
Hydration HC
CH3
CH3
HO
1,2-diCO -
HC C +H3C 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 bond
between the functionalized carbon. so, whilewecan usean acid derivative 3 or
 An aldehyde 5 forone half of the mol, weare forced to use a synthon o
f
unnatural polarity, the acyl anion 2 forother half.
1 2
3
4 HO 2
3
1 2
The hydroxy – ketone that could come from the acetylenic alcohol by hydration&
hence fromacetonewith the anion of acetyleneacting as the acyl anion equivalent.

R1
R2
OH
HO
FGI
electrophilic addition R1
R
witting
2 R1- CHO + Ph3P
+
R2
FGI
Br R2
Method from alkenes.
R1
R2
OH
H2N
R1
O
R2 H3C
CH3
C - N
1,2 dix FGI
electrophilic addition
starting
Many possible
material
Epoxide give rise to many 1,2 difunctionalised comp such as 6 with control over
stereochemistry . Reaction of theepoxide give the anti stereochemistry in 6 in contrast
Tothe syn stereochemistry in 1.

HO
HO
OH
NH
CH3
H3C
C - N
reductive amination
HO
CHO
OH
HO
HO
HO
Functionalisation
1,2 diCO
1
2
OH
3
CH3
α Functionalisation of carbonyl compound
Metaproterenol 1 is an adrenalineanalogue used as bronchodilater.
the might be inserted by reductive amination on the aldehyde 2 & this might a Α
–functionalisation of the available ketone .

R1
R2
O
O
CH2
-
O
+
H2C
+
R2
O
CH2
O
-
X
R2
O
R1
enolate
1,4 diCO
1
R1
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
asα - haloketone 5.

O
COOEt
CH
-
O
+
+
O
-
H
Br
H COOEt
1,4 diCO
1
2
H2C COOEt
3
4
5
Reaction of enol (ate)s with reagents fora2 synthons
A simple example would be the keto-ester 1 disconnect the bond at the
branchpoint & that suggest the synthon 2&3 . The reagent for 3 can be
bromoester5 butweshall need tochooseour enolateequivalent carefully .If
should not too basicas the marked proton in 5 between Br & COO2Et are rather
acidic.

H
N
O
O
Ph
Ph
HOOC COOH
Ph
NC COOH
Ph
COOH
+
CN
-
2 C - N
FGI
Conjugate addition of acyl anion equivalents
The anticonvulsant phensuximide 1 being an imide , comes from a
dicarboxylic 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
H2C
-
R2
O
+
R2
O
-
1
2
3
4
R2
OSiMe3
5
Direct addition of homoenolates
The same disconnection but of the opposite polarity requires some
acylating agent forsynthon this no problemaswe havevarious derivative
at our disposal but the nucleophilic synthon 3 or homoenolate, is
another matter.
There is no stabilisation of the anioneas drawn but if were
tocyclise tooxyanion 4, it would be rather morestable & there is
evidence trapping with silicon togive 5 .

synthesis of 1,2 & 1,4diCO comp byoxidative cleavage.
If we wanted to add bromketones 4 to enolate 3 to make the 1,4 dicarbonyl
compound 5 .Wecould not usea lithium enolate because it would be too basic
no such difficulties exist in the reaction of enolate with allylic halides such as 2
Anyenol equivalent will do as thereare noacidic hydrogen & allylic halides
aregood electrophilic for the SN2 reaction.
H C
3
CH3
CH3 CH2
O
H C
3
CH3
CH2 H3C
CH3
O
-
H3C
CH3
CH3
O
CH3
O
2
3
O
H3C
4
1 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 –alkene
from phosphonium salt 2 but the ketoaldehyde 3 would need protection , perhaps
as the acetal 4.
Reconnection

R
CH3
O
Ph3P+
+ OHC CH3
O
O
CH3
OHC
1
R
2
3
CH3
OH 4
witting
FGI
The problem is how to protect the ketone rather than the aldehyde & the answer is
Protectit when the aldehyde is not there .Reconnection to the alkeneachieves
this & the ketone can be made byreactionof someenolatewith allyl bromide.
O O
CH3
OHC
4 O
CH3
O
H2C
reconnect
O
CH3
2
CH2
Br
+ H3C CH2
-
FGI H C
C - C
O
Reconnection

O
O
O OHC
O
COOEt
COOEt
COOEt
O COOEt
Ph
reconnect
aldol
O
COOEt
COOEt
H3C
H2C
-
CH3
O
Br COOEt
Br COOEt
+
1
2 3
4
5
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 .
O
Thealdol disconnection 3 reveals methyl ketonewith 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 bespecificallycleaved by hydrogenation to give 4 .
Condensationwith unenolisable benzaldehyde is unambiguous & ozonedoes the rest.
Reconnection

H3C
CO2Bn
O
2(NaH)
2(Br COOEt)
Me
O
COOEt
BnO2C
COOEt
Me
O
COOEt
COOEt
Pd/C H2
O COOEt
Ph
PhCHO
base
COOEt
OHC
O
COOEt
COOEt
2
7
8
4
3
O3
Me2S
Reconnection

Two gr. C-C disconnection VI- 1,6 dicarbonyl compound
R1
R2
O
R1
CH2
+
O
H2C
-
R2
O
+
?
1
2
3
4
5
6
O
1
2 3
1,6 dicarbonyl compound disconnectin the middlewe 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 conjugateaddition .
Disconnecting elsewhere is no helpas the truedifficulty is that the two
carbonylgroup are too farapart for this approach.
Strategyof reconnection is needed the main strategy for the synthesis of 1,6 diCO
compound

O
CH3
CH3
H3C
CH3
O
CH3
CH3
H3C
CH3
O
CH3
CH3
H C
3
OHC1
2
3
4
5
6
aldol
6
7
8
R1
R2
O
1
2
3
4
5
6
O
R1
R2
weaken ?
R1
R2
4 5
1
Reconnect intramolecularly the marked atom C-1 & C-6 form a ring 4 &
the bond between these atom must be made weakerthan anyother bond in
the molecule Ironicallywecan do this by making it adouble bond 5
Bicyclic ketone made from the simpleenone that had to be made. Aldol
disconnection reveals the ketoaldehyde.

H3C
CH3
H3C
O
11
MeLi
H+H+
CH3
H3C
CH3
H3C
9
O3
Me2S
CH3
O
CH3
CH3
H C
3
OHC1
2
3
4
5
6
CH3
O
CH3
H3C
CH3
KOH
MeOH
synthesis
CH3
O
CH3
H3C
CH3
OHC1
2
3
4
5
6
CH3
H C
3
CH3
H3C
CH3
H3C
CH3
H3C
OH
H3C
CH3
H3C
O
8
9 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 .
Twogr. C-C disconnection VI- 1,6 dicarbonyl compound

H2C
+
CH2
O
R
O
R
O
HOOC
HOOC
R
O3
H2O2
H2C
1
2
3
4
The Diels –AlderRoute to 1,6 dicarbonyl compound
Normally we have to make the cyclohexene , need for oxidative cleavage & one o
f
the
best route to such is Diels – Alderreaction .Generalized examplewould be
ozonolysis of alkene .The product hasa 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
O
O
O
H H H
H
O
O
O
+
H2C
H2C
1
5
6
7
8
reconnection
1,6 diCO
OMe FGI
OMe
D-A
diester required for synthesis of the antibiotic pentalenolactone reconnecting
the ester gives thecyclohexene . We mustchange the two ether group into
carbonyl group & one starting material is , diels alder adduct of butadiene &
maleicanhydride .

O
O
O
H
H 7
OMe
H
H
6
MeOOC
MeOOC OMe
OMe
H
H
5
LiAlH4
NaH,Mel
OMe O3MeOH,H2O2
CH2N2
O
O
O
O
CH3
H C
3
COOH
COOH HOOC
CH3
HO
CH3
OH
COOH
HOOC
CH3
CH3
2(C - O )
ester
1,1 dix
9 10
O
11

The synthesis followed this pattern with the ether 6 being made
immediately after the reduction of 7 & the ester made with diazomethane
CH2N2 afteroxidative cleavage.
 The bicyclic double lactone used as precursor for all four heterocyclic ring i
n
synthesis of Vit. B12 .disconnectionof both lactones revealsa ketone .
COOH
COOH
Twogr. C-C disconnection VI- 1,6 dicarbonyl compound

COOH
HOOC
CH3
CH3
COOH
O
11
COOH
H C
3
H3C O
H2C
H2C
1
+
H3C
H3C CH2
COOH
reconnect
1,6 diCO
1
2
3 4
5
6
1
2
3
4
5
6
diels
alder
12
13

The ketone 11 in fact has 1-4,1-5 & 1-6 relationship & if we redraw in 11a t
o
see
1,6 relationship clearly Being careful to get the sterochemistry right,we can
reconnect to the cyclohexene 12 & hence , byDiels- Aldrdisconection , find
the reactivedienophil disconnection 13 .The methyl & czrboxylic
groupare cis in 12 & must becis 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
Cyclisation reaction
Synthesis of butanone by reaction of the primary alkyl chloride with MeNH2 was
likely to give a pooryield . The problem is that the product 2 is alsoa nucleophile &
will reactat similar rate with alkyl chloride asdoes MeNH2. The reaction is
intermolecular & so bimolecular.

MeHN Cl
N
+
H
N
H3C
CH3
base
Cl
..
MeHN
The very similar reaction of 4 gives exclusively the pyrrolidine 5.The reaction 6 is
now intramolecular a unimolecularcyclisation in fact & is greatly preferred to any
bimolecular processes.

H3C Br
CH3
Cl
CH3
O
H3C
H3C
Cl
OH
CH3
O
CH3
H3C
H3C
NaOH
Mg.OEt
1 2
3
CH3
CH2
H3C
CH3
4
Three membered ring
Thesimple formation of epoxide 3 by the action of peroxyacidssuch as
mCPBAon alkenes 4.
They can equally well be made by cyclisation of chloro- alcohols 2 as in the
Cornforthaddition of a Grignard reagent to an α-chloroketone & cyclisation in base.

Ph
N
Ph
NH
X
Ph
O
PhCHO +
C - N C - N
Reductive amination
aldol
1
2
3
O
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.

N
H
O
X
H3N
+
COOEt
Cl
-
NH2 O
OEt
N
H
O
base
1
NH2 O
2
3
4
1
Five membered ring
Lactam come from acid derivative . Thecompounds such as aminoesterare
not stableas the freeamine butare usually isolated as saltsuchas hydrochloride .
When treated with base, give the freeaminewhich promptlycyclise to the lactam.

S 2 C - S X X
HO OH Cl Cl
S
PCl3 Na2S
The unsaturated rings does have a double bond but it is not next to the
heteroatom. It is an allylicrather than avinylic sulfide so two disconnection
at the alcohol oxidation level suggest the doublyallylicstarting material

N
H
O
CH3
CH3
H3C
H3C
H3C
3
CH3
CH H3C
O
H3C
CH3
O
H3C
CH3
O
H3C
CH3
O
2( 1,3 diX)
2( C - N)
2 aldol
O
CH3
CH3
O
CH3
H3C CH3 H3C
3
CH3
CH H3C
O
N
H
O
H3C
H3C
O
CH3
CH3
NH3
CH3
CaCl2
CH3
Six membered ring
Intramolecular reaction comes in the synthesis of tetramethyl piperidine ,
removal of nitrogen with conjugateaddition of ammonia to the dienone opens
the possibility of adoubledisconnection to reveal three moleculeof acetone.

O
N
H
O
R
R O
NH2
COOR R OH
NH2
COOR
+ Cl
R
O +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 obviouslyan epoxide 4 adduct with
ammonia.

N
Pr
COOMe
Cl Cl
N
CH3
COOMe
Pr
Cl CH3
F
Pr
1
2 3
HN
4
MeOOC
+
aldol
SnAr
Seven membered ring
Thecompound 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 nucleophilicaromaticsubstitution workwell
on aryl fluorides with orthoor para electron – withdrawing gr. such as the aldehyde 3.

N
O
Pr
Cl CH3
F
3
HN
-
O OC
Pr
Cl
N
CH3
COOMe
Pr
2
HCl
NaOH
Mel,K2CO3
MeONa
6
(MeO)2CO
Pr
N
Cl
COOMe
1
Thecompound 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.

H
N
O
CH3
CONH2
NH2 a
C - N
amide
COOH
b
C - N
reductive amination
N
O
CF3
COOH
HO
CHO
COOH
COOH
H
+
F3C
NH2
1
2
3
The related benzazapinones 1 can be made by formation o C-N bonds in
Twodifferent ways. Amide formation from compound like 2 is not surprising
butreductiveamination between the amide nitrogen & thealdehyde in 3 is a
testament to the efficient of cyclisation even when a seven- membered ring is the
product.
The intermediate 1 asan intermediate in the synthesis of adrug for the treatment
of osteoporosis ,chosethe double disconnection because we had way making singal
enantiomers of the diacid 2

H3C CH3
O
H C
3
HO
H C
3
CH3
OH
CH3
H C
3
H C
3
O
CH3
CH3
Rearrangement in synthesis
R R
OH O
?
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
The typical pinacol formed from acetone is important because it
rearranges in acid togive a tertiary alkyl ketone known as ‘ pinacolone’ . The
A key step is methyl migration asone of the OH group is lost.

O
a b
OH
OH
O
a
OH
HO
b
Pinacol
FGI
The bestway todo thedisconnection is to reverserearrangement & there are
twoway todo this a & b .
Thediol can be made by pinacol dimerisation of cyclopentanonewhilediol
would be the productof dihydroxylation of the alkene.

O OH
Bu
OH
Bu
SoCl2
Pyridine
O
BuLi
H+H+
Mg/Hg
THF
HO
The pinacol dimerisation use to make the diol & the pinacol rearrangement
to make the spirocyclic ketone.

O O
Cl
COOR
O
H OR
O
-
Cl
RO
halogenation
RO
-
The Favorskii rearrangment
Halogenation of cyclohexanongives the α –chloroketonetreatment 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 oneof the weak C-C bonds in the three membered ring.

R1 R2
O
R1 R2
NO2
R1 R2
NH2
base,(o)
or H2O+TiCl33
or H2O + H2SO4
H2,Pd/C
1
2
3
Aliphatic Nitro compound in synthesis
H2N
CH3
NH2
CH3
2
O N
CH3 O2N
CH3 Cl
+
H C NO
3 2
CH3
FGI C - C
1 2 3
4
Few aliphatic nitro compound arewanted as target molecule in theirown
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 byvarious
form of hydrolysis
Reduction of nitro compound
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 bedisconnected next to the tertiary centerafter the amine are
changed to nitro compound 2 . 2-Nitro propane 4 is available.
NO2

H3C NO2
CH3
O2N
Cl O2N
CH3
NO2
CH3
2
H N
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
samestep
Aliphatic Nitro compound in synthesis

CH2 CH2
OH
CH
OH
O
+
-
C CH
FGI
FGI
C - C
1
2
3
4
5
Use of Acetylenes(Alkynes)
CH
OH
CH2
OH
CH2
HC CH
NaNH2,NH3
O
Lindar
H2,Pd/BaSO4
poison
KHSO4
Synthesisof dienes
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 forsynthon 5 replace thevinyl metal derivative.

O
N
H
O
3
F C
Cl OH
NH2
F3C
Cl
O
NH2
F3C
Cl
+ HC
HC
HC
X
HC
OH
C - N
C - O
C - C
C - C
FGI
1
2
3
4
4
5
6
Alkynescontain Anti –AIDS
Efavirenz
Reverse transcriptase inhibitor efavirenz 1 is oneof a new generation of
Anti-AIDS drug . Disconnectionof two structural C-O bonds reveal 2
that 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 groupwork well.
Use of Acetylenes(Alkynes)

H3C CH3
O
+
CH3
O
H2C
CH3
O
1
2
3
H2C
C - X
1,3 diX
X
OH
R
+
OH
H3C
O
X
O
+
R
Br
O
R
H2C
C - X
H2C
C - X
O
1,2 diX R
1,2 diX
4 5
6
R
7 8
9
R
1
OH
X
R
1
C
+
2
R
OH
R
1
2
R
O 1
R
X
O O
1 + 1
R C R
Cl
O
C - X
C - X
1,1 diX
1,1 diX
2
R
10
11 12 13
14
15
Reversal of polarity, Cyclisation
can not be made
Reversal of polarity-synthesis of epoxide & halocarbonyl compound
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 .Thesynthons for 1,3-diX
& 1,1-diX relationship could be turned into reagent 3,12 & 15 simply by using the
Natural electrophilic behaviour of thecarbonyl group . The synthons 5,8 for
1,2 –diX relationship could not be turned into reagentsoeasily: reagent 6 does not
Resemble synthon 5 whilesynthon 8 look very unstable & such intermediates

Br
O
Br
Br CH3
Br
+ MeCOCl
C-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
carboxylicacid ,is simple providing that we notice the directing effectof two group
on the benzen ring in 2 and disconnection with Friedel -Craft .
O
Thesynthesis is very straightforward : no brominationoccurs on the ring aswould
beexpected in the absenceof lewis acid .Enols reactwith brominewithout
the need of any atalysis.

H3C
OH + NR2
OH
H C
3
O
NR2
1 2
3
+
H
HO
NR
OH
R
O
R
O
N
H
4
NR
5 6
Cyclisation reaction
The rateof cyclisation to form 3, 5, 6 & 7 membered ringsaregreater than the rates
of corresponding biomecular reaction . This is kinetics but the smaller loss of entropy
Isalso a factor . We should notexpectagood yield in an acid – catalysed ether
formation from two alcohol . If the reaction worked at all ,we should get dimer of
eachalcohol aswell as the mixed ether 1.
But if reaction were acyclisation of diol then thing would brvery different .The rate
of the cyclisation will be much greatersoeven this unpromising reaction should
go well .and no regioselectivity problem would arises .If the side chain on nitrogen
weredifferent 4 weshould still get the same product 5 regardlessof which OH group
were protonated & which acted as the nucleophile . The parent compound 6
is morpholine & this unit is present in manydrugs 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 right If
wedisconnectedeither C-N bond of the pyrroleweget back to ketone & an amine
N
H
R
NH2
R
O
C - N
enamine
N
H
N
R
C - N
NH2
NH R
O
In disconnectionof pyrroles ,disconnectionof both C-N bond gives avery
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 beadded & treatment
with acid cyclise the ketone to the furan.
N
H
2
R
1
R
NH3 + R
1 2
R
O O
O
R1
H+
enamine
C - N
Aromatic heterocycles

If the 1,4 –dicarbonyl compound is unsymmetrical ,then it disconnectat branchpoint
with the idea of using ad’ reagent for BuCHO in conjugate addition to the enone.
Pyrroles-e.g. Clopirac
Disconnectionof two C-N bonds reveals the diketone available as ‘acetonylacetone’
& the simplearomaticamine .The synthesis is to mix the two together . This synthesis
makes N-substituted pyrrolesavailable

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 justa tautomer of the thioamides. Though
thioketones are unastable , thioamides arestable toextra conjugation.
Thiazoles
When there are two different heteroatom in a five –membered ring , question of
regioselectivity 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 sideof thealkene forcyclisation to be possible.

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 ketoneor the saturated carbon atom as can the nitrogen but sulfuris
excellentat SN2 reactions while nitrogen is betterat addition tocarbonyl so 1 & not 9
is product. No intermediates are isolated: onceeither 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 cyclicanhydride.

N
N
H
R2
R1
+ R1
O
R2
O R1
O
CH3
X
O
R2
+
2 C - N
imine
enamine
1,3 -diCO
1
NH2-NH2
2
3 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
Disconnectionof both C-N bonesof pyridinegives an ene-dione butalkene has to
becis forcyclisation to be possible & conjugated cis-enones are rather unstable
It is usually easierto remove the double bond to reveal the saturated 1,5 diketone
thatcan be made byconjugate addition of an enolateto enone
N
R1 R2
2 C - N
R1 O R2
O R1 O R2
O
R1 CH2
-
O
H2C
+
R2
O
+
O
H2C
R2
FGI
1,5 diCO
imine enamine

Treatmentof the diketone with ammoniagives the dihydropyridine that is very
easilyoxidized by avariety of oxidant to the pyridine itself.
A hydrogen from C-4 is very easily removed as the product is aromatic .
If you don’twant to be bothered with the oxidation, you can use hydroxylamine instated
of ammonia .The intermediate is now unstable & eliminates water very easily .
Oneof the two marked Hs at the C-4 is lost is lostasa proton with cleavage of theweak
N-O bond togive the pyridine & water.
R1 O R2
O
NH2OH
N
H
H
1
R R2
HO
N
R1 R2
- H2O
R1 R2
O O
NH3
N
H
H
H
1
R 2
O2(air) or Ce(IV)
R quinone or
bromine
N
R1 R2

The bicyclic pyridine gives the diketone by disconnection & FGA. Disconnection at the
branchpoint suggests some enolate equivalent of cyclohexanone & the enone.
Vinyl ketoneare unstable & we often prefer to use the Mannich base instead.
N O
Ph
O
CH
-
O
H2C
H2C Ph
+
1,5 diCO
2 C - N
imine enamine
Ph

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 ratherdrawas a pyrimidone .
Disconnection of two C-N bonds gives simple starting materials , available dimethyl
guanidine & acetoacetatederivative.
N
N
CH3
H3C
NMe2
O NMe2
O
C - O
ester N
N
CH3
H3C
NMe2
OH
NH
N
CH3
H3C
NMe2
O
HN NH2
NMe2
+
H C
3
O
COOH
CH3
amide enamine 2 C - N

NaOEt
Mel
H C
3
O
COOEt
CH3
H3C
COOEt
O
NH
N
CH3
H3C O
NMe2
Cl NMe2
N
N
O
CH3
H3C
NMe2
O NMe2
O
The synthesis used ethyl acetoacetate which was methylated & cyclised with
guanidine to gives Aphoxdirectly.

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.
N
H
R
R
O
NH2
?
C - N
enamine
1
2
NH
N
H3C R
N
H
R
HOAc
or ZnCl2
3 1

H3C CH3
O
PhNHNH2
NH
N
H3C R
HOAc
or ZnCl2
NH
H2C R
NH
R
NH
H
R
NH
NH2
4
3
NH
5
6
7

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 thataromatises to 7.
Cyclisationof the NH2 group onto the imine & lossof ammonia gives the indole.

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

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