100173431 solved-problems-in-advanced-organic-synthesis-for-csir-net-chemistry

V.Adityavardhan
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ADICHEMISTRY
SOLVED PROBLEMS IN ADVANCED ORGANIC SYNTHESIS
(FOR CSIR NET & GATE)
1st EDITION
SAMPLE COPY
Release date: 7th, May 2012
Last updated on: 7th July 2012
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Problem 1.1 (IISc 2011)
O
Ph
O
Ph
O
Ph
OH
Pha) b) c) d)
O
O
i) PhMgBr
ii) H+
?
Answer: c

V.Adityavardhan
2
O
O
i) PhMgBr
BrMgO
O
Ph
ii) H+
OH
O
Ph
::
H
+
::
O
+
Ph
O HH
::
O
Ph
-H2O
-EtOH
-H+
1,2 addition of Grignard reagent
acid catalysed and conjugate
 bond assisted removal of
OH group
final removal step of EtOH
Explanation
Thinkdifferent:
What willhappenif1,4 addition occurs?
O
O
O
O
Ph
1, 4 - addtion of PhMgBr
O
Ph
H+
-EtOH
i) PhMgBr
ii) H+
OH
O
Ph
Same product!So it might bethe actualmechanism? But slimchances. Why?Thepossible explanation
might golikethis:
i) thepositive charge on4thpositionis diminished dueto contribution ofp-electronsofadjacentethoxy
‘O’throughconjugation(+M effect).

V.Adityavardhan
3
ii) The enolate ion formis less stable due to -I effect of‘O’.
iii) We also know that: 1,2 addition is kineticallymore favorable than 1,4-addition incase ofGrignard
reagents. It is because the R group attached to Mg in GR is a hard nucleophile and prefers carbonyl
carbon with considerablepositive charge (hard electrophile).
And if this is the mechanism, the removalof ethanolmaygive another product, though less likely, as
shown below.
O
Ph
Now start arguing!
Web Resource:
http://www.adichemistry.com/organic/organicreagents/grignard/grignard-reagent-reaction-1.html
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Problem 1.2
The most appropriate set of reagents for carrying out the following conversion is:
O OHCl
a) i) EtMgBr; ii) HCl b) i) (C2
H5
)2
CuLi; ii) HCl
c) i) C2
H5
Li; ii) HCl d) i) HCl; ii) EtMgBr
Answer: d
Explanation:
1,4-additionofHClfurnishes 4-chlorobutanone, whichreactswithGrignard reagent to get the desired
product.
O HCl OCl EtMgBr H3O+
H+
CH2
OH
Cl-
OHCl
OHCl
mechanism
However, the yields maynot be satisfactorydue to side reactionthat is possible inthe second step
with Grignard reagent. It mayundergo Wurtz like coupling reactionwith -CH2
Clgroup.

V.Adityavardhan
4
OCl EtMgBr O
MgBrCl
What about other options?
Option - a :
O HClEtMgBr H3O+
OH Cl Cl
H+
-H2O
Cl-
Cl-
major product
* 1,2-additionoccurs withGrignard reagent, since the ethylgroup attached to Mg has considerable
positive charge and is a hard nucleophile. It prefers to attack 2nd carbon (hard electrophile).
* In the reaction ofallylic alcoholwithHCl, the Cl-
prefersto attack the allylic carbocation fromless
hindered end. Hence the major product is 1-chloro-3-methyl-2-pentene.
Option - b
O Et2CuLi H3O+
O HCl
Expecting aldol reaction
1,4-addition occurs withLithiumdiethylcuprate, since ethylgroup attached to copper is a soft
nucleophile and prefers carbonat 4thposition (soft electrophile).
Option-c :
The products are same as incase ofoption-a. Ethyllithiumalso shows 1,2 additionlike Grignard
reagent.
Problem 1.3 (CSIR DEC 2011)
Choose the correct option for M & N formed in reactions sequence given below.
O
1) PhMgBr
2) TsOH
1) BH3.SMe2
2) PCC
3) mCPBA
N
a)
M
N=M=
Ph
c)
b)
d)M=
Ph
M=
Ph
O
Ph
O
Ph
ON=
N=M=
N=
OH Ph
O
Ph
O
Ph
O
Answer: a

V.Adityavardhan
5
Explanation:
*Atertiaryalcoholis formed upon1,2 additionofPhMgBr and is dehydrated inpresence ofTosylic
acid.
O
PhMgBr H3O+
OH Ph Ph
TsOH
-H2O
* Thus formed product is subjected to hydroboration with BH3
.Me2
S complexto yield 2-
phenylcyclohexanol, an anti-Markonikov’sproduct, which is oxidized to a ketone inpresence ofPCC.
The keto compound is subjected to Baeyer Villiger oxidation withmCPBAto get a lactone. The
PhCH- group is migrated onto oxygenin preference to CH2
group.
Ph Ph
OH PCC
Ph
O O
Ph
O
BH3.SMe2 mCPBA
Anti Markonikov's
product
Baeyer Villger
oxidation
Ph-CH- group has more
migratory aptitude than
CH2 group
Problem 1.4 (CSIR JUNE 2011)
The majorproduct formed in the following transformation is:
O
Ph
a) c)b) d)
Answer: d
1) MeMgCl, CuCl
2) Cl
O
Ph
O
Ph
O
Ph
O
Ph
Explanation:
* The Grignard reagent reacts with CuClto give Me2
CuMgCl, anorganocopper compound also
known as Gilmanreagent that is added to the-unsaturated ketone in 1,4-manner.
Initiallycopper associateswith the double bond to give a complex, whichthenundergoes oxidative
additionfollowedbyreductive elimination.
Thus formed enolate ion acts asa nucleophile and substitutesthe Clgroup ofallylchloride.
The attackon allylchloride is done fromthe opposite side ofmore bulkyphenylgroup.
2 MeMgCl + CuCl Me2CuMgCl + MgCl2
O
-OMe
MeO
F3C O CF3
O O
TFAA
OOMe
MeO
O
CF3

V.Adityavardhan
6
Problem 6.1 (IISc 2009)
a) b) c) d)
Answer: a
CHO
HEtOOC
H
? major product
CHO
+S
COOEt
C6H6

CHO
HH
EtOOC O
HH
EtOOC
O
H
H
EtOOC
Explanation:
ThisisCorey-Chaykovskireaction. Since thesulfur ylideisstable,cyclopropanationoccursmajorly
through 1,4-additionroute. The product isa thermodynamic one.
The CHOand COOEt groups get trans positions in thecyclopropane ring. This occurs since they
tend to orient as far awayas possible during thecyclopropanations step to avoidsteric repulsion.
S
+
CH
-
COOEt
O
H
H
S
+
CH
O
-
H
COOEt
H
C
H
OH
EtOOC
H
+ S
slow
1, 4 addition
irreversible
+
H
H
CHOH
C
S
+
EtOOC H
+
Stereochemistry of
cyclopropanation step
CHO & COOEt groups orient
in space so as to minimize
repulsion. Hence they assume
trans postions to each other in
cyclopropane ring.
Thinkdifferent:
What willbe the product if1,2-addition occurs?
An epoxide is formed. It iskineticallyfavored product.
But this isminor product. Why?
Since 1,2-addition step is reversible, the expulsion of ylide from the intermediate is also more
likely.
H
O
S
+
CH
H
O
-
COOEt
+S
+
CH
-
COOEt faster
1,2 addtion
reversible C
O
C
EtOOC
H
H
+ S
However, the 1,4-additionstep is irreversible due to formation ofstronger sigma C-C bond, the
equilibrium moves more towards 1,4 addition intermediate and the final outcome is the formation of
cyclopropane ring as the major product.
Note: But whenunstable sulfur ylidesare used, the major product is epoxide.

V.Adityavardhan
7
Web Resource:
http://www.adichemistry.com/organic/namedreactions/coreychaykovsky/corey-chaykovsky-1.html
Problem 6.2
a) b) c) d)
Answer: b
? major product
CH3 H
O N
H
CO2H
S
+
-
Ph
O
,
CHCl3, 0 O
C
cat.
CH3
H
O
Ph
O
CH3
H
O
Ph
O
CH3
H
O
Ph
O
CH3
H
O
Ph
O
MAIL ME AT

V.Adityavardhan
8
Crotonaldehyde froms an iminium
ion with Indoline-2-carboxylic acid
Corey-Chaykovsky
reaction
Explanation
CH3 H
O
N
CO2H
H
: N+
CO2H
CH3
H
OH
HH
+
N+
CO2H
H
CH3
S(CH3)2
Ph
O
:
Ph
O
N+
CO2H
CH3
H
S(CH3)2
Ph
O
N
CO2H
CH3
H
Ph
O
N+
CO2H
CH3
H
OH2
:
Ph
O
N
CO2H
CH3
H
OH
Ph
O
CH3
H
O
-H2O
Regeneration of
Indoline-2-carboxylic acid
The iminium and
PhCO groups try
to avoid steric
interaction with
adjacent CH3
group during the
formation of
cyclopropane ring
and hence both of
them are oriented
trans to methyl
group.
-S(CH3)2
Both -CHO &
PhCO groups are
cis to each other
since they try to
avoid -CH3 during
the reaction.
1,4-addition of
stabilized sulfur
ylide and hence
the formation of
cyclopropane ring.
H
CH3
HN
C
S(CH3)2
H
CO2H
O
Ph

V.Adityavardhan
9
CH
COOMe
COOMe
HMeO
-
CH-
COOMe
COOMe
+ MeOH
CH-
COOMe
COOMe
H
O OH
COOMe
COOMe
COOMe
COOMe-H2O
COOMe
COOMe
HCOOMe
COOMe
ene
reaction
Redraw
H
CH3
CO2Me
CO2Me
H
CO2Me
CO2Me
H
CO2Me
CO2Me
Stereochemistry of Alder-ene reaction
Problem 16.1 (CSIR June 2011)
a) b) c) d)
Answer: c
major product?
OH
OH
OH
(2S)-butane-1,2,4-triol
p-TSA
O
+
OH
O O
OH
O O
O O
OH
O O
OH
Explanation:
* Butane-1,2,4-triolforms an acetal with3-pentanone to give a 1,3-dioxolane (a 5 membered ring) as

V.Adityavardhan
10
major product. The hydroxygroups onadjacent (1 & 2) positions take part inthis reaction.
Formation ofa 1,3-dioxane (a six membered ring) isomer as in options a & b is less preferred due to
steric factors.
* p-TSA, p-toluenesulfonic acid serves as acid catalyst.
O
H
+
O
OH
OH
H
:
O
OH
OH
OH
H
+
:
O
OH
OH
OH2
+
:
p-TSA
-H2O
O
OH
OH
C+
O
OH
OH
C+
:
redraw
O
O
OH
C
redraw
O
O
OH
Note: Work out the stereochemistryat chiralcarbon. In the optionc, the configuration at 2nd carbon is
retained and incidentallyit has S configurationasin the triol.
Problem 17.1 (CSIR Dec 2011)
The major products M and N in the following reaction sequence are:
Problem 35.2 (GATE 2010)
In the reaction sequence,
OH
OH
CH2OH
H
H
OH
i. Hg(OAc)2/MeOH
ii. NaBH4
(CH3)2CO
HCl
X Y
the majorproducts, X and Y, respectively, are:

V.Adityavardhan
11
OOH
OH
OH
MeO
O
O
O
MeO
OH
& OOH
OH
OH
MeO
& OO
O
OH
MeO
OOH
OH
OH
OMe
O
O
O
OH
OMe
& OOH
OH
OH
OMe
OO
O
OH
OMe
&
A) B)
C) D)
Answer: C
Explanation:
* Inthe solvo-oxymercuration, the Hg attacks fromthe less hindered side and the OMe approaches
fromthe top side and attacks the carbon-1, adjacent to ring oxygen, since the positive charge on this
carbon is stabilized due to +M effect ofring oxygen.
* Demercurationis achieved bysodiumborohydride to furnish adeoxysugar.
* Inthe finalstep, protectionofOH groups on4th and 6th carbons is achieved byusing acetone in
presence ofdry HCl.
However, a sixmembered cyclic acetalis formed instead offive membered one eventhough there is
diaxialinteractionfor methylgroups onisopropylidene moiety.
It is because ofinabilityofformationoftrans fusion of6/5membered ring that would be created when
the OH groups on 3rd and 4thcarbons, which are trans to each other, are involved in cyclic acetal
formation.
O
OH
HOH2C
OH
OH
OH
CH2OH
H
H
OH
Hg OAcAcO
O
OH
HOH2C
OH
Hg
+
OAc
O Me
H
:
OOH
OH
OH
OMe
AcOHg
OOH
OH
OH
OMe
O
O
O
OH OMe
NaBH4
(CH3)2CO
HCl
-OAc-
-H+
MAIL ME AT

V.Adityavardhan
12
I
+
Pd(OAc)2, PPh3
Et3N
1) 2) 3) 4)
I
Answer: 3
Explanation:
* It is a Heck reaction. 3-phenylcyclohexene is formed as the onlyproduct instead ofexpected 2-
phenylcyclohexene, due to requirement of synelimination ofhydride.
PPh3
Ph3P
Pd
oxidative addition
SYN
-hydride
elimination
reductive elimination
SYN
addition
PPh3
Ph
Ph3P
IPd
2+
I
PPh3
Ph
Ph3P
IPd
2+
PPh3
Ph3P I
Pd
2+Ph
H
PPh3
Ph3P
IPd
2+
H
Ph
IH
Problem 36.2
Choose the species that is not an intermediate in the following palladium catalyzed Heck
reaction.

V.Adityavardhan
13
Br
+
1) 2) 3) 4)
Answer: 2
O
O
O
O
Pd(PPh3)2
Et3N
Pd(PPh3)2
Br
O
O(Ph3P)2Pd
O
O
Pd(PPh3)2
Br
OO
Pd(PPh3)2
Br
Explanation:
PPh3
Ph3P
Pd
oxidative addition
SYN
-hydride
elimination
reductive elimination
SYN
addition
PPh3
Ph
Ph3P
BrPd
2+
Br
PPh3
Ph
Ph3P
BrPd
2+
PPh3
Ph3P
BrPd
2+
H
BrH
O
O
O
O
O
O
PPh3Ph3P
Br
Pd
2+
Ph
H
H
H
O
O
PPh3Ph3P
Br
Pd
2+
Ph H
H
H
Rotation of
C-C bond to give
more stable conformation
Ph O
O
Ph
O
O
PPh3Ph3P
Br
Pd
2+
H
-complex
-complex
The majorproduct formed on nitrationofN, N-dimethylaniline withconc. H2
SO4
andconc. HNO3

V.Adityavardhan
14
is:
NMe2
NO2
NMe2
NO2
NMe2
NO2
NO2
NMe2
NO2
A) B) C) D)
Answer: A
Explanation:
* The nitrationofN,N-dimethylaniline using 85% H2
SO4
gives 45%meta-nitro product and38% para-
nitro product alongwithN,N,N’,N’-tetramethylbenzidine.
* N,N-dimethylaniline is protonated in stronglyacidic mediumbyforming an N,N-dimethylanilinium
ion, PhNHMe2
+
.
Since NHMe2
+
group is deactivating towards electrophilic substitution due to -I effect, the nitration
occurs mostlyat meta position.
Why para product is also formed?
* Inthe reactionmixture, smallamount offree N,N-dimethylanilineis also present inequilibriumwithits
aniliniumsalt. Thisfree formis highlyreactive and gives para product and favors the equilibriumto the left
side.
N
MeMe
N
+ H
Me
Me
H+
present in less amount
but reacts faster to give
para isomer
present in more amount
but reacts slowly to give
meta isomer
* However, below 83% concentration ofH2
SO4
, meta substitution is seldomobserved and the
composition ofpara product and N,N,N’,N’-tetramethylbenzidine increases with dilutionofsulfuric
acid.
It is also interesting to note that, withdilution ofH2
SO4
, the major product is N,N,N’,N’-tetramethyl
benzidine rather thanthe para isomer.
Me2N NMe2
For example, 63% ofbenzidine derivative is formed when 74.7% ofH2
SO4
is used. This reaction
occurs throughionradical-radicalpair mechanism.
* Ortho product is also possible but formed in less amount due to steric factor.
In electrophilic aromatic substitution reactions, nitro group is meta-directing, because the
nitrogroup:
a) increases electron density at meta position.
b) increases electron density at ortho and para positions.
c) decreases electron density at meta position.
d) decreases electron density at ortho and para positions.

V.Adityavardhan
15
Answer: d
Explanation:
* Nitro group is electronwithdrawinggroup. It withdraws electrons both byinductive aswellas
mesomeric effect i.e., both -I &-M group.Among these, mesomeric effect has more effect on the
reactivity.
In the resonance forms that canbe written due to -M effect ofnitro group, the positive chargeis more
distributed onortho and para positions. Hence these positions become less reactive towards
electrophilic substitution, whencompared to meta position.
N
+
O
-
O
N
+
O
-
O
-
N
+
O
-
O
-
N
+
O
-
O
-
The positive charge is more concentrated on
ortho and para positions making these site less
reactive towards electrophilic substitution
* The comparisonofresonance forms ofWheland intermediates formed duringattack ofelectrophile,
E+
isillustrated below.
The Whelandintermediates formed, during ortho and para attacks are less stable, since one ofthe
contributing structure isleast stable due to presence ofpositive charges onadjacent carbon atoms.
MAIL ME AT

V.Adityavardhan
16
N
+
O
-
O
N
+
O
-
O
H
E
N
+
O
-
O
E
H
N
+
O
-
O
H E
N
+
O
-
O
H
E
N
+
O
-
O
H
E
N
+
O
-
O
H
E
N
+
O
-
O
E
H
N
+
O
-
O
E
H
N
+
O
-
O
E
H
N
+
O
-
O
H E
N
+
O
-
O
H E
N
+
O
-
O
H E
ortho attack
meta attack
para attack
Least stable
Least stable
E+
* The summaryofabove points are illustrated inthefollowing free energydiagram. Not onlythe
activation free energyrequired for metaattack is less, and also the Wheland intermediate formed is
more stable.
Transition
state
Wheland
intermediate
Freeenergy
Reaction coordinate
ortho & para attack
meta attack

V.Adityavardhan
17
O
H
+
OH OH
OH
H
OH
1,2-alkyl
shift
1,2-alkyl
shift
-H+
* However, the product given underoption-1 is also formed as a minor product.The mechanism
leading to formationofthis is also shownbelow.
O
H
+ OH
OH
H
OH
-H+
1,2-alkyl
shift
Why above one is a minor product?
In this case, less stable secondarycarbocation is formed inthe initialstep.
Problem 39.1
In the following reaction sequence, the major product is:
H
MeO
H
H
O
A) B)
C) D)
Answer: A
i) NH2OH, HCl
ii) ArSO2Cl, Py
iii) 
iv) H2O
NH
H
MeO
H
H
O
H
MeO
H
H
ONH
H
MeO
H
H
O
N
H
MeO
H
H
N
O

V.Adityavardhan
18
Explanation:
* The keto group is converted to following E-oxime andthen tosylated.
H
MeO
H
H
O
H
N
OH
H
N
OSO2Ar
S
O
Ar
O
Cl
NH2OH, HCl
E-Oxime
-HCl
Tosylation
Note: Onlypart ofthe reactant is shown.
* Thusformed product undergoes Beckmannrearrangement uponheating andsubsequent treatment
withwater to furnishanamide.
H
N
OSO2Ar
N
H
-
OSO2Ar
N
H
OH

NH
H
O
Beckmann rearrangement
H2O
-ArSO3H
Why not amide as given under option B is formed?
*Yes. It is also a possible product. It is formed due to migration ofmethylgroup when it is antito OH
group as inZ-oxime. However the migrationofmethylgroup is slowerthanthe ring andhence the this
amide is formedas minor product.
H
N
OH
H
ONH
Migration of methyl group occurs since it is anti to the OH group in Z-oxime.
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The intermediateAand the majorproduct B in the following reaction are:

V.Adityavardhan
19
1) 2)
3) 4)
A= an acyl cation B=
A= an acyl nitrene B=
Answer: 4
NH2
N3
O

A B
A= an acyl cation
A= an acyl nitrene
B=
B=
NH
O
N
H
N
H
O
NH
O
N
H
N
H
O
intermediate
Explanation:
* It is Curtius rearrangement. The acylazide loses dinitrogen, N2
to give anintermediate acylnitrene,
whichundergoesrearrangement ofarylgroup onto nitrogenand furnishes anisocyanate.
* The isocyanateis attacked bynucleophilic amine group to give aheterocyclic ring, 1,3-dihydro-2H-
benzimidazol-2-one.
NH2
N
O
N
N
-
NH2
N
O
N
N
NH2
N
O
Acyl nitrene
NH2
N
C
O
isocyanate
N
H
N
H
O
-N2
For the following reactions:A& B, the correct statement is:
Br
COOH
A)
Br
COOH
B)
KOBut
(excess) KOBut
(excess)
1)
2)
3)
4)
COOK
A gives B gives
COOK
A gives B gives
both A & B give
both A & B give
COOK
Answer: 2

V.Adityavardhan
20
Explanation:
* Incase ofreactionA, the most stable conformationofcyclohexane is withtert-butylgroup on
equatorialposition. Inthis conformation, there is a H atomantiperiplanar to the Brand hence undergo
E2
eliminationeasilybygiving cyclohexenewithcarboxylic group.
Br
O
-
O
H
-
OBu
t O
O
-
-HBr
* However, inreaction B, thecarboxylic group itselfis antiperiplanar to Br andhence the elimination
involves decarboxylation.
Br
O O
-
-CO2
-Br
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100173431 solved-problems-in-advanced-organic-synthesis-for-csir-net-chemistry

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