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1. 1
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ADICHEMISTRY
SOLVED PROBLEMS IN ADVANCED ORGANIC SYNTHESIS
(FOR CSIR NET & GATE)
1st EDITION
SAMPLE COPY
Release date: 7th, May 2012
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Last updated on: 7th July 2012
Current No. of problems solved: 108 (This will be updated again)
Current No. of pages: 104
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FOR COMPLETE DETAILS
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This book is intended for the aspirants of CSIR NET, SLET, APSET, GATE, IISc and other
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Problem 1.1 (IISc 2011)
ht
O
i) PhMgBr
?
O ii) H+
a) Ph b) Ph c) Ph d) Ph
O O O OH
Answer: c

2. 2
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Explanation
O
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O
1,2 addition of Grignard reagent
i) PhMgBr
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BrMgO Ph
.c
O
ii) H+
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+
H
:
HO Ph
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:
:
O acid catalysed and conjugate
 bond assisted removal of
:
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-H2O OH group
Ph
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:
H O H
:
+
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O
-EtOH final removal step of EtOH
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-H+
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Ph
O
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Think different:
What will happen if 1,4 addition occurs?
O OH O O
H+
i) PhMgBr
Ph Ph
-EtOH
ii) H+ O O Ph
O
1, 4 - addtion of PhMgBr
Same product! So it might be the actual mechanism? But slim chances. Why? The possible explanation
might go like this:
i) the positive charge on 4th position is diminished due to contribution of p-electrons of adjacent ethoxy
‘O’ through conjugation (+M effect).

3. 3
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ii) The enolate ion form is less stable due to -I effect of ‘O’.
iii) We also know that: 1,2 addition is kinetically more favorable than 1,4-addition in case of Grignard
reagents. It is because the R group attached to Mg in GR is a hard nucleophile and prefers carbonyl
carbon with considerable positive charge (hard electrophile).
And if this is the mechanism, the removal of ethanol may give another product, though less likely, as
shown below.
O
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Ph
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Now start arguing!
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Web Resource:
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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:
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O Cl OH
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a) i) EtMgBr; ii) HCl b) i) (C2H5)2CuLi; ii) HCl
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c) i) C2H5Li; ii) HCl d) i) HCl; ii) EtMgBr
Answer: d
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Explanation:
1,4-addition of HCl furnishes 4-chlorobutanone, which reacts with Grignard reagent to get the desired
product.
ht
H+
O HCl Cl O EtMgBr H3O+
Cl OH
Cl- mechanism
OH Cl OH
H2C
However, the yields may not be satisfactory due to side reaction that is possible in the second step
with Grignard reagent. It may undergo Wurtz like coupling reaction with -CH2Cl group.

4. 4
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Cl O EtMgBr O
MgBrCl
What about other options?
Option - a :
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major product
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O EtMgBr H3O+ OH HCl Cl Cl
H+
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Cl-
-H2O Cl-
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* 1,2-addition occurs with Grignard reagent, since the ethyl group attached to Mg has considerable
positive charge and is a hard nucleophile. It prefers to attack 2nd carbon (hard electrophile).
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* In the reaction of allylic alcohol with HCl, the Cl- prefers to attack the allylic carbocation from less
hindered end. Hence the major product is 1-chloro-3-methyl-2-pentene.
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Option - b
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O Et2CuLi H3O+ O HCl
Expecting aldol reaction
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1,4-addition occurs with Lithium diethyl cuprate, since ethyl group attached to copper is a soft
nucleophile and prefers carbon at 4th position (soft electrophile).
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Option-c :
The products are same as in case of option-a. Ethyl lithium also shows 1,2 addition like Grignard
reagent.
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Problem 1.3 (CSIR DEC 2011)
Choose the correct option for M & N formed in reactions sequence given below.
ht
O 1) BH3.SMe2
1) PhMgBr 2) PCC
M N
2) TsOH 3) mCPBA
Ph Ph HO Ph Ph
O
a) M= N= O b) M= N=
O O
Ph Ph Ph Ph
O
c) M= N= O d) M= N=
Answer: a

5. 5
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Explanation:
* A tertiary alcohol is formed upon 1,2 addition of PhMgBr and is dehydrated in presence of Tosylic
acid.
O HO Ph Ph
+
PhMgBr H3O TsOH
-H2O
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* Thus formed product is subjected to hydroboration with BH3.Me2S complex to yield 2-
phenylcyclohexanol, an anti-Markonikov’s product, which is oxidized to a ketone in presence of PCC.
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The keto compound is subjected to Baeyer Villiger oxidation with mCPBA to get a lactone. The
PhCH- group is migrated onto oxygen in preference to CH2 group.
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Baeyer Villger
Ph Ph Ph oxidation Ph
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BH3.SMe2 OH PCC O O
mCPBA
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Anti Markonikov's Ph-CH- group has more
product migratory aptitude than
CH2 group
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Problem 1.4 (CSIR JUNE 2011)
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The major product formed in the following transformation is:
O
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1) MeMgCl, CuCl
2) Cl
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Ph
O O
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O O
a) b) c) d)
Ph Ph Ph Ph
Answer: d
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Explanation:
* The Grignard reagent reacts with CuCl to give Me2CuMgCl, an organocopper compound also
known as Gilman reagent that is added to the -unsaturated ketone in 1,4-manner.
ht
Initially copper associates with the double bond to give a complex, which then undergoes oxidative
addition followed by reductive elimination.
Thus formed enolate ion acts as a nucleophile and substitutes the Cl group of allyl chloride.
The attack on allyl chloride is done from the opposite side of more bulky phenyl group.
2 MeMgCl + CuCl Me2CuMgCl + MgCl2
OMe O- OMe O
O
F 3C O CF 3
MeO CF 3
MeO O O
TFAA

6. 6
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NOTE THATTHIS IS A SAMPLE COPY ONLY. YOU NEED TO PURCHASE TO GET THE COMPLETE BOOK.
Problem 6.1 (IISc 2009)
C6H6
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S + ? major product
COOEt CHO

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a) b) c) O d) O
H CHO EtOOC CHO EtOOC EtOOC H
EtOOC H H H H H H
Answer: a
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Explanation:
This is Corey-Chaykovski reaction. Since the sulfur ylide is stable, cyclopropanation occurs majorly
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through 1,4-addition route. The product is a thermodynamic one.
The CHO and COOEt groups get trans positions in the cyclopropane ring. This occurs since they
tend to orient as far away as possible during the cyclopropanations step to avoid steric repulsion.
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O -
O H O
slow
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H H EtOOC
H 1, 4 addition H
+ - + H
S CH + S HC C + S
COOEt irreversible H
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COOEt
w.
H CHO
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CHO & COOEt groups orient
Stereochemistry of EtOOC H in space so as to minimize
H
cyclopropanation step repulsion. Hence they assume
C
H trans postions to each other in
+ + cyclopropane ring.
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S
Think different:
What will be the product if 1,2-addition occurs?
ht
An epoxide is formed. It is kinetically favored product.
But this is minor product. Why?
Since 1,2-addition step is reversible, the expulsion of ylide from the intermediate is also more
likely.
COOEt
COOEt faster
O 1,2 addtion - EtOOC O
+ - + O
S CH + S HC C + S
H
H reversible H C
H
However, the 1,4-addition step is irreversible due to formation of stronger 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 when unstable sulfur ylides are used, the major product is epoxide.

7. 7
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Web Resource:
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Problem 6.2
O
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cat. CO2H +
N S -
O , Ph
H
? major product
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H3C H
CHCl3, 0 OC
Ph H Ph H Ph H Ph H
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a) b) c) d)
O O O O O O O O
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CH3 CH3 CH3 CH3
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Answer: b
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ht

8. 8
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Explanation
CO 2H -H2O CO 2H
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N
:
+ N+
H H
H O HO
H
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H3C H
H3C
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Crotonaldehyde froms an iminium CO 2H S(CH3)2
N+
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ion with Indoline-2-carboxylic acid
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O
H
CH3 Ph
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The iminium and CO 2H
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:
PhCO groups try N
CO 2H to avoid steric
N H interaction with H
Corey-Chaykovsky
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adjacent CH3 Ph
CH3 reaction
O H group during the
formation of O S(CH3)2 1,4-addition of
H CH3 cyclopropane ring
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stabilized sulfur
Ph C -S(CH3)2
and hence both of ylide and hence
them are oriented the formation of
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S(CH3)2 trans to methyl CO 2H cyclopropane ring.
group. N+
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H
CH3
O
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Ph
CO 2H
N+
ht
: H
H2O CH3
O
Ph
Both -CHO & H CO 2H
PhCO groups are N
cis to each other O H Regeneration of
since they try to CH3 Indoline-2-carboxylic acid
O H O
avoid -CH3 during CH3
the reaction. Ph O
Ph

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