Solved Problems in Advanced Organic Synthesis CSIR NET

1

SOLVED PROBLEMS IN ADVANCED ORGANIC SYNTHESIS (FOR CSIR NET & GATE)

1st EDITION

Release date: 7th, May 2012 Last updated on: 11th Nov 2012 No. of problems solved: 200 No. of pages: 174

om

(This will be updated again)

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SAMPLE COPY

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This book is intended for the aspirants of CSIR NET, SLET, APSET, GATE, IISc and other University entrance exams. Most of the advanced level problems in organic synthesis from previous year question papers are solved and are thoroughly explained with mechanisms. It is a dynamic on-line version; updated frequently. You can purchase this book at different rates depending on your choice. PRICE DETAILS OPTION-1 Rs. 875/- (If you purchase at this rate, free updates are NOT available. However you can get free updates by contributing your knowledge on forum and by helping the author. For more details contact at the following mail id.)

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Problem 1.1

(IISc 2011)

O i) PhMgBr ? +

ii) H

O a)

Ph

b)

O Answer: c

Ph

c)

O

d)

Ph

O

Ph

OH

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ADICHEMISTRY

2

1,2 addition of Grignard reagent

Ph

ht tp V. :// ww Adi t w. ya ad va ich rdh em an ist ry .c

BrMgO

om

i) PhMgBr

O

ii) H+ :

+

H

HO

Ph

:

:

acid catalysed and conjugate  bond assisted removal of OH group

O

:

Ph

:

-H2O

H O H :

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O

O

+

-EtOH -H+ Ph

final removal step of EtOH

O

Think different: What will happen if 1,4 addition occurs?

i) PhMgBr O

ii) H+

O

O

OH

O

H+ Ph O

Ph O

-EtOH

Ph

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).

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Explanation O

om

Ph

Now start arguing!

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O

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Note: Updates & support through forums are only available to those who purchased this book from the author of this book at his site: http://www.adichemistry.com. Problem 1.2 The most appropriate set of reagents for carrying out the following conversion is: O

Cl

a) i) EtMgBr; ii) HCl c) i) C2H5Li; ii) HCl Answer: d

OH

b) i) (C2H5)2CuLi; ii) HCl d) i) HCl; ii) EtMgBr

Explanation: 1,4-addition of HCl furnishes 4-chlorobutanone, which reacts with Grignard reagent to get the desired product. H+

O

ClH2C

HCl

Cl

mechanism Cl OH

O

EtMgBr

H3O+

Cl

OH

OH

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.

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3 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.

4 O

EtMgBr

O MgBrCl

What about other options?

O

H3O+

EtMgBr

Cl

Cl

HCl

OH

om

major product

H+

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Option - a :

Cl-

-H2O

Cl-

* 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). * 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. Option - b O

H3O+

Et2CuLi

O

HCl

Expecting aldol reaction

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). Option-c : The products are same as in case of option-a. Ethyl lithium also shows 1,2 addition like Grignard reagent. NOTE THATTHIS IS A SAMPLE COPY ONLY. YOU NEED TO PURCHASE TO GET THE COMPLETE BOOK.

Problem 1.3 (CSIR DEC 2011) Choose the correct option for M & N formed in reactions sequence given below. O

1) BH3.SMe2 2) PCC

1) PhMgBr

N

M 2) TsOH

3) mCPBA Ph

Ph a)

M=

N=

Ph

HO Ph O

b)

M=

O

O Ph c)

M=

Answer: a

Ph N=

Ph O

O

N=

d)

M=

Ph O N=

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Cl

5

O PhMgBr

H3O

TsOH -H2O

om

* Thus formed product is subjected to hydroboration with BH3.Me2S complex to yield 2phenylcyclohexanol, an anti-Markonikov’s product, which is oxidized to a ketone in presence of PCC. 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.

Ph

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Ph

HO Ph +

Ph

Ph

OH

BH3.SMe2

Baeyer Villger oxidation Ph

O

PCC

O

mCPBA

O

Anti Markonikov's product

Ph-CH- group has more migratory aptitude than CH2 group

Problem 1.4 (CSIR JUNE 2011) The major product formed in the following transformation is: O

1) MeMgCl, CuCl Ph O a)

2)

Cl

O

O

O

b)

c)

Ph Answer: d

d)

Ph

Ph

Ph

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. 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

MeO

O O

CF 3 O

TFAA

MeO

CF 3

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Explanation: * A tertiary alcohol is formed upon 1,2 addition of PhMgBr and is dehydrated in presence of Tosylic acid.

6

(IISc 2009) C6H6 +

a)

H

CHO

EtOOC Answer: a

? major product

CHO

COOEt

 b)

EtOOC H

H

c)

CHO

d)

O

EtOOC

H

om

S

H

H

O

EtOOC

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Problem 6.1

H

H

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Explanation: This is Corey-Chaykovski reaction. Since the sulfur ylide is stable, cyclopropanation occurs majorly 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. O

H +

-

H

O

slow 1, 4 addition

-

H

H

+

S HC

S CH + COOEt

H

irreversible

O

EtOOC

C

H

+ S

H

COOEt

H CHO

Stereochemistry of cyclopropanation step

H

EtOOC

C

CHO & COOEt groups orient in space so as to minimize repulsion. Hence they assume trans postions to each other in cyclopropane ring.

H

H

+ + S

Think different: What will be the product if 1,2-addition occurs? 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 +

-

S CH

faster 1,2 addtion

O

COOEt +

S HC

+ H

reversible

O

EtOOC

-

O C

H

H

C

H

+ S

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

cat. N H

O

S

,

Ph

H

O

CH3 Answer: b

Ph

CHCl3, 0 OC Ph

H

b)

O

-

om

Ph a)

+

? major product

H

H3C

O

CO2H

Ph

H

ht tp V. :// ww Adi t w. ya ad va ich rdh em an ist ry .c

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Problem 6.2

c)

O

O

CH3

H

d)

O

O

O

CH3

O CH3

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8

+

H

om

H

H3C

ht tp V. :// ww Adi t w. ya ad va ich rdh em an ist ry .c

H3C

H H

HO

O

H

CO 2H

N+

Crotonaldehyde froms an iminium ion with Indoline-2-carboxylic acid

CO 2H

N+

The iminium and PhCO groups try CO 2H to avoid steric N H interaction with adjacent CH3 group during the H O formation of CH3 cyclopropane ring Ph C H and hence both of them are oriented S(CH3)2 trans to methyl group.

S(CH3)2 O

H

CH3 Ph

N

CO 2H

:

H Ph

CH3

O

S(CH3)2

-S(CH3)2

CO 2H

N+

Corey-Chaykovsky reaction 1,4-addition of stabilized sulfur ylide and hence the formation of cyclopropane ring.

H

CH3

O

Ph

CO 2H

N+

H

H2O

:

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:

N

-H2O

CO 2H

CH3

O Ph

Both -CHO & PhCO groups are cis to each other since they try to avoid -CH3 during the reaction.

H

CO 2H

N

O CH3 O Ph

Regeneration of Indoline-2-carboxylic acid

H H

O

CH3

O Ph

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Explanation

9

-

COOMe + MeOH CHCOOMe

COOMe HC COOMe

H

OH H

COOMe -H O 2

COOMe CHCOOMe

COOMe

COOMe

om

COOMe

Redraw

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O

ene

COOMe

COOMe

reaction

COOMe

H

COOMe

Stereochemistry of Alder-ene reaction

CO2Me

CO2Me

CO2Me CO2Me

CO2Me

H

CO2Me

H

H

CH3

Problem 16.1

(CSIR June 2011) OH

p-TSA

+

O

major product?

OH

OH (2S)-butane-1,2,4-triol

OH a) HO

b) HO O

O

c) O

O

OH d)

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 major product. The hydroxy groups on adjacent (1 & 2) positions take part in this reaction. Formation of a 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.

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MeO

10

OH

: OH

OH

+

O

OH + H

OH2

+

:

H

OH

O

-H2O OH OH

om

O

OH H p-TSA

ht tp V. :// ww Adi t w. ya ad va ich rdh em an ist ry .c O

C+

redraw OH

OH

O

OH

O

redraw

HO

:

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:O

O

O

O

C+

C

Note: Work out the stereochemistry at chiral carbon. In the option c, the configuration at 2nd carbon is retained and incidentally it has S configuration as in 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, CH2OH O

H H OH HO H

i. Hg(OAc)2/MeOH

(CH3)2CO

X

ii. NaBH4

Y

HCl

the major products, X and Y, respectively, are:

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OH

11 HO O

O &

HO

HO

O HO

MeO

MeO

HO

HO

Answer: C

O

HO

HO

O

O

HO

HO O

O &

HO

O

O

OMe

OMe

O

O MeO

D) O &

O

O &

HO MeO

HO

C)

HO

OMe

OMe

ht tp V. :// ww Adi t w. ya ad va ich rdh em an ist ry .c

Explanation: * In the solvo-oxymercuration, the Hg attacks from the less hindered side and the OMe approaches from the top side and attacks the carbon-1, adjacent to ring oxygen, since the positive charge on this carbon is stabilized due to +M effect of ring oxygen. * Demercuration is achieved by sodium borohydride to furnish a deoxy sugar. * In the final step, protection of OH groups on 4th and 6th carbons is achieved by using acetone in presence of dry HCl. However, a six membered cyclic acetal is formed instead of five membered one even though there is diaxial interaction for methyl groups on isopropylidene moiety. It is because of inability of formation of trans fusion of 6/5 membered ring that would be created when the OH groups on 3rd and 4th carbons, which are trans to each other, are involved in cyclic acetal formation. H

CH2OH O H H OH HO H

HOH2C

O

HOH 2C

-OAc-

O Me

O

+

-H

HO

HO HO

HO HO

HO

:

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O

HO

om

HO

B)

O OMe

HO

AcOHg

+

Hg

AcO Hg OAc

OAc

HO

NaBH4

HO

HO

O OMe

(CH3)2CO HCl

HO

O

O

O OMe

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A)

12

Pd(OAc)2, PPh3

+

Et3N 1)

2)

3)

4)

om

Answer: 3

Explanation: * It is a Heck reaction. 3-phenylcyclohexene is formed as the only product instead of expected 2phenylcyclohexene, due to requirement of syn elimination of hydride.

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I

I

PPh3

Pd

oxidative addition

Ph3P

HI

reductive elimination

Ph

PPh3

2+

Pd

I

Ph3P

PPh3

H

2+

Pd

I

Ph3P

PPh3

Ph

Ph

Ph3P

SYN -hydride elimination

H

Ph

Ph3P

PPh3

2+

Pd

I

SYN addition

2+

Pd

I

Problem 36.2 Choose the species that is not an intermediate in the following palladium catalyzed Heck reaction.

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I

13 O Pd(PPh3)2

O

+ O

Et3N

Pd(PPh3)2

(Ph3P)2Pd

O

2)

O

3) O Br

Answer: 2

Explanation:

Pd(PPh3)2

4)

O Pd(PPh3)2

O

O

om

1)

Br

O

ht tp V. :// ww Adi t w. ya ad va ich rdh em an ist ry .c

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Br

Br

HBr

PPh3

Pd

reductive elimination

O Ph

2+

H

Pd

oxidative addition

Ph3P

PPh3 Br

Ph3P

O

PPh3 2+

Pd

Ph

Br O

Ph3P

O

O

O

-complex

H

Ph

2+

Ph3P

Pd

PPh3

O

Br

SYN -hydride elimination

H

Ph H Ph3P

O

Ph Ph3P

O

H

2+

Pd

PPh3

Br

H

Ph3P Rotation of C-C bond to give more stable conformation

-complex PPh3

2+

Pd

Br

O

SYN addition

O

H

Ph

O

H

2+

Pd

PPh3

Br

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Problem 37.1 (GATE 2004) The major product formed on nitration of N, N-dimethylaniline with conc. H2SO4 and conc. HNO3

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Br

14

A)

NMe2

B)

NMe2 NO2

C)

NMe2 NO2

D)

NMe2

NO2

NO2

om

Answer: A

Explanation: * The nitration of N,N-dimethylaniline using 85% H2SO4 gives 45% meta-nitro product and 38% paranitro product along with N,N,N’,N’-tetramethyl benzidine. * N,N-dimethyl aniline is protonated in strongly acidic medium by forming an N,N-dimethyl anilinium ion, PhNHMe2+. Since NHMe2+ group is deactivating towards electrophilic substitution due to -I effect, the nitration occurs mostly at meta position.

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NO2

Why para product is also formed? * In the reaction mixture, small amount of free N,N-dimethylaniline is also present in equilibrium with its anilinium salt. This free form is highly reactive and gives para product and favors the equilibriumto the left side. Me

Me

Me

Me

N

+

N

H

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 of H2SO4, meta substitution is seldom observed and the composition of para product and N,N,N’,N’-tetramethyl benzidine increases with dilution of sulfuric acid. It is also interesting to note that, with dilution of H2SO4, the major product is N,N,N’,N’-tetramethyl benzidine rather than the para isomer. Me2N

NMe2

For example, 63% of benzidine derivative is formed when 74.7% of H2SO4 is used. This reaction occurs through ion radical-radical pair mechanism. * Ortho product is also possible but formed in less amount due to steric factor. Problem 37.2 (GATE 2004) In electrophilic aromatic substitution reactions, nitro group is meta-directing, because the nitro group: 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.

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is:

15

+

N

O

-

O

+

N

O

-

O

+

O

-

O

N

-

om

O

+

O

-

N

ht tp V. :// ww Adi t w. ya ad va ich rdh em an ist ry .c

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Explanation: * Nitro group is electron withdrawing group. It withdraws electrons both by inductive as well as mesomeric effect i.e., both -I & -M group. Among these, mesomeric effect has more effect on the reactivity. In the resonance forms that can be written due to -M effect of nitro group, the positive charge is more distributed on ortho and para positions. Hence these positions become less reactive towards electrophilic substitution, when compared to meta position.

The positive charge is more concentrated on ortho and para positions making these site less reactive towards electrophilic substitution

* The comparison of resonance forms of Wheland intermediates formed during attack of electrophile, E+ is illustrated below. The Wheland intermediates formed, during ortho and para attacks are less stable, since one of the contributing structure is least stable due to presence of positive charges on adjacent carbon atoms. WANT TO PURCHASE THE ENTIRE BOOK? TO KNOW THE METHOD OF PAYMENT MAIL ME AT

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Answer: d

16

+

-

O

+

O

-

N

N

O

H

H

E

E

+

O

-

O

O

+

N

N

-

H

H

E

E

O

+

O

om

Least stable

-

O

N

E+

O

-

O

O

-

O

O

-

ht tp V. :// ww Adi t w. ya ad va ich rdh em an ist ry .c +

N

meta attack

+

O

+

O

-

O

H

E

+

O

-

O

N

H

+

E

O

+

O

-

H E

O

N

H

E

-

N

H E

H E

N

para attack

+

O

N

N

H E

+

O

-

N

H

E

Least stable

* The summary of above points are illustrated in the following free energy diagram. Not only the activation free energy required for meta attack is less, and also the Wheland intermediate formed is more stable.

Transition state

Free energy

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ortho attack

O

ortho & para attack meta attack

Wheland intermediate Reaction coordinate

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O

17

O

shift

HO

HO

om

1,2-alkyl shift

-H+ HO

HO

H

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H

* However, the product given under option-1 is also formed as a minor product. The mechanism leading to formation of this is also shown below. H

-H+

1,2-alkyl

+

H

O

shift

HO

HO

HO

Why above one is a minor product? In this case, less stable secondary carbocation is formed in the initial step. Problem 39.1 In the following reaction sequence, the major product is:

O

i) NH2OH, HCl

H

H MeO

ii) ArSO2Cl, Py

H

iii)  iv) H2O

HN

A)

B)

H

H

H

MeO O N

H

C) H Answer: A

O

H

H

H

MeO

MeO

HN

O

H

D) H

H MeO

N O H

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1,2-alkyl +

18

Ar OH N

NH2OH, HCl

-HCl H

H

MeO

Cl

Tosylation

E-Oxime

H

om

H

O

O

H

OSO2Ar N

S

Note: Only part of the reactant is shown. * Thus formed product undergoes Beckmann rearrangement upon heating and subsequent treatment with water to furnish an amide.

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O

OSO 2Ar N

-

N

N

OH

HN

H2O



H

OSO 2Ar

O

-ArSO3H

H

H

H

Beckmann rearrangement

Why not amide as given under option B is formed? * Yes. It is also a possible product. It is formed due to migration of methyl group when it is anti to OH group as in Z-oxime. However the migration of methyl group is slower than the ring and hence the this amide is formed as minor product. N

OH

HN

O

H

H Migration of methyl group occurs since it is anti to the OH group in Z-oxime.

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Problem 52.1 (CSIR JUNE 2012) The intermediate A and the major product B in the following reaction are:

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Explanation: * The keto group is converted to following E-oxime and then tosylated.

19



N3

A intermediate

NH2

B H N

B=

2) A= an acyl cation

NH

N H H N

O 3) A= an acyl nitrene

B=

O

B=

4) A= an acyl nitrene

NH

Answer: 4

om

1) A= an acyl cation

B=

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O

O

N H

Explanation: * It is Curtius rearrangement. The acyl azide loses dinitrogen, N2 to give an intermediate acyl nitrene, which undergoes rearrangement of aryl group on to nitrogen and furnishes an isocyanate. * The isocyanate is attacked by nucleophilic amine group to give a heterocyclic ring, 1,3-dihydro-2Hbenzimidazol-2-one. O

O

-

N NH2

N

N

N

N

O

N

-N2

N

NH2

NH2

Acyl nitrene

N

C

O

NH2

isocyanate

H N O N H

Problem 53.1 (CSIR JUNE 2012) For the following reactions: A & B, the correct statement is: Br

A)

KOBut (excess)

Br

B)

COOH

COOH

1) A gives

2) A gives Answer: 2

KOBut (excess)

B gives

COOK

B gives

COOK

3) both A & B give

4) both A & B give

COOK

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O

20

-HBr O

O -

OBu

-

O

t

om

H

-

* However, in reaction B, the carboxylic group itself is antiperiplanar to Br and hence the elimination involves decarboxylation.

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Br O

Br

-CO2 -Br

O

O

-

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Explanation: * In case of reaction A, the most stable conformation of cyclohexane is with tert-butyl group on equatorial position. In this conformation, there is a H atom anti periplanar to the Br and hence undergo E2 elimination easily by giving cyclohexene with carboxylic group.