SOLVED PROBLEMS IN ADVANCED ORGANIC SYNTHESIS

1 The original copy of this book is available only from http://www.adichemistry.com/. Do not distribute this book without permisssion of author. ADICHEMISTRY SOLVED PROBLEMS IN ADVANCED ORGANIC SYNTHESIS (FOR CSIR NET & GATE) 1st EDITION SAMPLE COPY Release date: 7th, May 2012Updates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ Last updated on: 7th July 2012 Current No. of problems solved: 108 (This will be updated again) Current No. of pages: 104 om VISIT http://www.adichemistry.com/common/htmlfiles/csir-gate-chemistry.html .c FOR COMPLETE DETAILS em an ry This book is intended for the aspirants of CSIR NET, SLET, APSET, GATE, IISc and other ich rdh ist University entrance exams. Most of the advanced level problems in organic synthesis from previous year question papers are solved and thoroughly explained. It is a dynamic on-line version; updated frequently. If you are interested see for the current offer rate ad va below. This price includes future updates too. PRICE DETAILS tya Rs. 725/- (If you purchase at this rate, there will be free updates only up to 200 problems & this price may be revised on every update depending on the number of problems added.) ww Adi TO KNOW THE METHOD OF PAYMENT w. MAIL ME AT tp V. ADICHEMADI @ GMAIL.COM 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. :// NOTE THATTHIS IS A SAMPLE COPY ONLY. YOU NEED TO PURCHASE TO GET THE COMPLETE BOOK. 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 The original copy of this book is available only from http://www.adichemistry.com/. Do not distribute this book without permisssion of author. Explanation OUpdates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ O 1,2 addition of Grignard reagent i) PhMgBr om BrMgO Ph .c O ii) H+ em an ry + H : HO Ph ich rdh ist : : O acid catalysed and conjugate  bond assisted removal of : ad va -H2O OH group Ph tya : H O H : + ww Adi O -EtOH final removal step of EtOH w. -H+ tp V. Ph O :// ht 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 The original copy of this book is available only from http://www.adichemistry.com/. Do not distribute this book without permisssion of author. 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. OUpdates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ Ph om Now start arguing! .c Web Resource: em an ry http://www.adichemistry.com/organic/organicreagents/grignard/grignard-reagent-reaction-1.html ich rdh WANT TO PURCHASE THE ENTIRE BOOK? ist TO KNOW THE METHOD OF PAYMENT MAIL ME AT ADICHEMADI@GMAIL.COM ad va 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. tya Problem 1.2 The most appropriate set of reagents for carrying out the following conversion is: ww Adi O Cl OH w. a) i) EtMgBr; ii) HCl b) i) (C2H5)2CuLi; ii) HCl tp V. c) i) C2H5Li; ii) HCl d) i) HCl; ii) EtMgBr Answer: d :// 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 The original copy of this book is available only from http://www.adichemistry.com/. Do not distribute this book without permisssion of author. Cl O EtMgBr O MgBrCl What about other options? Option - a :Updates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ major product om O EtMgBr H3O+ OH HCl Cl Cl H+ .c Cl- -H2O Cl- em an ry ich rdh ist * 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). ad va * 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. tya Option - b ww Adi O Et2CuLi H3O+ O HCl Expecting aldol reaction w. 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). tp V. 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. 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 The original copy of this book is available only from http://www.adichemistry.com/. Do not distribute this book without permisssion of author. 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 -H2OUpdates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ * 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. om 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. .c Baeyer Villger Ph Ph Ph oxidation Ph em an ry BH3.SMe2 OH PCC O O mCPBA ich rdh O ist Anti Markonikovs Ph-CH- group has more product migratory aptitude than CH2 group ad va Problem 1.4 (CSIR JUNE 2011) tya The major product formed in the following transformation is: O ww Adi 1) MeMgCl, CuCl 2) Cl w. Ph O O tp V. O O a) b) c) d) Ph Ph Ph Ph Answer: d :// 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 The original copy of this book is available only from http://www.adichemistry.com/. Do not distribute this book without permisssion of author. NOTE THATTHIS IS A SAMPLE COPY ONLY. YOU NEED TO PURCHASE TO GET THE COMPLETE BOOK. Problem 6.1 (IISc 2009) C6H6Updates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ S + ? major product COOEt CHO  om a) b) c) O d) O H CHO EtOOC CHO EtOOC EtOOC H EtOOC H H H H H H Answer: a .c em an ry Explanation: This is Corey-Chaykovski reaction. Since the sulfur ylide is stable, cyclopropanation occurs majorly ich rdh ist 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. ad va O - O H O slow tya H H EtOOC H 1, 4 addition H + - + H S CH + S HC C + S COOEt irreversible H ww Adi COOEt w. H CHO tp V. 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. :// 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 The original copy of this book is available only from http://www.adichemistry.com/. Do not distribute this book without permisssion of author. Web Resource: http://www.adichemistry.com/organic/namedreactions/coreychaykovsky/corey-chaykovsky-1.html Problem 6.2 OUpdates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ cat. CO2H + N S - O , Ph H ? major product om H3C H CHCl3, 0 OC Ph H Ph H Ph H Ph H .c a) b) c) d) O O O O O O O O em an ry CH3 CH3 CH3 CH3 ich rdh ist Answer: b ad va WANT TO PURCHASE THE ENTIRE BOOK? tya TO KNOW THE METHOD OF PAYMENT MAIL ME AT ADICHEMADI@GMAIL.COM ww Adi 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. w. tp V. :// ht

8 The original copy of this book is available only from http://www.adichemistry.com/. Do not distribute this book without permisssion of author. Explanation CO 2H -H2O CO 2HUpdates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ N : + N+ H H H O HO H om H3C H H3C .c Crotonaldehyde froms an iminium CO 2H S(CH3)2 N+ em an ion with Indoline-2-carboxylic acid ry O H CH3 Ph ich rdh ist The iminium and CO 2H ad va : PhCO groups try N CO 2H to avoid steric N H interaction with H Corey-Chaykovsky tya adjacent CH3 Ph CH3 reaction O H group during the formation of O S(CH3)2 1,4-addition of H CH3 cyclopropane ring ww Adi stabilized sulfur Ph C -S(CH3)2 and hence both of ylide and hence them are oriented the formation of w. S(CH3)2 trans to methyl CO 2H cyclopropane ring. group. N+ tp V. H CH3 O :// 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 The original copy of this book is available only from http://www.adichemistry.com/. Do not distribute this book without permisssion of author. COOMe COOMe - MeO H HC CH- + MeOH COOMe COOMe O OHUpdates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ COOMe -H O COOMe H 2 COOMe CH- COOMe COOMe COOMe om Redraw .c em an ry ene COOMe COOMe ich rdh ist reaction COOMe COOMe H ad va Stereochemistry of Alder-ene reaction tya CO2Me CO2Me CO2Me ww Adi CO2Me CO2Me CO2Me w. H H H tp V. CH3 Problem 16.1 (CSIR June 2011) :// 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

10 The original copy of this book is available only from http://www.adichemistry.com/. Do not distribute this book without permisssion of author. 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. OHUpdates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ OH : OH OH :O OH OH H p-TSA O O om O + + H OH OH2 : + .c H -H2O em an ry OH ich rdh ist OH O C+ ad va redraw tya OH OH O OH HO redraw : O ww Adi O O O C C+ w. tp V. 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, ht CH2OH H O i. Hg(OAc)2/MeOH (CH3)2CO H X Y OH HCl ii. NaBH4 HO H the major products, X and Y, respectively, are: NOTE THATTHIS IS A SAMPLE COPY ONLY. YOU NEED TO PURCHASE TO GET THE COMPLETE BOOK.

11 The original copy of this book is available only from http://www.adichemistry.com/. Do not distribute this book without permisssion of author. A) HO B) HO HO O O O HO O & O HO O & O HO HO HO O MeO MeO MeO MeOUpdates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ C) HO D) HO HO O O HO O & O & O O O HO om HO HO HO O OMe OMe OMe OMe Answer: C .c Explanation: em an ry * 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 ich rdh ist 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 ad va 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. tya 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. ww Adi H w. CH2OH O Me HO HOH 2C : HOH2C O O + tp V. H O -OAc- -H O HO H HO HO HO OMe OH HO HO HO + AcOHg AcO Hg OAc Hg H :// OAc HO NaBH4 O (CH3)2CO HO O O ht HO OMe O HCl HO OMe WANT TO PURCHASE THE ENTIRE BOOK? TO KNOW THE METHOD OF PAYMENT MAIL ME AT ADICHEMADI@GMAIL.COM 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.

12 The original copy of this book is available only from http://www.adichemistry.com/. Do not distribute this book without permisssion of author. I Pd(OAc)2, PPh3 + Et3N I 1) 2) 3) 4)Updates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ Answer: 3 om Explanation: * It is a Heck reaction. 3-phenylcyclohexene is formed as the only product instead of expected 2- phenylcyclohexene, due to requirement of syn elimination of hydride. .c em an ry ich rdh ist I PPh3 Pd ad va Ph3P oxidative addition HI tya reductive elimination PPh3 2+ Ph Pd I ww Adi Ph3P w. PPh3 2+ H Pd I tp V. Ph3P PPh3 :// 2+ Ph Ph Pd I SYN Ph3P SYN -hydride addition H PPh3 ht elimination Ph 2+ Pd Ph3P I Problem 36.2 Choose the species that is not an intermediate in the following palladium catalyzed Heck reaction.

13 The original copy of this book is available only from http://www.adichemistry.com/. Do not distribute this book without permisssion of author. Br O O Pd(PPh3)2 O + O Et3N Br Br Pd(PPh3)2 (Ph3P)2Pd O O Pd(PPh3)2Updates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ 1) 2) 3) 4) O O Pd(PPh3)2 O O Br om Answer: 2 Explanation: .c em an ry Br ich rdh ist HBr PPh3 reductive elimination Pd oxidative addition ad va Ph3P PPh3 2+ O H Pd Br tya Ph3P PPh3 Ph O 2+ Ph Pd Br ww Adi Ph3P O O O O w. -complex tp V. Ph H 2+ Ph3P Pd PPh3 Br O O :// SYN -complex -hydride PPh3 elimination O 2+ H Ph Pd Br O Ph H Ph3P H ht O 2+ H Ph3P Pd PPh3 O SYN H H addition Br Ph 2+ Ph3P Pd PPh3 Rotation of C-C bond to give Br more stable conformation NOTE THATTHIS IS A SAMPLE COPY ONLY. YOU NEED TO PURCHASE TO GET THE COMPLETE BOOK. Problem 37.1 (GATE 2004) The major product formed on nitration of N, N-dimethylaniline with conc. H2SO4 and conc. HNO3

14 The original copy of this book is available only from http://www.adichemistry.com/. Do not distribute this book without permisssion of author. is: A) NMe2 B) NMe2 C) NMe2 D) NMe2 NO2 NO2 NO2Updates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ Answer: A NO2 NO2 om Explanation: * The nitration of N,N-dimethylaniline using 85% H2SO4 gives 45% meta-nitro product and 38% para- nitro 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 .c ion, PhNHMe2+. em an ry Since NHMe2+ group is deactivating towards electrophilic substitution due to -I effect, the nitration occurs mostly at meta position. ich rdh ist Why para product is also formed? * In the reaction mixture, small amount of free N,N-dimethylaniline is also present in equilibrium with its ad va anilinium salt. This free form is highly reactive and gives para product and favors the equilibriumto the left side. tya Me Me Me Me + H N N H+ ww Adi w. present in less amount present in more amount but reacts faster to give but reacts slowly to give tp V. para isomer 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 ht 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.

15 The original copy of this book is available only from http://www.adichemistry.com/. Do not distribute this book without permisssion of author. Answer: d 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 moreUpdates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ distributed on ortho and para positions. Hence these positions become less reactive towards electrophilic substitution, when compared to meta position. om - - - - - - - O O O O O O O O + + + + N N N N .c em an ry The positive charge is more concentrated on ortho and para positions making these site less ich rdh ist reactive towards electrophilic substitution * The comparison of resonance forms of Wheland intermediates formed during attack of electrophile, ad va 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. tya WANT TO PURCHASE THE ENTIRE BOOK? TO KNOW THE METHOD OF PAYMENT ww Adi MAIL ME AT ADICHEMADI@GMAIL.COM w. 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. tp V. :// ht

16 The original copy of this book is available only from http://www.adichemistry.com/. Do not distribute this book without permisssion of author. - - - - O O O O O O O O + + + + ortho attack N N N N H H H H E E E EUpdates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ Least stable om - O O + - - - - N O O O O O O O O .c + + + + E+ meta attack N N N N em an ry H H H H ich rdh E E E E ist ad va - - - - O O O O O O O O + tya + + + N N N N para attack ww Adi H E H E H E H E w. Least stable tp V. * 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 ht Free energy ortho & para attack meta attack Wheland intermediate Reaction coordinate NOTE THATTHIS IS A SAMPLE COPY ONLY. YOU NEED TO PURCHASE TO GET THE COMPLETE BOOK.

17 The original copy of this book is available only from http://www.adichemistry.com/. Do not distribute this book without permisssion of author. 1,2-alkyl + H shift O HO HO 1,2-alkylUpdates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ shift om -H+ HO HO H .c * However, the product given under option-1 is also formed as a minor product. The mechanism em an ry leading to formation of this is also shown below. ich rdh ist H 1,2-alkyl -H+ + O HO shift H HO HO ad va Why above one is a minor product? tya In this case, less stable secondary carbocation is formed in the initial step. Problem 39.1 ww Adi In the following reaction sequence, the major product is: w. O tp V. i) NH2OH, HCl H ii) ArSO2Cl, Py H H iii)  :// MeO iv) H2O HN O HN O ht A) H B) H H H H H MeO MeO O N H N H O C) D) H H H H MeO MeO Answer: A

18 The original copy of this book is available only from http://www.adichemistry.com/. Do not distribute this book without permisssion of author. Explanation: * The keto group is converted to following E-oxime and then tosylated. Ar OH OSO2Ar N S O O N O ClUpdates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ H NH2OH, HCl -HCl H H H H Tosylation om MeO E-Oxime Note: Only part of the reactant is shown. * Thus formed product undergoes Beckmann rearrangement upon heating and subsequent treatment .c with water to furnish an amide. em an ry OSO 2Ar ich rdh ist - N OSO 2Ar N N OH HN O  H2O ad va -ArSO3H H H H H Beckmann rearrangement tya 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 ww Adi 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. w. tp V. N HN O OH :// H H Migration of methyl group occurs since it is anti to the OH group in Z-oxime. WANT TO PURCHASE THE ENTIRE BOOK? ht TO KNOW THE METHOD OF PAYMENT MAIL ME AT ADICHEMADI@GMAIL.COM 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 52.1 (CSIR JUNE 2012) The intermediate A and the major product B in the following reaction are:

19 The original copy of this book is available only from http://www.adichemistry.com/. Do not distribute this book without permisssion of author. O N3  A B NH2 intermediate H O N 1) A= an acyl cation B= 2) A= an acyl cation B= OUpdates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ NH N H H O om N 3) A= an acyl nitrene B= 4) A= an acyl nitrene B= O NH N Answer: 4 H .c em an Explanation: ry * 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. ich rdh ist * The isocyanate is attacked by nucleophilic amine group to give a heterocyclic ring, 1,3-dihydro-2H- benzimidazol-2-one. ad va O - O O N N N N -N2 N N N tya N C O NH2 NH2 NH2 NH2 Acyl nitrene isocyanate ww Adi w. H tp V. N O N H :// Problem 53.1 (CSIR JUNE 2012) For the following reactions: A & B, the correct statement is: Br KOBut (excess) Br KOBut (excess) ht A) B) COOH COOH 1) A gives B gives 3) both A & B give COOK 2) A gives B gives 4) both A & B give COOK COOK Answer: 2

20 The original copy of this book is available only from http://www.adichemistry.com/. Do not distribute this book without permisssion of author. 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. Br O -HBrUpdates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ - O - H O - t O OBu om * However, in reaction B, the carboxylic group itself is antiperiplanar to Br and hence the elimination involves decarboxylation. .c Br em an ry -CO2 ich rdh -Br ist - O O ad va WANT TO PURCHASE THE ENTIRE BOOK? TO KNOW THE METHOD OF PAYMENT MAIL ME AT tya ADICHEMADI@GMAIL.COM 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. ww Adi w. tp V. :// ht

SOLVED PROBLEMS IN ADVANCED ORGANIC SYNTHESIS

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SOLVED PROBLEMS IN ADVANCED ORGANIC SYNTHESIS Document Transcript