1The 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 Updates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ http://www.adichemistry.com/common/htmlfiles/csir-gate-chemistry.html FOR COMPLETE DETAILS 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.) 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 O O a) Ph ht tp V. :// ww Adi tya w. ad va ich rdh em an ist ry .c TO KNOW THE METHOD OF PAYMENT MAIL ME AT VISIT ADICHEMADI @ GMAIL.COM (IISc 2011) i) PhMgBr ? ii) H + b) Ph c) Ph d) O Answer: c O O om Ph OH Release date: 7th, May 2012 Last updated on: 11th Nov 2012 No. of problems solved: 200 No. of pages: 174 (This will be updated again) 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 O Updates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ O i) PhMgBr BrMgO Ph 1,2 addition of Grignard reagent ii) H+ : H + : HO Ph -H2O Ph -EtOH -H+ Ph Think different: What will happen if 1,4 addition occurs? O i) PhMgBr O ii) H+ OH 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). ht tp V. :// ww Adi tya w. ad va ich rdh em an ist ry .c O : O acid catalysed and conjugate bond assisted removal of OH group : O + : H O H : final removal step of EtOH O O H+ Ph O -EtOH om O Ph 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. Updates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ O The original copy of this book is available only from http://www.adichemistry.com/. Do not distribute this book without permisssion of author. Now start arguing! Web Resource: http://www.adichemistry.com/organic/organicreagents/grignard/grignard-reagent-reaction-1.html WANT TO PURCHASE THE ENTIRE BOOK? TO KNOW THE METHOD OF PAYMENT MAIL ME AT 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 OH a) i) EtMgBr; ii) HCl c) i) C2H5Li; ii) HCl Answer: d Explanation: 1,4-addition of HCl furnishes 4-chlorobutanone, which reacts with Grignard reagent to get the desired product. H+ ClH2C 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. ht tp V. :// ww Adi tya w. ad va ich rdh em an ist ry .c
[email protected] b) i) (C2H5)2CuLi; ii) HCl d) i) HCl; ii) EtMgBr O HCl Cl O EtMgBr H3O+ Cl OH mechanism Cl OH OH om Ph 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 O EtMgBr H3O+ OH H+ -H2O ClHCl Cl ht tp V. :// ww Adi tya w. ad va ich rdh em an ist ry .c Et2CuLi H3O+ O HCl * 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 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) PhMgBr M 2) TsOH Ph a) M= 1) BH3.SMe2 2) PCC 3) mCPBA Ph N HO Ph O O b) M= N= N= Ph c) M= N= Ph O d) M= Ph N= Answer: a om Cl Cl- Ph O O Ph O 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 PhMgBr H3O + HO Ph TsOH -H2O Ph Updates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ Ph BH3.SMe2 Problem 1.4 (CSIR JUNE 2011) The major product formed in the following transformation is: O 1) MeMgCl, CuCl Ph O a) 2) Cl 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. 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 OMe OF 3C MeO O O O CF 3 MeO CF 3 ht tp V. :// ww Adi tya w. ad va ich rdh em an ist ry .c Ph Ph Baeyer Villger oxidation Ph mCPBA OH PCC O O O Anti Markonikov's product Ph-CH- group has more migratory aptitude than CH2 group O O O b) c) d) Ph Ph Me2CuMgCl + MgCl2 OMe O O TFAA om Ph * 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. 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 Updates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ (IISc 2009) C6H6 + S COOEt a) CHO b) ? major product EtOOC H H CHO c) EtOOC H om d) H O H C H EtOOC H CHO O H EtOOC H O H 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 S CH + COOEt + - ht tp V. :// ww Adi tya w. ad va ich rdh em an ist ry .c H slow 1, 4 addition irreversible O H S HC + Answer: a H EtOOC + S COOEt H H CHO Stereochemistry of cyclopropanation step EtOOC H H C H + + S CHO & COOEt groups orient in space so as to minimize repulsion. Hence they assume trans postions to each other in cyclopropane ring. 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 + - O + H faster 1,2 addtion reversible COOEt S HC + O - EtOOC C H H O 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 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 Updates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ cat. O H3C H N H CO2H , CHCl3, 0 OC O S + - Ph a) O ht tp V. :// ww Adi tya w. ad va ich rdh em an ist ry .c b) c) d) O O O O O O CH3 CH3 Ph H Ph H Ph H om Ph CH3 ? major product H O CH3 Answer: b WANT TO PURCHASE THE ENTIRE BOOK? TO KNOW THE METHOD OF PAYMENT MAIL ME AT
[email protected] 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. 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 Updates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ CO 2H H O H + -H2O HO H3C N H H3C N+ CO 2H N H Ph O -S(CH3)2 N+ H O Ph N+ H H2O O Ph Both -CHO & PhCO groups are cis to each other since they try to avoid -CH3 during the reaction. H O CH3 O Ph H O O Ph N H : The iminium and PhCO groups try CO 2H to avoid steric N H interaction with adjacent CH3 group during the H O formation of H CH3 cyclopropane ring Ph C and hence both of them are oriented S(CH3)2 trans to methyl group. ht tp V. :// ww Adi tya w. ad va ich rdh em an ist ry .c Crotonaldehyde froms an iminium ion with Indoline-2-carboxylic acid N+ CO 2H S(CH3)2 O H CH3 Ph CO 2H CH3 S(CH3)2 CO 2H CH3 CO 2H : CH3 CO 2H Regeneration of Indoline-2-carboxylic acid CH3 om Corey-Chaykovsky reaction 1,4-addition of stabilized sulfur ylide and hence the formation of cyclopropane ring. : H H 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 HC COOMe COOMe + MeOH CHCOOMe MeO - H O OH H COOMe CHCOOMe COOMe -H O 2 COOMe COOMe COOMe Updates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ Redraw Stereochemistry of Alder-ene reaction ht tp V. :// ww Adi tya w. ad va ich rdh em an ist ry .c COOMe ene COOMe reaction H CO2Me CO2Me CO2Me CO2Me H H CH3 Problem 16.1 (CSIR June 2011) OH + p-TSA O OH major product? OH (2S)-butane-1,2,4-triol OH a) HO O O b) HO O O c) O O d) O O om COOMe COOMe CO2Me CO2Me H OH 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. 10 The original copy of this book is available only from http://www.adichemistry.com/. Do not distribute this book without permisssion of author. OH OH OH :O OH H p-TSA O OH + H O OH2 + : OH OH Updates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ O : H + OH OH O O : O 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 H H OH HO H O i. Hg(OAc)2/MeOH ii. NaBH4 the major products, X and Y, respectively, are: NOTE THATTHIS IS A SAMPLE COPY ONLY. YOU NEED TO PURCHASE TO GET THE COMPLETE BOOK. ht tp V. :// ww Adi tya w. ad va ich rdh em an ist ry .c O C+ redraw OH OH OH redraw HO O O C C+ (CH3)2CO HCl X Y om -H2O 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 MeO HO O & O HO O MeO O HO B) HO MeO HO O & O O MeO HO O HO Updates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ C) HO HO HO O & OMe O HO O O D) HO OMe HO HO HO om O HO O OMe O O O & OMe O O OMe Answer: 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 ht tp V. :// ww Adi tya w. ad va ich rdh em an ist ry .c HOH2C O -OAcHOH 2C O : O Me -H + HO HO HO HO HO HO AcO Hg OAc Hg + AcOHg OAc HO NaBH4 HO HO O OMe (CH3)2CO HCl HO O OMe WANT TO PURCHASE THE ENTIRE BOOK? TO KNOW THE METHOD OF PAYMENT MAIL ME AT
[email protected] 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 Updates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ 1) Answer: 3 2) 3) 4) reductive elimination ht tp V. :// ww Adi tya w. ad va ich rdh em an ist ry .c I PPh3 Pd HI Ph3P Ph Pd 2+ 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. oxidative addition Ph3P PPh3 I H Pd 2+ Ph3P Ph Ph Pd SYN -hydride elimination Ph3P H PPh3 Ph Pd 2+ Ph3P I Problem 36.2 Choose the species that is not an intermediate in the following palladium catalyzed Heck reaction. om I PPh3 I 2+ PPh3 SYN addition 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 + O Br Pd(PPh3)2 Pd(PPh3)2 Et3N (Ph3P)2Pd O 3) O O Br O O Pd(PPh3)2 4) O Pd(PPh3)2 O Updates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ 1) 2) Answer: 2 O Ph ht tp V. :// ww Adi tya w. ad va ich rdh em an ist ry .c HBr PPh3 reductive elimination 2+ Explanation: Pd PPh3 Br Ph3P oxidative addition H Pd O Ph3P Ph om Br PPh3 2+ Br Pd Br O O O Ph3P O -complex Ph H Ph3P Pd 2+ PPh3 Br O O -complex PPh3 Br SYN -hydride elimination H O O Ph H Ph3P H Ph Ph3P Pd 2+ Pd 2+ PPh3 H O H O Br Rotation of C-C bond to give more stable conformation Ph Ph3P H SYN addition Pd 2+ PPh3 Br 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 NO2 C) NMe2 NO2 D) NMe2 NO2 Updates & 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 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 N + * 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. ht tp V. :// ww Adi tya w. ad va ich rdh em an ist ry .c N Me H H+ present in less amount but reacts faster to give para isomer present in more amount but reacts slowly to give meta isomer 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. om 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 more distributed on ortho and para positions. Hence these positions become less reactive towards electrophilic substitution, when compared to meta position. O O - Updates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ N + O - N + O - O - N + O - O - N * 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 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. ht tp V. :// ww Adi tya w. ad va ich rdh em an ist ry .c The positive charge is more concentrated on ortho and para positions making these site less reactive towards electrophilic substitution
[email protected] om + O - 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 ortho attack N + O - O N H E + O - O H E N + O - O N H E + O - H E Updates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ Least stable O ht tp V. :// ww Adi tya w. ad va ich rdh em an ist ry .c meta attack N + N + O - O O - O E+ N + O - om N + O O - O N + O - H E H E H E H E O para attack N + O - O N + O - O N + O - O N + O - H E H E H E 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 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 + O HO shift HO 1,2-alkyl shift Updates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ -H+ HO HO 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 1,2-alkyl shift -H+ H + O 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: H MeO A) MeO O N H C) H MeO Answer: A ht tp V. :// ww Adi tya w. ad va ich rdh em an ist ry .c HO HO HO O i) NH2OH, HCl H ii) ArSO2Cl, Py iii) iv) H2O H HN O HN H B) H H H MeO H H H D) H MeO H H om O N O 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 O OH N O NH2OH, HCl H H E-Oxime S O Cl -HCl OSO2Ar N Updates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ H H MeO OSO 2Ar N 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 Migration of methyl group occurs since it is anti to the OH group in Z-oxime. 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: ht tp V. :// ww Adi tya w. ad va ich rdh em an ist ry .c - 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. N OSO 2Ar N OH H2O HN H -ArSO3H H H H WANT TO PURCHASE THE ENTIRE BOOK? TO KNOW THE METHOD OF PAYMENT MAIL ME AT
[email protected] om O Tosylation H 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 NH2 A intermediate B= B O H N 2) A= an acyl cation B= N H O Updates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ 1) A= an acyl cation NH 3) A= an acyl nitrene Answer: 4 B= NH 4) A= an acyl nitrene B= om N C NH2 H N N H O H N 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 - N NH2 ht tp V. :// ww Adi tya w. ad va ich rdh em an ist ry .c N N O N N N O -N2 N NH2 NH2 Acyl nitrene O isocyanate O Problem 53.1 (CSIR JUNE 2012) For the following reactions: A & B, the correct statement is: Br KOBut (excess) Br KOBut (excess) A) B) COOH COOH 1) A gives B gives COOK 3) both A & B give 2) A gives Answer: 2 COOK B gives 4) both A & B give COOK 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 -HBr O H - Updates & online help through forums are only available to those who purchased this book from the author at http://www.adichemistry.com/ O OBu t - O * However, in reaction B, the carboxylic group itself is antiperiplanar to Br and hence the elimination involves decarboxylation. O 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. ht tp V. :// ww Adi tya w. ad va ich rdh em an ist ry .c Br -CO2 -Br O - WANT TO PURCHASE THE ENTIRE BOOK? TO KNOW THE METHOD OF PAYMENT MAIL ME AT
[email protected] om