SOLVED PROBLEMS IN ADVANCED ORGANIC SYNTHESIS

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This book is useful to those who are preparing for CSIR NET in chemistry. It helps you in solving problems in advanced organic synthesis. You can get more information and details of payment at http://www.adichemistry.com

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SOLVED PROBLEMS IN ADVANCED ORGANIC SYNTHESIS

  1. 1. 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. 2. 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. 3. 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. 4. 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. 5. 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. 6. 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. 7. 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. 8. 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. 9. 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

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