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V.Adityavardhan
http://www.adichemistry.com
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Updates&onlinehelpthroughforumsareonlyavailabletothosewhopurchasedthisbookfr...
V.Adityavardhan
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V.Adityavardhan
http://www.adichemistry.com
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V.Adityavardhan
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100173431 solved-problems-in-advanced-organic-synthesis-for-csir-net-chemistry

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100173431 solved-problems-in-advanced-organic-synthesis-for-csir-net-chemistry

  1. 1. V.Adityavardhan http://www.adichemistry.com 1 Updates&onlinehelpthroughforumsareonlyavailabletothosewhopurchasedthisbookfromtheauthorathttp://www.adichemistry.com/ Theoriginalcopyofthisbookisavailableonlyfromhttp://www.adichemistry.com/.Donotdistributethisbookwithoutpermisssionofauthor. ADICHEMISTRY SOLVED PROBLEMS IN ADVANCED ORGANIC SYNTHESIS (FOR CSIR NET & GATE) 1st EDITION SAMPLE COPY Release date: 7th, May 2012 Last updated on: 7th July 2012 Current No. of problems solved: 108 (This willbe updated again) Current No. of pages: 104 VISIT http://www.adichemistry.com/common/htmlfiles/csir-gate-chemistry.html FOR COMPLETE DETAILS This book is intended for the aspirants ofCSIR NET, SLET,APSET, GATE, IISc and other Universityentrance exams. Most ofthe advanced levelproblems inorganicsynthesis fromprevious year questionpapers are solved andthoroughlyexplained. It is a dynamic on-line version;updated frequently. Ifyou are interested see for the current offer rate below. This priceincludes future updates too. PRICE DETAILS Rs. 725/-(Ifyou purchase at this rate, there willbe free updates onlyup to 200 problems & this price maybe revised on everyupdate depending on the number ofproblems added.) TO KNOW THE METHOD OF PAYMENT MAIL ME AT ADICHEMADI @ GMAIL.COM Note: Updates &support through forums areonlyavailable to those who purchased this book fromthe author ofthis book at his site:http://www.adichemistry.com. NOTETHATTHISISASAMPLECOPYONLY.YOUNEEDTOPURCHASETOGETTHECOMPLETEBOOK. Problem 1.1 (IISc 2011) O Ph O Ph O Ph OH Pha) b) c) d) O O i) PhMgBr ii) H+ ? Answer: c
  2. 2. V.Adityavardhan http://www.adichemistry.com 2 Updates&onlinehelpthroughforumsareonlyavailabletothosewhopurchasedthisbookfromtheauthorathttp://www.adichemistry.com/ Theoriginalcopyofthisbookisavailableonlyfromhttp://www.adichemistry.com/.Donotdistributethisbookwithoutpermisssionofauthor. O O i) PhMgBr BrMgO O Ph ii) H+ OH O Ph :: H + :: O + Ph O HH :: O Ph -H2O -EtOH -H+ 1,2 addition of Grignard reagent acid catalysed and conjugate  bond assisted removal of OH group final removal step of EtOH Explanation Thinkdifferent: What willhappenif1,4 addition occurs? O O O O Ph 1, 4 - addtion of PhMgBr O Ph H+ -EtOH i) PhMgBr ii) H+ OH O Ph Same product!So it might bethe actualmechanism? But slimchances. Why?Thepossible explanation might golikethis: i) thepositive charge on4thpositionis diminished dueto contribution ofp-electronsofadjacentethoxy ‘O’throughconjugation(+M effect).
  3. 3. V.Adityavardhan http://www.adichemistry.com 3 Updates&onlinehelpthroughforumsareonlyavailabletothosewhopurchasedthisbookfromtheauthorathttp://www.adichemistry.com/ Theoriginalcopyofthisbookisavailableonlyfromhttp://www.adichemistry.com/.Donotdistributethisbookwithoutpermisssionofauthor. ii) The enolate ion formis less stable due to -I effect of‘O’. iii) We also know that: 1,2 addition is kineticallymore favorable than 1,4-addition incase ofGrignard reagents. It is because the R group attached to Mg in GR is a hard nucleophile and prefers carbonyl carbon with considerablepositive charge (hard electrophile). And if this is the mechanism, the removalof ethanolmaygive another product, though less likely, as shown below. O Ph 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 ADICHEMADI@GMAIL.COM Note: Updates &support through forums areonlyavailable to those who purchased this book fromthe author ofthis 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 OHCl a) i) EtMgBr; ii) HCl b) i) (C2 H5 )2 CuLi; ii) HCl c) i) C2 H5 Li; ii) HCl d) i) HCl; ii) EtMgBr Answer: d Explanation: 1,4-additionofHClfurnishes 4-chlorobutanone, whichreactswithGrignard reagent to get the desired product. O HCl OCl EtMgBr H3O+ H+ CH2 OH Cl- OHCl OHCl mechanism However, the yields maynot be satisfactorydue to side reactionthat is possible inthe second step with Grignard reagent. It mayundergo Wurtz like coupling reactionwith -CH2 Clgroup.
  4. 4. V.Adityavardhan http://www.adichemistry.com 4 Updates&onlinehelpthroughforumsareonlyavailabletothosewhopurchasedthisbookfromtheauthorathttp://www.adichemistry.com/ Theoriginalcopyofthisbookisavailableonlyfromhttp://www.adichemistry.com/.Donotdistributethisbookwithoutpermisssionofauthor. OCl EtMgBr O MgBrCl What about other options? Option - a : O HClEtMgBr H3O+ OH Cl Cl H+ -H2O Cl- Cl- major product * 1,2-additionoccurs withGrignard reagent, since the ethylgroup attached to Mg has considerable positive charge and is a hard nucleophile. It prefers to attack 2nd carbon (hard electrophile). * In the reaction ofallylic alcoholwithHCl, the Cl- prefersto attack the allylic carbocation fromless hindered end. Hence the major product is 1-chloro-3-methyl-2-pentene. Option - b O Et2CuLi H3O+ O HCl Expecting aldol reaction 1,4-addition occurs withLithiumdiethylcuprate, since ethylgroup attached to copper is a soft nucleophile and prefers carbonat 4thposition (soft electrophile). Option-c : The products are same as incase ofoption-a. Ethyllithiumalso shows 1,2 additionlike Grignard reagent. NOTETHATTHISISASAMPLECOPYONLY.YOUNEEDTOPURCHASETOGETTHECOMPLETEBOOK. Problem 1.3 (CSIR DEC 2011) Choose the correct option for M & N formed in reactions sequence given below. O 1) PhMgBr 2) TsOH 1) BH3.SMe2 2) PCC 3) mCPBA N a) M N=M= Ph c) b) d)M= Ph M= Ph O Ph O Ph ON= N=M= N= OH Ph O Ph O Ph O Answer: a
  5. 5. V.Adityavardhan http://www.adichemistry.com 5 Updates&onlinehelpthroughforumsareonlyavailabletothosewhopurchasedthisbookfromtheauthorathttp://www.adichemistry.com/ Theoriginalcopyofthisbookisavailableonlyfromhttp://www.adichemistry.com/.Donotdistributethisbookwithoutpermisssionofauthor. Explanation: *Atertiaryalcoholis formed upon1,2 additionofPhMgBr and is dehydrated inpresence ofTosylic acid. O PhMgBr H3O+ OH Ph Ph TsOH -H2O * Thus formed product is subjected to hydroboration with BH3 .Me2 S complexto yield 2- phenylcyclohexanol, an anti-Markonikov’sproduct, which is oxidized to a ketone inpresence ofPCC. The keto compound is subjected to Baeyer Villiger oxidation withmCPBAto get a lactone. The PhCH- group is migrated onto oxygenin preference to CH2 group. Ph Ph OH PCC Ph O O Ph O BH3.SMe2 mCPBA Anti Markonikov's product Baeyer Villger oxidation Ph-CH- group has more migratory aptitude than CH2 group Problem 1.4 (CSIR JUNE 2011) The majorproduct formed in the following transformation is: O Ph a) c)b) d) Answer: d 1) MeMgCl, CuCl 2) Cl O Ph O Ph O Ph O Ph Explanation: * The Grignard reagent reacts with CuClto give Me2 CuMgCl, anorganocopper compound also known as Gilmanreagent that is added to the-unsaturated ketone in 1,4-manner. Initiallycopper associateswith the double bond to give a complex, whichthenundergoes oxidative additionfollowedbyreductive elimination. Thus formed enolate ion acts asa nucleophile and substitutesthe Clgroup ofallylchloride. The attackon allylchloride is done fromthe opposite side ofmore bulkyphenylgroup. 2 MeMgCl + CuCl Me2CuMgCl + MgCl2 O -OMe MeO F3C O CF3 O O TFAA OOMe MeO O CF3
  6. 6. V.Adityavardhan http://www.adichemistry.com 6 Updates&onlinehelpthroughforumsareonlyavailabletothosewhopurchasedthisbookfromtheauthorathttp://www.adichemistry.com/ Theoriginalcopyofthisbookisavailableonlyfromhttp://www.adichemistry.com/.Donotdistributethisbookwithoutpermisssionofauthor. NOTETHATTHISISASAMPLECOPYONLY.YOUNEEDTOPURCHASETOGETTHECOMPLETEBOOK. Problem 6.1 (IISc 2009) a) b) c) d) Answer: a CHO HEtOOC H ? major product CHO +S COOEt C6H6  CHO HH EtOOC O HH EtOOC O H H EtOOC Explanation: ThisisCorey-Chaykovskireaction. Since thesulfur ylideisstable,cyclopropanationoccursmajorly through 1,4-additionroute. The product isa thermodynamic one. The CHOand COOEt groups get trans positions in thecyclopropane ring. This occurs since they tend to orient as far awayas possible during thecyclopropanations step to avoidsteric repulsion. S + CH - COOEt O H H S + CH O - H COOEt H C H OH EtOOC H + S slow 1, 4 addition irreversible + H H CHOH C S + EtOOC H + Stereochemistry of cyclopropanation step CHO & COOEt groups orient in space so as to minimize repulsion. Hence they assume trans postions to each other in cyclopropane ring. Thinkdifferent: What willbe the product if1,2-addition occurs? An epoxide is formed. It iskineticallyfavored product. But this isminor product. Why? Since 1,2-addition step is reversible, the expulsion of ylide from the intermediate is also more likely. H O S + CH H O - COOEt +S + CH - COOEt faster 1,2 addtion reversible C O C EtOOC H H + S However, the 1,4-additionstep is irreversible due to formation ofstronger 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 whenunstable sulfur ylidesare used, the major product is epoxide.
  7. 7. V.Adityavardhan http://www.adichemistry.com 7 Updates&onlinehelpthroughforumsareonlyavailabletothosewhopurchasedthisbookfromtheauthorathttp://www.adichemistry.com/ Theoriginalcopyofthisbookisavailableonlyfromhttp://www.adichemistry.com/.Donotdistributethisbookwithoutpermisssionofauthor. Web Resource: http://www.adichemistry.com/organic/namedreactions/coreychaykovsky/corey-chaykovsky-1.html Problem 6.2 a) b) c) d) Answer: b ? major product CH3 H O N H CO2H S + - Ph O , CHCl3, 0 O C cat. CH3 H O Ph O CH3 H O Ph O CH3 H O Ph O CH3 H O Ph O WANT TO PURCHASE THE ENTIRE BOOK? TO KNOW THE METHOD OF PAYMENT MAIL ME AT ADICHEMADI@GMAIL.COM Note: Updates &support through forums areonlyavailable to those who purchased this book fromthe author ofthis book at his site:http://www.adichemistry.com.
  8. 8. V.Adityavardhan http://www.adichemistry.com 8 Updates&onlinehelpthroughforumsareonlyavailabletothosewhopurchasedthisbookfromtheauthorathttp://www.adichemistry.com/ Theoriginalcopyofthisbookisavailableonlyfromhttp://www.adichemistry.com/.Donotdistributethisbookwithoutpermisssionofauthor. Crotonaldehyde froms an iminium ion with Indoline-2-carboxylic acid Corey-Chaykovsky reaction Explanation CH3 H O N CO2H H : N+ CO2H CH3 H OH HH + N+ CO2H H CH3 S(CH3)2 Ph O : Ph O N+ CO2H CH3 H S(CH3)2 Ph O N CO2H CH3 H Ph O N+ CO2H CH3 H OH2 : Ph O N CO2H CH3 H OH Ph O CH3 H O -H2O Regeneration of Indoline-2-carboxylic acid The iminium and PhCO groups try to avoid steric interaction with adjacent CH3 group during the formation of cyclopropane ring and hence both of them are oriented trans to methyl group. -S(CH3)2 Both -CHO & PhCO groups are cis to each other since they try to avoid -CH3 during the reaction. 1,4-addition of stabilized sulfur ylide and hence the formation of cyclopropane ring. H CH3 HN C S(CH3)2 H CO2H O Ph
  9. 9. V.Adityavardhan http://www.adichemistry.com 9 Updates&onlinehelpthroughforumsareonlyavailabletothosewhopurchasedthisbookfromtheauthorathttp://www.adichemistry.com/ Theoriginalcopyofthisbookisavailableonlyfromhttp://www.adichemistry.com/.Donotdistributethisbookwithoutpermisssionofauthor. CH COOMe COOMe HMeO - CH- COOMe COOMe + MeOH CH- COOMe COOMe H O OH COOMe COOMe COOMe COOMe-H2O COOMe COOMe HCOOMe COOMe ene reaction Redraw H CH3 CO2Me CO2Me H CO2Me CO2Me H CO2Me CO2Me Stereochemistry of Alder-ene reaction Problem 16.1 (CSIR June 2011) a) b) c) d) Answer: c major product? OH OH OH (2S)-butane-1,2,4-triol p-TSA O + OH O O OH O O O O OH O O OH Explanation: * Butane-1,2,4-triolforms an acetal with3-pentanone to give a 1,3-dioxolane (a 5 membered ring) as
  10. 10. V.Adityavardhan http://www.adichemistry.com 10 Updates&onlinehelpthroughforumsareonlyavailabletothosewhopurchasedthisbookfromtheauthorathttp://www.adichemistry.com/ Theoriginalcopyofthisbookisavailableonlyfromhttp://www.adichemistry.com/.Donotdistributethisbookwithoutpermisssionofauthor. major product. The hydroxygroups onadjacent (1 & 2) positions take part inthis reaction. Formation ofa 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. O H + O OH OH H : O OH OH OH H + : O OH OH OH2 + : p-TSA -H2O O OH OH C+ O OH OH C+ : redraw O O OH C redraw O O OH Note: Work out the stereochemistryat chiralcarbon. In the optionc, the configuration at 2nd carbon is retained and incidentallyit has S configurationasin 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, OH OH CH2OH H H OH i. Hg(OAc)2/MeOH ii. NaBH4 (CH3)2CO HCl X Y the majorproducts, X and Y, respectively, are: NOTETHATTHISISASAMPLECOPYONLY.YOUNEEDTOPURCHASETOGETTHECOMPLETEBOOK.
  11. 11. V.Adityavardhan http://www.adichemistry.com 11 Updates&onlinehelpthroughforumsareonlyavailabletothosewhopurchasedthisbookfromtheauthorathttp://www.adichemistry.com/ Theoriginalcopyofthisbookisavailableonlyfromhttp://www.adichemistry.com/.Donotdistributethisbookwithoutpermisssionofauthor. OOH OH OH MeO O O O MeO OH & OOH OH OH MeO & OO O OH MeO OOH OH OH OMe O O O OH OMe & OOH OH OH OMe OO O OH OMe & A) B) C) D) Answer: C Explanation: * Inthe solvo-oxymercuration, the Hg attacks fromthe less hindered side and the OMe approaches fromthe top side and attacks the carbon-1, adjacent to ring oxygen, since the positive charge on this carbon is stabilized due to +M effect ofring oxygen. * Demercurationis achieved bysodiumborohydride to furnish adeoxysugar. * Inthe finalstep, protectionofOH groups on4th and 6th carbons is achieved byusing acetone in presence ofdry HCl. However, a sixmembered cyclic acetalis formed instead offive membered one eventhough there is diaxialinteractionfor methylgroups onisopropylidene moiety. It is because ofinabilityofformationoftrans fusion of6/5membered ring that would be created when the OH groups on 3rd and 4thcarbons, which are trans to each other, are involved in cyclic acetal formation. O OH HOH2C OH OH OH CH2OH H H OH Hg OAcAcO O OH HOH2C OH Hg + OAc O Me H : OOH OH OH OMe AcOHg OOH OH OH OMe O O O OH OMe NaBH4 (CH3)2CO HCl -OAc- -H+ WANT TO PURCHASE THE ENTIRE BOOK? TO KNOW THE METHOD OF PAYMENT MAIL ME AT ADICHEMADI@GMAIL.COM Note: Updates &support through forums areonlyavailable to those who purchased this book fromthe author ofthis book at his site:http://www.adichemistry.com.
  12. 12. V.Adityavardhan http://www.adichemistry.com 12 Updates&onlinehelpthroughforumsareonlyavailabletothosewhopurchasedthisbookfromtheauthorathttp://www.adichemistry.com/ Theoriginalcopyofthisbookisavailableonlyfromhttp://www.adichemistry.com/.Donotdistributethisbookwithoutpermisssionofauthor. I + Pd(OAc)2, PPh3 Et3N 1) 2) 3) 4) I Answer: 3 Explanation: * It is a Heck reaction. 3-phenylcyclohexene is formed as the onlyproduct instead ofexpected 2- phenylcyclohexene, due to requirement of synelimination ofhydride. PPh3 Ph3P Pd oxidative addition SYN -hydride elimination reductive elimination SYN addition PPh3 Ph Ph3P IPd 2+ I PPh3 Ph Ph3P IPd 2+ PPh3 Ph3P I Pd 2+Ph H PPh3 Ph3P IPd 2+ H Ph IH Problem 36.2 Choose the species that is not an intermediate in the following palladium catalyzed Heck reaction.
  13. 13. V.Adityavardhan http://www.adichemistry.com 13 Updates&onlinehelpthroughforumsareonlyavailabletothosewhopurchasedthisbookfromtheauthorathttp://www.adichemistry.com/ Theoriginalcopyofthisbookisavailableonlyfromhttp://www.adichemistry.com/.Donotdistributethisbookwithoutpermisssionofauthor. Br + 1) 2) 3) 4) Answer: 2 O O O O Pd(PPh3)2 Et3N Pd(PPh3)2 Br O O(Ph3P)2Pd O O Pd(PPh3)2 Br OO Pd(PPh3)2 Br Explanation: PPh3 Ph3P Pd oxidative addition SYN -hydride elimination reductive elimination SYN addition PPh3 Ph Ph3P BrPd 2+ Br PPh3 Ph Ph3P BrPd 2+ PPh3 Ph3P BrPd 2+ H BrH O O O O O O PPh3Ph3P Br Pd 2+ Ph H H H O O PPh3Ph3P Br Pd 2+ Ph H H H Rotation of C-C bond to give more stable conformation Ph O O Ph O O PPh3Ph3P Br Pd 2+ H -complex -complex NOTETHATTHISISASAMPLECOPYONLY.YOUNEEDTOPURCHASETOGETTHECOMPLETEBOOK. Problem 37.1 (GATE 2004) The majorproduct formed on nitrationofN, N-dimethylaniline withconc. H2 SO4 andconc. HNO3
  14. 14. V.Adityavardhan http://www.adichemistry.com 14 Updates&onlinehelpthroughforumsareonlyavailabletothosewhopurchasedthisbookfromtheauthorathttp://www.adichemistry.com/ Theoriginalcopyofthisbookisavailableonlyfromhttp://www.adichemistry.com/.Donotdistributethisbookwithoutpermisssionofauthor. is: NMe2 NO2 NMe2 NO2 NMe2 NO2 NO2 NMe2 NO2 A) B) C) D) Answer: A Explanation: * The nitrationofN,N-dimethylaniline using 85% H2 SO4 gives 45%meta-nitro product and38% para- nitro product alongwithN,N,N’,N’-tetramethylbenzidine. * N,N-dimethylaniline is protonated in stronglyacidic mediumbyforming an N,N-dimethylanilinium ion, PhNHMe2 + . Since NHMe2 + group is deactivating towards electrophilic substitution due to -I effect, the nitration occurs mostlyat meta position. Why para product is also formed? * Inthe reactionmixture, smallamount offree N,N-dimethylanilineis also present inequilibriumwithits aniliniumsalt. Thisfree formis highlyreactive and gives para product and favors the equilibriumto the left side. N MeMe N + H Me Me 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 ofH2 SO4 , meta substitution is seldomobserved and the composition ofpara product and N,N,N’,N’-tetramethylbenzidine increases with dilutionofsulfuric acid. It is also interesting to note that, withdilution ofH2 SO4 , the major product is N,N,N’,N’-tetramethyl benzidine rather thanthe para isomer. Me2N NMe2 For example, 63% ofbenzidine derivative is formed when 74.7% ofH2 SO4 is used. This reaction occurs throughionradical-radicalpair 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 nitrogroup: 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. 15. V.Adityavardhan http://www.adichemistry.com 15 Updates&onlinehelpthroughforumsareonlyavailabletothosewhopurchasedthisbookfromtheauthorathttp://www.adichemistry.com/ Theoriginalcopyofthisbookisavailableonlyfromhttp://www.adichemistry.com/.Donotdistributethisbookwithoutpermisssionofauthor. Answer: d Explanation: * Nitro group is electronwithdrawinggroup. It withdraws electrons both byinductive aswellas mesomeric effect i.e., both -I &-M group.Among these, mesomeric effect has more effect on the reactivity. In the resonance forms that canbe written due to -M effect ofnitro group, the positive chargeis more distributed onortho and para positions. Hence these positions become less reactive towards electrophilic substitution, whencompared to meta position. N + O - O N + O - O - N + O - O - N + O - O - The positive charge is more concentrated on ortho and para positions making these site less reactive towards electrophilic substitution * The comparisonofresonance forms ofWheland intermediates formed duringattack ofelectrophile, E+ isillustrated below. The Whelandintermediates formed, during ortho and para attacks are less stable, since one ofthe contributing structure isleast stable due to presence ofpositive charges onadjacent carbon atoms. WANT TO PURCHASE THE ENTIRE BOOK? TO KNOW THE METHOD OF PAYMENT MAIL ME AT ADICHEMADI@GMAIL.COM Note: Updates &support through forums areonlyavailable to those who purchased this book fromthe author ofthis book at his site:http://www.adichemistry.com.
  16. 16. V.Adityavardhan http://www.adichemistry.com 16 Updates&onlinehelpthroughforumsareonlyavailabletothosewhopurchasedthisbookfromtheauthorathttp://www.adichemistry.com/ Theoriginalcopyofthisbookisavailableonlyfromhttp://www.adichemistry.com/.Donotdistributethisbookwithoutpermisssionofauthor. N + O - O N + O - O H E N + O - O E H N + O - O H E N + O - O H E N + O - O H E N + O - O H E N + O - O E H N + O - O E H N + O - O E H N + O - O H E N + O - O H E N + O - O H E ortho attack meta attack para attack Least stable Least stable E+ * The summaryofabove points are illustrated inthefollowing free energydiagram. Not onlythe activation free energyrequired for metaattack is less, and also the Wheland intermediate formed is more stable. Transition state Wheland intermediate Freeenergy Reaction coordinate ortho & para attack meta attack NOTETHATTHISISASAMPLECOPYONLY.YOUNEEDTOPURCHASETOGETTHECOMPLETEBOOK.
  17. 17. V.Adityavardhan http://www.adichemistry.com 17 Updates&onlinehelpthroughforumsareonlyavailabletothosewhopurchasedthisbookfromtheauthorathttp://www.adichemistry.com/ Theoriginalcopyofthisbookisavailableonlyfromhttp://www.adichemistry.com/.Donotdistributethisbookwithoutpermisssionofauthor. O H + OH OH OH H OH 1,2-alkyl shift 1,2-alkyl shift -H+ * However, the product given underoption-1 is also formed as a minor product.The mechanism leading to formationofthis is also shownbelow. O H + OH OH H OH -H+ 1,2-alkyl shift Why above one is a minor product? In this case, less stable secondarycarbocation is formed inthe initialstep. Problem 39.1 In the following reaction sequence, the major product is: H MeO H H O A) B) C) D) Answer: A i) NH2OH, HCl ii) ArSO2Cl, Py iii)  iv) H2O NH H MeO H H O H MeO H H ONH H MeO H H O N H MeO H H N O
  18. 18. V.Adityavardhan http://www.adichemistry.com 18 Updates&onlinehelpthroughforumsareonlyavailabletothosewhopurchasedthisbookfromtheauthorathttp://www.adichemistry.com/ Theoriginalcopyofthisbookisavailableonlyfromhttp://www.adichemistry.com/.Donotdistributethisbookwithoutpermisssionofauthor. Explanation: * The keto group is converted to following E-oxime andthen tosylated. H MeO H H O H N OH H N OSO2Ar S O Ar O Cl NH2OH, HCl E-Oxime -HCl Tosylation Note: Onlypart ofthe reactant is shown. * Thusformed product undergoes Beckmannrearrangement uponheating andsubsequent treatment withwater to furnishanamide. H N OSO2Ar N H - OSO2Ar N H OH  NH H O Beckmann rearrangement H2O -ArSO3H Why not amide as given under option B is formed? *Yes. It is also a possible product. It is formed due to migration ofmethylgroup when it is antito OH group as inZ-oxime. However the migrationofmethylgroup is slowerthanthe ring andhence the this amide is formedas minor product. H N OH H ONH Migration of methyl group occurs since it is anti to the OH group in Z-oxime. WANT TO PURCHASE THE ENTIRE BOOK? TO KNOW THE METHOD OF PAYMENT MAIL ME AT ADICHEMADI@GMAIL.COM Note: Updates &support through forums areonlyavailable to those who purchased this book fromthe author ofthis book at his site:http://www.adichemistry.com. Problem 52.1 (CSIR JUNE 2012) The intermediateAand the majorproduct B in the following reaction are:
  19. 19. V.Adityavardhan http://www.adichemistry.com 19 Updates&onlinehelpthroughforumsareonlyavailabletothosewhopurchasedthisbookfromtheauthorathttp://www.adichemistry.com/ Theoriginalcopyofthisbookisavailableonlyfromhttp://www.adichemistry.com/.Donotdistributethisbookwithoutpermisssionofauthor. 1) 2) 3) 4) A= an acyl cation B= A= an acyl nitrene B= Answer: 4 NH2 N3 O  A B A= an acyl cation A= an acyl nitrene B= B= NH O N H N H O NH O N H N H O intermediate Explanation: * It is Curtius rearrangement. The acylazide loses dinitrogen, N2 to give anintermediate acylnitrene, whichundergoesrearrangement ofarylgroup onto nitrogenand furnishes anisocyanate. * The isocyanateis attacked bynucleophilic amine group to give aheterocyclic ring, 1,3-dihydro-2H- benzimidazol-2-one. NH2 N O N N - NH2 N O N N NH2 N O Acyl nitrene NH2 N C O isocyanate N H N H O -N2 Problem 53.1 (CSIR JUNE 2012) For the following reactions:A& B, the correct statement is: Br COOH A) Br COOH B) KOBut (excess) KOBut (excess) 1) 2) 3) 4) COOK A gives B gives COOK A gives B gives both A & B give both A & B give COOK Answer: 2
  20. 20. V.Adityavardhan http://www.adichemistry.com 20 Updates&onlinehelpthroughforumsareonlyavailabletothosewhopurchasedthisbookfromtheauthorathttp://www.adichemistry.com/ Theoriginalcopyofthisbookisavailableonlyfromhttp://www.adichemistry.com/.Donotdistributethisbookwithoutpermisssionofauthor. Explanation: * Incase ofreactionA, the most stable conformationofcyclohexane is withtert-butylgroup on equatorialposition. Inthis conformation, there is a H atomantiperiplanar to the Brand hence undergo E2 eliminationeasilybygiving cyclohexenewithcarboxylic group. Br O - O H - OBu t O O - -HBr * However, inreaction B, thecarboxylic group itselfis antiperiplanar to Br andhence the elimination involves decarboxylation. Br O O - -CO2 -Br WANT TO PURCHASE THE ENTIRE BOOK? TO KNOW THE METHOD OF PAYMENT MAIL ME AT ADICHEMADI@GMAIL.COM Note: Updates &support through forums areonlyavailable to those who purchased this book fromthe author ofthis book at his site:http://www.adichemistry.com.

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