Baixe Manual soluções Paula Bruice Química Orgânica Cap07 e outras Notas de estudo em PDF para Química, somente na Docsity! Chapter 7 215 Solutions to Problems 1. a. 1. Ifstercoisomers are not included, 3 different monosubstituted compounds can be formed. Tf stereoisomers are included, four different monosubstituted compounds can be formed because the second listed compound has an asymmetric carbon. Br€=CC=CCH,CH, HC=CC CCHEH; HC=CC=CCH,CH,Br Br 2. If stereoisomers are not included, 2 different monosubstituted compounds can be formed. If stereoisomers are included, three different monosubstituted compounds can be formed because the first compound has a double bond that can have cis-trans isomers. BrCH=CHC=CCH=CH, CHo=CC=sCCH=CH, Br b. 1. Ignoring stereoisomers, 5 different disubstituted compounds can be formed. BrC=CC =ECHCHs BrC=CC=CCH,CH,Br HC=CC econcIBr Br Br Br HC=CC=CCcH, HC=CC=CCH,CHBr Br Br 2. Ignoring stereoisomers, 5 different disubstituted compounds can be formed. BICH=CCamCCH=CH, PH CHOmCC=eH; BrCH=CHC=CCH=CHBr Br Br CH=GCECO=CH, Brg=CHC=CCH=CH, Br Br Br € 1. Three of the five different disubstituted compounds have an asymmetric carbon. BrC=CC CG HCHa HC=CC =CCH,CHBr HC=CC =CqHCH;Br Br Br Br Each of the compounds with an asymmetric carbon can have the R or the S configuration. Therefore, 8 disubstituted products are possible: (2x3) + 2 compounds without an asymmetric carbon. 2. Two of the five disubstituted compounds do not have cis-trans isomers. CH,=CC=CC=CH, BrC=CHC=CCH=CH, dr dr Br Two of the five disubstituted compounds have a pair of cis-trans isomers. BrCH=CC=CCH=CH, BrCH =CHC=CC=CH, Br Br One of the five disubstituted compounds has two pairs of cis-trans isomers. BrCH=CHC=CCH=CHBr Therefore, 10 disubstituted compounds are possible. 216 Chapter 7 Ladenburg benzene is a better proposal. It would form 1 monosubstituted compound, three disubstituted compounds, and would not add Bro, all in accordance with what early chemists knew about the structure of benzene. H H Br Br H Br H H Br H H Br H H H H H Br H H H H H Br Dewar benzene is not in accordance with what early chomists knew about the structure of benzene, because it would form two monosubstituted compounds, 6 disubstituted compounds, and it would add Bro. Br Br Br Br Br DO Pr Br Br Br Sor Br. 7 Br Br a. From examination of the contributing resonance structures in Figure 7.4 on page 271 of the text, one can conclude that all the carbon-oxygen bonds in the carbonate ion should be the same length. b. Because the two negative charges are shared equally by three oxygens, each oxygen should have two thirds of a negative charge. a. 1,2,3,5, and6 1. é Õ (VB VE Ss = “q — Os | Om 8. Chapter 7 219 o | Í + b. CH;C-CH=CH, <—» CH;C=CH-CH, é | + CH;C—CH=CHCH, <—> CHC=CII-CHCH; more stable because the positive change in on a secondary carbon q e CH;CH-CH=CH, 7 à CcH;C=CHCH, <—> CH;C—CHCH, more stable because only in this compound is the negative charge delocalized *NH, NEL » d. CH;—C—NH, <-> CH-C=NH, more stable because the positive charge is on a N rather than on an O *QH qH * CH;—-C—NH, <> CH;-C=NH, a. Because nitrogen is less electronegative than oxygen, it is better able to stabilize the positive charge by resonance. A en + : CN+ + CH;O-CH, <—> CH,;0=CH, CH;NH“CH, <—> CH;NH=CH, more stable b. In order for electron delocalization to occur, the atoms that share the electrons must be in the same plane. The two tert-butyl groups prevent the positively charged carbon and the benzene ring from being in the same plane. Therefore, the carbocation cannot be stabilized by electron delocalization. ( Dam ai + more stable 220 Chapter 7 e. The compound with delocalized electrons is more stable than the compound in which all the electrons are localized. as + AN + + CH;0—CH,CH, cH;Ó-CH, «<> CH;0=CH, all the electrons are localized electrons are delocalized d. The compound with delocalized electrons is more stable than the compound in which all the electrons are localized. + + CHCH, CHCH; CHCHs O — Cf Os + electrons are delocalized all the electrons are localized e. The OCH; group destabilizes the carbocation in the first species and destabilizes it in the second species. QeHs DCH; + + CcH=C>CH, <—> CH,—C=CH, NS 4 The OCH; group destabilizes the carbocation by inductive electron withdrawal. .. + .. É + CH SEE, <—>— cnólém-cH=cn, «<—>» CH;0=CH—CH=CH, The OCH; group stabilizes the carbocation by resonance electron donation. 9. a. Solved in the text. b. The contributing resonance structures show that there are four sites that could react with HO”. ci N ci E cl ci = a <> << + e. The contributing resonance structures show that there are three sites that could react with a bromine radical. 10. 1. 12. 13. Chapter 7 221 d. The contributing resonancc structures show that there are two sites that could be protonated. CO — O o Solved in the text. In each case, the compound shown is the stronger acid because the negative charge that results when it loses a proton can be delocalized. Electron delocalization is not possible for the other compound in each pair. a CH;CH=CHOH —>» HÍ + CH;CH=CHO <->» CH,CHCH=0 Ú 7 b. cré-on —» HÍ + CHC-O <—»> CH;C=0 e CHCH=CHOH —>» H + CH;CH=CHO <->» CH;CHCH=0 - + d. CH,CH=CHNH, —>» H' + CH;CH=CHNH, <> CH;CHCH=NH, a. Ethylamine is a stronger base because when the lone pair on the nitrogen in aniline is protonated, it can no longer be delocalized into the benzene ring. b. Ethoxide ion is a stronger base because a negatively charged oxygen is a stronger base than a neutral nitrogen. e. Ethoxide ion is a stronger base because when the Pphenolate ion is protonated, the pair of electrons that is protonated can no longer be delocalized into the benzene ring. The carboxylic acid is the most acidic because its conjugate base has greater resonance stabilization than does the conjugate base of phenol. The alcohol is the least stable because, unlike the negative change on the conjugate base of phenol, the negative change on its oxygen atom cannot be delocalized. o Ota - Oro rom 224 Chapter 7 21. a. different compounds d. resonance contributors b. resonance contributors e. different compounds c. different compounds Notice that in the structures that are different compounds, both atoms and electrons have changed their locations. In contrast, in structures that are resonance contributors, only the electrons have moved. 22. a. - + 1. CH,CH=CHOCH, <> CH;CHCH=OCH, major minor 2. CH,NH, CH,NH, Jo o — É the two resonance contributors have the same stability 3. CH;CHCEN <——> CH,;CH=C=N minor major D— o the two resonance contributors have the same stability 5. + + + OCH; A OCHs OocH, OCH, OCH, Io C—É major minor minor minor major + 4º + 9 6. CH;—N o CH;—N No V o o the two resonance contributors have the same stability Chapter 7 225 o ! [+ CH;CH,COCH,CH, «es CH;CH,C=0CH,CH, major minor CH;CH=CHCH=CHCH, CH;CHCH=CHCH=CH, minor hM 2” major CH,CH=CHCHCH=CH, major c Il | + HCNHCH, <> — HC=NHCH, major minor + + CH;CH=CHCH, <—> CH;CHCH=CH, minor major f [ IN a A SN “0” Do o? Do o” So each of the three resonance contributors has the same stability o 1 - | I HCCH=CHCH, «<—>» HCCHCH=CH, <— > HC=CHCH=CH, minor minor major + + + + The five contributors are equally stable. o o” o” Ra — +47 +47 CH;CH-N <>" CHÉR-N O «—> cHcH=N minor o minor o major o / N 226 Chapter 7 15. The electrons can move in two different direction. They can move out of the benzene ring toward the alkene group; they can move into the benzene ring away from the alkyl group. x + + CH=CH, CHCH, CHCH, CHCH, CH=CH, major minor minor minor major CHCH, CHCH, CHCH, CH=CH, + + DO a <> minor minor minor major 16. + + + ci A ci cl cl ci major minor minor minor major H H l 7 1 17. CH;C—CH-CCH, <—> CH;C—CH=CCH, <—> CH;C=CH—CCH; minor major major or -. I 18. CH;COCH,CH, <—> — CH;=COCH,CH; minor major b. 2,4, 6, 11, and 13 23. ê a. CH,CHCH=CH, b CH; e q * This makes the greater This makes the greater This makes the greater contribution because the contribution because the contribution because the positive charge is on positive charge is on negative charge is on a secondary carbon. a tertiary carbon. an oxygen atom. 30. a. CH;CH,0” - N e. CH;CHCH,CCH; o d. CH,CHCH; ? e. CHG—CH CH; O f . N 31. Chapter 7 229 The stronger base is the less stable base of cach pair in Problem 29. Less stable because the negative charge cannot be delocalized. b. CH;CCHCH,CCH, Less stable because the negative charge can be delocalized onto only one carbonyl oxygen. Less stable because the negative charge cannot be delocalized. Less stable because neither of the nitrogen lone pairs can be delocalized. Less stable because the negative charge is delocalized onto a carbon. Less stable because the negative charge can be delocalized onto only one carbonyl oxygen. The carbocation leading to 1,1-dichloroethane is more stable than the carbocation leading to 1,2-dichloroethane because the positive charge on the intermediate leading to 1,1-dichloroethane is shared by carbon and chlorine. Since the more stable carbocation is the one that is easier to form, the final product of the reaction is 1,1-dichloroethane. HC=CH + .. HA CH,CH,-C: —+ GHocHci HCl a. +CI ci => H)C=CH-CI: 1,2-dichloroethane A + . CHCH-CI: —» — CHsCHCI qi a 1,1-dichloroethane 4 CH,CH=GI: 230 Chapter 7 32. The resonance contributors of pyrrole are more stable because the positive charge is on nitrogen. In furan, the positive charge is on oxygen which, being more electronegative, is less stable with a positive charge. ss <> ss <> «Ly <> (ja <> al N N N N N H H H H H (Ne ENS LD) e De dO o 9 9 o 2 33. A isthc most acidic because the electrons left behind when the proton is removed can be delocalized onto two oxygen atoms. B is the next most acidic because the electrons left behind when the proton is removed can be delocalized onto one oxygen atom. C is the least acidic because the electrons left behind when the proton is removed cannot be delocalized. mol Il Il N CH;CCH,CCH, > CH;CCH,CH,CCH, > CH;CCH,CH;CH,CCH; A B c 34. — Ofthe two possible carbocations that can be formed in reaction a, the more stable carbocation is the one formed by adding the electrophile to the sp? carbon bonded to the greater number of hydrogens. Itis more stable because the positive charge is shared by carbon and fluorine. the carbocation formed by adding the electrophile to the sp? carbon bonded to the greater number of hydrogens + = CH,CH, —F: the carbocation formed by not adding the electrophile to the sp? carbon bonded to the greater number of hydrogens Of the two possible carbocations that can be formed in reaction b, the more stable carbocation is the one formed by not adding the electrophile to the sp? carbon bonded to the greater number of hydrogens. The fluoro substituent is not in a position to help to stabilize the positive charge in either carbocation. In the carbocation formed by not adding the electrophile to the sp? carbon bonded to the greater number of hydrogens, the electron-withdrawing fluoro substituent is farther away from the positive charge. Chapter 7 231 CH;CHCF, the carbocation formed by adding the electrophile to the sp? carbon bonded to the greater number of hydrogens CH,CH,CF, the carbocation formed by not adding the electrophile to the sp? carbon bonded to the greater number of hydrogens 35. a. and d. oH Potential Energy o oH (o - o o a ais Progress of the Reaction O —d 0 d N T Ei o o