Manual soluções - cap.04 Paula Bruice

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Resolução de exercícios de química orgânica do capítulo 4 de Paula Bruice vol.01.

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Manual soluções - cap.04 Paula Bruice

  1. 1. Chapter 4 127 Solutions to Problems CH3 l 3- > > CH3CH2CH2çH2 b. The halogen atoms decrease the stability of the carbocation because, since they are more electronegative than a hydrogen, they are more effective than a hydrogen at withdrawing electrons away from the positively charged carbon. This increases the concentration of positive charge on the carbocation which makes it less stable. Because fluorine is more electronegatíve than chloiine, the Huorine-substituted carbocation is less stable than the chloiine-substituted carbocation. CH3ClJHCH2çH2 > cmencHzçnz > crgtltncnzçn¡ CH3 c¡ F 2. a. The bond orbitals of the carbon that is adjacent to the positively charged carbon are the bond orbitals that are available for overlap with the vacant p orbital. Because the methyl cation does not have a carbon adjacent to the positively charged carbon, there are no bond orbitals available for overlap with the vacant p orbital. b. An ethyl cation is more stable because it has three carbon-hydrogen bond orbitals available for overlap with the vacant p orbital, while a methyl cation does not have any carbon-hydrogen bond orbitals available for overlap with the vacant p orbital. 3. a. products b. reactants c. reactants d. products 4. CH3 CH3 a. cH3CH2çHCH3 c- @Ám e. @Br Br ? Hs ? Hs b. CH3CH3çCH3 d. CH3('3CH2CH2CH2CH3 f. CH3CH2('3HCH3 Br Br Br
  2. 2. 128 Chapter 4 $113 $H3 a_ CH¡ = CCH3 b. <: >_CH2CH= CH2 C- ©~C= CH2 d' <: >'= CHCH3 or <: >-CH2CH3 ? Hs a. : GHz In both a and b, the compound that is more highly regioselective is the one where the choice is between forming a tertiary carbocation or a primary carbocation. In the less regioselective compound, the choice is between forming a tertiary carbocation or a secondary carbocation, so the difference in the stability of the two possible carbocations is not as great. As long as the pH is greater than -2.5 and less than about 15, more than 50% of 2-propano1 would be in its neutral, nonprotonated form. Because when the pH = pKa, half the compound is in its acid form and half is in its basic form, at a pH less than - 2.5, more than half of the compound will be in its positively charge protonated form. At a pH greater than about 15, more than half of the compound will exist as the negatively charged anion. c. the neutral alcohol d. the second and third steps in the forward direction a. 3 transition states b. Zintermediates 3- CHacHzCHzçHcHa c CH3CH2CH2CH2Cl2HCH3 and CH3CH¡CH2(| IHCH2CH3 OH OH OH “i O” “" O6* OH
  3. 3. Chapter 4 129 10. a. ?Ha ? Hs ? Hs ? Hs 1. cH3<I: cH3 2. CH3CIZCH3 3. CH3(|3CH3 4. CH3(| '.'CH3 c¡ Br OH OCH3 b. The first step in all the reactions is addition of an electrophilic proton (H+) to the carbon of the CH¡ group. The tert-butyl carbocation is fomied as an intermediate in each of the reactions. c. The nucleophile that adds to the tert-butyl carbocation is different in each reaction. In reactions #3 and #4, there is a third step-a proton is lost from the group that was the nucleophile in the second step of the reaction. 3. (12113 ? H3 4_ ? H3 ([3113 CHgçcHg --> CH3C| ZCH3 + n30* CH§CH3 --> CH3(ECH3 + 113o* 63H OH 53cm OCH3 Hzõ_ H cH3§H H jÍ jf ll. a¡ Q + CH3OH --)› OCH3 (5113 + C| JH3 b. CH2=CCH3 + CH3OH À» CH3O'CCH3 CH3 + c. CH3CH= CHCH3 + CH3CH2OH _Hs_ CH3CH¡('2HCH3 or OCH2CH3 + CH2=CHCH2CH3 + CH3CH2OH A» cH3cH2çHcH3 or 0CH2CH3 + CH2=CH2 + CH3CH2CHCH3 43+ cHaCHzçHCHa (BH 0CH2CH3
  4. 4. 130 Chapter 4 d. CH3CH= CHCH3 + Hzo -›- CH3CHCH2CH3 or OH + CH2=CHCH2CH3 + Hzo _Ha- CH3(|2HCH2CH3 OH OH + e. c; + L» Ó + r. CH3CH2CH= CHCH2CH3 + Hzo -H-›- CH3CH2C| ZHCH2CHZCH3 OH 12. + CH3(|3HCH2CH2QH ? H3 _ _L *_ CH3C-CH2 -H-›- CH3CCH3 cmcncnzcnzpà CGH? , I I l 3 H CH¡ md @H3 C143 CH B: is any base that is $ present in the reaction mixture ? Hs HB* + cH3c¡: HcH2cH2o<| :cH3 CH3 CH3 13. Solved in the text.
  5. 5. Chapter 4 131 14. LZ-hydride B H** + shift + B; lr a. CH CHCH= CH -› CH CHCHCH -› CH CCH CH -› CH CCHQCH 3' 2 3' 3 3' 2 3 3' 3 CH3 CH3 CH¡ CH3 secondary ÍBTUBPY CH2 CH: Br CH3 b. + + . - c. CH3C|2HCH2CH= CH2 -H-› CH3(IJHCHZCHCH3 L CH3CHCH2CHCH3 CH3 CH3 CH3 BT d. GHz CH; Br CH¡ H+ + Br' É» É» CH CH 1,2-methyl CH CH Br 13 H+ + I 3 shift 13+ Br- I 3¡ e. CH = CHCCH -› CH CHCCH _› CH CHCCH _› CH CH-CCH 2 l 3 3 | 3 3 l 3 3 | 3 CH3 CH3 CH3 CH3 secondary tertiary f. CH3 cHa CH? , + Br_ c. ) H/ r QE I' “dl CH3 CH3 Ó e gb + Br 1-Bromo-3-methylcyclohexane and 1-bromo-4-methylcyclohexane will be obtained in approximately equal amounts because in each case the initially formed carbocation is secondary.
  6. 6. 132 15. 16. 17. 18. Chapter 4 C113 Addition of H* would form a CH &CH CH carbocation that could rearrange. 3 2 3 àr Addition of Br* forms a cyclic bromonium ion rather than a carbocation so there is no rearrangement. a. The first step in the reaction of ethene with Br; forms a cyclic bromonium ion, whereas the first step in the reaction of ethene with I-IBr forms a carbocation. b. If the bromide ion were to attack the positively charged bromine atom, a highly unstable compound (with a negative charge on carbon and a positive charge on bromine) would be formed. + m WB¡ co_ - 'Í' HC E + : Br: CHz-CHz-Br-Br 2 Notice that the electrostatic potential maps of the cyclic bromonium íons on p. 158 of the text show that the ring carbons are the least electron dense (most blue) atoms in the intermediate. The nucleophile that is present in greater concentration is more apt to collide with the carbocation intermediate. Therefore, if the solvent is a nucleophile, the major product will come from reaction of the solvent with the carbocation or cyclic bromonium (or chloronium) ion intermediate, because the concentration of the solvent is much greater than the concentration of the other nucleophile. (For example, in "a" the concentration of CH3OH is much greater than the concentration of ClÍ) (EH: CH3 a. ClCH¡Cl2CH3 and ClCH2(|2CH3 c. CH3CH2CHCH3 and CH3CH2C| HCH3 QCH; Cl OH Cl maJm' major b- CH3CHCH3 and CH3CHCH3 d- CH3CH2çHCH3 and CH3CH2CIIHCH3 I Br 0CH3 Br major major As elements, sodium and potassium achieve an outer shell of eight electrons by losing the single electron they have in the 3s (in the case of Na) or 4s (in the case of K) orbital, thereby becoming Na+ and K+. In order to form a covalent bond, they would have to regain electrons in these orbitals, thus losing the stability associated with having an outer shell of eight electrons and no extra electrons.
  7. 7. Chapter 4 133 19. Because chlorine is more electronegative than iodine, iodine will be the electrophile. Therefore, it will become attached to the sp? carbon that is bonded to the greater number of hydrogens. (i /1+ CH2CH2CH= CH2 + I-Cl _› CH3CH2CH-CH2_› CH3CH2(| IHCHZI k/ ,Ç-J; Cl 20. a. CH2CHCH2CH3 b. CH2CHCH2CH3 c. CH2CHcH2CH3 d. CH2CHCH2CH3 Br Br Br OH Br OCH2CH3 Br OCH3 21. Notice that the addition of water or the addition of an alcohol to an alkene can be carried out using an acid Catalyst (Section 4.5) or by mercuration/ demercuration (Section 4.8). H+ OCH-, CH3 a. -_-_› - CH3CH2OH OI' 1. Hg(O¡CCF¡)¡, CH¡CH2OH 2. NaBH4 b. CH2 H, CH3 CH3 CH; É* OH OI' H3., OH 2 H2O °r or 1. Hg(OAc)2, H2O, THF í-í-H 3- NaBHA 2. NaBH4 + c. CH3CH= CHCH3 __i-_› CH3CHCH2CH3 CH3CH2OH i or OCH2CH3 1. Hg(O¡CCF3)2, CH¡CH2OH 2. NaBH4
  8. 8. 134 Chapter 4 ? Hs H+ ? Ha d. CH¡ = CCHZCH3 -_›- cH3ccH2cH3 CH3OH l OCH¡ OI' 1. Hg(O¡CCF¡)2, CH¡OH 2. NaBH4 22. CH3 + CH¡ l,2-hydn'de CH3 CH3 I H I shift I Hzo | a. CH2=CHCHCH3 +- CH3CHCHCH3 ? ›- CH3CH2CCH3 -›- CH3CH2C|3CH3 + + OH CH¡ CH3 I 1. Hg(OAc)2, H2O, THF I b. cH2=cHcHcH3 44a- CH3CHCHCH3 2. NaBH4 | OH In a, a carbocation rearrangement is required to get the desired product from the given starting materia In b, the desired product must be obtained from a reaction that will not form a carbocation íntermediat 23. Because one mole of BH3 reacts with three moles of an alkene, one third of a mole of BH3 is needed to react with each mole of alkene. Therefore, two thirds of a mole of BH3 is needed to react with two moles of an alkene (in this case, l-pentene). 24. cm CH3 _ I a. CH3C= CHCH3 CH3CHC| jHCH3 2.HoÍHo, Ho 2 2 2 OH CH3 CH3 b_ 1. BH¡ OH 2. HOÍ H2O2, Hzo
  9. 9. Chapter 4 135 25. Addition of a proton to the carbon that is bonded to the greater number of hydrogens forms a carbocation intermediate. The alcohol group in the same molecule is the nucleophile that reacts with the carbocation. CIHB HQ* ? H3 H3C CH2=CCH2(l2HCH3 É» CH3-CCH¡C| ZHCH3 -›- H (QO/ CH: CH CH 3 o l 2 LJ I 2 CZ¡ OH = <.>. H sua HB* is any acid present in the solution H3C B: is any base present in the solution CH3 H3C O + HB+ CH¡ ' Br + Br- : -> CH3 CH¡ ° BT Br + HBr _'-)' + Br- 27. Because alkene A has the smaller heat of hydrogenation, it is more stable. 26. CH3 28. a_ CH2CH3 This alkene is the most stable bcacause it has the greatest number of alkyl substituents bonded to the Spz carbons. CH2CH3 b CH2CH3 ° a This alkene is the least stable bcacause it has the fewest number of alkyl substituents bonded to the sp2 carbons. CH2CH3 c CHZCH¡ ' This alkene has the smallest heat of hydrogenation because it is the most stable of the three alkenes. CH2CH3
  10. 10. 136 Chapter 4 29. H C / CH3 H3cx H CHzCHz CH3CH2 CHZCH¡ 3 *CH cH/ Cns / / / Cíc > Cíc > Cíc > / x / / x CH3CH2 H H H H H 2 trans substituents 2 cis substituents 2 CiS SUbSÚÍUCnÍS that cause greater steric strain CH3CH2CH2CH2 H / /C"= C H H one substituent 30. Solved in the text. 31. The reaction of 3-methylcyclohexene with HBr and peroxide would form approximately equal amounts of l-bromo-2-methylcyclohexane (the desired product) and l-bromo-B-methyl- cyclohexane because the bromine radical could add to either the l-position or the 2-position of the alkene since in both cases a secondary radical would be formed. Thus, only half as much of the desired product would be formed from 3-methylcyclohexene than from l-methyl- cyclohexene. CH: CH; CH; Br. Br _ . + Br im* im CH3 CH3 Br + Br l-bromo-Z-methyl- 1 -bromo-3-methyl- cyclohexane cyclohexane
  11. 11. Chapter 4 l 37 If 3-methylcyclohexene were treated with HBr in the absence of peroxide, little l-bromo- 2- methylcyclohexane (the desired product) would be formed because the secondary carbocation with a positive charge at the 2-position would rear-range to a more stable tertiary carbocation. CH: CH3 CH; CH3 H* + $135 *Br* *BF CH3 CH3 Br CHE] Br l-bromo-3-methyl- l-bromo-2-methyl- 1'bT°m°'1'm°thY¡* cyclohexane cyclohexane Cycbhexane minor 32. + a. CH3CH= CH2 . _.H_›- CH3CHCH3 CH3OH I OCH3 l|3r b. CH3CH2CH= CHCH3 + Br, --›- CH3CH2(|3HCHCH3 Br CH: CH2OH °' 1. BH¡ --_: -›- 2. HO , 11202, HZO GHz CHa CH3O CH3 d. H* or -->- CH3OH GHz CH3 Br CH3 e. HBr OI' ›
  12. 12. 138 Chapter 4 f H+ OCH2CH2CH3 ' Ô + CH3CH2CH20H -› Cr 33. Bl' ? H3 CH2CH3 . b. CH CCH CH C- a dcHzcHa 3¡ 2 3 @Br Br 34. electrophile nucleophile a. CH3CHCH3 + : õif -_› CH3(| ZHCH3 + . . w : ç_1: nucleophile electrophile __ b. CH3CH= CH --> CH3ÇHCH2BI= electrophile ? Hiu (ljH3 nucleophile _--› CH3(!2-(I)CH3 c_ CH3C| + + CH3QH CH3H CH3 35. ? Hs ? Hs a. CH3C= CHCH3 + HBr : -› CH3(l2-CH2CH3 Br ? Hs . ?Hs b. CH3C= CHCH3 + HBr Períwíí CH3CH-(|3HCH3 Br ? Ha d. BI' CH3
  13. 13. ?Hs ? Hs C. + HI a I m. CH3C= CHCH3 --_-›- CH3CH-(EHCH3 2.HO. H0' 2 2 OH ? Hs Cl Br ? H3 . ?H3 d. cH3c= cHcH3 + HI ENAdS_ CH3C-CH2CH3 I ? Ha ? Hs e. CH3C= CHCH3 + IC] : ›- CH3C-CHCH3 Cl 1 CH3 (E113 f- cH3c': =cHcH3 _ff- CH3CH-CH2CH3 ? Hs g. CH3C= CHCH3 + Br¡ + excess NaCl -a- CH3(lj-(IjHCH3 ? H3 1. Hg(OAc)2, HZO ? H3 h. CH3C= CHCH3 cH3ç-cH2cH3 OH CH3 CH3 i. cH3c= cHcH3 + HZO “Link CH3(I3-CH2CH3 ôn ? Hs ? Ha j- CH3C= CHCH3 + Br, __›. CH3C-CHCH3 CHZCIZ Ér 1|3r ? H3 H o ? H3 k. cH3c= cHcH3 + Br¡ -2-›- cH3c-çHcH3 H0 Br CH3 CH3 l. CH3Cl= CHCH3 + Br¡ 395- CH3C-CHCH3 CH3(| ) i3: ? H3 1. BH, ?H3 Chapter 4 139
  14. 14. 140 Chapter 4 CH CH i 3 l. Hg(O2CCF3)2, CH3OH l 3 n. CH3C= CHCH3 CH3Ç"CH2CH3 ' d 4 OCH3 36. ? H3 ? H3 ? H3 CH3CH= C(| :HCH2CH3 CH3C= CCH2CH2CH3 CH3CH= CHCH(| IHCH3 CH3 CH3 CH3 3,4-dimethyl-2-hexene Z3-ÓÍmCÍhYÍ-Ê-Í1CXCHC 4v5'dimeth)'l'z'hexene 2,3-Dimethyl-2-hexene is the most stable because it has the greatest number alkyl substituents bonded to the sp¡ carbons. Because it is the most stable, it has the smallest heat of hydrogenation. 4,5-Dimethyl-2-hexene has the fewest alkyl substituents bonded to the spz carbons. making it the least stable of the three alkenes. It, therefore, has the greatest heat of hydrogenation. 37. A H 1,2-hydride CH cHcH-CH + m SW* * 3 l - 2 H-Br e CH3C-CHCH3 --> CH3C| ZCHZCH3 CH; CH3 CH3 l: 1:3; lzlâi-Ê ir ir CHaÇHCHCHa CH3CCH2CH3 CH: CH3 38. 3- ? H3 b. CH2=CHCH2CH2CH2CH3 CH2=CHCH2CHCH3 CH3C= CCH3 5B CH ? H3 3 ' 3 CH2=CHCHCH2CH3 cH3(| jcH= -cH2 This compound is most stable. CH3 CHS It has 4 alkyl substituents bonded to the spz carbons. These compounds are the least stable. Each has only one alkyl substituem bonded to the sp? carbons.
  15. 15. Chapter 4 39. : õ:' : Õ i) II _ a. CH3-CÚCH3 à» CH3-C-CH3 + CH3O CH3 b, + Êú-Hz É* CH3CÉC_ + : NH3 c. CH3CH2-Br + ong? -› CH3CHZ-ÇZCH3 + Br' V 40. CH2$HCHZBT U Br CH2CH2CH2OH CH2CH2CH3 l' B“3 B M? ? 0/ 2. Hzoz, Ho' r? CHzfHCHg H+ CH2CH= CH2 CH2çHCH3 OH 'Hzo __›HBT Br + HB *ECÉOH Br CH2CH2CH2Br 2 pêroxide g H2O CH2ClIHCH2Br (j OH U OCH3 141
  16. 16. 142 Chapter 4 41. a. CH3-Cl b. CH3CH¡CH2 H c. CH3ICH3 d. Br-Br 42. a. Cl b. OCH¡ Br 43. b. CH3CH2CH2CH= CH2 _-›. CH¡ Cl c. à, r. CH3CH2CH= CHCH2CH3 é. peroxide N33¡ C- no reaction (ge, Cl uses a 3 sp3 orbital in bond formation, while Br uses a 4 sp3 orbital. Cl, therefore, forms a shorter and stronger bond with carbon. A primary radical is less stable than a secondary radical, so it is harder to break a bond that results in the formation of a primary radical than it is to break a bond that results in the formation of a secondary radical. A methyl radical is less stable than a primary radical, so it is harder to break a bond that results in the formation of two methyl radicals than it is to break a bond that results in the formation of a methyl radical and a primary radical. I forms a weaker bond than Br because I uses a 5 sp3 orbital in bond formation, while Br uses a 4 sp3 orbital. e. 8- without an OH acid Catalyst r c¡ f. h- Br Cl Cl CH2CH= CH2 d. H* . H2O HCl e. CH3CH2CH= CHCH2CH3 _gâ- 2 or CXCCSS CH3CH2CH= CHCH2CH3 _BL NaCl CXCCSS ¡. @com j. Cl
  17. 17. Chapter 4 44. C112 CH3 a. and b. N O. CHz CHzBr + HB, peroxide CH3 cHa , HB, &E; Ó/ B' c Yes CH: H3C Cl + HCl f* Õ CH3 H3C Cl d. Peroxide has no effect on the addition of HCl, so both compounds will give the same product (the product shown in c). 45. ? H3 en, a CH3C= CHCH2CH3 a. CH3CCH3 r. 0/0113 g- CH3CHCH2CH2 + d. CH3CH2CH2CHCH3 Cl b. CH3ÕHCH2CH3 c. CH3CHCH3 + 143
  18. 18. 144 Chapter 4 46. a. To determine their relative rates of hydration, the rate constant of each alkene is divided by the smallest rate constant of the series (3.51 x 10's). propene = 4.95 x 10'3/3.5l x l0'3 = 1.41 cis-Z-butene = 8.32 x lO'8/3.5l x l0'3 = 2.37 trans-Z-butene = 3.51 x l0'3/3.51 x l0'3 = l 2-methyl-2-butene = 2.15 x 104/351 x 10-8 = 6.12 x 103 2,3-dimethyl-2-butene = 3.42 x l0'4/3.51 x 10's = 9.74 x 103 b. Both compounds form the same carbocation, but since (Z)-2-butene is less stable than (E)-2-butene, (Z)-2-butene has a smaller free energy of activation. c. 2-Methyl-2-butene reacts faster because it forms a tertiary carbocation in the rate-limiting step, while cis-Z-butene forms a less stable secondary carbocation. d. Both compounds form tertiary carbocation intermediates. However, 2,3-dimethyl-2-butene has two sp¡ carbons that can react with a proton to form the tertiary carbocation whereas 2-methyl-2-butene has only one sp2 carbon that can react with a proton to form the tertiary carbocation. Therefore there will be more collisions with the proper Orientation that lead to a productive reaction in the case of 2,3-dimethyl-2-butene. 47. a H H b. Cl H c. Cl H / ¡ l IC= C / C=C IC: H c¡ H CH3 H CH¡ 48. Only a symmetrical alkene will form the same alkyl halide when it reacts with HBr in the presence of peroxides that it forms when it reacts with HBr in the absence of peroxides. ag b. There are many more symmetrical alkenes in the case of an alkene with an even number of carbon atoms. CH3CH2 / CH2CH3 H3C / CH3 CH3 / c=c / C=c H H HaC CH3 me CH CH3 CH; CH3 3 CH3CH2 / H c= c / H CH2CH3 CH3 CH3
  19. 19. 49. 50. 51. 52. Chapter 4 145 No, he should not follow the student's advice. Markovnikov's rule would indicate that the secondary carbocation is more stable than the primary carbocation. However because of the electron-withdrawing fluoro substituents in this compound, the primary carbocation is more stable than the secondary carbocation, since in the latter the positive charge is closer to the fluoro substituents. So the major product will be 1,1,l-trifluoro-3-iodopropane, not l, l,1-trif| uoro-2- iodopropane, the compound that would be predicted to be the major product by Markovnikov's rule. mw + a. CH3CH2CH= CH2 __› cmcnzcncns f» CH3CH2CHCH3 . . , H C? CH3 H F cH3§Hl CH3CH2CEHCH3 OCH3 b. the first step e. the sec-butylcation c. H* f. methanol d. l-butene a. Both l-butene and 2-butene react with HCl to form Z-chlorobutanc. b. Both alkenes form the same carbocation, but because 2-butene is more stable than l-butene, 2-butene has the greater free energy of activation. c. Because l-butene has the smaller free energy of activation, it will react more rapidly with HCl. d. Both compounds form the same carbocation, but since (Z)-2-butene is less stable, it will react more rapidly with HCl. a CH¡ CH3 CH3 CH3 óóóà b. l-Methylcyclopentene is the most stable. 4 alkenes c. Because 1-methylcyclopentene is the most stable, it would have the smallest heat of hydrogenation.
  20. 20. 146 Chapter 4 53. CH CH Í-_s CH3 _ Br CH3 a. C-CHz D + 3 + CHS _âr_ CH lj “A -+ -›-' 3 secondary CH3 CH; ___ CH¡ CH; :B: + . à ÚBr tertiary b. The initially formed carbocation is tertiary. c. The rearranged carbocation is secondary, which then undergoes another rearrangement to a more stable tertiary carbocation. d. The initially formed carbocation rearranges in order to release the strain in the four- membered ring. (A tertiary carbocation with a strained four-membered ring is less stable than a secondary carbocation with an unstrained five-membered ring. ) 54. It tells us that the first step of the mechanism is the slow step. If the first step is slow, the carbocation will react with water in a subsequent fast step, which means that the carbocation will not have time to lose a proton to reform the alkene. CH-CD --›-H+ É: -›-H2O _ 2 slow H CHDZ fast ? H-CHD? *OHZ
  21. 21. Chapter 4 147 If the first step were not the slow step, an equilibrium would be set up between the alkene and the carbocation, and because the carbocation could lose either H* or D* when it reformed the alkene, all the deuterium (D) would not be retained in the alkene. H+ ©-CH= CD2 _TE CH-CHD2 --› Õ-CH: HD + l + D+ @amem + H* 55. a. A proton adds to the alkene, forming a secondary carbocation, which undergoes a ring- expansion rearrangement to form a more stable tertiary carbocation. H3C CH= CH2 H3C CHCH3 CH3 CH3 + < Hzc . . H2O H", ___, _Hz_; .. H3C l CH3 HO H + H3C b. At first glance this appears to be a difficult mechanism, but examination of the reaction shows that the only electrophile available to the alkene adds to the spz carbon bonded to the most hydrogens (the one that results in the formation of the most stable carbocation). This is followed by the addition of the nucleophilic nitrogen to the other sp? carbon. CH3CH2CIÉ= CH2 + cH, =rÍ1=rÍ: -›- CH3CH2CCH2CH2N= N- -a- N CH3 U CH3 H3C
  22. 22. 148 Chapter 4 c. The weak Oxygen-Oxygen bond breaks homolytically, forming a radical that abstracts a hydrogen atom from HCCl3. The radical that is formed adds to the double bond. A n electron of the other double bond pairs with the unpaired electron, forming a bond that divides the ring. The new radical abstracts a hydrogen atom from HCCl3, and the propagation steps are repeated. RCÇ-“OR -? -> 2RO° RO-g-CCI¡ _a ROH + ›cc13 f) + -CC13 à» -: › CCl3 CCl3 r CÓ + 56. c¡ A Cl Cl ! Q . . _ l _ I _ a Cl-(E H + H9: 2+ Cl-C: :-› Cl-Ct + Cl Cl CC] 'f-HZO c¡ o c¡ c¡ I n . . - l I _ b. ci-c-'lc-fg: A» Cl-Cz' : › Cl-C: + c¡ c¡ Na* c1 +C02

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