Pré-visualização parcial do texto
Baixe Manual soluções Paula Bruice Química Orgânica Cap10 e outras Notas de estudo em PDF para Química, somente na Docsity!
282 Chapter 10 Solutions to Problems 1. It decreases the magnitude of the rate constant, which causes the reaction to be slower. 2. çã q 8 CH;CH,CH,CH;CH;Br > CH;CHCH,CH,Br > CH;CH>CHCH;Br > CHsCH>ÇBr CH; 3. a. Solved in the text. b. The SN2 reaction of (R)-2-bromobutane with hydroxide ion will form ($)-2-butanol. c. The SN2 reaction of (S)-3-chlorohexane and with hydroxide ion will form (R)-3-hexanol. d. The SN2 reaction of 3-iodopentane (it does not have an asymmetric carbon) with hydroxide ion will form 3-pentanol (it does not have an asymmetric carbon). 4. a. RO” because ROH is a weaker acid than RSH. b. RS” because it is less well solvated by water. 5. a. aprotic b. aprotic e. protic d. aprotic 6. Solved in the text. a. CH,CH,Br + HO” HO” is a stronger nucelophile than HO. b. CHsCHCH,Br + HO” The alkyl halide has less steric hindrance. CH; e. CH;CH,Cl + CH;S” CHg3S” is a stronger nucleophile than CH307 in a solvent that can form hydrogen bonds. d. CH;CH,Br + 7 Br” is a weaker base than CI” so Br” is a better leaving group. Chapter 10 283 Solved in the text. a. Reaction of an alkyl halide with ammonia gives a low yield of primary amine because as soon as the primary amine is formed, it can react with another molecule of alky] halide to form a secondary amine. b. The alkyl azide is not treated with hydrogen until after all the alkyl halide has reacted with azide ion. Therefore, when the primary amine is formed, there is no alkyl halide for it to react with to form a secondary amine. In each reaction, the conjugate acid of the leaving group is stronger than the conjugate acid of the nucleophile. That means the leaving group is the weaker base (better leaving group). HO Hci RC=CH HBr pKa = 15.7 pKa = 7.0 pka = 25 pKa = 9.0 HS HBr Hicaiiy Hr Pka = 70 ph = 80 pk = 91 pKa = -10.0 ROH HI N pKa = 15.5 pKa = -10.0 NH, HBr pKa = 94 pKa = 9.0 RSH HBr pKa = 10.5 PKa = 9.0 CH;CH;NE,CH, Hm NH; HCI pKa = 10.8 pKa = -10.0 PKa = 36 pKa = 7.0 a - a. CH,CH,0CH; e CH;CH,N(CH;); Br b. CH,CH,N; d. CH,CH,SCH,CH, - acet CH;CH;Cl + KI =» CHCHI + K'cr Q KÍ CY precipitates out in acetone, which drives the reaction to the right. CH; CHÇBr > CHÇHBr > CH;CH,CH,Br > CH,Br CH; CH; Chapter 10 285 17. qu o qa CH,=CHCCH, ——> CH,=CHÇCH, <> CH,CH=CCH; + CI” (6) | 9 9 cH;CO” CH;CO ca, CH CH;=CHÇCH; CH;CH=CCH, OCCH; CHsco (0) (0) 18. The trans isomer will be formed in greater yield because the departing bromide ion can block the approach of the incoming nucleophile to the side of the carbocation vacated by the bromide ion. 19. a. 1. The reactant does not have any asymmetric carbons. The Sy2 reaction proceeds with back-side attack. H Br H H cHO” Da a CH H CH; OCH, 2. The reactant does not have any asymmetric carbons. The Sy2 reaction proceeds with back-side attack. H oH H = H HO; [6] Sn2 H CH; CH; b. 1. H Br H H H ocH, CH;0H Snl + CH H CH; OCH; CH; H H oH H H H H 2 CH;0H 4 [o] Sul H OH 286 20. 24. 22. Chapter 10 Because the rate of an SN1 reaction is not affected by increasing the concentration of the nucleophile, while the rate of an Sn2 reaction is increased when the concentration of the nucleophile is increased, you first have to determine whether the reactions are Snl or SN2 reactions. a is an SN2 reaction because the configuration of the product is inverted compared with the reactant. b is an SN2 reaction because the reactant is a primary alkyl halide. cis an SN1 reaction because the reactant is a tertiary alkyl halide. Because they are SN2 reactions, a and b will go faster if the concentration of the nucleophile is increased. Because it is an SN1 reaction, the rate of c will not change if the concentration of the nucleophile is increased. trans-4-Bromo-2-hexene is more reactive in an Sn1 solvolysis reaction, because the carbocation that is formed is stabilized by electron delocalization. (It is a secondary allylic carbocation.) The carbocation formed by the other alkyl halide is less stable because the secondary carbocation cannot be stabilized by resonance. H, CH,CH H CH,CH. um Panic anaieo v E pars H [CHaCHs Sa ——» O SW <—> an CHiÇH cH.CH cHCH É Br a. CH;CH,CH,I T is a weaker base than Br, so” is a better leaving group. b. CH.OCH,CI The electron-withdrawing methoxy group increases the ê CH electrophilicity of the carbon bonded to the leaving group. 3 I e. CH;CH,CHBr This compound has less steric hindrance. Gta d. CH;CH,CHCH,Br A primary carbon is less sterically hindered than a secondary carbon. e ( N-cncrtr A primary carbon is less sterically hindered than a secondary carbon. f. (Sen Nucleophilic attack cannot occur on an sp2 hybridized carbon. g. CH;CH=CHCHCH, Nucleophilic attack cannot occur on an sp? hybridized carbon. I Br 288 Chapter 10 b. An SN2 reaction, because a good nucleophile is used. GHaCH>CH, TA O Eca, H;NÊ Even though ammonia is not negatively charged, it is a good nucleophile. Recall that an amine is a much stronger base than an alcohol. (The pKa of a protonated alcohol is - —2.5. The pKa of the ammonium ion is 9.4; see Appendix II in the text.) c. An SN2 reaction, because of the high concentration of a good nucleophile. OCH; d. An SNI reaction, because a poor nucleophile is used. Notice that the initially formed econdamy carbocation rearranges to a more stable tertiary carbocation. OCH; E OM cmo E Os The product does not have a chirality center. e. An SN2 reaction, because of the high concentration of a good nucleophile. GHoCHs ca Sem OCH, f. An SN] reaction, because a poor nucleophile is used. ÇH-CH CEC o Cru Hi >cH; cH” NH cH,O OcH, Chapter 10 289 26. a. Solved in the text. SNZ kol2-brozróbutane] [HOY b. h s Sn2 + Sul ko[2-bromabutane] [HO] + ky[2-brompbutane] Sn2 — 320x10º x 1x10? 3.20 x 10% 32 + 2 Sn2+Snl — 3.20x10% +1.50x10% 3.20 x 10* + 150 x 10? 153 , =2% 27. a. Increasing the polarity will decrease the rate of the reaction because the concentration of charge on the reactants is greater (the reactants are charged) than the concentration of charge on the transition state. ” Increasing the polarity will decrease the rate of the reaction because the concentration of charge on the reactants is greater (the reactants are charged) than the concentration of charge on the transition state. e. Increasing the polarity will increase the rate of the reaction because the concentration of charge on the reactants is less (the reactants are not charged) than the concentration of charge on the transition state. 28. a. CH,Br + HO ——» CH,OH + Br” HO” is a better nucleophile than H,O. b. CHI + HO —>» CHOH + I- IN is a better leaving group than CI” + e CH;Br + NH, —>» CH;NH, + Br NH, is a better nucleophile than H,O. a cuBr + Ho PMSO cmom + Br DMSO will not stabilize the nucleophile by hydrogen bonding. EtOH + e CH,Br + NH, ——>» CH;NH, + Br The more polar solvent will be more able to stabilize the transition state. 35. 36. 37. 38. 39. Chapter 10 291 The weaker base is the better leaving group. a. HO c. HS e Ir b. NH; d. HS” f. Br” - - Da a. HO ec HS e CH;NH, g. CH;CO b. CH;O” d. CH;CH,S” f. “CEN h. CHC=C” ú a. CH;CH;S” > CH;CHO > CH,CO e NH, > HO id (o > (9-0 d > B”> cr The pKa would increase (it would be a weaker acid) because of a decreased tendency to form a charged species in a less polar solvent. a. The Sn2 reaction proceeds with inversion of configuration. CH,CH,CH. a CH,CH,CH. pd cH,O I quot td Ci — >» nr E BS NºH HZ “och; CH; CH; (R)-2-bromopentane or (S)-2-pentanol CH,CH,CH; E CH,CH,CH; B cH cH;O 3 —> CcH;——0CH,; H H 292 Chapter 10 b. The SN1 reaction proceceds with racemization (either complete or partial). qproton º GHaCHCH; CH,CH,CH; Vim = EN + é. BY 'H HZ >oH HO” NºH Ce cH; CH; (R)-2-bromopentane or (S)-2-pentanol (R)-2-pentanol CH,CH,CHs . CH,CH,CH; CH,CH,CH; HO B CH; > cH—fon + HO—-CH, H H c. trans-1-Chloro-2-methylcyclohexane exists as a pair of enantiomers. The SN2 reaction proceeds with back-side attack. PP BL HC H H CH; HC OCH, CH CH; d. trans-1-Chloro-2-methylcyclohexane exists as a pair of enantiomers. The asymmetric carbon is lost when the carbocation that is formed in the SN1 reaction undergoes a 1,2-hydride shift. A LIT H CH; e. 3-Bromo-2-methylpentane exists as a pair of enantiomers. The asymmetric carbon is lost when the carbocation that is formed in the Sy1 reaction undergoes a 1,2-hydride shift. Hs GE CH; CH;CHÇHCH;CH, — CH;CHCHCH,CH; — CH;CCH;CH,CH; Br pres CH; CHGOHCHCH, OCH; 294 Chapter 10 43. Pres + HBÍ CH; rasa HO CHCH,;NH, i i 0H Mo), OH O L B: Fi em H, NÉ Hs soca -0,CCHCH,CH, OHOH OHOH 44. cH; à CHA Em cHo-H (O — or por e à HCc—ÀH CH, CH,CHs CH,CH; (2R,38) E cH; b. CH J o. a — cacem, o HOME PAN HA-cH, CH CH orar CHCH; CH; em A H—ocH a C— c==CH; or 3 HZ “em cH HC—-H CH;0 cs CH,CH, Chapter 10 295 cH a cm A 3 Nc—c=CH,CH; or H—0CH; Ergo H-cH; CH; CH É: (28,3R) CH,CH; e. GHaCHs GHaCHs CH,CH, (fot, C, + ns€ or H CH; + CH—-H Ed NO: E'y >em, Nm CH “e(cH,) (CH5)C G(CHsa (CH), paia o E,CH OCH,CH; CH;CH;O OCH,CH; OCH,CH; The alkyl halide dissociates to form a carbocation that undergoes a 1,2-methyl shift. Because the carbocation is planar, the methly groups with its pair of electrons can add either to the top or the bottom of the carbocation. Therefore, the asymmetric carbon in the product can have either the R or the S configuration. f. ( )-emocinen, 45. a. We can predict that this is an SN1 reaction, because acetate ion is a relatively weak nucieophile and the alkyl halide are sterically hindered. CH; cH; CH; CHGCH=CHCH,CH, — CH;CCH=CHCH,CH; <> CH;C=CHCHCH,CH; º ci o o + [cristo [cisto CH CH; CH;CCH=CHCH,CH, + CH;C=CHÇHCH CH, OCCH; OCCH,; | | (o) (o) Chapter 10 297 48. Br, a. a Br + HBr Br cH,0H b. E Br 5 OCH; + HBr Br, e (O) o (Om ES (om + um 49. u CH,CH; * CH; / cH;O--H Ni cÊ=CH, or 30 EVA H;Cc—H cH;CH, CH,CH; CH,CH; (RAS) CH,CH; b. CH, A cH,O—-—H “c—C=CH,CH, or A X HH; CH;CH, CHs CH,CH; (RAR) CH,CH e + 2CHs CHsCHA — c-=CH,CH, or H—oCH; Hurj N HA cd CH; nei (38,4R) CHCHs CH,CH d. dA Ca CCH, or oc, ad CH,CH, HC+-H 3 (5,48) CHLCH; 50. — Tetrahydrofuran can solvate a charge better because the floppy ethyl substituents of diethyl ether provide steric hindrance, making it difficult for the nonbonding electrons of the oxygen to approach the compound to be solvated. 298 Chapter 10 51. / DN H HÓ: 1 a" “ron | CH; H,0: tra Br — ——» O a HO: cH; 3 0H cH; + HO” a Ho! HO NA 
