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Solutions to Problems 1. b. ua fu CHsCH>CHCH, Br Ss CH, CHCHCHÇCR, Br CH, crdem,caci CH; Chapter 11 The alkene that is formed is more stable than the alkene that is formed from the other alkyl halide. Br is a better leaving group (weaker base) than CI. The alkene that is formed is more stable than the alkene that is formed from the other alkyl halide. The other alkyl halide cannot undergo an E2 reaction because it doesn't have any -hydrogens. The reaction of 2-bromo-2,3-dimethylbutane with sodium tert-butoxide forms two alkene elimination products. Because of steric hindrance, the least stable alkene (2,3-dimethyl-1-butene) is the one that is easier to make. 2-bromo-2,3-dimethylbutane Free Energy Hs ÇHs CHG—CHCH; Br (CH)sCO CH; CH; CH; CH; CH,=C—CHCH;, + CHC==CCH, 2,3-dimethyl-1-butene 2,3-dimethyl-2-butene 2-bromo-2,3-dimethylbutane 2,3-dimethyl-1-butene 2,3-dimethyl-2-butene ——»>» Progress of the Reaction 303 304 3. Chapter 11 a. CH;CH=CHCH; Removal of a -hydrogen from the most substituted carbon forms the most stable "alkene-like" transition state. b. CH,=CHCH,CH, Removal of a B-hydrogen from the least substituted carbon forms the most stable "carbanion-like" transition state. CH; e CH,C=CHCH,CH, Removal of a B-hydrogen from the most substituted carbon forms the most stable "alkene-like" transition state. d. CH;CH=CHCH=CH, ME B-hydrogen is removed that will lead to a conjugated alkene. & The f-hydrogen is removed that will lead to a conjugated alkene. CH; f. CH;CHCH=CHCH, Removal of a f-hydrogen from the least substituted carbon forms the most stable "carbanion-like” transition state. cH; a. CH,CHCHCH;CH; It forms the more stable alkene; the alkene has a greater Br number of substituents bonded to the sp? carbons. It forms is more stable alkene; the new double bond b. is conjugated with the double bond that is already there. Br e. CH4CH,CHCH,CH; It has four hydrogens that can be removed to form an Br alkene with two substituents on the sp2 carbons so it has a greater probability of an effective collision with the nucleophile than the other alkyl halide that has only two such hydrogens. d. ( )cinguencn, It forms the more stable alkene; the new double bond is br conjugated with the phenyl substituent. 306 Chapter 11 8. a. E2 beause a strong base is used. e. El because a weak base is used. CH;CH=CHCH; CH; CH;C=CCH; b. El because a weak base is used. du CH;CH=CHCH, 3 f. E2 because a strong base is used. c. El because a weak base is used. Gs CH; CH;CCH=CH, CH,C=CH, dm, d. E2 because a strong base is used. Notice that even though the same alkyl halide CH; is used in "e" and "f”, different products are obtained because the carbocation that forms CH;C=CH, under Sy1 conditions rearranges. 9; a. E2 beer bromide] [HO"] EZ+EL k[tert-buyyl bromide] [HO-] + ky[tert-by6yl bromide] E2 o 7.1x10% x 5.0 o 35.5 x 10% 2355 96 EZ+EL 71x10º x 50 + 15x10) 355x10%+ 1.5x10º 37 . = 96% b. E2 7.1x 10? x 2.5 x 10º 1.78x 107 178 12% == a = = 1.2% EZ+EL 71x10 x25x107+150x107 1.78x107 +150x 107 152 10. ms a. 1. CH;CH,CH=CCH; No stereoisomers are possible. 2. CH;CH, H No=s nd The major product has the bulkier group bonded to one / sp? carbon on the opposite side of the double bond from H O the bulkier group bonded the other sp? carbon. 3 CHCHo H The major product is the conjugated diene with the c=c" bulkier groups on opposite sides of the double bond. N / H CH=CH, b. In none of the reactions is the major product dependent on whether you started with the R or S enantiomer. 1. 12. Chapter 11 307 a. Solved in the text. e EÇHo H d. CH; Puma GEs H;C CH,CH,CH; b. CH;CH,CH=CCH, Elimination occurs when the H and Br to be eliminated are in axial positions. When Br is in an axial position in the cis isomer, it has an axial hydrogen on each of the adjacent carbons. The one bonded to the same carbon as the ethyl group will be more apt to be the one eliminated with Br because the product formed is more stable and, therefore, more easily formed than the product formed when the other H is eliminated with Br. When Br is in an axial position in the trans isomer, it has an axial hydrogen on only one adjacent carbon, and it is not the carbon that is bonded to the ethyl group. Therefore, a different product is formed. cis-1-bromo-2-ethylcyclohexane trans-1-bromo-2-ethylcyclohexane Br Br H H CH,CH; H <— mm <— + HBr H + HBr CH,CH, CHCH, 1-ethylcyclohexene 3-ethylcyclohexene Chapter 11 309 Cl + a SAE? ocn, OCH, CH; CH; Cl a SNVEL VEL + Tm shift cH, [7 [ore CH; cH; SE CH; 8a RIO; f. á + Cu H' Hr" E / NH cH;O CHCH; cHo CHCH; cHCH, “ocH, H,C H H H se j q ) He H H A C=C + c=€ c=c + C=c / " / x H CH, HC CH, H cH, Hx cH, major elimination product minor major elimination product minor 310 Chapter 11 15. Bristhe weakest base so it is the best Icaving group, and F is the worst leaving group. It is easier to break a C—H bond than a C—D bond, so the more B-hydrogens that are replaced by a deuterium, the slower the rate of the E2 reaction will be. CH; CH; cH; GD CD; CHÇ-Br > CHÇ-C1 > CHÇ—F > CHÇ-F > CDiÇ—F CH; CH; CH; CD; CD; 16. The rate-limiting step in an El reaction is carbocation formation. Because the proton is removed : . - + in a subsequent fast step, the difference in the rate of removal of an HÍ versus a D” would not be reflected in the rate constant. Therefore, the deuterium kinetic isotope effect would be close to 1. 17. Because CH3S” is a stronger nucleophile and weaker base than CH30”, the ratio of substitution (where CH3S” reacts as a nucleophile) to elimination (where CH3S” reacts as a base) will increase when the nucleophile is changed from CH30” to CH3S”. 18. In order to undergo an elimination reaction under E2 conditions, the substituents that are to be eliminated (H and Br) must both be in axial positions. Drawing the compound in the chair conformation shows that when Br is in an axial position, the adjacent hydrogens are in equatorial positions, so an elimination reaction can'”t take place. CH; CH; Br 19. a.1. noreaction b. 1. primarily substitution 2. no reaction 2. substitution and elimination 3. substitution and elimination 3. substitution and elimination 4. substitution and elimination 4. elimination 312 Chapter 11 A Br NaOH 2 - CH,CH,B d. CH;CH;CHCHCH,OH ay CHyCH;CH;CHCH;O Cir CH; CH; CHsCH,CH,CHCH,OCH,CH; or CH; CHsCH,CH)CHCH,Br CH; cHCHOH > CHCHO —— 2 + CHsCH,CHCHCH,0CH,CH; CH; Both methods can be used to synthesize the target molecule. The first method is preferred because ethyl bromide has less steric hindrance than does 1-bromo-2-methylpentane. 23. CH, CH; CH, CH; l CH,CH,0 | Í I CHÇ-Br o CHÇ-OH + CHÇ-0CHCH, + CHC=CH, CH, 2 CH; CH; 24. cH, CH; CH, à CHCHCHCH,CH; ET CH;CHCHCHCH; + CH;C=CHCH-CH; N: Br oH The SN2 reaction will form the substitution product with the inverted configuration compared to the configuration of the starting material. But, because the configuration of the starting material is not specified (we do not know if it is the R isomer, the S isomer, or a mixture of R and S), we cannot specify the configuration of the SN2 product. GH HO GH 12-hydride CH; CHsCHCHCH,CH; RT CH;CHÇHCH,CH; it CH;CCH,CH,CH; Br s | cH; GH I CH;C=CHCH;CH; + CHAÇCH,CH-CH 0H Chapter 11 313 cH, HC cH; Hc CH,CH db. CHCHCH>CH; ET Voc | Siga ER Br x H CH,CH; H CH; minor cH HO cH Lo dlilut po HC CH; CH;CH,CCH;CH,; Se CH;CH;CCH,CH, + C== I Sxl/EI I 4 x Br 0H H CH,CH, HC, CH,CH; Cc=— á H minor 3 25. Because an allene is less stable than an alkyne, it is harder to make. 26. a. Ho pr because it forms a six-membered ring, while the other compound would form a seven-membered ring. A seven-membered ring is more strained than a six-membered ring so the six-membered ring is formed more easily (see Table 2.9 on p. 96 of the text). b. HAN AN because it forms a five-membered ring, Br while the other compound would form a four-membered ring. A four-membered ring is more strained than a five-membered ring so the five-membered ring is formed more easily. c HO pr because it forms a seven-membered ring, while the other compound would form an eight-membered ring. An eight-membered ring is more strained than a seven- membered ring so the seven-membered ring is formed more easily; also, the Br and OH are less likely to be in the proper position relative to one another for reaction because there are more bonds around which rotation to an unfavorable conformation can occur in the compound that leads to the eight-membered ring (Section 24.6). Chapter 11 315 tertBuO d. CH;CH,CH,CH;Br E CH,CH;CH=CH, By CHCHCHCHBr | (Notice that a bulky base is used to Br encourage elimination over substitution.) . | NH, excess ) H,0 1. NH, CH;CH,CCH,CH,CH; «<< CH,;CH,C=CCH,CH, = — CH;CH,C=CH gtatdo quilot-dis H,SO, au dão ads 5. CH,CH,Br à ie? a CH=CH ter BuO A 2 e BrCH,CH,CH,CH,Br ——> CH,=CHCH=CH, — > excess (1,3-butadiene is both the diene and the dienophile) H | d/C Oda 29. 8. The product depends on the configuration of the reactant, which has not been specified. If the reactant is cis. H Ha CH, CY E b. C== e =. “O If the reactant is trans. CH; H cH; CH; c Na, e f. / N H CH,CH; 316 Chapter 11 30. a. b. e 31. a. b. CH; H “ee a O 8 CH; H CH; Clio, Fa CE==C e. No reaction because it is a primary alkyl halide. H cH; CHa, H / ' CH; C==C . / N H CH,CH; 1.3º>2º>1º 2. An El reaction is not affected by the strength of the base, but a weak base favors an El reaction by disfavoring an E2 reaction. 3. An El reaction is not affected by the concentration of the base, but a low concentration of a base favors an El reaction by disfavoring an E2 reaction. 4. An aprotic polar solvent favors an El reaction if the reactant is charged. A protic polar solvent favors an El reaction if the reactant is not charged. 1.3º>2º>1º 2. A strong base favors an E2 reaction. 3. A high concentration of a base favors an E2 reaction. 4. An aprotic polar solvent favors an E2 reaction if either of the reactants is charged. A protic polar solvent favors an E2 reaction if neither of the reactants is charged. 32. The predominant product is the elimination product because tertiary alkyl halídes react with nucleophiles to form an elimination product and little substitution product. CH Gs CH5ÇCH; + CHCHO —>— cH;C=CH, ci predominant product Rather than a tertiary alkyl halide and a primary alkoxide ion, he should have used a primary alkyl halide and a tertiary alkoxide ion. q q CH;CHCI + CHÇOS — CHAÇOCHCH; + ci” CH; CH; 318 3s. 36. Chapter 11 f. CHsCH A CH;CH, Es C=C + Cc= / N f N cH$ CH; cHy H minor a. ethoxide ion because elimination is favored by the bulkier base, and tert-butoxide ion is bulkier than ethoxide ion b. -SCN because elimination is favored by the stronger base, and “OCN is a stronger base than “SCN e. Br” because elimination is favored by the stronger base, and CI” is a stronger base than Br” d. CH;S” because elimination is favored by the stronger base, and CH30 is a stronger base than cH;S- The first compound has two axial hydrogens adjacent to Br, the second has one axial hydrogen adjacent to Br but it cannot form the more substituted (more stable) alkene that can be formed by the first compound. The last compound cannot undergo an E2 reaction because it does not have an axial hydrogen adjacent to Br. Br Br Br 37. 38. Chapter 11 319 a. a S,A Co CH,CH; NBS. A CHCH, (CH)co CH=CH, peroxide u A r Notice that a bulky base and heat are used in the last step to encourage elimination over substitution. 1.NH, b. CH;CH,CH=CH, Br CH;CH,CH,CH,Br + CH,CH,CH,CH,NH, peroxide 2.HO- After the Sn2 reaction with NH3, the solution is made basic so the final amine product is in its basic form. HOCH,CH,CH=CH, eronãE HOCH,CH,CH,CH;Br Dê» “OCH,CH,CH,CH,Br | (3 e d. O Es DM cH,0H dá Br OCH; NB CHCO CH,CH,CH=CH, road CHÇHCH=CH, Cinco CH,=CHCH=CH, Br ? CH, CH; GH GH; CH; CH; l CH(—CHCH, Ls cH=C—CHCH, + CHC=CCH, 2,3-dimethyl-1-butene 2,3-dimethyl-2-butene Br GHbCHs a. CH,;CH,CO Because it is the most sterically hindered base, l it gives the highest percentage of the 1-alkene. CH,CH; b. CH;CH,O. Because it is the least sterically hindered base, it gives the highest percentage of the 2-alkene. 40. 41. Chapterll 321 a. Br Br Big a ci Cl DD tert-Bul O ny Of Õ Es * fia o ci Br oH b. Br Br - > tert-BuO NBS, 4, 2 O hy Of O peroxide dao The number of atoms in the ring is given by n. Three and four membered rings have strain, so they are harder to make than 5 and 6 membered rings. The three-membered ring is formed faster than the four-membered ring because the compound leading to the three-membered ring has one less carbon-carbon single bond that can rotate to give a conformer in which the reacting groups are positioned too far from one another for reaction. Fai Br A [= 5 os e» H,N H5N 7 N CH, =cH, CA Even though the compound that forms the five-membered ring has one more bond that can rotate to give a conformer in which the reacting groups are positioned too far from one another for reaction than the compound that forms the four-membered ring, the five-membered ring is formed faster because it is relatively strain free. So lack of strain more than makes up for the lower probability of having the reacting groups in the proper position for reaction. Now the rate of the ring-forming reaction gets slower as the size of the ring being formed gets larger, because the reactant has more bonds that can rotate to give conformers in which the reacting groups are positioned too far from one another for reaction. 322 42. 43. 44. Chapter 11 In an E2 reaction, both groups to be eliminated must be in axial positions. When the bromine is in the axial position in the cis isomer, the tert-butyl substituent is in the more stable equatorial position. When the bromine is in the axial position in the trans isomer, the tert-butyl substituent is in the less stable axial position. Thus, elimination takes place via the most stable conformer in the cis isomer and via a less stable chair conformer in the trans isomer, so the cis isomer undergoes elimination more rapidly. o fx LE (CHa)3€ cis-1-bromo-4-tert-butylcyclohexane trans-1-bromo-4-tert-butylcyclohexane q cendugoes, O —» É (aco, E OCH CHEOCH; É 1— És io + Br Because the f-carbon from which the hydrogen is to be removed is bonded to only one hydrogen, the configuration of the reactant determines the configuration of the E2 elimination product. To determine the configuration of the product, convert the prospective formula of the Fischer projection to a staggered a Newman projection, in which the groups (H and Br) that are to be eliminated anti to one another. (Remember that the Fischer projection of the reactant shows the molecule in an eclipsed conformation, since both horizontal bonds are pointing toward the viewer.) Thus the elimination product formed from (28,38)-2-chloro-3-methypentane is the E isomer. 
