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334 Chapter 12 Solutions to Problems The relative reactivity would be: tertiary > primary > secondary. If secondary alcohols reacted by an SN2 mechanism, they would be less reactive than primary alcohols because they are more sterically hindered than primary alcohols. All four alcohols react by an Sy1 mechanism because they are either secondary or tertiary alcohols. “e a CH;CH;CHCH, ===5 CH;CHCHCH, —— CH;CH)CHCH; + HO 0H +0H H - p CH;CHoÇHCH; Br Db. CH É CH; CH; cr CH; By A 3 €1 oH = OH e” iá a (6| H + HO The carbocation that is initially formed in c and the carbocation that is initially formed in d rearrange in order to form more stable carbocations. ps HÓ q Hs e CH;C—CHCH, == CHC—ÇHCH; > CH;Ç-CHCH; + HO CH, OH CH;'0H CH; H secondary carbocation 1,2-methyl shift CH; º CH; CH,C-CHCH, «BL cHC-CHCH, a + Br CH, — CH; tertiary carbocation Chapter 12 335 H cH; QH cH; *0H CH; CHs d. CHCH, ' CHCH, CHCH, CH,CH; —»» + =—— ——— LZhydride shift - + HO tertiary — carbocation secondary carbocation | CH; CH,CH, ci The stronger the base, the more difficult it is to displace. Water is a stronger base by about 7 pKa units than Br”; (the pKa of HBr =-—9; the pKa of CH30H2* = -1.7). Br + cHÔH —L> CcHBr + HO H Ammonia is a stronger base by about 18 pKa units than Br”; (the pKa of HBr = — 9; the pKa CH3NH3* = 9.4). Thus, ammonia is far too basic to be displaced by Br. Br” + CH;NH, —A sy no reaction 4. Solved in the text. All the syntheses below are shown to be accomplished by converting the alcohol into a sulfonate ester and then treating the sulfonate ester with the desired nucleophile: a could also have been carried out by converting the alcohol directly into the alkyl bromide with HBr; b-f could have been carried out by converting the alcohol into an alkyl halide and then adding the desired nucleophile. Chapter 12 337 Recall that alkenes undergo electrophilic addition reactions, and the first step in an electrophilic addition reaction is addition of an electrophile to the alkene. The acid-catalyzed dehydration reaction is reversible because the electrophile (H+) needed for the first step of the reverse electrophilic addition reaction is available. Base-promoted elimination reaction of a hydrogen halide is not reversible because an electrophile is not available to react with the alkene. 1-Bromo-1-methylcyclopentane would be formed because Br- would react with the carbocation. Tf the carbocation were able to lose a proton before it reacts with Br”, HBr could add to the resulting alkene, thereby forming the same alkyl halide as would be formed by addition of Br to the carbocation. cH Br . 4 Bo CH, — G + cH,0H CH; CH; Do» O O 0H a. In order to synthesize an unsymmetrical ether (ROR') by this method, two different alcohols (ROH and R'OH) would have to be heated with sulfuric acid. Therefore, three different ethers would be obtained as products. Consequently, the desired ether would account for considerably less than half of the total amount of ether that is synthesized. H,SO, ROH + ROH ra ROR + ROR' + ROR b. It could be synthesized by a Williamson ether synthesis. (See Section 11.9 on page 425 of the text.) - CH;CH;B - cH;cH;cH;oH Ni» cHcHcHO SECHBL cp CHLCHLOCELCH, + Br 338 Chapter 12 10. GH cH l 3 a. CH,CH,CHCH,0H b. oH 11. & CHsCHhCHCHCH; cH,0H J oH CH; —» CH; + | À + HO + o — Ho HO: cH; a b. Because of the difficulty of forming a primary carbocation, dehydration is an E2 reaction. The alkene that results is protonated and the proton that is removed is the one that results in formation of the most stable alkene. o” CH a! (tl — + HOH H,0: EN H H » ———— (Pica (Dom + HO | Ho! + d 1 57 CH; 9 340 Chapter 12 Tn “d” and “f”, the reactant is a primary alcohol. Therefore, elimination of water takes place by an E2 mechanism. Because the dehydration reaction is being carried out in an acidic solution, the alkene that is formed initially is protonated to form a carbocation. The proton that is then lost from the carbocation is the one that results in formation of the most stable alkene. H,0: d. HySOg ) ao) Sa H = mi ——» P HO: cH,0H CH,-0H CH, CH, q + H;0” | vo (OL cH 3 CH; CH; I HSO4 | e CHCHCH-ÇCH; ==5 CHCHCH—ÇCH; OH CH; a OH CH; x cH; I CHsCH)CH—ÇCH; + HO CH; | ,2-methyl shift ” CHs CH; CH; HO" + = <—— CHSCILCH-CCH N cHcHS CH; cH; H,SO4 ” f. CH,CH,CH,CH,CH,0H === CH,CH,CH,CH;CH,0H — CH,CH,CH,CH=CH, | + HO" CHsCH, cH CH;CH N Nes pod 3 N psi HO! + VE + [e=€ =—— CH,CH,CH,CHCH; N H H H cH; + HO major product 2 13. 14. 15. Chapter 12 341 In the presence of excess HI, the alcohol that is formed when the ether is cleaved is subsequently converted into an alkyl iodide. CH,0CH,CH,CH, e CH; + CH;CH,CH,0H + CH;CH,CH;I + HO We saw that HCI does not cleave ethers, because CI” is not a strong enough nucleophile. Fis an even weaker nucleophile, so HF cannot cleave ethers. Therefore, ethers can be cleaved only with HBr or HI. Notice that if there is excess concentrated HI, the initially formed alcohol can be converted to an alkyl iodide in c, d, and f. a. Solved in the text. s Cleavage occurs by an SN1 pathway because the benzyl carbocation that is formed is relatively stable; I- will attack the benzyl carbocation. Drs no) e. Cleavage occurs by an Sw1 pathway because the benzyl carbocation that is formed is relatively stable; I- will attack the benzyl carbocation. CH,CH,CH,0H + Ed) d. HOCH;CH,CH,CH,CH; EL ICH,CH,CH,CH,CH;I + HO OH tautomerization [a CHI + Cr of Athee enol Of enol ketone f. Cleavage will occur by an Sl pathway because the tertiary carbocation that is formed is relatively stable; I- will attack the tertiary carbocation. io qo HI HOCH,CH,CH-CCH; —> ICH,CH>CHCCH; + HO I I Chapter 12 343 20. The carbocation leading to 1-naphthol can be stabilized without destroying the aromaticity of the intact benzene ring. The carbocation leading to 2-naphthol can be stabilized only by destroying the aromaticity of the intact benzene ring. therefore, the carbocation leading to 1-naphthol is more stable. oH oH + mei <> + VN carbocation that leads to 2-naphthol 21. without an NIH shift | —Ç + DOH with an NIH shift D D, 4ôm Com Q oH o SN Tê H NIH D H Hiog To sit + HO CH; CH; CH; 344 Chapter 12 22. — The epoxide opens in the direction that will give the most stable carbocation. The carbocation undergoes an NIH shift and, as a result of the NIH shift, both reactants form the same ketone intermediate. Because they form the same intermediate, they form the same products. The deuterium-containing product is the major product because in the last step of the reaction it is easier to break a carbon-hydrogen bond than a carbon-deuterium bond. NIH shift +0H E: H D D, Nam NIH e major so minor se! 23. a. Solvedin the text. b. The compound without the double bond in the second ring is more apt to be carcinogenic It opens to form a less stabie carbocation than the other compound because it can be stabilized by electron delocalization only if the aromaticity of the benzene ring is destroyed. Because the carbocation is less stable, it is formed more slowly, giving the carcinogenic pathway a better chance to compete with ring-opening. 9H — SK stable carbocation “02-00” more stable carbocation 346 Chapter 12 c. The two different carbocations formed by phenanthrenes II and IIT differ in stability. One carbocation is more stable than the other because it can be stabilized by resonance without disrupting the aromaticity of the adjacent ring. The more stable carbocation leads to the major product. HO oH major product minor product d. Phenanthrene oxide I is the most carcinogenic because it is the only one that opens to form a carbocation that cannot be stabilized without disrupting the aromaticity of the other ring(s). 26. a cH;cH,sH -HOS cmcH,s CHCHBr cH,CH;SCH,CH; + Br” b. The reaction cannot be done with methanethiol and tert-butyl bromide, because a tertiary alkyl halide would form an elimination product rather than a substitution product. Hs r je H,B ig CHÇSH HO, cuçs CHBL cyoscH, + Br CH; CH, CH; Chapter 12 347 e. The highest yield is obtained by having the less substituted of the two R groups of the thioether be the alkyl halide and the more substituted be the thiol. cH; CH; CH; l HO. 1 CH;CH,Br - CH,=CHCHSH ——» CH,=CHCHS ——> CH,=CHCHSCH,CH; + Br d. The synthesis must be done this way because the sp? carbon of the benzene ring cannot undergo backside attack. 27. — The first compound is too insoluble. The second compound is too reactive. The third compound is more soluble than the first because of the oxygen-containing substituent. The third compound is less reactive than the second because the lone pair can be delocalized into the benzene ring, so the lone pair is less apt to displace a chloride ion. 28. a. CH;CH,CH,CH,CH,0H b. ( )-cmemctom e ( cmemon (o) L 9 CH,CH; N CH,CH, a. CH,CH,MgBr H Esto Ep H Fuso, a O qi 29. Chapter 12 349 34. CH(CH,)CH,Br > CH(CHoCHyLi CUL (CH;CH,CH,CH;CH,)CuLi CHSCH ACHo)CHhBr | Hr “H CH(CH,), CH5)pCHa É — ” H H 35. 36. CH,CH,CH,C=CH 37. Dm + me po so + aà cre H=CH, 350 Chapter 12 38. jo N a. CH;CH,CH,OCCH; f. CH;CH>ÇH-ÇCHs cHO OH b. CH;CH,CH,CH,Br g. CH;CH=CHCH, e cmencmemo(/ 5 h. CHaÇHCH,CH,CI cH, CH; q d. CH;CH,ÇH-CCH; i. ( )rctencrom oH OCH; e Í CH=CH, j , CH,CH,CH, 39. a CH,CH,0H The other alcohol cannot undergo dehydration because its B-carbon is not bonded to a hydrogen. A secondary allylic carbocation is more stable than a secondary carbocation. oH HCl ,0H e A tertiary carbocation is more stable than a secondary carbocation. QH CHCH; The rate-limiting step in both dehydrations is carbocation formation: a secondary benzylic carbocation is more stable than a secondary carbocation and, therefore, easier to form. 352 Chapter 12 D Db o . OCH; Am + R-S=co —> men NE ES A d CACHE” Ag il CH sor sx cHcH NH . jo) inversion Db (6)slicdeuterios (S)-1-deuterio- I-propanol 1-methoxypropane or (o) O cHCH CH,CH; T qu tda cHO CH,CH; Hop + Rj —» RIO | D s iai D T OCH; N H (o) jo) H inversion H . (R)-1-deuterio- (S)-1-deuterio- 1-propanol 1-methoxypropane e. Since the desired product has the same configuration as the starting material, it can be synthesized using two consecutive reactions that each involves inversion of configuration. For example, treating the starting material with PBr, forms ($)-1-bromo-1-deuteriopropane followed by reaction of the alky halide with methoxide ion. An casier way to prepare the desired compound is to use reactions that don't break any of the bonds to the asymmetric carbon. D D Db a NaH ia CH Au caca —> cuca Nom cH;cH” NH oH o” OCH; (R)-1-deuterio- (R)-1-deuterio- 1-propanol 1-methoxypropane 42. 2-hexene and 3-hexene HySO, CH;CH,CH,CH,CH,CH,0H + CH;CH,CH,CH,CH=CH, N CH;CH,CH,CH=CHCH, = CH;CH,CH,CH,CHCH; + 2-hexene a + CH;CH,CH;CHCH;CH; —> CH;CH;CH=CHCH,CH; + CH;CH,CH,CH=CHCH; 3-hexene 2-hexene (cis and trans) + H' (cis and trans) 43. get a CH;CBr + CH;CH,0H cH; b. CHCHCH,OH + CHil CH; e CHCH, (Hs AR cH$ CH; d. CH ,CHa [= a CH CH; “A CH,CH,CH; Chapter 12 353 44. 2,3-Dimethyl-2-butanol will dehydrate faster because it is a tertiary alcohol, while 3,3-dimethyl- 2-butanol is a secondary alcohol. Es q CHsCH—CCHs CH;CH—CCH; CH; 0H OH CH; 2,3-dimethyl-2-butanol a tertiary alcohol 3,3-dimethyl-2-butanol a secondary alcohol 45. oH O e O é O NAM (0) b. cH; NS 1. À CHoaMeBr. BO, O! NAM 200 Hr Br CH=CH, e BN Pd(PPha)4 1. BH;/THF || + cH=cH, ainda art vo (CH3CH,);N 2.H,0,, HO” CH,CH,0H 
