Baixe Manual soluções Paula Bruice Química Orgânica Cap12 e outras Notas de estudo em PDF para Química, somente na Docsity! 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. AI four alcohols react by an Sx1 mechanism because they are either secondary or tertiary alcohols. “e à 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 N 3 Cl oH = OH q iá > cl 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 HÁ q Hs e CH;C—CHCH, == CHC—ÇHCH; ——> CHiÇ—CHCH; + HO CH, OH CH;'0H CH; H secondary carbocation 1,2-methyl shift cH; º CH; CH,C-CHCH. «EL CH;C—CHCH a 3 ço 3 Br CH, - CH; tertiary carbocation Chapter 12 335 H cH; 9H cH, *0H cH cH ! 31 3 à 8 d. CHCH, » CHCH, CHCH, CH,CH; mma Di + —<— ——— LEhydride shift . + HO tertiary carbocation secondary carbocation pr CH; CH,CH, ci 3. 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> cHBr + 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. 5. 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. 338 Chapter 12 10. GH cH I 3 a. CH,;CH,CHCH,0H b. oH 11. & CHsCHhCHCHCH; cH,0H JF oH CH; —» CH; + | q + HO + o — H, HO: cH; en 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 (lt — + HOH H,0: EN H H 3 ———— (Pic (Dom, + HO | Ho! + d N 57 CH 9 Chapter 12 339 o c. CH. Aê dj CH, QH VE + HO CAE CH; E CH; CH; CH; Br CN ai e CH; cH; ' 12. CH ESO GH cH; a. CH;CH;C—CHCH, = É CHsCH€—CHCH; — a CHsCHÇ—CHCH; OHCH; *QHCH; CH, + HO CH; CH; HO" + *c=d N cHcHS om; H 0H +0H b. i I * CHCH,CH; CHCH,CH, CHCH,CH, HySO, 2SOy — 5) or —» [meme + HO HO! + E S-cmen, cai [Dcmen, Loss of any one of four hydrogens leads to this product. "a oH c. 2-0» 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 ) a) RR H = mem —— >» P HO: CcH,0H CH,-0H CH, CH, q + Ho” | nO cH 3 CH; CH; I HSO4 I e CHCHCH-ÇCH; ==5 CHCHCH—ÇCH; OH CH; a OH CH; x CH; Í CHsCH)CH—ÇCH; + HO CH, | ,2-methyl shift a CHs, cH; CH; HO + =d <—— CHSCHLCH-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 CcH;CH . Nes poa 3 N A HO! + PRN + [e=€ =—— CH,CH,CH,CHCH; N H H H cH; + HO major product 2 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 E + VN carbocation that leads to 2-naphthol 21. without an NIH shift | —Ç + DOH with an NIH shift D D, 4ôm Con Q oH o SN A H NIH D H Hiog To it + 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 A 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. De” — SK stable carbocation “02” 00” more stable carbocation Chapter 12 345 Each arene oxide will open in the direction that forms the most stable carbocation. Thus, the arene oxide opens so the positive charge can be stabilized by electron delocalization from the methoxide group. *QCH; oc The arene oxide opens to form the most stable carbocation intermediate, which is the one where the positive charge is father away from the electron-withdrawing NO, group. oH o Ê OH , HO. NA “NS ÍNA o” So o” So “0” Fo more stable less stable a. Note e a bond joining two rings cannot be epoxidized. bes b. The epoxide ring in phenanthrenes T and III can open in two iss iene to give two different carbocations and, therefore, two different phenols. oH 348 Chapter 12 oH 9 CH CH; * 3 b. CH,MgBr H HsO, ja CH; CH; mam GY HSO, | H, << Pd/C 30. Solvedin the text. 31. All the reactions occur because, in each case, the reactant acid is a stronger acid than the product acid (methane, pK, = 50). 32. Because silicon is more electronegative than magnesium, transmetallation will occur. 4 CH;MeCl + Sil —> — (CHyySi + 4MgCh 33. Because the alkyl halide will undergo an elimination reaction instead of the alkyl group of the Gilman reagent substituting for the halogen. Chapter 12 349 34. CHCH,)CH,Br > CH(CHoCHyLi CUL (CH;CH,CH,CH;CH,)CuLi CHSXCH ACHo)CHhBr | Hr “H CH(CH,), CH5),CHa “L — =” H H 35. 36. CH,CH,CH,C=CH 37. (Dm + met po so + cnc H=CH, 350 Chapter 12 38. q N a. CH;CH,CH,0OCCH; f. CH;CH>ÇH-ÇCHs cHO OH b. CH;CH,CH,CH,Br g. CH;CH=CHCH; e cmencmeo(/ 5 h. CHaÇHCH,CH,CI cH, CH; q d. CH;CH,ÇH-CCH; i. ( )ctencrom OH OCH, e. , CH=CH, j 7 CH,CH,CHs, 39. CH,CH,0H The other alcohol cannot undergo dehydration because its B-carbon is not bonded to a hydrogen. oH b A secondary allylic carbocation is more stable than a a secondary carbocation. Hc. ,0H c A tertiary carbocation is more stable than a secondary carbocation. GH d. 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. 43. qts a CH;CBr + CH;CH,0H CH; b. CH;CHCH,OH + CHyl cH; e CH;CH, (Hs Pi cH$ CH; d. CH ,CHa E CH CH; 44. 2-butanol is a secondary alcohol. ia CHÇH—ÇCH; CH; OH 2,3-dimethyl-2-butanol a tertiary alcohol 45. E ne + CH=CH, CH=CH, PA(PPha)4 ——+» (CHCH,),N Chapter 12 353 CH,CH,CH; 2,3-Dimethyl-2-butanol will dehydrate faster because it is a tertiary alcohol, while 3,3-dimethyl- cH. [o CH;CH—CCH, OH CH, 3,3-dimethyl-2-butanol a secondary alcohol om - O Siam CH,CH,0H 1. BH;/THF Do ds 2.H,05, HO 354 Chapter 12 E Mgct CcH,CH,0H d. Mg, 1. ethylene oxide ——————» EO 2.H* pra CH,CH,CEN CH,CH,OTs e. CH;CH,C=CH NH cHCH,C=C” O + Pg 1. HC CH, CH;CH,C=CCH,CH,0H HB f. CHÇHCH,OH — CHsÇHCH,Br TS CHsÇHCH,MgBr CH; CH; 2 CH; o /N 1. HC cH, 2.H* CHsÇHCH;CH,CH,OH CH; 46. A CHÇHCH—CH, + CH0! — CH;CHCH—CH,OCH; ci No Cc | o /N . CcH;CH— CHCH,OCH, + Ci Chapter 12 355 47. a. Ifan NIH shift occurs, both carbocations will form the same keto-intermediate. Because it is about four times easier to break a C—H bond (ks) compared with a C—D bond (k4), about 80% of the deuterium will be retained. H. 0Hp HO p H Hp k k 2 no e NIH NIH mit DL A a SODA 4 H oH b. If an NIH shift does not occur, 50% of the deuterium will be retained because the epoxide can open equally easily in either direction, and ky is equal to ko. A | H oH HO SS 358 Chapter 12 0H MgBr CY PBr; —— o 1. 2x 2. Hº OE CH; CH; cH; I HSO, I HBr I e CHÇCHCHCH, SI CHC=CHCH CH, Sim CHCHÇHCH;CH; peroxide oH Br MgBr e cHcH,cH=cH, FÊ cH,CH;CH,CH;Br —U > CH;CH,CH,CH,Li peroxide a use some of thex cueso N productofthe firststepasthe + reagent for this step (CH;CH,CH,CH,),CuLi N q CH,CH,CH,CH,Br | CH;CH,CH,CH,CH,CH,CH,CH; Chapter 12 359 55. o) 207 /N ss (Co HC——cH, + HO: —>» | CH,CH,0H VA No) Su 105 HOCH,CH,0OCH,CH, (o N NE H,O - HOCH,CH,OCH,CH,0CH,CH,0H | «+ HOCH,CH,0CH,CH,0CH,CH,O + HO” 56. o cH;cH 2 2-ethyloxirane a. CH;CH>ÇHCH,0H OH 0.1 M HCl is a dilute solution of HClin water b. CH;CH,CHCH,OH OCH; e. CH;CH;CHCH,CH,CH; 0H d. CH;CH,CHCH,0H oH e CH;CH,ÇCHCH,OCH, 0H 57. Ethyl alcohol is not obtained as a product, because it reacts with the excess HI and forms ethyl iodide. cH,cmocHcH, —Hl — > CHCHjI CH;CH,0H my A CH;CH,I + HO 360 Chapter 12 58. a. w o pe CN A ———» CH30CH,CHCH,CH,CH;!Br q CH,CH,CH,Br O, cHOCH, (Y + Br b. 02) O A LN a ——» FÓCHnCHCH,CH;CH, Br CH,CH,CH,B ea J a dem cH;: | OCH, Du o e. The six-membered ring is formed by attack on the more sterically hindered carbon of the epoxide. Attack on the less sterically hindered carbon is preferred. 59. A=Mg D =ethylene oxide G =ethylene oxide B = diethyl ether E=H+ H=H+ C=CH3MgBr F=Na 60. Greg did not get any of the expected product, because the Grignard reagent reacted with the hydrogen of the alcohol group. Addition of HCI/H,0 protonated the alkoxide ion and opened the epoxide ring. HCl 0 HCl 0 HC OH 3 3 oH HC CH;MgBr + —>> CH + > HO OH O- 0H 64. 65. Chapter 12 363 a. OA , 10H Br—Br HO: HO st —— + HO o EA O + Br. +H,0 b. Only the trans isomers will be formed because the epoxide undergoes backside attack by methoxide ion. oH oH OCH, — CH,O, b is the fastest reaction; a is the slowest reaction In order to form the epoxide, the alkoxide ion must attack the backside of the carbon that is bonded to Br. This means that the OH and Br substituents must both be in axial positions. To be diaxial, they must be trans to each other. CB o a does not form an epoxide, because the OH and Br substituents are cis to each other. b and c can form epoxides because the OH and Br substituents are trans to each other. The rate of formation of the epxoide is given by K Key where k” is the rate constant for the substitution reaction, and Keg is the equilibrium constant for the equatorial/axial conformers. When the OH and Br substituents are in the required diaxial position, the large tert-butyl substituent is in the equatorial position in b and in the axial position in e. 364 Chapter 12 Br Br oH f oH b c Because the more stable conformer is the one with the large tert-butyl group in the equatorial position, the more stable conformer of b has the OH and Br substituents in the required diaxial position (Keq is large), while the OH and Br substituents in c are in the required diaxial position in its less stable conformer (Keq is small). Therefore, b reacts faster than e. Br Br H H (CHyC H C(CH;) OH H 0H b c 66. — The benzene ring can become attached to either of the sp2 carbons of the alkene, resulting in the formation of a pair of constitutional isomers. fa N > — oo 67. om oH :0H OH 0H (1 H;SO4 (1 SG CHÇ—C-CH; === CHÇ—C-CH; —> CHC-CH; + HO CH,CH; CHsCH; CHsCH; + CH;O cH;0H + Lol [ll HO” + CHC—C-CH; CH;C—C-CH, Í CH; CH; Chapter 12 365 68. DO o Hs Gl E a. CHsÇ—CH,0H CHÇ—CH,0H ——>» CH;C-CH,0H + H,O0 + oH *+0H . | q o CH;CH-CH ===> CH;C=CHOH + H,0' b. Dehydration of the primary alcohol group cannot occur because it cannot lose water via an El pathway, because a primary carbocation cannot be formed. It cannot lose water via an E2 pathway, because the f-carbon is not bonded to a hydrogen. 69. 9 H;SO, 7 qo CE + A —— >» + HO 0H 408) We Í | o *+0H ' =—— HO + iá 70. N(CHs) OH O au N(CH» OH + CHNHCH, —» + QU R,R-isomer $,S-isomer