有机化学课后习题答案南开大学版9醇和酚

1、 3第九章 醇和酚命名以下化合物：12-乙基-1-己醇23-甲基-2-丁烯醇3溴代对苯二酚5氯-7溴-2萘酚3，4-4甲基-5苯基-1戊烯-3醇完成以下反响式：ClClCl(1)CH3(2)(3)CH2OHHO(4)(5)CH2CH3OH(6)CH3CH3COCOCH3CH3(7)HBrIHHIBrHCH2CH3CH2CH3(8)OCH3用化学方法将以下混合物分别成单一组分。OCH3苯酚OHH+苯酚水层NaOHOHH2O油层环己醇OHO2NNO2NO2CH水层H+2，4，6-三硝基苯酚NaHCO3H2O3O N油层2NaHCO3H2O32NO2NO2用简洁的化学方法鉴别以下化合物。FeCl3检

2、验出苯酚，用Na 检出溴苯，然后用Lucas 试剂检验出伯、仲、叔醇。按与氢溴酸反响的活性挨次排列以下化合物。641352按亲核性强弱排列以下负离子：1432651OHOC H25+C H+OHOHCC HH+25C2H5C2H525C HH2OOHO25+CC HH+C C2H525C HOHOHC2H525OHOC HCC HH+C HH+6565C6H5H2O65C6H5C6H5写出以下反响可能的机理：+OH+C HCC HH+C H+2525H O25C HC H2C H25252C2H5C2H5C2H5H+C2H5+OH+H+H+H2O+C2H5 CHCH3H+ H2OC2H5 CH

3、CH3OH+OH+C2H5H+C2H5OHOH OHH+H2ONaBH4H+ CH3OH+CH3H+OOH+OHCH3OOHH+CH3OCH3O+AH+3CH +CO3OHOCH3或CHCHOHOCH3或CHCH3BCH3或CCH3OOHCH3OH或CCH3CHCH3C3CH3OH OHOH OH3CH3O或CH3CH3CH3CCH3CH3CH3完成以下转化：醇(1)CH2OHH2SO4Br2BrNaOH醇(2)HOH2SO4B2H6H2O2OH-CH3OH 及对映体H HCH3TsClOTsH HCH3C2H5ONa醇CH3OOH2OH+(3)OHPBr3Mg乙醚MgBrCOHCH2CH3(

4、4)(CH3)2NB2H6H2O2OH-(CH3)2NOHSarrett CrO3.Py(CH3)2NO以苯、甲苯、环己醇及四个碳以下的有机物为原料合成：(1)Br2BrFeMgBrH OHOOMg乙醚2H+HOOMg乙醚2HH SO24CH CH CH ClAlCl323CH2CH3ClC H ONa2 5醇C H ONa2 5醇Cl3hvClCH3CCHC H MgBr2 C H MgBr2 5OCCH3H O2H+OHO2NH SONaOHHNOCH324HO SCHO3O3HOCH33 HOCH3CH3Br2CH3MgBr2CH3HOMgH OFe乙醚H+HOMgH OH SOB HH

5、 OHO24CH32 62 2CH3HOOH-Mg(CH3)3CClMg乙醚(CH3)3CMgClClCHCH=CH22 (CH3)3CCH2CH=CH2CHH3HNi2CH3 CCH2CH2CH3CH3Ni CrON23CrON23B HH OOCH MgBrCH MgBr3H O2H+OHCHK Cr OH H SO24CH3CH32 62 2OH22 7OOH-H SO24OOHMgCH CCHH OCH3CH2CH2CH2MgBr乙醚Br332H+CH3CH2CH2CH2CCH3CH3PBr3CH3CH2CH2CH2CCH3CH3MgCH3CH2CHCH3Mg乙醚BrHCHOH O2H

6、+2CH3PBr3CH3CH2CHCH2BrCH3NaOH醇CH3CH2C=CH2CH3CH CHH2SO4HO SCH CHNaOHCH3CH2ClAlClCH3CH2ClAlCl3熔融Br3HOCH2CH3NBSHOCHCH3NaOH醇NaOCH=CH2CH2ClCH2OCH=CH2CH2ClH2SO4HO SHOCH3ICH OBr2Fe3熔融OHBr2FeOHCH3OBrMgCH OMgBr CH3CHO H2OCH3OCHCH3乙醚3H+OHOOA一对对映体 B COHOOC I2/NaOH发生碘仿反响，说明是甲基酮，见醛酮一章P353OTs(CH3)3COTs(CH3)3CHE2H

7、(CH3)3C及对映体E2 反响机理反响，所以速率与对甲苯磺酸酯、乙氧基负离子浓度均成正比。H(CH3)3COTs(CH3)3COTs C2H5ONaE1H(CH3)3C及对映体因稳定构象很难翻转，OTs 不能处于直立键，无法反式共平面消退，所以只能按E1 机理反响，故反响速度慢，且反响速率只取决于对甲苯磺酸酯的浓度。ACH3或OHCH3CH3OH CH3CH3BOHOHCH3CODCHECH3FCH33ClOH(HBr,ClOH(CH3Cl+)构型转化Br ( + )ClClOH (+)HBr, 邻基参与构型保持Br ( + )当卤素与羟基处于顺式时，卤素不能发生邻基参与，只能发生 SN2 反响，构型发生翻转，生成反式的卤代物，当卤素与羟基处于反式时可以邻基参与，构型保持，因而生成相 同的产物。OHH3C OHCHCH3H+3CH3HOCH3HOOHHOCH3H+CHOHHOCH3CH3CH3CH3CO重排为反式共平面重排，且发生在环上的反响，a e 键高。发生了溴的邻基参与反响。BrCH3CHCH3HOHBrH+CH3HCH3CCHOH2Br+CCH+CCH3HBr-HBrCH3HCH3CHBrS,RBrCH3BrH+Br+CCH3Br+CCH3HBr-S,RBrCH3HCCHCH3OHHCH3CCHOH2CH3HCH3CCHBrR, R+它们对映体的生成过程类