有机化学高鸿宾第四版答案第十章-醚和环氧

第十章醚和环氧化合物习题(一)写出分子式为CHO的所有醚及环氧化合物，并命名之。512解：CH3OCHCH2CH3CH3CH3OCH2CHCH3CH3CH3OCH2CH2CH2CH3甲基丁基醚甲基仲丁基醚甲基异丙基醚CH3CH3OCCH3CH3CH3CH2OCHCH3CH3CH3CH2OCH2CH2CH3甲基叔丁基醚乙基丙基醚乙基异丙基醚(二)完成下列反应式：OCHCH2CH3CH3Br+CH3CH2CHCH3ONa(1)CHCH3CH253CH(2)CH3CH2I+NaOC25CH3CH2OCHHCH3CH2CHCH2BrNaOHOHCH3CH2CHCH2O(3)(4)NH2H2NHONH3O+OHBrBrNH3(5)C(CH3)CH2NH2OHOHClCHCH2OCH2CHOCHCH2OH+Cl(6)CHCl3OH(三)在3,5-二氧杂环己醇OO的稳定椅型构象中，羟基处在a键的位置，试解释之。解：羟基处有利于形成具有五元环状结构的分子内氢键：于a键时，HOHOO或OOO(四)完成下列转变：OCOCH3CH2OCH3(1)OCOCH3Na+C2H5OHCH2OHCH2OCH3CH3I解：NaOHBrOH(2)及OPhK2CrOOH27O解：H2SO4OHOBrMgMgBrH2O/H+THFOC6H5CO3H(TM)Br及CH2CHCH3OHCH3CHCH3OH(3)H2SO4CF3CO3HCHCHCH3OHCH2=CHCH3CH2CHCH3O3解：CH2CHCH3BrMgMgBrH2O/H+CH2CHCH3OHOTHFCH2CH2CH2OCH2CH3CH2OHCHCH2OH及(4)3H2SO4CF3CO3HCHCH2OHCH2=CH2CH2CH23解：o170COHBrCH3CH2BrOCH2OHCH2ClSOCl2MgCH2MgClH2O/H+dryether(1)NaCH2CH2CH2OC2H5CH2CH2CH2OH(2)C2H5BrCHCHOH及32(5)OHSO4oCHCH2OHCH2=CH22解：3170CCF3CO3H+CH2=CH2O(五)推测下列化合物A~F的结构，并注明化合物E及F的立体构型。(1)Mg/纯醚Br2(1)CH2=CHCH2BrA(C4H8O)B(C4H8Br2O)(2)HCHO，(3)H3O+KOH25oCKOHC(C4H7BrO)DOBrCH2=CHCH2CH2OHBrCH2CHCH2CH2OHBr解：OA(C4H8O)B(C4H8BrO)2C(C4H7BrO)COOHH(1)LiAlH4/乙醚KOH,H2O(2)ClE(C3H7ClO)F(C3H6O)(2)H3O+CH3CH2OHHCH3CHCH2OCl解：CH3E(C3H7ClO)F(C3H6O)(六)选择最好的方法，合成下列化合物(原料任选)。(CH3)3CCHOCH(CH3)2CH3O(1)(2)解：(1)用烷氧汞化-脱汞反应制备：CH3CH3CCH3BrCH3CH33CHC=CH2H2OHBrMg干醚3CH3CHOCHCMgBrH+CH3CH3CH3CCH=CH2CH3CH3OHCH3CCHCHCH3P2O53(-H2O)(CF3COO)2Hg(CH3)2CHOHNaBH4(CH3)3CCHOCH(CH3)2CH3(CH3)3CCH=CH2OH-ClHCl+O(2)H2OH+OHNaONaOHOHH2,PtH2SO4O或者：(七)完成下列反应式，并用反应机理解释之。CH3(1)H2CCCH3CH3HOCH2CCH3OCH3CH3OHH2SO4(少量)OCH3HCCCH3CH3H2CCCH3CH3H2CCCH3OHCH3OHH+解：(1)2OOHCH3CH3HOCH2CCH3OCH3-H+H2CCCH3HOHOCH3CH3CH3CH3OHCH3ONa(少量)(2)H3CCCH2OH3CCCH2OCH3OHCH3HCCCH2CH3H3CCCH2OCH3OCH3H3CCCH2OCH3OHCH3OHCH3O-解：3OOOH2SO4,THF(3)OHOH解：OOHH+OHO-H+(八)一个未知物的分子式为C2H4O，它的红外光谱图中3600~3200c-1m和1800~1600cm-1处都没有峰，试问上述化合物的结构如何？解：C2H4O的不饱和度为1，而IR光谱又表明该化合物不含O-H