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C5H12O的所有醚及环氧化合物，并命名之。解：CH3OCH2CH2CH2CH3CH3OCHCH2CH3CH3CH3OCH2CHCH3CH3甲基丁基醚甲基仲丁基醚甲基异丙基醚CH3OCCH3CH3CH3CH3CH2OCH2CH2CH3CH3CH2OCHCH3CH3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：(1)Br+CH3CH2CHCH3ONaOCHCH2CH3CH3(2)CH3CH2I+CNaOHC2H5CH3COHC2H5CH3CH3CH2(3)CH3CH2CHCH2BrOHNaOHCH2CHOCH3CH2(4)ONH3NH2OHH2NHO+(5)BrONH3BrC(CH3)CH2NH2OH(6)CH2CHOHClCHCl3CH2OHCHCl+CH2CHO(三)在3,5-二氧杂环己醇OOOH的稳定椅型构象中，羟基处在a键的位置，试解释之。解：羟基处于a键时，有利于形成具有五元环状结构的分子内氢键：OOOH或OOOH(四)完成下列转变：(1)COCH3OCH2OCH3解：COCH3OCH2OCH3CH3INaOHNa+C2H5OHCH2OH(2)BrOHOPh及解：K2Cr2O7H2SO4OHOMgBrH2O/H+MgTHFOOHBrC6H5CO3HO(TM)(3)Br及CH2CHCH3OHCH3CHCH3OH解：CH3CHCH3OHCHCH2OCH3H2SO4CH2=CHCH3CF3CO3HCH2CHCH3OHMgBrH2O/H+MgTHFBrCHCH2OCH3(4)CH2OHCH3CH2OHCH2CH2CH2OCH2CH3及解：CH3CH2BrH2SO4170CoCF3CO3HCH3CH2OHCH2=CH2CH2CH2OHBrCH2ClH2O/H+CH2OHCH2MgClMgdryetherOSOCl2CH2CH2CH2OC2H5CH2CH2CH2OH(1)Na(2)C2H5Br(5)及CH3CH2OHO解：H2SO4170CoCH3CH2OHCH2=CH2OCH2=CH2+CF3CO3H(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)CH2=CHCH2Br(1)Mg/纯醚(2)HCHO，(3)H3O+Br2A(C4H8O)B(C4H8Br2O)KOHKOH25CoC(C4H7BrO)DO解：A(C4H8O)B(C4H8Br2O)C(C4H7BrO)CH2=CHCH2CH2OHBrCH2CHCH2CH2OHBrOBr(2)(1)LiAlH4/乙醚(2)H3O+KOH,H2OE(C3H7ClO)F(C3H6O)HCOOHCH3Cl解：E(C3H7ClO)F(C3H6O)HCH2OHCH3ClCCH2OHCH3(六)选择最好的方法，合成下列化合物(原料任选)。(1)(CH3)3CCHOCH(CH3)2CH3(2)O解：(1)用烷氧汞化-脱汞反应制备：CH3C=CH2CH3Mg干醚HBrCH3CCH3CH3BrCCH3MgBrCH3CH3CH3CHOH2OH+CCH3CHCH3CH3CH3OHCCH3CH=CH2CH3CH3P2O5(-H2O)(CH3)3CCH=CH2(CF3COO)2Hg(CH3)2CHOHNaBH4OH-(CH3)3CCHCH3OCH(CH3)2(2)O+OHClHClONaH2OH+Na或者：OHOHH2SO4H2,PtO(七)完成下列反应式，并用反应机理解释之。(1)CH2COCH3CH3H2SO4(少量)CH3OHCHOCH2CH3CH3OCH3解：(1)CH2COCH3CH3CH3OHH+CH2COCH3CH3HCH2COHCH3CH3CH2CHOCH3CH3OCH3H-H+CHOCH2CH3CH3OCH3(2)CCH2OH3CCH3CH3ONa(少量)CH3OHCCH2H3CCH3OCH3OH解：CCH2OH3CCH3CH3OHCH3O-CCH2OH3CCH3OCH3CCH2H3CCH3OCH3OH(3)H2SO4,THFOO解：OOHH+OHOHHOO-H+(八)一个未知物的分子式为C2H4O，它的红外光谱图中3600~3200cm-1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和C=C，所以该化合物分子中有一个环。其结构为CH2H2CO(九)化合物(A)的分子式为C6H14O，其1HNMR谱图如下，试写出其构造式。CHOCHCH3CH3CH3CH3ab