有机化学(第四版)+高鸿宾版+答案 联系客服

发布时间 : 星期五 文章有机化学(第四版)+高鸿宾版+答案更新完毕开始阅读0a035fb269dc5022aaea00fe

Br(2)

OH及K2Cr2O7H2SO4OPh

解:

OHO

BrMgTHFMgBrOH2O/H+OH

C6H5CO3HO (TM)

?(3)

Br及CH3CHCH3OHCH3CHCH3OHH2SO4?CH2CHCH3OHCF3CO3H

解:

CH2=CHCH3CH2OCHCH3

CH2CHCH3OH2O/H+BrMgTHFMgBrCH2CHCH3

OH(4)

CH2OH及CH3CH2OHH2SO4170 CoCH2CH2CH2OCH2CH3CF3CO3H

解:

CH3CH2OHCH2=CH2HBrCH2OCH2

CH3CH2BrMgdry etherCH2OHSOCl2CH2Cl(1) Na(2) C2H5BrCH2MgClCH2CH2CH2OC2H5OH2O/H+

CH2CH2CH2OH

(5)

及CH3CH2OHOH2SO4170 Co

解:CH3CH2OHCH2=CH2

+CH2=CH2?CF3CO3H

O(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=CHCH2BrKOH(1) Mg/纯醚(2) HCHO, (3) H3O+KOH?A(C4H8O)

Br2B(C4H8Br2O)

25 CoC(C4H7BrO)DOBr解:

CH2=CHCH2CH2OHA(C4H8O)BrCH2CHCH2CH2OHBrB(C4H8Br2O)E(C3H7ClO)KOH , H2OOC(C4H7BrO)F(C3H6O)

COOH(1) LiAlH/乙醚4ClH(2)

(2) H3O+CH3CH2OHClH解: CH3E(C3H7ClO)CH2OCCH3H

F(C3H6O)(六) 选择最好的方法,合成下列化合物(原料任选)。

(1)

(CH3)3CCHOCH(CH3)2CH3CH3HBr (2)

OCH3

解:(1) 用烷氧汞化-脱汞反应制备:

CH3CH3CCH3BrMg干醚CH3C=CH2CH3CMgBrCH3CH3CHOH2OH+

CH3OHCH3CCH3CHCH3(-HO)2P2O5CH3CH3CCH=CH2

NaBH4OH-CH3(CF3COO)2Hg(CH3)2CHOH(CH3)3CCH=CH2(2)

(CH3)3CCHOCH(CH3)2CH3

+HClClO ?H2OH+OHNaONaOH 或者:

H2 ,Pt?OHH2SO4?O

(七) 完成下列反应式,并用反应机理解释之。

CH3(1) H2CCCH3OCH3CH3OHH2SO4(少量)CH3HOCH2CCH3OCH3CH3H2CCOHCH3H2COHCH3CCH3CH3OH

解:(1) H2CCOCH3H+ CH3 H2CCCH3-H+CH3HOCH2CCH3 OCH3HOHOCH3CH3(2) H3CCOCH2CH3OHCH3ONa(少量)CH3H3CCCH2OH:

OCH3

CH3H3CCOCH2CH3O-CH3H3CCCH2OH2SO4,THFOCH3CH3OHCH3H3CCCH2OHOCH3

O(3) 解

O

OHOH+OHOH HO-H+O (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

H2CCH2O

(九) 化合物(A)的分子式为C6H14O,其1HNMR谱图如下,试写出其构造式。

CH3CH3a

CH3CHOCHCH3 b