有機(jī)化學(xué)第16章課后習(xí)題答案第二版王積濤等ppt課件

1、習(xí)題與解答習(xí)題與解答（2）.CHCO2C2H5ClCHCO2C2H5，（1）.CH3CCHCCH3CH3CH=CHCHCCH3OO，O解：CH3CH2CCHCCH3OCH3CH2-C=CH-CCH3OO-OCH3CH2-C-CH=CCH3O-OCH3CH=CHCHCCH3OCH3CH-CH=CH-CCH3OCH3CH=CH-CH=C-CH3O-堿性：CH3CH=CHCHCCH3OCH3CH2CCHCCH3OOEEEEE1. 寫(xiě)出下列負(fù)離子的共振結(jié)構(gòu)式。分別比較兩對(duì)子的堿性。解：CHCO2C2H5ClCH=C-OC2H5ClCH-COC2H5ClO-OCH-COC2H5OCH-COC2H5OCl

2、ClEE堿性：CHCO2C2H5ClCHCO2C2H5ECHCO2C2H5CH-COC2H5OCH-COC2H5OCH-COC2H5OCH=C-OC2H5O-2. 按生成烯醇式難易程度排列下列化合物。CH2CH3ABOOOOCOOCH=CH2COOCH2CH3AOOBOO解：CH=CH23. 3-環(huán)己烯酮在催化作用下與2-環(huán)己烯酮存在以下平衡， 寫(xiě)出它們相互轉(zhuǎn)化的機(jī)理OOH+解：OH+-H6OH6HO+H+-H4. 完成下列反應(yīng)式：（1）.C5H6COCH2CH3 + HCO2C2H5NaOC2H5H+C6H5C-CH-CHOCH3O(3).CO2H2LiNCH(CH3)22CH3IH2OCO

3、-Li+OLiCOHCH3O(4) .(CH3)2CHCHO+NCCH2CO2C2H5吡咯烷CH3CHCH=CCNCOC2H5OCH3(2).CH2=CHCOCH=CH2 + CH2(CO2C2H5)2NaOC2H5OCC2H5OCCOC2H5OO(5)COCH3C-CH3OO KOHH+CCOOCH2(8).OO+CH3CC6H5H+ONaOC2H5CH2CH2CH2-C-CH2-C-C6H5OHOO(6).C6H5CH2CHO+ClCHCO2C2H5ZnH+/H2OCH3C6H5CH-CH-COC2H5OH CH3O為Reformatsky反應(yīng)5. 2-甲基-3-丁酮酸乙酯在乙醇中用乙醇鈉

4、處理后加入環(huán)氧乙烷得到一個(gè)化合物。這 個(gè)化合物的分子式為C7H10O3 它的1HNMR數(shù)據(jù)為1.7(三重峰，2H),1.3(單峰， 3H),2.1(單峰，3H），3.9(三重峰，2H）。寫(xiě)出這個(gè)化合物的結(jié)構(gòu)式和反應(yīng)機(jī) 理。(7).C6H5CH2CHOClHCCO2C2H5CH3+NaOC2H5CH2CHCCOC2H5CH3OO解：CH3CCH-C-OC2H5CH3CH2CH2OONaOC2H5OCH3CCCCH2CH2OCH3OOOC2H5CH3-C-C-C-OC2H5CH3OOH-OC2H5CH3-C-CCH3OOCH2CH2OCH3-C-CCH3OOCH2CH2-O-CH-C-CHCH3O

5、CH2CH2O-O-OC2H5CH3CCCCH2CH2OCH3OOabcdC-OC2H5C-OC2H5C-OC2H56設(shè)計(jì)合成路線 把 牛兒醇轉(zhuǎn)化為下列化合物。 (1) (CH3)2C=CHCH2CH2C(CH3)=CHCH2CH2CO2C2H5(2) (CH3)2C=CHCH2CH2C(CH3)=CHCH2CH2COCH3(CH3)2C=CHCH2CH2C(CH3)=CHCH2OH 牛兒醇7. 由丙二酸二乙酯或乙酰乙酸乙酯為起始原料合成：(1).CH2=CHCH2CHCO2HCH3（1）SoCl2Mg/乙醚CH2CH2OH+CrO3-吡啶參P274Ag+(NH3)2C2H5OHH2SO4O(

6、CH3)2C=CHCH2CH2C(CH3)=CHCH2OH(CH3)2C=CHCH2CH2C(CH3)=CHCH2Cl(CH3)2C=CHCH2CH2C(CH3)=CHCH2-CH2CH2OH(CH3)2C=CHCH2CH2C(CH3)=CHCH2-CH2CHO(CH3)2C=CHCH2CH2C(CH3)=CHCH2CH2COC2H5(2). CH3CCH2COC2H5OONaOC2H5CH3CCH-CHOC2H5OOCH3-C-CH-C-OC2H5OOCH3-OHH+, -CO2OCH3(CH3)2C=CHCH2CH2C(CH3)=CHCH2ClCH2CH=C-CH2CH2-CH=C(CH3

7、)2CH3CCH2-CH2CH=C-CH2CH2CH=C(CH3)2CH2(COC2H5)2解：OOCH3-OHH+, -CO2CH2=CHCH2CHCO2HCH3 NaOC2H5 CH2=CHCH2Cl NaOC2H5 CH3ICH2=CH-CH2-C - (COC2H5)2CH2=CH-CH2-CH(COC2H5)2(2).CH2COHOO解：CH2COH+CH2(COC2H5)2OHOHC2H5OCOOOOOOO合成：C2H5OCH2=CHCH=ONaOC2H5OO+OOHC2H5OCONaOC2H5OOC2H5ONaOC2H5CH2(CO2C2H5)2OOC2H5OCH(COC2H5)

8、2O-OHH+, -CO2OCH2COOH(3).COOHCH2(COC2H5)2ONaOC2H5BrCH2BrCH2CH2CH2CH2CH2BrCH(COC2H5)2ONaOC2H5 C(COC2H5)2O-OHH+, -CO2COOH解：(5). CH3CCHCH2CH2CHCCH3CH3CH3OOO(4).CH3CCH2CH2CH2CO2HCH3CCH2COC2H5NaOC2H5CH2=CHCOC2H5CH3-C-CH-COC2H5CH2CHCOC2H5-OHH+, -CO2解：OOOOOOCH3CCH2-CH2CH2COHOO(6). CH3OO解：CH3OO CH3-CH-CH2CH

9、2-C-OHOHOOOO CH3-CCH2-CH2COH CH3CCH2CH(CO2C2H5)2OOO(5).CH3CH3O2CH3CCH2COC2H5NaOC2H5CH3I 2CH3CCHCOC2H5CH3NaOC2H5 2CH3C-C-COC2H5CH3OOBrCH2CH2BrCH3CC-CH2CH2-C - C-CH3OC2H5CH3OC2H5CH3OO-OHH+, -CO2CH3CH3解： CH3CCHCH2CH2CHCCH3 CH3CCHCH2CH2CHCCH3OOOOCH2(CO2C2H5)2NaOC2H5BrCH2CCH3CH3CCH2-CH(CO2C2H5)2NaBH4CH3C

10、H-CH2CH9CO2C2H5)2OOOH-OHH+, -CO2CH3OO CH(CO2C2H5)(7).CH2(CO2C2H5)2解：2NaOC2H5參P512- C -CO2C2H5CO2C2H5BrCH2CH2CH2BrCO2C2H5CO2C2H5LiAlH4H+CH2OHCH2OHPBr3CH2BrCH2BrCO2C2H5CO2C2H5CH2(CO2C2H5)22 NaOC2H5-OHH+, -CO2COOHCOOHEE8. 寫(xiě)出下列反應(yīng)機(jī)理：質(zhì)子遷移OHOCH2CH(CH3)2HO-H+HOCOCH2CH(CH3)2OO-OCH2CH(CH3)2OH質(zhì)子遷移OCH2CH(CH3)2O

11、HOH-H2OOCH2CH(CH3)2O(1).O+ (CH3)2CHCH2OHOCH2CH(CH3)2OO解：OOH+OOHHOCH2CH(CH3)2OOHOCH2CHCH3CH3H H2SO4苯E(2). C2H5O2C CH3 H+/H2OCH2CH2CO2C2H5OO NaOC2H5OO CH3解：OOC2H5O2CCH3-OC2H5OO-C2H5OCCH3OOC2H5C2H5OCOCH3C-O2H5OOOCH2CH2CO2C2H5C-OC2H5OCHCH3OCH2CH2CO2C2H5OC2H5O-CH3-OC2H5CH2CH2CO2C2H5CH3HOO-HOC2H5CH2CH2COC

12、2H5CH3OOH+CH2CH2COC2H5CH3OOOOOCH3-CCH2COC2H5OOOHCHONaOC2H5CCH2CCH3OCH3C-CH2COC2H5NaOC2H5三乙生成C 對(duì)上步反應(yīng)的產(chǎn)物進(jìn)行1，4-加成EtO-CO(3).+ H+/H2OOCO2H+O2CH3CCH2CO2C2H5 HH3CHH2O CO2 NaOC2H5 CH32C2H5OH解:OCH3CO2HOCH3OOCH3OOOOC-OEtCO2EtCO2EtEtO-CEtO-COOCHCCH3OCH-C-CH3OONaOC2H5相當(dāng)于羥醛縮合CH3COOC2H5OC2H5O-COH+/H2OCH3CO2HO+ CO

13、2 + 2C2H5OHC-OEtEtO-C9 .錳酮（menthone）為萜類化合物，其結(jié)構(gòu)符合異戊二烯規(guī)律，即萜一般 由異戊二烯碳骨架首尾相接而成。錳酮分子式為C10H18O，它可被下列方 法合成。解：CH3CH2C-OC2H5OC-OC2H5ONaOC2H5CH3CHC-OEtOC-OC2H5OCH3C-OEtOO3-甲基庚二酸二乙酯分子內(nèi)的酯縮合NaOC2H5(CH3)2CHICH3OC-OC2H5O(CH3)2HC(B)(A)-OHH+， -CO2O錳酮3-甲基庚二酸二乙酯錳酮1)NaOC2H52)H+/H2OA (C10H16O3)1)NaOC2H52)(CH3)CHIB1)OH-/

14、H2O2)H+/10 .根據(jù)下列反應(yīng)寫(xiě)出D的結(jié)構(gòu)式。ZnH+/H2OA(C11H22O3)CrO3NaOC2H5芐氯OH-/H2OH+，B(C11H20O3)C(C18H26O3) D (C15H22O)庚醛 + -溴代乙酸乙酯解：CH3(CH2)5CHO + BrCH2COC2H5 CH3(CH2)5CH-CH2COC2H5ZnH+/H2O參P479 Refoematsky反應(yīng)OHO(A)OOCrO3CH3(CH2)5CCH2COC2H5NaOC2H5芐氯OH-/H2OH+，(B)CH2(C)CH3(CH2)5CCH2CH2(D)OO CH3(CH2)5C-CH-COC2H511 .通過(guò)酯縮

15、合反應(yīng)合成：OOOCH2CH2CNOO(2)(4)（1） C6H5CH2CCH2C6H5(3) (CH3C)2CHCH2C6H5O解：(1)C6H5-CH2-C-CH2-C6H5 結(jié)構(gòu)對(duì)稱，由碳酸二乙酯與兩分子苯乙酸乙酯 縮合而成。OCH-COEtH + C2H5O-C -OC2H5 + HCH-COEtCH-C-CH-C-OEtOC-OEtO-HH+， -CO2CH2-C-CH2-2NaOC2H5OOOO(2)OOC2H5OC2H5OCCOC2H5COC2H5O由兩分子丁二酸二乙酯縮合而成加致活劑C2H5OCC2H5OOOOOOOOCOEtEtOC+EtOCEtOCEtOCCOEtCOEtC

16、OEt-2C2H5O-EtOCCOEt-OHH+， -CO2OO2NaOC2H5OOOOOOOOOOOOOOOC2H5ONaCH3COClNaOC2H5C6H5CH2ClCH3C-C-COC2H5OC-CH3OCH2-C-CH3O H+， ,CO2 -OH(CH3C)2CHCH2C6H5O CH3C-CH2COC2H5 CH3C-CH-COC2H5(3)(CH3C)2CHCH2C6H5O有丙酮結(jié)構(gòu)，可由三乙為原料進(jìn)行合成OO（4）CH2CH2CNOCOC2H5CH2CH2CNCOC2H5+ CH2=CHCNCOC2H5C-OC2H5NaOC2H5 H+OCOEtNaOC2H5 CH2=CHCN

17、己二酸二乙酯 Dieckmann縮合COEtCH2CH2CN H+， CO2 -OHOCH2CH2CNOOOOOOO解 ：（1）ONHNOHO+ NH2OH己內(nèi)酰胺12 .完成下列轉(zhuǎn)化： (1) C2H5OC(CH2)5CO2C2H5ONH(2)OCO2C2H5CO2C2H5CH2CH3OOOOC2H5OC(CH2)5COC2H5NaOC2H5 H+OCOC2H5 H+， ,CO2 -OH庚二酸二乙酯ONH2OHNOHONHBeckmann重排（2）該題是Robinson擴(kuò)環(huán)反應(yīng)之后再進(jìn)行Wittig加成。OCO2Et+ CH2=CH-C-CH2CH3ONaOC2H5共扼加成CO2EtCH3O

18、ONaOC2H5羥醛縮合CO2EtOPh3P=CH2CO2C2H5CH2CH3CH313 . 由“ 三乙” 、丙二酸二乙酯和不超過(guò)四個(gè)碳的原料合成： （1）CO2HCO2H(2)OO(3) CH3CHCHCH2C-CH3H3CCH2CH3OHOH(1) 2CH(CO2Et)2 2CH(CO2C2H5)2 H2C2NaOC2H5Br(CH2)4BrCH2-CH2-CH(CO2Et)2解：CH2-CH(CO2Et)2NaOC2H5CH(CO2Et)2C(CO2Et)2I2CH(CO2Et)2C(CO2Et)2NaOC2H5C(CO2Et)2C(CO2Et)2C-OEtOC-OEtOC-OEtOOEtCO H+， CO2 -OHCOOHCOOHO(2)OO添致活基OOCOEtEtOCOOCOEtEtOC+ EtO-C-C-OEt由戊二酸二乙酯與草