ZZM有機(jī)化學(xué)后題答案

1、有機(jī)化學(xué)（第四版，汪小蘭）課后題答案第二章飽和脂肪烴2.1a.2,4,4三甲基正丁基壬烷butyl2,4,4trimethylnonaneb. 正己烷hexanec.3,3二乙基戊烷3,3diethylpentaned.3甲基異丙基辛烷isopropyl3methyloctane e. 甲基丙烷（異丁烷）methylpropane (iso-butane)f. 2,2二甲基丙烷（新戊烷）2,2dimethylpropane(neopentane)g.3甲基戊烷methylpentaneh.甲基乙基庚烷5ethyl2methylheptane2.2a=b=d=e為 2,3,5三甲基己烷c=f為

2、2,3,4,5四甲基己烷2.3a.錯(cuò)，應(yīng)為 2,2二甲基丁烷Cb.c.d.f.e.錯(cuò)，應(yīng)為 2,3,5三甲基庚烷錯(cuò)，應(yīng)為 2,3,3三甲基戊烷h.2.5g.cbead2.63 種123ClClCla 是共同的2.72.8BrBrHHHHABBrHBrHHHCBrHHBrHHDBrBrHHHH2.10這個(gè)化合物為2.11穩(wěn)定性cab第三章不飽和脂肪烴3.1用系統(tǒng)命名法命名下列化合物a.2乙基丁烯ethyl1buteneb.2丙基己烯propyl1hexenec.3,5二甲基庚烯3,5dimethyl3heptened.2,5二甲基己烯2,5dimethyl2hexene3.2-2-有機(jī)化學(xué)（第四

3、版，汪小蘭）課后題答案a.b. 錯(cuò)，應(yīng)為1丁烯c.d.e.f.錯(cuò)，應(yīng)為2,3二甲基戊烯g.h.3.4錯(cuò)，應(yīng)為2甲基3乙基己烯下列烯烴哪個(gè)有順、反異構(gòu)？寫出順、反異構(gòu)體的構(gòu)型，并命名。答案： c , d , e ,f 有順反異構(gòu)Cc. C2H5HCCH2IH( Z )1碘2戊烯 ( E )1碘2戊烯CCCH2IHHC2H5d.CCCH(CH3)2HH3CHCCHCH(CH3)2H3CH( Z )4甲基2戊烯 ( E )4甲基2戊烯e.CCHCH( Z )1,3戊二烯HH3CCH2CCHCH( E )1,3戊二烯H3CHCH2f.CHC( 2Z,4Z )2,4庚二烯HH3CCCHC2H5HCHCH

4、H3CCCHHC2H5( 2Z,4E )2,4庚二烯CHCH3CHCCHHC2H5( 2E,4E )2,4庚二烯CHC( 2E,4Z )2,4庚二烯H3CHCCHC2H5H3.5完成下列反應(yīng)式，寫出產(chǎn)物或所需試劑a.CH3CH2CH=CH2H2SO4CH3CH2CH CH3OSO2OHb.(CH3)2C=CHCH3HBr(CH3)2C-CH2CH3Brc.d.CH3CH2CH=CH2CH3CH2CH=CH2CH3CH2CH2CH2OHCH3CH2CH-CH3OHe.+ CH3CH2CHOf.CH2=CHCH2OHCl2 / H2OClCH2CH-CH2OH1). BH32).H2O2 , OH

5、H2O / H+1). O3(CH3)2C=CHCH2CH3 2). Zn , H2O CH3COCH3OH3.61己烯正己烷Br2 / CCl4or KMnO4無反應(yīng)褪色正己烷1己烯3.7-3-HH有機(jī)化學(xué)（第四版，汪小蘭）課后題答案CH2=CHCH2CH3CH3CH=CHCH3或3.8將下列碳正離子按穩(wěn)定性由大至小排列：CH3CH3 CH3H3C C CH+CH3CH3H3C C CHCH3CH3H3C CCH3+CH2 CH23.9 寫出下列反應(yīng)的轉(zhuǎn)化過程：C=CHCH2CH2CH2CH=CH3CH3C+CH3CH3H3CH3CCH3CH3+C-CH2CH2CH2CH2CH=C+H_3.

6、10or3.11a.4甲基己炔methyl2hexyneb.2,2,7,7四甲基3,5辛二炔2,2,7,7tetramethyl3,5octadiyneCHCd.c.3.13a.LindlarcatH2C CH2b.HC CHNi / H2CH3CH3c.HC CH+ H2OH2SO4HgSO4CH3CHOd.HC CH + HClHgCl2CH2=CHCle.H3CC CHHgBr2HBrCH3C=CH2BrHBrBrBrCH3-C CH3f.H3CC CH +Br2CH3C=CHBrBrg.H3CC CH + H2OH2SO4HgSO4CH3COCH3HC CH + H2Lindlarca

7、tH3CC CH + H2h.H3CC CH+HBrHgBr2CH3C=CH2Bri.CH3CH=CH2HBr(CH3)2CHBr3.14-4-有機(jī)化學(xué)（第四版，汪小蘭）課后題答案-5-正庚烷1,4庚二烯Ag(NH3)2+灰白色1庚炔Br2 / CCl4褪色1,4庚二烯a.b.1庚炔2甲基戊烷2己炔1己炔Ag(NH3)2+無反應(yīng)灰白色無反應(yīng)正庚烷1,4庚二烯1己炔無反應(yīng)褪色無反應(yīng)正庚烷2己炔2甲基戊烷 Br2 / CCl42己炔2甲基戊烷a.CH3CH2CH2C CHHCl (過量)CH3CH2CH2C3.15ClCH3b.CH3CH2C CCH3 + KMnO4H+ClCH3CH2COOH+

8、 CH3COOHH2SO4HgSO4c.CH3CH2C CCH3 +H2OCH3CH2CH2COCH3+ CH3CH2COCH2CH3d.CH2=CHCH=CH2+CH2=CHCHOCHOe.CH3CH2C CH+HCNCH3CH2C=CH2CN3.16CH3CHCH2C CHH3C3.17AH3CCH2CH2CH2C CHBCH3CH=CHCH=CHCH33.18CH2=CHCH=CH2HBrCH3CH CH=CH2Br+CH3CH=CHCH2BrCH2=CHCH=CH22HBrCH3CHBrCH CH3 +BrCH3CHBrCH2 CH2BrCH2=CHCH2CH=CH2HBrCH3CHC

9、H2CH=CH2CH2=CHCH2CH=CH22HBrBrCH3CHCH2CH CH3Br第四章Br環(huán)烴4.1C5H10不飽和度=1有機(jī)化學(xué)（第四版，汪小蘭）課后題答案-6-a.b.c.d.環(huán)戊烷1甲基環(huán)丁烷順1,2二甲基環(huán)丙烷cyclopentane1methylcyclobutanecis1,2dimethylcyclopropane反1,2二甲基環(huán)丙烷 trans1,2dimethyllcyclopropanee.f.1,1二甲基環(huán)丙烷乙基環(huán)丙烷1,1dimethylcyclopropaneethylcyclopropane4.3a. 1,1二氯環(huán)庚烷1,1dichlorocyclohe

10、ptaneb. 2,6二甲基萘2,6dimethylnaphthalenec.1甲基異丙基1,4環(huán)己二烯1isopropyl4methyl1,4cyclohexadiened.對(duì)異丙基甲苯pisopropyltoluenee.chlorobenzenesulfonic acidf.氯苯磺酸CH3Clg.CH3CH3h.2chloro4nitrotoluenecis1,3dimethylcyclopentane4.4a.HBrNO22,3dimethyl1phenyl1pentene完成下列反應(yīng)：CH3CH3Brb.+Cl2高溫Clc.+Cl2ClClClCl+d.CH2CH3+ Br2FeBr

11、3BrC2H5 +BrC2H5e.CH(CH3)2+Cl2高溫C(CH3)2Cl有機(jī)化學(xué)（第四版，汪小蘭）課后題答案f.CH3O3Znpowder , H2OCHOOg.CH3H2SO4H2O ,CH3OHAlCl3h.i.+CH3CH2Cl2+ HNO3CH2CH3NO2+CH3NO2j.+ KMnO4H+COOHk.CH=CH2+Cl2CH2ClCHCl4.521.4.6CH3CH3CH3CH3+Br2FeBr3CH3CH3CH3CH3+BrBr+Br2FeBr3CH3CH3+CH3BrCH3+Br2FeBr3BrCH3CH3CH3CH3Br4.7b , d 有芳香性4.8-7-有機(jī)化學(xué)（

12、第四版，汪小蘭）課后題答案a.BCA 1,3環(huán)己二烯苯1己炔Ag(NH3)2+C灰白色無反應(yīng)A Br2 / CCl4B無反應(yīng)褪色BAb.AB環(huán)丙烷丙烯KMnO4無反應(yīng)褪色AB4.9a.b.c.e.f.ClCOOHNHCOCH3CH3NO2COCH3OCH3d.4.10NO2Bra.b.Br2FeBr3HNO3H2SO4BrNO2HNO3H2SO4Br2FeBr3BrO2Nc.CH32 Cl2FeCl3CH3ClCld.CH3+Cl2FeCl3CH3ClKMnO 4COOHCle.CH3KMnO 4COOHCl2FeCl3COOHCl-8-有機(jī)化學(xué)（第四版，汪小蘭）課后題答案-9-f.HNO3H

13、2SO4NO2CH3H3C2Br2FeBr 3BrCH3Brg.Br2FeBr 3CH3CH3BrKMnO4COOHBrHNO3H2SO4NO2COOHBrNO24.11可能為orororororor即環(huán)辛烯及環(huán)烯雙鍵碳上含非支鏈取代基的分子式為 C8H14O2 的各種異構(gòu)體，例如以上各種異構(gòu)體。4.12H3CC2H54.13BrBrA.4.14A.BrClBrBrBrClBrBrClBCD4.15Cl將下列結(jié)構(gòu)改寫為鍵線式。a.b.c.d.e.OOOHor有機(jī)化學(xué)（第四版，汪小蘭）課后題答案第五章旋光異構(gòu)5.4a.CH2CH2CH2CH3OHBrHCH2OHBrHHBrBrCH2CH3(R)

14、COOHBrCH2OHHb.COOHHOOCCH CHBrBrCOOH( meso- )CH2CH3(S)COOHHBrBrHCOOHCOOH( 2R,3R )BrHHBrCOOH( 2S,3S )c.COOHH3CCH CHBr BrHHCOOHBrBrBrBrCOOHHHHBrCOOHBrHCOOH( 2R,3S )COOH( 2S,3S )CH3( 2R,3R )BrHCOOHHBrd.CH3C=CHCOOHCH3CH3( 2S,3R )( 無 )5.6COOHHCH3CH2CH3(R)COOHH3CHCH2CH3(S)C5H10O25.7 =1C6H12A*CH2=CHCHCH2CH3

15、CH3BCH3CH2CHCH2CH3CH35.10(I)HCOOHOHCH3(R)紙面上旋轉(zhuǎn) 90oOHHOOCCH3H(S)離開紙面旋轉(zhuǎn)COOHHOHCH3(S)- 10 -( 對(duì)映異構(gòu) )( 對(duì)映異構(gòu) )HCNOHOH有機(jī)化學(xué)（第四版，汪小蘭）課后題答案第六章鹵代烴6.1CH3Bra. CH3CH-CHCH3b.CH3H3C CCH3CH2Ic.Brd.Cle.ClClf.2碘丙烷iodopropaneg.三氯甲烷 trichloromethane or Chloroformh.1,2二氯乙烷1,2dichloroethanei.氯丙烯chloro1propenej.1氯丙烯chloro1

16、propene6.4寫出下列反應(yīng)的主要產(chǎn)物，或必要溶劑或試劑a.C6H5CH2ClMgEt2OC6H5CH2MgClCO2C6H5CH2COOMgCl+H2OC6H5CH2COOHb.CH2=CHCH2Br+NaOC 2H5CH2=CHCH2OC2H5c.CH=CHBrCH2Br+AgNO 3EtOHr.tCH=CHBrCH2ONO 2+AgBrCH2=CHCHOCHOKOH-EtO HtOH-EHKO光照Brd.e.CH2ClCl+ Br2NaOHH2OBrBrCH2OHClC2H5OHCH3CH2CH2CH3 + BrMgOC 2H5f.g.CH3ClCH3CH2CH2CH2Br+MgEt

17、2OH2OSN2 歷程CH3CH2CH2CH2MgBrOHh.ClCH3+H2OSN1 歷程OHCH3CH3+OHCH3i.BrKOH-EtO Hj.CH2=CHCH2ClCH2=CHCH2CNk.(CH3)3Cl+ NaOHH2O(CH3)3COH6.5下列各對(duì)化合物按 SN2 歷程進(jìn)行反應(yīng),哪一個(gè)反應(yīng)速率較快?- 11 -有機(jī)化學(xué)（第四版，汪小蘭）課后題答案- 12 -a.(CH3)2CHI(CH3)3CClb.(CH3)2CHI(CH3)2CHClc.CH2ClCld.CH3CHCH2CH2BrCH3CH3CH2CHCH2BrCH3e.CH3CH2CH2CH2ClCH3CH2CH=CHC

18、l6.6將下列化合物按 SN1 歷程反應(yīng)的活性由大到小排列bca6.7(a) 反應(yīng)活化能(b) 反應(yīng)過渡態(tài)(c) 反應(yīng)熱放熱6.8ACH3CH2CHCH3BrBCH2=CHCH2CH3C.CH3CH=CHCH3(Z) and (E)6.9怎樣鑒別下列各組化合物？鑒別a,b,dAgNO3 / EtOHc. Br26.10CH3CHCH2BrCH3KOH-EtO HCH3C=CH2CH3H+H2OCH3CCH3CH3a.OHb.CH3CCH3CH3HBrCH3CCH3CH2BrBr 2HOBrCH3CCH3CH2Brd.Brc.OH Bre.6.11ABrCH2CH2CH3B.CH2=CHCH3C

19、.CH3CHBrCH3HH有機(jī)化學(xué)（第四版，汪小蘭）課后題答案第七章醇酚醚7.1命名下列化合物a. (3Z)戊烯醇 (3Z)penten1olb. 2溴丙醇 2bromopropanolc. 2,5庚二醇 2,5heptanediold. 4苯基2戊醇 4phenyl2pentanole. (1R,2R) 2 甲 基環(huán)己醇 (1R,2R) 2 methylcyclohexanolf. 乙二醇二 甲醚ethanediol 1,2 dimethyl etherg.(S) 環(huán)氧丙烷(S) 1,2 epoxypropaneh.間甲基苯酚mmethylphenoli. 1苯基乙醇1phenylethan

20、olj. 4硝基1萘酚4nitro1naphthol7.3完成下列轉(zhuǎn)化a.OHCrO3.Py2Ob.CH3CH2CH2OH濃 H2SO4 CH3CH=CH2Br2BrBrCH3CH-CH2KOH / EtOHc.ACH3CH2CH2OH(CH3)2CHB rCH3CH2CH2OCH(CH3)2+CH3CH=CH2HBrCH3CHCH3BrCH3C CHNaB.CH3CH2CH2OHHBrH+CH3CH=CH2CH3CH2CH2BrOHCH3CHCH3d.CH3CH2CH2CH2OHCH3CH2CH=CH2OHCH3CH2CHCH3OHOHe.濃 H2SO4SO3Hf.CH2=CH2稀 冷 KM

21、nO4CH2-CH22 OHOCH2CH2OCH2CH2OCH2CH2OHOH OHClCH2CH2OHHOCH2CH2OCH2CH2OCH2CH2OHH2O+Cl2g.CH3CH2CH=CH2CH3CH2CH2CH2OHh.ClCH2CH2CH2CH2OHNaOHONaOHClCH2CH2OHHOHH+NaONaCH3CHCH3CH3CH2CH2BrT.MHBrCH3CH2CH2CH2BrCH3CH2OHNaOH1) B2H6 , Et2O2) H2O2 , OHHOH+7.5下列化合物是否可形成分子內(nèi)氫鍵？寫出帶有分子內(nèi)氫健的結(jié)構(gòu)式。a , b , d 可以形成- 13 -H有機(jī)化學(xué)（第四

22、版，汪小蘭）課后題答案a,NOOOHcis NOOHOtransHONOOOHNOOb.OHOH2CH3CCH3d.OOH7.6寫出下列反應(yīng)的歷程OH+OH2bab+a+_H2OH_ H+_ H+7.7寫出下列反應(yīng)的產(chǎn)物或反應(yīng)物a.(CH3)2CHCH2CH2OH+HBr(CH3)2CHCH2CH2Brb.OH +HCl無水 ZnCl2Clc.OCH3CH2CH2OCH3+HI (過量)+d.OCH3+HI (過量)CH2CH2CH3e.(CH3)2CHBr+NaOC2H5(CH3)2CHOC2H5+CH3CH=CH2f.KMnO4OHCH3(CH2)3C CH3Og.CH3COOHCH CH

23、2CH3HIO4CH3COOH + CH3CH2CHOh.CH3OHOHHIO4CH3COCH2CH2CHOi.OHCH3+Br2OHCH3BrBrCCl4 , CS2 中單取代CH3(CH2)3CHCH3OHj.CH3(CH2)2CHOHCH2CH3分子內(nèi)脫水濃 H2SO4CH3CH2CH=CHCH2CH3+ CH3CH2CH2CH=CHCH37.8- 14 -H有機(jī)化學(xué)（第四版，汪小蘭）課后題答案- 15 -AOCH3B.OHC.CH3I第八章醛、酮、醌8.1a.異丁醛甲基丙醛methylpropanalisobutanalb.苯乙醛phenylethanalc.對(duì)甲基苯甲醛pmethyl

24、benzaldehyded.3甲基丁酮methylbutanonee. 2,4二甲基戊酮 2,4dimethyl3pentanonef.間甲氧基苯甲醛mmethoxybenzaldehydeg.h. BrCH2CH2CHOi.O甲基丁烯醛 3methyl2butenalOj.CCl3CH2COCH2CH3k.(CH3)2CCHOl.CH3CH2COCH2CHOm.CH=CHCHOn.C CH3Oo.丙烯醛 propenalp.二苯甲酮diphenyl Ketone8.3 寫出下列反應(yīng)的主要產(chǎn)物a.CH3COCH2CH3+H2NOHCH3CCH2CH3NOHb.Cl3CCHO+H2OCl3CCH

25、OHOHc.H3CCHOHOOCCOOH+KMnO4H+d.CH3CH2CHO稀 NaOHCH3CH2CH-CHCHOOHCH3e.C6H5COCH3+C6H5MgBrC6H5CH3C6H5C OMgBrH+H2OC6H5CH3C6H5C OHf.O+H2NNHC6H5NNHC6H5g.(CH3)3CCHO濃 NaOH(CH3)3CCH2OH(CH3)3CCOOH+h.O+ (CH3)2C(CH2OH)2無水 HClOOOi.+K2Cr2O7+HOOC(CH2)3COOHj.CHOKMnO4室溫COOHHHHg , H+2)- 16 -k.有機(jī)化學(xué)有機(jī)化學(xué)（第四版，汪小蘭）課后題答案（第四版，

26、汪小蘭）課后題答案OCl2 , H2OOHCOCH2ClC CH3COOH +CHCl3l.OC CH3+Cl2H+ HClCl+ClOHCH3m.n.CH2=CHCH2CH2COCH3CH2=CHCOCH3+ HBrCH3-CHCH2CH2COCH3BrCH2CH2COCH3o.CH2=CHCHO+ HCNNCCH2CH2CHO+CH2=CHCHCNOHp.C6H5CHO+CH3COCH3稀 NaOHOC6H5CHCH2C-CH3OH8.4a.ABCD丙醛丙酮丙醇異丙醇2,4二硝基苯肼有沉淀無沉淀Tollen 試劑I2 / NaOH沉淀無沉淀無沉淀黃色沉淀b.A戊醛B2戊酮C環(huán)戊酮Tolle

27、n 試劑沉淀A無沉淀BCI2 / NaOHCHI3無沉淀BC8.4完成下列轉(zhuǎn)化a.C2H5OHCrO3.(Py)2CH3CHOCH3CHOHCNCH3CHCOOHOHb.COCl無水 AlCl3COc.ONaBH 4OHd.HCCH+H2OCH3CHO稀 OHCH3CH=CHCHO H2 / NiCH3CH2CH2CH2OHe.CH3Cl2光CH2ClMgEt2OOHCH2CCH3CH3f.CH3CH=CHCHOOHOH 無水 HClCH3CH=CHCHOO稀 冷 KMn O4OHOOCH3CH-CHCHOH OHH3O+CH3CH-CHCHOOH OHg.CH3CH2CH2OHHBrCH3C

28、H2CH2BrMgEt2OCH3CH2CH2MgBrCH3CH2CH2CH2OHHCN+H2OH2OCH2MgCl 1) CH3COCH3+1) HCHO2) H+8.5ACH3CH2CHO + CH3MgBrBCH3CHO + CH3CH2MgBr8.6分別由苯及甲苯合成 2苯基乙醇H有機(jī)化學(xué)（第四版，汪小蘭）課后題答案- 17 -CH3Cl2CH2ClMgEt2OCH2CH2OHCH2MgCl 1) HCHO2) H+光照BrMgBrOBr2FeMgEt2OH+H2O8.8a. 縮酮b.半縮酮c.d半縮醛8.9b.H3COOH+CH3CHO + HOCH2CH2OH+OHa.OHOOHOH

29、OOHOHOc.H+OHOHHOHO+OHOHHOHO8.10A.CHCH3CHOHB.CHCH3H3CH3CH3CH3COCC.CCHCH3H3CH3C8.11OCH38.12ACH3CH-C-CH2CH3CH3OB.CHH3CH3CCHCH2CH3OHC.CH3CH3CCHCH2CH3D.CH3CH2CHOE.CH3COCH38.13BrBrHOOCCOOHOAOBOCODH- 18 -有機(jī)化學(xué)有機(jī)化學(xué)（第四版，汪小蘭）課后題答案（第四版，汪小蘭）課后題答案第九章 羧酸及其衍生物9.1 a. 2甲基丙酸 2Methylpropanoic acid(異丁酸 Isobutanoic acid

30、)b. 鄰羥基苯甲酸（水楊酸）oHydroxybenzoic acid c. 2丁烯酸 2Butenoic acidd3溴丁酸3Bromobutanoic acide. 丁酰氯Butanoyl Chloridef.丁酸酐Butanoic anhydrideg. 丙酸乙酯 Ethyl propanoateh.乙酸丙酯 Propyl acetatei. 苯甲酰胺Benzamidej. 順丁烯二酸Maleic acids.COt.H3CCOk.COOCH3COOCH3l.HCOOCH(CH3)2m.CH3CH2CONHCH39.29.3gabcfehd寫出下列反應(yīng)的主要產(chǎn)物a.Na2Cr2O7-H2

31、SO4COOHCOOH+COOHCOOHb.(CH3)2CHOH+COOCH(CH 3)2H3CCOClH3Cc.HOCH2CH2COOHLiAlH4HOCH2CH2CH2OHd.NCCH2CH2CN+H2ONaOHOOCCH2CH2COOH+HOOCCH2CH2COOHe.CH2COOHCH2COOHBa(OH) 2Of.CH3COCl+CH3無水 AlCl3CH3COCH3+CH3COCH3g.(CH3CO)2O+OHOCOCH 3h.CH3CH2COOC 2H5NaOC 2H5CH3CH2COCHCOOC 2H5CH3i.CH3COOC2H5+CH3CH2CH2OH+CH3COOCH 2

32、CH2CH3+C2H5OHj.CH3CH(COOH)2CH3CH2COOHk.COOH+HClClCOOH+ CO2H有機(jī)化學(xué)（第四版，汪小蘭）課后題答案- 19 -l.2HOCH2CH2OH+COOCH2CH2OOCm.COOHLiAlH4CH2OHCOOH +n.HCOOH+OHHCOOH+NaOC2H5Oo.p.NCH2CH2COOC2H5CH2CH2COOC2H5CONH2OHNCOOC2H5COO +q.CH2(COOC2H5)2+H2NCONH2HNNHNH3OOO9.4a.KmnO4 b.FeCl3 c.Br2 or KmnO4d.FeCl3 2,4-二硝基苯肼或 I2 / Na

33、OH9.5完成下列轉(zhuǎn)化：a.OCNOHH+COOHOHb.CH3CH2CH2BrCNCH3CH2CH2CNCH3CH2CH2COOHc.(CH3)2CHOHCrO3.(Py)2(CH3)2C O(CH3)2CCNH+(CH3)2CCOOHOHd.CH3CH3KMnO4COOHCOOHCOOHCOOHBa(OH)2OHOOOOOOe.(CH3)2C=CH2HBr(CH3)3CBrMgEt2O(CH3)3CMgBrf.CH3BrAlCl3CH3KMnO4COOHBr2FeCOOHBrg.HCCHH2OH+ , Hg2+CH3CHOKMnO4CH3COOHCH3CH2OHH+CH3COOC2H5HC

34、NH2OHCN1) CO2(CH3)3CCOOH2) H+ / H2OH2OH+H2OH3O1) EtONa濃 OHCH2CH2CH31) OHH ,H- 20 -h.有機(jī)化學(xué)（第四版，汪小蘭）課后題答案OHNO3HOOC(CH2)4COOHOi.CH3CH2COOHLiAlH4CH3CH2CH2OHHBrCH3CH2CH2BrMgEt2OCH3CH2CH2MgBrCH3CH2CH2CH2CH2OH CH3CH2CH2CH2COOH或H+CH3COCH2COOC2H5 CH3 CO CH COOC2H5CH3(CH2)3COOHj.CH3COOHCl2PCH2COOHClCNCH2COOHCN

35、EtOH+CH2(COOC2H5)2k.OOONH3CH2COONH4CH2CONH2(CH3CO)2Ol.m.CO2CH3OHCH3CH2COOH+H2OSOCl2COOHOHCH3CH2COClOHCOOHOOCCH3CH3CH2COOn.CH3CH(COOC2H5)2CH3CH2COOHO KMnO42) CH3CH2CH2Br+2) H+ 3)9.6己醇A已酸 B對(duì)甲苯酚 CNaHCO3 水溶液水相已酸鈉HCl已酸 B有機(jī)相已醇對(duì)甲苯酚NaOH水相 HCl酚鈉有機(jī)相 已醇 A酚CCOOHCOOH9.7COOHCOOHCOOHCOOH HOOCCOOH HOOCHCH3CH2COOHCO

36、OHHOOC(Z)易成酐 (E) 不易 (Z) 易成酐 (E) 不易成酐9.8A.CH2COOHCH2COOHB.OOOC.H2C COCH3H2C COCH3OOD.CH2CH2OHCH2CH2OHCH2COO有機(jī)化學(xué)有機(jī)化學(xué)（第四版，汪小蘭）課后題答案（第四版，汪小蘭）課后題答案第十章 取代酸n. 4-氧代戊酸（4oxopentanoic acid)10.1 m. 3-氯丁酸（3-Chlorobutanoic acid）10.2a.(A)CH3CH2CH2COCH2COOCH3OHCOOHCH3CHCOOH(B)(C)FeCl3ABC顯色不顯色Na2CO3A溶解，有氣體B無變化OH10.3

37、 寫出下列反應(yīng)的主要產(chǎn)物：a.b.CH3COCHCOOC2H5CH3CH3COCHCO2CH3CH2CO2CH3c.CH3CH2CHCOOHOHHCOOCHCCOCH3CH2CH2CH3d.COOCH3COCH3稀 H+1) 稀 OH-2) H+ , H2O濃 NaOHCH3COCH2CH3_ + CH3COO + CH3OHCH2COOOCH3 + CO2+ CH3OHOCe.OCH2CH2CH3COOHOCH2CH2CH3f.HOOCCH2COCCOOHCH3CH3CH3COCH(CH3)2NaOH-H2Og.h.CH3CH2CHCOOHClCH3CHCH2COOHCH3CH2CHCOON

38、aOHCH3CH=CHCOOHOOHOCH3NaOH-H 2OCH3CHCH2COONaCH2OHi.+ CO2j.OHCH3CH3CH2CCOOH稀 H2SO4k.稀 H2SO4+HCOOHOCH3CH2CCH3l.CH3CH2COCO2HCH3CHCOCO2HCH3CH3CH2CHOCH3CHCOOHCH3- 21 -CO2+ CO2) HCH2COCH3 2) H1) 稀 OH-H- 22 -COOHCOOHCOOHO有機(jī)化學(xué)有機(jī)化學(xué)（第四版，汪小蘭）課后題答案（第四版，汪小蘭）課后題答案OOm.n.OOOOo,CH3CH2COOH+Cl2PCH3CHCOOH1) NaOH,2) HCl

39、,Cl10.5 完成下列轉(zhuǎn)化：a.BrCH2(CH2)2CH2CO2HNaOHOOb.OCO2CH3EtOC2H5ClCH2COCH3OCO2CH3OCH2COCH3c.CH3COOHCH3CH2OH+CH3COOC 2H5NaOC 2H5CH3COCH2COOC2H5CH3COCCOOC2H5CH3CO+ ,1) NaOC 2H52) BrCH2CH2CH2Br1) OH-2) H+,d.CH3COOC2H5NaOC2H5CH3COCH2COOC2H5CH3CH3COCHCOOC2H5NaOC2H5COCH3CH3CCOOOC2H5CH3CHCOOC2H5CH3CHCOOHCH3CHCOOH

40、1) NaOC2H52) BrCH32) CH3CHCOOC2H5Br1)+_1) 濃OH ,2) HCH3CHCH2CH3CH3CHOCH3COO NH4NHCOCH3 H有機(jī)化學(xué)（第四版，汪小蘭）課后題答案- 23 -第十一章含氮化合物11.2a. 硝基乙烷b.p亞硝基甲苯c.N乙基苯胺d.對(duì)甲苯重氮?dú)滗逅猁}或溴化重氮對(duì)甲苯e.鄰溴乙酰苯胺f. 丁腈g. 對(duì)硝基苯肼h. 1,6己二胺i. 丁 二 酰 亞 胺j. 亞 硝 基 二 乙 胺k. 溴 化 十 二 烷 基 芐 基 二 甲 銨m. HOCH2CH2NH2n.l. (CH3)3N+CH2CH2OHOH(CH3)3N+CH2CH2OCOC

41、H3OHo. HOCH(OH)CH2NHCH3HOp.(CH3)2CHCH2NH2r.N,N-二甲基乙胺q.NHH2N-C-NH2HO11.4OHHOHHONH3CHH3CH3CH NH3Ca 中有三種氫鍵b. 中只有一種氫鍵H3CH3C N HHHH3CH3C NH11.5如何解釋下列事實(shí)？a. 因?yàn)樵谄S胺中，未與苯環(huán)直接相連，其孤對(duì)電子不能與苯環(huán)共軛，所以堿性與烷基胺基本相似。NH2H3C中，硝基具有強(qiáng)的吸電子效應(yīng)。 -NH2 中N上孤對(duì)電子更多地偏向苯環(huán)，所以與苯胺相比,其堿性更弱。b. O2N而NH2 中，甲基具有一定的給電子效應(yīng)，使 -NH2 中N上電子云密度增加，所以與苯胺相比,其

42、堿性略強(qiáng).11.6+_CH3CH(NH2)CH2CH3(+)酒石酸胺 (+)胺 (+)酸鹽酸鹽(+)(-)拆分,而后再分別加 NaOH 析出胺.物理性質(zhì)不同可11.9完成下列轉(zhuǎn)化：a.HNO3H2SO4NO2NH2b.NH2(CH3CO)2ONHCOCH3HNO3 O2N+NH2O2NFe + HClc.CH3COOHCH3COOHP2O5SOCl2(CH3CO)2OCH3COClNH3NH3CH3CONH2CH3CONH2CH3COOHNH3 +CH3CONH2d.2ClSOCl2NH3CH3CH(OH)CHCH3NH2CH3CHCH2CH3CH3CH2OHHClCH3CH2ClCrO3(P

43、y)2MgCH3CH2MgClEt2O1) CH3CH2MgCl+N Cl0-5 CN2 Cl0-5 C有機(jī)化學(xué)（第四版，汪小蘭）課后題答案NHOOKOHNKOOOOCH3NCHCH2CH3H+COOHCOOH+ CH3CHCH2CH3ClCH3CHCH2CH3NH2NH2(CH3CO)2ONHCOCH3NO2NH2NH2Br+-BrNH2+NBrNH2Ne.f.Fe + HClBr2 , FeNO2Fe + HClBrNaNO2+ HCl0弱 HCH3g.HNO3CH3Fe + HClCH3NaNO2+ HCl0N2+Cl-H3COH-NO2NOHH3CH2NHON11.10NHNNCH3a

44、.OCb.NOCH3Cc.H3CCH3+Nd.NCOCOOHH3C+NCH3CH2NCONHNCH3f.Ne.OSOg.NNOH3C+Nh.H3C+H2+N ClH_CH3NO2H_11.11- 24 -(CH3)3N+CH2CH2CH2CH3 + Cl+ Na + OHa.有機(jī)化學(xué)有機(jī)化學(xué)（第四版，汪小蘭）課后題答案（第四版，汪小蘭）課后題答案CH3NH2NHCH3COOHOH(A)(B)(D)NaOHCOOH(C)ABCDH3CSO2ClNaOHFeCl3ABDC不溶不溶可溶可溶顯色不顯色b.(CH3)3N HCl(A)(CH3CH2)4N+Br (B)AgNO3AgClAgBr黃白 AB

45、或者用 NaOHA分層 均相由于三甲胺 b. p. 3,可能逸出,也可能部分溶于 NaOH,所以用 AgNO3 作鑒別較好.11.12寫出下列反應(yīng)的主要產(chǎn)物：a.(C2H5)3N+ CH3CHCH3Br(C2H5)3N+ CHCH3CH3Brb.(CH3)3N+CH2CH2CH2CH3Cl + NaOH+ c.CH3CH2COClCH3NHCH3+ H3CCH3CH2CONCH3d.N(C2H5)2+HNO2N(C2H5)2NO11.13ABNH2CNH3CH3CSO2ClN使用 Hinsberg 反應(yīng).(注意分離提純和鑒別程序的不同)NHCH3CH3SO2NHSO2NCH3CH3 (D)CH

46、3 (E)蒸鎦除去C NaOH 水洗除 ECH3有機(jī)相CH3NSO2H3CCH3H+CH3NH 。HClOHNHCH3純(D)11.15將芐胺、芐醇及對(duì)甲苯酚的混合物分離為三種純的組分。- 25 -有機(jī)化學(xué)（第四版，汪小蘭）課后題答案- 26 -CH2NH2CH2OH(A)(B)CH3HO(C)NaOH 水溶液有機(jī)相水相ABC 的鈉鹽水相有機(jī)相HClA 的鹽酸鹽BANaOH稀 HClC再進(jìn)一步分別純化11.16 = 0飽和胺B 具有CH3CHC4H9OH(可進(jìn)行碘仿反應(yīng)）C(C6H12)KMnO4所以 C 為 CH3CH=CHCHCH3CH3倒推回去CH3COOH + CH3CHCOOHCH3

47、CH3BA6 2+2+1-152CH3CHCH2CHOHCH3CH3CH3CHCH2CHNH2CH3第十二章含硫和含磷有機(jī)化合物12.6 由指定原料及其它無機(jī)試劑寫出下列合成路線。a.CH3CH2CH2CH2OHHClCH3CH2CH2CH2ClNa2SCH3CH2CH2CH2SCH2CH2CH2CH3CH3CH2CH2CH2SO2CH2CH2CH2CH3b.CH3濃 H2SO4SO3HNO2CH3HNO3H2SO4H3CH3CNH2Fe + HCl H3CSO2NHH3CCH3KMnO4or CH3CO3HPCl3SO2ClH3C有機(jī)化學(xué)（第四版，汪小蘭）課后題答案第十三章碳水化合物13.1

48、1CHOCH2OHCHOCH2OHCH2OHOCH2OH13.12e. Tollen 試劑a. Bendict 試劑; b. I2c. I2d. Br2_H2O13.13寫出下列反應(yīng)的主要產(chǎn)物或反應(yīng)物：CHOa.NaOHH2OCH2OHOCHOCH2OHCH2OHOHOCH2OHAg(NH3)2+b.CH2OHCOOHCH2OHc.OHOH2COOCH3HOH2CCH3OH無水 HClOBr2-H2OCH2OHOCOOHd.( -麥芽糖)CH2OHOHOCH2OHOCH2OHOe.CH2OHOOOHOCH2OHOCOOH+Ag(NH3)2CH2OH( -纖維二糖）CH2OHOf.CHOCH2O

49、HHNO3COOHCOOH13.14- 27 -NaOHd.(CH3O)2SO4a.O2-HBr2C 2H 5無有機(jī)化學(xué)（第四版，汪小蘭）課后題答案- 28 -CHOCH2OHCH2OHOD- 甘露糖CH2OCH3OOCH3 OCH3OCH3CH2OHOAcOAcOCHOOAcOAcCH2OAcCOOHCH2OHCH2OHCH2OHCOOHCOOHf. NaBH4h. 催化氫化b. HNO3e. (CH3CO)2O5 HCOOH+1 HCHOlOHHC水j. HIO4c稀liHCOHOCH2CH3OCH2CH3+COOHCH2OHg.HCN , 再酸性水解CH2OHOOH+ C2H5OHCOO

50、HCH2OH13.15如 D- 葡萄糖呈五員環(huán)狀，則OCH 2 OHOCH 2OH(CH 3)2SO 4NaOHOCH 3OOCH 3OOCH 3CH 2OCH 3HNO 3COOHOCH 3CH 2 OCH 3COOHOCH 3H3COCOOH無水 HClCH 3OH稀 HCl+HOCH 2OCH 3H3COH 3CO OH OCH 3 OCH 3OHHHOCH 3OOCH 3CHOOCH 3OHOCH 3CH 2OCH 313.19a.D- 葡萄糖CH2OHOb. 不反應(yīng)c. 不反應(yīng)OHd.甲基化OCH3CH2OCH3OOCH3OOCH3OCH3OHOOn13.20ONH2OOHOHOHO

51、H2Ca.HOOHOH2CNH2OHOH2COHOnb._+NH3ClOH- 29 -有機(jī)化學(xué)有機(jī)化學(xué)（第四版，汪小蘭）課后題答案（第四版，汪小蘭）課后題答案第十四章 氨基酸、多肽與蛋白質(zhì)14.2COOHNCOOHOb.a.Oc.CH2-CH-COOH+OH NH3CH2-CH-COOOH NH2d. HOOC-CH2-CH-COOH+NH3OOC-CH2-CH-COONH2+NH2CH2CHCOOH+NH3CH2CHCOO NH214.3a.CH3CHCOOHNH2(A)(B)H2NCH2CH2COOHNH2 (C)NaOH可溶不溶(分層)ABC茚三酮反應(yīng)顯色不顯色ABb.NH2蘇氨酸絲氨酸

52、H3CCH-CHCOOHOH NH2HOCH2CHCOOHI2/NaOHCHI3無變化c.乳酸H3CCHCOOHOH丙氨酸H3CCHCOOHNH2茚三酮顯色不顯色14.4(CH3)2CHCH(NH 2)COOHH2N(CH 2)4CH(NH 2)COOHNH2主要存在形式(CH 3)2CHCH(NH 2)COO H2N(CH 2)4CHCOO IPIP5.969.74PH=8時(shí)PH=10時(shí)a.b.纈氨酸賴氨酸CH2-CHCOOHOH NH2c.絲氨酸IP5.68PH=1時(shí)CH2-CHCOOHOH +NH3HOOC(CH2)2CHCOOHHOOC(CH2)2CHCOOHNH2d.谷氨酸IP 3.

53、22PH=3時(shí)+NH314.5寫出下列反應(yīng)的主要產(chǎn)物a.CH3CHCO2C2H5+H2ONH2HClCH3CHCOOH+NH3Clb.CH3CHCO2C2H5+NH2(CH3CO)2OCH3CHCO2C2H5NHCOCH3c.CH3CHCONH2NH2+HNO2 (過量)CH3CHCOOHOHCH3CHCOOH + H3N CH2COOHNH3CH3 N(CH2CH3)3NH3有機(jī)化學(xué)（第四版，汪小蘭）課后題答案H2OH+d. CH3CHCONHCHCONHCH2COOHNH2 CH2CH(CH3)2+(CH3)2CHCH2CHCOOH+e.CH3CHCOOHNH2+ CH3CH2COClCH

54、3CHCOOHNHCOCH2CH3(CH3)2CHCH2CHCOOCH3NH2CH3CH2CH-CHCOOH+h. (丙氨酸)CH3CHCOOHNH2CH2CH(NH2)COOHBr2-H2OHOCH2CH(NH2)COOHi. 酪氨酸 HOHCHNNHCHCCCH3H3Cf. (亮氨酸) (CH3)2CHCH2CHCOOH + CH3OH (過量) HClNH2g. (異亮氨酸) CH3CH2CH-CHCOOH + CH3CH2I (過量)CH3 NH2 OOBrj. (丙氨酸)CH3CHCOOHNH2+ O2NNO2FNO2O2NBrCH3CHNHCOOHk. NH2CH2CH2CH2CH

55、2COOHHNOl.CH2COOHNH2.HCl+SOCl2CH2COClNH2.HCl14.6 =2n+2+N 數(shù) - 實(shí)際氫數(shù)23= 1 2+2+1-72屬氨基酸，三個(gè)碳，有旋光活性，應(yīng)為丙氨酸 CH3CHCOOHNH214.7P/ Cl2NH3CH3CHCH2 COOHCH3CH3CH-CHCOOHCH3 ClCH3CH-CHCOOHCH3 NH2如果在無手性條件下,得到的產(chǎn)物無旋光活性,因?yàn)樵诼却嵘傻哪且徊綗o立體選擇性.14.8三肽,N 端 亮氨酸,C 端甘氨酸.中性.14.10a絲氨酸-甘氨酸-亮氨酸，簡(jiǎn)寫為 ：絲-甘-亮b14.1114.12谷氨酸-苯丙氨酸-蘇氨酸，簡(jiǎn)寫為 ：

56、谷-苯丙-蘇此多肽含有游離的羧基,且羧基與 NH3 形成酰胺.丙-甘-丙 或 丙-丙-甘14.13精脯脯甘苯丙絲脯苯丙精- 30 -有機(jī)化學(xué)（第四版，汪小蘭）課后題答案第十五章類脂化合物15.3a. KOH,b.Br2 or KMnO4c.KOHd.Ca(OH)2e. Br2 or KMnO415.4寫出由三棕櫚油酸甘油酯制備表面活性劑十六烷基硫酸鈉的反應(yīng)式。O(CH2)7CH=CH(CH2)5CH3(CH2)7CH=CH(CH 2)5CH3CH2O COCHO CCH2O C(CH2)7CH=CH(CH 2)5CH3OKOHH+CH3(CH2)5CH=CH(CH2)7COOHH2CatCH3

57、(CH2)14COOHLiAlH4CH3(CH2)14CH2OHH2SO4CH3(CH2)14OSO3HNaOHCH3(CH2)14CH2OSO3Na15.5 卵磷脂結(jié)構(gòu)中既含親水基,又含有疏水基,因此可以將水與油兩者較好的相溶在一起。15.6下列化合物哪個(gè)有表面活性劑的作用？a 、d 有表面活性劑的作用。15.7O(CH2)7CH=CH(CH2)7CH3(CH2)14CH3(CH2)14CH3CH2O CO* CHO COCH2O C15.8OCH2O C(CH2)14CH3CH2O COCHO CO(CH2)14CH3(CH2)14CH3KOHH+3 CH3(CH2)14COOHLiAlH

58、4CH3(CH2)14CH2OHH+,CH3(CH2)14COO(CH2)15CH315.9腦苷脂是由神經(jīng)組織中得到的一種鞘糖脂。如果將它水解，將得到哪些產(chǎn)物？CH2OHOOHHOCH2CHNH2CHOHCHCH(CH2)12CH3HOCO(CH2)22CH3+15.10a(4)b(2)c(3)d(1)15.11a 反b 順c 順d 反e 反f 反15.13寫出薄荷醇的三個(gè)異構(gòu)體的椅式構(gòu)型（不必寫出對(duì)映體）。- 31 -CH3 H有機(jī)化學(xué)（第四版，汪小蘭）課后題答案H3COH薄荷醇H3COHH3COHOHCH312315.16CHCH2CHCH3CH3CH2CCH3CHCH3非萜類 應(yīng)為CHC

59、HH2CH2CHCCH3CH3CCH3CH315.17CHH2CH2CHCCH2CCH3CH3OHCCH315.18 利用雌二醇的酚羥基酸性, 用 NaOH 水溶液分離15.19a.完成下列反應(yīng)式：ClClb.BrHOc.CH=CH-COCH3d.OCOCH3OHe.OHOHCH3COCH3+HOOHCOOBr15.20+OCH3O+CH3O+- 32 -H+CH3O+CH3OCH2有機(jī)化學(xué)（第四版，汪小蘭）課后題答案- 33 -第十六章雜環(huán)化合物161命名或?qū)懗鼋Y(jié)構(gòu)a.OCOOHNNHNOHNb.HOc.NHCH3d.NNOHe.NCH3f.NCOOHg.SSO3Hh.糠醛i.噻唑3-甲基吲

60、哚8-羥基喹啉2-苯基苯并吡喃j.k.l.答案:a.2-呋喃甲酸b.2,6-二羥基嘌呤c.3-甲基吡咯d.5-羥基嘧啶e.N-甲基吡咯f.3-吡啶甲酸2-噻吩磺酸g.h.OCHOi.SNj.NHCH3k.NOHl.O16.2a. 維生素 A 萜類b. 維生素 B1,B2,B6,B12 雜環(huán)化合物c. 維生素 PP 雜環(huán)化合物d. 維生素 C單糖的衍生物e. 維生素 D 己三烯衍生物f. 維生素 K醌g. 葉酸雜環(huán)化合物16.4a. 可溶于酸b. 既可溶于酸又可溶于堿 c. 既可溶于堿又可溶于酸 e. 可溶于堿參閱教材 P293 .(教材 P294)核苷由核糖或脫氧核糖與嘌呤或嘧啶化合而成,核苷