ZZM有機(jī)化學(xué)后題答案

ZZM有機(jī)化學(xué)后題答案第1頁/共37頁《有機(jī)化學(xué)》（第四版，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－１－戊烯g.

h.3.4錯(cuò)，應(yīng)為2－甲基－3－乙基－２－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.C2H5

HC

CH2IH(Z)－1－碘－2－戊烯(E)－1－碘－2－戊烯CC

CH2IH

HC2H5d.CC

CH(CH3)2HH3C HCC

HCH(CH3)2H3C H(Z)－4－甲基－2－戊烯(E)－4－甲基－2－戊烯e.CC

HCH(Z)－1,3－戊二烯

HH3CCH2CC

HCH(E)－1,3－戊二烯H3C HCH2f.C

HC(2Z,4Z)－2,4－庚二烯

HH3CCCH

C2H5HC

HC

HH3CCCH

HC2H5(2Z,4E)－2,4－庚二烯C

HCH3C HCCH

HC2H5(2E,4E)－2,4－庚二烯C

HC(2E,4Z)－2,4－庚二烯H3C HCCH

C2H5H3.5完成下列反應(yīng)式，寫出產(chǎn)物或所需試劑．a(chǎn).CH3CH2CH=CH2H2SO4CH3CH2CHCH3OSO2OHb.(CH3)2C=CHCH3HBr(CH3)2C-CH2CH3Brc.d.CH3CH2CH=CH2CH3CH2CH=CH2

CH3CH2CH2CH2OHCH3CH2CH-CH3OHe.+CH3CH2CHOf.CH2=CHCH2OHCl2/H2OClCH2CH-CH2OH1).BH32).H2O2,OH－

H2O/H+

1).O3(CH3)2C=CHCH2CH32).Zn,H2OCH3COCH3

OH3.61－己烯 正己烷

Br2/CCl4orKMnO4無反應(yīng)褪色正己烷1－己烯3.7

-3-第2頁/共37頁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：CH3

CH3CH3H3CCCH ++>CH3

CH3H3CCCH CH3>

CH3H3CC CH3

+CH2CH23.9寫出下列反應(yīng)的轉(zhuǎn)化過程：C=CHCH2CH2CH2CH=CH3C H3C+CH3

CH3H3C H3CCH3

CH3+C-CH2CH2CH2CH2CH=C+

+H_3.10or3.11a.4－甲基－２－己炔４－methyl－2－h(huán)exyneb.2,2,7,7－四甲基－3,5－辛二炔2,2,7,7－tetramethyl－3,5－octadiyneCHCd.c.3.13a.Lindlar catH2CCH2b.HCCHNi/H2CH3CH3c.HCCH+H2OH2SO4HgSO4CH3CHOd.HCCH+HClHgCl2CH2=CHCle.H3CCCHHgBr2HBrCH3C=CH2BrHBrBr

BrCH3-CCH3f.H3CCCH+Br2CH3C=CHBrBrg.H3CCCH+H2OH2SO4HgSO4CH3COCH3HCCH+H2Lindlar catH3CCCH+H2h.H3CCCH+HBrHgBr2CH3C=CH2Bri.CH3CH=CH2HBr(CH3)2CHBr3.14

-4-第3頁/共37頁《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.CH3CH2CH2CCHHCl(過量)CH3CH2CH2C3.15

ClCH3b.CH3CH2CCCH3+KMnO4H+

ClCH3CH2COOH+CH3COOHH2SO4HgSO4c.CH3CH2CCCH3+H2OCH3CH2CH2COCH3+CH3CH2COCH2CH3d.CH2=CHCH=CH2+CH2=CHCHOCHOe.CH3CH2CCH+HCNCH3CH2C=CH2

CN3.16

CH3CHCH2CCH H3C3.17AH3CCH2CH2CH2CCHBCH3CH=CHCH=CHCH33.18CH2=CHCH=CH2HBrCH3CHCH=CH2Br+CH3CH=CHCH2BrCH2=CHCH=CH22HBrCH3CH BrCHCH3+BrCH3CH BrCH2CH2

BrCH2=CHCH2CH=CH2HBrCH3CHCH2CH=CH2CH2=CHCH2CH=CH22HBr

BrCH3CHCH2CHCH3

Br第四章Br

環(huán)烴4.1C5H10不飽和度Π=1第4頁/共37頁《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cyclopentane 1－methylcyclobutanecis－1,2－dimethylcyclopropane反－1,2－二甲基環(huán)丙烷trans－1,2－dimethyllcyclopropanee. f.1,1－二甲基環(huán)丙烷 乙基環(huán)丙烷

1,1－dimethylcyclopropaneethylcyclopropane4.3a.1,1－二氯環(huán)庚烷1,1－dichlorocycloheptaneb.2,6－二甲基萘2,6－dimethylnaphthalenec.1－甲基－４－異丙基－1,4－環(huán)己二烯1－isopropyl―4－methyl－1,4－cyclohexadiened.對異丙基甲苯p－isopropyltoluenee.２２－chlorobenzenesulfonicacidf.－氯苯磺酸

CH3

Clg.CH3CH3h.2－chloro－4－nitrotoluenecis－1,3－dimethylcyclopentane4.4a.HBr

NO22,3－dimethyl－1－phenyl－1－pentene

完成下列反應(yīng)：

CH3

CH3Brb.+Cl2高溫Clc.+Cl2ClClClCl+d.CH2CH3+Br2FeBr3BrC2H5+BrC2H5e.CH(CH3)2+Cl2高溫C(CH3)2Cl第5頁/共37頁《有機(jī)化學(xué)》（第四版，汪小蘭）課后題答案f.CH3O3Zn－powder,H2OCHOOg.CH3

H2SO4H2O,CH3

OHAlCl3h.i.

+CH3CH2Cl2

+HNO3

CH2CH3NO2+CH3NO2j.+KMnO4H+COOHk.CH=CH2+Cl2CH2ClCHCl4.52

1.4.6CH3CH3CH3CH3+Br2FeBr3CH3CH3CH3CH3+

BrBr+Br2FeBr3CH3CH3

+CH3BrCH3+Br2FeBr3

BrCH3

CH3CH3

CH3Br4.7b,d有芳香性4.8

-7-第6頁/共37頁《有機(jī)化學(xué)》（第四版，汪小蘭）課后題答案a.BCA1,3－環(huán)己二烯

苯1－己炔Ag(NH3)2+C灰白色無反應(yīng)ABr2/CCl4

B

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.Br2FeBr3

HNO3H2SO4Br NO2

HNO3H2SO4

Br2

FeBr3BrO2Nc.CH3

2Cl2FeCl3CH3ClCld.CH3+Cl2FeCl3CH3ClKMnO4COOHCle.CH3KMnO4COOHCl2

FeCl3

COOH Cl-8-第7頁/共37頁《有機(jī)化學(xué)》（第四版，汪小蘭）課后題答案-9-f.

HNO3H2SO4NO2CH3H3C2Br2FeBr3BrCH3Brg.Br2FeBr3CH3CH3BrKMnO4COOHBrHNO3H2SO4NO2

COOHBrNO24.11可能為orororor

oror……

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.BrClBrBrBrClBrBrClBCD4.15

Cl將下列結(jié)構(gòu)改寫為鍵線式。a.b.c.d.e.OO OHor第8頁/共37頁《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.CH2CH2CH2CH3OHBrHCH2OH BrHHBrBrCH2CH3

(R) COOHBrCH2OH Hb.COOHHOOCCHCH Br Br

COOH(meso-)CH2CH3

(S) COOH

HBrBrHCOOH

COOH(2R,3R)Br HHBr

COOH(2S,3S)c.COOHH3CCHCH BrBrHHCOOH Br BrBrBrCOOH H H

HBrCOOH Br H

COOH(2R,3S)

COOH(2S,3S)

CH3(2R,3R)Br HCOOH H Brd.CH3C=CHCOOH CH3

CH3

(2S,3R)(無)5.6COOHHCH3CH2CH3

(R)COOHH3CHCH2CH3(S)C5H10O25.7Π=1C6H12A

*CH2=CHCHCH2CH3

CH3BCH3CH2CHCH2CH3

CH35.10(I)HCOOH OHCH3

(R)紙面上旋轉(zhuǎn)90oOHHOOCCH3

H(S)離開紙面旋轉(zhuǎn)COOHHOH

CH3(S) -10-(對映異構(gòu))(對映異構(gòu))第9頁/共37頁HCNOHOH《有機(jī)化學(xué)》（第四版，汪小蘭）課后題答案第六章鹵代烴6.1CH3

Bra.CH3CH--CHCH3b.

CH3H3CCCH3CH2Ic.Brd.Cle.ClClf.2－碘丙烷２－iodopropaneg.三氯甲烷trichloromethaneorChloroformh.1,2－二氯乙烷1,2－dichloroethanei.３－氯－１－丙烯３－chloro－1－propenej.1－氯－１－丙烯１－chloro－1－propene6.4寫出下列反應(yīng)的主要產(chǎn)物，或必要溶劑或試劑a.C6H5CH2Cl

MgEt2OC6H5CH2MgClCO2C6H5CH2COOMgCl

+H2OC6H5CH2COOHb.CH2=CHCH2Br+NaOC2H5CH2=CHCH2OC2H5c.CH=CHBrCH2Br+AgNO3EtOH r.tCH=CHBrCH2ONO2+AgBrCH2=CHCHOCHO

KOH--EtOH tOH --E HKO光照Br

d.e.CH2ClCl+Br2

NaOHH2O

Br BrCH2OHClC2H5OHCH3CH2CH2CH3+BrMgOC2H5

f.g.CH3ClCH3CH2CH2CH2Br +

Mg Et2OH2OSN2歷程CH3CH2CH2CH2MgBr OH

－h(huán).ClCH3

+H2O

－SN1歷程OHCH3

CH3

+OH CH3i.BrKOH--EtOHj.CH2=CHCH2Cl－CH2=CHCH2CNk.(CH3)3Cl+NaOHH2O(CH3)3COH6.5下列各對化合物按SN2歷程進(jìn)行反應(yīng),哪一個(gè)反應(yīng)速率較快?

-11-第10頁/共37頁《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)2CHCl>c.CH2ClCld.CH3CHCH2CH2Br>CH3CH3CH2CHCH2BrCH3e.CH3CH2CH2CH2Cl>CH3CH2CH=CHCl6.6將下列化合物按SN1歷程反應(yīng)的活性由大到小排列b>c>a6.7(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--EtOHCH3C=CH2CH3H+H2OCH3CCH3CH3a.OH b.CH3CCH3CH3HBrCH3CCH3CH2BrBr2HOBrCH3CCH3CH2Brd.Brc.OHBr e.6.11ABrCH2CH2CH3B.CH2=CHCH3C.CH3CHBrCH3第11頁/共37頁HH《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)－３－戊烯醇(3Z)－３－penten－1－olb.2－溴丙醇2－bromopropanolc.2,5－庚二醇2,5－h(huán)eptanediold.4－苯基－2－戊醇4－phenyl－2－pentanole.(1R,2R)－2－甲基環(huán)己醇(1R,2R)－2－methylcyclohexanolf.乙二醇二甲醚ethanediol－1,2－dimethyletherg.(S)－環(huán)氧丙烷(S)－1,2－epoxypropaneh.間甲基苯酚m－methylphenoli.1－苯基乙醇1－phenylethanolj.4－硝基－1－萘酚4－nitro－1－naphthol7.3完成下列轉(zhuǎn)化a.OHCrO3.Py2Ob.CH3CH2CH2OH濃H2SO4CH3CH=CH2Br2BrBrCH3CH-CH2KOH/EtOHc.ACH3CH2CH2OH(CH3)2CHBrCH3CH2CH2OCH(CH3)2+CH3CH=CH2HBrCH3CHCH3BrCH3CCHNaB.CH3CH2CH2OHHBr

H+CH3CH=CH2CH3CH2CH2Br

OHCH3CHCH3d.CH3CH2CH2CH2OHCH3CH2CH=CH2

OHCH3CH2CHCH3OHOHe.

濃H2SO4

SO3Hf.CH2=CH2稀冷KMnO4CH2-CH22OHOCH2CH2OCH2CH2OCH2CH2OHOHOH ClCH2CH2OHHOCH2CH2OCH2CH2OCH2CH2OHH2O+Cl2g.CH3CH2CH=CH2CH3CH2CH2CH2OHh.ClCH2CH2CH2CH2OHNaOHO

NaOHClCH2CH2OHHOH H+Na

ONaCH3CHCH3CH3CH2CH2BrT.MHBrCH3CH2CH2CH2BrCH3CH2OH NaOH1)B2H6,Et2O2)H2O2,OH－HOH

+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-第12頁/共37頁H《有機(jī)化學(xué)》（第四版，汪小蘭）課后題答案a,NOOOHcis－NOOH

Otrans－H

O NOOOHNOOb.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)ⅠⅠⅠCH2CH2ⅠCH3Ⅰe.(CH3)2CHBr+NaOC2H5(CH3)2CHOC2H5+CH3CH=CH2f.KMnO4

OH－CH3(CH2)3CCH3

Og.CH3COOHCHCH2CH3HIO4CH3COOH+CH3CH2CHOh.CH3OH OHHIO4CH3COCH2CH2CHOi.OHCH3+Br2OHCH3BrBrCCl4,CS2中單取代CH3(CH2)3CHCH3

OHj.CH3(CH2)2CHOHCH2CH3分子內(nèi)脫水濃H2SO4CH3CH2CH=CHCH2CH3+CH3CH2CH2CH=CHCH37.8

-14-第13頁/共37頁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２－甲基丙醛２－methylpropanalisobutanalb.苯乙醛phenylethanalc.對甲基苯甲醛p－methylbenzaldehyded.3－甲基－２－丁酮３－methyl－２－butanonee.2,4－二甲基－３－戊酮2,4－dimethyl－3－pentanonef.間甲氧基苯甲醛m－methoxybenzaldehydeg.３－h(huán).BrCH2CH2CHOi.O

甲基－２－丁烯醛3－methyl－2－butenalOj.CCl3CH2COCH2CH3k.(CH3)2CCHOl.CH3CH2COCH2CHOm.CH=CHCHOn.CCH3Oo.丙烯醛propenalp.二苯甲酮diphenylKetone8.3寫出下列反應(yīng)的主要產(chǎn)物a.CH3COCH2CH3+H2N－OHCH3CCH2CH3NOHb.Cl3CCHO+H2OCl3CCHOHOHc.H3CCHOHOOCCOOH+KMnO4H+d.CH3CH2CHO稀NaOHCH3CH2CH-CHCHOOHCH3e.C6H5COCH3+C6H5MgBrC6H5

CH3C6H5COMgBrH+H2OC6H5

CH3C6H5COHf.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COOH第14頁/共37頁HHHg,H+2)-16-k.

《有機(jī)化學(xué)》（第四版，汪小蘭）課后題答案OCl2,H2O OH－COCH2ClCCH3COOH+CHCl3l.OCCH3+Cl2H++HClCl+ClOHCH3m.n.CH2=CHCH2CH2COCH3CH2=CHCOCH3

+HBr

CH3-CHCH2CH2COCH3BrCH2CH2COCH3o.CH2=CHCHO+HCNNCCH2CH2CHO+CH2=CHCHCNOHp.C6H5CHO+CH3COCH3稀NaOH

OC6H5CHCH2C-CH3

OH8.4a.ABCD丙醛丙酮丙醇 異丙醇2,4－二硝基苯肼有沉淀無沉淀Ａ ＢＣ ＤTollen試劑

I2/NaOH

沉淀 無沉淀 無沉淀黃色沉淀ＡＢＣ Ｄb.A戊醛B2－戊酮C環(huán)戊酮Tollen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)2CH3CHOCH3CHOH CNCH3CHCOOH OHb.COCl無水AlCl3COc.ONaBH4OHd.HCCH+H2OCH3CHO稀OH－CH3CH=CHCHOH2/NiCH3CH2CH2CH2OHe.CH3

Cl2

光CH2ClMgEt2OOHCH2CCH3

CH3f.CH3CH=CHCHOOHOH無水HClCH3CH=CHCHOO稀冷KMnO4

OH－OOCH3CH-CHCH OHOHH3O+CH3CH-CHCHO OHOHg.CH3CH2CH2OHHBrCH3CH2CH2BrMg Et2OCH3CH2CH2MgBrCH3CH2CH2CH2OHHCN

+H2OH2OCH2MgCl1)CH3COCH3 +1)HCHO2)H+8.5ACH3CH2CHO+CH3MgBrBCH3CHO+CH3CH2MgBr8.6分別由苯及甲苯合成2－苯基乙醇第15頁/共37頁H《有機(jī)化學(xué)》（第四版，汪小蘭）課后題答案-17-CH3Cl2CH2ClMgEt2OCH2CH2OHCH2MgCl1)HCHO2)H+光照

BrMgBrOBr2Fe

MgEt2OH+H2O8.8a.縮酮b.半縮酮c.d半縮醛8.9b.H3COOH+CH3CHO+HOCH2CH2OH+OHa.

OHOOHOHOOHOHOc.H+OHOHHOH O+OHOHHOH O8.10A.CHCH3CHOHB.CHCH3H3C H3CH3C H3COCC.CCHCH3H3C H3C8.11OCH38.12ACH3CH-C-CH2CH3

CH3OB.CHH3CH3CCHCH2CH3OHC.CH3CH3CCHCH2CH3D.CH3CH2CHOE.CH3COCH38.13BrBrHOOCCOOHOAOBOCOD第16頁/共37頁H-18-《有機(jī)化學(xué)》（第四版，汪小蘭）課后題答案

第九章羧酸及其衍生物9.1a.2－甲基丙酸2－Methylpropanoicacid(異丁酸Isobutanoicacid)b.鄰羥基苯甲酸（水楊酸）o－Hydroxybenzoicacidc.2－丁烯酸2－Butenoicacidd3－溴丁酸3－Bromobutanoicacide.丁酰氯ButanoylChloridef.丁酸酐Butanoicanhydrideg.丙酸乙酯Ethylpropanoateh.乙酸丙酯Propylacetatei.苯甲酰胺Benzamidej.順丁烯二酸Maleicacids.COt.H3CCOk.COOCH3COOCH3l.HCOOCH(CH3)2m.CH3CH2CONHCH39.29.3

g>a>b>c>f>e>h>d寫出下列反應(yīng)的主要產(chǎn)物a.Na2Cr2O7--H2SO4COOHCOOH+COOHCOOHb.(CH3)2CHOH+COOCH(CH3)2H3CCOClH3Cc.HOCH2CH2COOHLiAlH4HOCH2CH2CH2OHd.NCCH2CH2CN+H2ONaOH－OOCCH2CH2COO－H+HOOCCH2CH2COOHe.CH2COOHCH2COOHBa(OH)2Of.CH3COCl+CH3無水AlCl3CH3COCH3+CH3COCH3g.(CH3CO)2O+OHOCOCH3h.CH3CH2COOC2H5NaOC2H5CH3CH2COCHCOOC2H5CH3i.CH3COOC2H5+CH3CH2CH2OH+CH3COOCH2CH2CH3+C2H5OHj.CH3CH(COOH)2CH3CH2COOHk.COOH+HClClCOOH+CO2第17頁/共37頁H《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.NCH2CH2COOC2H5CH2CH2COOC2H5

CONH2OH－N

COOC2H5COO－+q.CH2(COOC2H5)2+H2NCONH2HNNHNH3OOO9.4a.KmnO4b.FeCl3c.Br2orKmnO4d.①FeCl3②2,4-二硝基苯肼或I2/NaOH9.5完成下列轉(zhuǎn)化：a.OCNOHH+COOHOHb.CH3CH2CH2BrCN－CH3CH2CH2CNCH3CH2CH2COOHc.(CH3)2CHOHCrO3.(Py)2(CH3)2CO(CH3)2CCNH+(CH3)2CCOOHOHd.CH3CH3KMnO4COOHCOOH COOHCOOHBa(OH)2OH OOOOOOe.(CH3)2C=CH2HBr(CH3)3CBrMgEt2O(CH3)3CMgBrf.CH3BrAlCl3CH3KMnO4COOHBr2FeCOOHBrg.HCCH

H2OH+,Hg2+CH3CHOKMnO4CH3COOHCH3CH2OH H+CH3COOC2H5HCNH2OHCN

1)CO2

(CH3)3CCOOH2)H+/H2OH2O

H+H2O第18頁/共37頁H3O1)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.CH3CH2COOHLiAlH4CH3CH2CH2OHHBrCH3CH2CH2BrMgEt2OCH3CH2CH2MgBrCH3CH2CH2CH2CH2OHCH3CH2CH2CH2COOH或

H+CH3COCH2COOC2H5CH3COCHCOOC2H5－CH3(CH2)3COOHj.CH3COOHCl2

PCH2COOHClCN－CH2COOHCNEtOH

+CH2(COOC2H5)2k.OOONH3CH2COONH4CH2CONH2(CH3CO)2Ol.m.

CO2CH3

OHCH3CH2COOH

+H2O SOCl2

COOH OHCH3CH2COClOH

COOH OOCCH3CH3CH2COOn.CH3CH(COOC2H5)2CH3CH2COOHOKMnO4

2)CH3CH2CH2Br+

－2)H+3)9.6己醇A已酸B對甲苯酚CNaHCO3水溶液水相已酸鈉HCl已酸B有機(jī)相已醇對甲苯酚NaOH水相HCl

酚鈉 有機(jī)相已醇A酚CCOOHCOOH9.7

COOHCOOHCOOHCOOHHOOCCOOHHOOCHCH3CH2COOHCOOH HOOC

(Z)易成酐(E)不易(Z)易成酐(E)不易成酐9.8A.CH2COOHCH2COOHB.OOOC.H2CCOCH3H2CCOCH3OOD.CH2CH2OHCH2CH2OH第19頁/共37頁CH2COO《有機(jī)化學(xué)》（第四版，汪小蘭）課后題答案

第十章取代酸n.4--氧代戊酸（4—oxopentanoicacid)10.1m.3--氯丁酸（3--Chlorobutanoicacid）10.2a.(A)CH3CH2CH2COCH2COOCH3

OH

COOHCH3CHCOOH(B)(C)FeCl3A

BC顯色不顯色Na2CO3A溶解，有氣體

B無變化

OH10.3寫出下列反應(yīng)的主要產(chǎn)物：a.b.CH3COCHCOOC2H5

CH3

CH3COCHCO2CH3

CH2CO2CH3c.CH3CH2CHCOOH OHHC OOCHCCOCH3CH2CH2CH3d.COOCH3COCH3稀H+1)稀OH-2)H+,H2O

濃NaOHCH3COCH2CH3

_ _ _+CH3COO+CH3OHCH2COO

OCH3+CO2+CH3OHOCe.OCH2CH2CH3

COOHOCH2CH2CH3f.HOOCCH2COCCOOH CH3CH3CH3COCH(CH3)2NaOH--H2Og.h.CH3CH2CHCOOH ClCH3CHCH2COOH

CH3CH2CHCOONa OHCH3CH=CHCOOHOOHOCH3NaOH--H2OCH3CHCH2COONa CH2OHi.+CO2j.OH

CH3CH3CH2CCOOH稀H2SO4k.稀H2SO4++HCOOH

OCH3CH2CCH3l.CH3CH2COCO2H CH3CHCOCO2HCH3CH3CH2CHOCH3CHCOOHCH3-21-

CO2+CO第20頁/共37頁2)HCH2COCH32)H1)稀OH-H-22-COOHCOOHCOOHO《有機(jī)化學(xué)》（第四版，汪小蘭）課后題答案

OOm. n.OOOOo,CH3CH2COOH+Cl2PCH3CHCOOH1)NaOH,2)HCl,

Cl10.5完成下列轉(zhuǎn)化：a.BrCH2(CH2)2CH2CO2HNaOHOOb.OCO2CH3

EtOC2H5ClCH2COCH3OCO2CH3OCH2COCH3c.CH3COOHCH3CH2OH

+CH3COOC2H5NaOC2H5CH3COCH2COOC2H5CH3COCCOOC2H5CH3CO+,

1)NaOC2H52)BrCH2CH2CH2Br1)OH-2)H+,d.CH3COOC2H5NaOC2H5CH3COCH2COOC2H5

CH3CH3COCHCOOC2H5NaOC2H5

COCH3CH3CCOOOC2H5CH3CHCOOC2H5CH3CHCOOHCH3CHCOOH1)NaOC2H52)BrCH32)CH3CHCOOC2H5

Br1)+

_1)濃OH,第21頁/共37頁2)HCH3CHCH2CH3CH3CHOCH3COONH4NHCOCH3H《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.對甲苯重氮?dú)滗逅猁}或溴化重氮對甲苯e.鄰溴乙酰苯胺f.丁腈g.對硝基苯肼h.1,6－己二胺i.丁二酰亞胺j.Ｎ－亞硝基二乙胺k.溴化十二烷基芐基二甲銨m.HOCH2CH2NH2n.l.[(CH3)3N+CH2CH2OH]OH－[(CH3)3N+CH2CH2OCOCH3]OH－o.HOCH(OH)CH2NHCH3HO p.(CH3)2CHCH2NH2r.N,N--二甲基乙胺q.

NHH2N-C-NH2

HO11.4OHH

OHHON

H3C HH3C

H3C HNH3CⅠⅢa中有ⅢⅡⅠ三種氫鍵b.中只有Ⅰ一種氫鍵H3CH3CNH H

ⅡH

H3CH3CN H11.5如何解釋下列事實(shí)？a.因?yàn)樵谄S胺中，Ｎ未與苯環(huán)直接相連，其孤對電子不能與苯環(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.HNO3

H2SO4NO2NH2b.NH2

(CH3CO)2ONHCOCH3

HNO3O2N+NH2O2NFe+HClc.CH3COOHCH3COOHP2O5SOCl2(CH3CO)2OCH3COCl

NH3NH3

CH3CONH2CH3CONH2CH3COOHNH3－+CH3CONH2d.2ClSOCl2

NH3CH3CH(OH)CHCH3

NH2

CH3CHCH2CH3CH3CH2OH HClCH3CH2Cl

CrO3(Py)2

Mg CH3CH2MgClEt2O1)CH3CH2MgCl

+第22頁/共37頁NCl－0-5CN2Cl0-5C《有機(jī)化學(xué)》（第四版，汪小蘭）課后題答案NHOOKOHNKOOOO

CH3NCHCH2CH3H+COOHCOOH+CH3CHCH2CH3

ClCH3CHCH2CH3NH2NH2(CH3CO)2ONHCOCH3NO2NH2NH2Br+-BrNH2+NBrNH2Ne.f.Fe+HClBr2,Fe NO2

Fe+HCl BrNaNO2+HCl

0弱HCH3g.HNO3CH3

Fe+HClCH3NaNO2+HCl0N2+Cl-H3COH--N

O2NOH H3CH2N HON11.10NHNNCH3a.OCb.NOCH3Cc.H3CCH3+Nd.NCOCOOHH3C

+NCH3CH2NCO×××××NHNCH3f.Ne.OSOg.NNOH3C+Nh.H3C+H2+NCl H－

_Ⅰ

_ ⅠCH3

NO2H_11.11

-24-第23頁/共37頁[(CH3)3N+CH2CH2CH2CH3]+Cl－+Na+OHa.

《有機(jī)化學(xué)》（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或者用NaOHA分層Ｂ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+CH3CHCH3

Br(C2H5)3N+CHCH3

CH3Br－b.[(CH3)3N+CH2CH2CH2CH3]Cl－+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)NHCH3

CH3SO2NHSO2NCH3CH3(D)CH3(E)蒸鎦除去CNaOH水洗除E

CH3有機(jī)相CH3NSO2H3C CH3H+CH3NH。HClOH－NHCH3

純Ａ(D)11.15將芐胺、芐醇及對甲苯酚的混合物分離為三種純的組分。

-25-第24頁/共37頁《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11.16Ω==0飽和胺B具有CH3CHC4H9

OH(可進(jìn)行碘仿反應(yīng)）C(C6H12)KMnO4所以C為CH3CH=CHCHCH3CH3倒推回去CH3COOH+CH3CHCOOH CH3

CH3BA6╳2+2+1-15 2CH3CHCH2CH OH CH3

CH3

CH3CHCH2CH NH2

CH3

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+HClH3CSO2NHH3CCH3

KMnO4orCH3CO3HPCl3SO2ClH3C第25頁/共37頁《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13.11CHOCH2OHCHOCH2OHCH2OH OCH2OH13.12e.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.NaOHH2OCH2OH OCHOCH2OH

CH2OH OH OCH2OHAg(NH3)2+b.CH2OH COOH CH2OHc.OHOH2COOCH3HOH2C

CH3OH無水HClOBr2--H2OCH2OHOCOOHd.(β--麥芽糖)CH2OH OH OCH2OH OCH2OH Oe.CH2OH OOOHOCH2OHOCOOH

+Ag(NH3)2

CH2OH(α-纖維二糖）CH2OH Of.CHOCH2OHHNO3COOHCOOH13.14

-27-第26頁/共37頁NaOHd.(CH3O)2SO4a.O2--HBr2C2H5無《有機(jī)化學(xué)》（第四版，汪小蘭）課后題答案-28-CHOCH2OH

CH2OH OD--甘露糖

CH2OCH3

O OCH3OCH3OCH3CH2OH OAcOAcOCHO OAc OAcCH2OAcCOOHCH2OHCH2OHCH2OHCOOHCOOH

f.NaBH4h.催化氫化b.HNO3e.(CH3CO)2O5HCOOH +1HCHOl

OH HC水j.HIO4

c稀l

iHC~OH~OCH2CH3~OCH2CH3+COOHCH2OH

g.HCN,再酸性水解CH2OH O~OH+C2H5OH

COOH CH2OH13.15

如D--葡萄糖呈五員環(huán)狀，則OCH2OHOCH2OH(CH3)2SO4

NaOHOCH3

OOCH3

OOCH3CH2OCH3HNO3COOH OCH3CH2OCH3COOHOCH3H3COCOOH無水HCl CH3OH稀HCl+HOCH2OCH3H3COH3CO~OH~OCH3~OCH3~OHHHOCH3O

OCH3CHO OCH3

OH OCH3CH2OCH313.19a.D--葡萄糖CH2OH Ob.不反應(yīng)c.不反應(yīng)OHd.甲基化OCH3CH2OCH3

OOCH3O

OCH3OCH3OHOOn13.20ONH2OO

HOHO HOH2Ca.

HOO HOH2C

NH2O

HOH2CO HOnb.

_+NH3Cl OH第27頁/共37頁-29-

《有機(jī)化學(xué)》（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)14.2COOHNCOOHO

b.－a.O－c.CH2-CH-COOH

+OHNH3

－CH2-CH-COOOHNH2d.HOOC-CH2-CH-COOH

+

NH3－OOC-CH2-CH-COO－

NH2+NH2CH2CHCOOH

+

NH3CH2CHCOO－

NH214.3a.CH3CHCOOH

NH2(A)(B)H2NCH2CH2COOH

NH2(C)NaOH可溶不溶(分層)ABC茚三酮反應(yīng)顯色不顯色A Bb.NH2蘇氨酸絲氨酸H3CCH-CHCOOH

OHNH2HOCH2CHCOOHI2/NaOHCHI3

無變化c.乳酸H3CCHCOOH

OH丙氨酸H3CCHCOOHNH2茚三酮顯色不顯色14.4(CH3)2CHCH(NH2)COOHH2N(CH2)4CH(NH2)COOHNH2

主要存在形式(CH3)2CHCH(NH2)COO－H2N(CH2)4CHCOO－IPIP5.96 9.74PH=8時(shí)PH=10時(shí)a.b.纈氨酸賴氨酸CH2-CHCOOHOHNH2c.絲氨酸IP5.68PH=1時(shí)CH2-CHCOOHOH+NH3HOOC(CH2)2CHCOOHHOOC(CH2)2CHCOOHNH2d..谷氨酸IP3.22PH=3時(shí)+NH314.5寫出下列反應(yīng)的主要產(chǎn)物a.CH3CHCO2C2H5+H2ONH2HClCH3CHCOOH+NH3Cl－b.CH3CHCO2C2H5+NH2(CH3CO)2OCH3CHCO2C2H5NHCOCH3c.CH3CHCONH2NH2+HNO2(過量)CH3CHCOOHOH第28頁/共37頁CH3CHCOOH+H3NCH2COOHNH3CH3N(CH2CH3)3NH3《有機(jī)化學(xué)》（第四版，汪小蘭）課后題答案H2OH+d.CH3CHCONHCHCONHCH2COOH

NH2CH2CH(CH3)2+++(CH3)2CHCH2CHCOOH+e.CH3CHCOOHNH2++CH3CH2COClCH3CHCOOHNHCOCH2CH3(CH3)2CHCH2CHCOOCH3

NH2CH3CH2CH-CHCOOH

+h.(丙氨酸)CH3CHCOOH

NH2CH2CH(NH2)COOHBr2--H2OHOCH2CH(NH2)COOHi.酪氨酸HOHCHNNHCHCCCH3H3Cf.(亮氨酸)(CH3)2CHCH2CHCOOH+CH3OH(過量)HCl

NH2g.(異亮氨酸)CH3CH2CH-CHCOOH+CH3CH2I(過量)

CH3NH2OO Brj.(丙氨酸)CH3CHCOOH

NH2+O2NNO2

FNO2O2NBr

CH3CHNHCOOHk.NH2CH2CH2CH2CH2COOHHNOl.CH2COOHNH2.HCl+SOCl2CH2COClNH2.HCl14.6Ω=2n+2+N數(shù)-實(shí)際氫數(shù)

23==1╳2+2+1-7 2

屬氨基酸，三個(gè)碳，有旋光活性，應(yīng)為丙氨酸CH3CHCOOH

NH214.7P/Cl2NH3CH3CHCH2COOH

CH3CH3CH-CHCOOH

CH3ClCH3CH-CHCOOH

CH3NH2

如果在無手性條件下,得到的產(chǎn)物無旋光活性,因?yàn)樵讦哩C氯代酸生成的那一步無立體選擇性.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，簡寫為：絲--甘--亮b14.1114.12

谷氨酸--苯丙氨酸--蘇氨酸，簡寫為：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-第29頁/共37頁《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,Δb.Br2orKMnO4c.KOHd.Ca(OH)2e.Br2orKMnO415.4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(CH2)5CH3CH2OC OCHOCCH2OC(CH2)7CH=CH(CH2)5CH3OKOHH+CH3(CH2)5CH=CH(CH2)7COOH

H2CatCH3(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。15.7

O

(CH2)7CH=CH(CH2)7CH3(CH2)14CH3

(CH2)14CH3

CH2OC O

*CHOC O CH2OC15.8

OCH2OC(CH2)14CH3CH2OC OCHOC O

(CH2)14CH3(CH2)14CH3KOHH+3CH3(CH2)14COOHLiAlH4CH3(CH2)14CH2OH

H+,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？CH2OH OOHHOCH2CHNH2

CHOHCHCH(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（不必寫出對映體）。

-31-第30頁/共37頁CH3H《有機(jī)化學(xué)》（第四版，汪小蘭）課后題答案H3COH薄荷醇H3COHH3COHOHCH312315.16CHCH2CHCH3CH3C

H2

CCH3

CHCH3非萜類∴應(yīng)為CHCHH2CH2CHCCH3CH3CCH3CH315.17CHH2CH2CHCCH2CCH3CH3OHC

CH315.18利用雌二醇的酚羥基酸性,用NaOH水溶液分離15.19

a.完成下列反應(yīng)式：

ClClb.BrHOc.CH=CH-COCH3d.OCOCH3OHe.OHOHCH3COCH3+HOOHCOOBr15.20+OCH3O+