華東理工大學(xué)有機(jī)化學(xué)課后答案

1、精品感謝下載載部分習(xí)題參考答案2-1順-1,2-二澳環(huán)己烷(6)1,7,7-三甲基二環(huán)2.2.1庚烷5-甲基螺3.4辛烷(9)(10)CH3CH3IIH3CC-C-CH3(11)(CH3)2CHCH2CH2CH32-4(3)>(2)>(5)>(1)>(4)2-6(1)正丙基(n-Pr-)(2)異丙基(i-Pr-)2-72-82-92-102-112-132-15(4)叔丁基(t-Bu-)甲基(Me-)(3)CH3CH2CH(CH3)2(4)(CH3)4C(3)、(6)等同;(2)、(5)構(gòu)造異構(gòu);(1)、(4)CH3ch3(3)異丁基(i-Bu-)(6)乙基(Et-)構(gòu)

2、象異構(gòu)(1)用Br2。因氫原子活性有差異，澳原子活性適中，反應(yīng)選擇性強(qiáng),(2)用Cl2。只有一種氫，氯原子反應(yīng)活性高。CH3CH2的穩(wěn)定性大于CH3,-易于生成。2,3-二甲基丁烷有四個典型構(gòu)象式,2,2,3,3-四甲基丁烷有二個;H3cH式為CH3由HhCHCH3CH3HH(2)H3c(順CH(CH3)2CH3<>一CH)>.CH(CH3)2<CH3CH(CH3)2CH(CH3)2CH3主要得到JBr。前者最穩(wěn)定的構(gòu)象C2H5Cl1CH3Cl2-16(4)>(2)>(3)>(1)4-4CH3(CH3)2CHC = OCH3IBrO= CHCH3(8)

3、OO(CH3)2CHCCHCH3、OH3cCH2BrBrMJA(10)BrOCH3CH3(11)H2cCn'nCOOCH34-5(1)乙烯基CH3CH2CH2CH=CH2烯丙基CH3CH2pH=CHCH3有順反異構(gòu)丙烯基CH3CH3CH3CH2C=ch2CH3CHCH3異丙烯基CH2=CHCH(CH3)2乙烯基(2) 2,3-二甲基-2-丁烯2-甲基-2-戊烯反-3-己烯順-3-己烯1-己烯(3) 2-甲基-1-丙烯快(形成叔碳正離子)快(形成叔碳正離子)4-6(1)親電加成反應(yīng)，中間體為碳正離子，有重排(CH3)2CHCH=CH2H+(CH3)2ChChCH3Br*(CH3)2CHC

4、HCH3Br重排4-B廠CH3C一CH2CH3*(CH3)2CCH2CH3CH3Br(2)甲醇與碳正離子結(jié)合；直接失去質(zhì)子而形成醒+_H八八CH2CH2+CH3OH*CH2CH2”CH2CH2BrBr+OCH3BrOCH3H(3)BrBrH+H'CH3Br+BrCH3+HCH3BrHHCH3CF3CHCH2CF3CHCH3CF3cH2cH2（穩(wěn)定）Cl"CF3cH2cH2clCH3OCH2CH2CH3OCHClCH3OCHC1CH3重排Cl4-7CHCHHCH3CH3OCHCH3（穩(wěn)定）ch3hHTCH2CH3CCH2CH2cCH2CH2CCH3CCH3CCH3CH3HCH3

5、cHHCCH3CH2CH2CH3hCH2CH2CH3CH3cCH3CH3CCH3或5-9（1）Br2/CCl4；Ag（NH3”NO3(2)順丁稀二酸酎;Ag(NH3)2NO35-11H3cH3cHCH3-CCH3H3cHCCH3CH3CH2CH2CHCH2CH2CH3Ch3H3cHHC1CC二CHCH3CH3H3cHHHHCe=eCH3H3cC-CHCHCH3HCH3C-CH5-12ACH3CHeCHCH3BCH3cHe三CCH2CH2CH3CH3cH3cH2cH2cOOHcH3chcOOHCh3CH3CHCCH3ch36-1(a)c3H7NO(b)C2H3OCl6-3CH3CH=CHCH2C

6、H2CH3a-CH2CH3m/z=84一CH2CH2CH3CH3CH=CHCH2CH3CHCHm/z=55重排上CH2CH=m/z=41ch26-4共軻，£(a)>db)6-5(a)227nm,(b)237nm,(c)227nm,(d)232nm6-6(a)>(b)6-8(2)(b)>(a)(a)>(b)30003100cm-1C=C-Hv,m28502950cm-1CH3,CH2v,s16201680cm-iC=Cv,mCH 31430cm-19101000cm-1C=CH6-9偶合常數(shù)不同，反式偶合常數(shù)較大7-97-107-14chch3ch3Er7-16

7、7-17CHaCHaOF7-188-11)1,4-二氯丁烷2)2-甲基-3-氯-6-澳-1,4-己二烯3) (R)-3-甲基-2-氯戊烷4) (S)-1-苯基-1-澳丙烷5) 4-甲基-3-澳苯磺酸6)3-對澳苯-1-丙烯CH3I8-51)(a)(CH3)3CBr>CH3CH2CHBr>CH3CH2CH2CH2Br2)CH3cH2cH2Br>(CH32CHCH2Br>(CH3)3CCH2BrCH3(b)CH3cH2cH2cH2Br>CH3cH2cHBr>(CH3)3cBr8-61)CH3CH2CH2CH2Br反應(yīng)較快。因甲基支鏈的空間位阻降低了Sn2反應(yīng)的速

8、度。2) (CH3)3Br反應(yīng)較快，為Sn1反應(yīng)。(CH3)2CHBr首先進(jìn)行Sn2反應(yīng)，但因水為8-88-98-108-118-158-179-39-41)2)3)9-51 )、弱的親核試劑，故反應(yīng)慢。3) SH-反應(yīng)快于OH-,因S的親核性大于O。4) (CH3)2CHCH2Br快，因?yàn)锽r比Cl易離去。D>C>A>B1) >4)>2)>3),Sn2反應(yīng)歷程A>B>C,Sn1反應(yīng)歷程對于Sn2反應(yīng)來說，新戊基鹵因空間位阻較大，難以進(jìn)行；對Sn1反應(yīng)來說，新戊基鹵是伯鹵烷，Sn1反應(yīng)困難。ch2=ch-chch3CH2CH-CHCH3CH2=C

9、H-CH=CH2BrBrBrBrABC1) A錯，澳應(yīng)加在第一個碳上。B錯，-OH的活潑H會與格式試劑反應(yīng)。2) B錯，叔鹵烷遇-CN易消除。3) B錯，與苯環(huán)相連的澳不活潑，不易水解。CH=CH-CH2CH34)產(chǎn)物為，共軻雙烯穩(wěn)定。3)2)1)加水分層，得有機(jī)層異丙觸，水層分儲得異丙醇加入氧化鈣吸水，過濾，蒸儲加入氫氧化鈉溶液分層，油層為苯甲醛，水層酸化，得苯酚2）、止醇、1,2丙一醇環(huán)己烷甲內(nèi)醒NaH2J無氣體放出r無氣體放出AH2'正丁醇1,2一丙二醇不變色»正丁醇Pb（OAc）4變色加1,2丙二醇環(huán)己烷冷濃硫酸A環(huán)己烷甲內(nèi)髓J不分層E.甲內(nèi)醍分層3）、盧卡斯試劑，茉

10、醇反應(yīng)快,苯甲醛和間甲苯酚不反應(yīng)，加入堿，則間甲苯酚溶于水,苯甲醛不溶。9-7CH3CH2I0CH39-10（A）、（B）、（C）、（D）的構(gòu)造式分別為：0HOCl66609-11（A）、（B）、（C）的構(gòu)造式分別為:OCH3OHCH3I9-12A)、(B)的構(gòu)造式分別為：(B)CH 3CH2CH(CH 3)CH 20H(A)(CH3)2CHCH2CH20H9-13（A）、（B）、（C）的構(gòu)造式分別為OHOHO(2) CH3-CH=CHt =(4) CH3CH2CH-SO 3Na1OH(6) CH3CH2CH-CH-CHO 11OH CH3(8) CH3cH2cH20H(10) CH3CH-C

11、HO1Br(12) CH3CH2CH=N-OHa -H10-4.(1)CH3CH2CH20H(3)CH3CH2CH2OH(5)CH3CH2CH-CN1OH(7)CH3CH2C+C-CHOCH3O(9)CH3CH2CH；O(11)CH3CH2COOH(13) CH3CH2CH=N-NHC6H510-5.(6)(7)(10)不能反應(yīng)，10-7.(1) CH3COCHO>CH3CHO>CH3COCH2CH3>(CH3)3CCOC(CH3)3(2) CH3COCCI3>C2H5COCH3(3) CICH2CHO>BrCH2CHO>CH2CHCHO>CH3CH2

12、CHO(4) CF3CHO>CH3CHO>CH3COCHCH2>CH3COCH310-10.共同點(diǎn)：在17001730cm-1左右有厥基的強(qiáng)吸收峰CH=CHCHO不同點(diǎn)：一的紅外光譜中在14501650cm-1出現(xiàn)24個苯環(huán)的特征峰，同時在3020cm-1以及1660cm-1附近出現(xiàn)烯煌的特征峰。10-11.CH 3cH 2cH 2cH=CCHOCH2CH 3CH 3cH 2cH 2cH=CCH 20HCH2CH3(1)(3)I ' ' -OC2H51./ 、OC2H5(CH3)2C-CHO(CH3”CCH0(CH3)22-CH(OC2H5)2(CH3)2C-

13、CH(OC2H5)29H-0HBrBrMgBrCH(CH3)21)NaOH/EtOH2)B2H6/H2O2QHNNH-C-NH2110(6)(12)CH2CH2CH3(CH3)2CHCHCH3(CH3)2CHCCH3(CH3)2CHCCH 3IIIIOH B：O C：OH10-14 . A:OCH3CH2-C-CHCH3I10-16.CH3OIICH3-C-CH210-17. A:OCH3OCH3-C-CH2C:OHOHICH3CH CH2D:10-18.OCH3 -CCH2cH(OCH3)210-19.1）烯醇式（平面結(jié)構(gòu)）酮式轉(zhuǎn)化。2）格氏試劑的大基團(tuán)難以接近厥基而奪取厥基“-H形成烯醇結(jié)

14、構(gòu)。3）順，反異構(gòu)，R2C=N-OH11-11）丙酸乙酯2）甲酸3）a-氨基己酰胺4）鄰氨基苯甲酸甲酯5）B-泰甲酸6）N,N-二乙基丙酰胺7）苯甲酰氯8）鄰苯二甲酰亞胺9）2-甲基丁烯二酸酎10)反-2-甲基環(huán)己基甲酸11)”-乙基乙酰乙酸乙酯12)2-環(huán)己酮基甲酸甲酯13)對乙?；郊姿嵋阴?4)甲酸節(jié)酯OCH2=C_C-OCH3ch3NHCOCH3(6)NH2cOOC2H5OhC.CHNCH，11-2II(8)och3Co(10)(12)CH3OC(11)1COill0+CH2CHbFOh11-3COOCH2CH2OHCoOCH2CH20H(13)OII。_c_CH3COOH(1) CH

15、3CH2C00H>CH3cH2cH2cH20H>ch3cH2cH2cHO>ch3cH20cH2cH3(2) CH3CH2COOH>CH3CONHCH3>CH3COOCH311-4(正)丁酸在水中溶解度大于丙酸甲酯，竣基中氫鍵作用11-51)竣酸質(zhì)子氫81013,與-OCH383.54.0很大區(qū)別2)談基特征吸收竣酸的羥基吸收峰更寬，常覆蓋C-H伸縮振動3)酸酎的厥基吸收波數(shù)更高17507800cm-1,且出現(xiàn)兩個峰11-6.(1)甲酸是液體，其余為固體甲酸 乙二酸 丙二酸 丁二酸AgNO3/NH3；：士雙酸11-7CH2C1Cl11III(2) CH3CCI>

16、;CH3C-O-CCH3OO> CH3C-OCH2CH3>CH3JNH2-ICO2T丙二酸11-8(1) HCOOCH3>CH3c02cH3>CH3c02c2H5>CH3c02cH(CH3)2>CH3c02c(CH3)3,C02CH3(2) O2NC02cH3>ClC02cH3>C、C02cH3>CH3O_產(chǎn)11-9COClICO2cH2cH3(2)CH3cH2CON(CH3)3_0_O8_CH3Br0CH3(5)H-C-0-LHCH3C2H50I0IIJrCH30OC2H5(6)IO0c2H50OH(7)1rCH2COOC2H5(8)HO

17、CH2CH2CH2CH2CH20H(13)11-10CH3CH2CHC0(10)CH3l|0CCHCH2CH3NH20000(9)OCH3cH2c_CHCOC2H5(11)CN(12)C2H5C2H5O&NHONHO(16)OCH(14)CH3cHCOC2H5C=oCOOC2H5O(15)O.11COOC2H5(1)OOClCH2c_OH>CH3c_OH>CH3CH2OH_O_OO2N_COH>_COHO2N-OH>O>CH3CNH2O>COH節(jié)CH20HO>CH3OCOH草酸丙二酸甲酸乙酸苯酚11-11(1)'CH3CII-A1C13

18、CH3OII-COOHBr21COOHIFeIBr(2)CH3Cl2I光”CH2ClCN-分CH2CNH3O+oCH2C-OHMg干醛(CH3)2C=CH2一HBrv.(CH3)2C-CH3Br1)C02(CH3)2CCH3(CH3)3CCOOHMgBr2)H301)NaOC2H500NaBH40H0(4)2CH3CH2CO2CH2CH3CH3CH2CHC+COCH2CH3-CH3cH2cHCH-COCH2cH32)HCH3Ch3HC=CHHCNCH2=CHCNH20,H+CH3OHH+-CH2=CHC00CH3O(6)CH3OH/H+NaBH4HOCO2CH3(8)(9)(10)11-121

19、1-13OA1C13CO2HOOH1HBrCH3CHCH3CH3OOMg/®0HZn/HgHClHCNCHOCH3CH2CH2COOH_Cl2/P(2)OH1)CO22)H3O*CH3OIICH3CHCOOHCH3OHCHCNCH3CH2CHCOOHIClCN-1)C2HgONaCH3COCH2COOC2H525+一OHH2O,HCH3OCHCOOHH2O/H+CH3CH2CHCOOHCOOHCH3COCHCO2C2H5-CH2CH32)C2H5BrOCH3-CHCH210%NaOHHa+OA4-CO1)C2H50NaCH(CO2c2H5)21)C2H5ONa2)CH2cH3BrCH

20、2CH3CH3coec02c2H5CH2CH3CH2(C02c2H5)2一2)BrCH2CHBrCfCH(CO2c2H5)2h3+o-CO2COOHCOOH1)NaOC2H5OCH3c(CH2)3COC2H5514CH2cCH2cH2cH2COC2H5H+OOC2H5-C2H5OH笫環(huán)己酮甲酸乙酯含有活潑亞甲基,在堿性條件下與冉伊不飽和酮，先發(fā)生1,4加成（Micheal反應(yīng)），再發(fā)生分子內(nèi)的羥醛縮合,生成環(huán)狀%3-不飽和酮（Robinsen增環(huán)反應(yīng)）。CO2c2H5NaOC2H5OII-OCOOC2H5NaOC2H5NaOC2H5COOC2H5COOC2H5COOC2H5C2H5OH_C2H

21、5O-OfOO-COOC2H5(3)a-鹵代酯在強(qiáng)堿作用下與酯、酮反應(yīng)生成a吊環(huán)氧竣酸酯(Darzen反應(yīng))。OIIClCH2cOEtICHCOOEtCHCOOEt11-14CH3CH2OHOgCH3COOHC2H50HACH3c02c2H5Br2/PZn/醛-BrZnCH2COOC2H5H+“-vCH2COOHOHReformastsky 反應(yīng)CH 2Clch2(co2ch2ch3)21CH3cH2ONa稀 OH-H 3+OA -CO2CH2CH2COOH分析O11CCH 3/.chMchO OII ICH3cCH2COC2H5+BrCH2cH2Br合成CH3CH2OHONaOC2H5BrC

22、H2CH2Br稀OH-O OC2H5OHNaOC2 H5CH3CO2c2H 5CH 3CCH 2coe 2H 5H+C2H50H+OOCH31.H3OOH-12/, co2 CH3C-CH2CH2CH2CH2CCH3OH*'OCCH3 分析：COOC 2H 5CHCH3COOC2H5COOC2H5己二酸COOC 2H 5CO2c2HCO2c2H合成：CO2HCO2HC2H50HH+CO 2c2H 5CO2c2H5NaOC2H5COOC2H5ch2= chCch3C2H50HCOOC2H5/ "IfO OOH-COOC2H5NaOC2H5CH3cH= PPh3COOC2H531

23、1-15A.CH3CH2COOCH2CH3;B.CH3CH2COCH(CH3)COOCH2CH3;C.CH3CH2COC(C2H5)(CH3)COOCH2CH3；D.CH3CH2COCH(CH3)CH2CH3。11-16BrCH 2CH2CH2BJNCCH 2CH 2CH 2CN ,HO2CCH2CH2CO2H ,11-17CH3COCH(CH3)OH?；衔锶茉贒2O中，一OH的質(zhì)子被D取代，8=3.75的峰消失。12-3.bp:B>C>A.12-4.A:b>a>cB:b>c>aC:d>a>c>bD:a>b>cE:b>a

24、F:a>b>c>d12-5.HNOHC CHOolc CHO1. OH -:n - BuNH 22. H3O -不能有效的制備Me3CNH2,Me3CCH2NH2,Et2NH，所以此法反應(yīng)為SN2歷程Me3CNH2是叔碳,新戊基位阻大，不利SN2反應(yīng)歷程，Et2NH對消除有利。p-CH3c6H4Br難發(fā)生親核取代，C=CX.12-6.PhSO2NHEtPhSOzNEtNa.PhSO2NEt2ABCPhS020H+Et2NH2DE12-7.A:CH3CH2CH2NH2,B:CH3CH2CH2NHEt,C:CH3cH2cH2MEtEt12-8.A:PhCH=NPh,B:PhCH2

25、NHPh,C:PhSO2NMe2,D:phCH2NHCHOCH3E+F：(CH3)2C=CH2+EtNMe2,G:PhCH2CH2NCH2CH3，CH3+H：PhCH2cH2NCH2CH3OH-,|：PhCH=CH2+NCH2cH3MeMeMeMeJ：1N(Me)2,K：I+N(Me)312-10.H2,Pd/BaSO4NH3C6H13COCI*C6H13CHOaC6H13CH2NH2H2/Pta. C6H13coel+C6H13cH2nh2C6H13CONHCH2c6H13LiAlH4C6H13CH2NHCH2c6H13b. C6H13COCINH3-C6H13CONH2MBr2»

26、C6H13NH2/,NaOHH2/PtC6H13CHO+C6H13NH2C6H13CH2NHC6H1312-11.a.CH3H3PO2CH3b.FeHClCH3HNO3H2SO4NO2CH3ClCH3Cl2FeCH3IClNO2Fe/HClNaNO2HCl”NO2(EtO)2OCH3Cl2水解FeNH2ClNHCOCH3CH3NaNO2H3PO2HClI'(b)ClN2Cl12-12.小混酸JO'n/HCl.NO2(CHS”。魯3.：.-NO2.I1no2水解NaNO2NaNO2.f*HCl.NaHCO3,Cu2O".INO212-13.a.EtNO2EtSnAc2O

27、HClNH2EtClClCl21.OHFe2.HNO23.CuClb.NH2EtCl2ClEtuhClnh2NaNO2H3PO2HClEtClCl12-14.h2c一ch2NH34h2c一ch2HOCH2cH2NHCH2cH20HP2O512-15.COOHCH3混酸NO2CH3KMnONO2SOCl2Et2NCH2CH2OH12-16.NH2H2so412-18.Ni/H2COOCH2CH2NEt2Et2NHTVO-E2NCH2CH2OHnh2NHSO3HHNO2NNMe2HSO3N二NPH57LN(CH3)2A：NH9卜NHB：NHCOCH3C：NO2OOIIIJII1NHN.HNH2OC

28、OCH3D:CH3CH2CH2COOHE：CH3CH2NH2+HCOOHF:CH3CH2CH2NHCH3G：-COOCH3H+I：CH2CN+AgIJ:不反應(yīng)AO2NCH3K：-NHfVNO2-12-23.用NaNO2/HCl分別反應(yīng)，a.在低溫時反應(yīng)可見有氮?dú)夥懦?；b.低溫時反應(yīng)生成重氮鹽，加入-蔡酚后得紅色偶氮化合物；c.生成黃色對亞硝基N,N-二甲基苯胺鹽酸鹽，中和后成為綠色固體；d.生成黃色油狀物。12-24.解答：加乙酰氯處理，二丁胺生成乙?；铮梢耘c三丁胺分離。廠丁rNO"HLUHHO CH2- N= N-OH h j /-11,H2O 1HOCT重排二一飛飛擴(kuò)壞11二

29、' 112-26.A可與HNO 2發(fā)生重氮化反應(yīng)，故其為芳1所以A為五取代芳伯胺。又根據(jù)其1HNMRA xz51 cc-t*HO CH2- N三 NHO CH25、一 N2上 J l1 hOo-JU1 一 ) 、一,'白胺，又它不能發(fā)生芳環(huán)上的親電取代反應(yīng)， 數(shù)據(jù)推測A的結(jié)構(gòu)為一(二:-NH2。從B的1HHO CH2NH2HO ch2- N-N=OHO CH2- N-N OH12-25.*NMR數(shù)據(jù)SH7.1(4H)看，B為二取代的芳胺，它不能發(fā)生重氮化反應(yīng)，其不是芳伯胺，再結(jié)合1HNMR數(shù)據(jù)，推測B的結(jié)構(gòu)為:CH3CH3CH 3IAg 2。，H 2Och3ch3CHCH3+

30、+NCH 3CH 3OH-CH3N(CH 3)212-27.1)OH3)OHCH20H0h 0>CH2OHOOh hOHOC2H5OPO3H22)4)OHCOOCH 3OCH2OHOHOHOHOHOHCH3I.Ag2)皿3N+H2/Pt13-1.1)3)HOH2CHOHOOHHOHI2CHOCH 20HOHOHoch3OH2)4)仁。與IICH20HOH HH(Ho2COHI Oj, HH 一 OH13-2.13-3.1)2)3)4)5)HC=NOHH-C-OHOH-C-HH_C-OHH-C-OHCh20HCOOHH-C-OHOH-C-HH-C-OHH-C-OHCh20HCH20H3CH

31、20HH_C-OHOH-&HH-C-OHH-C-OHCh20HC=NNHC6H5C=NNHC6H5OH-C-HH_LOHH-C-OHCh20H13-4.1）甲基葡萄糖甘葡萄糖果糖淀粉淀粉I2不顯色甲基葡萄糖甘葡萄糖果糖2)甲基葡萄糖甘葡萄糖果糖菲林試劑葡萄糖果糖,Br2/H20j褪色一L不褪色葡萄糖一果糖無沉淀-甲基葡萄糖甘麥芽糖乳糖蔗糖甘露糖菲林試劑麥芽糖乳糖甘露糖,苯朋姬耿測熔點(diǎn)糖豚確定不同的糖蔗糖13-5.因?yàn)樗嵝詶l件不能發(fā)生差向異構(gòu)化反應(yīng),故不能反應(yīng)；而堿性條件可以發(fā)生差向異構(gòu)化反應(yīng)，通過烯二醇結(jié)果轉(zhuǎn)化為酮糖（果糖），故可以反應(yīng)。13-6.無還原性，為糖甘。能被麥芽糖酶水解，故

32、為“一糖甘，因而，其結(jié)構(gòu)為CH20HOCH(CH3)2HOH13-7.CHOHO-C-HH-C-OHCH20H(A)CHOH-C-OHH-C-OHCH20H(B)13-8.HOHhcHOHH2OHCHOHOHHOHC1-4111HHOH(H0HHCNlOHHOH)H20HCH2OH+CNCOOHHOHCH2OH1HO-C-HHO-C-HHOII_HH2O.HOHCHcutHHHOHHOH一OHHOHCNCOOHH-C-OHH-C-OHCH2OHCH20H1)2)HO、.OC，H.,OC，HO.一OCH-C-OHHO-C-HH-C-OH溟水HO-C-H濃HNOH-C-OH3.HO-C-HH-C-OHH-C-OHH-C-OHH-C-OHCOOHH-C-OHCh2ohH-C-OHCH2OH3)4)HOCOO-OHOHOHCH20HCH20HC=OHO-C-HH-C-OHH-C-OHCH2OHCa2+COOHC=OCHOHOHHOHH2O2,Fe3+H0CH2OH-C02HOOHOHCH2OHCH20HCH2OH1H-C-OHHO-C-H1HO-C-HHO-C-H1Na-Hg-H-C-OH+H-C-OH1H2H-C-OHH-C-OH1CH2OHCH2OH13-9.CHOH-C-OHH-C-OHH-