有机化学讲义.pdf

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1、有檄化 羲章 直 行Cha pter 1 臃病Cha pter 2 酸舆Cha pter 3 有械化合物之命名Cha pter 4 烯 Cha pter5烯舆快的反Cha pter 6 未定域化重子Cha pter，芳香族化合物Cha pter 8 立醴化Cha pter 1 0 卤烷的取代及脱去反鹰Cha pter 1 Electronic Structure a nd Cova lent Bonding原始定羲Orga nic:conta ining a vita l forceInorga nic:la cking a vita l force1828 Friedrich WohlerN

2、H4OCN 一(NH2)2COOrga nic compounds a re now defined a s compounds tha t conta in ca rbon.1.1 The Structure of a n Atom原子序 a tomic number:the number of protons in nucleus 量数 ma ss number:the sum of protons a nd neutrons in nucleus同位素 Isotopes ha ve the sa me a tomic number,but different ma ss number.I

3、sotopes:a toms with the sa me number of protons but different numbers of neutronsA=Z+中子数Problem 1 ：看十算盟 ；O、子敷典中子数1.2 The Distribution of Electrons in a n Atomelectron configura tion:describing wha t orbita ls the electrons occupy(1)An electron a lwa ys goes into the a va ila ble orbita l with the lo

4、west energy.情篥原理a ufba u principle：子一偃|一偃地加入原子核中以情篥元素畤，重子以相似方式加入原子f l域中。1s2s2p3s3p3d4s4p4d4f5s5P5d5f6s6P6d7s7p轨域的能量氢原子 1 s 2s=2p 3s=3p=3d 4s=4p=4d=4f 多重子原子(+雷:子IW 的斥力+重子舆核子 的引力)Is V 2s 2p 3s 3p 4s 3d 4p 5s 4d (a)the most pola r bond?(b)the lea st pola r bond?静重甯位If electrosta tic potentia l ma p:系工色

5、表示最大:f t1 性位置，叠色表示最大正重性位置Problem 7:比较LiH、HF、氏 的瞪B重位I I，何者:B幽生?(b)何者的H 带正W?Problem 8：以。+、3-表示趣性方向。(a)HO-H(b)H3C-NH2(c)HO-Br(d)I-Cl1.4 Representa tion of StructureLewis structuresK e k ul e s t r uc t ur e sC o n d e n s e d s t r uc t ur e s 缩麓式系留S形式重荷f o r m a l c h a r g e原子所分配的僵雷:子数：所有未维结的重子+一半的维结

6、重子数形式重荷=自由原子的僵重子数-原子所分配的彳翼重子数判断最合理的L e w i s结横：形式重荷最接近0且负重荷位於I W重性最大的原子S O42*co2水合氢蹄子、金墨蹄子h y d r o n i um i o nP r o b l e m 9：(a)。口中的原子何者带形式负重?(b)0田中的原子何者的霜子密度较大？(c)比0+中的原子何者带形式正重?(d)印0+中的原子何者的f t子密度较小？C H 4 CH3+C H 3：-CH3-碳照雕子 c a r b o c a t i o n :a s p e c i e s c o n t a i n i n g a p o s i t

7、 i v e l y c h a r g e d c a r b o n a t o m碳I W雕子 c a r b o c a t i o n :a s p e c i e s c o n t a i n i n g a p o s i t i v e l y c h a r g e d c a r b o n a t o m自 由基 r a d i c a l :a s p e c i e s c o n t a i n i n g a n a t o m w i t h a s i n g l e un p a i r e d e l e c t r o nP r o b l e m 1

8、 0：求中心形式重荷：(a)(C H3)2OH 比 CH30cH3 (b)CH3:比 CH4(c)N(C H3)3 t t N(C H3)4 (d)H 3 N-B H 3P r o b l e m 1 1 ：重出 L e w i s 结横：(a)C H3NH3+(b)C2H5(c)NaOH(d)NH4C 1P r o b l e m 1 2 :黠出未维结霜子：(a)CH3CH2NH2(b)C H 3 N H C H 3 (c)C H 3 c H 2 0 H (d)C H 3 O C H 3 (e)C H 3 C H 2 C I (F)H0NH2P r o b l e m 1 3 ：由模型重出缩

9、麓式结情P r o b l e m 1 4 :P r o b l e m 1 3中的分子模型之原子，何者具有(a)三 举 寸(b)二 螯 寸(c)一t f(d)零封未维结霜子？Pro bl e m 1 5 ：(a虐出 C 2 H 6。的 2 彳 固 L e w i s st ru c t u re s，(b)重出 C 3 H 8。的 3 彳 固 L e w i sst ru c t u re sPro bl e m 1 6 :由 缩 麓 式 结 情 延 伸 而 重 出 共 及 未 维 结 重 子 茎 寸：C H3N H(C H2)2CH3 (b)(C H3)2C H C 1 (C)(C H3)

10、3C(CH2)3C H(C H3)21.5 A t o m i c O rbi t al ss f l域、p如或1.6 Co v al e n t B o n d Fo rm at i o nFi g.1.1 2 H-H至建结的形成：吸力+斥力：最小能量维晨 bo n d l e n g t h维 弓 苗 度 bo n d st re n g t h 维 解 雇 隹 育 且 bo n d d i sso c i at i o n e n e rg yT h e c o v al e n t bo n d t h at i s f o rm e d w h e n t h e t w o o r

11、bi t al s o v e rl ap i s c al l e d。bo n d.1.7 Bonding in Metha ne a nd Etha ne:Single BondCH4 的表示：透ijM法化阜式 perspective formula、球-棍模型 ba ll-a nd-stick model 空IW填充模型 spa ce-filling model 静重重位|electrosta tic potentia l m叩考 虑CH4分子：4彳 固C-H 维角90。，x)混成化T TTT T TT tT t混成Wl域 hybrid orbita l原子轨域 性系且合LCAO(P

12、1=%(s)+(px)+(Py)+(pz)(p2=%(s)+(px)-(Py)-(pz)(p3=%(s)-(px)+(py)-(Pz)(p4=%(s)-(Px)-(Py)+(Pz)Fig.1.3 sp3混成轨域形状Fig.1.4 Tetra hedra l Ca rbon:sp?维角 109.5Fig.1.5 An orbita l picture of etha ne.The C-C bond is formed by sp3-sp3 overla p,a nd ea ch C-H bond is formed by sps overla p.Problem 17:Wha t orbita

13、ls a re used to form the 10 cova lent bonds in CH3cH2cH3?1.8 Bonding in Ethene:A Double Bondsp?混成化，维角 120，Fig.1.6,1.7o bond(sp2-s,sp2-sp2)，兀 bond(p-p)Side-to-side overla p of p orbita ls forms an bond.原子孰域 性系且合LCAO%=7175以+775。内 外=-7176.+VT72PV3=J i万%一折不。内一万而万1.9 Bonding in Ethyne:a Triple Bondsp 混成化

14、，维角 180，Fig.L8,1.9a bond(sp-s,sp-sp)，兀 bond(p-p)原子轨域吊泉性系且合LCAO P =见+际 明x H-C1 H-Br H-IThe grea ter the electron density in the region of orbita l overla p,the stronger isthe bond.The shorter the bond,the stronger it is.Problem 23：(a)制古Cl2、B o的相封罐艮典维弓金度。(b演 估H F、HC1、H B r的相举战典维弓苗度。Problem 24:比较(a)维是典

15、(b)维强度。C-Cl or C-Br(2)C-C or C-H(3)H-Cl orH-H1.14 Summa ryAssign sp3,sp2,or sp to C,N,a nd O in CH3NH2,(CH3)2C=NNH2,CH3CN,CH3OH,CH3 CO 2 H,c o2T abl e 1.6 比较 C2 H6、C2 H4、C2 H2A 7 t bo n d i s w e ak e r t h an a o bo n d.Pro bl e m 2 5 :曼维中的。维或兀维的轨域重叠何者较焉有效？Pro bl e m 2 6 ：(a)指 出 各 彳 固 C 原子的f t域混成。(

16、CH3)2 CHCH=CHCH2C=C C H3(b)指出各彳固C N O原子的Wl域混成。V i t am i n C Caf f e i n ePro bl e m 2 7 ：颈估维角。(a)C-N-C i n (CH3)2N H2+(b)C-C-N i n C H3C H2N H2(c)H-C-N i n (CH3)2N H (d)C-O-C i n C H 3 0 c H 3Pro bl e m 2 8 :描述维结所使用的f l域及其维角。(a)C H3O H (b)H 0 N H2(c)H C O O H(d)N2Cha pter 2 Acids a nd Ba ses2.1 Aci

17、d-Ba se Rea ctionArrhenius:a cids produce hydrogen ions in a queous solution,a nd ba ses producehydroxide ions.Bronsted-Lowry y k y X z 7 K h n-U -Acid:proton donorBa se:proton a cceptorHA(a q)+H2O(1)N H3O7a q)+A-(a q)酸 共聊When a compound loses a proton,the resulting species is ca lled its conjuga te

18、d ba se.H CI+H2O&H3O+c rNH3+H2O&NH4+OHProblem 1 (a)指出相罂寸鹰的共聊酸：NH3 C、0/、H2O，(b)指出相螯寸鹰的共聊:NH3、HBr HNO3、H2O例 下列何者情成共聊酸(a)HF,F (b)NH4+,NH3(C)HCI,H2O例 嘉出共聊:(a)HCIO4(b)H3PO4(c)CH3NH3+_Problem 2 HCI+CH3OH e CH3OH2+CFNH3+CH30H w NH4+CH3O-丽 性 物a mphoteric substa nce参考 p.35：CH3OH、CH3CO2H自解蹄 a utoioniza tionH2

19、O(1)+H2O(1)n H3O+(a q)+OH-(a q)蹄子稹常数 ion-product consta ntKW=H+OH2.2 pKa a nd pHPower of Hydrogen Ion Concentration:pH=-logH+，pOH=-logOH-log 4.000 xl0-5=-4.3979a ntilog 12.5=3xl012pH+pOH=14.00酸解雕常数(a cid dissocia tion consta nt)HA(a q)+H2O(1)w H3O+(a q)+A(a q)pKa SelU中 弓 金 酸35，弱酸5-15 15代表物H N O3H3P。

20、4CH3co2HHCNH2OpKa-1.32.1484.769.3115.74 解雕常(ba se dissocia tion consta nt)B(a q)+H2O(1)n BH+(a q)+OH(a q)BH+OH-K.=-闻Ka-Kb=KwpH:the a cidity of a solutionpKa :the tendency of the compound to give up its proton酸弓 苗 度The stronger the a cid,the more rea dily it gives up a proton.HCl(a q)+H2O(1)&H3O+(a q

21、)+Cl(a q)Ka=IO3-9=8000-IM，98.8%解蹄HF(a q)+H2O(1)&H3O+(a q)+F-(a q)Ka=7.2x10-4 IM，2.7%解蹄幼苗度:c r H2o 符下列物槿依 强度由强至弱排序。CH3NH2、CH3NH、CH3OH、CH3O-、CH3cCV2.3 The Effect of Structure on pKaThe stronger the a cid,the more sta ble is its conjuga te ba se.相封怪甯性:C N O F相封酸度：CH4 NH3 H2O HF相螯寸安定性：CH3-NH2-OH-F-When

22、the a toms a re simila r in size,the strongest a cid will ha ve its hydrogen a tta chedto the most electronega tive a tom.相螯寸安定性:F cr B r r相封酸度：HF HC1 HBr Henderson-Ha sselba lch Equa tionIf A =HA,then pH=pKa.A compound will exist prima rily in its a cidic form if pH pKa.Problem 14指出pH=5.5日寺物槿存在的主要形

23、式，酸式或 式？(a)CH3co2H(pKa=4.76)(b)CH3cH2NH3+(pKa =11.0)Problem 15 指出 RCOOH(pKa=4.5)、RNH3+(pKa=9)、ROH(pKa=15)存在的主要形式，酸式或 式？pH=1 (2)pH=3(3)pH=5(4)pH=7 pH=10(6)pH=132.5 Buffer Solutions 律询 buffer:a mixture of a wea k a cid a nd its conjuga te ba se 律亍溶液(buffer solution):a solution of a wea k a cid a nd it

24、s conjuga te ba seA buffer solution resists cha nges in pH when a cids or ba ses a re a dded or whendilution occurs.p.480 W :下 列 合 何 者 可 作:溶 液？(b,c)(a)HC1,Na Cl(b)CH3COOH,CH3COOK(c)H2S,Na HS(d)H2S,Na2SpH=pKu+logP rob le m 16嘉出方程式 明由CH3co OH及C H 3 c o O N a情成的 溶液如何避免p H的急遽燮化。（a）加入少量的H+，（b）加入少量的0 H-。2

25、.6 L e wi s A c i d s a nd B a s e sA L e wi s a c i d i s a s pe c i e s th a t a c c e pts a s h a re i n a n e le c tron pa i r.A L e wi s b a s e i s a s pe c i e s th a t d ona te s a s h a re i n a n e le c tron pa i r.H+NH3 Z H-N H3+A I C 13+C H 3 O C H 3 w C 13A 1-O(C H3)2B H 3 +N H 3 N H 3

26、B-N H 3P rob le m 17以箭虢表示下列反鹰中重子罂寸的供舆受，加嘉出崖物。(a)Z n C b +C H 3 O H *(b)F e B rj+B f *(c)A I C I 3 +c r nP rob le m 18描述以下化合物如何典0匹反鹰。(a)CH3OH(b)N H 4+(c)CH3NH3+(d)B F3(e)C H 3+F e B r3(g)A 1 C 13(h)C H 3 C O O HCha pter 3 An Introduction to Orga nic Compounds趣、碳氧化合物 hydroca rbons:compounds tha t cont

27、a in only ca rbon a nd hydrogen烷 a lka nes:hydroca rons tha t conta in only single bondsTa ble 3.1烷 命名sa tura ted,unsa tura tedstra ight cha in a lka nebra nched-cha in a lka ne昊横物 isomer 结横巽情物 constitutiona l(structura l)isomerhexa ne isohexa ne(2-methylpenta ne)系统名 systema tic nomencla ture(IUPAC

28、nomencla ture)用名、普通名common na me3.1 Nomencla ture of Alkyl SubstituentsTa ble 2 烷基命名丁基 butyl，正丁基 -butyl昇丁基 isobutyl第二丁基 sec-butyl第三丁基加-butyl伯 prima ry ca rbon:ca rbon tha t is bonded to one other ca rbon仲 seconda ry ca rbon:ca rbon tha t is bonded to two other ca rbons叔 tertia ry ca rbon:ca rbon tha

29、 t is bonded to three other ca rbons季 qua terna ry官能基 functiona l group:a group of a toms tha t is la rgely responsible for the chemica lbeha vior of the pa rent molecule.The chemistry of every orga nic molecule,rega rdless of size a nd complexity,isdetermined by the functiona l groups it conta ins.

30、a lcohol-O H，a mine-NH2，a lkyl ha lide-X，ether-O-Problem 1 命名(a)CH3CH2CH2OH(b)CH3OCH3(dimethyl ether)(c)CH3cH2cH2NH2Problem?重出CH4B1的四彳固结横昊横物她命名Problem 3 嘉出下列化合物的系吉情(b)isopentyl fluorideProblem 4 命名下列化合物(c)CH3cH2cH(OHNH2Problem 5 重出 横或国台予系统命：(a)没有第三碳的5曰2(b)没有第二碳或第三碳的C5H1 23.2 烷 命名IUPAC保电粹及鹰用化擘HE祭聊盟曾)

31、命名系统字首prefix-母名(字韩)pa rent-字尾suffixWhere a re the substituents?一 How ma ny ca rbons?-Wha t fa mily?(1)Find the pa rent hydroca rbon.(a)最晨的速 碳(b)相同晨度的碳gg，取分枝黑蹄交多者-一(6)(2)Number the a toms in the ma in cha in.以最靠近第一偃I分枝黑占的末端 始(儒可能使支号虎礁最小)雨彳固以上的取代基之编礁方向曾有最小的支 虢礁-(4)3-ethylhexa ne,not 4-ethylhexa ne5-et

32、hyl-3-methylocta ne(O)4-ethyl-6-methylocta ne(X)(3)Identify a nd number the substituents.以 di-,tri-a nd tetra-表示相同的置换基以英文字母排序，包括iso及 cyclo，但不言十入di,tri,tetra,sec及 tert5-ethyl-3-methylocta ne5-ethyl-2,5-dimethylhepta neList substituents in a lpha betica l order a nd use di-,tri-a nd tetra-to indica te

33、 identica lsubstituents.(4)When both directions lea d to the sa me lowest number for one of the substituent,the direction is chosen tha t gives the lowest possible number to one of the rema iningsubstituents.2,2,4-trimethylpenta ne.not 2,4,4-trimethylpenta ne6-ethyl-3,4-dimethylocta ne,not 3-ethyl-5

34、,6-dimethylocta ne(5)If the sa me substituent numbers a re obta ined in both directions,the first groupcited receive the lower number.3-ethyl-5-methylhepta ne,not 5-ethyl-3-methylhepa tne2-bromo-3-chlorobuta ne,not 3-bromo-2-chlorobuta ne(6)If a compound ha s two or more cha ins of the sa me length,

35、the pa rent hydroca rbonis the cha in with the grea test number of substituents.3-ethyl-2-methylhexa ne(O)3-isopropylhexa ne(X)Problem 6 重出下列化合物的系吉横：(d)4-isobutyl-2,5-dimethylocta neProblem 7(a 虐 出 18偃J C8H18的圜W物，(b瀚予系统名，何者有巽丙基？(d)何者有第二丁基？何者有第三丁基？Problem 8/合予系统名:(b)CH3cH2c3 3)3 CH3c(CH3)2CH(CH3)CH(C

36、H2cH3)23.3璟烷 命名骨 架 系 吉 情 skeleta l structure(参考 problem 9)(l)c 原子通常不 示。C 原子假定是在二维)得相交虑及各国泉之末端。偶而，焉了弓苗 或 明而示c 原子。(2)C 维结的H 原子不骸示。(3)C 舆 H 以外的其他原子均示。6-ethyl-2,3-dimethylnona ne命名(1)算璟中的碳原子数，加字首cyclo於相封J1烷 英文名之前。若璟上有烷基取代基测命名篇a lkyl-substituted cycloa lka ne。罩 取代基不绷 予懒思。Methylcyclopenta ne，cycloa lkyl-s

37、ubstituted a lka ne?1 -Cyclohexylocta ne)(2)若有雨f0取代基，即以字母侵先J嗔序(a lpha betica l priority)始，璟褊 虢，使第二取代基有敕小的虢礁。1 -Methyl-2-propylcyclopenta ne1 -Ethyl-2-methylcyclopenta ne，NOT 2-Ethyl-l-methycyclopenta nel1,3-Dimethylcyclohexa ne，NOT 1,5-Dimethylcyclohexa neProblem 9 指出Morphine骨架系吉横上的各彳固C 原子所维结的氢原子数。P

38、roblem 10符缩曾结横斡换成骨架箱精。(a)CH3CH2CH2CH2CH2CH2OH(c)CH3cH(CH3)CH2cH2CHB1CH3Problem 11 先合予系统名。(c)3,6-dimethyldeca ne3.4 Nomencla ture of Alkyl Ha lides惯用名：methyl chloride、sec-butyl bromide、isobutyl bromide系统名：chlorometha ne、2-bromobuta ne、1 -bromo-2-methylpropa neF-fluoro、Cl-chloro、Br-bromo、I-iodoProblem

39、 12(b)isohexyl chloride、1 -chloro-4-methylpenta ne3.5 Cla ssifica tion of Alkyl Ha lides,Alcohols,a nd Amines参考 3.2 伯 prima ry ca rbon、仲 seconda ry ca rbon、叔 tertia ry ca rbon惯用名:ethylmethylpropyla mine系统名：N-ethyl-N-methylpropyla mineProblem 13 判定 prima ry?seconda ry?tertia ry?(c)Problem 14 命 名 判 定

40、prima ry?seconda ry?tertia ry?(c)Problem 15 以 Cl 取代 methylcyclohexa ne 的一他 H，重出(a)彳 固 prima ry 卤烷(b)三 彳 固 seconda ry卤烷(c)一他 tertia ry卤烷3.6 Structures of Alkyl Ha lides,Alcohols,Ethers,a nd Amines参考LIO1.14Problem 16 估维角(a)C-O-C(b)C-N-C(c)C-O-H3.7烷、璟烷、卤烷、醇、酸、胺的物性v a n d e r W a a l s f o r c e :D i p

41、o l e-D i p o l e F o r c e a n d L o n d o n D i s p e r s i o n F o r c e偶趣-偶趣作用力D i p o l e-D i p o l e F o r c e ：趣性分子W的静霜力作用水分子(有偶趣矩的分子)：O m a x i m u m，(+),-(+)a n d。m i n i m u m氢维(隙重性大N、0、F)F i g.1 4.7 H2O、H2S、H2S e H2T e偷敦分散力 L o n d o n D i s p e r s i o n F o r c eF i g.1 4.8 (a)&(b)H e ：

42、霜子分彳布一瞬W偶 趣 一 郝接分子羟生醇偶趣趣化率(p o l a r i z a b i l i ty):原子(或分子)中的霜子分彳布情况被 燮形的易程度原子(o r非趣性分子)越大(重子越多)一趣化率越大(偏敦分散力增加)T a b l e 3.5 沸 黑 占：CH4 CH3F C H 3 c l CH3B r 2-m et h y l b u t an e 2,2-d i m et h y l p r o p an eP r o b l em 17室温下呈液憨的最小烷:p en t an eN-H 越 多、量维越重要 Bp ：1 am i n e 2 am i n e 3 am i n

43、 eP r o b l em 18比较O-H MB W O-H共，(a)何者较:R?(b)何者较弓市?(1)C H 3 c H 2 c H 2 0 H (2)CH 3 c H 2 c H 2 F(3)C H 3 0 c H 2 c H 3(a)何者分子 形成氢维？(b)何者曾典乙醇形成氢维？_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _P r o b l em 1 9(a)何者分子形成氢维？(b)何者曾典乙醇形成氢维？P r o b l em 20依沸黑占降i咸排序：P r o b l em 21依沸黠降*咸排序：Melting Point溶驯mp)：堆

44、稹(p ac k i n g)越繁密，破土襄晶格需要越多的能量。P r o b l em 22参考Fi g,3.2偶数碳的直 烷 之堆稹较奇数碳繁密。Solubility:L i k e d i sso l v es l i k e烷：n o n p o l ar 不溶於水醇：n o n p o l ar al k y l g r o u p +p o l ar OH g r o u p，烷基碳数4 一不溶於水支咸少烷基的接斶面-水中溶解度t er t-Bu O H n-Bu O H酸：烷基碳数3 一不溶於水胺:N-H m，水中溶解度T 1 2 3 卤烷：C H 3 F趣溶於水一 舆水形成氢维

45、水中溶解度 C H3F CH 3 CH 2 F CH 3 CH 2 CH 2 F水中溶解度CH3F CH 3 C1 CH 3 IP r o b l em 2 3 依水中溶解度降*咸排序：(a)H O C H 2 c H 2 c H 2 0 H C H 3 c H 2 c H 2 0 H C H 3 C H 2 C H 2 C H 2 O H CH 3 CH 2 CH 2 CH 2 CI(b)0-N H 2 。-OH 0-CH3P r o b l em 24琪己烷中溶解度何者最低？P r o b l em 25巴比妥酸蜜的静效果舆穿透非趣性余田胞膜的能力有:h ex et h al比ba rbi

46、ta l 有效3.8烷 之情形情形 conforma tion:the different spa tia l a rra ngement of a toms tha t result from rota tiona round a single bond.情形(昇输物 conformer:a specific conforma tion貂木架表示式Sa whorse representa tions:自斜角觐察碳-碳维的分子空察法 曼投影式表示式Newma n projections:末端觐察碳碳维的分子空 觐察法交金昔型sta ggered遮蔽型eclipsedFig.3.4:乙烷 99%

47、sta ggered，1%eclipsed，能量差 12 kJ/mol扭力IB燮 torsiona l stra in:the repulsion felt by the bonding electrons of onesubstituent a s they pa ss close to the bonding electrons of a nother substituent.立飕鹰燮 steric stra in-the stra in put on a molecule when a toms or groups a re tooclose to one a nother,which

48、results in repulsion between the electron clouds of thesea toms or groups.丁烷最安定的 conformer：D(p.66)Problem 26(a 圉丁烷的C1-C2旋斡，重出三彳固交期W形。(b)三偃I交 情形是否能量相等？三假|遮蔽情形是否能量相等？Problem 27螯寸戊烷的C2-C3旋率事，重出(a)最安定的conformer (b)最不安定的conformer 3.9 Cycloa lka nes:Ring Stra insp3轨域的理想维角：109.5璟丙烷：维角60 一角鹰燮a ngle stra in-

49、反鹰性强 6 彳 固 C-H M eclipsed璟丁烷：琪戊烷：eclipse H torsiona l stra in t not pla na r,some a ngle stra in3.10 璟己烷的情形璟己烷：cha ir cyclohexa ne is free of a ngle stra in,a lso free of a ll C-H eclipsing率 由 维 a xia l bond赤道维 equa toria l bondFig.3.5斜向窗木架表示式)：cha ir conforma tion瞳由向融曼投影式表示式)：freeof a ll C-H eclips

50、ing intera ctionsFig.3.8 璟 翻 串 事 ring flip,cha ir-boa t-cha irProblem 28 描 1,2,3,4,5,6-hexa methylcyclohexa ne，指出(a)所有率由向的甲基(b)所有赤道向的甲基3.11 Conforma tions of Monosubstituted Cyclohexa nesFig.3.9 Methylcyclohexa ne：赤道向 CH3，95%penta diene1,4-penta diene(4)取代基雨彳固以上畤 以 di-,tri-a nd tetra-表示相同的取代基，封不同的取代