第 7,8 章烯烴和炔烴 : 性質, 命名, 合成及加成反應

Transcription

1 第 7,8 章烯烴和炔烴 : 性質, 命名, 合成及加成反應 C 3 Ethene Propene Ethyne sp 2 sp 一 ) 烯烴的 E, Z 命名系統 : 3 C C 3 C 3 cis-2-butene 3 C trans-2-butene

2 cis- trans- 系統對某些烯烴類化合物, 不能給予明確的命名, 例如 : Cl F 在 (E)-, (Z), 命名系統中, 先確立雙鍵同一碳上基團的優先順序, 再比較兩個雙鍵碳原子上的基團, 優先基團在同側為 Z, 反向為 E:

3 cis 或 trans 類的烯烴亦可以 Z,E 系統來命名 : 課堂練習,page 289:

4 二 ) 烯烴的相對穩定性 1)cis 與 trans 的立體異構物相比較 由於 cis 異構物有較大的 crowing, 故較 trans 異構物不穩定 2) 烯烴的穩定性 : 烯烴雙鍵上的烷基取代越多, 烯烴的穩定性就越大 課堂練習,page 291:

5 烯烴穩定性的測量 : 3) 環烯烴的穩定性 : 小於五個碳的環烯烴只能以 cis 的形式存在

6 三 ) 烯烴的製備 1) 鹵代烷烴的 E2 消除反應 secondary and tertiary halides 一般來說比較適用於 E2 反應 當所用的 alkyl halides 為一級時, 大的鹼必須使用 高濃度的強鹼通常用於反應當中 (NaOEt,t-BuONa) 3 C C C 3 C 2 5 O - Na+, C 2 5 O 3 C C 3 C C 3 C 2 5 O - Na+, C 2 5 O (C 3 ) 3 C - Na+, (C 3 ) 3 CO

7 A) Zaitsev s rule: 當在消除反應中有兩個可能的烯烴產物時, 在用以小體積鹼的情況下 ; 有較多烷基取代的烯烴的形成佔優勢

8 The trisubstituted alkene-like transition state will be most stable and have the lowest ΔG Kinetic control of product formation: When one of two products is formed because its free energy of activation is lower and therefore the rate of its formation is higher This reaction is said to be under kinetic control 課堂練習,page 295:

9 Bulky bases such as potassium tert-butoxide have difficulty removing sterically hindered hydrogens and generally only react with more accessible hydrogens (e.g. primary hydrogens): ofmann rule

10 B) E2 反應中的立體化學 : 四個原子 (-C-C-L) 必須在同一平面上 ; 被稱為 coplanarity. 其中,anti coplanar 比 syn coplanar 更佔優勢 : Base R R L R R Anti coplanar orientation is preferred because all atoms are staggered 在環狀 halides 中, 離去基團, 被鹼所攻擊的氫必須處在 diaxial 的位置 :

11 符合 Zaitsev's rule

12 igher activation energy 認真做課堂練習,page 298: 2) 烯烴的第二種製備方法 : 醇的酸催化脫水反應 Elimination

13 7.34 Explain: Base Reason 1: the bromide doesn't occupy the axial position Reason 2: the double bond can not be formedonthebridgehead

14 此類反應所需的酸的濃度以及溫度同常取決於 alcohol 底物的結構 : a)1 o alcohol: 最不容易發生反應 ; 通常需要強酸高熱條件 b)2 o alcohol: 比 1 級醇容易一些 c)3 o alcohol: 最容易發生酸催化脫水反應

15 一級和二級醇在脫水反應的同時通常伴有重排反應的發生 : A) 3 o,2 o 脫水反應的機制 : E1

16 Rate-determining step

17 The developing positive charge is most effectively delocalized in the transition state by hyperconjugation 課堂練習,page 300:

18 B) 1 o 脫水反應的機制 : E2 Primary alcohols cannot undergo E1 dehydration because of the instability of the carbocation-like transition state in the 2nd step Rate-determining step

19 C) 伴隨著某些二級 alcohol 酸催化脫水反應 (E1) 而發生的重排反應

20 In the third step the less stable 2 o carbocation rearranges by shift of a methyl group with its electrons (a methanide). This is called a 1,2 shift. The removal of a proton to form the alkene occurs to give the Zaitzev (most substituted) product as the major one.

21 課堂練習,page 305:

22 D) 伴隨著某些一級 alcohol 酸催化脫水反應 (E2) 而發生的重排反應

23 3) 烯烴的第三種製備方法 : 炔烴的氫化及還原反應 A) 製備炔烴的幾種方法 : a) Dihydrohalogenation of vic-dibromides to form alkynes: 課堂練習,page 308:

24 O PCl 5 NaN 2 Cl Cl NaN 2 Cl NaN 2 2 NaN 2 b) Replacement of the acetylene hydrogen atom of terminal alkynes: The relative acidity of acetylenic hydrogens in solution 認真做課堂練習,page 309:

25

26 此類反應適用於一級 alkyl halides. 對於 2,3 級 alkyl halides 來講 ;E2 反應佔優勢 : 課堂練習,page 310:

27 B) 由炔烴轉化為烯烴的方法 a) 氫化加成反應

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29

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31 Exercise:Page : Correct the following names: a) Change 3-trans-pentene into 2-trans-pentene b) Change 1,1-dimethylethene into 2-methylpropene d) Change 4-methylcyclobutene 3-methylcyclobutene e) C 3 3 C Cl Change 3-Chloro-2-butene into 2-chloro-2-butene c) f) hange 2-methylcyclohexene into 1-methylcyclohexene Change 5,6-Dichlorocyclohexene into 3,4-dichlorocyclohexene 7.19: Write the structural formula for each of the following:

32 a) g) C 3 b) h) Cl c) i) d) e) j) k) 3 C Cl C 3 F l) C 3 f) F F 7.20: Write the three-dimensional formulas: C 3

33 a) (E, 4S)-4-bromo-2-hexene (Z, 4S)-4-bromo-2-hexene (E, 4R)-4-bromo-2-hexene (Z, 4R)-4-bromo-2-hexene 7.21: Give the IUPAC names:

34 a) d) (Z)-3,5-Dimethyl-2-hexene 1-sec-Butyl-2-methylcyclohexene b) Cl -Chloro-3-methyl-1-cyclopentene c) e) Cl (Z,R)-3-chloro-hex-4-en-1-yne f) 6-methyl-3-heptyne 2-Pentyl-1-heptene

35 7.22: Outline a synthesis of propene form each of the following: a) Cl O - Na + O b) Cl EtO - Na +, EtO c) O Conc. 2 SO 4, 180 o C

36 d) O 3 PO 4 e) NaN 2 f) 2, Pd/CaCO 3

37 7.23: Outline a synthesis of cyclopentene: a) EtO - Na +, EtO Cl b) Cl EtO - Na +, EtO Cl Cl 2 /Pd EtO - Na +, EtO c) O 3 PO : Starting with ethyne, outline the synthesis of each of the following:

38 a) 1) NaN 2 2) C 3 I C 3 b) 1) NaN 2 2) c) 1) NaN 2 2) C 3 I C 3 1) NaN 2 2) C 3 I 3 C C 3 d) 3 C C 3 (from question c) 2,Pd/CaCO 3 3 C C 3 e) 3 C C 3 (from question c) Li, N 3 3 C C 3

39 f) 1) NaN 2 2) C 2 C 2 C 3 g) C 2 C 2 C 3 1) NaN 2 2) C 3 I 3 C C 2 C 2 C 3 h) 3 C C 2 C 2 C 3 2, Pd/CaCO 3 (From question g) i) 3 C C 2 C 2 C 3 Na, N 3 (From question g) j) 1) NaN 2 2) C 2 C 3 1) NaN 2 2) 3 C 2 C C 2 C 3

40 7.26: Explain: A - C 3 + O - 2 O C 3 C 3 base C 3

41 7.27: Outline the synthesis of phenylethyne form each of the following: a) b) Ph NaN 2, N 3 Ph NaN 2, N 3 Ph c) 2, CCl 4 NaN 2, N 3 Ph d) O PCl 5 Cl Cl NaN 2, N 3 Ph

42 7.28: Arrange the acidity in decreasing order: O > > > 7.29: Write the products:

43 a) C 3 NaOEt, EtO C 3 + C 3 C 3 C 3 (major) C 3 (minior) b) C 3 NaOEt, EtO C 3 C 3 C 3 C 3 C 3 c) NaOEt, EtO + C 3 C 3 (major) C 2 (minior) 3 C d) NaOEt, EtO + e) NaOEt, EtO (major) Et (minior) + (major) (minior) f) NaOEt, EtO

44 7.30: Write the products: a) C 3 C 3 t-buona, t-buo C 3 C 3 (major) + C 3 (minior) C 3 b) C 3 t-buona, t-buo C 3 3 C C 3 3 C C 3 c) 3 C d) C 3 t-buona, t-buo t-buona, t-buo C 3 3 C (major) C C (minior) + e) (minior) t-buona, t-buo + (major) (minior)

45 7.32: Arrange the reactivity towards the acid-catalyzed dehydration: O > O > O 7.33: Give the products:

46 a) O + b) + + (hydride migration) O (major) (minor) c) O + + d) O + (major) + (major) 3 C O e) + C 3 C 3 major C 3 minor

47 四 ) 烯烴的加成反應 A) 烴雙鍵的親核性以及親電試劑 (electrophilic reagents; electrophiles) 一般來說, 反應均為放熱反應 (exothermic), 故有利於產物的形成 (one π and σ two σ). 第一類非離子型加成反應 : 氫化反應 第二類離子型加成反應 : 與親電試劑的加成反應 :

48

49 烯烴由於鬆散的 π 電子的存在, 故可以作為親核試劑, 而受到親電試劑 (electrophiles) 的進攻 親電試劑 : +, 2, Lewis acids, Ag +, etc.

50

51 B) Addition of hydrogen halides: a) 反應活性 :I>>Cl>F b) 反應的選擇性 : 3 C C 3 3 C C 3 Drawing the mechanism of the formation of these two compounds

52 3 C 3 C 3 C 3 C Major C 3 3 C 3 C C 2 little is formed c) Markovnikov'rule: 在 X 與部分不對稱烯烴的反應中 ; + 總是加在含氫原子較多的碳原子上 d) Markovnikov'rule 的理論解釋 : 正碳離子越穩定, 它所形成的速度就越快, 所對應產物的形成就越多 Rate-determining step

53 d) Markovnikov'rule 的現代理論 : 在部分離子型不對稱烯烴的加成反應中 ; 形成穩定的正碳離子的反應方向佔優勢 認真做課堂練習,page 335: d) regioselectivity: Stereoselective reaction

54 regioselectivity: 當反應理論上可以產生兩種以上不同的 constitutional isomers, 而實際上只產生一種產物或某一種產物佔優勢時, 此類反應被稱為具有區域選擇性 e) Addition of hydrogen halides 一般不具有立體選擇性 :

55 C) Addition of 2SO4: Sulfate 進一步水解成醇 ( 第一種方法 ): 課堂練習,page 338:

56 D) 製備 alcohol 的第二種方法 : 酸催化水解反應 The + must exist in the form of hydronium ion in water solution a) rate-determining step: 介紹 reversible reaction 的概念

57 b): The mechanism is the reverse of that for dehydration of an alcohol: ydration is favored by addition of a small amount of acid and a large amount of water 認真做課堂練習,page : rearrangement

58 E) 製備 alcohol 的第三種方法 :Oxymercuration of alkenes 討論此反應的機制 :

59 (compared with the previous methods, rearrangement seldom occur, see page 342) 課堂練習,page 342: 3 C C 3 C 3 C C 2 3 O+, 2 O F) 製備 alcohol 的第四種方法 :ydroboration-oxidation ( 硼氫化 - 氧化 ) Anti-Markovnikov: 與前三種由烯烴轉化為醇的方法相比較 ; 此方法是反 Markovnikov's rule, 即產物中的氫原子是加在含氫較少的碳上 Reagent: B26 or B3.TF 3 C O C 3 C 3 C C 3? I? 3 O + g(oac) 2, 2 O NaB 4, O -? Stereoselectivity And regioselectivity

60 反應機制 : 第一步 : 硼氫加成 (ydroboration) syn-addition The boron atom attached to the less substituted carbon atom of double bond: Anti-Markovnikov; due to steric factors and electronic factors. 3 C 3 C When the reaction is completed: B C3

61 第二步 : 氧化反應 : When the reaction is completed: O 第三步 : 水解反應 : O B O O O 3 + B O O

62 Regioselectivity: Comparing the following reactions: 3 C 3 C C C 3 3 O+, 2 O 3 C 3 C O C 2 C 3

63 Stereoselectivity: C 3?? C 3 O

64 認真做課堂練習,page : G):Addition of bromine and chlorine: C 3 C=CC 3 Cl 2,CCl 4 C 3 CCC 3 Cl Cl ow to explain the formation of trans compounds? (R,R) (S,S)

65 a) The mechanism of the halogenation: 8 7 = +1 SN2: attack from the back side

66 b) The stereochemistry of the addition of halogens to alkene (R,R) (S,S) racemic mixture of trans-1,2-dibromocyclopentane enantiomers are formed stereospecific reactions: 從一種特定的立體異構物的起始物得到特定的立體異構物的產物的反應 注意 stereospecific 反應與 stereoselective 反應的不同含義

67 stereospecific reactions 之實例

68

69 Exercise in page 354

70 ):alohydrin formation:

71

72 2, 2 O 2 O O + Racemic O Symmetrical operation

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74 I): 烯烴的其他反應 : a) 與卡賓 (carbene,r 2 C:) 的加成反應 : 高活性 R 2 R 3 R 1 R 4 C 2 : Syn addition R 1 R 2 R 3 R 4

75 KOC(C 3 ) 3, CCl 3 3 C Cl Cl C 3 KOC(C 3 ) 3, C 3 C 3, KOC(C 3 ) 3

76 Simmons-Smith reaction: 3 C 3 C 2 C ZnI I 3 C C 3

77 b) 烯烴的氧化反應 3 C O O Os O O 3 C O Os O O O NaSO 3, 2 O 3 C O O syn NaSO 3, 2 O O Os O O O O O Os O O O O

78

79 3C C 3 C 3 C 3 OsO 4, NaSO 3, 2 O or KMnO 4, O -, 2 O, cold OsO 4, NaSO 3, 2 O 3 C meso O O (R) (S) C 3 O C 3 C 3 Meso 3C OsO 4, NaSO 3, 2 O 3 C C 3 C 3 O O (R) (R) racemic O 3 C (S) 3 C (S) O O C 3 (R) O O (S)

80 c) 烯烴的 Oxidative Cleavage of Alkenes

81

82 Exercise 365: Ozonolysis

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84 五 ) 炔烴的加成反應

85 Markovnikov

86 J): 烯烴的親核加成反應的生化反應實例

87

88 8.27 給出產物結構 : a) I b) 2, Pt I c) Dilute 2 SO 4 即 3 O+, 2 O O d) Cold concentrated 2 SO 4 OSO 3 e) Cold concentrated 2 SO 4 then 2 O, heat OSO 3 O f) g), Al 2 O 3 2, CCl 4

89 h) 2, CCl 4 Then KI in acetone I i) j) 2 in 2 O Cl, Al 2 O 3 I O k) Cold dilute KMnO 4 O - O Cl i) m) h) i) j) O O 3, then Zn O OAc + OsO 4, NaSO 3 2 O O O KMnO 4, O - O heat, 3 O + O g(oac) 2, 2 O O then NaB 4, O - B 3, 2 O 2 O- O O + O O

90 8.28 給出產物結構 : a) I b) 2, Pt I c) Dilute 2 SO 4 即 3 O+, 2 O O d) Cold concentrated 2 SO 4 OSO 3 e) Cold concentrated 2 SO 4 then 2 O, heat O f), Al 2 O 3 g) 2, CCl 4 +

91 I I h) i) 2, CCl 4 Then KI in acetone 2 in 2 O I O + + I j) Cl, Al 2 O 3 O k) i) m) Cold dilute KMnO 4 O - O 3, then Zn OAc OsO 4, NaSO 3 2 O O O CO CO Cl O O h) KMnO 4, O - heat, 3 O + CO 2 CO 2 g(oac) 2, 2 O i) then NaB 4, O - O j) B 3, 2 O 2 O- O

92 8.29 給出產物結構 : a) 1 mole 2 b) 1 mole c) 2 mole d) 2 (in excess) e) 2, Pd/CaCO 3

93 f) NaN 2,C 3 I 3 C g) NaN 2 (C 3 ) 3 C

94 8.29 給出產物結構 : a) one equivalent 3 C C 3 3 C C 3 b) two equivalent 3 C C 3 3 C C 3 c) 2 one equivalent 3 C C 3 3 C C 3 d) 2 two equivalent 3 C C 3 3 C C 3 e) 2, NiB(P-2) 3 C 3 C C 3 syn addition C 3

95 f) Cl, one molar 3 C 3 C C 3 Cl C 3 g) Li, N 3 3 C C 3 3 C C 3 h) 3 C C 3 2,excess i) 2, two molar 3 C C 3 j) KMnO 4, O - 3 C C 3 then 3 O + 2C 3 CO 2

96 j) 3 C C 3 O 3, OAc 2C 3 CO 2 k) NaN 2, N 3 3 C C 3 NR 8.31 從下列化合物合成 1-butyne a) 2, CCl 4 NaN 2 Cl t-buo - K + b) Following a) c) Cl NaN 2

97 Cl d) Cl NaN 2 e) NaN 2 C 3 C 2 (1 eq) 8.32 從 2-methylpropene 合成下列化合物

98 a) 3 O+, 2 O O b) c) d) Cl F Cl F

99 e) Cl 2, 2 O O Cl f) B 3, 2 O 2, O - O 8.38 給出產物結構 : O O3, Zn, OAc O + O O O + O

100 8.46 給出產物結構 ; 注明立體結構 3 C Et OsO 4 a) NaSO 3, 2 O 3 C Et O O (R) (S) O O 3 C (S) Et (R) b) 3 C Et OsO 4 NaSO 3, 2 O 3 C Et O O (R) (R) O O 3 C (S) Et (S)

101 c) 3 C Et 2, CCl 4 3 C Et (R) + (R) Et 3 C (S) (S) d) 3 C Et 2, CCl 4 3 C Et (R) + (S) Et 3 C (S) (R)