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1 第 6 章 Ionic Reactions-Nucelophilic Substitution and Elimination Reaction of Alkyl Halides ( 鹵代烷烴的親核取代反應及消除反應 ) 1) 鹵代烷烴 C-X 鍵的極化 (polarized): 由於 C 原子與 X 原子之間電負性的差異, 造成了 C-X 鍵的極化, 使得 C 帶 partial positive charge, X 帶 partial negative charge. 此現象決定了鹵代烷烴的化學反應性.

2 以 CH 3 X 為例, 從 F I, C-X 鍵的鍵長增加, 極化度降低, 鍵能減小

3 Organic halides 包括 :alkyl halides; vinilic halides and aryl halides H Vinyl halideh H X CH 2 Cl 2 : dichloromethane; CH 3 Cl: chloroform; CCl 4 : carbon tetrachloride

4 2) 有機反應中重要的一類反應 - 親核取代反應 (Nucleophilic substitution Reactions) 以 RX 為例 : Nu: - 被稱為 nucleophiles( 親核試劑 ), 它為具有未成對電子的負電離子或分子在反應中進攻底物的正電中心, 取代底物 (substrates) 中的離去基團 (leaving groups).

6 Exercise in page 242: Rewrite the reactions as net ionic reactions: CH 3 I + CH 3 CH 2 O - CH 3 OCH 2 CH 3 + I - I - + CH 3 CH 2 Br CH 3 CH 2 I + Br - H + CH 3 OH + (CH 3 ) 3 CCl (CH 3 ) 3 COCH 3 + Br - CH 3 OH (CH 3 ) 3 COCH 3 + CH 3 OH 2 +

7 L - 被稱為 leaving group ( 離去基 ), 好的離去基在離開底物後, 應為為穩定的弱鹼離子. 3) S N 2(Bimolecular nucleophilic substitution) 親核取代反應 A) 親核取代反應可能的途徑 : a) Nu: - + RX R X - Nu NuR + X - 在新鍵形成的同時, 舊鍵亦發生斷裂 (S N 2) 或者是 : b) RX R + + X - Nu: - + R + NuR RX 先發生斷裂產正碳離子, 再受到親核試劑的攻擊 (SN1)

8 B) S N 2 反應的機制的探討 : a) 從反應的初始速度 (initial rate) 來證明反應機制 : Second order reaction 反應的初始速度與兩種反應物的濃度成正比, 間接證明反應是由兩種分子碰撞而發生 (second order).

9 b) Sir Christopher Ingold 所提出的 SN2 反應機制 : 反應的過渡態 : 反應過程中所經歷的一種能量高而不穩定的狀態, 它所對應的時間僅為 S 左右反應的過渡態不是反應的中間體 (intermediates).

10 c) SN2 反應能量之探討 Exergonic reaction: negative ΔG o (products favored) An energy barrier is evident because a bond is being broken in going to the transition state (which is the top of the energy barrier). The difference in energy between starting material and the transition state is the free energy of activation (ΔG ). The difference in energy between starting molecules and products is the free energy change of the reaction,δg o.

11 Endergonic reaction: positive ΔG o (products not favored)

12 d) SN2 反應立體化學之探討 :

13 hbrtinkingifthismoleculeisthechiralmolecu- le?i 課堂練習 : a) Page 250: I

14 b) Page 251: a) In a SN2 reaction, based on the configuration of the product, deduce the configuration of the start material: H Cl CH 3 Et (R)-2-Chlorobutane OH - H CH 3 OH Et (S)--2-Butanol b) In a SN2 reaction H Cl CH 3 I - H CH 3 I Et (R)-2-Chlorobutane Et (S)--(-)-2-Iodobutane H CH 3 I Et (R--(+)-2-Iodobutane

15 4) 另一類親核取代反應 : SN1(Unimolecular nucleophilic substitution) 產物的生成速度與 tert-butyl chloride 的濃度成正比, 而與 OH- 的濃度無關可以據此推論,OH - 並沒有參與到影響反應速度的過渡態當中此類反應被稱之為 SN1(Unimolecular nucleophilic substitution) A) SN1 反應的 Mechanism:

16 Intermediate 1 Intermediate 2 在多階段的反應中, 最慢的反應過程通常決定整個反應的速度, 它被稱之為 rate-determining step; ΔG 為最大. In the case above k1<<k2 or k3, and the first step is rate determining

17 B) 正碳離子 (carboncations) 的結構及穩定性 A carbocation has only 6 electrons, is sp 2 hybridized and has an empty p orbital

18 Hyperconjugation: stabilizes the carbocation by donation of electrons from an adjacent carbon-hydrogen or carbon-carbon σ bond into the empty p orbital( 電子從 C-H 或 C-Cσ 軌道分散到鄰近的未充滿 p 軌道, 可使得 p 軌道上的正電荷分散, 降低體系的能量, 穩定正碳離子 )

19 C) SN1 反應立體化學之探討 : 有光學活性的反應物經過化學反應後生成外消旋的產物, 稱為 : Racemization.

20 課堂練習 :a) Page 257:

21 (CH 3 ) 3 C + CH 3 CH 3 OH + OH CH 3 D) solvolysis: 鹵代烷烴與溶劑分子發生的親核取代反應 hydrolysis methanolysis

5) 決定 SN1 和 SN2 反應途徑的因素 a) 底物結構 (the structure of the substrate) SN2 Reactions steric effect In SN2 reactions alkyl halides show the following general order of reactivity: Steric hinderance:

23 5) 決定 SN1 和 SN2 反應途徑的因素 a) 底物結構 (the structure of the substrate) SN2 Reactions steric effect In SN2 reactions alkyl halides show the following general order of reactivity: Steric hinderance: the spatial arrangement of the atoms or groups at or near a reacting site hinders or retards a reaction. In tertiary and neopentyl halides, the reacting carbon is too sterically hindered to react.

24 SN1 reactions 決定 SN1 反應的關鍵因素是看是否可以形成穩定的正碳離子 Generally, a) for alkyl halides, only tertiary halides undergo SN1 reactions because only they can form relatively stabilized carbocations b) allylic halides and benzylic halides can also undergo SN1 reactions easily. Explain (b)?

25 The Hammond-Leffler Postulate The transition state for an exergonic reaction looks very much like starting material The transition state for an endergonic reaction looks very much like product Generally the transition state looks most like the species it is closest to in energy

26 課堂練習 :Page 261 b) 親核試劑的濃度與強度 (concentration and strength of nucleophile): SN1 反應並不受親核試劑濃度及強度的影響對 SN2 反應來講 : 1) Rate is directly proportional to the concentration of nucleophile; 2) Stronger nucleophiles react faster: 規則 1: A negatively charged nucleophile is always more reactive than its neutral conjugate acid 規則 2: When comparing nucleophiles with the same nucleophilic atom, nucleophilicities parallel basicities: Note: the necleophility and basicity are not measure in the same way (OH - and CN - ).

27 c) 溶剂对 SN2 反应的影响 1) In Protic solvent: having hydrogen attached to a strongly electronegtive element (H 2 O, ROH etc) 溶劑分子通過氫鍵聚集在親核試劑周圍, 降低它和親核性規則 3: 原子半徑較小的親核試劑易形成較強的氫鍵, 而且較大的原子在質子型溶劑中極化性較強, 較易於提供電子進行親核進攻, 為此 : 在 protic solvent 中 :realative nucleophilicity:

28 2) In Aprotic solvent: relative nucleophilicity: 課堂練習 :Page 264 3) 在 Protic solvent 中,SN1 的反應速度會大大增強, 因為溶劑對中間體正碳離子的穩定作用課堂練習 :Page 265

29 d) 離去基團的性質 : 在 SN1 和 SN2 反應中, 好的離去基團應該是穩定的陰離子 ( 弱鹼 ) 或中性小分子規則 4: Leaving group ability of halides: This trend is opposite to basicity (see Table in page 105): 其他好的離去基團 : Triflate ion: CF 3 SO 3-,

30 規則 5: 強鹼極少可能作為好的離去基團課堂練習 :Page 266

31 Summary S N 1 vs. S N 2 In both types of reaction alkyl iodides react the fastest because of superior leaving group ability

32 SN1, SN2 反應小結 : 1) 一般來講, 能形成穩定正碳離子的底物 (tertiary halides), 親核性較弱的 nucleophiles, 以及 polar protic solvents 有利於 SN1 反應的發生. Such as solvolysis reactions of tertiary halides. 2) 立體位阻小的底物 (methyl halides and primary halides), 親核性較強的 nucleophiles, polar aprotic solvent, 以及高濃度的 nucleophiles 則有利於 SN2 反應的發生. 3) 離去基團的影響對 SN1 和 SN2 反應相同, 即穩定的弱鹼離子為好的離去基

33 Functional Group Transformations using SN2 Reactions: Stereochemistry can be controlled in SN2 reactions

34 生化反應中 SN2 反應實例

37 課堂練習 :Page 270:Starting with (S)-2-bromobutane, outline syntheses of Each of the following compounds: Et H Br CH 3 ONa, CH 3 OH H 3 CO Et H H 3 C CH 3 7) 鹵代烷烴的消除反應 (elimination reaction) 為製備烯烴的主要方法 Explain what is α carbon and what is β carbon

38 消除反應所用強鹼的製備 : 1,2-elimination

departure of the leaving group The orientation of the

39 a)e2 消除反應的反應的機制 E 2 : E 2 reaction involves concerted removal of the proton, formation of the double bond, and departure of the leaving group The orientation of the hydrogen being removed and the leaving group is in the same plane.

40 b) E2 versus SN2: 兩種反應均在高濃度的 nucleophiles 下容易發生規則 1: 當底物為一級鹵代烷烴, 在小體積強鹼條件下,SN2 反應佔優勢規則 2: 當底物為二級鹵代烷烴, 在強鹼條件下,E2 反應佔優勢

41 Br 想一想此化合物會不會發生消去反應? 規則 3: 當底物為三級鹵代烷烴, SN2 反应几乎不会发生在加熱強鹼條件下, 消除反應占絕對優勢????? 規則 4: 加熱條件有利於消除反應 (both E1 and E2) 的發生規則 5: 體積大的鹼有利於 E2 反應的發生

42 CH 3 O - + CH 3 (CH 2 ) 15 CH 2 CH 2 Br CH 3 OH, 65 o C CH 3 (CH 2 ) 15 CH 2 CH 2 OCH 3 + CH 3 (CH 2 ) 15 CH=CH 2 99% 1% 規則 6: 低極化的強鹼 (NH 2-,CH 3 O -, etc) 有利於 E2 反應的發生 ; 而高極化的弱鹼 (Cl -, CH 3 CO 2- ) 則有利於 SN2 反應的發生

43 c)e1 消除反應及其反應機制 The E1 reaction competes with the SN1 reaction and likewise goes through a carbocation intermediate.

44 d) E1 versus SN1: 加熱及強鹼有利於 E1 反應的進行課堂練習 : 認真討論 cases on Page 278

45 課堂練習 6.13 Show how you might use a nucleophilic substitution reaction of propyl bromide to synthesize each of the following compounds. a) Br OH -, H 2 O OH g) Br N(CH 3 ) 3 N Br - b) Br NaI, EtOH I h) Br NaCN CN DMF c) Br EtONa, EtOH OEt i) Br H 2 S HS d) Br CH 3 SH SCH 3 e) Br CH 3 CO - 2 Na + CO 2 CH 3 DMF f) Br NaN 3 N 3 EtOH 6.14 Which alkyl halides would you expect to react more rapidly by an SN2 reaction

46 a) CH 3 CH 2 CH 2 Br > Br c) Cl > Cl > b) CH 3 CH 2 CH 2 I CH 3 CH 2 CH 2 Cl d) > Cl Cl e) Cl > Br 6.15 Which SN2 reaction of each pair would you expect to react more rapidly in protic solvent? Explain a) 1) >2) b) 2) > 1) c) 2) > 1) d) 2 > 1) the concentration of nucleophile is higher 原子半徑較小的親核試劑易形成較強的氫鍵原子半徑較小的親核試劑易形成較強的氫鍵 6.16 Which SN1 reaction of eeach pair would you expect to react more rapidly in protic solvent? Explain

47 a) 2) >1): Br - is better leaving group b) 1) > 2): CH 3 OH is weaker base (see page 105) c) 2) >1): concentration d) about the same; e) 1) >2) 6.17 synthesize the following compounds: a) CH 3 Br ) CH 3 CH 2 Br c) CH 3 Br d) CH 3 CH 2 Br NaI, EtOH NaI, EtOH NaOH, H 2 O NaOH, H 2 O CH 3 I CH 3 OH CH 3 CH 2 I CH 3 CH 2 OH H 2 S e) CH 3 Br CH 3 SH H 2 S ) CH 3 CH 2 Br CH 3 CH 2 SH NaCN, DMF g) CH 3 Br CH 3 CN NaCN, DMF h) CH 3 CH 2 Br CH 3 CH 2 CN NaOCH 3, CH 3 OH i) CH 3 Br CH 3 OCH 3 NaOCH 3, CH 3 OH j) CH 3 CH 2 Br CH 3 CH 2 OCH a) Strong base can not be as leaving group. b) same as above c) same as above d)??? e) CH3O - is a strong base f) the formation of NH4 + and CH3OH k) Br O - K +

48 6.19 Which compound is more suitable for preparing styrene by dehydrohalogenation using KOH in ethanol? Br Br Br SN2 HO Br OH -, EtOH E2

49 6.20 Which of the following reaction is more suitable for the preparation of CH3OCH(CH3)2 Answer: 2) Is better than 1), because 對於 1) 來說 ; 消除反應佔優勢 write the major product: a) Br CH 3 CH 2 O -, CH 3 CH 2 OH, 50 o C SN2 OCH 3 b) Br (CH 3 ) 3 CO -, (CH 3 ) 3 OH, 50 o C E2 c) Br CH 3 O -, CH 3 OH, 50 o C E2 d) Br (CH 3 ) 3 CO -, (CH 3 ) 3 OH, 50 o C E1 Cl e) (H 3 C) 3 C I -, 50 o C, acetone SN2 (H 3 C) 3 C I Cl CH 3 OCH 3 f) (H 3 C) 3 C 25 o C, CH 3 OH SN1 (H 3 C) 3 C OCH 3 + (H 3 C) 3 C CH 3

50 e) Cl CH 3 O -, 50 o C, CH 3 OH E2 f) CH 3 CO - 2, CH 3 CO 2 H Cl SN2 O e) H 3 C Et H Br (R) OH -, 25 o C SN2 HO CH 3 H Et (S) O f) Cl CH 3 (S) CH 3 OH, 25 o C SN1 H 3 CO CH 3 + racemic H 3 C OCH 3 f) (S) Br H I -, 50 o C CH 3 OH SN1 I (R) SN2 H I

51 6.23 Explain: Br In SN2 reaction, Br is more than 10 time faster than because of the less steric hindrance Br Br CH 3 CH 2 O -, CH 3 CH 2 OH CH 3 CH 2 O -, CH 3 CH 2 OH OCH 2 CH 3 major product major product 對 isobutyl bromide 來講,SN2 反應會受到部分立體阻礙, 故 E2 反應在強鹼條件下佔優勢 6.24 Answer: SN Answer: in protic solvent; nucleophicility: CH3NH - >CH3NH2 (C6H5)3P>(C6H5)3N HS - >H2S CH3O - >CH3CO2 - CH3SH>CH3OH H2O>H3O + NH3>NH4 + OH - >CH3CO2 -

52 6.33 Give the structure of the products: H F NaI ( 1 mol) I F Br (1 mol) H H H Cl (1 mol) Cl NaI ( 1 mol) I Cl Br Br + + Na - S S - Na + S S Cl OH NaH Cl O - O H 3 C H NaNH 2 H 3 C C C - CH 3 I H 3 C CH 3

PowerPoint 簡報

PowerPoint 簡報第 15 章苯類化合物的有機反應 1) 苯環與親電試劑 (electrophile) 發生的取代反應 (electrophilic aromatic substitution) Br 2, C 4 Br Br a) Halogenation F 2 與苯環反應相當劇烈, 故通常得到多取代的產物 Step 1: Step 2: sp2 sp3 sp2 Arenium 是中間體 (intermediate)