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1 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

3 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:

4 Step 2: sp2 sp3 sp2 Arenium 是中間體 (intermediate)

5 Step 3:

6 b) Nitrition: HN 3, H 2 S 4 N 2 + H HS 4 - Step 1: generation of electrophile H S H + H N - H S - + H N H - H2 + N Step 2: Electrophilic addition:

7 N H N - Step 3: Regeneration of aromatic system: H 2 H N 2 + H3 + N -

8 c) Sulfonation: Fuming sulfuric acid H S H + H S H H S - + H H S H S + H 3 +

10 d) Friedel-Craft Alkylation 在苯環上引入烷基 + Al 3 + HF 解釋反應機制 + HF + H BF 3

12 Page 673; 解釋反應機制 : + HF Why not

13 e) Friedel-Craft Acylation 反應在苯環上引入酰基 (acyl group: RC) Al 3 + R + H R Explain the mechanism H 3 C + H 3 C Al 3 CH 3 + H 3 C H

14 H 2

15 Page 675; drawing the mechanism: H 3 C + H 3 C Al 3 CH 3 + H 3 C H

16 Drawing the mechanism

17 Zn/ Hg

18 設計合成途徑 H CH 3

19 + Al 3 H S 2 2 H Al 3 Hg/Zn

20 f) Friedel-Craft 反應的應用所受到的限制親電試劑的重排

21 Page 677; 給出主產物並解釋反應機制 + Al 3 + H HF

22 + Al 3 Hg/Zn Page 677; 已知起始物合成下列化合物 + Al 3 Hg/Zn H S 2 + Al 3

23 g) 取代基效應 : i) 當苯環上有以下取代基時, 反應活性降低

24 ii) 多重烷基化反應的發生 iii) Polyacylation does not occur because the acyl group deactivates the aromatic ring to further substitution

25 在苯類化合物的親核取代反應中, 若取代基的存在使得取代苯比苯本身更為 active, 則此類取代基被稱為 activating groups (ortho-para directors); 反之則稱為 deactivating group (meta director). 比苯的反應更容易進行, 故 CH 3 為 activating group 反應速度只有苯的 10-4, 故 N 2 為 deactivating group

27 鹵族元素取代基 (Br, ) 為一特例 ; 它們是 deactivating group, 但可使反應發生在 ortho-para 位

32 Page 681 exercise: Predict the major producs: CH 3 fume H 2 S 4 CH 3 S 3 H + CH 3 S 3 H C 2 H C 2 H H 2 S 4, HN 3 N 2 N 2 N 2 Br 2, FeBr 3 Br H H H CH 3 C + Al 3

33 Halo groups are ortho-para directors but are also deactivating * The electron-withdrawing inductive effect of the halide is the primary influence that deactivates haloaromatic compounds toward electrophilic aromatic substitution **The electron-donating resonance effect of the halogen's unshared electron pairs is the primary ortho-para directing influence

34 g) 取代反應之應用實例 ( ): When designing a synthesis of substituted benzenes, the order in which the substituents are introduced is crucial 如何由苯製備下列化合物 : Br N 2 Br Br N 2 N 2 Br Br Br Br 2, FeBr 3 H 2 S 4, HN 3 + N 2 N 2 N 2 N 2 H 2 S 4, HN 3 Br 2, FeBr 3 Br

35 如何由甲苯製備下列化合物 : C 2 H C 2 H N 2 C 2 H N 2 N 2

36 如何由甲苯製備下列化合物 : C 3 Br C 3 C 3 Br Br CH 3 CH 3 CH 3 Br2, Fe3 Br + Br 2, light 2, light C 3 Br C 3 Br CH 3 C 3 C 3 2, light Br2, Fe3 Br

37 ? CH 2 CH 3 2, Al 3 Zn/Hg

38 The concept of the protecting group

39 h) rientation in disubstituted benzene: i) When two substituents are present on the ring initially, the more powerful activating group generally determines the orientation of subsequent substitution ii) rtho-para directors determine orientation over meta directors iii) Substitution does not occur between meta substituents due to steric hindrance Exercise 703

40 Exercise on page 703 H CN S 3 H CF 3 CH 3 N 2

41 2) Benzylic 的自由基和正碳離子反應

43 a) 自由基反應 Drawing the mechanisms of the above reactions

44 2, C 4, light major product 2 2 H H H H H H

45 Br NBS, C 4, light Explain major product b) Benzylic 的正碳離子反應

46 3) 脫水反應 Dehydration of the alcohol below yields only the more stable conjugated alkenyl benzene 4) 氧化反應

48 5)Benzylic halide 的 SN2, SN1 反應

50 6) 苯類化合物的還原反應 Birch reduction

51 15.26: give the major product: a) b) 2, Fe 3 2, Fe F F c) 2, Fe 3 C 2 H C 2 H d) 2, Fe 3 N 2 N 2 e) 2, Fe 3 e) 2, Fe 3 + f) 2, Fe 3 + g) Et Et Et 2, Fe 3 +

52 15.27: give the major product: a) NHCCH 3 H 2 S 4, HN 3 NHCCH 3 d) C 2 H H 2 S 4, HN 3 C 2 H 2 N 2 N b) CCH 3 H 2 S 4, HN 3 C 2 CH 3 2 N e) H 2 S 4, HN 3 c) C 2 H H 2 S 4, HN 3 C 2 H N 2 N : Predict the major product(s):

53 a) H f) H 2 ( 1 molar equivalent) Pt Br b) CH 3 - Na + c) HA g) KMn4, H-, heat H 3 + H heat H d) e) Peroxide HBr HA, H 2 H Br

54 15.29: From benzene, synthesize the following compounds: a) Br FeBr 3 b) FeBr 3 ) Zn/Hg Fe 3 d) Zn/Hg Fe 3

55 e) 2, Fe 3 HA N 2 N 2 H 2 S 4, HN 3 H2S4, HN3 h) N 2 i) H 2 S 4, HN 3 N 2 Br 2, FeBr 3 N 2 Br Br 2, FeBr 3 j) H 2 S 4, HN 3 Br 2 N Br 2 Fe 3 j) CH 3 Br, FeBr 3 KMn 4, H -, heat H 2 C H 3 +

56 k) 2 Fe 3 H 2 S 4 H 3 S l) 2 Fe 3 H 2 S 4, HN 3 2 N N 2 m) H 2 S 4 H2 S 4, HN 3 S 3 H S 3 H

酸檸檬？鹼檸檬？?食物的酸鹼性.doc

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