1. REAKSI ELIMINASI

2. suatu reaksi dimana substrat kehilangan unsur dari molekul kecil seperti HCl, H 2 O atau Cl 2 selama reaksi pembentukan produk REAKSI ELIMINASI - HCl C C H Cl C C

3. REAKSI ELIMINASI Alkil halida + basa kuat dan panas Alkohol + asam kuat dan panas Kehilangan HCl Kehilangan H 2 O    DUA CONTOH E2 E1

4. + HCl - HCl ELIMINASI LAWAN DARI ADISI Kondisi basa + panas Kondisi asam interkonversi dari alkil halida dan alkena ALKIL HALIDA ALKENA C H 3 C H 2 C H C H 3 C l C H 3 C H C H C H 3 HCl pekat NaOH + panas

5. + H 2 O - H 2 O Kondisi asam kuat + panas Kondisi asam encer 2-6 M H 2 SO 4 H 2 SO 4 pekat interkonversi dari alkohol dan alkena ALKOHOL ALKENA ELIMINASI LAWAN DARI ADISI C H 3 C H C H C H 3 C H 3 C H 2 C H C H 3 O H

6. BASA KUAT

7. BASA KUAT NaOH water Basa pelarut yang diijinkan KOH water, MeOH, EtOH NaOR ROH (same R group) NaNH 2 NH 3 (liq) -33 o C Digunakan untuk reaksi eliminasi Halida (RX) tidak larut dalam air, tetapi larut dalam alkohol, Karena itu KOH atau NaOR dalam alkohol sering digunakan

8. ALKIL HALIDA + BASA KUAT + PANAS E2

9. SUMMARY TO DATE BASA KUAT Dibutuhkan REGIOSELECTIVE Mengikuti aturan Zaitsev Rule ) - membentuk alkena yang lebih tersubstitusi STEREOSPECIFIC  -H dan X harus ANTI-COPLANAR REAKSI ELIMINASI ALKIL HALIDA Panas Selalu dibutuhkan

10. MEKANISME SERENTAK = HANYA SATU TAHAP Semua ikatan terputus dan terbentuk tanpa pembentukan suatu intermediet Eliminasi Bimolekular Basa kuat alkil halida E 2

11. Orde Reaksi Molekularitas Tahap Penentu Kecepatan Keadaan Transisi Kompleks Teraktivasi Jumlah eksponen dari konsentrasi pada persamaan kecepatan Jumlah spesi yang ada pada tahap penentu kecepatan Tahap paling lambat dalam rangkaian reaksi Energi tertinggi dalam profil energi suatu reaksi Spesi yang ada pada keadaan transisi

12. REAKSI SERENTAK E2 Satu Tahap – tanpa Intermediet

13. mekanisme Kompleks teraktivasi   ELIMINASI E2 Serentak: semua terjadi pada satu waktu, tanpa suatu intermediet C H C H B r H B :

14. Reaksi Serentak (Satu Tahap) product starting material Keadaan Transisi TS Energi Aktivasi Ea Panas Reaksi  H E N E R G Y

15. C C H Br R R H H O CH 3 : .. .. .. .. .. : : Serangan basa pada hydrogen-  memulai reaksi sp 3 sp 3 Ketika elektron ini memasuki orbital tetangga, maka proses ini akan mendorong pasangan ikatan keluar The critical event is the removal of the  -H.

16. C C H Br R R H H O CH 3 .. .. .. : : .. . . Pembentukan ikatan rangkap dan Kehilangan atom bromida 2p 2p

17. REAKSI ADALAH ELIMINASI-  Gugus fungsi terikat Ke karbon-  .  -carbon  -carbon Hydrogen -  terikat Ke karbon-  . Karena hydrogen-  hilang dalam reaksi ini Maka disebut Eliminasi-  . Reagent = Basa kuat

18. C C C l H B: Basa mengambil HYDROGEN-  C C C l H B : : .. : : .. .. MEKANISME

19. REGIOSELECTIVITY ALKIL HALIDA + BASA KUAT (E2)

20. APA YANG TERJADI JIKA LEBIH DARI SATU HYDROGEN-  ?   ’ MANA YANG HILANG ?

21. ELIMINASI ADALAH REGIOSELECTIVE   ’  -H  ’ -H Produk Mayor adalah Yang punya energi terendah 2-bromobutana 2-butena 1-butena major product 81 % minor product 19 %

22. major product minor product   ’’  ’  ’’ =    ’ MORE REGIOSELECTIVITY identical to  -product  ’’ 1-chloro-1-methylcyclohexane 1-methylcyclohexene methylenecyclohexane 1-methylcyclohexene 3 kemungkinan Kehilangan hydrogen- 

23. -27.8 -25.4  H +H 2 +H 2 kcal mole METHYLCYCLOHEXENE ISOMERS Both are hydrogenated to the same product therefore their energies may be compared. methylcyclohexane RECALL

24. ATURAN SAYTZEV Reaksi yang memberikan alkena yang lebih tersubstitusi (energi terendah) sebagai produk mayor

25. monosubstitusi trisubstitusi disubstitusi decreasing energy tetrasubstitusi Peningkatan substitusi ISOMER LEBIH TERSUBSTITUSI LEBIH STABIL ISOMER ALKENA Perbedaan posisi dari ikatan rangkap cis trans 1,1- 1,2- 1,2- H H R H H H R R R H R H H R R H R H R R R R R R

26. major product 81 % minor product 19 % major product minor product APLIKASI ATURAN ZAITZEV DISUBSTITUSI TRISUBSTITUSI DISUBSTITUSI MONOSUBSTITUSI 2-bromobutane 1-chloro-1-methylcyclohexane

27. STEREOKIMIA REAKSI ELIMINASI ALKYL HALIDE + STRONG BASE (E2)

28. STEREOKIMIA TWO EXTREME POSSIBILITIES FOR THE ELIMINATION PROCESS C C H Cl syn elimination C C H Cl anti elimination tidak umum terjadi sering terjadi H H Cl Cl anti-coplanar

29. 1-Bromo-2-methylcyclohexane major product Zaitsev The expected result (naïve) : minor The result actually depends on the stereochemistry of the starting material ( cis or trans ). drawn flat without stereochemistry

30. cis The elimination needs to have H and Br anti-coplanar plus a small amount of major product (Zaitsev) Br CH 3 THE CIS STEREOISOMER The other chair won’t work. Why? trans shown on next slide

31. trans only product Br CH 3 THE TRANS STEREOISOMER The other chair won’t work. Why? no methylcyclohexene is formed

32. ALKIL HALIDA + BASA LEMAH (SOLVOLYSIS) E1 Pemindahan hydrogen-  menjadi sukar tanpa basa kuat, Sehingga reaksi berjalan denngan mekanisme berbeda

33. Reaksi Eliminasi E1(dua tahap) + : X slow fast rate = k [RX] karbokation 3 o > 2 o > 1 o Bekerja baik dalam pelarut polar Pembentukan ion unimolecular Tahap 1 Tahap 2 Basa lemah

34. starting material product Ea 1 Ea 2  H intermediate TS 2 TS 1 PROFIL ENERGi two step reaction E N E R G Y step 1 step 2 carbocation slow E1

35. STEREOSPECIFICITY

36. C X C H C X C H C C H C C H + + Carbocation is sp 2 hybridized ( planar ) and can react from either side. These two carbocations are equivalent by rotation and by symmetry. anti syn REAKSI E1 TIDAK STEREOSPECIFIC rotation rate of C-C rotation = 10 10 to 10 12 / sec ( karbokation terbuka adalah palanar dan bisa berputar) Elimination can be either syn or anti .

37. REGIOSELECTIVITY

38. REAKSI E1 ADALAH REGIOSELECTIVE MENGIKUTI ATURAN ZAITSEV major minor 0.001 M KOH / EtOH tertiary trisubstitusi (stereochemistry is not a problem as in E2) disubstitusi very dilute base Zaitsev

39. MEKANISME ELIMINASI LAIN

40. 3 jenis reaksi eliminasi : E2 E1 E1cb serentak halogen first proton second proton first halogen second carbocation carbanion

41. PERBEDAAN ANTARA E1 DAN E2

42. [RX] constant, [B] increasing Rate rate = k 1 [RX] E1 rate = k 2 [RX] [B] E2 KECEPATAN DENGAN PENINGKATAN KONSENTRASI BASA second order first order E1 terjadi pada Konsentrasi basa rendah E2 pada konsentrasi basa lebih tinggi

43. [RX] constant, [Base] increasing Rate PENGARUH KONSENTRASI BASA PADA REAKSI E1/E2 secondary RX, k’ tertiary RX, k’’ primary RX, k k 1 [RX] E1 For E1 elimination : k’’ (tertiary) > k’ (secondary) > k (primary). k 2 [RX] [B] E2 1 1 1 At high base concentration E1 never has a chance. At low base concentration E2 is nonexistent

44. secondary RX, k’ tertiary RX, k’’ primary RX, k [RX] constant, [B] Rate k 1 [RX] E1 k 2 [RX] [B] E2 For E2 elimination : line slopes k 2 differ for 1 o ,2 o ,3 o . Different substrates react at different rates, primary secondary tertiary 1 1 2 k 2 k’ k’’ 2 PENGARUH KONSENTRASI BASA PADA REAKSI E1/E2

45. Obviously for E1 which forms a carbocation intermediate kecepatan : tertiary > secondary > primary > methyl But this same order holds for E2 also. STRUKTUR SUBSTRAT R-C-X R R R-C-X R H R-C-X H H primary secondary tertiary tertiary has more  -hydrogens EtO - more opportunites for reaction

46. Jika basa kuat ada pada konsentrasi tinggi, atau substrat adalah halida 1 0 , reaksi berjalan E2 KAPAN MEKANISME E1 TERJADI ??? E1 terjadi hanya: 1) Pada konsentrasi basa rendah 2) Dengan solvolisis (pelarut adalah basa) 3) Dengan substrat tersier dan beresonasi (alkil halida)

47. Mekanisme E2 Mekanisme E1 Basa kuat Kons basa tinggi. Basa lemah Kons basa rendah ALKIL HALIDA + BASA solvolysis Harus membentuk karbokation baik or Membutuhkan anti-coplanar (pelarut basa) stereospecific regioselective not stereospecific regioselective

48. EXAMPLES N a O E t E t O H E t O H 6M K O H 0.01 M E2 E1 rate = k [RBr] rate = k [RBr] [OEt] C H B r C H 3 C H B r C H 3 C H C H 2 C H C H 3 + .. :O E t ..

49. PENATAAN ULANG KARBOKATION

50. JENIS PENATAAN ULANG KARBOKATION Perpindahan metil Perpindahan hidrogen Perpindahan fenil Gugus pindah dengan elektron ikatannya Pergeseran 1,2-metil Pergeseran 1,2-hydrogen Pergeseran 1,2-phenyl

51. methyl shift protonation loss of H 2 O loss of H + Penataan ulang – pergesaran 1,2-METIL H 2 SO 4

52. Kenapa terjadi penataan ulang ? Ion sekunder Ion tersier Karbokation menata ulang Untuk mencapai energi terendah Energi karbokation 3 o < 2 o < 1 o < CH 3 + energy decrease lowest highest

53. Gugus mana yang pindah ? tertiary benzylic Gugus yang memberikan Karbokation terbaik yang Akan pindah yes secondary benzylic secondary no no Kompetisi antara H atau Me atau Ph ? BINGO !

54. DO YOU HAVE A CARBOCATION? CAN YOU FORM A BETTER CARBOCATION ? STOP E X R E M E R E G N A D STOP - LOOK - THINK ALWAYS EVALUATE FOR A REARRANGEMENT

55. REAKSI HOFMANN E2

56. ATURAN HOFMANN When you have a bulky leaving group like -N(CH 3 ) 3 + the least-substituted alkene will be the major product. BULKY = Branched at the first atom attached to the chain OTHER GROUPS FOLLOW THE ZAITSEV RULE trimethyl ammonium dimethyl sulfonium chain chain Big is not the same as bulky.

57. HOFMANN ELIMINATION Hofmann found that when the leaving group was -N(CH 3 ) 3 + E2 elimination reactions gave the least-substituted alkene. 95% 5% 31% 69% Hofmann Zaitsev

58. 31% 69% 30% 70% 48% 52% 87% 13% 98% 2% HOFMANN ZAITSEV EFFECT OF INCREASING SUBSTITUENT BULK ( cis + trans ) E2 Big is not the same as bulky.

59. C C C C C H 3 X H H HOFMANN ZAITSEV cis trans ( all H equivalent ) bulky groups cause crowding and give Hofmann products ANALYSIS OF 2-SUBSTITUTED PENTANE ELIMINATIONS less crowding in this area of the molecule

60. cis trans ZAITSEV PRODUCTS HOFMANN PRODUCT N(CH 3 ) 3 CH 2 CH 2 CH 3 H N(CH 3 ) 3 CH 3 CH 2 CH 3 H CH 3 CH 2 CH 3 N(CH 3 ) 3 H most steric crowding no steric crowding less steric crowding H H H H H H H WOULD MAKE WOULD MAKE NOT FORMED FORMED

61. BULKY BASES ALSO INCREASE HOFMANN PRODUCT 81% 19% 47% 53% ZAITSEV HOFMANN bulky base t -butoxide methoxide

62. BULKY  -SUBSTITUENTS  80% 20% 79% 21% 14% 86% ZAITSEV HOFMANN NaOEt NaOEt NaOEt    What constitutes bulky? NO NO YES t-butyl is bulky ! a methyl group is not bulky even two or three are not bulky

63. crowded less crowded THE ELIMINATION MOVES TO A LESS CROWDED REGION 86% 14% spacer REACTIVE CONFORMATIONS crowding crowding H H CH 3 Br C CH 3 CH 3 CH 3 CH 3 C H H CH 3 H Br CH 2 CH 3 CH 3 CH 3

64. EXAMPLES

65. Br CH 3 CH 3 CH 3 N(CH 3 ) 3 + I - CH 2 Br CH 3 CH 2 CH 3 NaOEt EtOH /  KOH EtOH /  NaOtBu tBuOH /  + ~90% ~90% ~60/40% Zaitsev Hofmann NORMAL BULKY LEAVING GROUP BULKY BASE HOW THE VARIOUS FACTORS AFFECT THE OUTCOME Bulky base alone not as effective as bulky leaving group Prototypical “Hofmann” elimination Bromine is big, not bulky

66. Br CH 3 CH 3 N(CH 3 ) 3 + I - CH 2 NaOtBu tBuOH /  BULKY BASE & LEAVING GROUP ~100% Hofmann tBu CH 3 NaOEt EtOH /  BULKY  -SUBSTITUENT HOW THE VARIOUS FACTORS AFFECT THE OUTCOME ( CONTINUED ) tBu either cis or trans to Br - same result H CH 2 tBu + no double bond here Double Whammy ! Favors Hofmann products Bulky base + bulky leaving group Use a bulky base here and ...

67. E2 REACTIONS DEVIATE FROM THE ZAITSEV RULE 1. If the favored  -hydrogen can’t achieve anti-coplanar geometry. 2. If the double bond would form at a bridgehead. 3. If there is a bulky leaving group. 4. If there is a bulky base. 5. If there is a bulky  -substituent. (see next slides)

68.  -ELIMINATION occurs when the substrate has NO  -HYDROGENS on these carbon atoms

69. ELIMINATIONS SUMMARY

70. K.I.S.S. alkyl halide + strong base + heat = E2 alkyl halide + solvent + heat (solvolysis) = E1 alcohol + strong acid + heat = E1 (acid assisted) typical situation for E1cb H next to C=O (easy to remove) X = strong base (difficult to break bond) Only E1 reactions have rearrangements (carbocations) Only E2 reactions require anti-coplanar  -hydrogens THE MOST BASIC STUFF

71. E1cb E2 E1 E1 E1 acid assisted strong strong weak base base base “ solvolysis” Zaitsev if stereochem allows Zaitsev Zaitsev Zaitsev stereospecific anti-coplanar alkyl halides alcohols special special case - not common acidic neutral carbocation rearrangements concerted stepwise - carbocation requires: acidic H and poor leaving group stepwise - carbanion not stereospecific  -elim. if no  -H Hofmann if bulky groups THE BIG PICTURE

72. THERE IS A RANGE OF DIFFERENT MECHANISMS FOR   -ELIMINATION REACTIONS SUMMARY

73. E1cb E2 E1 E1 E1 acid assisted strong strong weak base base base “ solvolysis” Zaitsev if stereochem allows Zaitsev Zaitsev Zaitsev stereospecific anti-coplanar alkyl halides alcohols special special case - not common acidic neutral carbocation rearrangements concerted stepwise - carbocation COMPARISON OF  -ELIMINATION MECHANISMS requires: acidic H and poor leaving group stepwise - carbanion not stereospecific

74. K.I.S.S. alkyl halide + strong base + heat = E2 alkyl halide + solvent + heat (solvolysis) = E1 alcohol + strong acid + heat = E1 (acid assisted) typical situation for E1cb H next to C=O (easy to remove) X = strong base (difficult to break bond) Only E1 reactions have rearrangements (carbocations) Only E2 reactions require anti-coplanar  -hydrogens