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06/04/2015 1 REAKSI ELIMINASI  PELEPASAN MOLEKUL YZ DARI ATOM-ATOM C BERDAMPINGAN DLM SUATU MOLEKUL PEREAKSI.  DEHIDROHALOGENASI DAN DEHIDRASI.

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Presentasi berjudul: "06/04/2015 1 REAKSI ELIMINASI  PELEPASAN MOLEKUL YZ DARI ATOM-ATOM C BERDAMPINGAN DLM SUATU MOLEKUL PEREAKSI.  DEHIDROHALOGENASI DAN DEHIDRASI."— Transcript presentasi:

1 06/04/ REAKSI ELIMINASI  PELEPASAN MOLEKUL YZ DARI ATOM-ATOM C BERDAMPINGAN DLM SUATU MOLEKUL PEREAKSI.  DEHIDROHALOGENASI DAN DEHIDRASI

2 06/04/20152 DEHIDROHALOGENASI ALKIL HALIDA DEHIDROHALOGENASI ALKIL HALIDA PELEPASAN/ PENARIKAN HX DARI ATOM-2 C BERDAMPINGAN DLM SEBUAH ALKIL HALIDAPELEPASAN/ PENARIKAN HX DARI ATOM-2 C BERDAMPINGAN DLM SEBUAH ALKIL HALIDA DILAKUKAN DG BASA KUATDILAKUKAN DG BASA KUAT MISALNYA: CH 3 ONa, C 2 H 5 ONa DAN (CH 3 ) 3 COKMISALNYA: CH 3 ONa, C 2 H 5 ONa DAN (CH 3 ) 3 COK

3 06/04/20153 CONTOH DEHIDROBROMINASI ALKIL BROMIDA

4 06/04/20154 DEHIDRASI ALKOHOL DEHIDRASI ALKOHOL PELEPASAN/ PENARIKAN H 2 O DARI ATOM-2 C BERDAMPINGAN DLM SEBUAH ALKOHOLPELEPASAN/ PENARIKAN H 2 O DARI ATOM-2 C BERDAMPINGAN DLM SEBUAH ALKOHOL DILAKUKAN DG ASAM KUATDILAKUKAN DG ASAM KUAT MISALNYA: H 2 SO 4 DAN H 3 PO 4MISALNYA: H 2 SO 4 DAN H 3 PO 4

5 06/04/20155 MEKANIME REAKSI ELIMINASI 1.REAKSI ELIMINASI BIMOLEKULER (E2) r = k [R-X].[:B - ] r = k [CH 3 CHBrCH 3 ].[C 2 H 5 O - ] 2.REAKSI ELIMINASI UNIMOLEKULER (E1) r = k [R-X] r = k [(CH3) 3 C-Cl] r = k [(CH3) 3 C-Cl]

6 06/04/20156 REAKSI E2  DEHIDROBROMINASI ISOPROPIL BROMIDA DG LARUTAN NATRIUM ETOKSIDA DLM ETANOL  

7 06/04/20157 MEKANISME REAKSI E2 1.BASA MENYERANG AT. H  DARI ARAH BERLAWANAN DG X (Br) 2.AT. X (Br)  PERGI DR ARAH BERLAWANAN DG BASA SBG ION X - (Br - ) 3.AT-2 C  DAN  MEMBENTUK IKATAN RANGKAP MENGHASILKAN ALKENA

8 06/04/20158 REAKSI E1  DEHIDROBROMINASI t-BUTIL BROMIDA DG LARUTAN NATRIUM ETOKSIDA DLM ETANOL

9 06/04/20159 MEKANISME REAKSI E1 1.ALKIL HALIDA MENGALAMI IONISASI MENGHASILKAN ION KARBONIUM 2.BASA MENYERANG AT. H  3.AT C  MEMBENTUK IKATAN RANGKAP DENGAN AT. C  MENGHASILKAN ALKENA

10 06/04/ KOMPETISI S N 2 DAN E2 PD HALIDA 3 o KENAIKAN T TERJADI REAKSI E2

11 06/04/ KERUAHAN BASA PEREAKSI 1.KERUAHAN BASA PEREAKSI MENDORONG REAKSI ELIMINASI

12 06/04/ KEBASAAN 1.BASA LEMAH SPT Cl -, CH 3 COO -, Br -, I - MENDORONG S N 2 2.BASA KUAT: C 2 H 5 O -, OH -, NH 2 - MENDORONG E2

13 06/04/ KOMPETISI S N 2 DAN E2 1.REAKSI S N 2 MELIBATKAN BASA LEMAH, BASA SEDERHANA DLM PELARUT POLAR, DAN SUHU RENDAH 2.REAKSI E2 MELIBATKAN BASA KUAT, BASA MERUAH, DLM PELARUT POLAR, DAN SUHU TINGGI.

14 06/04/ BAGAIMANA KOMPETISI S N 1 DAN E1 1.REAKSI E1 MELIBATKAN KARBOKATION STABIL, BASA LEMAH, PELARUT POLAR DAN SUHU TINGGI 2.REAKSI S N 1 MELIBATKAN KARBOKATION STABIL, BASA LEMAH, PELARUT POLAR DAN SUHU RENDAH

15 06/04/ ATURAN ZAITSEV 1.DEHIDROBROMINASI 2-BROMO-2-METILBUTANA DG NATRIUM ETOKSIDA 2.REAKSI ELIMINASI TERJADI DG MEMBERIKAN ALKENA YG LEBIH TERSUBSTITUSI.   

16 06/04/ BUKTI LAIN ATURAN ZAITSEV 1.DEHIDROKLORINASI NEOMENTIL KLORIDA DAN MENTIL KLORIDA DG NATRIUM ETOKSIDA

17 06/04/ ANTI ZAITSEV 1.DEHIDROBROMINASI 2-BROMO-2- METILBUTANA DG KALIUM t-BUTOKSIDA 2.ELIMINASI DG BASA MERUAH MEMBERIKAN ALKENA YG KURANG TERSUBSTITUSI.  

18 06/04/ BAGAIMANA KOMPETISI S N 1 DAN E1 1.REAKSI E1 MELIBATKAN KARBOKATION STABIL, BASA LEMAH, PELARUT POLAR DAN SUHU TINGGI 2.REAKSI S N 1 MELIBATKAN KARBOKATION STABIL, BASA LEMAH, PELARUT POLAR DAN SUHU RENDAH

19 06/04/ PROBLEM 1 WHEN CIS -1-BROMO-4-tert-BUTYLCYCLOHEXANE IS TREATED WITH SODIUM ETHOXIDE IN ETHANOL, IT REACT RAPIDLY. THE PRODUCT IS 4-tert- BUTYLCYCLOHEXENE. UNDER THE SAME CONDITIONS, TRANS-1-BROMO-4-tert- BUTYLCYCLOHEXANE REACTS VERY SLOWLY. WRITE CONFORMATIONAL STRUCTURE AND EXPLAIN THE DIFFERENCE IN REACTIVITY OF THIS CIS-TRANS ISOMERS.WHEN CIS -1-BROMO-4-tert-BUTYLCYCLOHEXANE IS TREATED WITH SODIUM ETHOXIDE IN ETHANOL, IT REACT RAPIDLY. THE PRODUCT IS 4-tert- BUTYLCYCLOHEXENE. UNDER THE SAME CONDITIONS, TRANS-1-BROMO-4-tert- BUTYLCYCLOHEXANE REACTS VERY SLOWLY. WRITE CONFORMATIONAL STRUCTURE AND EXPLAIN THE DIFFERENCE IN REACTIVITY OF THIS CIS-TRANS ISOMERS.

20 06/04/ REAKSINYA

21 06/04/ JAWAB PROBLEM 1 PD ISOMER CIS TERDAPAT 2 AT H  DG POSISI AKSIAL, BEGITU JUGA Br BERADA PD POSISI AKSIAL SHG REAKSI E2 BERLANGSUNG LEBIH CEPAT.PD ISOMER CIS TERDAPAT 2 AT H  DG POSISI AKSIAL, BEGITU JUGA Br BERADA PD POSISI AKSIAL SHG REAKSI E2 BERLANGSUNG LEBIH CEPAT. PD ISOMER TRANS BAIK AT H  MAUPUN Br YG AKAN BEREAKSI DLM POSISI EQUATORIAL, SEHG MENYEBABKAN REAKSI E2 SANGAT LAMBAT (SULIT).PD ISOMER TRANS BAIK AT H  MAUPUN Br YG AKAN BEREAKSI DLM POSISI EQUATORIAL, SEHG MENYEBABKAN REAKSI E2 SANGAT LAMBAT (SULIT).

22 06/04/ REAKSINYA

23 06/04/ PROBLEM 2 A). When cis-1-bromo-2-methylcyclohexane undergoes an E2 reaction, two product (cycloalkenes) are formed. What are these two cycloalkenes, and which would you expect to be the major product? Write conformational structures showing how each is formed.A). When cis-1-bromo-2-methylcyclohexane undergoes an E2 reaction, two product (cycloalkenes) are formed. What are these two cycloalkenes, and which would you expect to be the major product? Write conformational structures showing how each is formed. B) When trans-1-bromo-2-methylcyclohexane reacts in an E2 reaction, only one cycloalkene is formed. What is this product? Write conformational structures showing why is the only product.B) When trans-1-bromo-2-methylcyclohexane reacts in an E2 reaction, only one cycloalkene is formed. What is this product? Write conformational structures showing why is the only product.

24 06/04/ KESIMPULAN CH 3 X METIL RCH 2 X (1 o ) (R) 2 CHX (2 o ) (R) 3 CX (3 o ) REAKSI BIMOLEKULER S N 1/E1/E2 MEMBERIKAN REAKSI S N 2 MEMBERI S N 2, KECUALI DG BASA MERUAH MEMBERI REAKSI E2 MEMBERIKAN REAKSI S N 2 DG BASA LEMAH DLM SOLVOLISIS BEREAKSI S N 1/E1, PD T RENDAH S N 1 DOMINAN MEMBERIKAN REAKSI E2 DG BASA KUAT BASA KUAT REAKSI E2 DOMINAN

25 06/04/ A BIOLOGICAL NUCLEOPHILIC SUBSTITUTION REACTION: BIOLOGICAL METHYLATION 1.THE CELLS OF LIVING ORGANISMS SYNTHESIZE MANY OF THE COMPOUNDS THEY NEED FROM SMALLER MOLECULES. 2.A NUMBER OF REACTIONS TAKE PLACE IN THE CELLS OF PLANTS AND ANIMALS THAT INVOLVE THE TRANSFER OF A METHYL GROUP FROM AN AMINO ACID CALLED METHIONINE TO SOME OTHER COMPOUND.

26 06/04/ SOME OF THE COMPOUNDS THAT GET THEIR METHYL GROUP FROM METHIONINE

27 06/04/ METHYL TRANSFER MECHANISM

28 06/04/ BIOSYNTHESIS OF CHOLINE

29 06/04/ DEHIDRASI ALKOHOL 1.PELEPASAN H 2 O DR ALKOHOL-2 MELALUI PEMANASAN DG ASAM KUAT 2.ASAM BRONSTED: H 2 SO 4 DAN H 3 PO 4 ASAM LEWIS: ALUMINA (Al 2 O 3 ) DLM INDUSTRI 3.REAKSI DEHIDRASI ALKOHOL MENUNJUKKAN BBP KARAKTERISTIK PENTING.

30 06/04/ KONDISI REAKSI 1.ALKOHOL PRIMER MEMERLUKAN ASAM PEKAT DAN SUHU TINGGI

31 06/04/ CAMPURAN ALKENA 1.BBP ALKOHOL TERDEHIDRASI MEMBERIKAN HASIL CAMPURAN ALKENA 2.MENGIKUTI ATURAN ZAITZEV

32 06/04/ PENATAAN ULANG BBP ALKOHOL 1 o DAN 2 o MENGALAMI PENATAAN ULANG SELAMA DEHIDRASIBBP ALKOHOL 1 o DAN 2 o MENGALAMI PENATAAN ULANG SELAMA DEHIDRASI

33 06/04/ MEKANISME UMUM DEHIDRASI ALKOHOL TERKATALISIS ASAM MENGIKUTI MEKANISME E1MENGIKUTI MEKANISME E1

34 06/04/ MEKANISME PENATAAN ULANG PENATAAN ULANG SELAMA DEHIDRASI DARI 3,3- DIMETIL-2-BUTANOLPENATAAN ULANG SELAMA DEHIDRASI DARI 3,3- DIMETIL-2-BUTANOL

35 06/04/ FINAL PRODUCT

36 06/04/ BEBERAPA PERGESERAN DLM PENATAAN ULANG ION KARBONIUM

37 06/04/ DEBROMINATION OF VICINAL DIBROMIDES 1.VIC-DIBROMIDES UNDERGO THE LOSS OF A MOLECULE OF BROMINE (Br 2 )WHEN THEY ARE TREATED WITH A SOLUTION OF SODIUM IODIDE IN ACETON OR MIXTURE OF ZINC DUST IN ETHANOL. 2.DEBROMINATION BY SODIUM IODIDE TAKE PLACE BY AN E2 MECHANISM.

38 06/04/ THE MECHANISM OF DEBROMINATION

39 06/04/ PROBLEM 1 Although ethyl bromide and isobutyl bromide are both primary halides, ethyl bromide undergoes S N 2 reactions more than ten times faster than isobutyl bromide. When each compound is treated with a strong base (CH 3 CH 2 O - ), isobutyl bromide gives a greater yield of elimination products than substitution products, whereas with ethyl bromide this behavior is reversed. What factor accounts for these results?Although ethyl bromide and isobutyl bromide are both primary halides, ethyl bromide undergoes S N 2 reactions more than ten times faster than isobutyl bromide. When each compound is treated with a strong base (CH 3 CH 2 O - ), isobutyl bromide gives a greater yield of elimination products than substitution products, whereas with ethyl bromide this behavior is reversed. What factor accounts for these results?

40 06/04/ PROBLEM 2 Consider the reaction of I - with CH 3 CH 2 Cl. (a). Would you expect the reaction to be S N 1 or S N 2? The rate constant for the reaction at 60 o is 5x10 -5 liter mole -1 sec -1. (b). What is the reaction rate if [I - ] = 0.1 mole liter -1 and [CH 3 CH 2 Cl] = 0.1 mole liter -1 ? (c). If [I - ] = 0.1 mole liter -1 and [CH 3 CH 2 Cl] = 0.2 mole liter -1 ? (d). If [I - ] = 0.2 mole liter -1 and [CH 3 CH 2 Cl] = 0.1 mole liter -1 ? (e). If [I - ] = 0.2 mole liter -1 and [CH 3 CH 2 Cl] = 0.2 mole liter -1 ?Consider the reaction of I - with CH 3 CH 2 Cl. (a). Would you expect the reaction to be S N 1 or S N 2? The rate constant for the reaction at 60 o is 5x10 -5 liter mole -1 sec -1. (b). What is the reaction rate if [I - ] = 0.1 mole liter -1 and [CH 3 CH 2 Cl] = 0.1 mole liter -1 ? (c). If [I - ] = 0.1 mole liter -1 and [CH 3 CH 2 Cl] = 0.2 mole liter -1 ? (d). If [I - ] = 0.2 mole liter -1 and [CH 3 CH 2 Cl] = 0.1 mole liter -1 ? (e). If [I - ] = 0.2 mole liter -1 and [CH 3 CH 2 Cl] = 0.2 mole liter -1 ?

41 06/04/ PROBLEM 3 When tert-butyl bromide undergoes S N 1 hydrolisis, adding a “common ion” (i.e NaBr) to the aqueous solution has no effect on the rate. On the other hand when (C 6 H 5 ) 2 CHBr undergoes S N 1 hydrolisis, adding NaBr retards the reaction. Given that the (C 6 H 5 ) 2 CH + cation is known to be much more stable than the (CH 3 ) 3 C + cation, provide an explanation for the different behavior of the two compounds.When tert-butyl bromide undergoes S N 1 hydrolisis, adding a “common ion” (i.e NaBr) to the aqueous solution has no effect on the rate. On the other hand when (C 6 H 5 ) 2 CHBr undergoes S N 1 hydrolisis, adding NaBr retards the reaction. Given that the (C 6 H 5 ) 2 CH + cation is known to be much more stable than the (CH 3 ) 3 C + cation, provide an explanation for the different behavior of the two compounds.


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