REAKSI ELIMINASI

20/04/23

1

REAKSI ELIMINASIREAKSI ELIMINASI

PELEPASAN MOLEKUL YZ DARI ATOM-PELEPASAN MOLEKUL YZ DARI ATOM-ATOM C BERDAMPINGAN DLM SUATU ATOM C BERDAMPINGAN DLM SUATU MOLEKUL PEREAKSI.MOLEKUL PEREAKSI. DEHIDROHALOGENASI DAN DEHIDROHALOGENASI DAN DEHIDRASIDEHIDRASI

C C

Y Z

C C + YZ

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• DEHIDROHALOGENASI ALKIL DEHIDROHALOGENASI ALKIL HALIDAHALIDA

• PELEPASAN/ PENARIKAN HX DARI PELEPASAN/ PENARIKAN HX DARI ATOM-2 C BERDAMPINGAN DLM ATOM-2 C BERDAMPINGAN DLM SEBUAH ALKIL HALIDASEBUAH ALKIL HALIDA

• DILAKUKAN DG BASA KUATDILAKUKAN DG BASA KUAT

• MISALNYA: CHMISALNYA: CH33ONa, CONa, C22HH55ONa DAN ONa DAN (CH(CH33))33COKCOK

2 C2H5OH + 2 Na 2 C2H5O-Na+ + H2

CH3OH + NaH CH3O-Na+ + H2

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CONTOHCONTOH

DEHIDROBROMINASI ALKIL DEHIDROBROMINASI ALKIL BROMIDABROMIDAH3C

CH Br

H3C

C2H5ONa

(C2H5OH, 55oC)

H2C C

H

CH3

+ NaBr + C2H5OH

H3C

C Br

H3C

H3C

C2H5ONa

(C2H5OH, 25oC)

H2C C

CH3

+ NaBr + C2H5OH

CH3

79%

91%

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• DEHIDRASI ALKOHOLDEHIDRASI ALKOHOL

• PELEPASAN/ PENARIKAN HPELEPASAN/ PENARIKAN H22O DARI ATOM-O DARI ATOM-2 C BERDAMPINGAN DLM SEBUAH 2 C BERDAMPINGAN DLM SEBUAH ALKOHOLALKOHOL

• DILAKUKAN DG ASAM KUATDILAKUKAN DG ASAM KUAT

• MISALNYA: HMISALNYA: H22SOSO44 DAN H DAN H33POPO44

H3C-CH2-CH2

OH

H2SO4 PEKAT

170oCH3C-CH2 CH2 + H2O

OH H3PO4 85%

165-170oC+ H2O

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MEKANIME REAKSI MEKANIME REAKSI ELIMINASIELIMINASI

1.1. REAKSI ELIMINASI BIMOLEKULER REAKSI ELIMINASI BIMOLEKULER (E2)(E2)

r = k [R-X].[:Br = k [R-X].[:B--]]

r = k [CHr = k [CH33CHBrCHCHBrCH33].[C].[C22HH55OO--]]

2.2. REAKSI ELIMINASI REAKSI ELIMINASI UNIMOLEKULER (E1)UNIMOLEKULER (E1)

r = k [R-X] r = k [R-X]

r = k [(CH3)r = k [(CH3)33C-Cl] C-Cl]

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REAKSI E2 REAKSI E2

DEHIDROBROMINASI ISOPROPIL DEHIDROBROMINASI ISOPROPIL BROMIDA DG LARUTAN NATRIUM BROMIDA DG LARUTAN NATRIUM ETOKSIDA DLM ETANOL ETOKSIDA DLM ETANOL

C2H5O:-

H

H2C C CH3

Br

H H

H2C C CH3

Br-

HC2H5O:-

..

C C

H

CH3

H

H

+ Br- C2H5OH+

PROPENAISOPROPIL BROMIDA

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MEKANISME REAKSI E2MEKANISME REAKSI E2

1.1. BASA MENYERANG AT. H BASA MENYERANG AT. H DARI ARAH DARI ARAH BERLAWANAN DG X (Br)BERLAWANAN DG X (Br)

2.2. AT . X (Br) AT . X (Br) PERGI DR ARAH BERLAWANAN PERGI DR ARAH BERLAWANAN DG BASA SBG ION XDG BASA SBG ION X- - (Br(Br--))

3.3. AT-2 C AT-2 C DAN DAN MEMBENTUK IKATAN MEMBENTUK IKATAN RANGKAP MENGHASILKAN ALKENARANGKAP MENGHASILKAN ALKENA

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REAKSI E1REAKSI E1

DEHIDROBROMINASI t-BUTIL BROMIDA DG DEHIDROBROMINASI t-BUTIL BROMIDA DG LARUTAN NATRIUM ETOKSIDA DLM LARUTAN NATRIUM ETOKSIDA DLM ETANOL ETANOL

1. H3C

CH3C

H3C

Br CH3C+H3C

H3C

+ Br-

2. C2H5O:-

CH3C+CH2

H3CH

H2C CCH3

CH3

+ C2H5O:H

2-METILPROPENA

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MEKANISME REAKSI E1MEKANISME REAKSI E1

1.1. ALKIL HALIDA MENGALAMI ALKIL HALIDA MENGALAMI IONISASI MENGHASILKAN IONISASI MENGHASILKAN ION KARBONIUMION KARBONIUM

2.2. BASA MENYERANG AT. H BASA MENYERANG AT. H

3.3. AT C AT C MEMBENTUK MEMBENTUK IKATAN RANGKAP DENGAN IKATAN RANGKAP DENGAN AT. C AT. C MENGHASILKAN MENGHASILKAN ALKENAALKENA

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KOMPETISI SKOMPETISI SNN2 DAN E22 DAN E21. C2H5ONa + H3CCH2Br

ETANOL

T=550CC2H5OCH2CH3 + H2C CH2

SN2= 90% E2 = 10%

2.

(1o)

+ NaBr

C2H5ONa +

H3C

H3C

CH BrETANOL

T=550C

C2H5O-CH-CH3 + H2C CH-CH3

SN2= 21% E2 = 79%

+ NaBr

CH3

(2o)

3. C2H5ONa +

H3C

H3C

C BrETANOL

T=250CC2H5O-C-CH3 + H2C C

SN2= 9% E2 = 91%

+ NaBr

CH3

(3o)

H3C

CH3

CH3

CH3

T=550C E2 = 100%

PD HALIDA 3o KENAIKAN T TERJADI REAKSI E2

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KERUAHAN BASA KERUAHAN BASA PEREAKSIPEREAKSI

1.1. KERUAHAN BASA PEREAKSI KERUAHAN BASA PEREAKSI MENDORONG REAKSI ELIMINASIMENDORONG REAKSI ELIMINASI

CH3O:- + CH3(CH2)15CH2CH2

Br

METANOL

DIREFLUX

CH3(CH2)15CH2CH2

CH3(CH2)15CH2 CH2

OCH399%

1%

+

CH3(CH2)15CH2CH2

Br

t-BUTILALKOHOL

40o

CH3(CH2)15CH2CH2

CH3(CH2)15CH2 CH2

OC-CH315%

85%

+CH3-C-O:-

CH3

CH3

CH3

CH3

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KEBASAANKEBASAAN

1.1. BASA LEMAH SPT ClBASA LEMAH SPT Cl--, CH, CH33COOCOO--, Br, Br--, I, I- -

MENDORONG SMENDORONG SNN22

2.2. BASA KUAT: CBASA KUAT: C22HH55OO--, OH, OH--, NH, NH22--

MENDORONG E2MENDORONG E2

CH3COO:- + CH3-CH

CH3

Br

C2H5O:-

CH3-CH

CH3

O-CCH3

O

+ Br:-

100%

CH3-CH

CH3

Br+CH2=CH

CH3+ Br:-

79%

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KOMPETISI SKOMPETISI SNN2 DAN E22 DAN E2

1.1. REAKSI SREAKSI SNN2 MELIBATKAN 2 MELIBATKAN BASA LEMAH, BASA BASA LEMAH, BASA SEDERHANA DLM PELARUT SEDERHANA DLM PELARUT POLAR, DAN POLAR, DAN SUHU RENDAHSUHU RENDAH

2.2. REAKSI E2 MELIBATKAN BASA REAKSI E2 MELIBATKAN BASA KUAT, BASA MERUAH, DLM KUAT, BASA MERUAH, DLM PELARUT POLAR, DAN PELARUT POLAR, DAN SUHU SUHU TINGGITINGGI..

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BAGAIMANA KOMPETISI SBAGAIMANA KOMPETISI SNN1 DAN 1 DAN E1E1

1.1. REAKSI E1 MELIBATKAN REAKSI E1 MELIBATKAN KARBOKATION STABIL, BASA KARBOKATION STABIL, BASA LEMAH, PELARUT POLAR DAN LEMAH, PELARUT POLAR DAN SUHU TINGGISUHU TINGGI

2.2. REAKSI SREAKSI SNN1 MELIBATKAN 1 MELIBATKAN KARBOKATION STABIL, BASA KARBOKATION STABIL, BASA LEMAH, PELARUT POLAR DAN LEMAH, PELARUT POLAR DAN SUHU RENDAHSUHU RENDAH

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ATURAN ZAITSEVATURAN ZAITSEV

1.1. DEHIDROBROMINASI 2-BROMO-2-DEHIDROBROMINASI 2-BROMO-2-METILBUTANA DG NATRIUM ETOKSIDAMETILBUTANA DG NATRIUM ETOKSIDA

2.2. REAKSI ELIMINASI TERJADI DG REAKSI ELIMINASI TERJADI DG MEMBERIKAN ALKENA YG LEBIH MEMBERIKAN ALKENA YG LEBIH TERSUBSTITUSI.TERSUBSTITUSI.

C2H5O:-

CH3-CH C Br

CH2

H

CH3

HA

B

A

B CH3CH2C

CH2

CH32-METIL-1-BUTENA

CH3CH C

CH3

CH3

2-METIL-2-BUTENA

2-BROMO-2-METILBUTANA

(31%)

(69%)

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BUKTI LAIN ATURAN BUKTI LAIN ATURAN ZAITSEVZAITSEV1.1. DEHIDROKLORINASI NEOMENTIL KLORIDA DAN DEHIDROKLORINASI NEOMENTIL KLORIDA DAN

MENTIL KLORIDA DG NATRIUM ETOKSIDAMENTIL KLORIDA DG NATRIUM ETOKSIDA

H3CCl

H

HH

H

H

CH(CH3)2

C2H5O:-

C2H5O:H

MENTILKLORIDA2-MENTEN

H3C CH(CH3)2

Cl

H

HH

H

H

C2H5O:-

AB

A. 3-MENTEN (78%)

B. 2-MENTEN (22%)

CH(CH3)2

CH(CH3)2

H3C

CH(CH3)2

(100%)

NEOMENTILKLORIDA

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ANTI ZAITSEVANTI ZAITSEV

1.1. DEHIDROBROMINASI 2-BROMO-2-DEHIDROBROMINASI 2-BROMO-2-METILBUTANA DG KALIUM t-BUTOKSIDAMETILBUTANA DG KALIUM t-BUTOKSIDA

2.2. ELIMINASI DG BASA MERUAH ELIMINASI DG BASA MERUAH MEMBERIKAN ALKENA YG KURANG MEMBERIKAN ALKENA YG KURANG TERSUBSTITUSI.TERSUBSTITUSI.

CH3-C-O:-

CH3-CH C Br

CH2

H

CH3

HA

B

A

B CH3CH2C

CH2

CH32-METIL-1-BUTENA

CH3CH2 C

CH3

CH3

2-METIL-2-BUTENA

2-BROMO-2-METILBUTANA

(72,5%)

(27,5%)

H3C

H3C

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BAGAIMANA KOMPETISI SBAGAIMANA KOMPETISI SNN1 DAN 1 DAN E1E1

1.1. REAKSI E1 MELIBATKAN REAKSI E1 MELIBATKAN KARBOKATION STABIL, BASA KARBOKATION STABIL, BASA LEMAH, PELARUT POLAR DAN LEMAH, PELARUT POLAR DAN SUHU TINGGISUHU TINGGI

2.2. REAKSI SREAKSI SNN1 MELIBATKAN 1 MELIBATKAN KARBOKATION STABIL, BASA KARBOKATION STABIL, BASA LEMAH, PELARUT POLAR DAN LEMAH, PELARUT POLAR DAN SUHU RENDAHSUHU RENDAH

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PROBLEM 1PROBLEM 1

• WHEN WHEN CISCIS -1-BROMO-4-tert-BUTYLCYCLOHEXANE IS -1-BROMO-4-tert-BUTYLCYCLOHEXANE IS

TREATED WITH SODIUM ETHOXIDE IN ETHANOL, IT TREATED WITH SODIUM ETHOXIDE IN ETHANOL, IT

REACT RAPIDLY. THE PRODUCT IS 4-tert-REACT RAPIDLY. THE PRODUCT IS 4-tert-

BUTYLCYCLOHEXENE. UNDER THE SAME BUTYLCYCLOHEXENE. UNDER THE SAME

CONDITIONS, CONDITIONS, TRANSTRANS-1-BROMO-4-tert--1-BROMO-4-tert-

BUTYLCYCLOHEXANE REACTS VERY SLOWLY. BUTYLCYCLOHEXANE REACTS VERY SLOWLY.

WRITE CONFORMATIONAL STRUCTURE AND EXPLAIN WRITE CONFORMATIONAL STRUCTURE AND EXPLAIN

THE DIFFERENCE IN REACTIVITY OF THIS THE DIFFERENCE IN REACTIVITY OF THIS CISCIS-TRANS -TRANS

ISOMERS.ISOMERS.

20/04/23 20

REAKSINYAREAKSINYA

H

H

H Br

H

HH

H

H

H

H

C2H5O:-

H

H

H

Br

H

H

H

H

H

H

H

CIS

TRANS

SULIT

20/04/23 21

JAWAB PROBLEM 1JAWAB PROBLEM 1

• PD ISOMER CIS TERDAPAT 2 AT H PD ISOMER CIS TERDAPAT 2 AT H DG DG POSISI AKSIAL, BEGITU JUGA Br BERADA POSISI AKSIAL, BEGITU JUGA Br BERADA PD POSISI AKSIAL SHG REAKSI E2 PD POSISI AKSIAL SHG REAKSI E2 BERLANGSUNG LEBIH CEPAT.BERLANGSUNG LEBIH CEPAT.

• PD ISOMER TRANS BAIK AT H PD ISOMER TRANS BAIK AT H MAUPUN MAUPUN Br YG AKAN BEREAKSI DLM POSISI Br YG AKAN BEREAKSI DLM POSISI EQUATORIAL, SEHG MENYEBABKAN EQUATORIAL, SEHG MENYEBABKAN REAKSI E2 SANGAT LAMBAT (SULIT).REAKSI E2 SANGAT LAMBAT (SULIT).

20/04/23 22

REAKSINYAREAKSINYA

H

H

H Br

H

HH

H

H

H

H

C2H5O:-

H

H

H

Br

H

H

H

H

H

H

H

CIS

TRANS

SULIT

20/04/23 23

PROBLEM 2PROBLEM 2• A). When A). When ciscis-1-bromo-2--1-bromo-2-

methylcyclohexane undergoes an E2 methylcyclohexane undergoes an E2 reaction, two product (cycloalkenes) are reaction, two product (cycloalkenes) are formed. What are these two cycloalkenes, formed. What are these two cycloalkenes, and which would you expect to be the and which would you expect to be the major product? Write conformational major product? Write conformational structures showing how each is formed.structures showing how each is formed.

• B) When B) When transtrans-1-bromo-2--1-bromo-2-methylcyclohexane reacts in an E2 methylcyclohexane reacts in an E2 reaction, only one cycloalkene is formed. reaction, only one cycloalkene is formed. What is this product? Write What is this product? Write conformational structures showing why is conformational structures showing why is the only product.the only product.

20/04/23 24

KESIMPULANKESIMPULANCHCH33XX

METILMETIL

RCHRCH22XX

(1(1oo))

(R)(R)22CHXCHX

(2(2oo))

(R)(R)33CXCX

(3(3oo))

REAKSI BIMOLEKULERREAKSI BIMOLEKULER SSNN1/E1/E21/E1/E2

MEMBERIKAMEMBERIKAN REAKSI N REAKSI SSNN22

MEMBERI MEMBERI SSNN2, 2, KECUALI KECUALI DG BASA DG BASA MERUAH MERUAH MEMBERI MEMBERI REAKSI E2REAKSI E2

MEMBERIKAMEMBERIKAN REAKSI N REAKSI SSNN2 DG BASA 2 DG BASA LEMAHLEMAH

DLM DLM SOLVOLISIS SOLVOLISIS BEREAKSI BEREAKSI SSNN1/E1, PD T 1/E1, PD T RENDAH SRENDAH SNN1 1 DOMINAN DOMINAN

MEMBERIKAMEMBERIKAN REAKSI E2 N REAKSI E2 DG BASA DG BASA KUATKUAT

BASA KUAT BASA KUAT REAKSI E2 REAKSI E2 DOMINANDOMINAN

20/04/23 25

A BIOLOGICAL NUCLEOPHILIC A BIOLOGICAL NUCLEOPHILIC SUBSTITUTION REACTION: BIOLOGICAL SUBSTITUTION REACTION: BIOLOGICAL

METHYLATIONMETHYLATION

1.1. THE CELLS OF LIVING ORGANISMS THE CELLS OF LIVING ORGANISMS

SYNTHESIZE MANY OF THE COMPOUNDS SYNTHESIZE MANY OF THE COMPOUNDS

THEY NEED FROM SMALLER MOLECULES.THEY NEED FROM SMALLER MOLECULES.

2.2. A NUMBER OF REACTIONS TAKE PLACE IN A NUMBER OF REACTIONS TAKE PLACE IN

THE CELLS OF PLANTS AND ANIMALS THE CELLS OF PLANTS AND ANIMALS

THAT INVOLVE THE TRANSFER OF A THAT INVOLVE THE TRANSFER OF A

METHYL GROUP FROM AN AMINO ACID METHYL GROUP FROM AN AMINO ACID

CALLED METHIONINE TO SOME OTHER CALLED METHIONINE TO SOME OTHER

COMPOUND.COMPOUND.

20/04/23 26

SOME OF THE COMPOUNDS THAT GET SOME OF THE COMPOUNDS THAT GET THEIR METHYL GROUP FROM THEIR METHYL GROUP FROM

METHIONINEMETHIONINE

-OOC-CH-CH2-CH2-S

NH3+

CH3

METHIONINE

CH3

N+ CH2-CH2-OH

CH3

CH3

CH-CH2NH CH3HO

HOCHOLINE

ADRENALINE

N

CH3N NICOTINE

20/04/23 27

METHYL TRANSFER MECHANISMMETHYL TRANSFER MECHANISM

-OOCCH-CH2CH2S:

CH3

NH2

+ CH2

O P O P O P OH

OHOHOH

O O O

O ADENINE

OH OH

HH H

H

LEAVING GROUP

NUCLEOPHILE

-OOCCH-CH2CH2S

CH3

NH2

CH2 O ADENINE

OH OH

HH H

H

-:O P O P O P OH

OHOHOH

O O O+

TRIPHOSPHAT ION

S-ADENOSYMETHIONINE

ACTIVE METHYL TRANSFERER

20/04/23 28

BIOSYNTHESIS OF CHOLINEBIOSYNTHESIS OF CHOLINE

-OOCCH-CH2CH2S

CH3

NH3+

CH2 O ADENINE

OH OH

HH H

H +

S-ADENOSYMETHIONINE

ACTIVE METHYL TRANSFERER

CH3

N+ CH2-CH2-OH

CH3

CH3

CHOLINE

CH3

:N CH2-CH2-OH

CH3

N,N-DIMETHYLAMINOETHANOL

-OOCCH-CH2CH2S CH2 O ADENINE

OH OH

HH H

H+

NH3+

..

20/04/23 29

DEHIDRASI ALKOHOLDEHIDRASI ALKOHOL

1.1. PELEPASAN HPELEPASAN H22O DR ALKOHOL-2 O DR ALKOHOL-2

MELALUI PEMANASAN DG ASAM MELALUI PEMANASAN DG ASAM

KUATKUAT

2.2. ASAM BRONSTED: HASAM BRONSTED: H22SOSO44 DAN H DAN H33POPO44

ASAM LEWIS: ALUMINA (AlASAM LEWIS: ALUMINA (Al22OO33) DLM ) DLM

INDUSTRIINDUSTRI

3.3. REAKSI DEHIDRASI ALKOHOL REAKSI DEHIDRASI ALKOHOL

MENUNJUKKAN BBP KARAKTERISTIK MENUNJUKKAN BBP KARAKTERISTIK

PENTING.PENTING.

20/04/23 30

KONDISI REAKSIKONDISI REAKSI1.1. ALKOHOL PRIMER MEMERLUKAN ALKOHOL PRIMER MEMERLUKAN

ASAM PEKAT DAN SUHU TINGGIASAM PEKAT DAN SUHU TINGGICH3-CH2-CH2-OH

H2SO4 PKT

170oC

CH3-CH CH2

PROPIL ALKOHOL PROPENA

OH 85% H3PO4

170oC+ H2O

SIKLOHEKSANOL SIKLOHEKSENA (80%)

+ H2O

CH3

H3C C

CH3

OH20% H2SO4

85oC

+ H2OH3C C

CH3

CH2t-BUTIL ALKOHOL

2-METILPROPENA (84%)

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CAMPURAN ALKENACAMPURAN ALKENA1.1. BBP ALKOHOL TERDEHIDRASI BBP ALKOHOL TERDEHIDRASI

MEMBERIKAN HASIL CAMPURAN MEMBERIKAN HASIL CAMPURAN ALKENAALKENA

2.2. MENGIKUTI ATURAN ZAITZEVMENGIKUTI ATURAN ZAITZEV

H3C-CH2-CH2-CH2

OH

H2SO4 PKT

170oCC C

H3C

CH3H

H

C C

H3C CH3

H H

+ + C C

CH2CH3H

H H

TRANSCIS

UTAMACH3

OH

2-METIL-SIKLOHEKSANOL

170oC

85% H3PO4

CH3 CH3

1-METILSIKLOHEKSENA 3-METILSIKLOHEKSENA

20/04/23 32

PENATAAN ULANG PENATAAN ULANG

• BBP ALKOHOL 1BBP ALKOHOL 1oo DAN 2 DAN 2oo MENGALAMI MENGALAMI PENATAAN ULANG SELAMA DEHIDRASIPENATAAN ULANG SELAMA DEHIDRASI

CH3

H3C C CH CH3

H3C OH

3,3-DIMETIL-2-BUTANOL

..

+

CH3

H3C

C C

1) H3C

CH3

H2C C

CH3

CHCH3

CH3

2,3-DIMETIL-2-BUTENE 2,3-DIMETIL-1-BUTENE

(80%) (20%)

20/04/23 33

MEKANISME UMUM MEKANISME UMUM DEHIDRASI ALKOHOL DEHIDRASI ALKOHOL TERKATALISIS ASAMTERKATALISIS ASAM

• MENGIKUTI MEKANISME E1MENGIKUTI MEKANISME E1

STEP 1: C C OH....

H

+ H3O :+ C C O+-H..

H

+ H2O :

H(FAST)

STEP 2: C C O+-H..

H

H

+ H2O :C C +

H

(SLOW)

+ H2O :C C +

H

STEP 3:..

+ H3O :+C C (FAST)

20/04/23 34

MEKANISME PENATAAN ULANGMEKANISME PENATAAN ULANG

• PENATAAN ULANG SELAMA DEHIDRASI PENATAAN ULANG SELAMA DEHIDRASI DARI 3,3-DIMETIL-2-BUTANOLDARI 3,3-DIMETIL-2-BUTANOL

CH3

H3C C CH CH3

H3C OH

3,3-DIMETIL-2-BUTANOL

..

+ H3O+CH3

H3C C CH CH3

H3C O+H2..

+ H2O

1)

2) CH3

H3C C CH CH3

H3C O+H2..

CH3

H3C C C+ CH3

H3C

H2O+

H

3)CH3

H3C C C+ CH3

H3C H

CH3

H3C C C CH3

H3C H

+

CH3

H3C C C CH3

H3C H

+

(2o)

(3o)

20/04/23 35

FINAL PRODUCTFINAL PRODUCT

H CH2 C+ C CH3

H

H3C CH3

A) B) A)

B)

H2C C CHCH3

H3C CH3

MINOR PRODUCT

C C

CH3

CH3

H3C

H3C

MAJOR PRODUCT

20/04/23 36

BEBERAPA PERGESERAN DLM BEBERAPA PERGESERAN DLM PENATAAN ULANG ION PENATAAN ULANG ION

KARBONIUM KARBONIUM CH3

H3C-C CH+ CH3

CH3

CH3

H3C-C+ C CH3

CH3H

(2o)(3o)

PERGESERAN METIDA

H3C-CH2CH C+H2

H

H3C-CH2C+H CH2

H

(2o)(1o)

PERGESERAN HIDRIDA

CH+ CH3

CH3+ CH3

CH3H(2o) (3o)

20/04/23 37

DEBROMINATION OF VICINAL DEBROMINATION OF VICINAL DIBROMIDESDIBROMIDES

1.1. VICVIC-DIBROMIDES -DIBROMIDES UNDERGO THE LOSS UNDERGO THE LOSS OF A MOLECULE OF OF A MOLECULE OF BROMINE (BrBROMINE (Br22)WHEN )WHEN THEY ARE TREATED THEY ARE TREATED WITH A SOLUTION WITH A SOLUTION OF SODIUM IODIDE OF SODIUM IODIDE IN ACETON OR IN ACETON OR MIXTURE OF ZINC MIXTURE OF ZINC DUST IN ETHANOL. DUST IN ETHANOL.

2.2. DEBROMINATION BY DEBROMINATION BY SODIUM IODIDE SODIUM IODIDE TAKE PLACE BY AN TAKE PLACE BY AN E2 MECHANISM.E2 MECHANISM.

H3C CH CH2

Br Br

H3C CH2 CH

Br

BrVIC GEM

20/04/23 38

THE MECHANISM OF THE MECHANISM OF DEBROMINATIONDEBROMINATION

I:-

CH CH2H3C

Br

Br

CH CH2H3C

+ IBr + :Br-

I:- + IBr I2 + Br:-

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PROBLEM 1PROBLEM 1

• Although ethyl bromide and isobutyl Although ethyl bromide and isobutyl bromide are both primary halides, ethyl bromide are both primary halides, ethyl bromide undergoes Sbromide undergoes SNN2 reactions more 2 reactions more than ten times faster than isobutyl than ten times faster than isobutyl bromide. When each compound is bromide. When each compound is treated with a strong base (CHtreated with a strong base (CH33CHCH22OO--), ), isobutyl bromide gives a greater yield of isobutyl bromide gives a greater yield of elimination products than substitution elimination products than substitution products, whereas with ethyl bromide products, whereas with ethyl bromide this behavior is reversed. What factor this behavior is reversed. What factor accounts for these results?accounts for these results?

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PROBLEM 2PROBLEM 2

• Consider the reaction of IConsider the reaction of I-- with CH with CH33CHCH22Cl. Cl. (a). Would you expect the reaction to be (a). Would you expect the reaction to be SSNN1 or S1 or SNN2? The rate constant for the 2? The rate constant for the reaction at 60reaction at 60oo is 5x10 is 5x10-5-5 liter mole liter mole-1-1 sec sec-1-1. . (b). What is the reaction rate if [I(b). What is the reaction rate if [I--] = 0.1 ] = 0.1 mole litermole liter-1 -1 and [CHand [CH33CHCH22Cl] = 0.1 mole Cl] = 0.1 mole literliter-1-1? (c). If [I? (c). If [I--] = 0.1 mole liter] = 0.1 mole liter-1 -1 and and [CH[CH33CHCH22Cl] = 0.2 mole literCl] = 0.2 mole liter-1-1? (d). If [I? (d). If [I--] = ] = 0.2 mole liter0.2 mole liter-1 -1 and [CHand [CH33CHCH22Cl] = 0.1 mole Cl] = 0.1 mole literliter-1-1? (e). If [I? (e). If [I--] = 0.2 mole liter] = 0.2 mole liter-1 -1 and and [CH[CH33CHCH22Cl] = 0.2 mole literCl] = 0.2 mole liter-1-1? ?

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PROBLEM 3PROBLEM 3

• When When terttert-butyl bromide undergoes S-butyl bromide undergoes SNN1 1

hydrolisis, adding a “common ion” (i.e hydrolisis, adding a “common ion” (i.e NaBr) to the aqueous solution has no NaBr) to the aqueous solution has no effect on the rate. On the other hand effect on the rate. On the other hand when (Cwhen (C66HH55))22CHBr undergoes SCHBr undergoes SNN1 1

hydrolisis, adding NaBr retards the hydrolisis, adding NaBr retards the reaction. Given that the (Creaction. Given that the (C66HH55))22CHCH++cation cation

is known to be much more stable than is known to be much more stable than the (CHthe (CH33))33CC++ cation, provide an cation, provide an

explanation for the different behavior of explanation for the different behavior of the two compounds.the two compounds.