Transcript Document 7743152

REAKSI ELIMINASI
PELEPASAN MOLEKUL YZ DARI ATOMATOM BERDAMPINGAN DLM SUATU
MOLEKUL PEREAKSI.
 DEHIDROHALOGENASI DAN DEHIDRASI

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Y
Z
C
C
C
C
+
YZ
1
• DEHIDROHALOGENASI ALKIL HALIDA
• PELEPASAN/ PENARIKAN HX DARI ATOM-2
C BERDAMPINGAN DLM SEBUAH ALKIL
HALIDA
• DILAKUKAN DG BASA KUAT
• MISALNYA: CH3ONa, C2H5ONa DAN
(CH3)3COK
2 C2H5OH + 2 Na
CH3OH
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+ NaH
2 C2H5O-Na+ + H2
CH3O-Na+
+ H2
2
• DEHIDRASI ALKOHOL
• PELEPASAN/ PENARIKAN H2O DARI ATOM-2 C
BERDAMPINGAN DLM SEBUAH ALKOHOL
• DILAKUKAN DG ASAM KUAT
• MISALNYA: H2SO4 DANH3PO4
H3C-CH2-CH2
OH
OH
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H2SO4 PEKAT
170oC
H3PO4 85%
165-170oC
H3C-CH2
CH2 + H O
2
+ H2O
3
CONTOH
DEHIDROBROMINASI ALKIL BROMIDA
H3C
C2H5ONa
C
H
Br
H3C
H3C
H3C
C
H3C
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(C2H5OH, 55oC)
H
H2C
C2H5ONa
Br
(C2H5OH, 25oC)
C
+ NaBr + C2H5OH
CH3
79%
H2C
CH3
C
+ NaBr + C2H5OH
CH3
91%
4
MEKANIME REAKSI ELIMINASI
1.
REAKSI ELIMINASI BIMOLEKULER (E2)
r = k [R-X].[:B-]
r = k [CH3CHBrCH3].[C2H5O-]
2. REAKSI ELIMINASI UNIMOLEKULER (E1)
r = k [R-X]
r = k [(CH3)3C-Cl]
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REAKSI E2
 DEHIDROBROMINASI ISOPROPIL BROMIDA
DG LARUTAN NATRIUM ETOKSIDA DLM
ETANOL
C2H5O:H

H2C
H

C
C2H5O:- H
CH3
Br
H2C
H
C
.. Br
H
CH3
H
C
H
C + Br- + C2H5OH
CH3
PROPENA
ISOPROPIL BROMIDA
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MEKANISME REAKSI E2
1. BASA MENYERANG AT H  DR 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
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REAKSI E1
 DEHIDROBROMINASI t-BUTIL BROMIDA DG
LARUTAN NATRIUM ETOKSIDA DLM ETANOL
1.
H3C
H3C
C
Br
C2H5O:-
CH2
H
H3C
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C+ CH3 + Br-
H3C
H3C
2.
H3C
C+ CH3
H2C
CH3
C
+ C2H5O:H
CH3
2-METILPROPENA
8
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
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KOMPETISI SN2 DAN E2
1.
ETANOL
C2H5ONa + H3CCH2Br
T=550C
(1o)
H3C
2.
C2H5ONa +
H3C
H3C
CH
Br
ETANOL
T=550C
(2o)
3. C2H5ONa +
Br
H3C C
H3C o
(3 )
ETANOL
T=250C
C2H5OCH2CH3 +
SN2= 90%
CH3
CH2 + NaBr
E2 = 10%
C2H5O-CH-CH3 + H2C
CH-CH3 + NaBr
E2 = 79%
SN2= 21%
CH3
CH3
C2H5O-C-CH3 + H2C
CH3
SN2= 9%
T=550C
H2C
C
+ NaBr
CH3
E2 = 91%
E2 = 100%
PD HALIDA 3o KENAIKAN T TERJADI REAKSI E2
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KERUAHAN BASA PEREAKSI
1.
KERUAHAN BASA PEREAKSI MENDORONG
REAKSI ELIMINASI
CH3O:- + CH (CH ) CH CH METANOL
3
2 15 2 2
Br
DIREFLUX
CH3(CH2)15CH2CH2 OCH3
99%
+
CH3(CH2)15CH2 CH2
1%
CH3
CH3-C-O:CH3
CH3(CH2)15CH2CH2
Br
t-BUTIL
ALKOHOL
40o
CH3
CH3(CH2)15CH2CH2 OC-CH3
15%
CH3
+
CH3(CH2)15CH2 CH2
85%
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KEBASAAN
1.
2.
BASA LEMAH SPT Cl-, CH3COO-, Br-, IMENDORONG SN2
BASA KUAT: C2H5O-, OH-, NH2- MENDORONG E2
CH3
CH3COO:- +
C2H5O:-
CH3-CH
CH3
Br
CH3
+
CH3-CH
Br
O
CH3-CH O-CCH3 + Br:
100%
CH3
+ Br:-
CH2=CH
79%
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KOMPETISI SN2 DAN E2
1. REAKSI SN2 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.
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ATURAN ZAITSEV
1.
DEHIDROBROMINASI 2-BROMO-2-METILBUTANA
DG NATRIUM ETOKSIDA
C2H5O:A
H
CH3-CH

H
CH2
B
B

Br
CH3
2-BROMO-2-METILBUTANA
2.
(31%)
CH3
2-METIL-1-BUTENA
 CH2
C
CH3CH2C
CH3
A
CH3CH
C
(69%)
CH3
2-METIL-2-BUTENA
REAKSI ELIMINASI TERJADI DG MEMBERIKAN
ALKENA YG LEBIH TERSUBSTITUSI.
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BUKTI LAIN ATURAN ZAITSEV
1.
DEHIDROKLORINASI NEOMENTIL KLORIDA
DAN MENTIL KLORIDA DG NATRIUM ETOKSIDA
H3C
Cl
H
H
C2H5O:H
H
H3C
H
CH(CH3)2
H
MENTILKLORIDA
C2H5O:-
2-MENTEN (100%)
C2H5O:B
H A
H
H3C
H
H
H
CH(CH3)2
CH(CH3)2
A. 3-MENTEN (78%)
CH(CH3)2
Cl
NEOMENTILKLORIDA
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CH(CH3)2
B. 2-MENTEN (22%)
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ANTI ZAITSEV
1.
2.
DEHIDROBROMINASI 2-BROMO-2METILBUTANA DG KALIUM t-BUTOKSIDA
ELIMINASI DG BASA MERUAH MEMBERIKAN
ALKENA YG KURANG TERSUBSTITUSI.
H3C
CH3-C-O:H3C
A
H
H
CH2
B
B
 CH
2
CH3-CH
C


2-BROMO-2-METILBUTANA
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(72,5%)
CH3
2-METIL-1-BUTENA
Br
CH3
CH3CH2C
CH3
A
CH3CH2
C
(27,5%)
CH3
2-METIL-2-BUTENA
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BAGAIMANA KOMPETISI SN1 DAN E1
1. REAKSI E1 MELIBATKAN
KARBOKATION STABIL, BASA
LEMAH, PELARUT POLAR DAN
SUHU TINGGI
2. REAKSI SN1 MELIBATKAN
KARBOKATION STABIL, BASA
LEMAH, PELARUT POLAR DAN
SUHU RENDAH
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PROBLEM 1
• WHEN CIS -1-BROMO-4-tert-BUTYLCYCLOHEXANE IS
TREATED WITH SODIUM ETHOXIDE IN ETHANOL, IT
REACT RAPIDLY. THE PRODUCT IS 4-tertBUTYLCYCLOHEXENE. UNDER THE SAME
CONDITIONS, TRANS-1-BROMO-4-tertBUTYLCYCLOHEXANE REACTS VERY SLOWLY.
WRITE CONFORMATIONAL STRUCTURE AND EXPLAIN
THE DIFFERENCE IN REACTIVITY OF THIS CIS-TRANS
ISOMERS.
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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 TRANS BAIK AT H  MAUPUN
Br YG AKAN BEREAKSI DLM POSISI
EQUATORIAL, SEHG MENYEBABKAN
REAKSI E2 SANGAT LAMBAT (SULIT).
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REAKSINYA
H
Br
H
HH
H
H
H
H
H
H
CIS
C2H5O:-
H
H
HH
TRANS
H
H
H
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H
H
Br
SULIT
H
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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.
• 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.
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KESIMPULAN
CH3X
RCH2X
(R)2CHX
METIL
(1o)
(2o)
REAKSI BIMOLEKULER
SN1/E1/E2
MEMBERI
SN2, KECUALI
DG BASA
MERUAH
MEMBERI
REAKSI E2
MEMBERIKAN
REAKSI SN2 DG
BASA LEMAH
DLM
SOLVOLISIS
BEREAKSI
SN1/E1, PD T
RENDAH SN1
DOMINAN
MEMBERIKAN
REAKSI E2 DG
BASA KUAT
BASA KUAT
REAKSI E2
DOMINAN
MEMBERIKAN
REAKSI SN2
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(R)3CX
(3o)
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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.
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SOME OF THE COMPOUNDS THAT GET
THEIR METHYL GROUP FROM METHIONINE
CH3
-OOC-CH-CH2-CH2-S
NH3+
CH3
CH3
N+
CH3
CH2-CH2-OH
CHOLINE
HO
METHIONINE
HO
CH-CH2NH
CH3
ADRENALINE
N
N
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CH3
NICOTINE
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METHYL TRANSFER MECHANISM
OH
NUCLEOPHILE
O
P
O
-OOCCH-CH2CH2S: +
NH2
P
O
CH3
OH
H
H
OH
O
OH
NH2
OH
CH2
O
H
H
OH
LEAVING GROUP
H
CH3
-OOCCH-CH2CH2S
OH
P
O
O
ADENINE
CH2
O
H
OH
ADENINE -:O
+
H
H
OH
P
O
OH
O
P
OH
O
O
P
OH
O
TRIPHOSPHAT ION
S-ADENOSYMETHIONINE
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ACTIVE METHYL TRANSFERER
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BIOSYNTHESIS OF CHOLINE
CH3
-OOCCH-CH2CH2S
NH3+
CH2
O
H
H
H
OH
CH3
ADENINE
H
+
:N
CH2-CH2-OH
CH3
OH
N,N-DIMETHYLAMINOETHANOL
S-ADENOSYMETHIONINE
ACTIVE METHYL TRANSFERER
..
-OOCCH-CH2CH2S
NH3+
CH2
O
H
H
H
OH
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ADENINE
CH3
H
+
CH3
N+
CH3
CH2-CH2-OH
CHOLINE
OH
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DEHIDRASI ALKOHOL
1.
PELEPASAN H2O DR ALKOHOL-2
MELALUI PEMANASAN DG ASAM KUAT
2. ASAM BRONSTED: H2SO4 DAN H3PO4
ASAM LEWIS: ALUMINA (Al2O3) DLM
INDUSTRI
3. REAKSI DEHIDRASI ALKOHOL
MENUNJUKKAN BBP KARAKTERISTIK
PENTING.
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KONDISI REAKSI
1.
ALKOHOL PRIMER MEMERLUKAN ASAM
PEKAT DAN SUHU TINGGI
H2SO4 PKT
CH3-CH2-CH2-OH
170oC
PROPIL ALKOHOL
OH
H3C
C
85% H3PO4
170oC
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+
H2O
SIKLOHEKSENA (80%)
CH3
20% H2SO4
OH
CH3
t-BUTIL ALKOHOL
H2O
PROPENA
SIKLOHEKSANOL
CH3
CH2 +
CH3-CH
85oC
H3C
C
+
H2O
CH2
2-METILPROPENA (84%)
28
CAMPURAN ALKENA
1.
BBP ALKOHOL TERDEHIDRASI
MEMBERIKAN HASIL CAMPURAN ALKENA
H3C-CH2-CH2-CH2
OH
H C
H2SO4 PKT 3
C
170oC
H H C
3
C
C +
CH3 H
H
TRANS
CH3
OH 85% H PO
3 4
170oC
UTAMA
CH3
CH3 H
C
+
H
CH2CH3
C C
H
H
CIS
CH3
2-METIL-SIKLOHEKSANOL
3-METILSIKLOHEKSENA
2. MENGIKUTI 1-METILSIKLOHEKSENA
ATURAN ZAITZEV
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PENATAAN ULANG
• BBP ALKOHOL 1o DAN 2o MENGALAMI
PENATAAN ULANG SELAMA DEHIDRASI
CH3
1)
H3C
H3C C CH CH3
H3C OH
..
3,3-DIMETIL-2-BUTANOL
C
H3C
C
+
H2C
C
CHCH3
CH3
2,3-DIMETIL-2-BUTENE
(80%)
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CH3 CH3
CH3
2,3-DIMETIL-1-BUTENE
(20%)
30
MEKANISME UMUM DEHIDRASI
ALKOHOL TERKATALISIS ASAM
• MENGIKUTI MEKANISME E1
STEP 1:
C
C
..
+
OH
+
H
O
:
3
..
C
H
H
STEP 2:
C
C
C
O+-H + H2O :
..
H
(SLOW)
C
O+-H
..
C+
+ H2O :
H
H
STEP 3:
H
(FAST)
C
..
+
C + H2O :
(FAST)
C
C
+ H3O :+
H
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MEKANISME PENATAAN ULANG
• PENATAAN ULANG SELAMA DEHIDRASI DARI 3,3DIMETIL-2-BUTANOL
CH3
1)
H3C
C
CH
H3C
OH
..
CH3 + H O+
3
H3C
CH3
C CH
H3C
CH3 + H2O
+
O
.. H2
3,3-DIMETIL-2-BUTANOL
2)
H3C
CH3
C CH
H3C
CH3
H3C C C+ CH3
H3C H (2o)
CH3
+
O
.. H2
H3C
C
H3C
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H2O
CH3
CH3
3)
+
C+ CH3
H
CH3
+
C
C
H3C
H
H3C
CH3
+
H3C
C
H3C
C
CH3
H (3o)
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FINAL PRODUCT
H3C
A)
H
B) H
CH2
C+
H3C
C
A)
CH3
H2C
C
CH3
CHCH3
MINOR PRODUCT
CH3
H3C
B)
CH3
C
H3C
C
CH3
MAJOR PRODUCT
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BEBERAPA PERGESERAN DLM
PENATAAN ULANG ION KARBONIUM
PERGESERAN METIDA
CH3
H3C-C
CH+
CH3
CH3
H3C-C+
CH3
CH3
CH3 H
(3o)
(2o)
PERGESERAN HIDRIDA
H
H3C-CH2CH
C+H2
(1o)
H3C-CH2C+H
CH+
(2o)
CH3
H
CH2
(2o)
CH3
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C
+
CH3
CH3
H
(3o)
34
DEBROMINATION OF VICINAL DIBROMIDES
1.
2.
VIC-DIBROMIDES
UNDERGO THE LOSS OF
A MOLECULE OF
BROMINE (Br2)WHEN
THEY ARE TREATED
WITH A SOLUTION OF
H3C
SODIUM IODIDE IN
ACETON OR MIXTURE
OF ZINC DUST IN
ETHANOL.
DEBROMINATION BY
SODIUM IODIDE TAKE
PLACE BY AN E2
MECHANISM.
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Br
CH
CH2
Br
Br
VIC
H3C
CH2 CH
Br
GEM
35
THE MECHANISM OF DEBROMINATION
I:Br
CH
H3C
I:-
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CH2
Br
+ IBr
CH
CH2 + IBr + :Br-
H3C
I2 + Br:-
36
PROBLEM 1
•
Although ethyl bromide and isobutyl bromide
are both primary halides, ethyl bromide
undergoes SN2 reactions more than ten times
faster than isobutyl bromide. When each
compound is treated with a strong base
(CH3CH2O-), 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?
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PROBLEM 2
•
Consider the reaction of I- with CH3CH2Cl. (a).
Would you expect the reaction to be SN1 or SN2?
The rate constant for the reaction at 60o is 5x10-5
liter mole-1 sec-1. (b). What is the reaction rate if
[I-] = 0.1 mole liter-1 and [CH3CH2Cl] = 0.1 mole
liter-1? (c). If [I-] = 0.1 mole liter-1 and
[CH3CH2Cl] = 0.2 mole liter-1? (d). If [I-] = 0.2
mole liter-1 and [CH3CH2Cl] = 0.1 mole liter-1?
(e). If [I-] = 0.2 mole liter-1 and [CH3CH2Cl] = 0.2
mole liter-1?
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PROBLEM 3
• When tert-butyl bromide undergoes SN1 hydrolisis,
adding a “common ion” (i.e NaBr) to the aqueous
solution has no effect on the rate. On the other
hand when (C6H5)2CHBr undergoes SN1
hydrolisis, adding NaBr retards the reaction. Given
that the (C6H5)2CH+cation is known to be much
more stable than the (CH3)3C+ cation, provide an
explanation for the different behavior of the two
compounds.
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