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ALKYL HALIDES –
ELIMINATION REACTIONS
ALKYL HALIDES UNDERGO ELIMINATION OF HX
WHEN TREATED WITH BASE. THE PRODUCTS ARE
ALKENES.
CH3
CH3
CH3
C
Br
KOH, EtOH
heat
CH3
CH3
+ HBr
C
CH2
ELIMINATION REACTIONS USUALLY REQUIRE FORCING
CONDITIONS, I.E. HEAT AND STRONG BASE.
THE ELIMINATION REACTIONS WHICH ALKYL HALIDES UNDERGO
ARE KNOWN AS I,2-ELIMINATIONS OR  ELIMINATIONS.
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ALKYL HALIDES: ELIMINATION REACTIONS
CH3
CH3

C
CH3
Br
heat
CH3

KOH, EtOH
CH3
+ HBr
C
CH2

The elements of H-X are lost from neighboring
carbon atoms and a C=C is formed. The head
carbon of the alkyl halide is termed  (“alpha”)
and the carbon atom or atoms next to it are
designated  (“beta”).
The halogen atom is lost from the  carbon, and
the hydrogen from one of the  carbons.
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ALKYL HALIDES: ELIMINATION REACTIONS
THE TWO MOST IMPORTANT MECHANISMS
BY WHICH ALKYL HALIDES UNDERGO
ELIMINATION REACTIONS ARE:
1. THE E1 MECHANISM (UNIMOLECULAR);
2. THE E2 MECHANISM (BIMOLECULAR).
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ELIMINATION REACTIONS OF ALKYL HALIDES:
THE UNIMOLECULAR MECHANISM (E1)
(a)
R
R

R C

C
H
R
slow, r.d.s.

R C
B
H
R
X

C

C
R
+ X
R
H
R
R
(b)
R

C
R
fast
R
R
C
+ B-H
C
R
R
The slow, rate determining step entails one species
– the alkyl halide.
The rate of the reaction = k[alkyl halide]
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Note the carbocation intermediate
ELIMINATION REACTIONS OF ALKYL HALIDES:
THE UNIMOLECULAR MECHANISM (E1)
A carbocation intermediate is formed when alkyl halides
undergo elimination via the E1 (unimolecular) mechanism.
3o alkyl halides are likely to lose HX via this mechanism.
For t-butyl bromide in aqueous alcoholic KOH:
H
(a)

H C
CH3

C
H
Br
HO

H C

H C
H

C
CH3
+ Br
CH3
H
CH3
H
H
(b)
slow, r.d.s.
H

C
CH3
fast
H
C
C
CH3
+ H-O-H
CH3
CH3
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ELIMINATION REACTIONS OF ALKYL HALIDES:
THE BIMOLECULAR MECHANISM (E2)
B
R
H
R

C

C
R
X
R
R
R
C
C
R
R
+ B-H + X
THIS IS A CONCERTED REACTION.
BOND FORMATION AND BOND BREAKING TAKE
PLACE SIMULTANEOUSLY.
THE RATE DETERMINING STEP ENTAILS THE BASE AND
THE ALKYL HALIDE.
RATE = k[alkyl halide][base]
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THE BIMOLECULAR MECHANISM (E2)
A VERY IMPORTANT FEATURE
For an alkyl halide to undergo elimination via the
E2 mechanism, the H and X groups must be anti to
each other and be in the same plane with each
other and the carbon atoms to which they are
R
attached.
B
H
R  R
R

X
R
R
C
C
R + B-H + X
R
THE ELEMENTS OF H-X MUST BE
ANTIPERIPLANAR.
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OTHER ASPECTS OF E1 AND E2 REACTIONS
1.
2.
THE DISTINCTION BETWEEN THE E1 AND E2
MECHANISMS IS NOT AS CLEAR AS THE
DISTINCTION BETWEEN THE SN1 AND SN2
MECHANISMS.
3O AND 2O ALKYL HALIDES WILL ELIMINATE H-X
VIA BOTH THE E1 AND E2 MECHANISMS.
THE ELIMINATION OF H-X FROM 1O ALKYL
HALIDES TAKES PLACE VIA THE E2 MECHANISM
ONLY.
FOR BOTH E1 AND E2 MECHANISMS, THE RATES
FOLLOW THE TREND:
3O R-X > 2O R-X > 1O R-X (do not react via E1)
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OTHER ASPECTS OF E1 AND E2 REACTIONS
3.
FOR MANY ALKYL HALIDES, THERE ARE TWO
POSSIBLE ELIMINATION PRODUCTS.
THE 3O ALKYL HALIDE BELOW HAS THREE 
CARBONS; TWO ARE IDENTICAL METHYL (CH3)
GROUPS, AND THE THIRD IS A METHYLENE (CH2)
GROUP.
H
H C
H
CH3 H
C
C CH3

Br
H
LET US EXAMINE THE ELIMINATION OF H-Br FROM
THIS COMPOUND VIA THE E1 MECHANISM.
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ELIMINATION PRODUCTS: E1 MECHANISM
Two products can result from the loss of H-Br
STEP I
H
H C
H
CH3
H
C
C CH3

Br
H
slow, r.d.s
CH3
H
C
H C
H
H
C CH3

H
carbocation
STEP 2
H
H C
H
OH
CH3
C
H
C CH3

H
H
H C
H
 CH3
C
C CH3

H
and/or
HO
H
H C
H
+ Br
+ H2O
more substituted alkene
more stable
SAYTZEFF PRODUCT

CH3
H
H
C
C CH3

C
H
H

CH3
C H
C CH3

+ H2O
H
less substituted alkene
less stable
HOFMANN PRODUCT
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ELIMINATION PRODUCTS: E2 MECHANISM
THE 2O ALKYL HALIDE SHOWN BELOW HAS TWO 
CARBONS WHICH ARE NOT IDENTICAL.
ONE IS A METHYL (CH3) GROUP AND THE OTHER IS A
METHYLENE (CH2) GROUP.
LET US EXAMINE THE ELIMINATION OF H-Br FROM THIS
COMPOUND VIA THE E2 MECHANISM.
H
H3C C
H
H
H
C
C H

Br
H
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ELIMINATION PRODUCTS: E2 MECHANISM
TWO PRODUCTS CAN FORM VIA THE E2 MECHANISM
HO
H
H3C C
H
H

C
H
Br
H
CH3
C H

C
H 
 H
C
+ H2O
+ Br
CH3
more substituted alkene
more stable
SAYTZEFF PRODUCT
and/or
OH
H
H3C C
H
H

C
H
Br
H
C H


CH3CH2
H

C
C
H
H
less substituted alkene
less stable
HOFMANN PRODUCT
+ H2O
+ Br
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ELIMINATION PRODUCTS:HOFMANN VS. SAYTZEFF
THE PROPORTION OF THE LESS SUBSTITUTED ALKENE
(HOFMANN PRODUCT) CAN BE INCREASED BY USING A VERY
BULKY BASE. TWO EXAMPLES OF BULKY BASES ARE SHOWN
CH3
O K
H3C C
potassium t-butoxide
K
CH3
t-BuO
CH2CH3
CH3CH2 C
O K
CH2CH3
potassium 3-ethyl-3-pentoxide
Et
Et C
Et
O K
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ELIMINATION PRODUCTS: HOFMANN VS. SAYTZEFF
BULKY BASES INCREASE THE PROPORTION OF THE
LESS SUBSTITUTED ALKENE (HOFMANN PRODUCT)
FORMED IN ELIMINATION REACTIONS.
CH3
H

C C H
H3C C

CH3
H
Cl
H
Et
Et C O K
Et
H
 CH
3
H3C C C
C H
CH3

H
substituted alkene
97% less
HOFMANN PRODUCT
The H’s on the less substituted  carbon are more sterically
accessible to the base than are the H’s on the more substituted 
carbon. When the base is very bulky, then the H’s on the less
substituted  carbon are almost exclusively removed, and the less
substituted (Hofmann) alkene product predominates.
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ELIMINATION PRODUCTS: HOFMANN VS. SAYTZEFF
STERIC ACCESSIBILITY OF THE  H AFFECTS THE
OUTCOME OF ELIMINATION REACTIONS.
If the H on the  carbon whose elimination leads
to the more substituted alkene is very crowded,
then the proportion of the
CH3H CH3
less substituted alkene product H C C C C
3
 C H

will be high.
CH3 H
CH3H

H3C C C
CH3 H
CH
3H

C C H Et O Na

H
Br
H
less substituted alkene
HOFMANN PRODUCT
+
CH3
very
crowded
more
accessible
MAJOR
CH
3
H
C
C H MINOR

H
H3C C 
C
CH3
H
more substituted alkene
SAYTZEFF PRODUCT
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SUBSTITUTION VERSUS ELIMINATION: SN1 VS E1
When substitution reactions are carried out
on 3o alkyl halides (SN1 reactions), products
of elimination (alkenes) are almost inevitably
formed.
Let us consider the the following reaction.
CH2CH3
CH3CH2 C CH2CH3 + H2O
Br
CH2CH3
CH3CH2 C CH2CH3 + HBr
OH
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SUBSTITUTION VERSUS ELIMINATION: SN1 VS E1
CH2CH3
CH3CH2 C CH2CH3 + H2O
Br
CH2CH3
CH3CH2 C CH2CH3 + HBr
OH
In this reaction the carbocation intermediate, once it is
formed, can lose a proton by reaction with as weak a base as
H2O to give appreciable quantities of the alkene (elimination)
product.
CH3
H2O
H
CH3CH2
H3C
C H
C
C
C
CH2CH3
CH3CH2
H
CH2CH3
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SUBSTITUTION VERSUS ELIMINATION: E2 VS SN2
IT IS EASIER TO CREATE CONDITIONS WHICH
FAVOR THE E2 MECHANISM OVER THE SN2
MECHANISM, OR VICE VERSA.
VERY STRONG BASE
(ETHOXIDE AS OPPOSED TO HYDROXIDE)
RELATIVELY NON-POLAR SOLVENTS
(E.G. ETHANOL IN PREFERENCE TO WATER)
HIGHER TEMPERATURES, WILL FAVOR THE
E2 MECHANISM OVER THE SN2 MECHANISM.
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ORGANOMETALLIC COMPOUNDS
COMPOUNDS IN WHICH A METAL IS DIRECTLY BONDED
TO CARBON ARE KNOWN AS

ORGANOMETALLIC COMPOUNDS.

C
M
THE METAL-CARBON BOND IS
POLARIZED AS SHOWN.
METALS ARE LESS ELECTRONEGATIVE THAN CARBON;
LARGER DIFFERENCES IN ELECTRONEGATIVITY
BETWEEN THE METAL AND CARBON INCREASE THE
IONIC CHARACTER OF THE METAL-CARBON BOND.
IONIC CHARACTER OF METAL CARBON BONDS FOLLOWS
THE TREND
Na > Li > Mg > Al > Zn > Cd > Hg
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ORGANOMETALLIC COMPOUNDS
ALKYL DERIVATIVES OF ALMOST ALL
METALS HAVE BEEN PREPARED.
THESE ARE NAMED AS “ALKYLMETALS”
(CH3)2Hg
DIMETHYLMERCURY
(liquid; bp 92 oC; neurotoxin; environmental contaminant)
(CH3CH2)4Pb TETRAETHYLLEAD
(liquid; bp ~ 220 oC; toxic; formerly used as a gasoline
additive)
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GRIGNARD REAGENTS
ALKYLMAGNESIUM HALIDES, R-Mg-X, ARE
KNOWN AS GRIGNARD REAGENTS.
GRIGNARD REAGENTS ARE PREPARED BY
REACTING ALKYL HALIDES WITH EXCESS
MAGNESIUM METAL IN DRY ALCOHOL-FREE
DIETHYL ETHER OR TETRAHYDROFURAN
(THF). DIETHYL ETHER AND THF ARE
SOLVENTS.
CH3CH2 O CH2CH3
diethyl ether
O
tetrahydrofuran
(THF), a cyclic ether
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GRIGNARD REAGENTS
PREPARATION
R-X + Mg  R-Mg-X
(radical mechanism)
Ease of formation follows the trends shown below
R-I > R-Br > R-Cl.
CH3X > C2H5X >C3H7X
Grignard reagents are usually closely associated
with two molecules of the ethereal solvent in which
Et
they have been prepared.
O Et
R
Mg
X
O Et
Et
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