ALKYLHALIDE
By Mrs. Azduwin Khasri
23rd October 2012
ELIMINATION REACTIONS OF
ALKYLHALIDES
ELIMINATION REACTION
• Elimination reactions involve the loss of elements from the starting
material to form a new  bond in the product.
• The product of elimination reaction is an Alkene
E1 AND E2 REACTION
E1
E-Elimination
1-Unimolecular
E2
E-Elimination
2-Bimolecular
E2 REACTION
Hydroxide cannot act as a nucleophile in this reaction
because of the bulky tertiary halide. Rather, hydroxide
acts as a base and abstracts a proton.
MECHANISM OF E2 REACTION
The removal of a proton and a halide ion is called
dehydrohalogenation
MECHANISM OF E2 REACTION
Carbon attached to halogen
Carbon adjacent to α-carbon
An E2 reaction is also called a b-elimination or
a 1,2-elimination reaction
The weaker the base, the better it is as a leaving group
The Regioselectivity of the E2 Reaction
2 structurally
β-carbon
base
2 product produce
The major product of an E2 reaction is the most stable
alkene
Reaction coordinate diagram for the E2 reaction
2-Butene formed faster
than 1-butene
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The Zaitsev Rule
The more substituted alkene product is obtained when a
proton is removed from the b-carbon that is bonded
to the fewest hydrogens
2-β-Hydrogens
3-β-Hydrogens
Disubstituted
Monosubstituted
The more substituted alkene is not always the most stable
alkene
Conjugated alkene products are preferred over the
more substituted alkene product:
Do not use Zaitsev’s rule to predict the major product
If the alkylhalide has a double bond or a benzene ring.
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The anti-Zaitsev Rule (Hoffmann Product)
Bulky Alkyl halide
Zaitsev product
Hofmann product
Bulky bases affect the product distribution resulting in the
Hofmann product, the least substituted alkene
Less sterically
hindered
Major product- Most
stable product
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Another exception to Zaitsev’s rule
When the halogen is fluorine-The major product of the
E2 reaction of alkyl Fluoride is the less substituted
alkene.
STRONGEST BASEPOOREST LEAVING
GROUP
The Fluoride ion does not have as strong a propensity
to leave as another halide ion.
Consider the elimination of 2-fluoropentane…
A carbanion-like transition state
Major productLess substituted
alkene
Carbocation vs Carbanion
Carbocation stability 3° > 2° > 1°
Carbanion stability 1° > 2° > 3°
CLASS EXERCISE 1-(E2)
Which of the alkyl halides is more reactive in an
E2 reaction?
Producing more
stable alkene
Br better leaving
group
Producing more
stable alkene
CLASS EXERCISE 2-(E2)
Give the major elimination product obtained
from an E2 reaction of the following alkyl
halides with hydroxide ion:
ANSWER:
E1 REACTION
• First order elimination reaction
• Must have at least two step
Mechanism of E1 reaction
Alkylhalide
dissociates,forming
carbocation
A Base removes a
proton from β-carbon
How does a weak base like water remove a proton from
an sp3 carbon?
1) The presence of a positive charge greatly reduces the
pKa
2) Hyperconjugation weakens the C-H bond by electron
density
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The Regioselectivity of the E1 Reaction
Fewest β-Hydrogen (According to Zaitsev rule)
The major product in an E1 reaction is generally
the more substituted alkene
Reaction coordinate diagram for the E1 reaction
Because the first step is the rate-determining step, the rate of an
E1 reaction depends both on the ease with which the carbocation
is formed and how readily the leaving group leaves
E1 AND E2 REACTION
• Major product generally the most stable
alkene
• Tertiary alkyl halides are the most reactive
and the primary alkyl halides are the least
reactive.
• For alkyl halide with the same alkyl group,
alkyl iodide are the most reactive and alkyl
fluoride are the least reactive.
Because the E1 reaction forms a carbocation
intermediate, we need to consider carbocation
rearrangement
Competition Between E2
and E1 Reactions
1° Carbocation are too
unstable to be formed in
E1 reaction
An E2 is favored by a high concentration of strong base and an
aprotic polar solvent
An E1 is favored by a weak base and a protic polar solvent
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Stereochemistry of the E2 Reaction
The bonds to the eliminated groups (H and X) must be
in the same plane
Parallel on
the same
side
Syn elimination
Parallel on
the
opposite
side
Anti elimination
The anti elimination is favored over the syn elimination
Consider the stereoselectivity of the E2 reaction
Antielimination
Synelimination
(E)- : the higher priority
groups are on opposite
sides of the double
bond.
(Z)- : the higher priority
groups are on the same
side of the double
bond.
The alkene with the bulkiest groups on opposite sides of the
double bond will be formed in greater yield, because it is the more
stable alkene
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Antielimination
Synelimination
When only one hydrogen is bonded to the b-carbon, the
major product of an E2 reaction depends on the structure
of the alkene
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Stereochemistry of the E1 Reaction
The major stereoisomer obtained from an E1 reaction is
the alkene in which the bulkiest substituents are on
opposite sides of the double bond
Both syn and anti elimination can occur in an E1 reaction,
both E and Z formed
In contrast,E2 forms both E and Z only if the β-carbon is bonded
to 2 Hydrogen.
If β-carbon bonded to only 1 Hydrogen,E2 form only 1 product
because anti elimination favored.
Competition Between
Substitution and Elimination
Alkyl halides can undergo SN2, SN1, E2, and E1
1) HO- is called nu in a substitution reaction (Attack carbon) and a
base in elimination reaction (removes proton).
2) Decide whether the reaction conditions favor SN2/E2 or SN1/E1
•SN2/E2 reactions are favored by a high concentration of a good
nucleophile/strong base
•SN1/E1 reactions are favored by a poor nucleophile/weak base
3) Decide how much of the product will be the substitution product
and how much of the product will be the elimination product
SN2/E2 conditions
36
A bulky alkyl halide or a sterically hindered nucleophile
encourages elimination over substitution
37
A strong or a bulky base encourages elimination over
substitution
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High temperature favors elimination over
substitution:
Why? Because elimination is entropically favorable.
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Tertiary alkyl halides undergo only elimination under
SN2/E2 conditions:
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SN1/E1 conditions
The elimination reaction favored at higher temperatures.
Primary alkyl halides do not form carbocations; therefore
they cannot undergo SN1 and E1 reactions.
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ASSIGNMENT 1
(SUBSTITUTION AND ELIMINATIONALKYLHALIDE)
Submit on 6th November (Tuesday)
-The End-