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ALKYLHALIDE By Mrs. Azduwin Khasri 23 rd October 2012.
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Transcript of ALKYLHALIDE By Mrs. Azduwin Khasri 23 rd October 2012.
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ALKYLHALIDE
By Mrs. Azduwin Khasri
23rd October 2012
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ELIMINATION REACTIONS OF ALKYLHALIDES
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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
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E1 AND E2 REACTION
E1E-Elimination1-Unimolecular
E2E-Elimination2-Bimolecular
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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.
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MECHANISM OF E2 REACTION
The removal of a proton and a halide ion is called dehydrohalogenation
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An E2 reaction is also called a b-elimination or a 1,2-elimination reaction
MECHANISM OF E2 REACTION
Carbon attached to halogen
Carbon adjacent to α-carbon
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The weaker the base, the better it is as a leaving group
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The Regioselectivity of the E2 Reaction
2 structurally β-carbon base 2 product produce
The major product of an E2 reaction is the most stablealkene
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Reaction coordinate diagram for the E2 reaction
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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 bondedto the fewest hydrogens
2-β-Hydrogens 3-β-Hydrogens
Disubstituted Monosubstituted
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The more substituted alkene is not always the most stable alkene
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Conjugated alkene products are preferred over the more substituted alkene product:
Do not use Zaitsev’s rule to predict the major productIf the alkylhalide has a double bond or a benzene ring.
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Bulky bases affect the product distribution resulting in the Hofmann product, the least substituted alkene
Hofmann productZaitsev product
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The anti-Zaitsev Rule (Hoffmann Product)Bulky Alkyl halide
Less sterically hindered
Major product- Most stable product
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Another exception to Zaitsev’s rule
STRONGEST BASE-POOREST LEAVING GROUP
When the halogen is fluorine-The major product of the E2 reaction of alkyl Fluoride is the less substituted alkene.
The Fluoride ion does not have as strong a propensity to leave as another halide ion.
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Consider the elimination of 2-fluoropentane…
A carbanion-like transition stateMajor product-Less substituted alkene
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Carbocation vs Carbanion
Carbocation stability 3° > 2° > 1°
Carbanion stability 1° > 2° > 3°
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CLASS EXERCISE 1-(E2)Which of the alkyl halides is more reactive in an
E2 reaction?
Br better leaving group
Producing more stable alkene
Producing more stable alkene
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Give the major elimination product obtained from an E2 reaction of the following alkyl halides with hydroxide ion:
CLASS EXERCISE 2-(E2)
ANSWER:
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E1 REACTION
• First order elimination reaction• Must have at least two step
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Mechanism of E1 reaction
Alkylhalide dissociates,forming
carbocationA Base removes a
proton from β-carbon
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How does a weak base like water remove a proton froman 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
The major product in an E1 reaction is generally the more substituted alkene
Fewest β-Hydrogen (According to Zaitsev rule)
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Reaction coordinate diagram for the E1 reaction
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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
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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.
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Because the E1 reaction forms a carbocationintermediate, we need to consider carbocation rearrangement
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Competition Between E2 and E1 Reactions
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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1° Carbocation are too unstable to be formed in
E1 reaction
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Stereochemistry of the E2 ReactionThe bonds to the eliminated groups (H and X) must bein the same plane
The anti elimination is favored over the syn elimination
Parallel on the same
side
Parallel on the
opposite side
Syn elimination Anti elimination
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Consider the stereoselectivity of the E2 reaction
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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(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.
Anti-elimination
Syn-elimination
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Anti-elimination
Syn-elimination
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When only one hydrogen is bonded to the b-carbon, themajor product of an E2 reaction depends on the structureof the alkene
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Stereochemistry of the E1 ReactionThe major stereoisomer obtained from an E1 reaction isthe alkene in which the bulkiest substituents are onopposite 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.
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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
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SN2/E2 conditions
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A bulky alkyl halide or a sterically hindered nucleophileencourages elimination over substitution
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A strong or a bulky base encourages elimination oversubstitution
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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
Primary alkyl halides do not form carbocations; thereforethey cannot undergo SN1 and E1 reactions.
The elimination reaction favored at higher temperatures.
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ASSIGNMENT 1(SUBSTITUTION AND ELIMINATION-
ALKYLHALIDE)
Submit on 6th November (Tuesday)
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-The End-