ALKYLHALIDE. By Mrs. Azduwin Khasri 23 rd 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.
By Mrs. AzduwinKhasri
23rd October 2012
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.
The removal of a proton and a halide ion is called dehydrohalogenation
An E2 reaction is also called a b-elimination or a 1,2-elimination reaction
Carbon attached to halogen
Carbon adjacent to α-carbon
2 product produce
2 structurally β-carbon
The major product of an E2 reaction is the most stable
2-Butene formed faster than 1-butene
The more substituted alkeneproduct is obtained when a
proton is removed from the b-carbon that is bonded
to the fewest hydrogens
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.
Bulky Alkyl halide
Bulky bases affect the product distribution resulting in the Hofmann product, the least substituted alkene
Less sterically hindered
Major product- Most stable product
When the halogen is fluorine-The major product of the E2 reaction of alkyl Fluoride is the less substituted alkene.
STRONGEST BASE-POOREST LEAVING GROUP
The Fluoride ion does not have as strong a propensity to leave as another halide ion.
Major product-Less substituted alkene
A carbanion-like transition state
Carbocation stability 3° > 2° > 1°
Carbanion stability 1° > 2° > 3°
Which of the alkyl halides is more reactive in an E2 reaction?
Producing more stable alkene
Br better leaving group
Producing more stable alkene
Give the major elimination product obtained from an E2 reaction of the following alkyl halides with hydroxide ion:
A Base removes a proton from β-carbon
an sp3 carbon?
1) The presence of a positive charge greatly reduces the
2) Hyperconjugation weakens the C-H bond by electron
Fewest β-Hydrogen (According to Zaitsev rule)
The major product in an E1 reaction is generally the more substituted alkene
E1 reaction depends both on the ease with which the carbocation
is formed and how readily the leaving group leaves
intermediate, we need to consider carbocation
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
The bonds to the eliminated groups (H and X) must be
in the same plane
Parallel on the opposite side
Parallel on the same side
The anti elimination is favored over the 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
major product of an E2 reaction depends on the structure
of the alkene
The major stereoisomer obtained from an E1 reaction is
the alkene in which the bulkiest substituentsare on
opposite sides of the double bond
Bothsyn 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.
Substitution and Elimination
Alkyl halides can undergo SN2, SN1, E2, and E1
3) Decide how much of the product will be the substitution product
and how much of the product will be the elimination product
encourages elimination over substitution
Why? Because elimination is entropically favorable.
The elimination reaction favored at higher temperatures.
Primary alkyl halides do not form carbocations; therefore
they cannot undergo SN1 and E1 reactions.
Submit on 6th November (Tuesday)