Elimination Reactions / Carbon Carbon Double bond formation Reactions
In an Elimination Reaction, Groups are eliminated from a Reactant.
For example, when an alkyl halide undergoes an elimination reaction, the halogen (X) is removed from one carbon and a proton is removed from an adjacent carbon.
A double bond is formed between the two carbons from which the atoms are eliminated.
Therefore, the product of an elimination reaction is an alkene (Olefin) i.e. C=C.
1.1) Introduction to C=C formation:
The formation of carbon - carbon double bonds is important in organic synthesis, not only for the obvious reason that the compound being synthesized may contain a double bond, but also because formation of the double bond allows the introduction of a wide variety of functional groups.
1.2) β-Elimination reactions:
One of the most commonly used methods for forming carbon - carbon double bonds is by β-Elimination reactions of the types.
where X = e.g. OH, OCOR, halogen, OSO2R, +NR3, etc.
Included among these reactions are acidcatalysed dehydrations of alcohols, solvolytic and base-induced eliminations from alkyl halides or sulfonates and the Hofmann elimination from quaternary ammonium salts.
They proceed by both E2 (elimination bimolecular) and E1 (elimination unimolecular) mechanisms.
A third mechanism, involving initial proton abstraction as the rate-determining step, followed by loss of X- is termed E1cB.
2) Saytzeff's Rule & Hofmann's Rule
It is an elimination reaction.
According to Saytzeff's rule , during dehydration, more substituted alkene (olefin) is formed as a major product, since greater the substitution of double bond greater is the stability of alkene.
It is found in general that acid-catalysed dehydration of alcohols and other E1 eliminations
In this type of elimination reaction the least substituted olefin is generally formed as a major product. This is called the Hofmann's Rule.
All such reactions bear charged leaving groups like -NR3 + or -SR2+and involve strong bases.
In E2 elimination both products are possible i.e. Saytzeff's & Hofmann's product.
3) E1 (Elimination Unimolecular):
Elimination reaction that alkyl halides can undergo is an E1 elimination.
The reaction of tert-butyl bromide with water to form 2-methylpropene is an example of an E1 reaction; "E" stands for elimination and "1" stands for unimolecular.
The major products generally the more substituted alkene. In this elimination reaction product always forms as "E" i.e. Trans product.
An E1 reaction is a first-order elimination reaction because the rate of the reaction depends only on the concentration of the alkyl halide.
rate = k[alkyl halide]
Only the alkyl halide is involved in the rate-determining step of the reaction.
Therefore, there must be at least two steps in the reaction.
When two elimination products can be formed in an E1 reaction, the major products generally the more substituted alkene.
Due to more stable transition state for the formation of internal alkene it undergo elimination with Saytzeff's rule to form internal double bond.
Relative reactivities of alkyl halides in an E1 reaction = relative stabilities of carbocations:
Relative reactivities of alkyl halides in an E1 reaction:
Because the E1 reaction forms a carbocation intermediate, the carbon skeleton can rearrange before the proton is lost, if rearrangement leads to a more stable carbocation.
In the following reaction, the initially formed secondary carbocation undergoes a 1,2-hydride shift to form a more stable secondary allylic cation:
Once carbocation formed then it will get rearrange to form anti double bond and internal double bond.
4) E2 (Elimination Bimolecular):
Elimination reaction that alkyl halides can undergo is an E1 elimination.
The reaction of The reaction of tert-butyl bromide with hydroxide ion is an example of an E2 reaction; "E" stands for elimination and "2" stands for bimolecular.
The product of an elimination reaction is an alkene.
The major product of the reaction depend on starting material and used base and their steric hindrance to form more or less substituted alkene.
In this elimination reaction product may be "E" or "Z".
The rate of an E2 reaction depends on the concentrations of both tert-butyl bromide and hydroxide ion. It is, therefore, a Second-Order Reaction
rate = k[alkyl halide][base]
The rate law tells us that both tert-butyl bromide and hydroxide ion are involved in the rate-determining step of the reaction.
The following mechanism agrees with the observed second-order kinetics:
E2 reaction is a concerted, one-step reaction: The proton and the bromide ion are removed in the same step, so no intermediate is formed.
4.1) Mechanism of the E2 Reaction :
The carbon to which the halogen is attached is called the -carbon.
A carbon adjacent to the α-carbon is called a β -carbon.
Because the elimination reaction is initiated by removing a proton from a β-carbon, an E2 reaction is sometimes called a β- elimination reaction.
It is also called a 1,2-elimination reaction because the atoms being removed are on adjacent carbons. ( by any base)
In a series of alkyl halides with the same alkyl group, alkyl iodides are the most reactive and alkyl fluorides the least reactive in E2 reactions because weaker bases are better leaving groups.
4.2) The Regioselectivity of the E2 Reaction
2-bromobutane has two structurally different β-carbons from which a proton can be removed.
So when 2-bromobutane reacts with a base, two elimination products are formed: 2-butene and 1-butene.
This E2 reaction is regioselective because more of one constitutional isomer is formed than the other.
What is the regioselectivity of an E2 reaction? In other words, what are the factors that dictate which of the two elimination products will be formed in greater yield?
To answer this question, we must determine which of the alkenes is formed more easily—that is, which is formed faster.
The reaction coordinate diagram for the E2 reaction of 2-bromobutane is shown in below figure :
- In the transition state leading to an alkene, the C-H and C-Br bonds are partially broken and the double bond is partially formed (partially broken and partially formed bonds are indicated by dashed lines), giving the transition state an alkene-like structure.
- Because the transition state has an alkene-like structure, any factors that stabilize the alkene will also stabilize the transition state leading to its formation, allowing the alkene to be formed faster.
- The difference in the rate of formation of the two alkenes is not very great.
Consequently, both products are formed, but the more stable of the two alkenes will be the major product of the reaction.
a) If bases are smaller then the forming product is Saytzeff's product :
b) Relative reactivities of alkyl halides in an E2 reaction:
Because elimination from a tertiary alkyl halide typically leads to a more substituted alkene than does elimination from a secondary alkyl halide, and elimination from a secondary alkyl halide generally leads to a more substituted alkene than does elimination from a primary alkyl halide, the relative reactivities of alkyl halides in an E2 reaction are as follows:
c) If there is Double bond in starting material
The conjugated alkene is the more stable alkene even though it is not the most substituted alkene.
The major product of each reaction is, therefore, the conjugated alkene because, being more stable, it is more easily formed
d) If Bases are bulky then Hoffman product:
Sterically bulky and the approach to the alkyl halide is sterically hindered, the base will preferentially remove the most accessible hydrogen.
In the following reaction, it is easier for the bulky tert-butoxide ion to remove one of the more exposed terminal hydrogens, which leads to formation of the less substituted alkene.
Because the less substituted alkene is more easily formed, it is the major product of the reaction.
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