Elimination reaction in Organic Compounds involves the removal of certain groups from the reactant molecule usually leading to the formation of an unsaturated compound as a product.
For example the reaction between an alcohol and a mineral acid at high temperature is an elimination reaction leading to formation of an alkene along with the elimination of a molecule of water.
The elimination reaction between an alkyl halide and a base that leads to the formation of an alkene can follow three major pathways. They are named E1, E2 and E1CB. In the name E stands for elimination (as expected) and the numbers 1 and 2 stand for unimolecular and bimolecular respectively. In E1CB, CB stands for conjugate base.
As we see in the above example, atoms from adjacent carbon atoms are removed; E1 and E2 reactions are often referred to as beta-elimination.
Let us talk about the elimination reaction through E2 mechanism.
Elimination reaction through E2 mechanism is a concerted reaction i.e. the reaction happens in a single step. Both the base and the alkyl halide take part in that step. As this reaction happens in only one step that step is the rate determining step and as both the base and the alkyl halide are involved in the rate determining step, the rate of the reaction will depend on the concentrations of both the alkyl halide and the base. So, this reaction is a second order reaction.
The mechanism of E2 elimination is shown below.
The base abstracts the hydrogen in the beta position to the leaving group and starts the reaction. At the same time a pi-bond is formed between the adjacent carbon atoms and also the halide is eliminated. A bond between the base and the hydrogen starts forming; the bond between the hydrogen and carbon starts breaking; a pi-bond between the adjacent carbon atoms start forming; the bond between the carbon and halogen starts breaking. This reaction goes through a transition state in which the bonds that start breaking become weaker and weaker and eventually break completely while the bonds that start forming become stronger and stronger and eventually become completely formed bonds.
The transition state of the reaction is shown below (the dotted lines represent a bond that is either being formed or being broken).
There are three important things related to a E2 elimination reaction.
1. The beta – hydrogen and the halogen should be on the same plane.
As seen in the mechanism a pi-bond is formed between two adjacent carbon atoms during the reaction at the same time when the C – H bond and C – Br bond are breaking. A pi-bond is formed by the sideways overlap of two p-orbitals. For maximum overlap and thereby form a strong bond both the p-orbitals should be parallel to each other on the same plane. So, the beta-hydrogen and the halogen should be on the same plane. This is necessary because the sp3 orbital of the carbon bonded to the hydrogen and the sp3 orbital of the carbon bonded to the halogen become the p-orbitals that overlap to form the pi-bond.
2. The beta-hydrogen and the halogen should preferably be anti-coplanar
Not only the beta-hydrogen and the halogen should be on the same plane they should preferably be anti (opposite sides) to each other than syn (same side). If elimination happens in anti-coplanar conformation it is called anti-elimination. This is necessary to avoid steric and electronic repulsions in the transition state which increase the energy of the transition state. The rate of the reaction depends on the energy of the transition state. The higher the energy of the transition state the difficult it is for it to be formed and slower the rate of the reaction. Any factor increasing the energy of the transition state will slow the reaction down. However, in molecules where the beta-hydrogen and the halogen are syn-coplanar due to the geometry of the molecule, syn-elimination happens.
3. More substituted alkene is formed as the major elimination product
The stability of an alkene depends on the number of alkyl substitutions present on the carbon atoms forming the pi-bond or more precisely, more the number of alpha-hydrogens on an alkene more stable it will become due to hypeconjugation. During elimination reactions, the substrates may have two different sets of beta-hydrogens, one leading to the formation of less susbtituted alkene and one leading to the formation of more substituted alkene. The more substituted alkene is formed as the major product as it is more stable. The product of the E2 elimination reaction resembles the transition state from which it is formed. The transition state leading to a more substituted alkene will be less energetic than the transition state leading to a less substituted alkene.
Apart from the three things mentioned above there are other interesting features too of E2 elimination such as the stereochemical outcome, competition with E1 and SN2 reactions etc.
For example the reaction between an alcohol and a mineral acid at high temperature is an elimination reaction leading to formation of an alkene along with the elimination of a molecule of water.
The elimination reaction between an alkyl halide and a base that leads to the formation of an alkene can follow three major pathways. They are named E1, E2 and E1CB. In the name E stands for elimination (as expected) and the numbers 1 and 2 stand for unimolecular and bimolecular respectively. In E1CB, CB stands for conjugate base.
As we see in the above example, atoms from adjacent carbon atoms are removed; E1 and E2 reactions are often referred to as beta-elimination.
Let us talk about the elimination reaction through E2 mechanism.
Elimination reaction through E2 mechanism is a concerted reaction i.e. the reaction happens in a single step. Both the base and the alkyl halide take part in that step. As this reaction happens in only one step that step is the rate determining step and as both the base and the alkyl halide are involved in the rate determining step, the rate of the reaction will depend on the concentrations of both the alkyl halide and the base. So, this reaction is a second order reaction.
The mechanism of E2 elimination is shown below.
The base abstracts the hydrogen in the beta position to the leaving group and starts the reaction. At the same time a pi-bond is formed between the adjacent carbon atoms and also the halide is eliminated. A bond between the base and the hydrogen starts forming; the bond between the hydrogen and carbon starts breaking; a pi-bond between the adjacent carbon atoms start forming; the bond between the carbon and halogen starts breaking. This reaction goes through a transition state in which the bonds that start breaking become weaker and weaker and eventually break completely while the bonds that start forming become stronger and stronger and eventually become completely formed bonds.
The transition state of the reaction is shown below (the dotted lines represent a bond that is either being formed or being broken).
There are three important things related to a E2 elimination reaction.
1. The beta – hydrogen and the halogen should be on the same plane.
As seen in the mechanism a pi-bond is formed between two adjacent carbon atoms during the reaction at the same time when the C – H bond and C – Br bond are breaking. A pi-bond is formed by the sideways overlap of two p-orbitals. For maximum overlap and thereby form a strong bond both the p-orbitals should be parallel to each other on the same plane. So, the beta-hydrogen and the halogen should be on the same plane. This is necessary because the sp3 orbital of the carbon bonded to the hydrogen and the sp3 orbital of the carbon bonded to the halogen become the p-orbitals that overlap to form the pi-bond.
2. The beta-hydrogen and the halogen should preferably be anti-coplanar
Not only the beta-hydrogen and the halogen should be on the same plane they should preferably be anti (opposite sides) to each other than syn (same side). If elimination happens in anti-coplanar conformation it is called anti-elimination. This is necessary to avoid steric and electronic repulsions in the transition state which increase the energy of the transition state. The rate of the reaction depends on the energy of the transition state. The higher the energy of the transition state the difficult it is for it to be formed and slower the rate of the reaction. Any factor increasing the energy of the transition state will slow the reaction down. However, in molecules where the beta-hydrogen and the halogen are syn-coplanar due to the geometry of the molecule, syn-elimination happens.
3. More substituted alkene is formed as the major elimination product
The stability of an alkene depends on the number of alkyl substitutions present on the carbon atoms forming the pi-bond or more precisely, more the number of alpha-hydrogens on an alkene more stable it will become due to hypeconjugation. During elimination reactions, the substrates may have two different sets of beta-hydrogens, one leading to the formation of less susbtituted alkene and one leading to the formation of more substituted alkene. The more substituted alkene is formed as the major product as it is more stable. The product of the E2 elimination reaction resembles the transition state from which it is formed. The transition state leading to a more substituted alkene will be less energetic than the transition state leading to a less substituted alkene.
Apart from the three things mentioned above there are other interesting features too of E2 elimination such as the stereochemical outcome, competition with E1 and SN2 reactions etc.
