Bimolecular Elimination Reactions

Bimolecular Elimination Reactions (E₂Reaction)

The elimination reaction in which the rate of reaction is dependent on both the concentration of substrate and reagent, i.e. kinetically second order. Since, the reaction follows second order kinetics so, takes place in a single step. That means the leaving group (LG) and proton from β-carbon eliminates simultaneously.
Rate ∝ [Substrate] [Base]
Rate = k [Substrate] [Base]

Mechanism of Bimolecular Elimination (E₂) Reactions

The E₂ mechanism is a single-step reaction process. During the reaction, the base attacks the β-carbon's hydrogen atom and removes, simultaneously carbon-carbon double bond forms by departure of the leaving group (nucleophile) from α-carbon. In the transition state, the β C-H and α C-LG bonds are stretched on the attack of the reagent with the incipient π-bond formation.

The energy of the transition state is least when the two leaving groups, the α and β carbons and the attacking base are coplanar in the transition state. The two leaving groups should be trans to each-other (i.e., in an anti-periplanar geometry) for effective π-bond formation. In case of alkenes, where double bond cannot rotate, elimination is difficult when the two leaving groups are cis- to each-other.

Conditions for Bimolecular Elimination Reactions

1. Branching at α- and β- carbons of the substrate for stability of the olefin.
2. Strong base of high concentration required to break a strong C-H bond.
3. Solvent of low polarity i.e. polar aprotic solvent (usually DMF and DMSO used) required because polar aprotic solvents do not interact with the base (no hydrogen bonding) and leaving the base naked and more reactive.
4. Low temperature.
5. Good leaving group (weak bases are good leaving group).

Stereoselectivity of Bimolecular Elimination Reactions

E₂ reactions are stereoselective that means two stereoisomers are possible but one of them is formed in excess.

Two stereoisomers are formed – a cis and a trans alkene in the above example. The trans alkene is the major product because of higher stability of the trans isomer.

Stereospecificity of Bimolecular Elimination Reactions

If there is only one β-hydrogen in the substrate and no choice of β-hydrogen for the elimination in that case the structure of the product depends on the structure of the reactant.

In the reaction given below, only the Z alkene is formed even though it is less stable than the E alkene, because, there is no choice of β-hydrogen elimination.

Trick to Predict the Product of Stereospecific E₂

1. If the β-hydrogen and leaving group (LG) are trans to each-other (180°), eliminate them and draw a double bond between α- and β-carbon as per zig-zag.

2. If the β-hydrogen and leaving group (LG) are cis to each-other, flip the zig-zag and draw the alkene.

Remember
☛ If there are two β hydrogens, it is stereoselective.
☛ If there is only one β hydrogen, it is stereospecific.

Examples of Bimolecular Elimination Reactions

E₂ Reaction in Cyclohexane Derivatives

In order for E₂ reaction to occur, the leaving group and the hydrogen must be anti to each other. In cyclohexane derivatives, this requires the leaving group and the hydrogen to be trans and the ring must be in the conformation where both of these groups are axial. In the give example, the conformation that has the chlorine axial also has the methyl and isopropyl (i-pr) groups axial. The conformation is less stable than its ring flip (ring-inversion) conformation. As a result, E₂ reaction is slow because an indivisual molecule spends only a small amount of time in the required conformation with the chlorine axial. Only one hydrogen is axial to chlorine, so the reaction proceeds a single alkene product.

In the example given below, the conformation that has the chlorine axial has the methyl and isopropyl groups equitorial. This conformation is more stable than its ring flip conformation. As a result, E₂ reacton is about 40 times faster. Both the axial hydrogens are anti to the chlorine. So, reaction proceeds to two alkene products. In accordance with Saytzeff's rule, the more highly substituted alkene resulting from the blue colored hydrogen is the major product.

Also read E₁ Reaction | E_1cB Reaction | E_i Reaction