What is E2 Reaction

E2 reaction is a type of elimination reaction. A beta-hydrogen and a leaving group are removed from a compound (substrate) to form a double bond in a single concerted step. The removal usually takes place in the presence of a strong base that removes the proton. The rate of reaction is proportional to the concentration of both the substrate and the base. Therefore, a good leaving group is required because it is involved in the rate-determining step. In addition, the leaving group must be coplanar in order to form a pi bond. The E2 reaction is common in alkyl halide and is similar to the SN2 reaction ^[1-7].

Characteristics of E2 Reaction

• Also known as bimolecular elimination
• One step removal mechanism process (concerted)
• Carbon-leaving group and carbon-hydrogen bonds break off simultaneously
• Carbon goes from sp3 to sp2 hybridization state
• Reaction rate is proportional to both the substrates and the base – second-order kinetics (R = k [substrate][base])
• Better leaving group leads to faster rates
• Alkoxide (RO^–) and hydroxide (OH^–) bases are generally used
• Polar aprotic solvents increase the rate
• Both hydrogen and leaving group must be on the same plane in order for the double bond to form (coplanar)
• Stereoselective and regioselective

Regioselectivity and Stereochemistry of E2 Reaction

Regioselectivity

In an E2 reaction, there can be more than one product. However, the most stable product is favored. Therefore, according to Zaitsev’s rule, the product with the most substituted carbon in the C=C pi bond will be thermodynamically the most stable (Zaitsev product).

The reactivity of the carbocation towards the E2 reaction is as follows:

Tertiary > Secondary > Primary > Methyl

Stereochemistry

The bond connecting the leaving group to the alpha-carbon and the bond connecting the hydrogen to the beta-carbon must be in the same plane. This orientation allows for the proper overlapping of the two carbon orbitals and facilitates the formation of the pi-bond. Furthermore, the hydrogen and the leaving group must be in anti-position to each other, i.e., they must be in an anti-periplanar geometry. This orientation allows the transition state of the reaction to be stable and lower in energy. Therefore, overall, the hydrogen and the leaving must be in anti-coplanar conformation. Because of the anti-coplanar conformation, one stereoisomer (E isomer) will be produced preferably over the other ^[8].

Examples of E2 Reaction ^[1-7]

1. Dehydrohalogenation of Alkyl Halides

1. In the presence of potassium ethoxide (C₂H₅O^– K⁺) in ethanol (C₂H₅OH), isobutyl bromide (C₄H₉Br) converts into isobutene (C₄H₈).

C₄H₉Br + C₂H₅O^– K⁺ → C₄H₈ + C₂H₅OH + KBr

2. In the presence of potassium ethoxide (C₂H₅O^– K⁺) in ethanol (C₂H₅OH), 2-bromopropane (C₃H₇Br) gives propene (C₃H₆).

C₃H₇Br +→ C₃H₆ + C₂H₅OH + KBr

2. Dehydrohalogenation of Cyclic Compound

In the presence of alcoholic potassium hydroxide (KOH/C₂H₅OH), 2-methyl-1-chlorocyclohexane (C₇H₁₃Cl) gives 1-methylcyclohexene (C₇H₁₂).

C₇H₁₃Cl + + C₂H₅O^– K⁺ → C₇H₁₂ + C₂H₅OH + KBr

Mechanism of E2 Reaction

The mechanism of the E2 reaction is a single-step concerted process with one transition state. First, the base removes the proton from the beta-carbon, and the electron pair in the beta-carbon forms a pi-bond with the alpha carbon. Then, finally, the leaving group comes off from the alpha-carbon with an electron pair ^{[1-7, 9-12]}.

Reactivity of Carbocation

The reactivity of the carbocation towards the E2 reaction is as follows:

Tertiary > Secondary > Primary > Methyl