E2 reactions: Mechanism, Stereochemistry, and Examples

E2 reactions, as the name indicates, are bimolecular, single-step β-elimination reactions. Bond breaking and formation occur simultaneously in a single transition state. E2 reactions are regioselective and the Zaitsev rule is followed. It is favored in a polar aprotic solvent such as acetone and dimethyl sulfoxide as the attacking base gets solvated by strong hydrogen bonds in polar protic solvents (such as water) and becomes less effective.

E2 reactions are analogous to bimolecular nucleophilic substitution reactions(S_N2). That is the reason why they compete with each other. In addition, these reactions follow second-order kinetics, such that they depend upon the concentration of both the base and alkyl halide.

Rate = k [substrate][Base] → (1)

Additionally, in the E2 reaction, the stereochemistry of the substrate greatly influences the rate of reaction. The stereochemistry can be determined by the relative orientation of the hydrogen atom and the leaving group on the adjacent carbon atom. If the substituents are anti-periplanar i.e., at an angle of 180º to each other in the same plane, the reaction will proceed at a relatively faster rate.

E2 Reaction Mechanism

E2 is a bimolecular (second-order) elimination reaction. It consists of a single-step mechanism. In this single step, the base attacks a substrate to abstract a proton of β-carbon, and a leaving group, leaves from alpha-carbon simultaneously, to form a double bond. This reaction typically occurs with strong bases such as hydroxide or alkoxide ions.

The E2 reaction mechanism is given by the general equation given below:

Where,

B is the attacking base and X is the leaving group.

Removal of a proton, H⁺.
Formation of (pi) 𝝅- bond.
The departure of the leaving group.

In the E2 reactions, the base eliminates a β-hydrogen and halide simultaneously from carbon and forms a double bond along with conjugate acid and halide. At the completion of the response, the C=C double bond and water molecule are completely formed, with the removal of the leaving group.

Orientation of double bond

The orientation of the double bond in the E2 reaction can be determined by the relative stability of possible alkenes. If a substrate has a β-hydrogen on only one carbon atom, there is no doubt about double bond positions and the reaction is regiospecific. On the other hand, if the substrate if two β-hydrogens are available, then the product formation is regioselective and follows Zaitsev’s rule.

If the substrate contains an uncharged leaving group then the double bond will preferably go in such a direction to highly substituted products can be produced. However, if a substrate contains a charged leaving group the double bond will go toward the least substituted carbon (Hofmann rule).

Competition of S_N2 and E2 reaction mechanism

E2 reactions compete with SN2 reactions. Both these reactions lead to entirely different products. There is always competition between them because both are closely related in the mechanism. A specific substrate responds to changes in reaction conditions and favors either E2 or S_N2 reactions. Both are bimolecular reactions and the relative rate of their reaction depends upon the carbonation arrangement. However, the presence of a strong base always favors the E2 reaction even with the tertiary substrate.

Stabilty of carbocation

The solvent also affects the reaction mechanism, if the solvent behaves as a nucleophile the reaction would proceed through the S_N2 pathway whereas, if the solvent behaves as a base then the reaction would proceed through the E2 pathway. Protic solvents are capable of H-bonding favor E2 elimination reactions. Moreover, the temperature effect is also essential to determine the reaction, whether E2 or SN2. For instance, high temperature favors E2, whereas low temperature favors SN2 reaction.

Stereochemistry of E2 reactions

E2 reactions are stereoselective reactions. The stereochemistry of E2 reactions depends on the number of β-hydrogen. When there are two β-hydrogens attached to that carbon atom from which H is eliminated then the E2 reaction will be stereoselective(not stereospecific) and yield more stable products. When only one β-hydrogens is attached to alkyl halide it gives a stereospecific isomer in such cases, only one conformation is possible which is not more stable.

Example of stereoselective E2 reactions

stereoselective e2 reaction

Example of stereospecific E2 reaction

stereospecific e2 reaction

Examples of E2 reactions

Dehydrohalogenation of alkyl halides

Alkyl halides undergo elimination to produce alkenes. The dehydrohalogenation of alkyl halides is a β -elimination reaction. it involves the loss of hydrogen and a halide from an alkyl halide. During Dehydrohalogenation Alkyl halides react with a strong base, such as sodium ethoxide.

E2 reaction of Cyclic compounds

Key Takeaways

E2 elimination is a bimolecular elimination reaction.
It is a single-step reaction.
In the E2 reaction mechanism, the carbon-hydrogen and carbon-leaving group (C-X) bonds break to form a new double bond.
Secondary and tertiary substrate gives an E2 elimination reaction.
E2 elimination reaction is a second-order reaction.
The rate of reaction depends on the concentration of substrate and base.
Strong bases are required for the E2 elimination reaction.
The Stereochemistry of E2 reactions depends on the number of β-hydrogen.
In Stereoselective E2 Reactions the trans-alkene product will usually be favored.
E2 reactions are regioselective.
The polar aprotic solvent is suitable.

Related Resources

Concepts Berg

Is the E2 stereoselective or stereospecific?

The E2 reaction is stereoselective. It gives stable stereoisomers as major products.

Why are trans products mostly produced in excess in the E2 elimination reaction?

During the E2 elimination reaction, two stereoisomers cis and a trans alkene are formed. Trans alkene is the major product due to its higher stability.

What are the similarities between E2 and S_N 2 reactions?

Similarities between E2 and S_N 2 reactions are given below:

Both E2 and S_N2 reactions are bimolecular.
They both have a single-step mechanism.
Both follow second-order kinetics.
They required a base to proceed with their reactions.
Both reactions are common in the primary and secondary structures of organic compounds.
They required a strong leaving group.

What are the differences between E2 and S_N 2 reactions?

The key differences between E2 and S_N2 reactions are as follows:

E2 is an elimination reaction, whereas S_N2 is a substitution reaction.
E2 reactions require a strong base whereas S_N2 reactions require a strong nucleophile.
In the E2 reaction alkenes are formed as a product, on the other hand, S_N2 substituted products are formed.
High temperature favors E2, whereas low temperature favors S_N2.

What product do we obtain as a result of the E2 reaction?

Zaitsev products(highly substituted alkene) are formed as a result of the E2 reaction.

What factors favor E2 over S_N 2 reactions?

The following factors favor E2 over S_N2

High temperature
strong base
Polar aprotic solvent

Which compound will undergo E2 reaction?

E2 reactions proceed with secondary and tertiary alkyl halides, a strong base is necessary with a primary halide. The mechanism should be a single-step reaction with one transition state.

What is the effect of the base on E2 reactions?

The rate of E2 reaction depends on strength of the base. E2 reactions are regioselective and favor the formation of Zaitsev products. The base appears in the rate equation, hence the rate of the E2 elimination reaction increases as the strength of the base increases.

Why are E2 reactions irreversible?

E2 reaction is irreversible because they are kinetically controlled reactions.

References

12th edition of Organic Chemistry by T.W Graham Solomons, Craig B. Fryhle, and Scott A. Synder.
A textbook of organic chemistry by M.Younas.
7th edition of March’s Advanced Organic Chemistry by Michael B. Smith
Lecture 13