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Elimination - E2 Mechanism, Rate, Energy Profile Diagram, Stereochemistry, Practice Problems and FAQ

Elimination - E2 Mechanism, Rate, Energy Profile Diagram, Stereochemistry, Practice Problems and FAQ

Imagine a beauty pageant contest in which contestants from various countries participated. In the finale, contestants from five countries are left and are made to stand in a line on

 a stage. In the second last round, contestants from the Philippines and France get eliminated such that those from Venezuela, South Africa and India, come closer. The ultimate aim of the contest is to filter the best suitable candidate. 

Similarly, in some organic reactions, atoms from two adjacent carbon atoms are eliminated to form a stronger bond containing product. 

Depending upon the mechanism of how two atoms are eliminated, Majorly they are of three types, E1, E2, E1cB. Let's study the E2 mechanism in detail.

TABLE OF CONTENTS

  • Elimination
  • Mechanism of Elimination 
  • E2 Reaction
  • Reactions That Follow E2 Mechanism
  • Rate of the Reaction in E2 Mechanism
  • Energy Profile Diagram for E2 Elimination Reaction
  • Stereochemistry of E2 Mechanism
  • Practice Problems
  • Frequently Asked Questions - FAQs

Elimination Reactions

In elimination, two atoms/substituents are removed from the adjacent atoms ( and positions) to form a new multiple bond. It is also known as elimination.

For example, when a haloalkane is made to react with a base, , in the presence of heat (, the elimination of the halogen (from the and proton ( from the takes place to give the corresponding alkene.

Mechanism of -elimination

Generally, the base ( ) abstracts a proton ( ) from the followed by the removal of halide ion from the . This can be understood with the help of the mechanism given below.

One of the important examples of elimination is dehydrohalogenation. In dehydrohalogenation, hydrogen from the and a halogen from the get eliminated when heated with an alcoholic solution of

Based on the fact that either abstraction of a proton or the removal of halogen is the first step or there will be simultaneous removal, elimination reactions can be classified into three types namely, E1 E2 and E1cB.

E2 Reaction

Generally, two groups/atoms depart simultaneously from adjacent carbons along with the proton being abstracted by a base. Example-Dehydohalogenation of alkyl halide in presence of base,

Reactions That Follow E2 Mechanism

1. Dehydrohalogenation of alkyl halide

As the name dehydrohalogenation suggests, halogen and hydrogen are removed. The reaction occurs in the presence of a strong base like . E2 reaction occurs in one step through a transition state. Let's consider a reaction of 2-chloro-2-methylpropane with in the presence of heat. It is a one-step process. 

Firstly, abstracts a proton from the followed by the simultaneous removal of to give. During the abstraction of the proton by the base and the removal of the leaving group, a transition state is formed. A partial double-bond character is observed in the transition state.

E2 reaction is a concerted reaction because the bond formation and the bond-breaking steps take place simultaneously. A concerted reaction is one in which the abstraction of a proton by a base and the departure of the leaving group occur simultaneously.

Rate of the Reaction in E2 Mechanism

Rate of reaction ∝

As the E2 mechanism is a single-step reaction hence, the rate of the reaction depends on the concentration of both the molecules involved. Thus, E2 elimination is a second-order reaction i.e. bimolecular reaction. The stability of alkene governs the rate of E2 reaction.

Example: Order of stabilities of alkenes will be:

Greater the stability of the alkene formed, more will be the rate of reaction.

Energy Profile Diagram for E2 Elimination Reaction

Let’s take the example of . It undergoes E2 elimination in the presence of methoxide ion such that two products are obtained, 2-butene and 1-butene. 

According to the Saytzeff rule, 2-butene will be the major product. In the E2 reaction of , removal of proton and takes place simultaneously. The removal of leads to the formation of two products. 2-butene being the stable alkene will have lesser energy than the relatively unstable alkene 1-butene. 

If the base is bulky or when the leaving group in E2 is poor, Hofmann rule is followed. It is just the opposite of the Saytzeff rule. According to the Hofmann rule, the least substituted alkene will be formed as the major product. Let’s understand the two cases where the Hofmann rule will be followed.

Case 1: When the base is bulky

Case 2: When the leaving group in E2 is poor

Stereochemistry of E2 Mechanism

E2 reaction is an anti-elimination reaction.

The transition state of E2 consists of an H atom and a leaving group on the substrate aligned in a plane. The alignment can be of two types: Anti-planar and syn-planar.

1. Anti-planar: The H and X atoms are oriented on the opposite sides in the plane of the molecule.

2. Syn-planar: The H and X atoms are oriented on the same side in the plane of the molecule.

Based on these two ways in which the and bonds align, elimination reactions can be of two types. 

1. Syn-elimination: If the substituents are removed from the same side of the bond, the reaction is known as syn-elimination.

2. Anti-elimination: If the substituents are removed from the opposite sides of the bond, the reaction is known as anti-elimination.

The base removes the proton from the alkyl halide that is oriented anti to the leaving group, and leaving group leaves-all in one concerted step. 

Practice Problems

Q1. Reaction conditions leading to Hofmann product is/are

  1. When the base is bulkier
  2. When the leaving group in E2 is poor
  3. When the base is weak 
  4. Both A and B

Answer: D

Solution: In the bimolecular elimination reactions (E2), the Hofmann product is obtained as the major product under two conditions:

  1. When the base is bulkier
  2. When the leaving group in E2 is poor

So, option D) is the correct answer.

Q2. More alkylated alkene is formed predominantly if the base is , while less alkylated alkene is obtained majorly when base is used. Which of the following statements is correct? 

  1. is a bulky base, so the Hofmann product is formed as the major product. 
  2. is a less bulky base, so the Saytzeff product is formed as the major product. 
  3. is a bulky base, so the Saytzeff product is formed as the major product. 
  4. Both A and B

Answer: D

Solution: is a bulkier base than . Bulky bases give product according to the Hofmann rule, which is the less substituted alkene. On the other hand, a more substituted alkene is obtained according to the Saytzeff rule if the base is less bulky.

So, option D) is the correct answer.

Q3. The reaction ofon reaction with alcoholic KOH produces: 

Answer: D

Solution: The reaction of on reaction with is an E2 elimination reaction. In an E2 reaction, the base removes the proton from the alkyl halide that is oriented anti to the leaving group ( ), and the leaving group leaves simultaneously. The removal of the proton by the base, the removal of the leaving group and the formation of the double bond all happen in one concerted step. The product formed is.

So, option D) is the correct answer.

Q4. The major product of the following reaction is:

A)



 

B)

C)

D)

Answer: B

Solution: Since the base is not bulky, the Saytzeff alkene is the major product formed.

So, option B) is the correct answer.

Frequently Asked Questions - FAQs

Q1. How to know whether the major product formed is according to the Hoffman or Saytzeff rule in elimination?
Answer: In the case of E1 elimination, the reaction occurs in presence of a weak base like , and a more substituted alkene will be formed as the major product according to the Saytzeff rule. On the other hand, in the case of E2 elimination, if the reaction occurs in the presence of a bulky base or poor leaving group is present, the least substituted product will be formed as the major product, according to the Hofmann rule.

Q2. Why is the least substituted alkene formed in the presence of a bulky base?
Answer: Bulky base abstracts the proton from the least hindered side. So, it abstracts the proton from that carbon which has a maximum number of hydrogen atoms, thus giving the least substituted alkene as the major product.

Q3. Give some examples of a poor leaving group?
Answer: Some examples of poor leaving groups are trialkyl ammonium salt () and dialkyl sulfonium salt ().

Q4. Other than dehydrohalogenation reaction, give one more example of E2 elimination reaction?
Answer. Dehalogenation of dihalide is another example of E2 elimination.

Related Topics

Electrophiles and Nucleophiles Types of Reactions
Nucleophilic Substitution Reaction-SN2 Nucleophilic Substitution Reaction-SNi, SNNGP
Substitution vs Elimination Factors Affecting SN1 and SN2
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