It is common for some organic compounds to have double bonds. Rather than having only a single bond as an alkane, it is common for the molecule to be an alkene.
Double bonds can be created by organic synthesis. One way to do this is by means of an elimination reaction. By reacting with a basic reagent, it is possible to create a compound with a double bond. This is called an elimination reaction.
There are three major types of elimination reactions, two of which are the E1 and E2 reactions. The E2 reaction is particularly important, and these synthetic reactions can be used to create alkenes.
So what kind of synthetic reaction is an elimination reaction? We will explain the reaction mechanism and how it works.
Table of Contents
- 1 In the Elimination Reaction, a Double Bond Is Made from an Alkane: β Elimination
- 2 Reaction Mechanism of Alkenes by E1 Reaction
- 3 The E2 Reaction Is a One-Step Reaction and the Nucleophile Is Involved in the Reaction Rate
- 4 E1 and E2 Reactions That Compete with SN1 and SN2 Reactions
- 5 E1cB Reaction, in Which Conjugation Causes Anion Production
- 6 Understand the Reaction Mechanism and Learn the Differences in Elimination Reactions
In the Elimination Reaction, a Double Bond Is Made from an Alkane: β Elimination
What is an elimination reaction is a synthetic reaction that creates a double bond, as I mentioned earlier. It is a frequently used technique for making alkenes from alkanes.
Why is it called an elimination reaction, even though it is a synthetic reaction to make a double bond? It’s because the substituents in the molecule leave to create a double bond.
In the elimination reaction, the hydrogen is pulled out of the molecule by a basic substance. Not only that, but the leaving groups attached to the adjacent carbons are leaved. As a result, a double bond is formed. These synthetic reactions in organic chemistry are elimination reactions.
*B stands for Base and L for Leaving group.
If the carbon atom which hydrogen is eliminated is the α-position, the next carbon atom is in the β-position. This reaction is called β-elimination because the leaving group in the β-position becomes leaved.
There Are Three Types of Elimination Reactions
Note that there are three types of elimination reactions. It is common that the hydrogen atoms are withdrawn by the basic substances and leaved while forming a double bond. However, the reaction mechanisms are different for each type of reaction. There are three types of reactions as follows
- E1 reaction: first, the leaving group is leaved and then the hydrogen atoms are pulled out.
- E2 reaction: simultaneous leaving of a leaving group and withdrawal of a hydrogen atom.
- E1cB reaction: first the hydrogen atoms are pulled out, followed by the leaving of the leaving groups.
Each reaction differs in terms of when the hydrogen atoms are pulled out.
They are all the same in terms of creating a double bond. But because the reaction mechanism is different, you have to learn to clearly separate them in organic chemistry.
Reaction Mechanism of Alkenes by E1 Reaction
Let’s start by checking the E1 reaction. In order for the E1 reaction to occur, the first thing to start is for the leaving group to break away. It looks like the following.
As the leaving group (L) leaves the molecule, it gives rise to a carbocation as an intermediate; in the E1 reaction, the speed of the reaction depends on how quickly the carbocation is produced. Think of the reaction rate as dependent on the carbocation, not on the strength of the basicity or the concentration of the reagent.
As soon as the carbocation is formed, the hydrogen is withdrawn to make a double bond; in the E1 reaction, it is a two-step elimination reaction in which an intermediate is formed and then a double bond is formed.
Stability of the Carbocation Is Important in the E1 Reaction
Since the carbocation is formed first in the E1 reaction, how stable the carbocation can be has a significant effect on the reaction rate.
There is no doubt that the carbocation is an unstable intermediate. However, if the intermediate is not stable to some extent, the intermediate will not be produced in the first place. It is precisely because the intermediate carbocation is stable that the E1 reaction proceeds.
These carbocations have a certain order of stability. The order is as follows.
- Tertiary > Secondary > Primary
No primary carbocation or methyl carbocation is produced. This is because they are very unstable. On the other hand, if it is a tertiary carbocation, it is produced spontaneously. For secondary carbocation, they do occur in a small amount.
You think of the E1 reaction, you should understand that it is mainly tertiary carbocation. The E1 reaction proceeds at the carbocation, which has three alkyl chains attached to it.
Saytzeff Rule to Obtain Synthetic Compounds
What kind of double bond is formed in the compound obtained in the E1 reaction? We explained earlier that alkenes are formed when the hydrogen is extracted. However, the position at which the hydrogen is withdrawn differs, which results in the formation of the following two types of compounds.
Also, if you look at the molecules, one has two hydrogen atoms and the other has six hydrogen atoms. Therefore, in terms of probability, the probability of a chemical reaction occurring is higher with six hydrogen atoms.
Moreover, with two hydrogen atoms, the base has to penetrate into the molecule in order to pull out the hydrogen atom. On the other hand, with six hydrogen atoms, the base can easily pull out the bare hydrogen atoms that exist outside the molecule. Therefore, it might seem that the reaction proceeds in the six-atom part.
In fact, however, the E1 reaction proceeds with two hydrogen atoms. Why does this happen?
There is a natural law of organic chemistry called the Saytzeff rule. The Saytzeff rule refers to the law of synthesis that ensures that when an alkene is made, there are many substituents in the molecule. As a result, alkenes with many substituents are synthesized.
The reason why the Saytzeff rule takes preference is that the transition state of the intermediate tends to be more stable. An alkene with more substituents requires less activation energy than an alkene with fewer substituents. The synthesis of multisubstituted alkenes is preferred due to the lower energy required for the reaction.
The Reason Why Multisubstituted Alkenes Are Stable is Hyperconjugation
Why do alkenes with more substituents tend to be more stable? The reason for this is rarely explained in a straightforward manner.
The content gets a little more difficult, so you can just remember, in the elimination reaction, multisubstituted alkenes are formed by the Saytzeff rule. So it is not a problem to skip this part. But if you want to know the reason in more detail, you can learn about the mechanism by which multisubstituted alkenes become stable.
In the stability of alkenes, the following is in order.
In the Saytzeff rule, this order of stability is important. One of the reasons for this order is the hyperconjugation. So what is hyperconjugation?
It is generally known that electrons are more easily stabilized when they are distributed over a wide range. All students of organic chemistry understand that if you can write a conjugated structure, it will be stable. This is because electrons can exist in many places due to delocalization.
If there is a double bond, there will always be a π bond there. A single bond is a σ-bond, and for these σ-bonds, the π-bond is the bond in its perpendicular state.
Even if there is a double bond, if it is a hydrogen atom that is attached to a carbon atom, it cannot interact with the carbon atom.
On the other hand, what if not a hydrogen atom but an alkyl chain is attached to the carbon atom as a substituent? In this case, the C-H bond and the pi bond of the double bond are parallel. As a result of the parallelization, they share each other’s electrons weakly.
This is hyperconjugation. The electrons are not distributed over a wide area by writing a resonance structure, as in a conjugate structure. It is just that the orbits are parallel, and the electrons are weakly shared, which is hyperconjugation. Due to the effect of hyperconjugation, it is more stable in alkenes with many substituents.
For these reasons, the Saytzeff rule is followed when creating a double bond by an elimination reaction.
In the Case of a Bulky Base, a Steric Hindrance Leads to Hofmann Elimination
By the way, some elimination reactions ignore the Saytzeff rule. Instead of making many substituents, the elimination reaction proceeds in such a way that it makes fewer substitutions. This is called Hofmann elimination.
When does a Hofmann elimination occur and not the Saytzeff rule? It occurs when a bulky nucleophile (a base with a large sterically-hindered) is used. Bulky bases include, for example, tert-butoxide.
Unlike common bases, tert-butoxide is highly sterically hindered by the addition of three methyl groups to the base. When using these bulky nucleophiles (bases), it is difficult for the reaction to occur according to the Saytzeff rule. As explained earlier, the hydrogen atoms inside the molecule have to be pulled out.
Because of the steric hindrance, the hydrogen atoms inside the molecule cannot be pulled out. On the other hand, hydrogen atoms bonded to the outside of a molecule can be pulled out without any problem, even from a bulky nucleophile.
Normally, we use reagents with low steric hindrance as bases. For this reason, the Saytzeff rule is followed. On the other hand, if a base with a large sterically-hindered base is used, it is possible to make alkenes with fewer substituents by Hofmann elimination.
The E2 Reaction Is a One-Step Reaction and the Nucleophile Is Involved in the Reaction Rate
On the other hand, how does the E2 reaction work? In the E2 reaction, the following things occur simultaneously.
- A nucleophilic agent (base) attacks a hydrogen atom.
- The leaving group is leaving.
In the E1 reaction, the first thing that starts to happen is that the leaving group is leaved. In the E2 reaction, on the other hand, these reactions occur simultaneously to create a double bond. Therefore, the following reactions occur at once.
The stability of the carbocation is not relevant in the E2 reaction because no carbocation is formed. The E2 reaction occurs at all alkyl groups, tertiary, secondary and primary. Since there are no intermediates, the E2 reaction is a one-step elimination reaction.
Also, since the reaction occurs in a single step, the rate of the E2 reaction involves both the nucleophilic reagent (basic compound) and the reacting molecules.
Anti-Elimination (Antiperiplanar) Causes the Synthetic Reaction to Proceed
What is the key to the E2 reaction? It’s stereochemistry.
If a new double bond is created in a compound with an alkyl chain attached to it, then two types of compounds are created. These are so-called E/Z isomers, and there are two types: cis and trans. For this, which of the following compounds is produced in the E2 reaction?
In order to understand this, you must learn two things
- The E2 response occurs in antiperiplanar.
- Anti-elimination occurs in a placement with less steric hindrance.
Let’s start by checking out the antiperiplanar. Considering the Newman projection, we can consider the following two conformations in stereochemistry.
When a hydrogen atom and a bromine atom (a leaving group) are in the same position, it is called a synperiplanar. It is an overlapping form, and in this case the energy state is high and unstable.
On the other hand, if the hydrogen atom and the bromine atom are on opposite sides, it is called an antiperiplanar. In this case, the atoms do not overlap each other and are not in a high energy (unstable) state. Moreover, only in the antiperiplanar state are the orbits perfectly parallel.
For this reason, E2 elimination proceeds in the antiperiplanar state. This is called anti-elimination.
Trans (E) Is Obtained to Reduce Steric Hindrance
And while we understand that the E2 reaction is anti-elimination, does it produce either cis (Z) or trans (E)? In this regard, trans (E) is the main one produced among the E/Z isomers.
Considering the Newman projection, in the cis (Z), the methyl groups are located next to each other. As a result, the energy is higher due to steric repulsion.
In contrast, what about trans (E)? In the case of trans, the methyl group is on the opposite side. Therefore, there is no steric repulsion and it is easy for elimination reaction to occur in this state. Let’s understand that in the elimination reaction, the main product of the trans (E) is obtained by anti-elimination.
Cyclohexane Undergoes an Elimination Reaction at the Axial Position
Understanding the stereochemistry in these elimination reactions will help us to understand the elimination mechanism of the E2 reaction in cyclohexane. Cyclohexane, also known as a cyclic compound, can take on two conformations. They are axial and equatorial.
The one in the horizontal position is the equatorial. On the other hand, the one in the upper or lower position is the axial. Of these, it is in the axial that the E2 reaction takes place.
Equatorial does not allow for an antiperiplanar state. In other words, the E2 reaction does not occur. Therefore, the synthetic reaction always proceeds in an axial.
In the case of cyclic compounds such as cyclohexane, the Saytzeff rule may not apply. For example, the following case is an example.
When making the synthesis involving cyclohexene, there is only one hydrogen atom in the compound shown in the above figure that will be in the antiperiplanar position when it becomes an axial. As a result, a compound is synthesized that does not follow the Saytzeff rule.
E1 and E2 Reactions That Compete with SN1 and SN2 Reactions
So far, we have identified the E1 and E2 reactions. In the E1 and E2 reactions, a double bond is created by the use of a nucleophile (base).
However, when nucleophiles are used, they can cause not only elimination reactions but also nucleophilic substitution reactions. The SN1 and SN2 reactions are known to be nucleophilic substitution reactions. In fact, both elimination reactions and nucleophilic substitution reactions often occur at the same time.
In such cases, how can we distinguish when each synthetic reaction occurs? In addition to the E1 and E2 reactions, the SN1 and SN2 reactions must also be identified. These reactions can be summarized in the following table.
|Strong nucleophile||SN2||SN2||E1 or E2 or SN1|
|Strong base||E2 (SN2)||E2 (SN2)||E2|
|–||SN1 or E1||SN1 or E1|
The important parts of the table above are noted in blue. When trying to predict which reaction will occur, focus on the blue text.
The reaction conditions will change which reaction proceeds. This section checks why different nucleophiles are used in different synthetic reactions.
SN1 or SN2 Reactions Occur When There Is High Nucleophilicity and Little Steric Hindrance
When considering the reagents to be reacted with, the SN1 or SN2 reactions occur with reagents that are highly nucleophilic and have little steric hindrance. In the SN1 and SN2 reactions, the nucleophile must approach and attack the interior of the molecule as follows.
Therefore, smaller molecules (reagents with less steric hindrance) will cause SN1 or SN2 reactions.
For strongly nucleophilic compounds, nucleophilic substitution reactions take precedence. These highly nucleophilic compounds include the following
In the case of these nucleophiles, the nucleophilic substitution reaction is the main process.
A Bulky Nucleophile or Strong Base Can Be Used in the E2 Reaction
On the other hand, when using a bulky nucleophile, the nucleophile is not able to get close to the inside of the molecule. As a result, an E2 reaction occurs.
We have already explained what happens in the case of bulky nucleophiles. In general, the E2 reaction follows the Saytzeff rule. However, due to the presence of steric hindrance, the reaction proceeds by Hofmann elimination when reagents such as tert-butoxide are used.
Even in the absence of such a bulky base, the E2 reaction is more likely to proceed when a strong base is used.
If the base is strong, the nucleophilic nature is also likely to be high. Therefore, the SN2 reaction also occurs at the same time. However, in general, strong bases attack hydrogen atoms. Rather than attacking the carbon atom to cause a nucleophilic substitution reaction, they prefer to pull the hydrogen atom out and make it stable.
Also, in a nucleophilic substitution reaction, it has to penetrate into the molecule and attack the carbon atom. It is easier to attack the bare hydrogen atoms that exist outside the molecule. Therefore, the stronger the base, the more likely it is to cause an E2 reaction rather than a nucleophilic substitution reaction.
Especially for tertiary alkyl groups, the SN2 reaction does not occur due to steric hindrance. Therefore, only the E2 reaction occurs.
It is important to understand that the stronger the base used, and the higher the steric hindrance of the reaction compound, the more elimination reactions will occur.
However, if the molecule of the strong base reagent to be used is small and the steric hindrance of the reacting compound is small, the SN2 reaction may take precedence over the E2 reaction. For example, sodium methoxide (CH3O–) and sodium ethoxide (CH3CH2O–) are strong bases, but very small molecules.
In this case, not only the strength of the base, but also all factors such as the steric hindrance of the reaction compound, temperature, and the state of the compound become relevant.
E1 and SN1 Reactions Occur at Weak Bases
On the other hand, a weak base may be used for synthesis. Weak bases include water, alcohol (methanol, ethanol) and acetic acid. These are also called protic solvents. Under these conditions, the E1 and SN1 reactions occur.
In the E2 and SN2 reactions, it is common to use reagents that are strongly nucleophilic (strongly basic). However, in the E1 and SN1 reactions, the basicity of the nucleophile is not related to the rate of the reaction. Rather, the reactivity is related to the amount of carbocation produced.
Therefore, the E1 and SN1 reactions proceed even when the base is weak. In addition, because the hydrogen atoms in the carbocation are very high acidic, the hydrogen atoms can be pulled out of the carbocation without any problem, even from a weak base.
-Both E1 and SN1 Reactions Occur
However, E1 and SN1 reactions cannot be separated. Let’s assume that if we utilize a protic solvent and react with a weak base, both the E1 and SN1 reactions will proceed to give two compounds. For example, the following.
It is possible to change the likelihood of the E1 and SN1 reactions to occur depending on the reaction conditions, for example, when the temperature is increased, the E1 reaction is given priority. However, the fact that the two reactions proceed at the same time does not change.
E1cB Reaction, in Which Conjugation Causes Anion Production
Most of what we have discussed so far is the elimination reaction. In addition to this, we should also try to understand the E1cB reaction. The E1cB reaction is that the hydrogen atoms are first pulled out by the base to form an anion.
Under what circumstances does the E1cB reaction occur? Let’s understand this as when a carbonyl group (-CO) is present near the leaving group. For example, the following.
In carbonyl groups, the acidity of the adjacent carbon (α carbon) is known to be high. This is because the hydrogen atoms bound to the α-carbon are pulled out by the base, allowing the following reaction to be written.
In short, think of the hydrogen next to the carbonyl group as being easily pulled out by the base to become an anion.
After becoming an anion in this way, the leaving group is released as the electrons return. As a result, a double bond is formed.
The cb in the E1cB reaction stands for conjugate base. The base pulls out hydrogen (proton) and does not immediately form a double bond. Once the anion is formed as a conjugate base. This reaction is called the E1cB reaction because it begins with deprotonation (the removal of a hydrogen atom).
The rate-limiting step in the E1cB reaction is the step where the leaving group is released. The proton next to the carbonyl group is highly acidic and can be easily extracted by a base. The rate of the reaction then depends on how quickly the leaving group is released as the electrons return.
Incidentally, when an anion-stabilizing functional group is present, such as a carbonyl group, the E1cB reaction is always preferred. It is important to understand that under all conditions, the E1cB reaction occurs over the E1, E2, SN1 and SN2 reactions.
Understand the Reaction Mechanism and Learn the Differences in Elimination Reactions
Two very important synthetic reactions in organic chemistry are the SN1 and SN2 reactions. However, once you have learned about these nucleophilic substitution reactions, in many cases you will be learning about elimination reactions.
There are many similarities between nucleophilic substitution reactions and elimination reactions. And as we have discussed, these reactions often occur in both cases. So we have to determine under what conditions the reaction takes precedence.
Also, even in the same elimination reaction, different nucleophiles (bases) will change the compounds formed. Normally, the reaction is by the Saytzeff rule, but it can also be a Hofmann elimination.
Once you understand the differences between these reaction mechanisms, the reaction conditions, and the reagents to be used, you will be able to synthesize alkenes at any time by learning about the three types of elimination reactions and understanding how the reactivity changes depending on the reagent (nucleophile) used.