Key Techniques for Effective Alkene Preparation
Alkenes are the hydrocarbons that contain a double bond between any two adjacent or adjoining carbon atoms. The structure of alkene is such that the number of hydrogen atoms is twice the number of carbon atoms. Alkenes form a homologous series, hence, the general formula of an alkene is stated as CnH2n. The simplest alkene which has one double bond in its structure is ethene, C2H4. Alkenes have important industrial uses, and also play an important role in our everyday lives.
There are various methods to prepare an alkene. The most commonly used ones are explained below.
Methods of Preparation of Alkenes
Preparation of Alkenes From Alkynes
Alkynes are hydrocarbons containing a triple bond between any two adjoining carbon atoms. Alkenes can be prepared from alkynes by carrying out hydrogenation in the presence of palletised charcoal. The charcoal which is used in this reaction has been moderately deactivated. Lindlar catalyst is palladium on calcium carbonate which has been deactivated by lead acetate to stop further hydrogenation. These alkenes will have a cis- form. To obtain a trans- alkene, these alkynes should be made to react with sodium in liquid ammonia.
(Δ, Lindlar catalyst/ Pd/C)
CH≡ CH + H2 —----------------> CH2=CH2
Ethyne Ethene
(Pd/C)
RC≡ CH + H2 ---------> RCH=CH2
Alkyne Alkene
(Pd/C)
CH3-C≡CH + H2 ------------> CH3-CH=CH2
Propyne Propene
Preparation of Alkenes From Alkyl Halides
In order to form alkenes from alkyl halides, dehydrohalogenation is performed. Alkyl halides have to be heated in the presence of alcoholic KOH. Alcoholic KOH is obtained when potassium hydroxide is dissolved in alcohol. As the reaction takes place, one molecule of halogen acid is eliminated, and a double bond is formed. The rate at which the reaction will proceed will be determined by the alkyl group and the attached halogen group. This reaction is also called beta elimination, as the halo group is removed from the alpha carbon while the hydrogen atom is removed from the beta carbon.
CH3 – CH2X ------> CH2=CH2 (alcoholic KOH, -HX)
(X = Cl, Br, I) (ethene)
The rate of the reaction is determined by the alkyl group and the type of halogen atom. In this case, the order is iodine > bromine > chlorine. According to the alkyl group, the reaction rate is tertiary > secondary > primary.
Preparation of Alkenes From Vicinal Dihalides
In vicinal dihalides, two halogen groups are attached to two adjoining carbon atoms in a compound. In geminal dihalides, the two halogens are attached to the same carbon atom.
When this dihalide undergoes a reaction with zinc or sodium iodide in acetone, the halogen groups attach to form a compound with zinc or sodium, leading to the formation of a double bond between those two carbon atoms. This reaction is called dehalogenation.
CH2Br - CH2Br + NaI ---------> CH2=CH2 +I2 + 2NaBr ( acetone)
CH2Br - CH2Br +Zn ----------> CH2=CH2 + ZnBr2
β Elimination Processes
In which two atoms on adjacent carbon atoms are removed, resulting in the production of a double bond, are commonly used to make alkenes.
(Image will be uploaded soon)
Alcohols are dehydrated, alkyl halides are dehydrohalogenation, and alkanes are halogenated as part of the process.
Dehydration of Alcohol
When an alcohol molecule is heated in the presence of a strong mineral acid, a molecule of water is removed from the molecule. The adjacent carbon atoms that have lost the hydrogen ion and the hydroxide group form a double bond.
(Image will be uploaded soon)
The steps in this dehydration reaction's mechanism are as follows.
Protonation of the alcohol is the first step.
(Image will be uploaded soon)
This step is a simple acid‐base reaction, which results in the formation of an oxonium ion, a positively charged oxygen atom.
Dissociation of the Oxonium Ion.
(Image will be uploaded soon)
The oxonium ion dissociates into a carbocation, a positively charged carbon atom that is an unstable intermediate.
Deprotonation of the Carbocation.
(Image will be uploaded soon)
The carbocation's positively charged end carbon attracts electrons in the overlap region, which bonds it to the adjacent carbon. The carbon becomes slightly positive as a result of the electron movement, which attracts electrons in the overlap regions of all other atoms bonded to it. As a result, the hydrogen on the carbon becomes slightly acidic, allowing it to be removed as a proton in an acid-base reaction.
Zaitsev Rule
An alcohol dehydration reaction, in which hydrogen atoms are lost from two different carbons on the carbocation, may be able to create a double bond in some cases. The more highly substituted alkene, that is, the alkene with the most substituents on the carbon atoms of the double bond, is always the major product, according to the Zaitsev rule. The following products are formed during the dehydration reaction of 2butanol.
(Image will be uploaded soon)
According to the Zaitsev rule, the main product is 2butene. It is worth noting that each carbon atom in the double bond of 2butene has one methyl group attached to it. In the case of 1butene, one carbon atom of the double bond has one substituent (the ethyl group), while the other has none.
Carbocation Rearrangement.
Rearrangement of carbocations In alcohol dehydration, the carbocation may rearrange to form more stable arrangements. The dehydration of 2 methyl 3 pentanols, for example, results in the formation of three alkenes. The mechanism of the reaction reveals that the extra compound is formed as a result of the carbocation intermediate being rearranged.
(Image will be uploaded soon)
The intermediate carbocation is rearranged to form the 2 methyl1 pentene molecule.
(Image will be uploaded soon)
A Hydride Ion (H: - ) Movement
results in the formation of a more stable carbocation. Carbocations, like carbon atoms, are classified as primary, secondary, and tertiary. A primary carbocation has one alkyl group attached to it, a secondary carbocation has two alkyl groups attached to it, and a tertiary carbocation has three alkyl groups attached to it.
(Image will be uploaded soon)
Alkyl groups can theoretically "push" electrons away from themselves. This is known as the inductive effect. The more alkyl groups that "push" electrons toward a positively charged carbon atom, the more stable the intermediate carbocation. This increase instability is due to charge density delocalization. A charge on an atom causes it to be stressed. The more the stress is distributed across the molecule, the lower the charge density on any one atom becomes, reducing stress. The ion becomes more stable as the stress decreases. Thus, tertiary carbocations have three alkyl groups on which to delocalize the positive charge, whereas secondary carbocations only have two alkyl groups on which to delocalize the positive charge. Secondary carbocations are more stable than primary carbocations for the same reason.
Alkyl groups, in reality, do not "push" electrons away from themselves, but rather have electrons removed from them. When an atom gains a positive charge and transforms into an ion, its electronegativity changes. The location of the overlap region relative to each carbon atom in the original bond between two carbon atoms is determined in part by the electronegativity of the two atoms. The overlap region moves closer to the more electronegative, positively charged carbon atom as the electronegativity of one of the carbon atoms increases due to ion formation. This electron density rearrangement results in a partial positive charge on the neighboring carbon. The charge gained by the second carbon atom equals the charge lost by the fully charged carbon atom. As a result, the charge is delocalized across the two carbons.
Alkyl Halide Dehydrohalogenation
Another elimination reaction is the dehydrohalogenation of alkyl halides, which involves the loss of hydrogen and a halide from an alkyl halide (RX). Normally, dehydrohalogenation is accomplished by reacting the alkyl halide with a strong base, such as sodium ethoxide.
(Image will be uploaded soon)
This reaction also follows the Zaitsev rule, so the major product in the reaction of 2chlorobutane with sodium ethoxide is 2butene.
(Image will be uploaded soon)
The following mechanism governs dehydrohalogenation reactions.
1. An acid-base reaction occurs when a strong base removes a slightly acidic hydrogen proton from an alkyl halide.
2. Electrons from the broken hydrogen carbon bond are drawn to the slightly positive carbon atom connected to the chlorine atom. The halogen atom breaks free as these electrons approach the second carbon, resulting in the formation of the double bond. This mechanism is depicted in the diagram below.
(Image will be uploaded soon)
Fun Facts About Alkenes
Alkenes can exist in all three states: solid, liquid, and gas. The first three alkenes are gaseous, the next fourteen are in the liquid state. As the molar mass increases further, they exist in the solid-state.
Alkenes are not soluble in water due to the existence of weak van der Waal forces.
Alkenes, however, are soluble in organic solvents such as benzene and acetone.
Higher the molar mass of a given alkene, greater will be its boiling point. Hence, the boiling points of higher alkenes are quite high.
The functional group attached to the alkene is the determinant of its polarity.
Being unsaturated in nature, alkenes are highly reactive compounds. Most of these reactions will take place at the site of the double bond, that is at the two carbon atoms between which the double bond is placed. They can easily undergo addition and oxidation reactions.
FAQs on Preparation of Alkenes: Complete Methods for Students
1. What are the main methods for preparing alkenes according to the Class 11 Chemistry syllabus?
As per the NCERT syllabus for Class 11, there are four primary methods to prepare alkenes:
- From Alkynes: Through controlled or partial hydrogenation of alkynes.
- From Alkyl Halides: By heating them with an alcoholic solution of potassium hydroxide (KOH) in a process called dehydrohalogenation.
- From Vicinal Dihalides: By reacting them with zinc dust, which removes the two halogen atoms.
- From Alcohols: By acidic dehydration, which involves heating an alcohol with a strong acid like concentrated sulphuric acid.
2. Can you explain Saytzeff's rule with a simple example?
Saytzeff's rule helps predict the major product in certain elimination reactions. It states that when forming an alkene, the more substituted alkene (the one with more alkyl groups attached to the double-bonded carbons) will be the main product because it is more stable. For example, when 2-bromobutane undergoes dehydrohalogenation, it can form But-2-ene and But-1-ene. But-2-ene is the major product (around 80%) because it is more substituted and therefore more stable.
3. Why is a more substituted alkene generally more stable?
A more substituted alkene is more stable primarily due to a phenomenon called hyperconjugation. The alkyl groups attached to the double-bonded carbons have C-H sigma bonds that can overlap with the empty pi-antibonding orbital of the double bond. This overlap delocalises the electrons, spreading the charge over a larger area and lowering the overall energy of the molecule. More alkyl groups mean more hyperconjugation and greater stability.
4. How does the dehydration of an alcohol lead to the formation of an alkene?
The dehydration of an alcohol is an elimination reaction where a molecule of water is removed. When an alcohol is heated with a strong protic acid, like concentrated sulphuric acid (H₂SO₄) or phosphoric acid (H₃PO₄), the acid protonates the -OH group, turning it into a good leaving group (-OH₂⁺). This group leaves as water, and a proton is removed from an adjacent carbon atom, resulting in the formation of a carbon-carbon double bond, which creates the alkene.
5. What is the key difference between preparing alkenes from alcohols versus from alkyl halides?
The main difference lies in the reagents used and the group that is eliminated. To prepare an alkene from an alcohol, you perform an acid-catalysed dehydration, where a molecule of water (H₂O) is removed. In contrast, preparing an alkene from an alkyl halide requires a base-induced dehydrohalogenation, typically using an alcoholic base like KOH, where a hydrogen halide molecule (like HCl or HBr) is eliminated.
6. How can we get a specific isomer (cis or trans) when preparing an alkene from an alkyne?
Yes, the choice of catalyst during the partial hydrogenation of an alkyne allows us to control the stereochemistry of the resulting alkene.
- To get a cis-alkene, a catalyst called Lindlar's catalyst (partially deactivated palladium) is used.
- To get a trans-alkene, the reduction is carried out using sodium metal in liquid ammonia, a reaction known as the Birch reduction.
7. What are some real-world applications of alkenes that make their preparation so important?
The preparation of alkenes is fundamental to many industries because they are key starting materials. Their most significant application is in the production of polymers. For example:
- Ethene is used to make polythene, one of the world's most common plastics, used for packaging, bags, and bottles.
- Propene is used to make polypropylene, a versatile plastic used in containers, car parts, and textiles.
- Alkenes are also used to artificially ripen fruits and synthesise other important chemicals like ethanol and ethylene glycol (antifreeze).
8. Is it ever possible to form the less substituted alkene as the major product instead of the Saytzeff product?
Yes, it is possible to favour the formation of the less substituted alkene, also known as the Hofmann product. This typically happens when a sterically bulky base (a large, physically big base) is used in the elimination reaction. The large base finds it difficult to access the more sterically hindered hydrogen needed to form the Saytzeff product, so it preferentially removes a more accessible, less hindered hydrogen, leading to the less stable, less substituted alkene as the major product.
