[Chemistry Class Notes] on Preparation of Alkenes Pdf for Exam

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. 

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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. 

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The steps in this dehydration reaction’s mechanism are as follows.  

Protonation of the alcohol is the first step. 

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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. 

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The oxonium ion dissociates into a carbocation, a positively charged carbon atom that is an unstable intermediate. 

Deprotonation of the Carbocation. 

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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. 

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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. 

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The intermediate carbocation is rearranged to form the 2 methyl1 pentene molecule. 

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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. 

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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. 

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This reaction also follows the Zaitsev rule, so the major product in the reaction of 2chlorobutane with sodium ethoxide is 2butene. 

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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. 

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Fun Facts About Alkenes 

1. 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.  

2. Alkenes are not soluble in water due to the existence of weak van der Waal forces. 

3. Alkenes, however, are soluble in organic solvents such as benzene and acetone.  

4. Higher the molar mass of a given alkene, greater will be its boiling point. Hence, the boiling points of higher alkenes are quite high.  

5. The functional group attached to the alkene is the determinant of its polarity.  

6. 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.