Preparation of Alkenes

Alkenes are hydrocarbons with at least one carbon-carbon double bond. These alkenes form basic raw materials of academic and industrial chemistry, ranging from chromatic colors applied on synthetic fibers to structural parts in modern polymers. By the presence of a double bond, alkenes are rather reactive. On this basis, their applicability ranges from simple reagents to important intermediates for the synthesis of complex organic compounds. Imagine the synthesis of a rigid, yet supple plastic container or synthesizing a critical pharmaceutical compound.

This Story also Contains

1. Alkenes
2. Methods of Alkene Preparation
3. Dehydration of Alcohols:
4. Dehydrohalogenation of Alkyl Halides
5. Dehydrohalogenation of Haloalkanes with Strong Bases
6. Some Solved Examples
7. Summary

Alkenes

Alkenes are hydrocarbons that contain at least one carbon-carbon double bond, $\mathrm{C}=\mathrm{C}$. It is this very double bond that to a large extent defines an alkene and gives unique chemical properties and reactivities. Alkenes have the general formula $\mathrm{C}_{\mathrm{n}} \mathrm{H}_2$ where 'n' refers to the number of carbon atoms. Because of this double bond, rotation about the bond is restricted and a plane is formed, making alkenes comparatively more reactive than alkanes. This reactivity is utilized in a broad scope of reactions, such as addition, polymerization, and oxidation, which makes alkenes rather useful in synthetic and industrial chemistry.

Alkynes on partial reduction with a calculated amount of dihydrogen in the presence of palladised charcoal partially deactivated with poisons like sulphur compounds or quinoline give alkenes. Partially deactivated palladised charcoal is known as Lindlar’s catalyst. Alkenes thus obtained have cis geometry. However, alkynes on reduction with sodium in liquid ammonia form trans alkenes.

Methods of Alkene Preparation

Dehydration of Alcohol by Conc.$\mathrm{H}_2 \mathrm{SO}_4$

It is the process of removal of water from alcohols using concentrated sulfuric acid.
The mechanism takes place according to Saytzeff's Rule: more substituted alkene is formed, or Hoffmann's Rule: less substituted alkene is formed.

Saytzeff's rule

This rule states that in dehydrohalogenation reactions, the preferred product is always that alkene which is most stable or in other words which has more number of $\alpha$-hydrogen atoms.

Hoffmann's rule

This rule states that the alkene formed would be the least stable as the major product or in other words that alkene would be formed which has the least number of $\alpha$-hydrogen atoms.

Dehydration by $\mathrm{Al}_2 \mathrm{O}_3$

Since the reagent used is $\mathrm{Al}_2 \mathrm{O}_3$ thus the Saytzeff's rule will be applied and E₂ elimination will take place and no carbocation will form. When ethanol is passed over heated aluminium oxide then ethene is formed as the final product. The reaction occurs as follows:

$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{OH} \xrightarrow{\mathrm{Al}_2 \mathrm{O}_3} \mathrm{CH}_2=\mathrm{CH}_2+\mathrm{H}_2 \mathrm{O}$

Dehydration by ThO₂

Since the reagent used is ThO₂, thus the Hoffmann's rule will be applied E₂ elimination will take place and no carbocation will form. The reaction occurs as follows:

$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}-\mathrm{OH}-\mathrm{CH}_3 \xrightarrow{\mathrm{ThO}_2} \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}=\mathrm{CH}_0$

Dehydration of Alcohols:

Alcohols undergo dehydration when allowed to react with concentrated acids in the presence of heat.

This reaction can be used to dehydrate all three types of alcohol viz. Primary, secondary, and tertiary alcohols. Some examples are given below:

It is to be noted that the dehydration usually occurs via the Unimolecular elimination reaction $(E_1)$ and involves a carbocation intermediate which can undergo rearrangement via hydride or alkyl shift and also undergo ring expansion for suitable substrates where the ring strain can be released. A drawback of this reaction is that a mixture of alkenes can be obtained due to the involvement of carbocation intermediates. Saytzeff’s alkene which is the more stable alkene is usually obtained as a major product.

Consider the examples given below in which the carbon skeleton changes due to carbocation rearrangement and ring expansion respectively.

Case of Methyl shift

The mechanism of the reaction is given below:

Case of Ring expansion

The mechanism of the reaction is given below

Dehydration of Al₂O₃ and ThO₂:
It uses catalysts like Aluminium oxide and thorium dioxide
It is applied industrially since it is highly efficient and selective

Dehydrohalogenation of Alkyl Halides

The removal of hydrogen halide, HX, from alkyl halide with a strong base
It mainly undergoes the E₂ elimination mechanism Dehalogenation of Vicinal Halides
The removal of the halogens on nearby carbon atoms in the presence of reducing agents like

When vicinal dihalides are heated with Zn dust or NaI/Acetone, an alkene having the same number of carbon is obtained. This reaction is known as dehalogenation. The reaction occurs as follows:

Mechanism

For example:

Dehydrohalogenation of Haloalkanes with Strong Bases

Secondary and tertiary alkyl halides undergo dehydrohalogenation on reaction with a strong base to form Alkenes. The reaction is an elimination reaction. It is to be noted that primary haloalkanes form ether by Williamson’s synthesis of Ethers. Some examples of the reaction are given below

This reaction is an example of $\beta$ elimination in which a $\beta-$ hydrogen is eliminated along with a halogen at the $\alpha$ carbon. The reaction occurs in a concerted mechanism and anti-elimination takes place as shown below.

If there are different types of$\beta$ hydrogen present in the substrate then usually the Saytzeff’s alkene is obtained as a major product. Please recall that Saytzeff’s alkene is the more substituted alkene having a greater number of $\alpha$ hydrogens or greater alkylation around the double bond.

However, in cases where bulky bases are used, the reaction usually takes place by the extraction of the least hindered hydrogen atom and often less substituted alkenes are obtained as a major product. Steric hindrance thus plays an important role in the reaction.

There is an anomaly shown in the reaction when Fluorine is present as the leaving group in the haloalkane and usually less substituted alkene is produced as a major product. This is explained by the poor leaving group ability of Fluorine and the reaction proceeds by a significant anionic character in the transition state.

The dehydrohalogenation occurs in an anti periplanar fashion and the hydrogen and the halogen should be in an anti orientation.

Zinc dust Wittig's reaction:

The phosphonium ylides are reacted with carbonyl compounds to form alkenes.
Known to exhibit stereoselectivity for the formation of certain alkenes.

In this reaction, methylene triphenyl phosphorane or phosphorous ylide is treated with a carbonyl compound to prepare an alkene. There are two important components of this reaction:

• A carbonyl compound
• A species known as "ylide". The "ylide" is a species with opposite charges on adjacent atoms.

This reaction is named after George Wittig who was awarded the Nobel prize for this work in 1979. A principal advantage of alkene synthesis by the Wittig reaction is that the location of the double bond is fixed, in contrast to the mixtures often produced by alcohol dehydration.

Mechanism

For example:

Pyrolysis of Quaternary Ammonium Salts:

Thermal decomposition of quaternary ammonium salts for the generation of alkenes.

Cope's Reaction:

It involves the thermal decomposition of tertiary amine oxides for the formation of alkenes and hydroxylamines.

Cope's reaction
When a tertiary amine oxide bearing one or more beta hydrogens is heated, it is converted to an alkene. The reaction is known as Cope elimination or Cope reaction. The net reaction is 1,2-elimination hence the name Cope elimination.

For example:

In Cope's elimination, the least hindered beta H is eliminated and Hoffman alkene is formed

Pyrolysis of Esters

When esters are heated in the presence of liquid N₂ and glass wool, the alkyl part of the ester converts into the respective alkene while the alkanoate part of the ester converts into the respective acid.

For example:

In pyrolysis of esters, the least hindered beta H is eliminated and Hoffman alkene is formed

Pyrolysis of Esters:

It generates alkenes through the thermal decomposition of esters.

Pyrolysis of quaternary ammonium salts follows the Hoffmann elimination. This means the less stable alkene will form. In this reaction, an amine reacts with 3 moles of methyl iodide and forms quaternary ammonium salt. Now heating this salt with moist Ag₂O or AgOH will form alkene.

The reaction occurs as follows:

$\mathrm{R}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NH}_2 \xrightarrow{\mathrm{CH}_3 \mathrm{I}} \mathrm{R}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NH}-\mathrm{CH}_3+\mathrm{HI} \xrightarrow{\mathrm{CH}_3 \mathrm{I}} \mathrm{R}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{N}-\left(\mathrm{CH}_3\right)_2+\mathrm{HI}$

$\xrightarrow{\mathrm{CH}_3 \mathrm{I}} \mathrm{R}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{N}^{+}-\left(\mathrm{CH}_3\right)_3 \xrightarrow[A \mathrm{AgOH}]{\text { Moist } \mathrm{Ag}_2 \mathrm{O} \text { or }} \mathrm{R}-\mathrm{CH}=\mathrm{CH}_2+\mathrm{N}-\left(\mathrm{CH}_3\right)_3$

Recommended topic video on (Preparation of Alkenes)

Some Solved Examples

Example 1

Question:
Conversion of alkyne to cis-alkene can be achieved using the reagent:

1.${ }_1 \mathrm{H}_2$, Lindlar's catalyst
2 . $\mathrm{H}_2 / \mathrm{Ni}_i$
3. $\mathrm{LiAlH}_4$
4${ }_4 \mathrm{Na} / \mathrm{liq} . \mathrm{NH}_3$

Solution:

As we have learned,

Preparation of alkene from alkyne -

Alkynes on partial reduction with a calculated amount of H₂ in the presence of Pd with charcoal give alkenes.

$\mathrm{H}_2$, Lindlar's Catalyst when reacts with alkyne gives cis-alkene while $\mathrm{Na}^{\prime}$, liq. $\mathrm{NH}_3$ gives trans-alkene..

Therefore, option (1) is correct.

Example 2

Question:
The hydrocarbon which cannot be reduced to an alkene in reaction with sodium in liquid ammonia is:

1. $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CCH}_2 \mathrm{CH}$
2. $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CCH}_2 \mathrm{CH}_2 \mathrm{CH}_3$
3.$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}$
4.$\mathrm{CH}_3 \mathrm{C} \equiv \mathrm{CCH}_3$

Solution:
$
\mathrm{CH}_3 \mathrm{CH}_2-\mathrm{C} \equiv \mathrm{CH} \xrightarrow[\Delta]{\mathrm{Na}^{-} / \mathrm{Liq}_4 \mathrm{NH}_3} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{C}^{-} \mathrm{Na}^{-}
$
It is a terminal alkyne, having acidic hydrogen.
Note: Solve it as a case of terminal alkynes, otherwise all alkynes react with Na in liq. $\mathrm{NH}_3$ :
Alkynes having terminal $\equiv \mathrm{C}-\mathrm{H}$ react with Na in liquid $\mathrm{NH}_3$ to yield $\mathrm{H}_2$

Therefore, option (3) is correct.

Example 3

Question:

The reagent needed for converting

is

1. Catalytic Hydrogenation
2. $\mathrm{H}_2$ / Lindlar's catalyst
3. $\mathrm{Li} / \mathrm{NH}_2$
4. $\mathrm{LiAlH}_4$

Solution:

$\mathrm{Li} /$ liq. $\mathrm{NH}_3$converts alkynes into trans alkenes.

Therefore, option (3) is correct.

Summary

Alkenes are hydrocarbons that contain at least one carbon-carbon double bond. There exist a number of ways for their preparation, each method having a different set of advantages and applications. These include Dehydration Dehydrohalogenation Dehalogenation Wittig's reaction Pyrolysis.