In this explainer, we will learn how to describe addition reactions of alkenes and predict what products are formed.
Alkenes are generally considered to be more reactive than alkanes because they contain an electron-rich carbon–carbon double bond (). The carbon–carbon double bond reacts with molecules and ions that have a full or partial positive electrostatic charge.
Alkenes are unsaturated hydrocarbon molecules that contain at least one carbon–carbon double bond ().
Electrophiles are chemical species that can accept a pair of electrons. Electrophiles ordinarily have a full or partial positive electrostatic charge, and they are attracted to areas of high electron density. Positively charged ions and polar molecules are good examples of electrophiles that are attracted to and can react with areas of high electron density. Some diatomic gas molecules can also be classed as electrophiles because they can generate a temporary partial positive charge when they interact with an area of high electron density.
Alkene molecules react with electrophile substances during addition reactions. The reactants combine together during the addition reaction, and they make a single type of product molecule. Alkene molecules can undergo different types of addition reactions. Alkene molecules can undergo hydrogenation, hydrohalogenation, halogenation, and hydration-type addition reactions.
Definition: Addition Reaction
Addition reactions are a type of reaction that happens when two or more reactant molecules combine together to make a single type of molecular product.
Hydrogen gas is combined together with an alkene molecule during hydrogenation reactions. The addition reaction produces a single type of saturated hydrocarbon product molecule. The simplest example would perhaps be the reaction between ethene and hydrogen gas that produces an ethane product.
The following equation shows a slightly more complex example of a hydrogenation reaction. The equation shows how 2-methylpropene molecules can be combined with hydrogen gas molecules to make the 2-methylpropane product.
2-Methylpropene does not generally react with hydrogen under normal conditions, but it will react with hydrogen at temperatures between and if it is combined with a platinum metal or nickel metal catalyst.
Chemists realized a long time ago that they could use hydrogenation reactions to make margarine from vegetable oils. The chemists realized that they could bubble hydrogen gas through a vegetable oil substance to saturate most or all of its carbon–carbon double bonds. The hydrogenation reaction produced a firmer and more viscous margarine product. Chemists then realized that they could speed up these hydrogenation reactions if they mixed vegetable oils with a finely divided nickel–aluminum alloy catalyst that is called Raney nickel catalyst.
Alkene molecules can also be combined with diatomic halogen gas reactant molecules to make a single dihalogenoalkane product. The following equation shows how ethene molecules can be combined with diatomic chlorine molecules during an addition reaction to make the 1,2-dichloroethane product.
Chemists do not need to use high temperature or pressure to make alkene molecules react with halogen molecules. Alkene molecules can react with chlorine (), bromine (), or iodine () molecules at room temperature and atmospheric pressure. Alkene molecules can also be reacted with fluorine () molecules at room temperature and atmospheric pressure, but this reaction is slightly more complex and cannot be classed as a simple addition reaction.
Bromine water (a mixture of and ) has a characteristic orange color, and this orange color is lost when bromine water is combined with a hydrocarbon substance that contains at least one carbon–carbon double bond. The orange color is lost as all of the dissolved bromine molecules react with the carbon–carbon double bonds of the liquid hydrocarbon substance. This reaction is usually known as the bromination reaction or the bromine test.
The bromination reaction can be used to determine if an unknown hydrocarbon substance should be classed as an alkene or an alkane. The bromine water will retain its characteristic orange color if the hydrocarbon is an alkane, and it will lose its characteristic orange color if the hydrocarbon is an alkene. The following reaction equation shows how the bromination reaction can be used to convert an ethene reactant into a 1,2-dibromoethane product:
There are other solutions that can be used to test for the presence of a carbon–carbon double bond. Chemists sometimes use a solution of bromine and carbon tetrachloride () molecules to determine if an unknown organic compound contains a carbon–carbon double bond. The mixture loses its characteristic dark reddish-orange color if the unknown organic substance has at least one carbon–carbon double bond. The mixture retains its characteristic dark reddish-orange color if the unknown substance does not have any carbon–carbon double bonds.
Example 1: Identifying What Happens When Bromine Water Is Combined with Propene
Which of the following occurs upon the addition of bromine water to propene?
- The color of bromine disappears with the formation of 1,2-dibromopropane.
- The color of bromine does not change and no reaction occurs.
- The color of bromine disappears with the formation of 1-bromopropene.
- The color of bromine disappears with the formation of 1-bromopropane.
- The color of bromine disappears with the formation of 1,3-dibromopropane.
Propene is an alkene molecule that has the structural formula. Bromine molecules can be combined with alkene molecules during an addition reaction. The bromine molecules are added across the carbon–carbon double bond. One bromine atom ends up being bonded to each one of the carbon atoms that contains the carbon–carbon double bond.
Bromine water is a dark orange solution that contains bromine and water molecules. Bromine water loses its characteristic orange color as its bromine molecules are consumed through reactions with carbon–carbon double bonds. Bromine water will react with a sample of propene because propene molecules contain a carbon–carbon double bond.
The color of bromine will disappear as propene is reacted with bromine molecules to produce a 1,2-dibromopropane product. We can use these statements to determine that option A is the correct answer for this question.
The Baeyer test can also be used to test for the presence of unsaturated hydrocarbons like alkenes. The Baeyer test occurs when a reagent that contains potassium permanganate is reacted with an unknown compound. Potassium permanganate has a characteristic deep purple color, and this color is lost when the potassium permanganate is reacted with an unsaturated compound.
The potassium permanganate acts as an oxidizing agent during the Baeyer test. The reaction can end up producing different types of diols or carbonyl molecules and even carbon dioxide gas. The composition of the product molecules is determined from the location of the carbon–carbon double bond, the length of the product molecules, and the concentration and temperature of the potassium permanganate solution.
The Baeyer test is used less frequently than the bromination test to determine if a compound is an alkene because the Baeyer solution can react with other functional groups on the molecule as well as any alkene groups and be decolorized.
The following reaction equation shows how ethene molecules can be reacted with cold, dilute, acidic permanganate. The  symbol is used to represent the oxygen atoms that are provided by the manganate ions. The reaction is a simplification for a more complicated reaction scheme that is beyond the scope of this explainer.
Alkene molecules are relatively reactive substances, and they can also be reacted with hydrogen halide molecules to make a single type of molecular product molecule.
The following reaction equation shows how a hydrogen iodide molecule can be combined with an ethene molecule to make the iodoethane product. The hydrogen iodide molecule adds across the carbon–carbon double bond.
It is easy to determine the type of molecular product that will be formed when a small symmetric alkene combines with a hydrogen halide molecule, but it can be more challenging to determine what type of product molecule will be formed when a hydrogen halide reacts with a larger unsymmetrical alkene molecule. The following reaction equation shows how a single hydrogen iodide molecule can be reacted with a single propene molecule to make two different types of molecular products.
Markovnikov’s rule can be used to determine which type of molecular product will be made when a hydrogen halide molecule reacts with a medium-to-large sized unsymmetrical alkene molecule such as propene or pent-2-ene.
Definition: Markovnikov’s Rule
Markovnikov’s rule states that the major product is the one where acidic hydrogen is added to carbon with the greatest number of hydrogen substituents.
Propene has one carbon–carbon double bond that is contained between the central carbon atom and one of the terminal carbon atoms. Markovnikov’s Rule can be applied to understand that the hydrogen atom must bond with the terminal carbon atom. Markovnikov’s Rule can similarly be applied to determine that the iodine atom must add to the central carbon atom of propene. Propene molecules will primarily form the 2-iodopropane product, not the 1-iodopropane product, when they are combined with a hydrogen iodide reactant.
Example 2: Determining What Product Is Formed When But-1-ene Is Reacted with Hydrogen Chloride
Consider the reaction of but-1-ene with :
What major product is formed from this reaction?
But-1-ene is an unsymmetrical alkene. An unsymmetrical alkene has a carbon–carbon double bond nearer to one end of the molecule than the other. Hydrogen halide molecules can be reacted with an unsymmetrical but-1-ene molecule to potentially make different types of product molecules. The chlorine atom can end up being bonded to either one of the carbon atoms that contains the carbon–carbon double bond. Markovnikov’s rule can be used to determine which type of product molecule will primarily be formed when any hydrogen chloride molecules are combined with unsymmetrical but-1-ene molecules.
Markovnikov’s rule states that the hydrogen atom of the or molecule adds to the carbon atom of the alkene molecule that has the greatest number of covalently bonded hydrogen atoms. Markovnikov’s rule can be used to determine that the hydrogen atom will add to the terminal carbon atom that contains the carbon–carbon double. The chlorine atom must add to the other carbon atom that contains the carbon–carbon double bond. This line of reasoning can be used to determine that 2-chlorobutane is the major product of this chemical reaction.
Example 3: Determining What Product Is Produced When the But-2-ene Hydrocarbon Is Combined with Hydrogen Chloride Reactants
The reaction of with but-2-ene is shown:
Why can this reaction only possibly produce a single product?
The but-2-ene hydrocarbon can be combined with a hydrogen chloride reactant to produce a chloroalkane product that has the chemical formula. We can visualize this reaction by designating each carbon atom a red-colored number. The following figure shows how the but-2-ene looks if we designate each carbon atom a red-colored number that ranges from one to four.
The hydrogen atom can bond with either the second (2) or third (3) carbon atom, and the chlorine atom would then have to bond with the other carbon atom that contains the carbon–carbon double bond. There are two different reactions that can occur, but the product will be the 2-chlorobutane molecule in either instance. The following figure shows how the but-2-ene molecule would form a 2-chlorobutane molecule regardless of how the hydrogen and chlorine atoms combined with the carbon–carbon double bond.
We can conclude that the reaction can only produce one type of molecular product because the but-2-ene molecules are symmetrical about their central carbon–carbon double bond ().
The final addition reaction we will investigate in this explainer is hydration. In a hydration reaction, water in the form of steam is added across a carbon–carbon double bond to produce an alcohol product. The following image shows how ethene can be combined with steam during a hydration reaction to make an ethanol product.
Chemists use a phosphoric acid catalyst () and a high temperature and pressure to speed up this hydration reaction and promote the formation of the ethanol product.
Sulfuric acid ( or ) can similarly be used to initiate an indirect hydration reaction. The hydration reaction occurs as an alkene molecule is reacted with the sulfuric acid catalyst to produce an alkyl hydrogen sulfate intermediate. The following equation shows how ethene molecules can be reacted with a sulfuric acid catalyst to make the ethyl hydrogen sulfate intermediate.
The ethyl hydrogen sulfate can then be hydrolyzed with water molecules to produce a more desirable ethanol product. The following equation shows how the sulfuric acid catalyst is regenerated as the ethyl hydrogen sulfate intermediate is converted into an ethanol product.
This reaction scheme is usually condensed down to the following much-simpler and easier-to-understand single-line equation.
Example 4: Identifying Which Type of Product Molecule Is Formed When Ethene Is Reacted With Water in the Presence of a Sulfuric Acid Catalyst
Fill in the blank: The addition of sulfuric acid to ethene, followed by dilution with water and distillation, can be used to produce .
- ethylene glycol
- ethanoic acid
Hydration is an example of an addition reaction. There are two types of hydration reactions that can be used to make an alcohol product. Alcohol products can be produced through direct hydration reactions when ethene molecules are reacted with steam in the presence of a strong acid catalyst.
Alcohol products can also be made through indirect hydration reactions when a sulfuric acid catalyst is used to produce intermediate alkyl hydrogen sulfate substances. The intermediate alkyl hydrogen sulfate substance can then be reacted with water to make a more desirable type of alcohol product molecule. Ethene molecules can, for example, be reacted with a sulfuric acid catalyst to produce the ethyl hydrogen sulfate intermediate. This intermediate can then be reacted with water molecules to produce a more desirable ethanol product. The ethanol product can be retrieved and extracted at the end of this multistep reaction process through distillation processes. These statements can be used to determine that option D is the correct answer for this question.
- Hydrogenation, hydrohalogenation, halogenation, and hydration reactions are all examples of addition reactions.
- Addition reactions are a type of reaction that happens when two or more reactant molecules combine together to make a single type of molecular product.
- The carbon–carbon double bond of any reactant alkene molecule is destroyed during an addition reaction.
- Alkenes can be combined with hydrogen during hydrogenation reactions to make a single type of saturated hydrocarbon product molecule.
- Alkenes can be combined with either diatomic halogen or hydrogen halide molecules during halogenation or hydrohalogenation reactions to make a single type of molecular product molecule.
- The halogenation reactions of alkenes can take place at room temperature and atmospheric pressure.
- The bromination and Baeyer tests can be used to determine if an unknown organic compound is an unsaturated hydrocarbon.
- Markovnikov’s rule can be used to determine the primary product that is produced as hydrogen halides are combined with unsymmetrical alkene molecules.
- The hydration reactions of alkenes are conducted with a strong acid catalyst because the catalyst can be used to produce an intermediate that readily transforms into more desirable alcohol product molecules.