If you remove two atoms or groups of atoms from a molecule, it forms a new product. This type of chemical reaction is called an elimination reaction. The name itself makes sense, as you are essentially eliminating a part of the molecule. Understanding elimination reactions is important in chemistry, as it can help predict and explain how certain reactions occur.
An elimination reaction can occur between the hydroxide ion, , and a halogenoalkane. This reaction produces water, a halide ion, and an alkene. Remember that a halogenoalkane is a hydrocarbon containing one or more halogen atoms. An example is chloromethane, , shown below.
The hydroxide ion acts as a base by accepting a proton, , from the halogenoalkane to form water.
A base is simply a proton acceptor. One of the adjacent carbon atoms then loses a halide ion in order to achieve a stable structure. This is also known as dehydrohalogenation. You’ll find the general mechanism below, using to represent the halogen.
We’ve included a labelled diagram to help you understand the mechanism.
During an elimination reaction, the hydroxide ion attacks a hydrogen atom with its lone pair of electrons. The hydrogen atom is always connected to a carbon atom next to the C-X bond. This causes the bonded pair of electrons in the C-H bond to form a C=C double bond. Meanwhile, the halogen takes the pair of electrons from the C-X bond and leaves as a leaving group.
A leaving group is a part of a molecule that breaks off during a chemical reaction. When the bond between the leaving group and parent molecule is broken, the bonding pair of electrons goes to the leaving group. To carry out this reaction, use either potassium hydroxide or sodium hydroxide in hot, ethanolic conditions. The sodium or potassium ion reacts with the halide ion to create either a sodium or potassium halide. Remember, this type of elimination occurs in ethanolic, not aqueous, conditions. It's called an E2 mechanism because two species are involved in the initial slow part of the reaction. An E1 mechanism is more complicated and also possible.
For elimination to occur, there must be a hydrogen on an adjacent carbon to the carbon bonded to the halogen. This sounds more complicated than it is! The easiest way to see if a halogenoalkane is suitable is to draw the molecule, then follow these simple steps:
Circle the C-X bond. Identify the adjacent carbons. See if any of these adjacent carbons contain a C-H bond.
If they do, your halogenoalkane can react! Identifying suitable halogenoalkanes. Notice that the carbon containing the C-X bond is also bonded to an R group.
For example, an elimination reaction could occur between a hydroxide ion and 2-bromobutane, but not with 1-bromo-2,2-dimethylpropane. We’ve drawn these molecules out below, labelling the steps we described above to make the explanation clearer:
Depending on the halogenoalkane used, we can form multiple different alkenes in an elimination reaction. This is because there could be multiple carbons with a C-H bond adjacent to the C-X bonded carbon, and so various hydrogen atoms can be attacked. Some of these alkenes may be stereoisomers.
Stereoisomers are molecules that have the same structural formula but different spatial arrangements of atoms. For example, the reaction between 2-chlorobutane and potassium hydroxide can produce three different alkenes:
For more information about stereoisomers, see Isomerism.
Different halogenoalkanes have varying levels of reactivity. For instance, iodopropane will react more quickly with hydroxide ions than chloropropane. This is because the C-I bond has a lower bond enthalpy compared to the C-Cl bond. Iodine is a bigger atom than chlorine, so the bonded pair of electrons involved in the C-X bond are farther away from the nucleus. This results in weaker attraction between the nucleus and the electrons, making the bond easier to break.
Elimination reactions and nucleophilic substitutions are two types of reactions that can occur with halogenoalkanes. However, the reaction conditions can be adjusted to promote one over the other. While some substitution will always happen during an elimination reaction and vice versa, the following table highlights the differences and similarities between these reactions. We will delve deeper into the favored conditions, favored halogenoalkane, and the role of the hydroxide ion later.
The favored conditions for either elimination or nucleophilic substitution reactions are influenced by the proportion of ethanol to water in the solvent and the temperature. Warm conditions are suitable for nucleophilic substitution, while hot conditions favor elimination. A more aqueous solution favors nucleophilic substitution, while a more ethanolic solution favors elimination. Using concentrated sodium or potassium hydroxide also promotes elimination.
The type of halogenoalkane used as a reactant also affects the dominant reaction. Tertiary halogenoalkanes, which have three alkyl groups attached to the carbon with the C-X bond, have more opportunities for elimination because they have more carbons with hydrogens adjacent to the C-X bonded carbon. Therefore, using tertiary halogenoalkanes favors elimination, while using primary halogenoalkanes favors nucleophilic substitution. It's essential to understand the favored conditions and favored halogenoalkanes to predict the dominant reaction accurately. It's also crucial to learn what your exam board expects you to know about reaction conditions. Some exam boards are content with aqueous solution for nucleophilic substitution reactions and ethanolic solution for elimination, while others require a more in-depth understanding, as described above.
The hydroxide ion plays different roles in elimination and nucleophilic substitution reactions, but this doesn't affect the dominant reaction type. In elimination, the hydroxide ion acts as a base by accepting a proton from one of the carbons adjacent to the carbon with the C-X bond. In contrast, in nucleophilic substitution, the hydroxide ion acts as a nucleophile by donating its lone pair of electrons to the partially charged carbon atom, leading to the substitution of the halogen out of the molecule.
As an example, 2-chloropropane undergoes elimination in hot ethanolic sodium hydroxide, where hydrogens on both carbons 1 and 3 can be attacked. However, the alkenes produced are identical, so no isomers are produced. The products in both cases are propene, water, and the chloride ion, which reacts with sodium to form sodium chloride.
Another example of an elimination reaction with a halogenoalkane is the elimination of 2-bromo-2-methylbutane using potassium hydroxide. This produces 2 different alkenes:
Elimination reactions involve the removal of two atoms or groups of atoms from a molecule to form a new product. One example of an elimination reaction is the dehydration of alcohols to form alkenes. In the case of elimination reactions with hydroxide ions and halogenoalkanes, they occur in hot, ethanolic conditions, and result in the production of water, a halide ion, and an alkene. This process is also referred to as dehydrohalogenation and utilizes an E2 mechanism. In elimination reactions, the hydroxide ion acts as a base, whereas in nucleophilic substitution, it acts as a nucleophile. Altering the reaction conditions can favor either elimination or nucleophilic substitution, and the favored halogenoalkane also plays a role in determining the dominant reaction type. It's important to understand these concepts to predict the outcome of a reaction accurately.
What is an elimination reaction?
An elimination reaction is a reaction in which two atoms, or groups of atoms, are removed from a molecule to form a new product.
Why are elimination reactions important?
Elimination reactions are important because they transform saturated compounds into unsaturated ones by forming a C=C double bond.
Are condensation reactions a type of elimination reaction?
Condensation reactions are elimination reactions, in which the two atoms removed react to form water. Condensation reactions are also known as dehydration reactions, and include the dehydration of alcohols to form an alkene and water.
What is the difference between substitution and elimination reactions?
In substitution reactions, an atom or group of atoms on a molecule is swapped for another atom or group of atoms. In elimination reactions, two atoms or groups of atoms are removed from a molecule.
