Elimination Reactions

Elimination reactions with halogenoalkanes

An elimination reaction can occur between the hydroxide ion, \(: OH^-\), and a halogenoalkane. This reaction produces water, a halide ion, and an alkene.

Fig. 2 - Chloromethane, an example of a halogenoalkane

The hydroxide ion acts as a base by accepting a proton, \(H^+\), from the halogenoalkane to form water.

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 \(X\) to represent the halogen.

Fig. 3 - The general mechanism for elimination reactions involving halogenoalkanes

We’ve included a labelled diagram to help you understand the mechanism.

Fig. 4 - The mechanism for elimination

1. The hydroxide ion attacks a hydrogen atom with its lone pair of electrons. Notice which carbon atom this hydrogen atom is attached to - it is always a carbon atom adjacent to the C-X bond.
2. The bonded pair of electrons from the C-H bond now becomes part of a C=C double bond.
3. The halogen takes the pair of electrons from the C-X bond and is expelled as a leaving group.

The reactant is either potassium hydroxide,\(KOH\) , or sodium hydroxide,\(NaOH\) . The reaction is carried out under reflux in hot, ethanolic conditions, and the sodium or potassium ion reacts with the halide ion to form either a sodium or potassium halide.

Study tip: Make sure that you remember that elimination occurs in ethanolic, NOT aqueous conditions.

Suitable halogenoalkanes

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:

1. Circle the C-X bond.
2. Identify the adjacent carbons.
3. See if any of these adjacent carbons contain a C-H bond.

If they do, your halogenoalkane can react!

Fig. 5 - Identifying suitable halogenoalkanes. Notice that the carbon containing the C-X bond is also bonded to an R group. This may also be suitable for elimination

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:

Fig. 6 - Finding suitable halogenoalkanes for elimination

Products

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.

For example, the reaction between 2-chlorobutane and potassium hydroxide can produce three different alkenes:

Fig. 7 - Elimination with 2-chlorobutane

Reactivity of the halogenoalkane

Some halogenoalkanes are a lot more reactive than others. Iodopropane, for example, will react faster with hydroxide ions than chloropropane. This is because the C-I bond has a lower bond enthalpy than the C-Cl bond. Iodine is a larger atom than chlorine and thus the bonded pair of electrons involved in the C-X bond are further from the nucleus. This means that there is weaker attraction between the nucleus and the electrons and the bond is easier to break.

Elimination reactions vs nucleophilic substitution

When carrying out an elimination reaction, some substitution will always occur, and vice versa. However, the reaction conditions can be controlled to favour one or the other of the reactions. The following table summarises the similarities and differences between the elimination and nucleophilic substitutions of halogenoalkanes. Favoured conditions, favoured halogenoalkane, and the role of the hydroxide ion will be explored further below.

Fig. 8 - A table comparing nucleophilic substitution with elimination

Favoured conditions

We stated above that elimination requires ethanolic potassium hydroxide. In actual fact, the solvent used is roughly a 50:50 split of ethanol and water. You might remember that halogenoalkanes are insoluble in water (see Nucleophilic Substitution Reactions), so a small amount of ethanol is needed to dissolve them. This is the case for both elimination and nucleophilic substitution reactions. However, the exact temperature and proportion of ethanol to water in the solvent influences the dominant reaction:

• Warm conditions favour nucleophilic substitution, whereas hot conditions favour elimination.
• A more aqueous, i.e., diluted solution favours nucleophilic substitution, whereas a more ethanolic solution favours elimination.
• Using more concentrated sodium or potassium hydroxide also favours elimination.

Study tip: Learn what your exam board wants you to know concerning reaction conditions. Some exam boards are happy with aqueous solution for nucleophilic substitution reactions and ethanolic solution for elimination, whereas some want you to explore the conditions in more depth, as given above.

Favoured halogenoalkane

The type of halogenoalkane used as a reactant affects the dominant reaction. Tertiary halogenoalkanes contain three alkyl groups attached to the carbon with the C-X bond, whereas primary halogenoalkanes have at most one. This means that tertiary halogenoalkanes have more opportunities for elimination, simply because they have more carbons with hydrogens adjacent to the C-X bonded carbon. Therefore, using tertiary halogenoalkanes favours elimination, whereas using primary halogenoalkanes favours nucleophilic substitution.

Fig. 9 - The type of halogenoalkane used has an effect on which reaction is favoured

The role of the hydroxide ion

The hydroxide ion takes different roles in the elimination and nucleophilic substitution, although this does not affect the dominant reaction type. In elimination, the hydroxide ion acts as a base whereas in nucleophilic substitution it acts as a nucleophile.

• A base is a proton acceptor. In elimination, the hydroxide ion takes a proton from one of the carbons adjacent to the carbon with the C-X bond.
• A nucleophile is an electron pair donor. In nucleophilic substitution, the hydroxide ion donates its lone pair of electrons to the partially charged carbon atom, causing the halogen to be substituted out of the molecule.

Examples of elimination reactions with halogenoalkanes

2-chloropropane undergoes elimination in hot ethanolic sodium hydroxide as shown below. Although hydrogens on both carbons 1 and 3 can be attacked, the alkenes produced are identical, so we don’t produce any isomers. In both cases, the products are propene, water, and the chloride ion, which reacts with sodium to form sodium chloride:

Fig. 10 - The elimination of 2-chloropropane to form propene, water, and the chloride ion

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:

Fig. 11 - The elimination of 2-bromo-2-methylbutane

Other common elimination reactions

Another type of elimination reaction is the dehydration of alcohols to form alkenes. (See Elimination Reactions of Alcohols.)