Halogenoalkanes undergo substitution reactions in the presence of a nucleophile such as the hydroxide ion, OH–, but most good nucleophiles are also great bases.
Bases are defined as proton acceptors (remember that a proton is essentially a H+ ion). Hydroxide ions readily accept protons / hydrogen ions using a lone pair of electrons to make a dative covalent bond to the hydrogen.
When a halogenoalkane reacts with a base, an elimination reaction occurs forming an alkene. It would seem that substitution reactions are in competition with elimination reactions.
As with substitution reactions, there are two mechanisms for an elimination reaction – E2 and E1 (2 stands for bimolecular, 1 for unimolecular). Primary halogenoalkanes only react via the E2 mechanism (and it’s slow), secondary and tertiary halogenoalkanes react faster via the E2 mechanisms and also via the E1 mechanism.
E2 reaction mechanism
The rate of elimination depends on the concentration of both the halogenoalkane and the base.
- The hydroxide ion is acting as a base, attacking a hydrogen atom on the β-carbon atom (this is the carbon atom adjacent to the carbon of the C-Br bond). At the same time, the C-Br bond breaks.
- There is a short lived transition state with a number of partial bonds – the negative charge is spread over the intermediate from the oxygen to the bromine.
- A double bond forms between the ⍺ and β carbon atoms and we have an alkene as the main product.
E1 reaction mechanism
The rate of elimination depends only on the concentration of the halogenoalkane.
The more stable the intermediate carbocation in the E1 mechanism, the faster the rate of this reaction. The simplest way to stabilise it is to replace the hydrogen atoms bonded to the carbon of the C-Br bond with electron donating alkyl groups, which explains why tertiary halogenoalkanes readily react using this mechanism but primary halogenoalkanes only react using the E2 mechanism.
How do I know whether a halogenoalkane will undergo substitution or elimination?
It will depend on a number of factors, but generally speaking …
Substitution is more likely to occur if we have a small strong nucleophile such as a hydroxide ion that can easily attack the 𝛅+ carbon of the C-X bond in a primary halogenoalkane (SN2) or the positively charged carbon of the tertiary carbocation (SN1).
Elimination is more likely to occur if we have a strong base such as the tert-butoxide ion which is too big to attack the 𝛅+ / positive carbon and finds it far easier to attack a hydrogen atom on the β-carbon (large anions are better bases than they are nucleophiles).
Elimination mechanisms have a higher activation energy than substitution because we have to break two bonds (C-H and C-Br) rather than one, which means that elimination reactions need higher temperatures.
We can also alter the odds of one reaction happening over another by changing the solvent – an aqueous solution of sodium hydroxide (solvent is water) at room temperature favours a substitution reaction whereas refluxing a halogenoalkane with concentrated sodium hydroxide in ethanol favours an elimination reaction.