10.5 – Halogenoalkanes
The halogens are the seventh group on the periodic table, including elements such as fluorine, chlorine, bromine and iodine.
Halogenoalkanes have a halogen bonded to the skeleton of the alkane. They have the general formula:
Where X is the halogen.
10.5.1 – Describe, using equations, the substitution of halogenoalkanes with sodium hydroxide
Halogens have greater electronegativity than carbon, causing halogenoalkanes to have a polar bond. This causes nucleophiles (species with a non-bonding pair of electrons such as OH–, H2O, NH3 and CN–) to be attracted to the carbon atom, allowing for substitution reaction to occur.
The curly arrows in the diagram below show the movement of electron pairs. The halogen is also called the leaving group.
The product of a substitution reaction with hydroxide is an alcohol. For example, the reaction between bromoethane (a primary halogenoalkane) and warm dilute sodium hydroxide:
On the other hand, the reaction between 2-bromo-2-methylpropane (a tertiary halogenoalkane) and warm dilute sodium hydroxide solution takes place according to the equation:
10.5.2 – Explain the substitution reactions of halogenoalkanes with sodium hydroxium in terms of SN1 and SN2 mechanisms
Heterolytic fission is when one of the species takes the shared electron pair, creating two oppositely charged ions.
X : X → X+ + X:–
Primary Halogenoalkanes
These have one alkyl group attached to the carbon bonded to the halogen, and undergo SN2 reactions. The hydrogen atoms that are also attached to the carbon are small and do not provide much protection for the halogen. This leaves it open to attack from a nucleophile.
The rate expression for this is: rate = k[C2H5Br][OH–]
The transition state formed during the reaction means that the rate is determined by both reactants. The slowest step involves both the halogenoalkane and the hydroxide ion, making it a bimolecular reaction. The transition state is when the carbon is weakly bonded to the halogen and the hydroxide.
Secondary Halogenoalkanes
These have two alkyl groups attached to the carbon bonded to the halogen, and can undergo both SN1 and SN2 reactions, and there is no way of determining which will happen – they may use one or the other, or a combination of the two.
Tertiary Halogenoalkanes
These have three alkyl groups attached to the carbon bonded to the halogen and undergo SN1 reactions. The alkyl groups are larger and protect the halogen from attack by nucleophiles, called steric hindrance.
This is a two-step reaction, happening according to the equations:
Since the first step is the slowest, it is the rate-determining step. There is only one molecule involved in this step; hence the overall reaction is SN1. The halogenoalkane splits heterolytically, with the halogen taking the bonding electron pair and forming a halide ion. The alkane now has a positive charge, called a carbocation. This means that the rate
expression is: rate = k[C(CH3)3Br]
Once the carbocation is formed, it is available to bond with the nucleophile. The surrounding alkyl groups keep the carbocation stable long enough for the second step to occur by donating electrons, or a positive inductive effect.