If you are not sure what a haloalkane is then have a look on the nomenclature page.
Haloalkanes contain a polar bond on the functional group, because the halogen is electronegative so becomes δ- and the carbon is positive. This means it is susceptible to nucleophillic attack by ions.
The stength of the C - Halogen bond also influences the rate of substitution. The C-F bond recquires a quite a bit (484 kJmol-1) of energy to break, however, the C-Br bond is more easy to break so the rate of reaction is faster.
The mechanism below uses the notation :Nu- to refer to the nucleophile which could either be OH- or NC-.
The mechanism below shows what happens with ammonia and a haloalkane.
The bromine ion and NH4 join to make NH4Br, and the final product is an amine
In the first mechanism the OH ion replaces the Br and forms an alcohol. However, the OH- can also act as a base, where an elimination reaction occurs, as outlined below.
So how do you decide whether substitution or elimination is going to happen. There are several factors that will determine what will happen. Have a look at the table below to find them out.
|Haloalkane Structure||Primary haloalkanes prefer substitution
Secondary halalkanes will do both at the same rate
Tertiary haloalkanes prefer elimination.
|Base Strength||Elimination is more likely as the strength of the base increases.|
|Temperature||Higher reaction temperatures make elimination happen more.|
If you are not sure about the differences between primary, secondary and tertiary then look at the page alcohols as the same rules apply.