Alcohols are formed when halogenalkanes are warmed with aqueous solutions of sodium hydroxide or potassium hydroxide. You are probably familiar with potassium and sodium hydroxide from any work you have done on acids and alkali where the sodium hydroxide and potassium hydroxide would have been used as a base (a H+ acceptor). However the hydroxide ion (OH-)can also behave as a nucleophile. There are many similarities between nucleophiles and bases, for example they both possess lone pairs of electrons for one. Normally strong bases are good nucleophiles and vice versa. However with sodium and potassium hydroxide we can alter the reaction conditions to "make it" behave as a base or a nucleophile.
If a halogenalkane is warmed with sodium or potassium hydroxide that has been prepared by dissolving the base in water to form an aqueous solution of sodium or potassium hydroxide then the hydroxide ion in this case tends to act as a nucleophile, as shown in the image below. Nucleophilic substitution using halogenalkanes and hydroxide ions results in the formation of alcohols.
However if the potassium or sodium hydroxide is dissolved in hot ethanol, to form an alcoholic solution of the base then a different type of reaction takes place. This reaction is called an elimination reaction and here the hydroxide ion acts as a base (H+ acceptor) and not a nucleophile. This elimination reaction occurs as a one step process with no intermediates being formed, there are a few key features of this elimination reaction you should be familar with:
In the example above using 2-bromopropane, a symmetrical
halogenalkane molecule it would not have made any difference to the
final product if I had removed a hydrogen atom from the carbon atom to the right of the
halogenalkane molecule as opposed to the one on the
left which is shown above. In either case the final product would still have been propene.
However if I had used an unsymmetrical halogenalkane then more than one product would have been obtained. As an example consider the products of the reflux reaction of 2-bromobutane with hot alcoholic potassium hydroxide (recall that these conditions are ideal for an elimination reaction).
It is a bit more tricky to predict the products of elimination reactions than with nucleophilic substitution reactions. However what we can say is that elimination reactions generally give mixtures of products. Consider the base induced elimination reaction of 2-bromobutane,with alcoholic potassium hydroxide solution as shown below. Now the hydroxide ion (the base) can in theory remove any of the hydrogen atoms attached to a carbon atom adjacent to the leaving group (the halogen), this means that all the blue hydrogens and all the brown hydrogens in the 2-bromobutane molecule can be removed by the basic hydroxide ion. This will lead to a mixture of products as shown.
The reaction may give a mixture of products but we can say with confidence that the more substituted alkene will be the
product of the reaction, so in the example above the but-2-ene is the alkene with the more alkyl groups attached
to the C=C so
it will be the main product of this particular reaction.
However as well as elimination reactions taking place it is highly likely that some substitution reactions will also take place and this will result in additional products. Elimination reaction work "best" with tertiary halogenalkanes where the reaction produces almost exclusively alkenes, however elimination reactions generally do not take place with primary halogenakanes, substitution reactions take place instead. Secondary halogenalkanes will give a mixture of elimination and substitution products - not ideal really!
However the reaction conditions also play a part in deciding whether the reaction will be a nucleophilic substitution or an elimination reaction: