In class you may have dehydrated an alcohol, for example ethanol using the apparatus shown below. Here the ethanol is soaked into to ceramic wool which is packed tightly into the bottom of a boling tube. A catalyst of pumice stone or aluminium oxide can be used, simply heat the catalyst using a gentle heating flame from a Bunsen burmer then pass the flame over the wool soaked in ethanol. This heat will vapourise the ethanol and as it passes over the catalyst it will be dehydrated to form the alkene ethene. Ethene is insoluble in water and so can be collected in a boiling tube held over a trough of water.
We can represent his dehydration as:
One of the most useful elimination reactions is the acid catalysed dehydration of alcohols using concentrated phosphoric or sulfuric acids. Phosphoric acid is the better reagent to use here as it in not an oxidising acid like sulfuric and this leads to fewer unwanted side reactions. This dehydration reaction results in the formation of alkenes. The mechanism of the dehydration of an alcohol can be thought as occurring in three steps:
This acid catalysed dehydration of an alcohol works best with tertiary alcohol. Secondary alcohols can be made to react but harsher conditions are required and primary alcohols are the slowest to react with high temperatures and the use of acids with high concentrations are both required to dehydrate a primary alcohol. The reasons for this should be familar to you, the reaction mechanism takes place through a carbocation intermediate and if you recall tertiary carbocations are more stable than secondary carbocation ions which are more stable than primary carbocations. This basically means that much more energy is required to get a primary carbocation to form than is required to get a tertiary carbocation to form.
In the example given above the alcohol propan-2-ol was dehydrated, this molecule is symmetrical and the dehydration of it can lead to the formation of only one product- propene. However in the example below the unsymmetrical tertiary alcohol 2-methylbutan-2-ol is dehydrated. This time there is more than one possible product. If you study the mechanism shown below you will see that at step 3, the loss of a proton or hydrogen ion (H+) there are two possiblities. Loss of one of the purple coloured hydrogen atoms or one of the blue coloured hydrogen atoms will lead to two different products. These two products will not be formed in equal amounts. Zaitsev's rule can be used to predict which product will be the major one and which the minor one. Zaitsev's rule basically says that the more substituted the alkene the more stable it will be. So here we have a di- and a tri-substituted alkene, so according to Zaitsev's rule the major product will be the trisubstituted alkene. This is outlined below:
You should take care with the products of these dehydration reactions, often the products will shown geometric (E/Z) isomerism and its a favourite exam question to name the three products of a dehydration reaction, just remember that 2 will be the E/Z iosomers.