Dehydration of Alcohols
Alcohol upon reaction with protic acids tends to lose a molecule of water to form alkenes. These reactions are known as dehydration of alcohols. It is an example of an elimination reaction. Its rate varies for the primary, secondary and tertiary alcohols. This variation of rate can be attributed to the stability of carbocation generated. Since the carbocation is most stable in the case of tertiary alcohols, the rate of dehydration is highest for tertiary alcohols in comparison to secondary and primary alcohols. Dehydration of alcohols follows a three-step mechanism:
- Formation of protonated alcohol
- Formation of carbocation
- Formation of alkenes
General dehydration reaction of alcohols can be seen as,
Mechanism of dehydration of alcohols:
Dehydration of alcohols can follow E1 or E2 mechanism. For primary alcohols, elimination reaction follows E2 mechanism while for secondary and tertiary alcohols elimination reaction follows E1 mechanism. Generally, it follows a three-step mechanism. The steps involved are explained below:
1. Formation of protonated alcohol:
In this step, the alcohol is acted upon by a protic acid. Due to the lone pairs present on oxygen atom it acts as a Lewis base. Protonation of alcoholic oxygen takes place which makes it a better leaving group. It is a reversible step which takes place very quickly.
2. Carbocation formation:
In this step, the C-O bond breaks generating a carbocation. This step is the slowest step in the mechanism of dehydration of an alcohol. Hence, the formation of carbocation is considered as the rate determining step.
3. Alkene formation:
This is the last step in dehydration of alcohols. Here the proton generated is eliminated with the help of a base. The carbon atom adjacent to the carbocation breaks the existing C-H bond to form C=C. Thus, an alkene is formed.
For a detailed discussion on dehydration of alcohols, please download Byju’s- The Learning App.
Practise This Question