Dehydration of Cyclohexanol – Preparation of an Alkene CHM 220
INTRODUCTION
Earlier this semester you reacted an alcohol (n-butanol) with a Hydro-halide acid (Hydro-bromic Acid) to
generate an alkyl halide. This reaction occurred through the SN2 pathway. In this experiment you will
react an alcohol (cyclohexanol) utilizing an acid catalyst (sulfuric acid) to “dehydrate” the alcohol and
form an alkene (cyclohexene).
Alcohols are frequently converted into the desired alkene using an acid catalyzed Elimination reaction.
The term “dehydrate” means to remove water and is used to identify the nature of the atoms/molecules
eliminated to form the double bond of the alkene in our product. Primary (1O) alcohols require strong
acids and significant amounts of heat (up to 180O C or more). Secondary alcohols (2O) are somewhat
easier to dehydrate requiring slightly lower temperatures, where Tertiary (3O) alcohols will undergo
dehydration at room temperature or only slightly above that.
The dehydration of cyclohexanol follows the E1 mechanistic pathway. The dehydration reaction
involved three steps. First to occur is the protonation of the alcohol by the acid, in the process the
Hydroxyl- group is converted from a poor leaving group to a good leaving group. This then result is the
formation of a 2O carbocation intermediate in the second step. Finally, we generate an alkene following
the loss of a proton adjacent to the carbocation.
OH
H
2
SO
4
OH2
++HSO
4
OH2
+H
2
O
H
+HSO
4
H
2
SO
4
+
In each case these reactions are reversible and are under conditions of equilibrium. Since each step is in
equilibrium we must make use of Le Chatelier’s principle to drive the reaction in the desired direction.
The simplest method of doing this is to remove the products of the reaction as they are formed. In this
experiment you will be distilling the product from the reaction mixture as it is formed.
