8.5 : 알켄의 산 촉매 수화

While the direct addition of water to an alkene leads to no reaction, in the presence of an acid, an alcohol is obtained by the net addition of a hydroxyl group and a hydrogen across the double bond.

Since strong acids, like sulfuric acid, dissociate completely in an aqueous solution, the acid participating in the reaction is the hydronium ion.

The first step is the slow protonation of the less substituted end of the double bond to form a more substituted carbocation.

In the second step, water, present in excess, acts as the nucleophile and attacks the carbocation to give the oxonium ion.

Lastly, water, with a pK_a of 15.7, acts as a base and deprotonates the more acidic oxonium ion with a pK_a of approximately -2, to yield the final product. The process is called acid-catalyzed, as a hydronium ion that initiates the reaction is recovered at the end.

The hydration of alkenes and the dehydration of alcohols are in equilibrium with each other, and the equilibrium position depends on the concentration of water and temperature.

According to Le Chatelier’s principle, the system in equilibrium adjusts to reduce the changes employed.

For instance, in the reaction of 2-methylpropene, water is present on the left side. The use of a dilute acid, which has more water, shifts the equilibrium to form more alcohol.

Alternatively, concentrated acid, which has less water, reverses the equilibrium to form the alkene and water.

Addition reactions are also temperature-dependent. Here, the enthalpy term is negative because the bond formation is exothermic. The entropy term, a product of a negative entropy change and a positive temperature value, is positive.

At low temperatures, the entropy term becomes smaller compared to the enthalpy term. Therefore, the Gibbs free energy becomes negative and the equilibrium constant being greater than one favors the formation of the alcohol.

In contrast, at higher temperatures, the entropy term dominates the enthalpy term, and the change in Gibbs free energy becomes positive. Consequently, the equilibrium constant becomes less than one, favoring the formation of the alkene.