Question

What is the detailed mechanism for the dehydration of 3-methyl-3-pentanol using sulfuric acid?

A) Protonation of the hydroxyl group initiates water removal, forming a carbocation intermediate. Subsequent deprotonation leads to the final product, 3-methyl-3-pentene.

B) Sulfuric acid reacts directly with 3-methyl-3-pentanol, causing the elimination of water and the immediate formation of 3-methyl-3-pentene.

C) Oxygen from sulfuric acid attacks the hydroxyl group, resulting in the expulsion of water and the creation of 3-methyl-3-pentene.

D) Sulfuric acid acts as a spectator in the reaction, merely providing a medium for heat, which induces the spontaneous dehydration of 3-methyl-3-pentanol.

Answer 1

The dehydration of 3-methyl-3-pentanol using sulfuric acid involves the protonation of the hydroxyl group to form a carbocation, followed by the removal of a β-hydrogen by a water molecule, resulting in the creation of 3-methyl-3-pentene.

Step-by-step explanation:

The detailed mechanism for the dehydration of 3-methyl-3-pentanol using sulfuric acid begins with the protonation of the hydroxyl group, which initiates the removal of water to form a carbocation intermediate. This protonation occurs because sulfuric acid is a strong dehydrating agent with a strong affinity for water.

In the next step, water acts as a base, removing a β-hydrogen (a hydrogen from the β-carbon) and causing an elimination reaction that results in the formation of the double bond of the alkene, 3-methyl-3-pentene.

Sulfuric acid serves as a catalyst because its conjugate base, the bisulfate ion (HSO4-), is a weak nucleophile and does not partake in substitution reactions. Instead, it facilitates the dehydration. Water, being a better nucleophile and present in excess, wins in the subsequent proton removal step, eventually leading to the creation of the alkene product.