explain how 2_hexanolis prrpared by acid -catalyzed hydration of an appropriate alkane showing the reaction mechanism as well

1 answer

To prepare 2-hexanol via acid-catalyzed hydration, we need to start with an appropriate alkene as our precursor. A common choice is to use 1-hexene, which is the simplest alkene that will yield 2-hexanol through this reaction. The process involves the following steps:

Step 1: Formation of Carbocation

  1. Protonation of Alkene: The first step in the acid-catalyzed hydration reaction involves the addition of a proton (H⁺) from an acid (commonly sulfuric acid, H₂SO₄) to the double bond of 1-hexene. This generates a carbocation intermediate.

    The reaction can be represented as: \[ \text{C}6\text{H}{12} + \text{H}^+ \rightarrow \text{Carbocation (C}6\text{H}{13}^+\text{)} \]

    The carbocation formed is a secondary carbocation because it is located at the carbon that becomes the second carbon in the 2-hexanol.

Step 2: Nucleophilic Attack

  1. Nucleophilic Attack by Water: Water (H₂O) then acts as a nucleophile and attacks the carbocation, leading to the formation of an alcohol.

    The reaction can be written as follows: \[ \text{C}6\text{H}{13}^+ + \text{H}_2\text{O} \rightarrow \text{C}6\text{H}{13}\text{OH} + \text{H}^+ \]

    Here, we are forming 2-hexanol (C₆H₁₄O). The proton is regenerated in this step, making the acid catalyst available for further reactions.

Step 3: Deprotonation

  1. Deprotonation: The final step involves the deprotonation of the alcohol to yield the final 2-hexanol product.

    This can be represented as: \[ \text{C}6\text{H}{13}\text{OH} \xrightarrow{\text{Base}} \text{C}6\text{H}{13}\text{O}^- + \text{H}^+ \]

Summary Reaction

Overall, the acid-catalyzed hydration of 1-hexene yields 2-hexanol as follows: \[ \text{C}6\text{H}{12} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{C}6\text{H}{14}\text{O} \]

Mechanism Summary

  1. Protonation of the alkene forms a carbocation intermediate.
  2. Water attacks the carbocation, forming an alcohol.
  3. Deprotonation of the alcohol gives the final product, 2-hexanol.

This mechanism follows Markovnikov's rule, where the more substituted carbocation is formed, leading to the formation of 2-hexanol.