Question

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

Answers

GPT-4o mini
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

2. **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

3. **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.