The dehydration of isopropyl alcohol (also known as isopropanol or 2-propanol) to produce propene involves the removal of a water molecule (H₂O) from the alcohol. This reaction typically requires an acid catalyst and can be done through the following general mechanism:
Steps for Dehydration of Isopropyl Alcohol to Form Propene:
-
Protonation of the Alcohol: The dehydration reaction begins with the protonation of the hydroxyl group (-OH) of isopropyl alcohol by an acid catalyst (like sulfuric acid, H₂SO₄). This increases the electrophilicity of the hydroxyl group and makes the hydroxyl group a better leaving group.
\[ \text{(CH}_3\text{)}_2\text{CHOH} + \text{H}^+ \rightarrow \text{(CH}_3\text{)}_2\text{C(OH)}^+ \text{H}_2 \]
-
Formation of a Carbocation: Once the hydroxyl group is protonated, it can leave as water (H₂O), forming a carbocation intermediate. For isopropyl alcohol, this will yield a secondary carbocation at the second carbon.
\[ \text{(CH}_3\text{)}_2\text{C(OH)}^+ \text{H}_2 \rightarrow \text{(CH}_3\text{)}_2\text{C}^+ + \text{H}_2\text{O} \]
-
Elimination of a Proton: The next step is the elimination of a β-hydrogen (a hydrogen atom from an adjacent carbon) from the carbocation intermediate. This step results in the formation of a double bond between the two carbons, producing propene.
\[ \text{(CH}_3\text{)}_2\text{C}^+ \rightarrow \text{C}(\text{C}_1\text{H}_3)(\text{C}_2\text{H}) + \text{H}^+ \]
-
Formation of Propene: The final product is propene (C₃H₆).
Overall reaction: \[ \text{(CH}_3\text{)}_2\text{CHOH} \xrightarrow{\text{H}^+} \text{C}_3\text{H}_6 + \text{H}_2\text{O} \]
Summary:
- The reaction requires an acid catalyst (such as sulfuric acid).
- The reaction first involves protonation, forming a better leaving group.
- Next, the hydroxyl group leaves, forming a carbocation.
- Finally, a β-hydrogen is eliminated to form propene.
This dehydration reaction is an example of an elimination reaction, where a molecule of water is removed and a double bond is formed.
1 answer