The dehydration of isopropyl alcohol (also known as 2-propanol) to propene occurs through an acid-catalyzed elimination reaction, commonly known as an E1 or E2 mechanism, depending on the conditions run. Here's an overview of the general mechanism:
Mechanism
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Protonation of Alcohol: The reaction typically begins with the protonation of the hydroxyl (-OH) group of isopropyl alcohol by an acid (usually sulfuric acid or phosphoric acid), converting the alcohol into a better leaving group (water).
\[ \text{(CH}_3\text{)}_2\text{CHOH} + \text{H}^+ \rightarrow \text{(CH}_3\text{)}_2\text{C(OH}_2\text{)}^+ \]
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Formation of Carbocation: After protonation, the -OHâ‚‚ group can leave, resulting in the formation of a carbocation. In the case of isopropyl alcohol, the carbocation formed is a secondary carbocation.
\[ \text{(CH}_3\text{)}_2\text{C(OH}_2\text{)}^+ \rightarrow \text{(CH}_3\text{)}_2\text{C}^+ + \text{H}_2\text{O} \]
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Elimination of a Hydrogen Atom: Next, a base (which can be the conjugate base of the acid used, or another base present in the system) abstracts a proton from a carbon adjacent to the carbocation. This leads to the formation of a double bond between the carbocation carbon and the adjacent carbon atom.
\[ \text{(CH}_3\text{)}_2\text{C}^+ + \text{Base} \rightarrow \text{C}_3\text{H}_6 + \text{H}^+ \]
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Formation of Propene: The final product is propene, which is an alkene, as a result of the elimination of the water molecule and formation of the double bond.
Summary Reaction
The overall reaction can be summarized as follows:
\[ \text{(CH}_3\text{)}_2\text{CHOH} \xrightarrow{\text{Acid}} \text{C}_3\text{H}_6 + \text{H}_2\text{O} \]
Conclusion
The dehydration of isopropyl alcohol to form propene is a fundamental reaction in organic chemistry and is widely used in various industrial applications, including the production of propene, which is a key building block in the synthesis of many chemicals and plastics.
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