CBSE · Class 12 · Chemistry · Alcohols, Phenols and EthersExplain the mechanism of acid-catalyzed dehydration of alcohols to form alkenes with proper steps. Also, explain why the reaction follows Saytzeff's rule.

Mechanism of Acid-Catalyzed Dehydration of Alcohols

\nThe acid-catalyzed dehydration of alcohols to form alkenes is an important chemical transformation in organic chemistry. This reaction typically involves secondary and tertiary alcohols and proceeds via an E1 elimination mechanism. The overall reaction involves the loss of a water molecule from the alcohol molecule in the presence of an acid catalyst such as concentrated sulfuric acid ($H_2SO_4$) or phosphoric acid ($H_3PO_4$) at elevated temperatures.

Step 1: Protonation of Alcohol\nIn the first step, the oxygen atom of the alcohol molecule uses its lone pair of electrons to accept a proton ($H^+$) from the acid catalyst. This results in the formation of a protonated alcohol, also known as an alkyloxonium ion. This step is fast and reversible. Protonation converts the poor leaving group ($-OH$) into a much better leaving group, which is a water molecule ($$-OH_2^+$).

Step 2: Formation of Carbocation (Rate-Determining Step)\nIn the second step, the carbon-oxygen bond of the protonated alcohol breaks heterolytically, releasing a neutral water molecule and leaving behind a positively charged carbon species known as a carbocation. This step is slow and is the rate-determining step of the entire reaction mechanism. The stability of the carbocation formed determines the ease of dehydration; therefore, tertiary carbocations form most readily, followed by secondary and primary carbocations.

Step 3: Elimination of a Proton to Form Alkene\nIn the final step, a base (such as a water molecule or bisulfate ion) removes a proton ($H^+$) from a carbon atom adjacent to the positively charged carbocation (the $\beta$-carbon). The electrons from the $\text{C-H}$ bond shift to form a double bond between the $\alpha$ and $\beta$ carbons, generating the final neutral alkene and regenerating the acid catalyst.

Why the Reaction Follows Saytzeff's Rule\nSaytzeff's rule states that in elimination reactions, the major product is the alkene that contains the greater number of alkyl groups attached to the doubly bonded carbon atoms (i.e., the more substituted alkene). During dehydration, when there is a possibility of forming more than one alkene, the more substituted alkene is more stable due to hyperconjugation and inductive effects. Consequently, the transition state leading to the more stable alkene has lower activation energy, making it the major product.