Question

What do you understand by dehydration reaction? Write the mechanism of dehydration reaction of alcohol.

Hint:

Dehydration of alcohols generally follows the Saytzeff's Rule: the more substituted alkene is the major product because it is more stable. Always check whether the mechanism proceeds via E1 (carbocation intermediate) or E2 (concerted elimination), depending on the nature of the alcohol.

Answer 1

Dehydration reaction: A dehydration reaction is a chemical reaction in which a molecule of water (\(H_2O\)) is eliminated from the reacting molecule. In the case of alcohols, when heated with a strong acid catalyst (such as conc. H$_2$SO$_4$ or H$_3$PO$_4$), alcohols undergo dehydration to form alkenes.
\[ R-CH_2-CH_2-OH \;\xrightarrow[\;\;]{\text{conc. H$_2$SO$_4$, heat}} \; R-CH=CH_2 + H_2O \] This reaction follows the mechanism of E1 elimination for secondary and tertiary alcohols and E2 elimination for primary alcohols.
Mechanism of Dehydration of Alcohol (E1 for 2$^\circ$ or 3$^\circ$ alcohols):

Step 1: Protonation of alcohol.
The lone pair of electrons on the oxygen atom of the alcohol attacks a proton (\(H^+\)) from the acid catalyst, converting the hydroxyl group into a better leaving group (water). \[ R-CH(OH)-R' + H^+ \;\rightarrow\; R-CH(OH_2^+)-R' \]

Step 2: Formation of carbocation.
The protonated alcohol loses a water molecule, forming a carbocation intermediate. \[ R-CH(OH_2^+)-R' \;\rightarrow\; R-C^+-R' + H_2O \]

Step 3: Elimination of proton.
A base (often \(HSO_4^-\)) abstracts a $\beta$-hydrogen atom from the carbon adjacent to the carbocation, leading to the formation of a double bond (alkene). \[ R-C^+-CH_2 \;\rightarrow\; R-CH=CH_2 + H^+ \] Overall reaction: \[ \text{Alcohol} \;\xrightarrow[\;\;]{\text{conc. H$_2$SO$_4$},\;170^\circ C} \;\text{Alkene} + H_2O \]