Acid-catalysed hydration of an alkene takes place through which reaction intermediate?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Carbocation
An alkene adds water in the presence of an acid catalyst to give an alcohol. For propene:
\( \displaystyle \ce{CH3CH=CH2 + H2O} \xrightarrow{\ce{H+}} \ce{CH3-CH(OH)-CH3} \)
The reaction goes in three steps:
Protonation: the π electrons of the double bond attack \( \ce{H3O+} \) (an electrophilic attack), and a carbocation forms: \( \ce{CH3CH=CH2 + H3O+} \rightleftharpoons \ce{(CH3)2CH^+ + H2O} \)
Nucleophilic attack: water attacks the carbocation: \( \ce{(CH3)2CH^+ + H2O} \rightleftharpoons \ce{(CH3)2CH-OH2^+} \)
Deprotonation: water removes a proton to give the alcohol: \( \ce{(CH3)2CH-OH2^+ + H2O -> CH3-CH(OH)-CH3 + H3O+} \)
Why the other options are incorrect
Option (2) Carbanion — Incorrect. A carbanion carries a negative charge on carbon; here the alkene gives its electrons to \( \ce{H+} \), so the carbon becomes positive, not negative.
Option (3) Free radical — Incorrect. A free radical has an unpaired electron and forms when a bond breaks evenly; in acid-catalysed hydration the electrons move in pairs to form ions.
Option (4) Carbene — Incorrect. A carbene is a neutral carbon with only two bonds; no such species forms when an alkene is protonated.
NCERT (2026-27 edition):
7.4.1 Preparation of Alcohols – “Mechanism
The mechanism of the reaction involves the following three steps: Step 1: Protonation of alkene to form carbocation by electrophilic attack of H3O+.”