Home › JEE Advanced › Chemistry › Alcohols, Phenols and Ethers ›  The figure shows Step I of the acid-catalysed hydration of an alkene (e.g. propene). Which intermediate forms after H₃O⁺ attacks the alkene?
AA carbocation on the more-substituted carbon (Markovnikov)
BA carbanion on the more-substituted carbon
CA carbene
DA free radical on the terminal carbon
Answer & Solution
Correct answer: A. A carbocation on the more-substituted carbon (Markovnikov)
Acid-catalysed hydration starts with protonation of the alkene by H₃O⁺. The proton attaches to the carbon that bears more hydrogens, leaving a **carbocation** on the more substituted carbon (Markovnikov's rule, stability driven).
Step II then has water attack that carbocation; Step III deprotonates to give the alcohol. The full chain is electrophilic addition by a polar (ionic) mechanism — no radicals, no carbenes.
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