Explain the preparation of alcohols by acid-catalyzed hydration of alkenes with its mechanism.
(N/A) Preparation: Alkenes react with water $(HOH)$ in the presence of an acid catalyst (dilute $HCl$ or $H_2SO_4$) to form alcohols.
If the alkene is unsymmetrical,the addition follows 'Markovnikov's rule',meaning the $-OH$ group attaches to the more substituted carbon and the $-H$ atom attaches to the less substituted carbon of the double bond.
$(b)$ Mechanism: The acid-catalyzed hydration of alkenes is an electrophilic addition reaction occurring in three steps.
Step-$1$: Protonation of the alkene by an electrophile $H_3O^+$ (or $H^+$) to form a more stable carbocation. This is the slow and rate-determining step as the $\pi$-bond of $C=C$ breaks.
Step-$2$: Nucleophilic attack of water $(H_2\ddot{O}:)$ on the carbocation $(M)$ to form a protonated alcohol.
Step-$3$: Deprotonation to form the final alcohol product.
If the alkene is unsymmetrical,the addition follows 'Markovnikov's rule',meaning the $-OH$ group attaches to the more substituted carbon and the $-H$ atom attaches to the less substituted carbon of the double bond.
$(b)$ Mechanism: The acid-catalyzed hydration of alkenes is an electrophilic addition reaction occurring in three steps.
Step-$1$: Protonation of the alkene by an electrophile $H_3O^+$ (or $H^+$) to form a more stable carbocation. This is the slow and rate-determining step as the $\pi$-bond of $C=C$ breaks.
Step-$2$: Nucleophilic attack of water $(H_2\ddot{O}:)$ on the carbocation $(M)$ to form a protonated alcohol.
Step-$3$: Deprotonation to form the final alcohol product.
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