Write the mechanism (using curved arrow notation) when propene react with water in the presence of acid and forms 2-propanol.
Answers
Answer:
Propene reacts with water in the presence of a dilute, strong acid to produce propanol. The dilute, strong acid does not take place in the reaction itself.
The net addition of water to alkenes is known as hydration. The result involves breaking the pi bond in the alkene and an OH bond in water and the formation of a C-H bond and a C-OH bond.
Alkenes undergo an addition reaction with water in the presence of a catalyst to form an alcohol. This type of addition reaction is called hydration. The water is added directly to the carbon – carbon double bond.
The addition of water to an alkene in the presence of a catalytic amount of strong acid leads to the formation of alcohols (hydroxy‐alkanes). This reaction proceeds via a standard carbocation mechanism and follows the Markovnikov rule.
Answer:
Explanation:
CH3-CH=CH2 + H2SO4 —→CH3-CH(OSO3H) CH3
CH3CH(OSO3H) CH3 + H2O----->CH3-CH(OH)CH3 +H2SO4.
Explanation:
mechanism for the acid-catalyzed addition of water
addition of the electrophile
addition of the nucleophile
CH3-CH=CH2 + H2SO4 —→CH3-CH(OSO3H) CH3
CH3CH(OSO3H) CH3 + H2O----->CH3-CH(OH)CH3 +H2SO4.
Propene on acid catalyzed hydration first gives isopropyl hydrogen sulphate which on boiling with water gives propan -2-ol.
Explanation: The active site on the propene molecule is the double bond (C=C). In the presence of a dilute strong acid, water will add across the double bond in propene (propylene) to produce a mixture of alcohols: propan-1-ol (1-propanol) and propan-2-ol (2-propanol).
