The reaction of propene with H_3O^{+} proceed | vedclass

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Question

(C) The reaction of propene $(CH_3-CH=CH_2)$ with $H_3O^{+}$ (acid-catalyzed hydration) follows Markovnikov's rule.In the first step,a proton $(H^{+})

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$CH_3-CH(O^{+}H_2)-CH_3$

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(C) The reaction of propene $(CH_3-CH=CH_2)$ with $H_3O^{+}$ (acid-catalyzed hydration) follows Markovnikov's rule.In the first step,a proton $(H^{+})$ adds to the terminal carbon to form the more stable secondary carbocation $(CH_3-C^{+}H-CH_3)$.In the second step,water $(H_2O)$ acts as a nucleophile and attacks the carbocation to form the oxonium ion intermediate $CH_3-CH(O^{+}H_2)-CH_3$.

The reaction of propene with $H_3O^{+}$ proceeds through which of the following intermediates?

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