The reaction of propene with HOCl proceeds vi | vedclass

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Question

(B) The reaction of propene $(CH_3-CH=CH_2)$ with $HOCl$ involves the formation of a chloronium ion intermediate.$HOCl$ dissociates to provide an elec

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$Cl^{+}$ in the first step

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(B) The reaction of propene $(CH_3-CH=CH_2)$ with $HOCl$ involves the formation of a chloronium ion intermediate.$HOCl$ dissociates to provide an electrophilic $Cl^{+}$ species.In the first step,the $\pi$-bond of the alkene attacks the electrophilic $Cl^{+}$ to form a cyclic chloronium ion intermediate.This is followed by the nucleophilic attack of $\overset{\Theta}{O}H$ on the more substituted carbon atom of the chloronium ion to yield the final product,$1-chloro-2-propanol$ (or $2-chloro-1-propanol$ depending on regioselectivity,but the electrophile is $Cl^{+}$).Therefore,the reaction proceeds via the addition of $Cl^{+}$ in the first step.

The reaction of propene with $HOCl$ proceeds via the addition of

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