The reaction of propene with HOCl (Cl_2 + H_2 | vedclass

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Question

(D) The reaction of propene with $HOCl$ involves the electrophilic attack of $Cl^+$ on the double bond. The $Cl^+$ ion attacks the double bond to form

Vedclass · Accepted Answer

$CH_3-CH^+-CH_2-Cl$

Vedclass · Answer

(D) The reaction of propene with $HOCl$ involves the electrophilic attack of $Cl^+$ on the double bond. The $Cl^+$ ion attacks the double bond to form a cyclic chloronium ion intermediate,which is then opened by the nucleophilic attack of $H_2O$ or $OH^-$. However,in terms of carbocation character during the ring opening,the positive charge is stabilized on the more substituted carbon atom. The intermediate formed is $CH_3-CH^+-CH_2-Cl$.

The reaction of propene with $HOCl$ $(Cl_2 + H_2O)$ proceeds through the intermediate $:$

Similar Questions

The major product of the following reaction is $:$

An alkene $A$ on reaction with $O_{3}$ and $Zn-H_{2}O$ gives propanone and ethanal in equimolar ratio. Addition of $HCl$ to alkene $A$ gives $B$ as the major product. The structure of product $B$ is:

How many products will be formed in the reaction shown below?
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Reason : $trans-$Compound in $trans$ addition forms two types of stereoisomers.

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