For the reaction CH_3-CH=CH_2 + HOCl to A,the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The addition of $HOCl$ to propene follows Markovnikov's rule.$HOCl$ dissociates as $HO^{\delta-} - Cl^{\delta+}$.The electrophile $Cl^{+}$ attacks

Vedclass · Accepted Answer

$CH_3-CH(OH)-CH_2Cl$

Vedclass · Answer

(B) The addition of $HOCl$ to propene follows Markovnikov's rule.$HOCl$ dissociates as $HO^{\delta-} - Cl^{\delta+}$.The electrophile $Cl^{+}$ attacks the double bond to form a cyclic chloronium ion intermediate.The nucleophile $OH^{-}$ then attacks the more substituted carbon atom $(C-2)$ to open the ring.Thus,the reaction is: $CH_3-CH=CH_2 + HOCl 	o CH_3-CH(OH)-CH_2Cl$.Therefore,Option $B$ is the correct answer.

For the reaction $CH_3-CH=CH_2 + HOCl \to A$,the product $A$ is:

Similar Questions

Write about ozonolysis of alkene.

Explain the dehalogenation of vicinal dihalides with an example.

The final product $'C'$ is?

$CH_3-C(CH_3)_2-CH=CH_2 \xrightarrow[H_2O_2/OH^{-}]{B_2H_6/THF} (Z)$ Major product

The reaction of $2-$methylbuta$-1,3-$diene with $O_3$ followed by $Zn-H_2O$ gives:

Vedclass Test Series

Exam Paper Generator

Online Exam Module