The predominant product formed when 3-methyl- | vedclass

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Question

(C) The reaction of $3$-methyl-$2$-pentene $(CH_3-CH=C(CH_3)-CH_2-CH_3)$ with $HOCl$ proceeds via the formation of a cyclic chloronium ion intermediat

Vedclass · Accepted Answer

$CH_3-CH_2-C(OH)(CH_3)-CH(Cl)-CH_3$

Vedclass · Answer

(C) The reaction of $3$-methyl-$2$-pentene $(CH_3-CH=C(CH_3)-CH_2-CH_3)$ with $HOCl$ proceeds via the formation of a cyclic chloronium ion intermediate.$HOCl$ dissociates to provide $Cl^+$ (electrophile) and $OH^-$ (nucleophile).The $Cl^+$ ion attacks the double bond to form a cyclic chloronium ion.The nucleophile $(OH^-)$ then attacks the more substituted carbon atom $(C3)$ due to its higher carbocation-like character in the transition state,following Markovnikov's rule.Therefore,the major product is $3$-chloro-$3$-methylpentan-$2$-ol,which corresponds to the structure $CH_3-CH(Cl)-C(OH)(CH_3)-CH_2-CH_3$.

The predominant product formed when $3$-methyl-$2$-pentene reacts with $HOCl$ is:

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