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(B) The reaction proceeds via the formation of a carbocation intermediate.When $H^+$ attacks the double bond of $CH_3CH=CH_2$,it forms either a $2^\ci

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$B$

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(B) The reaction proceeds via the formation of a carbocation intermediate.When $H^+$ attacks the double bond of $CH_3CH=CH_2$,it forms either a $2^\circ$ carbocation $(CH_3CH^+CH_3)$ or a $1^\circ$ carbocation $(CH_3CH_2CH_2^+)$.The $2^\circ$ carbocation is more stable than the $1^\circ$ carbocation due to the inductive effect and hyperconjugation.Therefore,the $I^-$ ion attacks the $2^\circ$ carbocation to form $CH_3CHICH_3$ $(B)$ as the major product.This is in accordance with Markovnikov's Rule,which states that the negative part of the addendum adds to the carbon atom of the double bond having a lesser number of hydrogen atoms.

Which of the products will be the major product in the reaction given below?
$CH_3CH=CH_2 + HI \to CH_3CH_2CH_2I_{(A)} + CH_3CHICH_{3(B)}$

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Which of the products will be the major product in the reaction given below?$CH_3CH=CH_2 + HI \to CH_3CH_2CH_2I_{(A)} + CH_3CHICH_{3(B)}$