CH_3-CH=CH_2 xrightarrow{IBr} text{Product } | vedclass

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Question

(A) The reaction of an alkene with an interhalogen compound like $IBr$ follows Markovnikov's rule. In $IBr$,the electronegativity of $Br$ $(2.96)$ is

Vedclass · Accepted Answer

$CH_3-CH(Br)-CH_2I$

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(A) The reaction of an alkene with an interhalogen compound like $IBr$ follows Markovnikov's rule. In $IBr$,the electronegativity of $Br$ $(2.96)$ is higher than that of $I$ $(2.66)$,so the bond is polarized as $I^{\delta+} - Br^{\delta-}$. The electrophilic part is $I^{\delta+}$,which attacks the double bond to form a more stable carbocation. $CH_3-CH=CH_2 + I^{\delta+} \rightarrow CH_3-CH^+-CH_2I$. Then,the nucleophilic $Br^{\delta-}$ attacks the carbocation to form the final product: $CH_3-CH(Br)-CH_2I$.

$CH_3-CH=CH_2 \xrightarrow{IBr} \text{Product } (A) \text{ is}$

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