What is the product of the reaction: {H_2}C = | vedclass

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(B) The reaction of an alkene with $HBr$ in the presence of peroxide follows the anti-Markovnikov addition rule (Kharasch effect or peroxide effect).I

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$Br-CH_2-CH_2-(CH_2)_6-CH_3$

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(B) The reaction of an alkene with $HBr$ in the presence of peroxide follows the anti-Markovnikov addition rule (Kharasch effect or peroxide effect).In this reaction,the $Br^-$ radical attacks the less substituted carbon atom of the double bond.The reaction proceeds as follows:${H_2}C = CH-(CH_2)_6-CH_3 + HBr \xrightarrow{peroxide} Br-CH_2-CH_2-(CH_2)_6-CH_3$.Therefore,the product is $1-bromodecane$.

What is the product of the reaction: ${H_2}C = CH{(CH_2)_6}CH_3 \xrightarrow[peroxide]{HBr} ?$

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