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(D) The reaction of an alkene with $HBr$ in the presence of peroxide follows the anti-Markovnikov addition rule.In this reaction,the $H$ atom adds to

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$1,4-$bis(bromomethyl)cyclohexane

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(D) The reaction of an alkene with $HBr$ in the presence of peroxide follows the anti-Markovnikov addition rule.In this reaction,the $H$ atom adds to the carbon with fewer hydrogen atoms (or the more substituted carbon),and the $Br$ atom adds to the carbon with more hydrogen atoms.Given the reactant $H_2C=C_6H_{10}=CH_2$ (which is $1,4-$dimethylenecyclohexane),the addition of $HBr$ occurs at both terminal double bonds.Applying the anti-Markovnikov rule to both sides,the $Br$ atom attaches to the terminal $CH_2$ group,converting it into a $CH_2Br$ group.Thus,the product formed is $BrCH_2-C_6H_{10}-CH_2Br$,which is $1,4-$bis(bromomethyl)cyclohexane.

What is the product of the following reaction?
$H_2C=C_6H_{10}=CH_2 + HBr \xrightarrow{\text{peroxide}} ?$

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What is the product of the following reaction?$H_2C=C_6H_{10}=CH_2 + HBr \xrightarrow{\text{peroxide}} ?$