Give the major product of the reaction:

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(B) The reaction is the electrophilic addition of $HBr$ to an alkene. According to Markovnikov's rule,the hydrogen atom of $HBr$ adds to the carbon at

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$2-$bromo$-3-$methylheptane

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(B) The reaction is the electrophilic addition of $HBr$ to an alkene. According to Markovnikov's rule,the hydrogen atom of $HBr$ adds to the carbon atom of the double bond that already has more hydrogen atoms,while the bromine atom adds to the more substituted carbon atom. The starting material is $3-methylhept-1-ene$. Upon protonation of the double bond,a carbocation is formed at the $C2$ position,which is a secondary carbocation. However,the $C3$ position is a tertiary carbon. $A$ hydride shift from $C2$ to $C3$ is not possible here as the double bond is at the terminal position. Wait,let's re-examine the structure: The reactant is $3-methylhept-1-ene$. Addition of $H^+$ to $C1$ gives a carbocation at $C2$. This is a secondary carbocation. Attack of $Br^-$ at $C2$ gives $2-bromo-3-methylheptane$ as the major product.

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