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(A) The reaction involves the electrophilic addition of $HI$ to the alkene $CH_2=CH-CH_2-I$.According to Markovnikov's rule,the proton $(H^+)$ adds to

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$CH_3-CHI-CH_2-I$

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(A) The reaction involves the electrophilic addition of $HI$ to the alkene $CH_2=CH-CH_2-I$.According to Markovnikov's rule,the proton $(H^+)$ adds to the carbon atom with more hydrogen atoms,and the iodide ion $(I^-)$ adds to the more substituted carbon atom.Step $1$: The double bond attacks the $H^+$ of $HI$ to form the more stable carbocation,$CH_3-CH^+-CH_2-I$.Step $2$: The iodide ion $(I^-)$ then attacks the carbocation to form $CH_3-CHI-CH_2-I$.Since $HI$ is in excess,the reaction stops at the geminal diiodide product,$CH_3-CHI-CH_2-I$.

Predict the major product for the following reaction:
$CH_2=CH-CH_2-I \xrightarrow{HI \text{ (Excess)}} ?$

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Predict the major product for the following reaction:$CH_2=CH-CH_2-I \xrightarrow{HI \text{ (Excess)}} ?$