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(A) The reaction involves the electrophilic addition of $Br_2$ to the $C=C$ double bond in the presence of $MeOH$ as a nucleophilic solvent. This is a

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(A) The reaction involves the electrophilic addition of $Br_2$ to the $C=C$ double bond in the presence of $MeOH$ as a nucleophilic solvent. This is a classic bromomethoxylation reaction. The $Br_2$ molecule forms a cyclic bromonium ion intermediate with the alkene. Then,the nucleophilic solvent $MeOH$ attacks the more substituted carbon of the bromonium ion (following Markovnikov-like regioselectivity due to the stability of the developing positive charge),leading to the anti-addition of $Br$ and $OMe$ across the double bond. Therefore,the product is the one where $Br$ and $OMe$ are added across the double bond,which corresponds to the structure in option $A$.

The major product obtained in the following conversion is

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$CH_3-CH_2-CH_2-Br \xrightarrow{KOH} CH_3-CH=CH_2 + KBr + H_2O$

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