`HBr` reacts with `CH_(2)=CH-OCH_(3)` under anhydrous conditions at room temperature to give:

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Answer :

`HBr` reacts with `CH_(2)=CH-OCH_(3)` under anhydrous conditions at room temperature to give: A. `H_(3)C-CHBr- ... (3)OH` D. `BrCH_(2)-CH_(2)-OCH_(3)`

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