$HBr$ reacts with $CH_2 = CH - OCH_3$ under anhydrous conditions at room temperature to give :

Question

( A ) $BrCH_2CHO$ and $CH_3OH$
( B ) $H_3C - CHBr - OCH_3$
( C ) $CH_3CHO$ and $CH_3Br$
( D ) $BrCH_2 - CH_2 - OCH_3$