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

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

HB react with CH_2 =CH -OCH_3 under anhydrous conditions at room temperature to give

HBr reacts with CH_2 = CH - OCH_3 under anhydrous conditions at 25^@C to give

Hydrocarbon 'A' reacts with KMnO_4 under neutral conditions at room temperature to form only a diketone, CH_3CH_2underset(O)underset(||)C-underset(O)underset(||)C"CH"_2CH_2CH_3 . The hydrocarbon is

Benzene reacts with CH_(3)COCl in the presence of anhydrous AlCl_(3) to give

CH_(2)=CH(CH_(2))_(2)COOH and HBr reacts in presence of peroxide to give