The product (C) in the below mentioned reaction is: CH 3 - CH 2 - CH 2 - Br → ∆ KOH alc A → HBr B → KOH aq ∆ C

Step-by-step solution

When haloalkane or alkyl halide with a β hydrogen is heated with alcoholic solution of KOH , elimination of a hydrogen atom from β carbon and halogen atom from α - carbon occurs as a result, alkene is formed as product. Since β hydrogen atom is involved in the elimination reaction, it is often called β elimination. An alkene is formed. This alkene then on reaction with HBr undergoes addition reaction to give bromopropane, which on reaction with aqueous KOH gives alcohol, propan - 2 - ol.