What is the product of the following reaction: CH₃CH₂CH₂Br + KOH (alcoholic) → ?
Correct: B
An alcoholic solution of KOH promotes elimination (E2) reactions. CH₃CH₂CH₂Br is a primary alkyl halide, and in the presence of a strong base like KOH in ethanol, it will undergo an elimination reaction to form the alkene, CH₃CH=CH₂ (propene).