What is the product when CH₃CH₂CH₂Br reacts with KOH in methanol?
Alcoholic KOH (in methanol) promotes elimination (E₂) in CH₃CH₂CH₂Br , forming CH₃CH=CH₂ (propene).
Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Elimination Reactions and Reaction with Metals and Other Reactions