What is the product when CH₃CH₂CH₂Cl reacts with NaNH₂ in liquid ammonia?
NaNH₂ in liquid ammonia (strong base) promotes elimination (E₂) in CH₃CH₂CH₂Cl , forming CH₃CH=CH₂ (propene).
Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Polyhalogen Compounds - Chloroform Iodoform DDT Freons
