Which halide forms CH₃CH₂CH₂CH=CH₂ when treated with alcoholic KOH at high temperature?
Elimination (E₂) of CH₃CH₂CH₂CH₂CH₂Br (1-bromopentane) with alcoholic KOH forms 1-pentene ( CH₃CH₂CH₂CH=CH₂ ).
Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Elimination Reactions and Reaction with Metals and Other Reactions
