What is the product when (CH₃)2CHCH₂CH₂Br reacts with alcoholic KOH at low temperature?
Elimination (E₂) of (CH₃)2CHCH₂CH₂Br with alcoholic KOH forms (CH₃)2C=CHCH₃ (2-methyl-2-butene) per Zaitsev's rule.
Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Elimination Reactions and Reaction with Metals and Other Reactions