JEE Main Chemistry question
-Propyl bromide on treating with alcoholic KOH produces [K CET 2008]
Answer: (B)
Solution
Alcoholic KOH is a dehydrohalogenating (elimination) reagent: CHCHCHBr + alc. KOH → CHCH=CH (propene) + HBr + KBr.
-Propyl bromide on treating with alcoholic KOH produces [K CET 2008]
Alcoholic KOH is a dehydrohalogenating (elimination) reagent: CHCHCHBr + alc. KOH → CHCH=CH (propene) + HBr + KBr.