Question

n-propylbromide on treatment with ethanolic potassium hydroxide produces

• A. Propane
• B. Propene
• C. Propyne
• D. Propanol

Answer 1

The correct option is B Propene

When n-propylbromide is made to undergo reaction with ethanolic potassium hydroxide i.e. KOH, there is formation of the carbon-carbon double bond and removal of bromine and the product formed is the unsaturated compound i.e. alkene as per saytzeff rule. Apart from alkene, there is formation of potassium bromide and water as well.
$\underset{\text{n-Propyl bromide}}{C{H}_{3}C{H}_{2}C{H}_{2}Br}\underset{-HBr}{\overset{\text{ethanolic KOH}}{\to }}\underset{\text{Propene}}{C{H}_{3}CH=C{H}_2}+KBr+H_{2}O$
Hence, the correct answer is option (b).