Question

Identify the product formed when \( CH_3{-}CH_2{-}Br \) reacts with alcoholic \(KOH\).

• A. \(CH_3CH_2OH\)
• B. \(CH_2=CH_2\)
• C. \(CH_3CHO\)
• D. \(CH_3COOH\)

Hint:

Remember:

• Aqueous \(KOH\) → Substitution (alcohol formation)
• Alcoholic \(KOH\) → Elimination (alkene formation)

This is a common rule in reactions of alkyl halides.

Answer 1

The Correct Option is B

Concept: Alkyl halides react differently with \(KOH\) depending on the solvent:

• Aqueous \(KOH\) → Substitution reaction producing alcohol.
• Alcoholic \(KOH\) → Elimination reaction producing an alkene.

Alcoholic \(KOH\) promotes dehydrohalogenation, where a hydrogen atom and a halogen atom are removed from adjacent carbon atoms, forming a double bond.
Step 1: {Identify the reactant.} The given compound is \[ CH_3{-}CH_2{-}Br \] which is ethyl bromide, a primary alkyl halide.
Step 2: {Apply elimination with alcoholic \(KOH\).} In the presence of alcoholic \(KOH\), the reaction proceeds via elimination of:

• \(H\) from one carbon
• \(Br\) from the adjacent carbon

\[ CH_3CH_2Br \xrightarrow[\text{alc.}]{KOH} CH_2 = CH_2 + KBr + H_2O \]
Step 3: {Identify the product.} The elimination forms the alkene \[ CH_2 = CH_2 \] which is ethene.