Question

Identify the major product formed when Ethyl Bromide reacts with alcoholic KOH.

• A. Ethanol
• B. Ethene (Ethylene)
• C. Diethyl ether
• D. Acetaldehyde

Hint:

Alcoholic KOH favors elimination (E2) producing alkenes, while aqueous KOH favors substitution producing alcohols.

Answer 1

The Correct Option is B

Concept: Alkyl halides react with alcoholic KOH mainly through the elimination reaction (E2 mechanism). In this reaction:
• A hydrogen atom is removed from the \(\beta\)-carbon
• The halogen atom leaves from the \(\alpha\)-carbon
• A double bond is formed Thus an alkene is produced.

Step 1: Write the reaction. \[ CH_3CH_2Br + KOH(\text{alc}) \rightarrow CH_2=CH_2 + KBr + H_2O \]

Step 2: Identify the product. The elimination of HBr from ethyl bromide forms \[ CH_2 = CH_2 \] which is Ethene (Ethylene). \[ \boxed{\text{Ethene}} \]