Step 1: Understanding the Question:
The question asks for the major organic product of the reaction between ethyl bromide (\(CH_3CH_2Br\)) and alcoholic potassium hydroxide (KOH). The key is to recognize the role of the reagent, alcoholic KOH.
Step 2: Key Formula or Approach:
The reaction of an alkyl halide with a base can proceed via two competing pathways: substitution or elimination. The conditions determine the major product.
- Alcoholic KOH: A strong, non-nucleophilic base in a non-polar solvent. It favors the elimination reaction (dehydrohalogenation) to form an alkene.
- Aqueous KOH: Provides hydroxide ions (\(OH^-\)) which are strong nucleophiles in a polar solvent. It favors the substitution reaction (SN2) to form an alcohol.
Step 3: Detailed Explanation:
Since the reagent is alcoholic KOH, an elimination reaction will be the major pathway. This reaction is also known as a \(\beta\)-elimination or dehydrohalogenation.
1. The reactant is Ethyl Bromide (\(CH_3-CH_2-Br\)). The carbon atom bonded to Br is the \(\alpha\)-carbon, and the adjacent methyl carbon is the \(\beta\)-carbon.
2. The strong base (alkoxide ion, \(RO^-\), formed from KOH and alcohol) abstracts a hydrogen atom from the \(\beta\)-carbon.
3. The C-H bond electrons shift to form a double bond between the \(\alpha\) and \(\beta\) carbons.
4. The bromide ion (\(Br^-\)) is expelled as a leaving group from the \(\alpha\)-carbon.
The overall reaction is:
\[
\underset{\text{Ethyl Bromide}}{CH_3-CH_2-Br} + \text{KOH (alcoholic)} \longrightarrow \underset{\text{Ethene}}{CH_2=CH_2} + KBr + H_2O
\]
The major organic product is \(CH_2=CH_2\), which is named Ethene (common name: Ethylene).
Step 4: Final Answer:
The major product is Ethene (Ethylene).