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

Step 1: Understanding the Question:
The reaction involves an alkyl halide (ethyl bromide) reacting with potassium hydroxide.
The solvent (alcohol) significantly determines whether the reaction proceeds via substitution or elimination.
Step 2: Key Formula or Approach:
1. \(\beta\)-Elimination Reaction (Dehydrohalogenation): Occurs when an alkyl halide is heated with alcoholic KOH.
2. Aqueous KOH usually leads to nucleophilic substitution (\(S_N2/S_N1\)) forming alcohols.
3. Alcoholic KOH acts as a strong base, favoring the removal of a \(\beta\)-hydrogen and the leaving group (\(Br^-\)).
Step 3: Detailed Explanation:
The reactant is \(CH_3-CH_2-Br\) (Ethyl bromide).
In the presence of alcoholic \(KOH\), the \(OH^-\) ion acts as a base rather than a nucleophile.
It attacks the hydrogen atom on the \(\beta\)-carbon (the \(CH_3\) group).
The electrons from the \(C-H\) bond shift to form a double bond between the two carbons, while the bromine atom leaves as a bromide ion.
Reaction:
\[ CH_3CH_2Br + KOH (\text{alc.}) \rightarrow CH_2=CH_2 + KBr + H_2O \]
The resulting product is ethene (an alkene).
Step 4: Final Answer:
The product formed is \(CH_2 = CH_2\).