Question

CH₃CH₂Br → (aq. KOH) → A → (PCl₅) → B
Final product B:

• A. CH₃CH₂OH
• B. CH₃CH₂Cl
• C. CH₃CH₂Br
• D. CH₃CHO
• E. CH₃CH₂COOH

Hint:

Always distinguish between \textbf{aqueous} KOH (forms alcohol) and \textbf{alcoholic} KOH (forms alkene via elimination).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This is a two-step organic transformation involving nucleophilic substitution. Aqueous KOH converts alkyl halides to alcohols, and PCl₅ converts alcohols to alkyl chlorides.
Step 2: Key Formula or Approach:
1. $R-X + \text{aq. KOH} \rightarrow R-OH + KX$
2. $R-OH + \text{PCl}_5 \rightarrow R-Cl + \text{POCl}_3 + \text{HCl}$
Step 3: Detailed Explanation:
1. Step 1: Ethyl bromide ($\text{CH}_3\text{CH}_2\text{Br}$) reacts with aqueous KOH. The hydroxyl group (OH⁻) replaces the bromine atom. \[ \text{CH}_3\text{CH}_2\text{Br} + \text{KOH(aq)} \rightarrow \text{CH}_3\text{CH}_2\text{OH (A)} \] Product A is Ethanol.
2. Step 2: Ethanol reacts with $\text{PCl}_5$. The OH group is replaced by a Chlorine atom. \[ \text{CH}_3\text{CH}_2\text{OH} + \text{PCl}_5 \rightarrow \text{CH}_3\text{CH}_2\text{Cl (B)} + \text{POCl}_3 + \text{HCl} \] Product B is Ethyl chloride (Chloroethane).
Step 4: Final Answer:
The final product B is CH₃CH₂Cl.