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 problem outlines a two-step synthetic pathway. The first step involves treating an alkyl halide with aqueous potassium hydroxide, a standard condition for a nucleophilic substitution reaction that yields an alcohol. The second step involves reacting that alcohol with a phosphorus chloride reagent, which is a common method for converting alcohols back into alkyl halides, specifically alkyl chlorides.
Step 2: Key Formula or Approach:
{Analytical Note: The chemical structures provided in the question (like 'CHCHBr') lack standard numerical subscripts. Based on standard organic chemistry curriculum, these most likely represent ethyl derivatives, where 'CHCH' stands for the ethyl group \(\text{CH}_3\text{CH}_2-\). We will proceed with this logical assumption.}
Reaction 1: Alkyl Halide + aq. KOH \(\rightarrow\) Alcohol (via \(S_N2\) or \(S_N1\) mechanism)
Reaction 2: Alcohol + \(\text{PCl}_3\)/\(\text{PCl}_5\) \(\rightarrow\) Alkyl Chloride
Step 3: Detailed Explanation:
- Step 1: Identifying Product A
Assuming the starting material 'CHCHBr' is ethyl bromide (\(\text{CH}_3\text{CH}_2\text{Br}\)).
When ethyl bromide is treated with aqueous KOH, the hydroxide ion (\(\text{OH}^-\)) acts as a strong nucleophile and substitutes the bromide ion (\(\text{Br}^-\)), which is a good leaving group.
\[ \text{CH}_3\text{CH}_2\text{Br} + \text{KOH (aq)} \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{KBr} \]
Thus, intermediate product A is ethanol (\(\text{CH}_3\text{CH}_2\text{OH}\)). In the abbreviated notation of the question, this corresponds to 'CHCHOH'.
- Step 2: Identifying Final Product B
The intermediate A, ethanol (\(\text{CH}_3\text{CH}_2\text{OH}\)), is now reacted with a phosphorus chloride reagent, simply denoted as '(PCl)'. This typically implies phosphorus trichloride (\(\text{PCl}_3\)) or phosphorus pentachloride (\(\text{PCl}_5\)). Both serve the same primary function: replacing the hydroxyl group (\(-\text{OH}\)) with a chlorine atom (\(-\text{Cl}\)).
Using \(\text{PCl}_5\) as an example:
\[ \text{CH}_3\text{CH}_2\text{OH} + \text{PCl}_5 \rightarrow \text{CH}_3\text{CH}_2\text{Cl} + \text{POCl}_3 + \text{HCl} \]
The final product B is ethyl chloride (\(\text{CH}_3\text{CH}_2\text{Cl}\)). Translating this back into the specific, albeit typo-ridden, notation style of the options, it becomes 'CHCHCl'.
Comparing this deduction with the given options, we find it perfectly matches option (B).
Step 4: Final Answer:
The final product B is CHCHCl.