Question

Identify `A` in the following reaction:
\[ A \xrightarrow[\text{pyridine, reflux}]{\text{SOCl}_2} B \xrightarrow[\Delta]{\text{KCN (alcoholic)}} \text{Propane nitrile} \]

• A. Ethanol
• B. Propane
• C. 1-Chloropropane
• D. Propan-1-ol

Hint:

Alcohols react with \(\text{SOCl}_2\) to form alkyl chlorides, and alkyl chlorides react with alcoholic KCN to give nitriles with one additional carbon atom.

Answer 1

The Correct Option is A

Step 1: Analyze the first reaction.
In the first step, compound \(A\) reacts with thionyl chloride \((\text{SOCl}_2)\) in the presence of pyridine under reflux conditions. This reaction converts an alcohol into the corresponding alkyl chloride. Hence, compound \(A\) must be an alcohol.
Step 2: Analyze the second reaction.
The alkyl chloride \(B\) reacts with alcoholic potassium cyanide \((\text{KCN})\) on heating. In this reaction, the cyanide ion replaces the halide ion via nucleophilic substitution, forming a nitrile. The product formed is propane nitrile, which contains three carbon atoms.
Step 3: Carbon chain analysis.
Since the nitrile product has three carbon atoms, the alkyl halide \(B\) must be chloroethane, which upon reaction with KCN gives propionitrile (propane nitrile). Therefore, the original alcohol \(A\) must be ethanol.
Step 4: Evaluation of options.
(A) Ethanol: Correct, as it forms chloroethane with \(\text{SOCl}_2\), which then gives propane nitrile with KCN.
(B) Propane: Incorrect, as alkanes do not react with \(\text{SOCl}_2\).
(C) 1-Chloropropane: Incorrect, as it is already a halide and not the starting compound \(A\).
(D) Propan-1-ol: Incorrect, as it would lead to butane nitrile, not propane nitrile.
Step 5: Conclusion.
Thus, the compound \(A\) is correctly identified as ethanol.