When nitriles are treated with strong bases and acid, they undergo hydrolysis to form carboxylic acids, which can then undergo halogenation at the alpha position (C-2) to give alpha-halocarboxylic acids.
Step 1: Analyze the reaction with \( \text{OH}^-/\text{H}_2\text{O} \).
The given compound is an alkylnitrile, \( \text{CH}_3\text{CH}_2\text{C}\equiv\text{N} \). When treated with a strong base like \( \text{OH}^- \), the nitrile group undergoes hydrolysis, forming an amide (in this case, propionamide) with the reaction:
\[
\text{CH}_3\text{CH}_2\text{C}\equiv\text{N} + \text{OH}^- + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{C(O)NH}_2
\]
Step 2: Analyze the reaction with \( \text{H}^+/O^2- \).
When the amide undergoes acid hydrolysis, it forms the corresponding carboxylic acid, \( \text{CH}_3\text{CH}_2\text{COOH} \). In this case, the product is propanoic acid. Step 3: Analyze the reaction with \( \text{Red P}/\text{Cl}_2 \).
When the carboxylic acid reacts with \( \text{Red P}/\text{Cl}_2 \), a substitution reaction occurs, where the hydrogen atom at the alpha position (C-2) of the carboxylic acid group is replaced with a chlorine atom. This forms 2-chloro propanoic acid. Step 4: Conclusion.
Thus, the major product is 2-chloro propanoic acid. The correct IUPAC name is 2-chloro propanoic acid. Final Answer: 2-chloro propanoic acid
