Question

Methyl propanoate on hydrolysis with dilute NaOH forms a salt that on further acidification with conc. HCl forms

• A. A
• B. B
• C. C
• D. D

Hint:

The process of ester hydrolysis with NaOH is called saponification and produces a carboxylate salt and alcohol. Upon acidification with HCl, the carboxylate salt is converted into the free acid.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the final organic product of a two-step process: base-catalyzed hydrolysis (saponification) of an ester followed by acidification.
Step 2: Detailed Explanation:
1. Starting Material: Methyl propanoate (\( \text{CH}_3\text{CH}_2\text{COOCH}_3 \)).
2. Step 1 (Hydrolysis with NaOH): The ester is cleaved by the base to produce sodium propanoate and methanol.
\[ \text{CH}_3\text{CH}_2\text{COOCH}_3 + \text{NaOH} \longrightarrow \text{CH}_3\text{CH}_2\text{COONa} + \text{CH}_3\text{OH} \] The salt formed is sodium propanoate.
3. Step 2 (Acidification with HCl): The carboxylate salt reacts with acid to regenerate the carboxylic acid.
\[ \text{CH}_3\text{CH}_2\text{COONa} + \text{HCl} \longrightarrow \text{CH}_3\text{CH}_2\text{COOH} + \text{NaCl} \] The final product is propanoic acid (\( \text{C}_2\text{H}_5\text{COOH} \)).
Step 3: Final Answer:
The final product is propanoic acid, which is shown in structure (C).