Question:
Methyl propanoate on hydrolysis with dil NaOH forms a salt that on further acidification with conc. HCl forms
Methyl propanoate on hydrolysis with dil NaOH forms a salt that on further acidification with conc. HCl forms
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Alkaline hydrolysis gives the sodium carboxylate; acidification liberates the free acid with the same carbon chain.
Updated On: Nov 13, 2026
- \(\mathrm{CH}_{3}-\mathrm{COOH}\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
An ester on alkaline hydrolysis gives the sodium salt of the carboxylic acid and an alcohol.
Step 2: Hydrolysis
\(\text{CH}_3\text{CH}_2\text{COOCH}_3+\text{NaOH}\rightarrow\text{CH}_3\text{CH}_2\text{COONa}+\text{CH}_3\text{OH}\).
Step 3: Acidification
\(\text{CH}_3\text{CH}_2\text{COONa}+\text{HCl}\rightarrow\text{CH}_3\text{CH}_2\text{COOH}+\text{NaCl}\).
Step 4: Identify the product
The product is propanoic acid, \(\text{C}_2\text{H}_5\text{COOH}\).
Final Answer:
The product is propanoic acid. \[ \boxed{\text{(C) }\text{C}_2\text{H}_5\text{COOH (propanoic acid)}} \]
An ester on alkaline hydrolysis gives the sodium salt of the carboxylic acid and an alcohol.
Step 2: Hydrolysis
\(\text{CH}_3\text{CH}_2\text{COOCH}_3+\text{NaOH}\rightarrow\text{CH}_3\text{CH}_2\text{COONa}+\text{CH}_3\text{OH}\).
Step 3: Acidification
\(\text{CH}_3\text{CH}_2\text{COONa}+\text{HCl}\rightarrow\text{CH}_3\text{CH}_2\text{COOH}+\text{NaCl}\).
Step 4: Identify the product
The product is propanoic acid, \(\text{C}_2\text{H}_5\text{COOH}\).
Final Answer:
The product is propanoic acid. \[ \boxed{\text{(C) }\text{C}_2\text{H}_5\text{COOH (propanoic acid)}} \]
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