Question:
Identify the product ' B ' in the following sequence of reactions.
$\text{Methylpropanoate} \xrightarrow[\text{dil. NaOH}]{\Delta} \text{A} \xrightarrow[\text{Conc. HCl}]{\text{H}^+} \text{B}$
Identify the product ' B ' in the following sequence of reactions.
$\text{Methylpropanoate} \xrightarrow[\text{dil. NaOH}]{\Delta} \text{A} \xrightarrow[\text{Conc. HCl}]{\text{H}^+} \text{B}$
$\text{Methylpropanoate} \xrightarrow[\text{dil. NaOH}]{\Delta} \text{A} \xrightarrow[\text{Conc. HCl}]{\text{H}^+} \text{B}$
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Ester + Base $\rightarrow$ Salt. Salt + Acid $\rightarrow$ Carboxylic Acid. It's a two-step route to "undo" an esterification.
Updated On: May 14, 2026
- sodium propanoate
- propanone
- propanal
- propanoic acid
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The Correct Option is D
Solution and Explanation
Step 1: Concept
This sequence involves the basic hydrolysis of an ester (saponification) followed by acidification of the resulting salt.
Step 2: Meaning
Esters react with bases to form a salt of the carboxylic acid and an alcohol. Acidification of the carboxylate salt restores the parent carboxylic acid.
Step 3: Analysis
1. $\text{CH}_3\text{CH}_2\text{COOCH}_3$ (Methylpropanoate) reacts with NaOH to form $\text{CH}_3\text{CH}_2\text{COONa}$ (Sodium propanoate, A) and methanol. 2. $\text{CH}_3\text{CH}_2\text{COONa}$ (A) reacts with HCl to form $\text{CH}_3\text{CH}_2\text{COOH}$ (Propanoic acid, B) and NaCl.
Step 4: Conclusion
The final product B is propanoic acid. Final Answer: (D)
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