Methyl propanoate on hydrolysis with dilute NaOH forms a salt that on further acidification with conc. HCl forms
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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.
Step 1: Understand the reaction of methyl propanoate with NaOH.
Methyl propanoate is an ester. When esters react with sodium hydroxide (NaOH), they undergo hydrolysis. The hydrolysis of an ester with a base results in the formation of a salt and alcohol. This reaction is known as saponification. The product formed here would be the sodium salt of propanoic acid, i.e., sodium propanoate. Step 2: Hydrolysis equation of methyl propanoate.
The chemical equation for the hydrolysis of methyl propanoate with NaOH is:
\[
\text{CH}_3\text{COO}\text{CH}_3 + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{CH}_3\text{OH}
\]
The sodium salt formed is sodium acetate and methanol. Step 3: Acidification with concentrated HCl.
On further acidifying the sodium salt (sodium propanoate) with concentrated hydrochloric acid (HCl), the sodium ion is replaced by a proton (H+), and the result is the formation of propanoic acid. Step 4: Final product.
The final product after acidification is propanoic acid, \( \text{C}_2\text{H}_5\text{COOH} \), which is acetic acid. This corresponds to option (2).
