Question

Identify the product ' B ' in the following sequence of reactions.
$\text{Methylpropanoate} \xrightarrow[\text{dil. NaOH}]{\Delta} \text{A} \xrightarrow[\text{Conc. HCl}]{\text{H}^+} \text{B}$

• A. sodium propanoate
• B. propanone
• C. propanal
• D. propanoic acid

Hint:

Ester + Base $\rightarrow$ Salt. Salt + Acid $\rightarrow$ Carboxylic Acid. It's a two-step route to "undo" an esterification.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The chemical reaction sequence presented describes the classic alkaline hydrolysis of an ester molecule, followed by a subsequent acidification step to recover the neutral free carboxylic acid.
Step 2: Key Formula or Approach:
Approach: Determine the intermediate salt 'A' formed during ester saponification, and then predict the final product 'B' when this salt is protonated by a strong acid.
Step 3: Detailed Explanation:
1. Reaction 1 (Alkaline Hydrolysis): The starting material is methyl propanoate, an ester with the structure $\text{CH}_3\text{CH}_2\text{COOCH}_3$. When an ester is heated with an aqueous strong base like dilute $\text{NaOH}$, it undergoes saponification to form the sodium salt of the parent carboxylic acid and an alcohol byproduct. \[ \text{CH}_3\text{CH}_2\text{COOCH}_3 + \text{NaOH} \xrightarrow{\Delta} \text{CH}_3\text{CH}_2\text{COO}^-\text{Na}^+ \, (\text{A}) + \text{CH}_3\text{OH} \] The major organic intermediate 'A' containing the acyl group is sodium propanoate. 2. Reaction 2 (Acidification): The basic sodium salt of the acid (A) is then treated with an excess of strong acid, here concentrated $\text{HCl}$. This step protonates the carboxylate anion, liberating the neutral free carboxylic acid. \[ \text{CH}_3\text{CH}_2\text{COO}^-\text{Na}^+ + \text{HCl} \longrightarrow \text{CH}_3\text{CH}_2\text{COOH} \, (\text{B}) + \text{NaCl} \] The final isolated product 'B' is propanoic acid.
Step 4: Final Answer:
The final product B is propanoic acid.