Question: $\text{CH}_3\text{CH}_2\text{CHO}$ and $\text{CH}_3\text{CH}_2\text{COOH}$ can be distinguished by:

Step 1 (Governing Principle): [0.5 Mark]
The Sodium Bicarbonate ($\text{NaHCO}_3$) test is a characteristic chemical test used to identify carboxylic acids. Carboxylic acids react with sodium bicarbonate to evolve carbon dioxide gas, which is observed as brisk effervescence. Aldehydes do not undergo this acid-base reaction.

The chemical reaction for the carboxylic acid is:

$$\text{CH}_3\text{CH}_2\text{COOH} + \text{NaHCO}_3 \rightarrow \text{CH}_3\text{CH}_2\text{COONa} + \text{H}_2\text{O} + \text{CO}_2 \uparrow$$

Step 2 (Step-by-Step Deduction): [0.25 Mark]

• For $\text{CH}_3\text{CH}_2\text{COOH}$ (Propanoic acid): Being an acid, it reacts with $\text{NaHCO}_3$ to release $\text{CO}_2$ gas (brisk effervescence).
• For $\text{CH}_3\text{CH}_2\text{CHO}$ (Propanal): Being an aldehyde, it lacks the acidic proton required to react with $\text{NaHCO}_3$; hence, no effervescence is observed.

Step 3 (Final Concluded Answer): [0.25 Mark]
The correct option is (A) Sodium bicarbonate test.

💡 EL10 Concept Insight: Think of the Sodium Bicarbonate test as an "Acid Detector"—because carboxylic acids are acidic, they "fizz" (release bubbles of gas) when they meet baking soda, while aldehydes are neutral and just sit there quietly without reacting.