Question:easy

An aryl carboxylic acid on treatment with sodium hydrogen carbonate liberates a gaseous molecule. Identify the gas molecule liberated.

Show Hint

Carboxylic acids react with sodium bicarbonate to release \(CO_2\) gas. This reaction is a standard laboratory test for identifying the \(-COOH\) group.
Updated On: Jun 22, 2026
  • \(H_2\)
  • \(CO_2\)
  • \(CO\)
  • \(O_2\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Identify the functional group.
An aryl carboxylic acid carries the $-COOH$ group attached to an aromatic ring, for example benzoic acid $C_6H_5COOH$.
Step 2: Recall the acidity of carboxylic acids.
Carboxylic acids are acidic enough to react with weak bases such as sodium hydrogen carbonate, $NaHCO_3$.
Step 3: Write the general reaction.
\[ RCOOH + NaHCO_3 \rightarrow RCOONa + H_2O + CO_2. \]
Step 4: Spot the gas released.
The unstable carbonic acid first formed breaks up into water and carbon dioxide, so the gas that bubbles out is $CO_2$.
Step 5: Note the significance of this test.
Brisk effervescence of $CO_2$ with $NaHCO_3$ is a standard laboratory test that distinguishes carboxylic acids from weaker acids like phenols, which do not react.
Step 6: Conclude.
Therefore the gas liberated is
\[ \boxed{CO_2} \]
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