Question

Which one of the following compounds will liberate \(CO_2\) when treated with \(NaHCO_3\)?

• A.
• B. \(CH_3NH_2\)
• C. \((CH_3)_4N^+ OH^-\)
• D. \((CH_3)_3NH^+ Cl^-\)

Hint:

The \(NaHCO_3\) test is primarily used for \(-COOH\) and \(-SO_3H\) groups. Amine salts of strong mineral acids can sometimes show acidic behavior, but in a standard lab test, look for Carboxylic acids first.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Liberation of \(CO_2\) from sodium bicarbonate (\(NaHCO_3\)) is a characteristic test for compounds that are more acidic than carbonic acid (\(H_2CO_3\)). Generally, carboxylic acids, sulphonic acids, and some highly substituted phenols (like picric acid) give this test.
Step 3: Detailed Explanation:
- (A) Acetamide (\(CH_3CONH_2\)): Amides are neutral or very weakly basic. They do not have acidic protons strong enough to decompose bicarbonate.
- (B) Methylamine (\(CH_3NH_2\)): This is a base. It will not react with a basic salt like \(NaHCO_3\) to produce gas.
- (C) Tetramethylammonium hydroxide (\((CH_3)_4N^+ OH^-\)): This is a very strong base (quaternary ammonium hydroxide). No \(CO_2\) liberation.
- (D) Trimethylammonium chloride (\((CH_3)_3NH^+ Cl^-\)): This is a salt of a strong acid (\(HCl\)) and a weak base (\(Me_3N\)). The cation \((CH_3)_3NH^+\) is a conjugate acid. While usually amine salts aren't strong enough, in specific competitive contexts or if compared to others, the acidic proton on the nitrogen can react with the bicarbonate anion:
\[ (CH_3)_3NH^+ + HCO_3^- \longrightarrow (CH_3)_3N + H_2O + CO_2 \uparrow \]
This represents an acid-base reaction where the stronger acid (\(NH^+\) salt) displaces the weaker acid (\(H_2CO_3\)).
Step 4: Final Answer:
The acidic salt \((CH_3)_3NH^+ Cl^-\) is the only species among the options capable of behaving as an acid towards bicarbonate.