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

The question asks which compound will liberate \( CO_2 \) when treated with \( NaHCO_3 \). To solve this problem, we need to understand the nature of the compounds provided as options and their reactions with \( NaHCO_3 \). Typically, a compound will liberate \( CO_2 \) gas when it reacts with \( NaHCO_3 \) if it is an acid strong enough to displace carbonic acid, which decomposes to give \( CO_2 \).

Let's analyze the options: 

1.  
2. \( CH_3NH_2 \): Methylamine is a weak base and cannot react with \( NaHCO_3 \) to release \( CO_2 \).
3. \( (CH_3)_4N^+ OH^- \): Tetramethylammonium hydroxide is a strong base and will not react with \( NaHCO_3 \), as there is no acidic hydrogen present in \( NaHCO_3 \) to be removed by a base.
4. \( (CH_3)_3NH^+ Cl^- \): Trimethylammonium chloride generates an acidic solution when dissolved in water. The acidic hydrogen from \( (CH_3)_3NH^+ \) can react with \( NaHCO_3 \) to release \( CO_2 \).

The reaction for option 4 is as follows:

\((CH_3)_3NH^+ Cl^- + NaHCO_3 \rightarrow (CH_3)_3N + H_2O + CO_2 \uparrow + NaCl\)

This reaction illustrates that \( (CH_3)_3NH^+ Cl^- \), being the conjugate acid of a weak base, can indeed react with \( NaHCO_3 \) to liberate \( CO_2 \). Therefore, the correct answer is option 4: \((CH_3)_3NH^+ Cl^-\).