Animated Solution for Chemistry - Organic Chemistry: Which one of the following compounds will liberate ${\text{CO}}_2$, when treated with ${\text{NaHCO}}_3$?

When you are asked which compound will liberate carbon dioxide (${\text{CO}}_2$) gas upon treatment with sodium bicarbonate (${\text{NaHCO}}_3$), you are essentially being asked to identify the acidic compound among the given choices.

The sodium bicarbonate test is a standard laboratory procedure used to detect the presence of an acidic functional group. The underlying principle is a simple acid-base proton transfer. For a compound to liberate ${\text{CO}}_2$, it must be capable of donating a proton (${\text{H}}^{+}$) to the bicarbonate ion (${\text{HCO}}_3^{-}$).

When the bicarbonate ion accepts a proton, it forms carbonic acid (${\text{H}}_2{\text{CO}}_3$).

Let's systematically evaluate the acid-base nature of the options provided:

1. $({\text{CH}}_3{)}_3\stackrel{\oplus }{N}{\text{OH}}^{\ominus }$: This is a quaternary-like structure, but specifically, it represents a hydroxide salt of an amine, which is inherently a base. 2. ${\text{CH}}_3{\text{CONH}}_2$: This is acetamide. While it has hydrogen atoms attached to nitrogen, amides are generally neutral in aqueous solutions because the lone pair on the nitrogen atom is delocalized into the adjacent carbonyl group via resonance. 3. ${\text{CH}}_3{\text{NH}}_2$: This is methylamine, a primary aliphatic amine. It is a classic weak base.

None of the above three compounds have the acidic strength required to donate a proton to the bicarbonate ion.

Now, let's look at the first option: $({\text{CH}}_3{)}_3\stackrel{\oplus }{N}{\text{HCl}}^{\ominus }$ (Trimethylammonium chloride).

This compound is a salt formed from the neutralization of a weak base (trimethylamine) and a strong acid (hydrochloric acid). According to the principles of ionic equilibrium, the salt of a weak base and a strong acid undergoes cationic hydrolysis in water, making the resulting solution acidic.

The trimethylammonium ion ($({\text{CH}}_3{)}_3\stackrel{\oplus }{N}\text{H}$) acts as a Brønsted acid. It is perfectly capable of donating its extra proton to the bicarbonate ion.

The chemical reaction proceeds as follows:

The carbonic acid (${\text{H}}_2{\text{CO}}_3$) formed in this step is highly unstable under standard conditions. It rapidly decomposes into water and carbon dioxide gas:

This rapid release of ${\text{CO}}_2$ gas is observed as brisk effervescence, confirming that $({\text{CH}}_3{)}_3\stackrel{\oplus }{N}{\text{HCl}}^{\ominus }$ is the correct answer.