Which one of the following compounds will liberate CO$$_2$$, when treated with NaHCO$$_3$$?

We know that sodium hydrogencarbonate, written as $$\text{NaHCO}_3$$, is a salt of a weak acid $$\text{H}_2\text{CO}_3$$ (carbonic acid) and a strong base $$\text{NaOH}$$. Whenever an acid that is stronger than $$\text{H}_2\text{CO}_3$$ comes in contact with $$\text{NaHCO}_3$$, an acid-base reaction occurs in which $$\text{HCO}_3^-$$ is converted to $$\text{H}_2\text{CO}_3$$, and the unstable $$\text{H}_2\text{CO}_3$$ immediately decomposes to give carbon dioxide gas:

First write the general sequence of steps:

$$\text{H}^+ + \text{HCO}_3^- \;\longrightarrow\; \text{H}_2\text{CO}_3$$

$$\text{H}_2\text{CO}_3 \;\longrightarrow\; \text{H}_2\text{O} + \text{CO}_2\uparrow$$

So, out of the given options we have to identify the compound that can really supply the necessary $$\text{H}^+$$ (i.e. behaves as an acid in water) more strongly than carbonic acid.

Let us analyse each option in turn.

Option A: $$\text{CH}_3\text{NH}_3^+ \text{Cl}^-$$   (methyl­ammonium chloride)
This is a salt formed from a weak base $$\text{CH}_3\text{NH}_2$$ (methylamine) and a strong acid $$\text{HCl}$$. In aqueous solution the cation $$\text{CH}_3\text{NH}_3^+$$ behaves as a conjugate acid and releases a proton:

$$\text{CH}_3\text{NH}_3^+ \;\rightleftharpoons\; \text{CH}_3\text{NH}_2 + \text{H}^+$$

The liberated $$\text{H}^+$$ can react with $$\text{HCO}_3^-$$ exactly as outlined above. Writing the full ionic equation with $$\text{NaHCO}_3$$ we obtain

$$\text{CH}_3\text{NH}_3^+\,\text{Cl}^- + \text{Na}^+\,\text{HCO}_3^- \;\longrightarrow\; \text{CH}_3\text{NH}_2 + \text{Na}^+\,\text{Cl}^- + \text{H}_2\text{CO}_3$$

and then

$$\text{H}_2\text{CO}_3 \;\longrightarrow\; \text{H}_2\text{O} + \text{CO}_2\uparrow$$

Because a gas is evolved, brisk effervescence of $$\text{CO}_2$$ will be observed.

Option B: $$\text{CH}_3\text{NHO}^-$$ (methyloxime anion) is actually the conjugate base of hydroxylamine derivative; it does not donate a proton, hence it cannot supply $$\text{H}^+$$ to drive the above reaction with $$\text{NaHCO}_3$$.

Option C: $$\text{CH}_3\text{-CO-NH}_2$$ (acetamide) is a very weak acid (essentially neutral) because the -NH2 group is not able to donate a proton readily; therefore it fails to react with $$\text{NaHCO}_3$$.

Option D: $$\text{CH}_3\text{NH}_2$$ (methylamine) is itself a base, so instead of giving $$\text{H}^+$$ it would accept protons; there is no possibility of evolving $$\text{CO}_2$$ from $$\text{NaHCO}_3$$.

Among all choices, only the salt $$\text{CH}_3\text{NH}_3^+ \text{Cl}^-$$ is sufficiently acidic to protonate $$\text{HCO}_3^-$$ and liberate $$\text{CO}_2$$.

Hence, the correct answer is Option A.