Question

When propionic acid is treated with aqueous sodium bicarbonate, $C{O_2}$ is liberated. The C from $C{O_2}$ comes from:
A. Methyl group
B. Carboxylic acid group
C. Methylene group
D. Bicarbonate

Answer 1

The important characteristics test for the identification of acid is the sodium bicarbonate test. A litmus test is also used to identify the acidity of the compounds. When litmus paper turns red from blue, then the compound is acidic and when the litmus paper turns blue from red, then the compound is basic.

Complete step by step answer:
When propionic acid reacts with sodium bicarbonate, it liberates carbon dioxide. The reaction can be written as follows,
$C{H_3}C{H_2}COOH + NaHC{O_3} \to C{H_3}C{H_2}COONa + C{O_2} + {H_2}O$
Generally, the acidity of the compounds can be easily determined by the removal of a proton from that compound. The ease of removal of a proton from the given compound will indicate that the compound is highly acidic. Acid tends to dissociate to form acid anion and proton. Thus, when sodium bicarbonate approaches propionic acid, then propionic acid readily loses its protons and sodium ion from sodium bicarbonate will get attached to it. And also, during the reaction, carbon dioxide is evolved and it symbolizes that the compound is acid. The evolution of carbon dioxide is not from acid instead, it is from sodium bicarbonate. Propionic acid group tends to dissociate as $C{H_3}C{H_2}CO{O^ - }$ and ${H^ + }$. Thus, it won’t dissociate to give carbon dioxide. Sodium bicarbonate evolves carbon dioxide when it reacts with any organic acids.

So, the correct answer is Option C.

Note: Propionic acid acts as a preservative in food products mainly in bread and other baked items since it resists the growth of mould and bacteria. If propionic acid is used inappropriately, then it will become toxic and may cause several health problems such as severe burns and blindness.