Propanal reacts with methanal according to the following sequence:

$$\mathrm{Propanal+Methanal\xrightarrow{(i)\ dil.\ NaOH}\xrightarrow{(ii)\ \Delta}\xrightarrow{(iii)\ NaCN}\xrightarrow{(iv)\ H_3O^+}B}$$

The molecular formula of product $$B$$ is:

$$C_5H_8O_3$$

Identify the correct statement regarding product $$B$$.

Propanal undergoes crossed aldol condensation with methanal in the presence of dilute $$NaOH$$ to form a $$\beta$$-hydroxy aldehyde.

On heating, dehydration occurs to give the $$\alpha,\beta$$-unsaturated aldehyde:

$$CH_2=C(CH_3)CH_2CHO$$

Addition of $$HCN$$ generated from $$NaCN/H_3O^+$$ across the carbonyl group forms a cyanohydrin.

Acidic hydrolysis converts the $$-CN$$ group into a $$-COOH$$ group, giving an $$\alpha$$-hydroxy carboxylic acid with molecular formula:

$$C_5H_8O_3$$

The carbon bearing the $$-OH$$ group becomes a new chiral center.

Since the carbonyl group is planar, cyanide attacks from both faces with equal probability.

Hence, the product is obtained as a racemic mixture.

The presence of the $$-COOH$$ group causes the compound to react with saturated $$NaHCO_3$$, liberating carbon dioxide.

$$RCOOH+NaHCO_3\rightarrow RCOONa+CO_2+H_2O$$

Hence, the correct answer is:

$$\boxed{\text{c}}$$