Identify the reaction sequence

Two aliphatic aldehydes $P$ and $Q$ react in aqueous $K_2CO_3$.

• Aqueous $K_2CO_3$ is a mild base.
• Two aldehydes in base commonly undergo aldol condensation/addition if one has an $\alpha$-hydrogen.
• The product $R$ then reacts with $HCN$, so $R$ must still contain a carbonyl group.
• Thus, $R$ is a $\beta$-hydroxy aldehyde/ketone formed by aldol addition.

What happens on treatment with $HCN$?

Carbonyl compounds react with $HCN$ to form cyanohydrins:

So compound $S$ is the cyanohydrin of $R$.

Effect of acidification and heating of cyanohydrin

On acid hydrolysis, the nitrile group $(-CN)$ converts to carboxylic acid $(-COOH)$:

Therefore, acidification and heating of $S$ converts the cyanohydrin into an $\alpha$-hydroxy carboxylic acid derivative.

Reverse analysis of the final product

Since the final product is obtained from cyanohydrin hydrolysis, its skeleton must come from a carbonyl compound $R$ of the form:

Hence, $S$ must be the cyanohydrin corresponding to that carbonyl compound.

How is $R$ formed from two aliphatic aldehydes?

The classic crossed aldol combination consistent with later cyanohydrin formation is:

• $P =$ acetaldehyde, $CH_3CHO$
• $Q =$ formaldehyde, $HCHO$

Aldol addition gives:

This is $3$-hydroxypropanal.

Formation of $S$ from $R$

Cyanohydrin formation on the aldehyde group of $HOCH_2CH_2CHO$ gives:

So,

Hydrolysis of $S$

This matches the expected product type from the question.

Conclusion

Therefore, the correct structure of $S$ is the cyanohydrin:

which corresponds to Option D.

Comparison with stored answer

Stored correct answer: D

My derived answer: D

Hence, they agree.