Identify the reaction sequence

The sequence is:

$P + Q \xrightarrow[\text{aq. }K_2CO_3]{} R \xrightarrow{HCN} S \xrightarrow[\Delta]{H^+} \text{given product}$

• In aqueous $K_2CO_3$, two aliphatic aldehydes undergo aldol condensation to form a $\beta$-hydroxy carbonyl compound or, after dehydration, an $\alpha,\beta$-unsaturated aldehyde/ketone.
• Treatment with $HCN$ gives a cyanohydrin.
• Acidic hydrolysis of the nitrile group followed by heating converts $-CN$ into $-COOH$.

Work backward from the final step

If $R$ reacts with $HCN$, then $R$ must contain a carbonyl group. Hence $R$ is an aldehyde or ketone.

Since $S$ on hydrolysis gives the shown product, $S$ must be the cyanohydrin of $R$.

General cyanohydrin formation:

$R_2C=O + HCN \rightarrow R_2C(OH)CN$

Hydrolysis:

$R_2C(OH)CN \xrightarrow{H^+,\Delta} R_2C(OH)COOH$

Therefore, the final product must be an $\alpha$-hydroxy acid, and the immediate precursor $R$ is obtained by replacing the $-COOH$ carbon in the product by $-CHO$ or $>C=O$ origin through cyanohydrin chemistry.

Nature of $R$ from aldol reaction

Since $P$ and $Q$ are two aliphatic aldehydes reacting in base, the most likely product $R$ is the aldol addition product:

$\beta\text{-hydroxy aldehyde}$

Among typical possibilities, the aldol product of acetaldehyde and formaldehyde is:

$HOCH_2CH_2CHO$

This is 3-hydroxypropanal.

Check whether this fits the later steps

Cyanohydrin formation on 3-hydroxypropanal:

$HOCH_2CH_2CHO + HCN \rightarrow HOCH_2CH_2CH(OH)CN$

Hydrolysis of nitrile:

$HOCH_2CH_2CH(OH)CN \xrightarrow{H^+,\Delta} HOCH_2CH_2CH(OH)COOH$

This matches the required type of final product.

Conclusion

Thus, the compound $R$ is the aldol product:

$\boxed{HOCH_2CH_2CHO}$

which corresponds to Option A.

Comparison with stored answer

Stored correct answer = A.

My derived answer = A.

Hence, they agree.