JEE Chemistry 2025 Question with Solution

Given: A match-the-following question on named reactions for preparation of aldehydes.

Find: The correct correspondence between List-I and List-II.

From the solution text:

• (A) $\mathrm{RCN} \xrightarrow[(ii)\, H_3O^+]{(i)\, SnCl_2, HCl} \mathrm{RCHO}$ is the Stephen reaction, so (A) \rightarrow (IV).
• (B) Conversion of benzoyl chloride to benzaldehyde in presence of $\mathrm{H_2/Pd-BaSO_4}$ is Rosenmund reduction, so (B) \rightarrow (III).
• (C) Oxidation of toluene using $\mathrm{CrO_2Cl_2/CS_2}$ followed by hydrolysis gives benzaldehyde; this is Etard reaction, so (C) \rightarrow (I).
• (D) Formylation using $\mathrm{CO}$ and $\mathrm{HCl}$ in presence of anhydrous $\mathrm{AlCl_3/CuCl}$ is Gatterman-Koch reaction, so (D) \rightarrow (II).

Therefore, the correct matching is (A)-(IV), (B)-(III), (C)-(I), (D)-(II). This corresponds to Option A.

Note: the extracted options do not contain this exact matching verbatim, but the solution explicitly concludes this correspondence and states the correct option as A.

Given: Four reactions in List-I and four named reactions in List-II.

Find: Which option gives the correct match.

1. For (A), reduction of a nitrile to an aldehyde using $\mathrm{SnCl_2/HCl}$ followed by hydrolysis is the Stephen reaction.
2. For (B), selective reduction of an acyl chloride to an aldehyde using $\mathrm{H_2/Pd-BaSO_4}$ is Rosenmund reduction.
3. For (C), side-chain oxidation of toluene to benzaldehyde using chromyl chloride reagent is the Etard reaction.
4. For (D), introduction of a formyl group into benzene using $\mathrm{CO}$ and $\mathrm{HCl}$ with Lewis acid catalyst is the Gatterman-Koch reaction.

Hence,

So the defensible answer from the solution is Option A.