Match the LIST-I with LIST-II.
| LIST-I (Name Reaction) | LIST-II (Transformation) | ||
| A. | Rosenmund reduction | I. | Benzonitrile to benzaldehyde |
| B. | Etard reaction | II. | Benzene to benzaldehyde |
| C. | Gatterman-Koch reaction | III. | Benzoyl chloride to benzaldehyde |
| D. | Stephen reaction | IV. | Toluene to benzaldehyde |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-III, B-IV, C-II, D-I
Benzaldehyde can be made from four different starting materials. Each named reaction starts from a compound at a different oxidation level: an acid chloride is reduced, toluene is oxidised, benzene is formylated and a nitrile is partly reduced.
A. Rosenmund reduction — This matches III. Benzoyl chloride to benzaldehyde. An acyl chloride is hydrogenated over palladium on barium sulphate. The catalyst is partly poisoned so that the aldehyde is not reduced further to an alcohol.
\( \displaystyle \ce{C6H5COCl + H2} \xrightarrow{\ce{Pd-BaSO4}} \ce{C6H5CHO + HCl} \)
B. Etard reaction — This matches IV. Toluene to benzaldehyde. Chromyl chloride oxidises the methyl group to a chromium complex, which is hydrolysed to the aldehyde.
\( \displaystyle \ce{C6H5CH3} \xrightarrow[\text{(ii) }\ce{H3O^+}]{\text{(i) }\ce{CrO2Cl2}\text{, }\ce{CS2}} \ce{C6H5CHO} \)
C. Gatterman-Koch reaction — This matches II. Benzene to benzaldehyde. Carbon monoxide and HCl in the presence of anhydrous \( \ce{AlCl3} \)/CuCl put a –CHO group directly on the ring.
\( \displaystyle \ce{C6H6} \xrightarrow[\text{anhyd. }\ce{AlCl3}/\ce{CuCl}]{\ce{CO},\ \ce{HCl}} \ce{C6H5CHO} \)
D. Stephen reaction — This matches I. Benzonitrile to benzaldehyde. Stannous chloride and HCl reduce the nitrile to an imine, which is hydrolysed to the aldehyde.
\( \displaystyle \ce{C6H5CN} \xrightarrow[\text{(ii) }\ce{H3O^+}]{\text{(i) }\ce{SnCl2}\text{, }\ce{HCl}} \ce{C6H5CHO} \)
Shortcut: A–III appears only in Options (2) and (4), and C–II then leaves only Option (2).