Match List-I with List-II
| LIST-I | LIST-II |
| A. Hoffmann Bromamide degradation | I. X2/Red phosphorus |
| B. Hell-Volhard-Zelinsky reaction | II. Br2 + NaOH |
| C. Rosenmund reduction | III. SnCl2 + HCl |
| D. Stephen reaction | IV. H2, Pd-BaSO4 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
These are named reactions of carbonyl compounds and their derivatives: the Rosenmund and Stephen reactions reduce an acyl chloride and a nitrile to an aldehyde; the Hell-Volhard-Zelinsky reaction halogenates a carboxylic acid at the α-carbon; the Hoffmann bromamide degradation turns an amide into an amine with one carbon less.
A. Hoffmann Bromamide degradation — This matches II. Br2 + NaOH. An amide loses its carbonyl carbon to give a primary amine:
\( \ce{RCONH2 + Br2 + 4NaOH -> RNH2 + Na2CO3 + 2NaBr + 2H2O} \)B. Hell-Volhard-Zelinsky reaction — This matches I. X2/Red phosphorus. An acid with an α-hydrogen is halogenated at the α-carbon:
\( \ce{RCH2COOH} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{X2}\text{/Red P}} \ce{RCH(X)COOH} \)C. Rosenmund reduction — This matches IV. H2, Pd-BaSO4. An acyl chloride is reduced to an aldehyde over a poisoned catalyst:
\( \ce{RCOCl + H2} \xrightarrow{\ce{Pd-BaSO4}} \ce{RCHO + HCl} \)D. Stephen reaction — This matches III. SnCl2 + HCl. A nitrile is reduced to an imine, which is hydrolysed to an aldehyde:
\( \ce{RCN + SnCl2 + HCl -> RCH=NH} \xrightarrow{\ce{H3O+}} \ce{RCHO} \)