Match List - I and List - II.
List - I :
(a) R-COCl $\to$ R-CHO
(b) R-CH2-COOH $\to$ R-CH-Cl-COOH
(c) R-CONH2 $\to$ R-NH2
(d) R-COCH3 $\to$ R-CH2-CH3
List - II :
(i) Br2/NaOH
(ii) H2/Pd-BaSO4
(iii) Zn(Hg)/Conc. HCl
(iv) Cl2/Red P, H2O
Show Hint
- (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
- (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
- (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
- (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
The Correct Option is D
Solution and Explanation
Step 1: (a) R-COCl → R-CHO: This is the Rosenmund Reduction. The reagent used is Hydrogen in the presence of Palladium supported on Barium Sulphate (\(H_2/Pd\text{-}BaSO_4\)). The \(BaSO_4\) "poisons" the catalyst to stop the reduction at the aldehyde stage. So, (a)-(ii).
Step 2: (b) R-CH2-COOH → R-CH-Cl-COOH: This is the Hell-Volhard-Zelinsky (HVZ) Reaction. Carboxylic acids with \(\alpha\)-hydrogens react with \(Cl_2\) or \(Br_2\) in the presence of red phosphorus to form \(\alpha\)-halo carboxylic acids. So, (b)-(iv).
Step 3: (c) R-CONH2 → R-NH2: This is the Hoffmann Bromamide Degradation. It converts an amide into a primary amine with one less carbon atom using \(Br_2\) and \(NaOH\). So, (c)-(i).
Step 4: (d) R-COCH3 → R-CH2-CH3: This is the Clemmensen Reduction. It reduces the carbonyl group (\(>C=O\)) of aldehydes or ketones to a methylene group (\(>CH_2\)) using Zinc amalgam and concentrated \(HCl\). So, (d)-(iii).
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