Question

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

• A. (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
• B. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
• C. (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
• D. (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)

Hint:

To remember these: {R}osenmund = {R}eduction to Aldehyde; {H}VZ = {H}alogen at $\alpha$-position; {H}offmann = {H}omes in on reducing the carbon chain length; {C}lemmensen = {C}arbonyl to Alkane in acid.

Answer 1

The Correct Option is D

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).