Question

Match List-I with List-II 

Choose the correct answer from the options given below: 
 

• (A) - (III), (B) - (II), (C) - (IV), (D) - (I)
• (A) - (I), (B) - (II), (C) - (III), (D) - (IV)
• (A) - (I), (B) - (II), (C) - (IV), (D) - (III)
• (A) - (III), (B) - (IV), (C) - (I), (D) - (II)

Hint:

For diazonium salt reactions: \textbf{Sandmeyer} uses a cuprous \textbf{S}alt (CuX). \textbf{Gatterman} is the "poor man's Sandmeyer" and just uses copper powder. For oxidation of toluene: CrO\(_2\)Cl\(_2\) \(\rightarrow\) Etard reaction. For amide to amine degradation: Br\(_2\)/NaOH \(\rightarrow\) Hofmann Bromamide.

Answer 1

The Correct Option is A

Step 1: Overview: This question tests your knowledge of four named reactions in aromatic organic chemistry. Step 2: Reaction Breakdown: (A) Benzene diazonium chloride (from aniline, NaNO\(_2\)/HCl) reacts with Cu/HCl to form chlorobenzene. This is the Gatterman reaction. Therefore, (A) matches (III). (B) A similar reaction to (A), but using cuprous salt (Cu\(_2\)Cl\(_2\)/HCl). This converts a diazonium salt to a halobenzene; it is the Sandmeyer reaction. So, (B) matches (II). (C) Conversion of an amide (benzamide) to a primary amine (aniline) with fewer carbons, using bromine and sodium hydroxide (Br\(_2\)/NaOH). This is the Hofmann Bromamide degradation reaction. So, (C) matches (IV). (D) Oxidation of a methyl group on an aromatic ring (toluene) to an aldehyde (benzaldehyde) using chromyl chloride (CrO\(_2\)Cl\(_2\)) in a non-polar solvent (CS\(_2\)). This is the Etard reaction. Hence, (D) matches (I). Step 3: Answer: The correct matching is: (A)-(III), (B)-(II), (C)-(IV), (D)-(I). This corresponds to option (1).