JEE MainChemistryAmines
Consider the following reaction sequence: Benzamide Br ₂, KOH [P] [ 273-278 K ] NaNO ₂, HCl [Q] KI [R] Na, dry ether [S] Identify the products [P], [R], and [S] respectively.
Options
- A[P] is Aniline; [R] is Iodobenzene; [S] is Biphenyl
- B[P] is Benzylamine; [R] is Benzyl iodide; [S] is 1,2-Diphenylethane
- C[P] is Aniline; [R] is Chlorobenzene; [S] is Biphenyl
- D[P] is Aniline; [R] is Iodobenzene; [S] is Benzene
Correct answer
A. [P] is Aniline; [R] is Iodobenzene; [S] is Biphenyl
Step-by-step solution
Step 1: Benzamide undergoes Hoffmann bromamide degradation when treated with Br ₂ and KOH. The carbonyl group is lost, yielding a primary aromatic amine with one less carbon atom. Thus, [P] is aniline. Step 2: Aniline reacts with NaNO ₂ and HCl at 273-278 K to undergo diazotization, forming benzene diazonium chloride [Q]. Step 3: Benzene diazonium chloride reacts with KI. The nucleophilic substitution of the diazonium group yields iodobenzene [R]. Step 4: Iodobenzene undergoes the Fittig reaction when treated with