An aromatic compound (A) on treatment with aqueous ammonia and heating forms compound (B) which on heating with Br2 and KOH forms a compound (C) of molecular formula C6H7N.
Identify (C) from the given reaction.
Answer & explanation
Correct answer: option 4
The correct answer is option 4- None of the above
-
A is likely benzoyl chloride or benzoic acid, which on reacting with aqueous ammonia and heating gives benzamide (B).
-
Compound B (benzamide) when heated with Br₂ and KOH undergoes Hofmann bromamide reaction, which removes the carbonyl carbon and converts the amide into an amine.
-
So, C = aniline (C₆H₅NH₂), which matches the molecular formula C₆H₇N.
-
-
Option 1 (Benzaldehyde): This is an aldehyde (C6H5CHO). The reaction Br2/KOH with an amide specifically produces an amine, not an aldehyde.
-
Option 2 (Benzamide): This is Compound (B), the intermediate. You must complete the second step of the reaction to find (C).
-
Option 3 (diprop-2-ynylamine): While this has the formula C6H7N, it is an aliphatic (non-aromatic) secondary amine. Since the starting material (A) was aromatic, the product must retain the benzene ring.
- Option 4 ( None of above ) is CORRECT