Consider the above reaction, the compound 'A' is :

Solution

Chemistry ma’am · 3 steps

Hofmann puts NH₂ exactly where C=O was, so read A backwards from the final aniline.

1. Identify the steps. Br₂/NaOH on C₈H₈ClNO is the Hofmann bromamide reaction. It turns an amide Ar-CONH₂ into Ar-NH₂ and loses one carbon as CO₂. So the middle compound is an amide, and A, which gives it with NH₃, is an acid chloride.

   Ar-COCl →(NH₃) Ar-CONH₂ →(Br₂/NaOH) Ar-NH₂

2. Map the positions. The NH₂ lands on the same ring carbon that held the -COCl group. In the product, taking NH₂ as C1, the CH₃ is at C3 and Cl is at C4. So A must have -COCl at C1, CH₃ at C3 and Cl at C4: 4-chloro-3-methylbenzoyl chloride.

   Option	Group at C1	CH₃ position	Cl on ring
   (a)	-COCl	C2	C4
   (b)	-COCl	C4	C3
   (c)	-COCl	C3	C4
   (d)	-CHO	CH₂Cl at C3	C4

3. So. Check the formula: C₆H₃(Cl)(CH₃)COCl is C₈H₆Cl₂O, and its amide is C₈H₈ClNO. Compound A is option (c).

The idea to keep. Hofmann bromamide swaps -CONH₂ for -NH₂ on the same carbon, with one carbon fewer. Keep every other substituent exactly where it is when you work backwards.

Answer from the NTA final answer key. Solution by PrepWiser.