Consider the above reaction, the compound 'A' is

The final step uses $$\mathrm{Br_2/NaOH}$$, which are the reagents for the Hoffmann bromamide degradation.

This reaction converts a primary amide $$\mathrm{(-CONH_2)}$$ into a primary amine $$\mathrm{(-NH_2)}$$ with loss of the carbonyl carbon.

Since the final product is $$\mathrm{4\text{-}chloro\text{-}3\text{-}methylaniline}$$, the intermediate must be $$\mathrm{4\text{-}chloro\text{-}3\text{-}methylbenzamide}$$.

The first step involves reaction with ammonia $$\mathrm{(NH_3)}$$, which converts an acid chloride $$\mathrm{(-COCl)}$$ into an amide $$\mathrm{(-CONH_2)}$$.

Therefore, compound $$\mathrm{A}$$ must be $$\mathrm{4\text{-}chloro\text{-}3\text{-}methylbenzoyl\ chloride}$$.

Its molecular formula $$\mathrm{C_8H_6Cl_2O}$$ matches the given formula.

Correct Answer: $$\mathrm{4\text{-}chloro\text{-}3\text{-}methylbenzoyl\ chloride}$$