Consider the following reaction,

What is A in the above reaction?

The final product is 4-bromophenylacetic acid.

$$Br-C_6H_4-CH_2COOH$$

Step 3 involves acidic hydrolysis using $$H_2O/H^+/\Delta$$. Nitriles undergo hydrolysis under these conditions to form carboxylic acids.

$$R-C\equiv N \xrightarrow{H_2O/H^+,\Delta} R-COOH$$

Therefore, the compound before hydrolysis must be 4-bromophenylacetonitrile.

$$Br-C_6H_4-CH_2CN$$

Step 2 involves treatment with $$CN^-$$, which undergoes nucleophilic substitution and replaces a halogen atom.

$$R-CH_2Cl \xrightarrow{CN^-} R-CH_2CN$$

Hence, the precursor for this step must be 4-bromobenzyl chloride.

$$Br-C_6H_4-CH_2Cl$$

Step 1 involves $$Cl_2/h\nu$$, which carries out free-radical substitution at the benzylic position.

$$Ar-CH_3 \xrightarrow{Cl_2/h\nu} Ar-CH_2Cl$$

Therefore, the starting compound $$A$$ must be 4-bromotoluene.

The reaction sequence is

$$Br-C_6H_4-CH_3 \xrightarrow{Cl_2/h\nu} Br-C_6H_4-CH_2Cl$$

$$Br-C_6H_4-CH_2Cl \xrightarrow{CN^-} Br-C_6H_4-CH_2CN$$

$$Br-C_6H_4-CH_2CN \xrightarrow{H_2O/H^+,\Delta} Br-C_6H_4-CH_2COOH$$

Hence, compound $$A$$ is 4-bromotoluene.

Therefore, the correct answer is Option (C).