[P] on treatment with $$Br_2/FeBr_3$$ in $$CCl_4$$ produced a single isomer $$C_8H_7O_2Br$$ while heating [P] with sodalime gives toluene. The compound [P] is:

The molecular formula of the monobrominated product is $$C_8H_7O_2Br$$.
During bromination one hydrogen atom is replaced by bromine, so the parent compound [P] must be $$C_8H_8O_2$$.

Heating [P] with sodalime (NaOH + CaO) carries out decarboxylation, i.e. $$R-COOH \rightarrow R-H + CO_2$$.
Because the decarboxylation product is toluene ($$C_7H_8$$), [P] must be an aromatic carboxylic acid containing one extra carbon compared to toluene. Therefore [P] is some isomer of $$CH_3-C_6H_4-COOH$$ (methyl-benzoic acid, also called toluic acid).

The three possible isomers are:
  • o-methylbenzoic acid (2-toluic acid)
  • m-methylbenzoic acid (3-toluic acid)
  • p-methylbenzoic acid (4-toluic acid)

We now use the bromination data. Electrophilic bromination in $$CCl_4/FeBr_3$$ is governed by the directing effects of the substituents present:

• $$-COOH$$ is a strong deactivating, meta-directing group.
• $$-CH_3$$ is an activating, ortho/para-directing group.

Case 1: o-methylbenzoic acid
  • The ring positions 3, 4, 5, 6 are all chemically nonequivalent and at least two of them are activated for substitution, so a mixture of brominated isomers would be obtained.

Case 2: m-methylbenzoic acid
  • Again several positions (2, 4, 5, 6) differ in reactivity, giving more than one monobromo product.

Case 3: p-methylbenzoic acid
  • Place $$-COOH$$ at position 1 and $$-CH_3$$ at position 4.
  • Both groups favour substitution at positions 3 and 5:
    - For $$-COOH$$ these are its meta positions.
    - For $$-CH_3$$ these are its ortho positions.
  • Positions 3 and 5 are equivalent by symmetry, so only one distinct monobromo isomer can form.

Thus only p-methylbenzoic acid satisfies the statement “produced a single isomer $$C_8H_7O_2Br$$”.

Therefore, the compound [P] is p-methylbenzoic acid (4-methylbenzoic acid).

Option A which is: p-methylbenzoic acid.