An organic compound A (C$$_6$$H$$_6$$O) gives dark green colouration with ferric chloride. On treatment with CHCl$$_3$$ and KOH, followed by acidification gives compound B. Compound B can also be obtained from compound C on reaction with pyridinium chlorochromate (PCC). Identify A, B and C.

The compound $$\mathrm{C_6H_6O}$$ gives a dark green colour with neutral $$\mathrm{FeCl_3}$$, indicating the presence of a phenolic $$\mathrm{-OH}$$ group. Hence, compound $$\mathrm{A}$$ is $$\mathrm{Phenol}$$.

Phenol reacts with $$\mathrm{CHCl_3/KOH}$$ followed by acidic workup through the $$\mathrm{Reimer\text{-}Tiemann\ Reaction}$$. This introduces a formyl group $$\mathrm{(-CHO)}$$ mainly at the ortho position, forming $$\mathrm{o\text{-}Hydroxybenzaldehyde\ (Salicylaldehyde)}$$.

Thus, compound $$\mathrm{B}$$ is salicylaldehyde.

Compound $$\mathrm{B}$$ is also formed by oxidation using $$\mathrm{PCC}$$. Since PCC oxidises primary alcohols into aldehydes, compound $$\mathrm{C}$$ must contain a $$\mathrm{-CH_2OH}$$ group at the ortho position.

Therefore, compound $$\mathrm{C}$$ is $$\mathrm{o\text{-}Hydroxybenzyl\ Alcohol}$$.

Correct Option: $$\mathrm{A}$$