Match List - I with List - II.

Choose the correct answer from the options given below:

In reaction A, phenol on heating with $$Zn$$ dust undergoes reduction. Zinc removes the oxygen atom as $$ZnO$$, converting phenol into benzene. Therefore, A matches with Benzene, i.e., $$A \rightarrow (III)$$.

In reaction B, phenol reacts with $$CHCl_3/NaOH$$ followed by acidification. This is the Reimer-Tiemann reaction, which introduces a formyl group $$(-CHO)$$ at the ortho position of phenol to form salicylaldehyde. Therefore, $$B \rightarrow (I)$$.

In reaction C, phenol reacts with $$CO_2/NaOH$$ followed by acidification. This is the Kolbe-Schmitt reaction, in which a carboxyl group $$(-COOH)$$ is introduced at the ortho position to form salicylic acid. Therefore, $$C \rightarrow (II)$$.

In reaction D, phenol reacts with concentrated $$HNO_3$$. Due to the strongly activating nature of the $$-OH$$ group, nitration occurs at all available ortho and para positions, producing 2,4,6-trinitrophenol (picric acid). Therefore, $$D \rightarrow (IV)$$.

Thus, the correct matching is:

$$A \rightarrow (III)$$

$$B \rightarrow (I)$$

$$C \rightarrow (II)$$

$$D \rightarrow (IV)$$

Hence, the correct answer is:

$$A-(III),\ B-(I),\ C-(II),\ D-(IV)$$