For the reaction sequence given below, the correct statement(s) is(are)

The given sequence of reagents fits very well with the following set of transformations:

$$\text{Anisole (methoxy-benzene)} \xrightarrow[\text{reflux}]{\text{excess HI}} \underset{X}{\text{Phenol}} \xrightarrow[\text{heat}]{CHCl_3/KOH} \underset{Y}{\text{Salicylaldehyde}} \xrightarrow[\;]{\text{alk. } KMnO_4\; \text{or } K_2Cr_2O_7/H^+} \underset{Z}{\text{Salicylic acid}}$$

Step-1 (Formation of X)
HI cleaves the C-O bond in anisole by an $$S_N1/S_N2$$ mechanism. The aromatic ring retains the -OH group while the alkyl part leaves as $$CH_3I$$. Hence $$X = C_6H_5OH$$ (phenol), an oxygen-containing compound.

Step-2 (Formation of Y)
Phenol on heating with $$CHCl_3$$ in strongly basic medium (Reimer-Tiemann reaction) gives an ortho-formyl derivative, salicylaldehyde ($$2\!-\!HO\!-\!C_6H_4CHO$$). Therefore $$Y$$ is also an oxygen-containing compound (it has both -OH and -CHO groups).

Step-3 (Formation of Z)
The -CHO group of salicylaldehyde is readily oxidised by alkaline $$KMnO_4$$ or $$K_2Cr_2O_7/H^+$$ to the -COOH group, giving salicylic acid ($$2\!-\!HO\!-\!C_6H_4COOH$$). $$Z$$ is a carboxylic acid, not an amine.

Checking the given statements

• Option A: “Both X and Y are oxygen containing compounds.”   Phenol (X) and salicylaldehyde (Y) each contain oxygen, so this statement is true.

• Option B: “Y on heating with $$CHCl_3/KOH$$ forms isocyanide.”   The carbylamine test (formation of isocyanide) requires a primary amine.   Y is an aldehyde, not an amine, so no isocyanide is produced.   Statement B is false.

• Option C: “Z reacts with Hinsberg’s reagent.”   Hinsberg’s test is specific for primary and secondary amines.   Z is a carboxylic acid, hence it does not react with benzenesulfonyl chloride.   Statement C is false.

• Option D: “Z is an aromatic primary amine.”   Z is salicylic acid, not an amine at all.   Statement D is false.

Thus, the only correct statement is:

Option A which is: Both X and Y are oxygen containing compounds.