Two products X and Y are formed in the following reaction sequence.
$\text{C}_6\text{H}_6 + \text{CH}_3\text{Cl} \xrightarrow{\text{Anhydr. AlCl}_3} W \xrightarrow[\text{warm}]{\text{dil. HNO}_3 + \text{dil. H}_2\text{SO}_4} X + Y$
The suitable method that can be used for the separation of products X and Y is :

The reaction of benzene with methyl chloride in the presence of anhydrous $\text{AlCl}_3$ is a Friedel-Crafts alkylation, which yields toluene (W).

$\text{C}_6\text{H}_6 + \text{CH}_3\text{Cl} \xrightarrow{\text{Anhydr. AlCl}_3} \text{C}_6\text{H}_5\text{CH}_3$

Nitration of toluene with a mixture of $\text{HNO}_3$ and $\text{H}_2\text{SO}_4$ yields a mixture of ortho-nitrotoluene and para-nitrotoluene as the major products (X and Y).

$\text{C}_6\text{H}_5\text{CH}_3 \xrightarrow[\text{warm}]{\text{dil. HNO}_3 + \text{dil. H}_2\text{SO}_4} \text{o-nitrotoluene} + \text{p-nitrotoluene}$

The boiling point of o-nitrotoluene is $222^\circ\text{C}$ and that of p-nitrotoluene is $238^\circ\text{C}$. Since the difference in their boiling points is small, they are separated by fractional distillation.

Answer: Fractional distillation