Identity $\mathrm{Z}$ in the following reaction
${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}\stackrel{{\mathrm{PBr}}_3}{⟶}\mathrm{X}\stackrel{\mathrm{alc}. \mathrm{KOH}}{⟶}\mathrm{Y}\underset{\left(\mathrm{ii}\right){\mathrm{H}}_2\mathrm{O}, \mathrm{heat}}{\overset{\left(\mathrm{i}\right) {\mathrm{H}}_2{\mathrm{SO}}_4, \mathrm{RT}}{\to }}\mathrm{Z}$

${\mathrm{PBr}}_3$ converts the ethanol into the ethyl bromide. The compound $\mathrm{X}$ is ethyl bromide. After that the alcoholic potassium hydroxide converts the ethyl bromide into the ethene. So, the compound $\mathrm{Y}$ is ethene. After that ethene gives the electrophilic addition reaction and formed the ethanol as a product. So, the compound $\mathrm{Z}$ is ethanol.

Reaction :-

${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}\stackrel{{\mathrm{PBr}}_3}{⟶}\left(\mathrm{X}\right){\mathrm{CH}}_3{\mathrm{CH}}_2-\mathrm{Br} \stackrel{\mathrm{Alc}. \mathrm{KOH}}{\to }\left(\mathrm{Y}\right){\mathrm{CH}}_2={\mathrm{CH}}_2\underset{\left(\mathrm{ii}\right){\mathrm{H}}_2\mathrm{O}, \mathrm{heat}}{\overset{\left(\mathrm{i}\right) {\mathrm{H}}_2{\mathrm{SO}}_4, \mathrm{RT}}{\to }}\left(\mathrm{Z}\right){\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}$

Hence, the compound $\mathrm{Z}$ in the given reaction is ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}$.