${\mathrm{C}}_2{\mathrm{H}}_4\underset{Al{\mathrm{Cl}}_3}{\stackrel{\mathrm{HCl}}{⟶}}\mathrm{A}\stackrel{\mathrm{KOH}(\mathrm{aq}.)}{\to }\mathrm{B}\frac{\text{ Conc }{\mathrm{H}}_2{\mathrm{SO}}_4}{{140}^0\mathrm{C}}\mathrm{C}$

What is the final product?

Given:- It is given to find the final product in the given reaction.

Explanation:-

The reaction can be completed as below.

${\mathrm{C}}_2{\mathrm{H}}_4\underset{Al{\mathrm{Cl}}_3}{\stackrel{\mathrm{HCl}}{⟶}}{\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{Cl}\stackrel{\mathrm{aq}. \mathrm{KOH}}{\to }{\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{OH}$

${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{OH} \underset{140°\mathrm{C}}{\overset{\mathrm{Con} {\mathrm{H}}_2{\mathrm{SO}}_4}{\to }} {\mathrm{C}}_2{\mathrm{H}}_5-\mathrm{O}- {\mathrm{C}}_2{\mathrm{H}}_5$

Hence, the correct answer is option (B) $C_2{H}_5-O-C_2{H}_5$.