${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Cl}+\mathrm{Na}\underset{-\mathrm{NaCl}}{\stackrel{\text{ dry ether }}{⟶}}\mathrm{A}\text{. }$ A on monochlorination gives how many isomers ?

The given reaction can be represented as below.

$2{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Cl}+2\mathrm{Na} \stackrel{\mathrm{dry} \mathrm{ether}}{\to }{\mathrm{C}}_4{\mathrm{H}}_{10}+2\mathrm{NaCl}$

In the above reaction, n-butane is obtained as a product (A). The monochlorination of n-butane gives 4 isomers.

Hence, A on monochlorination gives 4 isomers .

Hence, the correct answer is option (D) 4.