${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Cl}+\mathrm{Na}\underset{-\mathrm{NaCl}}{\to }\mathrm{A}$  on monochlorination gives how many isomers ?

$$\begin{gathered} {\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{Cl}+\mathrm{Na}\stackrel{\mathrm{ether}}{\to }\underset{\mathrm{butane}}{{\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_2-{\mathrm{CH}}_3} \\ \underset{\mathrm{butane}}{{\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_2-{\mathrm{CH}}_3}\stackrel{{\mathrm{Cl}}_2/\mathrm{h\nu }}{\to }\underset{\left(1\right)}{ {\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{Cl}}+ \\ \underset{\left(2\right)}{ {\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{C}*\mathrm{H}\left(\mathrm{Cl}\right)-{\mathrm{CH}}_3} \\ * = \mathrm{assymetric} \mathrm{carbon} \end{gathered}$$

Total no. of isomers=1+2=3