Chemistry - Halogen Derivative Question with Solution | TestHub

${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Br}+{\mathrm{Na}}^{\ast }⟶{\mathrm{CH}}_3{\mathrm{CH}}_2^{\ast }+\mathrm{NaBr}$ Intermediate free radical ${\mathrm{CH}}_3{\mathrm{CH}}_2^{\ast }$ combines to form ${\mathrm{CH}}_3{\mathrm{CH}}_2-{\mathrm{CH}}_2{\mathrm{CH}}_3$ (as a main product) and also ${\mathrm{CH}}_2=={\mathrm{CH}}_2$ and ${\mathrm{CH}}_3{\mathrm{CH}}_3$ by disproportion $${\mathrm{CH}}_3{\mathrm{CH}}_2^{\ast }+{\mathrm{CH}}_3{\mathrm{CH}}_2^{\ast }\to {\mathrm{CH}}_3{\mathrm{CH}}_2-{\mathrm{CH}}_2{\mathrm{CH}}_3$$ n-butane $${\mathrm{CH}}_3{\mathrm{CH}}_2^{\ast }+{\mathrm{CH}}_3{\mathrm{CH}}_2^{\ast }⟶{\mathrm{CH}}_2={\mathrm{CH}}_2+{\mathrm{CH}}_3-{\mathrm{CH}}_3$$ (disproportion reaction)