Chemistry - Halogen Derivative Question with Solution | TestHub

Sodium ethoxide (${\text{C}}_2{\text{H}}_5{\text{O}}^{-}{\text{Na}}^{+}$) acts as both a strong base and a strong nucleophile.

Isopropyl chloride (${\text{CH}}_3{\text{-CHCl-CH}}_3$) is a secondary alkyl halide.

When a secondary alkyl halide reacts with a strong base like sodium ethoxide, elimination (E2) competes with substitution (S${}_N$2). Due to the steric hindrance of the secondary alkyl halide and the strength of the base, dehydrohalogenation (elimination) dominates over substitution.

Thus, the major product is an alkene (propene), while the substitution product (ethyl isopropyl ether) is formed only as a minor product.

Reaction:
${\text{CH}}_3{\text{-CHCl-CH}}_3+{\text{C}}_2{\text{H}}_5{\text{O}}^{-}\to {\text{CH}}_3{\text{-CH=CH}}_2+{\text{C}}_2{\text{H}}_5\text{OH}+{\text{Cl}}^{-}$

Answer: ${\text{CH}}_3{\text{-CH=CH}}_2$