An alkyl halide with `beta`-hydrogen atoms on reaction with a base or a nucleophile has two competing paths. Substitutions (`S_(N^(1))` and `S_(N^(2))`) and elimination (`E_(1)` and `E_(2)`). The path adopted by them depends upon the nature of the alkyl halide, strength and size of the base/nucleophile and reaction conditions. The bulkiers nucleophile prefers to act as a base and abstracts a proton rather than approaching a tetravalent carbon atom (due to steric reasons) and vise versa. A primary alkyl halide can react by any of the four mechanism `(S_(N^(2)), S_(N^(1)), E_(2)` and `E_(1))` depending upon the stability or the intermediate carbocation or the substituted alkene formed and the reaction conditions and tertiary alkyl halide. The three possible paths `(S_(N^(1)), E_(1)` and `E_(2))`
2-Bromopropane is separately heated with aq `CH_(3)CO_(2)Na` or with `CH_(3)CH_(2)ONa//CH_(3)CH_(2)OH`. The major product obtained in each case respectively are

Options

A. Isopropyl acetate, propene
B. Propane, isopropylethyl ether
C. Isopropyl acetate, isopropyl ethyl ether
D. Propene in both the case

Answer

D

Explanation

`CH_(3)COO^(-)` is a weak base and hence brings about substitutions while `C_(2)H_(5)O^(-)` is a strong base and hence bring about eliminations.