When ethyl iodide and n-propyl iodide are allowed to react with sodium in the presence of dry ether, the number of alkanes that would be produced is

Options

A. One alkane
B. Two alkanes
C. Four alkanes
D. Three alkanes

Answer

D

Explanation

`C_(2)H_(5)I +2Na + C_3H_7I overset(dry)underset(ether) rarr underset("Pentane")(C_(2)H_(5)-C_(3)H_(7)) + 2NaI.`
`C_(2)H_(5)I + 2Na + C_(2)H_(5)I overset(dry)underset(ether) rarr underset("Butane")(C_(2)H_(5)-C_(2)H_(5))+2NaI`
`C_(3)H_(7)I + 2Na +C_3H_(7)I overset(dry)underset(ether)rarr underset("Hexane")(C_3H_7-C_3H_7) + 2NaI`