Question

When ethy iodide and n propyl iodide are allowed to react with na in dry ether the?

Answer 1

It’s an example of a coupling reaction - known as the Wurtz reaction, after the French chemist who first observed & documented it.[1]

The alkyl halide molecules are cleaved, with the halide forming the equivalent sodium salt whilst the alkyl groups couple together in a nucleophilic substitution reaction to form a longer chain alkane (butane in this case). The ether merely acts a solvent & reaction medium.

2 CH3CH2I + 2Na → C4H10 + 2NaI

Answer 2

CH3CH2I+CH3CH2CH2I−→−−EtherNaCH3CH2CH2CH3+CH3CH2CH2CH2CH3+CH3CH2CH2CH2CH2CH3CH3CH2I+CH3CH2CH2I→EtherNaCH3CH2CH2CH3+CH3CH2CH2CH2CH3+CH3CH2CH2CH2CH2CH3

So we can observe that when the reactants Ethyl iodide and n-propyl iodide are allowed to undergo the Wurtz reaction the possible alkenes formed are butane, pentane, and hexane. When (CH3CH2I)(CH3CH2I) undergoes self-addition reaction it forms (CH3CH2CH3CH3)(CH3CH2CH3CH3) butane. When (CH3CH2CH2I)(CH3CH2CH2I) undergoes self-addition reaction it forms (CH3CH2CH3CH2CH2CH3)(CH3CH2CH3CH2CH2CH3) hexane. Now when Ethyl iodide and n-propyl iodide are allowed to undergo cross-addition reaction they form another alkane (CH3CH2CH3CH2CH3)(CH3CH2CH3CH2CH3) pentane. So, the alkanes formed during the reaction are butane, pentane, and hexane are formed. Propane is the alkane that is not formed in the reaction.

hope it helps..