Question

Isopropyl chloride and ethyl chloride both react with Na in presence of dry ether. How many products are obtained?

Answer 1

Hint: In Wurtz reaction, two alkyl halide molecules are coupled in presence of sodium metal in anhydrous ether or tetrahydrofuran to form a new carbon bond and thus by giving a symmetrical alkane.

Complete answer:
In the question it is given that we have to find the product when isopropyl chloride is going to react with ethyl chloride in the presence of sodium and dry ether.
Let us know what an alkyl halide is.
Alkyl halides (also known as haloalkanes) are compounds in which one or more hydrogen atoms in an alkane have been replaced by halogen atoms. Alkyl halides fall into different classes depending on how the halogen atom is positioned on the chain of carbon atoms. Chloro ethane is also known as Ethyl Chloride. The chemical formula of ethyl chloride is $C_2{H}_{5}Cl$.
When an alkyl halide reacts with Na in presence of dry ether, then free radical alkyl forms along with halide ion because Na is a metal and provides a free electron.
$$2R-X+2Na\stackrel{DryEther}{\to }R-R+2NaCl$$
Thus Isopropyl chloride and ethyl chloride reacts with Na in presence of dry ether and gives 2,3-dimethyl butane.
 

Note:
Methane cannot be synthesized via the Wurtz reaction, since products of an organic coupling reaction must have at least two carbon atoms. The Wurtz coupling method generally fails when tertiary alkyl halides are used.