What happens when 1 - chloropropane is treats with Na metal in presence of dry ether

2 molecules of 1-chloropropane react with 2 molecules of Na in the presence of dry ether to give hexane and 2NaCl.2(Cl)-CH2CH2CH3 + 2Na in the presence of dry ether = CH3CH2CH2-CH3CH2CH3 + 2NaCl. This happens because homolytic cleavage takes place between Cl and the attached carbon thus producing free radicals. The Cl attaches with Na making NaCl. The same happens with the other symmetric compound. Thus 2 free radicals are generated on the 2 different carbon atoms. The free radicals combine to form a bigger hydrocarbon which is double that of the initial hydrocarbon taken I.e. propane to hexane. By this method, methane cannot be formed because this method gives hydrocarbon having double the number of carbon atoms s from initial but methane consists of only 1 carbon atom. Thank you!