Distributing three chlorine substituents among the carbons of a propane skeleton gives five distinct positional (constitutional) isomers: 1,1,1-; 1,1,2-; 1,1,3-; 1,2,2-; and 1,2,3-trichloropropane. (Note: the 1,1,2-isomer additionally has a chiral center and so exists as a pair of enantiomers, but at the level of counting structural isomers, as this question intends, the answer is 5.)