Propane molecule on chlorination under photochemical condition gives two di-chloro products, " x " and " y ". Amongst " x " and " y ", " x " is an optically active molecule. How many tri-chloro products (consider only structural isomers) will be obtained from " x " when it is further treated with chlorine under the photochemical condition?

Propane chlorination gives two dichloro products x and y, and x is optically active. Possible dichloropropane structural isomers are 1,1-dichloropropane: CHCl$$_2$$CH$$_2$$CH$$_3$$; 1,2-dichloropropane: CH$$_2$$ClCHClCH$$_3$$ (which has a chiral center at C2); 1,3-dichloropropane: ClCH$$_2$$CH$$_2$$CH$$_2$$Cl$$; and 2,2-dichloropropane: CH$$_3$$CCl$$_2$$CH$$_3$$. Only 1,2-dichloropropane is optically active, so x = CH$$_2$$ClCHClCH$$_3$$.

Further chlorination of CH$$_2$$ClCHClCH$$_3$$ can replace a hydrogen at three distinct positions. Substitution at C1 gives CHCl$$_2$$CHClCH$$_3$$ (1,1,2-trichloropropane); substitution at C2 yields CH$$_2$$ClCCl$$_2$$CH$$_3$$ (1,2,2-trichloropropane); and substitution at C3 gives CH$$_2$$ClCHClCH$$_2$$Cl (1,2,3-trichloropropane). Although C3 initially has three equivalent hydrogens, chlorination at any one of them leads to the same product.

In total, there are 3 structural trichloro isomers obtainable from x.

The correct answer is Option D: 3.