A hydrocarbon of molecular mass of 72 g mol^(-1) gives a single monochloro derivative and two dichloro derivatives on photochlorination. Give the structures of the hydrocarbon.

A hydrocarbon of molecular mass of 72 g mol^(-1) gives a single monoch

1.	A hydrocarbon of molecular mass of 72 g mol^(-1) gives a single monochloro derivative and two dichloro derivatives on photochlorination. Give the structures of the hydrocarbon.
Answer» Solution :The hydrocarbon having molecular mass of 72 g `mol^(-1)` must be `C_(5)H_(12)`. Since on photochlorination, it gives a single monochloro derivative, THEREFORE, all the twelve HYDROGENS in this hydrocarbon are equivalent. Thus, the hydrocarbon must be neopentane or 2,2-dimethylpropane, it forms only one monochloro derivative and two dichloro derivatives as shown below: `underset("2,2-Dimethylpropane (All the 12 H's are equivalent)")(CH_(3)-underset(CH_(3))underset(\|)overset(CH_(3))overset(\|)(C)-CH_(3))""underset("1-Chloro-2,2-dimethylpropane (Monochloro derivative)")(CH_(3)-underset(CH_(3))underset(\|)overset(CH_(3))overset(\|)(C)-CH_(2)Cl)""underset("1,3-Dichloro-2,2-dimethylpropane (Dichloro derivative)")(CH_(3)-underset(CH_(3))underset(\|)overset(CH_(2)Cl)overset(\|)(C)-CH_(2)Cl)""underset("1,1-Dichloro-2,2-dimethylpropane (Dichloro derivative)")(CH_(3)-underset(CH_(3))underset(\|)overset(CH_(3))overset(\|)(C)-CHCl_(2))`.