A hydrocarbon `(72 g cdot^(-1))` reacts with chlorine in the presence of light to produce one monochloro and two di-chloro derivaties. Write down the structure of the hydrocarbon.

Explanation

The molecular formula of the hydrocarbon having molecular mass `72g cdot mol^(-1) is C_(5)H_(12)`. Since, it produces one monochloro and two dichloro derivatives while reacting with `Cl_(2)` in the presence of light, all the H atoms of this hydrocarbon are equivalent. Therefore, the hydrocarbon is neopentane, `(CH_3)_(4)C`.
`CH_(3)-underset("Neopentane")underset(CH_3)underset(|)overset(CH_3)overset(|)(C)CH_(3)underset(hv)overset(Cl_(2))(rarr)CH_(3)-underset("monochloro derivative")underset(CH_3)underset(|)overset(CH_3)overset(|)(C)CH_(2)Cl`
`CH_(3)-underset("Neopentane")underset(CH_3)underset(|)overset(CH_3)overset(|)(C)CH_(3)underset(hv)overset(Cl_(2))(rarr)CH_(3)-underset("Two dichloro derivate")(underset(CH_3)underset(|)overset(CH_3Cl)
overset(|)(C)CH_(2)Cl+`CH_(3)-underset(CH_3)underset(|)overset(CH_3)overset(|)(C)-CHCl_(2))`.