Arrange each set of compounds in the order of increasing boiling points.
(i) Bromomethane, bromoform, chloromethane, dibromomethane
(ii) 1-chloropropane, iso-propyl chloride, 1-chlorobutane
(i) For the same alkyl group, boiling point increases with the size of the halogen atom. Thus, boiling point of bromomethane is higher than that of chloromethane. Further, the boiling point ’ increases as the number of halogen atoms increases. Thus, the boiling point of bromoform with three Br atoms is highest followed by dibromomethane with two Br atoms while that of bromomethane containing only one Br atom is lowest. Therefore CH3CI (chloromethane) < CH3Br (bromomethane) < CH2Br2 (dibromomethane) <CHBr3 (bromoform).
(ii) For the same halogen, boiling point increases with increase in size of the alkyl group. Thus, the boiling point of 1-chlorobutane is higher than those of
1- chloropropane and’isopropyl chloride. Further, the boiling point decreases as the branching increases. Thus, the boiling point of chloropropane is higher than isopropyl chloride.
Therefore, the arrangement of the given compounds in the order of increasing boiling points is as follows: (CH3 )2CHC1 (wo-propyl chloride or
2- chloropropane) < C1CH2CH2CH3
(1-chloropropane) < C1CH2CH2CH2CH3 (1-chlorobutane).