Question

Arrange the following compounds in the increasing order of their boiling points:

Butane, 1-Bromobutane, 1-Iodobutane, 1-Chlorobutane 

Answer 1

Butane < 1-Chlorobutane < 1-Bromobutane < 1-Iodobutane

Reason: 

Butane \[\ce{C4H10}\] has only London dispersion forces, as it is nonpolar and the smallest molecule among the given compounds. 
1-Chlorobutane \[\ce{C4H9Cl}\] has a polar bond \[\ce{(C - Cl)}\], which increases its boiling point compared to butane due to the presence of dipole-dipole interactions in addition to London dispersion forces.
Similarly, 1-bromobutane also has a polar bond \[\ce{(C - Br)}\], leading to dipole-dipole interactions. Bromine is larger than chlorine, so the molecule might have slightly stronger dispersion forces as well.
1-Iodobutane \[\ce{(C4H9I)}\] has the largest halogen atom (iodine), so it will have stronger London dispersion forces compared to the previous compounds due to its larger electron cloud.