The boiling point depends on the degree of INTERMOLECULAR force......the degree of interaction between individual molecules....
#"n-hexane"# versus #"2,2-dimethylbutane"#; here dispersion forces operate, and longer, linear molecules are able to interact more effectively. And thus #"n-hexane"# has a normal boiling point of #68# #""^@C#, versus #"2,2-dimethylbutane"#, normal boiling point of #50# #""^@C#. The same thing would occur if we attempted individual lengths of string from a pile of long lengths, versus a pile of short lengths. The shorter lengths would be easier to separate, because they would not get tangled.
For methylamine, and fluoromethane, hydrogen bonding operates in the former, but not in the fluorocarbon. And thus #"methylamine"# has a normal boiling point of #-6# #""^@C#, versus #H_3CF#, normal boiling point of #-78.4# #""^@C#.
For #"methanol"#, #"b.p."# #64.7# #""^@C#, and #"ethanol"#, #"b.p."# #78.4# #""^@C#, hydrogen bonding is the intermolecular force that operates strongly. And thus the boiling points of these alcohols are vastly elevated with respect to the boiling points of their parent alkanes....(#"methane"#, #"b.p"# #-161.5# #""^@C#; #"ethane"#, #"b.p"# #-89# #""^@C#). Even with methane and ethane, we can see the effects of dispersion forces with respect to a 2 carbon chain.
Note that the data, the volatilities, SHOULD have been supplied with this question. No-one is expected to remember boiling points (save that of water, and ethanol, and methanol, and benzene, and toluene, all of which I can remember!).
