In most textbooks, highly-branched isomers are said to boil lower than less-branched because being more compact they have lower surface areas and therefore weaker van der Waals attraction. But the most-branched octane, 2,2,3,3 tetramethylbutane (CH3)3C-C(CH3)3, boils higher than 'iso-octane' (CH3)3CCH2CH(CH3)2 which is less branched.
I suspect this is because there is a competing entropy effect. The more branched the alkane, the fewer degrees of internal freedom in the molecule, so it gains less entropy on liberation during vaporisation. This will tend to raise its boiling-point, as increase in entropy favours any process, especially an endothermic one like vaporisation. In the case of (CH3)3C-C(CH3)3 this entropy effect outweighs the lower enthalpy of vaporisation due to compact structure and weaker van der Waals forces.
This would seem to be confirmed by the fact that cycloalkanes always boil higher than the open-chain forms; the cycloalkane is constrained in internal motion and gains less entropy on liberation. Rigid, symmetrical molecules like adamantane and dodecahedrane, which essentially gain no greater freedom of internal motion on vaporisation, boil even higher than straight-chain molecules of similar molecular weight.
Can anybody confirm or refute this suggestion?