Question 11.8: Arrange the alkanes in each set in order of increasing boili...
Arrange the alkanes in each set in order of increasing boiling point.
(a) Butane, decane, and hexane
(b) 2-Methylheptane, octane, and 2,2,4-trimethylpentane
Strategy
The compounds in each set are alkanes, and the only forces of attraction between alkane molecules are very weak London dispersion forces. As the number of carbons in a hydrocarbon chain increases, London dispersion forces between chains increase; therefore, boiling point also increases (Section 5.7A). For alkanes that are constitutional isomers, the strength of London dispersion forces between molecules depends on shape. The more compact the shape, the weaker the intermolecular forces of attraction and the lower the boiling point.
(a) All three compounds are unbranched alkanes. Decane has the longest carbon chain, the strongest London forces between its molecules, and the highest boiling point. Butane has the shortest carbon chain and the lowest boiling point.
(b) These three alkanes are constitutional isomers with the molecular formula C8H18. 2,2,4-Trimethylpentane is the most highly branched isomer and therefore has the smallest surface area and the lowest boiling point. Octane, the unbranched isomer, has the largest surface area and the highest boiling point.
