Explaining an Anomalous Boiling Point in a Homologous Series
A class compares boiling points of four liquids: butan-1-ol (bp 118 °C), butan-2-ol (bp 100 °C), 2-methylpropan-2-ol (bp 82 °C) and butan-1-amine (bp 78 °C). All four have similar molar mass (73–74 g mol⁻¹). A student claims that 'the more branched the molecule, the lower the boiling point, because branching reduces hydrogen bonding'. The teacher points out that butan-1-amine has no O–H bond yet boils at almost the same temperature as 2-methylpropan-2-ol, and asks the class to separate the effects of hydrogen bonding from dispersion forces and molecular shape.
1. Evaluate the student's explanation. Which parts are correct and which are wrong?
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The trend within the butanol isomers is real: butan-1-ol > butan-2-ol > 2-methylpropan-2-ol. But the explanation is partly wrong. Branching does not remove hydrogen bonding; all three butanols have one O–H group and can donate and accept hydrogen bonds. What branching changes is molecular shape: a more spherical molecule has smaller surface area, so London dispersion forces between molecules are weaker. The primary alcohol also packs more efficiently, giving stronger dispersion contacts.
So the correct statement is: hydrogen bonding is present in all three butanols and is the dominant intermolecular force; the differences among isomers come mainly from dispersion forces, which depend on surface contact area and shape, not from loss of hydrogen bonding.
2. Explain why butan-1-amine boils at 78 °C, close to 2-methylpropan-2-ol, despite having no O–H bond.
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Butan-1-amine has N–H bonds. Nitrogen is more electronegative than carbon and hydrogen, so the N–H bond is polar and the lone pair on nitrogen can accept a hydrogen bond. Amines therefore form N–H···N hydrogen bonds, which are weaker than O–H···O hydrogen bonds because N is less electronegative than O, so the hydrogen bond is less strong.
However, butan-1-amine is a straight-chain molecule with greater surface area than the compact, roughly spherical 2-methylpropan-2-ol. Its dispersion forces are stronger. The two effects partly cancel: weaker hydrogen bonding but stronger dispersion forces, giving a similar boiling point (78 °C versus 82 °C). This shows that boiling point is not controlled by one force alone but by the sum of all intermolecular interactions plus molecular shape.