MCAT Organic Chemistry Question — Boiling Points
- Mar 31, 2017
- MCAT Question of the Day
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Ethyl benzene forms London dispersion forces and has a molecular weight of 106.2 g/mol. Benzaldehyde has a molecular weight of 106.1 g/mol and forms dipole-dipole forces in addition to London dispersion forces. Lastly, benzyl alcohol has a molecular weight of 108.1 g/mol and forms hydrogen bonds, a special type of dipole-dipole interaction, as well as London dispersion forces.
Since all of the molecular weights are approximately the same, the intermolecular forces of each compound will determine the boiling point. Hydrogen bonds (dipole-dipole) in benzyl alcohol are stronger than dipole-dipole forces in benzaldehyde, and all dipole-dipole forces are stronger than London dispersion forces. Therefore, benzyl alcohol (BP: 205.3oC) > benzaldehyde (BP: 178.1oC) > ethyl benzene (BP: 136.0oC), or III > II > I. Based on these comparisons, B is the correct answer.
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