To determine which compound has the lowest boiling point among the given options, we need to consider the molecular structure, intermolecular forces, and functional groups present in each compound. The options are:
- \(\text{CH}_3 - \text{O} - \text{CH}_2 - \text{CH}_3\) (Diethyl ether)
- \(\text{CH}_3 - \text{COOH}\) (Acetic acid)
- \(\text{CH}_3 - \text{CH}_2 - \text{CH}_2 - \text{CH}_3\) (Butane)
- \(\text{CH}_3 - \text{CH}_2 - \text{CH}_2 - \text{OH}\) (Propanol)
Let's analyze each option:
- Diethyl Ether (\(\text{CH}_3 - \text{O} - \text{CH}_2 - \text{CH}_3\)): It has weak dipole-dipole interactions due to the ether linkage and lacks hydrogen bonding.
- Acetic Acid (\(\text{CH}_3 - \text{COOH}\)): It forms strong hydrogen bonds due to its carboxylic acid group, resulting in a higher boiling point.
- Butane (\(\text{CH}_3 - \text{CH}_2 - \text{CH}_2 - \text{CH}_3\)): A nonpolar hydrocarbon with weak van der Waals forces, leading to a lower boiling point.
- Propanol (\(\text{CH}_3 - \text{CH}_2 - \text{CH}_2 - \text{OH}\)): It can form hydrogen bonds due to the hydroxyl group, which increases the boiling point.
Conclusion:
Among the given compounds, Butane (\(\text{CH}_3 - \text{CH}_2 - \text{CH}_2 - \text{CH}_3\)) contains only weak van der Waals forces (London dispersion forces) because it is a nonpolar hydrocarbon. This results in the lowest boiling point compared to the other compounds, which have stronger intermolecular forces due to dipole interactions or hydrogen bonding. Hence, Butane will have the lowest boiling point.