Based on the type or types of intermolecular forces, predict the substance in each pair that has the higher boiling point: (a) propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) or \(n\) -butane \(\left(\mathrm{C}_{4} \mathrm{H}_{10}\right),(\mathbf{b})\) diethyl ether $\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{3}\right)\( or 1 -butanol \)\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right)$ (c) sulfur dioxide \(\left(\mathrm{SO}_{2}\right)\) or sulfur trioxide \(\left(\mathrm{SO}_{3}\right)\), (d) phosgene \(\left(\mathrm{Cl}_{2} \mathrm{CO}\right)\) or formaldehyde \(\left(\mathrm{H}_{2} \mathrm{CO}\right)\). Look up and compare the normal boiling points and nor-

In this step, we need to identify the types of intermolecular forces (IMFs) present in each substance. The types of IMFs are London dispersion forces, dipole-dipole interactions, and hydrogen bonding.

After identifying the types of IMFs present in each substance, we will predict which substance has a higher boiling point based on the strength of these forces. The stronger the intermolecular forces, the higher the boiling point. (a) \(\mathrm{C}_{3} \mathrm{H}_{8}\) (propane) vs. \(\mathrm{C}_{4} \mathrm{H}_{10}\) (butane): - Both molecules are nonpolar, so they only have London dispersion forces. - Butane has a larger molecular size, so it has stronger London dispersion forces. - Prediction: Butane has a higher boiling point due to stronger London dispersion forces. (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCH}_{2}\mathrm{CH}_{3}\) (diethyl ether) vs. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) (1-butanol): - Diethyl ether has an oxygen in it, but it does not have a hydrogen bond. - 1-butanol has an alcohol group, which allows for hydrogen bonding. - Prediction: 1-butanol has a higher boiling point due to hydrogen bonding. (c) \(\mathrm{SO}_{2}\) (sulfur dioxide) vs. \(\mathrm{SO}_{3}\) (sulfur trioxide): - Both molecules have dipole-dipole interactions due to their polar nature. - Sulfur trioxide has a higher molecular weight and more oxygens, which gives it stronger dipole-dipole interactions. - Prediction: Sulfur trioxide has a higher boiling point due to stronger dipole-dipole interactions. (d) \(\mathrm{Cl}_{2} \mathrm{CO}\) (phosgene) vs. \(\mathrm{H}_{2} \mathrm{CO}\) (formaldehyde): - Both molecules have dipole-dipole interactions due to their polar nature. - Phosgene has larger molecular weight and chlorine atoms, which have a higher electronegativity that results in stronger dipole-dipole interactions. - Prediction: Phosgene has a higher boiling point due to stronger dipole-dipole interactions.

In this final step, look up the normal boiling points of the substances and compare them to verify the predictions made in Step 2. If there are differences, try to account for them by analyzing the molecular structure of the substances further.