Which type of intermolecular force accounts for each of these differences? (a) Acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO},\) boils at $56^{\circ} \mathrm{C} ;\( dimethyl sulfoxide or DMSO, (CH \)\left._{3}\right)_{2}$ SO, boils at \(189^{\circ}\) C. (b) \(\mathrm{CCl}_{4}\) is a liquid at atmospheric pressure and room temperature, whereas \(\mathrm{CH}_{4}\) is a gas under the same conditions. \((\mathbf{c})\) \(\mathrm{H}_{2} \mathrm{O}\) boils at \(100^{\circ} \mathrm{C}\) but \(\mathrm{H}_{2} \mathrm{~S}\) boils at \(-60^{\circ} \mathrm{C}\). (d) 1 -propanol boils at \(97^{\circ} \mathrm{C}\), whereas 2 -propanol boils at \(82.6^{\circ} \mathrm{C}\).

Acetone and DMSO have different boiling points due to the dipole-dipole interactions. Both molecules have a polar bond, which means that there is an unequal distribution of electrons between the participating atoms. Acetone has a polar C=O bond, and DMSO has a polar S=O bond. The dipole-dipole interaction in DMSO is stronger than in acetone because sulfur is more polarizable and has a larger size than the carbon atom in acetone. This leads to a stronger intermolecular attraction in DMSO, hence its higher boiling point.

CCl4 and CH4 have different states at room temperature: liquid and gas, respectively. The main intermolecular force responsible for this difference is London dispersion forces. These forces are caused by the temporary fluctuations in electron distribution around the molecules. CCl4 has a larger molar mass and more electrons than CH4, leading to a greater dispersion force, which causes the CCl4 to be a liquid at room temperature, while CH4 remains a gas.

H2O has a much higher boiling point than H2S due to the presence of hydrogen bonding. In H2O, the highly electronegative oxygen atoms attract the hydrogen electrons, leaving the hydrogen atoms electron-deficient. This induces a partial positive charge on the hydrogen atoms and partial negative charge on the oxygen atoms. When H2O molecules come close to each other, these charges lead to the formation of hydrogen bonds, which are much stronger than the dipole-dipole interactions present in H2S. This results in H2O having a significantly higher boiling point than H2S.

The difference in boiling points between 1-propanol and 2-propanol can be attributed to hydrogen bonding. Both molecules have O–H groups that can form hydrogen bonds, but the positioning of this group in the molecular structure affects the extent of hydrogen bonding the molecules can form. In 1-propanol, the OH group is located at an end of the molecule, which allows it to form more hydrogen bonds with adjacent molecules. In 2-propanol, the OH group is located at the middle of the molecule, which results in weaker hydrogen bonding interactions due to steric hindrance from surrounding methyl groups. This causes 1-propanol to have a higher boiling point than 2-propanol due to the stronger hydrogen bonding interactions.