Question

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 \(\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)\), (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 \(\left(\mathrm{H}_{2} \mathrm{CO}\right)\).

Short answer

The substances with higher boiling points are (a) n-butane, (b) 1-butanol, (c) sulfur dioxide, and (d) phosgene.

Step-by-step solution

(a) Analysis of propane (C3H8) and n-butane (C4H10)

Propane and n-butane are both nonpolar hydrocarbons. As hydrocarbons, the only type of intermolecular forces present are London dispersion forces. The strength of the London dispersion forces depends on the size and surface area of the molecule. n-butane has a larger molecular size and surface area than propane, which means it should have stronger London dispersion forces, resulting in a higher boiling point for n-butane.

(b) Analysis of diethyl ether (CH3CH2OCH2CH3) and 1-butanol (CH3CH2CH2CH2OH)

Diethyl ether has a polar C-O bond, while 1-butanol has a polar O-H bond in addition to the polar C-O bond. Diethyl ether has dipole-dipole interactions, while 1-butanol has hydrogen bonding, which is stronger than the dipole-dipole interactions in diethyl ether. This leads to a higher boiling point for 1-butanol.

(c) Analysis of sulfur dioxide (SO2) and sulfur trioxide (SO3)

Sulfur dioxide (SO2) is a polar molecule and has dipole-dipole interactions. Sulfur trioxide (SO3), on the other hand, is a nonpolar molecule (despite having polar bonds) as the bond dipoles cancel each other out due to the symmetrical structure, and it has London dispersion forces. Dipole-dipole interactions are stronger than London dispersion forces, so sulfur dioxide should have a higher boiling point than sulfur trioxide.

(d) Analysis of phosgene (Cl2CO) and formaldehyde (H2CO)

Phosgene is a polar molecule because of the presence of polar C-Cl and C-O bonds and so it will exhibit dipole-dipole interactions. Formaldehyde is also a polar molecule due to polar C-O and C-H bonds, and will have dipole-dipole interactions. The boiling point will depend on the strength of these interactions. The electronegativity difference on Cl-C and C-O bonds produces a stronger dipole in phosgene than the C-O and C-H bonds in formaldehyde. Since phosgene has stronger dipole-dipole interactions, it will have a higher boiling point than formaldehyde. In conclusion, the substance with the higher boiling point in each pair is: a) n-butane b) 1-butanol c) sulfur dioxide d) phosgene

Key concepts

London Dispersion Forces

When it comes to understanding why certain substances have higher boiling points than others, London dispersion forces offer key insights. These forces are the weakest type of intermolecular attraction that occur between all molecules, whether polar or nonpolar.

Named after the physicist Fritz London, these forces are caused by temporary dipoles forming when the electrons within a molecule are unevenly distributed at any given moment. This in turn induces a momentary dipole in nearby molecules. The larger and more electrons a molecule has, the stronger these London dispersion forces will be, simply because there are more electrons that can create dipoles.

For example, in comparing propane and n-butane, n-butane has a higher boiling point due to its larger size, which corresponds to a greater surface area for these temporary dipoles to work across.

Dipole-Dipole Interactions

Polar molecules have something extra: dipole-dipole interactions. These occur when the positive end of one polar molecule is attracted to the negative end of another. It's like molecular magnets aligning themselves with their opposites.

These interactions are significantly stronger than London dispersion forces but less potent than hydrogen bonds. For instance, when considering molecules like diethyl ether and 1-butanol, we see that both molecules are polar. However, due to the presence of an additional O-H bond, 1-butanol has both the dipole-dipole interactions and the capability to form hydrogen bonds, thereby giving it a higher boiling point compared to diethyl ether, which can only engage in dipole-dipole interactions.

Hydrogen Bonding

Hydrogen bonding is the star attraction among intermolecular forces when it comes to raising boiling points. This strong interaction occurs when a hydrogen atom bonded to a highly electronegative atom such as nitrogen, oxygen, or fluorine is attracted to another electronegative atom in a different molecule.

These bonds are particularly strong due to the high electronegativity of the atoms involved, which attracts the hydrogen atom powerfully. Substances capable of hydrogen bonding, like water, typically have much higher boiling points than those that don't.

In the case of 1-butanol, its ability to form hydrogen bonds due to its O-H group is the main reason for its higher boiling point compared to diethyl ether, which lacks this attribute.

Molecular Polarity

Molecular polarity is essentially the measure of how evenly electrical charges are distributed across a molecule. A molecule is considered polar if it has a distinct separation of electrical charges, meaning one end is more negative and the other end more positive.

The geometry of the molecule plays a crucial role in determining its polarity. If the structure of the molecule is such that the dipole moments of the bonds do not cancel each other out, the molecule is polar. Polar molecules interact through stronger forces like dipole-dipole interactions and can participate in hydrogen bonding if they contain hydrogen atoms bonded to highly electronegative atoms.

Take sulfur dioxide and sulfur trioxide, for example. Sulfur dioxide is polar due to its bent shape, resulting in unequal distribution of charge and therefore a higher boiling point through dipole-dipole interactions, as opposed to sulfur trioxide, which, despite having polar bonds, is nonpolar because of its symmetrical shape.