Problem 28 - Chemistry The Central Science In Si Units

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 \(\left(\mathrm{H}_{2} \mathrm{CO}\right)\).

Step-by-Step Solution

Answer

(a) n-butane, (b) 1-butanol, (c) \(\mathrm{SO}_{2}\), (d) phosgene.

Step 1: Analyze Propane vs. n-Butane

Propane (\(\mathrm{C}_{3}\mathrm{H}_{8}\)) and n-butane (\(\mathrm{C}_{4}\mathrm{H}_{10}\)) are both nonpolar hydrocarbons, and their primary intermolecular forces are London dispersion forces. London dispersion forces increase with molar mass and surface area. Since n-butane has a larger molar mass and more surface area than propane, it will have stronger dispersion forces and a higher boiling point.

Step 2: Analyze Diethyl Ether vs. 1-Butanol

Diethyl ether (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OCH}_{2}\mathrm{CH}_{3}\)) can engage in dipole-dipole interactions due to the polar \(-O-\) group, but it lacks hydrogen bonding. 1-Butanol (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{OH}\)) has an \(\text{-OH}\) group that allows for hydrogen bonding, the strongest type of intermolecular force in these examples. As a result, 1-butanol will have a higher boiling point than diethyl ether.

Step 3: Analyze Sulfur Dioxide vs. Sulfur Trioxide

Sulfur dioxide (\(\mathrm{SO}_{2}\)) is a polar molecule and can exhibit dipole-dipole interactions. Sulfur trioxide (\(\mathrm{SO}_{3}\)) is a nonpolar molecule and only exhibits London dispersion forces. Dipole-dipole interactions are stronger than dispersion forces; thus, \(\mathrm{SO}_{2}\) has a higher boiling point than \(\mathrm{SO}_{3}\).

Step 4: Analyze Phosgene vs. Formaldehyde

Phosgene (\(\mathrm{Cl}_{2}\mathrm{CO}\)) is a polar molecule with dipole-dipole interactions, primarily due to the presence of chlorines. Formaldehyde (\(\mathrm{H}_{2}\mathrm{CO}\)) is also polar but is smaller in size. Phosgene's larger size allows for stronger van der Waals forces, in addition to dipole-dipole interactions. As a result, phosgene has a higher boiling point than formaldehyde.

Key Concepts

Boiling Point

When discussing the boiling point, it’s all about the temperature at which a liquid turns into vapor. The boiling point is directly related to the strength of the intermolecular forces within a substance. Why is this important? A high boiling point indicates that stronger forces hold the molecules together, making them harder to break apart into a gaseous state.

In simpler terms, think of it as trying to pull apart two pieces of magnetic material. The stronger the magnetic force, the more effort it requires. Similarly, stronger intermolecular forces mean a higher boiling point. Understanding boiling points helps predict the behavior of substances under different conditions. It also offers clues about a substance's intermolecular forces, which can help determine which type of substance will win in a boiling point "face-off".

London Dispersion Forces

London dispersion forces are one of the fundamental intermolecular attractions in chemistry. They're the weakest type but are present in all molecules, whether polar or nonpolar. These forces arise due to temporary fluctuations in electron density within molecules.

What happens is that sometimes, electrons gather more on one side of a molecule than the other, causing a temporary dipole. This slight change in electron distribution creates an attraction between neighboring molecules.

• These forces increase as the size or mass of molecular structure increases.
• Larger molecules have more electrons, leading to stronger dispersion forces.
  
  In our example, n-butane has stronger dispersion forces than propane due to its larger size. Hence, n-butane has a higher boiling point because these temporary interactions are more significant, holding the molecules more strongly together.

Hydrogen Bonding

Hydrogen bonding represents a special, stronger type of dipole-dipole interaction. It occurs when hydrogen is bonded to highly electronegative elements like oxygen, nitrogen, or fluorine. What makes them unique? It's the direct involvement of hydrogen with these electronegative atoms.

Hydrogen atoms tend to carry a partial positive charge while attracting the partial negative charge of another nearby electronegative atom. This creates a robust intermolecular force responsible for high boiling points in substances like water.

• As seen with 1-butanol, the presence of an -OH group allows for hydrogen bonding.
• This strong bond gives 1-butanol a higher boiling point compared to diethyl ether, which lacks such interactions.
  
  Hydrogen bonds play a crucial role in biochemical systems and numerous other applications.

Dipole-Dipole Interactions

Dipole-dipole interactions occur between polar molecules, where one end of the molecule is slightly positive, and the other end is slightly negative. These interactions are stronger than London dispersion forces but weaker than hydrogen bonds.

Here's how it works: The positive end of one polar molecule is attracted to the negative end of another, creating a net attraction.

• In our exercise, sulfur dioxide (SO₂) displays dipole-dipole interactions due to its bent molecular shape creating asymmetry. On the contrary, sulfur trioxide (SO₃) is a symmetrical, nonpolar molecule.
• Thus, SO₂ has a higher boiling point since dipole-dipole interactions are stronger than the London dispersion forces found in SO₃.
  
  Understanding dipole-dipole interactions helps us comprehend why some substances have different boiling points, even if they're similar in size.