Question

Consider the compounds butanoic acid, pentanal, \(n\) -hexane, and 1 -pentanol. The boiling points of these compounds (in no specific order) are \(69^{\circ} \mathrm{C}, 103^{\circ} \mathrm{C}, 137^{\circ} \mathrm{C}\), and \(164^{\circ} \mathrm{C}\). Match the boiling points to the correct compound.

Short answer

The correct boiling points for the given compounds are: Butanoic acid: \(164^{\circ}\mathrm{C}\), Pentanal: \(103^{\circ}\mathrm{C}\), n-Hexane: \(69^{\circ}\mathrm{C}\), and 1-Pentanol: \(137^{\circ}\mathrm{C}\).

Step-by-step solution

Identify intermolecular forces in each compound

* Butanoic acid (C4H8O2): Since it is an acid (COOH group), it can form hydrogen bonds, dipole-dipole interactions, and van der Waals forces (London dispersion forces). * Pentanal (C5H10O): It has an aldehyde group (H-C=O), so it features dipole-dipole interactions and van der Waals forces. * n-Hexane (C6H14): It is an alkane, so it only has weak van der Waals forces. * 1-Pentanol (C5H12O): Since it is an alcohol (OH group), it can form hydrogen bonds, dipole-dipole interactions, and van der Waals forces.

Rank the compounds by intermolecular force strength

We know that hydrogen bonding is the strongest intermolecular force, then comes dipole-dipole interaction, and then van der Waals forces. With this information, we can rank the compounds from strongest intermolecular forces to weakest: 1. Butanoic acid (Hydrogen bonding; strongest) 2. 1-Pentanol (Hydrogen bonding) 3. Pentanal (Dipole-dipole interaction) 4. n-Hexane (Van der Waals forces; weakest)

Match boiling points based on intermolecular force strength

Now let's match the boiling points (in ascending order) to the compounds according to their intermolecular force strength: 1. n-Hexane: \(69^{\circ}\mathrm{C}\) (weakest intermolecular forces) 2. Pentanal: \(103^{\circ}\mathrm{C}\) (dipole-dipole interaction) 3. 1-Pentanol: \(137^{\circ}\mathrm{C}\) (hydrogen bonding) 4. Butanoic acid: \(164^{\circ}\mathrm{C}\) (strongest intermolecular forces) So, the correct boiling points are as follows: * Butanoic acid: \(164^{\circ}\mathrm{C}\) * Pentanal: \(103^{\circ}\mathrm{C}\) * n-Hexane: \(69^{\circ}\mathrm{C}\) * 1-Pentanol: \(137^{\circ}\mathrm{C}\)

Key concepts

Intermolecular Forces

Intermolecular forces are the forces that exist between molecules. These forces determine many physical properties, such as boiling points, melting points, and solubilities. Understanding these forces can help us predict how substances will behave in different conditions.

There are several types of intermolecular forces including:

• Hydrogen bonding: Very strong forces occurring in molecules containing hydrogen bonded to fluorine, oxygen, or nitrogen.
• Dipole-dipole interactions: Occur between molecules with permanent dipole moments.
• Van der Waals forces: Weak forces that arise from temporary fluctuations in electron distribution, present in all molecules.

Recognizing which type of force is present in a compound is crucial for understanding its properties, such as boiling point.

Hydrogen Bonding

Hydrogen bonding occurs when hydrogen is directly bonded to an electronegative atom such as oxygen, nitrogen, or fluorine. This strong type of intermolecular force greatly affects the boiling point of compounds because it requires more energy to break these bonds.

For example, in the original exercise, both butanoic acid and 1-pentanol exhibit hydrogen bonding. Butanoic acid has a carboxyl group (COOH) which allows it to form very strong hydrogen bonds, contributing to its high boiling point of \(164^{\circ} \mathrm{C}\). Similarly, 1-pentanol can form hydrogen bonds due to its hydroxyl (OH) group, resulting in a boiling point higher than compounds that lack hydrogen bonding.

Dipole-Dipole Interactions

Dipole-dipole interactions are a type of intermolecular force that occurs between molecules with polar bonds. These interactions happen between the positive end of one polar molecule and the negative end of another. They are stronger than van der Waals forces but weaker than hydrogen bonds.

In our case, pentanal shows dipole-dipole interactions due to its aldehyde group (H-C=O). The polar nature of the carbon-oxygen double bond leads to a partial charge which results in dipole-dipole forces. This middle-level force contributes to the boiling point of pentanal being \(103^{\circ} \mathrm{C}\), higher than non-polar molecules like n-hexane but lower than those with hydrogen bonding.

Van der Waals Forces

Van der Waals forces are the weakest of all intermolecular forces, yet they are present in every molecule. These forces result from the temporary attractions between molecules due to electron cloud shifts, leading to fleeting dipoles.

In the example, \(n\)-hexane displays van der Waals forces as it lacks polar groups or hydrogen bonds. This makes its boiling point the lowest among the given substances, at \(69^{\circ} \mathrm{C}\). Despite being weak, van der Waals forces are essential for understanding the behavior of non-polar compounds, such as many alkanes like \(n\)-hexane, under various temperature and pressure conditions.