Question

Arrange these compounds in order of increasing boiling point. Values in \({ }^{\circ} \mathrm{C}\) are 0,11 , and 97 . (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\)

Short answer

Question: Arrange the following compounds in order of increasing boiling points: (a) CH3CH2CH2OH, (b) CH3CH2-O-CH3, (c) CH3CH2CH2CH3. Answer: The order of the compounds with increasing boiling points is: (c) < (b) < (a), with boiling points of 0°C, 11°C, and 97°C, respectively.

Step-by-step solution

Identify the types of intermolecular forces

There are three main types of IMFs: London dispersion forces (LDF), dipole-dipole forces, and hydrogen bonding. LDF is present in all molecules, whereas dipole-dipole forces and hydrogen bonding are present in polar molecules and specific molecules containing O-H, N-H, or F-H bonds, respectively.

Identify the IMFs present in each compound

(a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\): The OH group makes this molecule polar, so it has dipole-dipole forces and LDF. Moreover, since it has an O-H bond, hydrogen bonding is also present. (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{3}\): The O atom makes this molecule polar, so it has dipole-dipole forces and LDF. However, since there's no O-H bond, it lacks hydrogen bonding. (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\): This molecule is nonpolar, so only LDF is present.

Compare the strengths of the intermolecular forces

Hydrogen bonding is the strongest IMF, followed by dipole-dipole forces, and LDF is the weakest. So, we can expect the compound with hydrogen bonding to have the highest boiling point, followed by the compound with dipole-dipole forces, and then the compound with only LDF.

Match the boiling points to the compounds based on the IMF strength

The given boiling points are 0, 11, and 97 \({}^{\circ} \mathrm{C}\). Since compound (a) has hydrogen bonding, it should have the highest boiling point, 97 \({}^{\circ} \mathrm{C}\). Since compound (b) has dipole-dipole forces, it should have the medium boiling point, 11 \({}^{\circ} \mathrm{C}\). Compound (c), which has only LDF, should have the lowest boiling point, 0 \({}^{\circ} \mathrm{C}\).

Arrange the compounds in order of increasing boiling points

Based on our analysis, the order of the compounds with increasing boiling points is: (c) < (b) < (a), with boiling points of 0 \({}^{\circ} \mathrm{C}\), 11 \({}^{\circ} \mathrm{C}\), and 97 \({}^{\circ} \mathrm{C}\), respectively.

Key concepts

Boiling Point

The boiling point of a compound is the temperature at which its liquid form turns to gas. It's a critical property, indicating how strong the forces are between molecules. The stronger these forces are, the more energy (or higher temperature) is required for molecules to escape into the gas phase.
This makes boiling point a key indicator of intermolecular force strength.

• When a compound has strong intermolecular forces, it needs more heat to break these forces, thus having a high boiling point.
• For weaker intermolecular forces, a lower amount of heat suffices to separate the molecules, resulting in a lower boiling point.

By arranging compounds based on their boiling points, we can deduce the relative strength of the intermolecular forces present.

Hydrogen Bonding

Hydrogen bonding occurs when hydrogen is bonded to strongly electronegative atoms such as oxygen, nitrogen, or fluorine. This type of bonding is much stronger than other types of intermolecular forces due to the significant dipole created by the hydrogen's positive charge and the electronegative atom's negative charge.

• In our example, compound (a), \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\), exhibits hydrogen bonding because it has an O-H bond.
• This strong bond requires more heat to break, explaining why compounds with hydrogen bonding often have higher boiling points compared to those with other intermolecular forces.

Understanding hydrogen bonding helps in predicting and explaining why certain substances have unique properties, such as water's unusually high boiling point.

Dipole-Dipole Forces

Dipole-dipole forces are a type of intermolecular force present in polar molecules. They arise from the attraction between the positive end of one polar molecule and the negative end of another. These forces are stronger than London dispersion forces but weaker than hydrogen bonds.

• Compound (b), \(\mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{3}\), has dipole-dipole forces due to its polar nature.
• Even though it doesn't have hydrogen bonds, the permanent dipoles resulting from its molecular structure mean it still has moderate intermolecular attractions.

Dipole-dipole forces help in determining the boiling point of polar molecules, generally leading to higher boiling points than nonpolar compounds due to these relatively stronger forces.