Question

The following pairs of compounds have about the same molar mass. Predict which compound in each pair has the higher boiling point. (a) \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{~F}\)

Short answer

(a) Ethanol (\(\mathrm{CH}_{3} \mathrm{CH}_{2}\mathrm{OH}\)) has a higher boiling point than dimethyl ether. (b) Propylamine (\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\)) has a higher boiling point than 1-fluoropropane.

Step-by-step solution

Identify Intermolecular Forces

To predict boiling points, consider the types of intermolecular forces present in each compound. Generally, stronger intermolecular forces correspond to higher boiling points.

Analyze Part (a) Compounds

For (a), compare \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) (dimethyl ether) and \(\mathrm{CH}_{3} \mathrm{CH}_{2}\mathrm{OH}\) (ethanol). Ethanol can form hydrogen bonds due to the presence of an OH group, which are strong intermolecular forces. Dimethyl ether can form dipole-dipole interactions but not hydrogen bonds, making its forces weaker than ethanol's.

Conclude Part (a) Result

Since ethanol can form hydrogen bonds and dimethyl ether cannot, ethanol (\(\mathrm{CH}_{3} \mathrm{CH}_{2}\mathrm{OH}\)) has a higher boiling point than dimethyl ether (\(\mathrm{CH}_{3} \mathrm{OCH}_{3}\)).

Analyze Part (b) Compounds

For (b), compare \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) (propylamine) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{~F}\) (1-fluoropropane). Propylamine can form hydrogen bonds because it has an NH group, whereas 1-fluoropropane can engage in dipole-dipole interactions due to the electronegativity of fluorine but cannot form hydrogen bonds.

Conclude Part (b) Result

Since propylamine can form hydrogen bonds, which are generally stronger than the dipole-dipole interactions in 1-fluoropropane, propylamine (\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\)) has a higher boiling point than 1-fluoropropane (\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{~F}\)).

Key concepts

Intermolecular Forces

Understanding intermolecular forces is essential for predicting the boiling points of substances. These forces are attractions that occur between molecules. They influence how molecules behave in terms of sticking together or pulling apart, ultimately affecting their physical properties. Different types of intermolecular forces vary in strength, determining how much energy is required for molecules to separate during boiling.
There are several types of intermolecular forces to consider:

• Dispersion forces: Present in all molecules, these are the weakest forces and increase with molecular size.
• Dipole-dipole interactions: Occur between polar molecules with permanent dipoles. These are stronger than dispersion forces.
• Hydrogen bonding: The strongest among the van der Waals forces, occurring when hydrogen is bonded to highly electronegative atoms like oxygen, nitrogen, or fluorine.

Recognizing the type of intermolecular forces in a compound allows for the prediction of its boiling point.

Hydrogen Bonding

Hydrogen bonding plays a significant role in determining the boiling points of molecules. It is an exceptionally strong type of dipole-dipole interaction. Hydrogen bonds form when hydrogen is covalently bonded to nitrogen, oxygen, or fluorine, as these elements have high electronegativity values. This leads to a large dipole, creating a strong attraction between molecules.
For example, ethanol (\(\mathrm{CH_{3}CH_{2}OH}\)) and propylamine (\(\mathrm{CH_{3}CH_{2}CH_{2}NH_{2}}\)) both contain groups capable of hydrogen bonding due to the presence of OH and NH groups, respectively. This makes these compounds capable of forming strong intermolecular attractions, elevating their boiling points.
Whenever hydrogen bonding is an option in a compound, anticipate higher boiling points compared to similar molecules lacking this feature.

Dipole-Dipole Interactions

Dipole-dipole interactions arise in polar molecules, where the positive end of one molecule is attracted to the negative end of another. These forces are stronger than dispersion forces but weaker than hydrogen bonds. Polar molecules have uneven charge distributions, creating a permanent dipole moment.
In the original exercise, dimethyl ether (\(\mathrm{CH_{3}OCH_{3}}\)) and 1-fluoropropane (\(\mathrm{CH_{3}CH_{2}CH_{2}F}\)) both have dipole-dipole interactions due to molecular polarity. However, their boiling points are lower than their counterparts capable of hydrogen bonding.
The strength and presence of dipole-dipole interactions can help estimate boiling behaviors when hydrogen bonds are absent.

Molar Mass Comparison

While intermolecular forces primarily dictate boiling points, molar mass also plays a role. Generally, compounds with larger molar masses have higher boiling points. This occurs because larger molecules have more electrons, leading to increased dispersion forces.
However, when comparing compounds of similar molar masses, the type of intermolecular force becomes more crucial. For instance, both comparative pairs in the exercise have similar molar masses—making factors like hydrogen bonding and dipole-dipole interactions more significant in determining the boiling points.
Ultimately, while molar mass provides a baseline, the presence and strength of intermolecular forces primarily account for difference in boiling points between similarly sized molecules.