Question

Explain why \(\mathrm{CH}_{3} \mathrm{CH}_{3}\) has a boiling point of \(-88.6^{\circ} \mathrm{C}\), but the boiling point of \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) is \(-6.3^{\circ} \mathrm{C}\).

Short answer

The boiling point of \(\mathrm{CH}_{3}\mathrm{NH}_{2}\) is higher than that of \(\mathrm{CH}_{3} \mathrm{CH}_{3}\) because \(\mathrm{CH}_{3}\mathrm{NH}_{2}\) has stronger intermolecular forces due to dipole-dipole forces and particularly hydrogen bond compared to \(\mathrm{CH}_{3} \mathrm{CH}_{3}\) which only has London dispersion forces.

Step-by-step solution

Identify the Molecular Type for both Compounds

When examining these molecules, it's found that \(\mathrm{CH}_{3} \mathrm{CH}_{3}\) is a nonpolar molecule while \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) is a polar molecule.

Determine the Intermolecular Forces

Since \(\mathrm{CH}_{3} \mathrm{CH}_{3}\) is nonpolar, it has London dispersion (LD) forces. On the other hand, \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) is a polar molecule, and hence, in addition to the LD forces, it also has dipole-dipole forces. Additionally, because nitrogen is present in \(\mathrm{CH}_{3} \mathrm{NH}_{2}\), it can form hydrogen bonds, a specific and much stronger type of dipole-dipole forces.

Compare Intermolecular Forces and Boiling Points

The strength of the intermolecular forces determines the boiling point of a substance. The stronger the forces, the more energy that is needed to break them, thereby raising the boiling point. Hence, \(\mathrm{CH}_{3} \mathrm{CH}_{3}\) having only LD forces has a lower boiling point, \(-88.6^{\circ} \mathrm{C}\) than \(\mathrm{CH}_{3}\mathrm{NH}_{2}\) which has LD forces, dipole-dipole forces and hydrogen bonds, therefore boils at a higher temperature, \(-6.3^{\circ} \mathrm{C}\).

Conclusion

It can be concluded that the greater number and stronger types of intermolecular forces in \(\mathrm{CH}_{3}\mathrm{NH}_{2}\) are responsible for its higher boiling point compared to \(\mathrm{CH}_{3} \mathrm{CH}_{3}\).

Key concepts

Molecular Polarity

Molecular polarity is a term used to describe how unevenly electrons are distributed across a molecule. This uneven distribution of electron density gives one end of the molecule a slight negative charge, and the other end a slight positive charge, creating what is known as a dipole. Polar molecules, like \texttt{CH}\(_3\)\texttt{NH}\(_2\), have a significant difference in electronegativity between the atoms, leading to a dipole moment. Nonpolar molecules, such as \texttt{CH}\(_3\)\texttt{CH}\(_3\), have a more even distribution of electrons, resulting in no overall dipole moment.

Polarity is a key factor in determining the kinds of intermolecular forces a molecule can participate in, which in turn affects properties like boiling points. In the given exercise, the polarity of \texttt{CH}\(_3\)\texttt{NH}\(_2\) allows it to engage in stronger intermolecular forces compared to those in nonpolar \texttt{CH}\(_3\)\texttt{CH}\(_3\), resulting in a higher boiling point.

London Dispersion Forces

London dispersion forces, also known as induced dipole-induced dipole interactions, are the weakest type of intermolecular force. They are present in all molecules, whether polar or nonpolar, including noble gases and nonpolar molecules like \texttt{CH}\(_3\)\texttt{CH}\(_3\). These forces result from the momentary fluctuations in the electron distribution within molecules, which produce a temporary dipole.

Even though they are weak, London dispersion forces become stronger with larger and heavier atoms or molecules because larger electron clouds are more easily distorted. In the case of \texttt{CH}\(_3\)\texttt{CH}\(_3\), London dispersion forces are the only type of intermolecular attraction present due to its nonpolar nature, contributing to its relatively low boiling point.

Hydrogen Bonds

Hydrogen bonds are a special type of dipole-dipole interaction and are much stronger than London dispersion forces. These bonds occur when hydrogen is attached to a highly electronegative atom such as nitrogen, oxygen, or fluorine, and it comes into close proximity with a lone pair on another electronegative atom.

In \texttt{CH}\(_3\)\texttt{NH}\(_2\), the hydrogen atoms bonded to nitrogen create a significant polarity due to the large electronegativity difference between nitrogen and hydrogen. As a result, \texttt{CH}\(_3\)\texttt{NH}\(_2\) is able to form strong hydrogen bonds with neighboring molecules, accounting for its much higher boiling point compared to \texttt{CH}\(_3\)\texttt{CH}\(_3\). This is a classic demonstration of how hydrogen bonding can dramatically affect the physical properties of a substance.

Dipole-Dipole Interactions

Dipole-dipole interactions are intermolecular forces that occur between two polar molecules. Unlike London dispersion forces that occur in all molecules, dipole-dipole interactions specifically require a molecule to be polar, possessing a positive and a negative end.

This type of force is stronger than London dispersion forces but weaker than hydrogen bonds. When molecules like \texttt{CH}\(_3\)\texttt{NH}\(_2\) are in close proximity, the positive end of one molecule is attracted to the negative end of another molecule, creating an intermolecular attraction that affects the boiling point. Due to these interactions, polar molecules generally have higher boiling points than nonpolar ones with similar molecular weights. As shown in our example, the polar \texttt{CH}\(_3\)\texttt{NH}\(_2\) molecules can engage in dipole-dipole interactions along with hydrogen bonding, contributing to their higher boiling point over nonpolar \texttt{CH}\(_3\)\texttt{CH}\(_3\) which can only display London dispersion forces.