Chapter 11: Problem 26

Rationalize the difference in boiling points in each pair: (a) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{O}\left(-23^{\circ} \mathrm{C}\right)$ and $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \quad\left(78^{\circ} \mathrm{C}\right),$ (b) $\mathrm{CO}_{2}\left(-78.5^{\circ} \mathrm{C}\right)$ and $\mathrm{CS}_{2}\left(46.2^{\circ} \mathrm{C}\right),(\mathbf{c}) \mathrm{CH}_{3} \mathrm{COCH}_{3}\left(50.5^{\circ} \mathrm{C}\right)$ and $\mathrm{CH}_{3} \mathrm{COOH}\left(101^{\circ} \mathrm{C}\right)$

Short Answer

Expert verified

The differences in boiling points can be rationalized by the intermolecular forces present in each pair of compounds: (a) Ethanol has a higher boiling point than dimethyl ether due to the presence of hydrogen bonds in addition to dipole-dipole interactions. (b) Carbon disulfide has a higher boiling point than carbon dioxide due to stronger van der Waals forces, resulting from its larger, more polarizable atoms. (c) Acetic acid has a higher boiling point than acetone due to the presence of hydrogen bonds in addition to dipole-dipole interactions.

Step by step solution

For pair (a), we have dimethyl ether $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{O}$ and ethanol $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$. Both of these compounds are polar and thus have dipole-dipole interactions. However, ethanol also has an -OH group, which can form hydrogen bonds. Dimethyl ether does not have any -OH groups and hence cannot form hydrogen bonds. #Step 2: Explain the difference in boiling points for pair (a)#

Boiling occurs when molecules have enough energy to overcome the intermolecular forces holding them together in the liquid state. Since ethanol can form hydrogen bonds, which are stronger than dipole-dipole interactions, it requires more energy to break these interactions and boil. This is why ethanol has a higher boiling point ($78^{\circ} \mathrm{C}$) than dimethyl ether ($-23^{\circ} \mathrm{C}$). #Step 3: Identify intermolecular forces in pair (b)#

For pair (b), we have carbon dioxide $\mathrm{CO}_{2}$ and carbon disulfide $\mathrm{CS}_{2}$. Both of these are linear molecules, and due to their symmetry, they have no net dipole moment and thus no dipole-dipole interactions. However, carbon disulfide has larger, more polarizable sulfur atoms, which results in stronger van der Waals (London dispersion) forces compared to carbon dioxide. #Step 4: Explain the difference in boiling points for pair (b)#

Since $\mathrm{CS}_{2}$ has stronger van der Waals forces due to its larger, more polarizable atoms, it requires more energy to overcome these forces to boil. This is why carbon disulfide has a higher boiling point ($46.2^{\circ} \mathrm{C}$) than carbon dioxide ($-78.5^{\circ} \mathrm{C}$). #Step 5: Identify intermolecular forces in pair (c)#

For pair (c), we have acetone $\mathrm{CH}_{3} \mathrm{COCH}_{3}$ and acetic acid $\mathrm{CH}_{3} \mathrm{COOH}$. Both of these compounds are polar and have dipole-dipole interactions. However, acetic acid has an -OH group, which can form hydrogen bonds, while acetone does not have any -OH groups and hence cannot form hydrogen bonds. #Step 6: Explain the difference in boiling points for pair (c)#

Since acetic acid can form hydrogen bonds, which are stronger than dipole-dipole interactions, it requires more energy to break these interactions and boil. This is why acetic acid has a higher boiling point ($101^{\circ} \mathrm{C}$) than acetone ($50.5^{\circ} \mathrm{C}$).