Problem 138 - General Chemistry

Question

Consider the following two compounds: \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} \quad\) 1-pentanol \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3} \quad\) hexane a. What are the different types of intermolecular forces that exist in each compound? b. One of these compounds has a normal boiling point of \(69^{\circ} \mathrm{C}\), and the other has a normal boiling point of \(138^{\circ} \mathrm{C} .\) What is the normal boiling point of hexane? Explain. C. One of these compounds has a viscosity of \(0.313 \mathrm{~g} /(\mathrm{cm} \cdot \mathrm{s})\) and the other has a viscosity of \(2.987 \mathrm{~g} /(\mathrm{cm} \cdot \mathrm{s}) .\) Assign viscosities to 1 -pentanol and to hexane. d. Consider the compound \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\). Where do you think the boiling point of this compound might lie: above both 1 -pentanol and hexane, intermediate between these two compounds, or below both of these two compounds?

Step-by-Step Solution

Answer

a. 1-pentanol has hydrogen bonds, dipole-dipole forces, and London forces; hexane only has London forces. b. Hexane: 69°C. c. Hexane: 0.313 g/(cm s), 1-pentanol: 2.987 g/(cm s). d. Boiling point intermediate.

Step 1: Identify Intermolecular Forces in Compounds

**For 1-pentanol (\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\)**):**- **Hydrogen bonding**: Due to the presence of an \(-OH\) group, it can form hydrogen bonds.- **Dipole-dipole interactions**: The molecule is polar because of the hydroxyl group.- **London dispersion forces**: Present in all molecular compounds due to the electron clouds.**For hexane (\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\)**):**- **London dispersion forces**: Hexane is nonpolar and only has dispersion forces.

Step 2: Determine the Boiling Point of Hexane

Since hexane only has London dispersion forces, while 1-pentanol has stronger hydrogen bonding and dipole-dipole interactions, hexane is expected to have a lower boiling point. Therefore, hexane's boiling point is \(69^{\circ} \mathrm{C}\).

Step 3: Assign Viscosities to Compounds

1-pentanol is expected to have higher viscosity than hexane due to hydrogen bonding, which creates more internal resistance to flow. Thus, 1-pentanol has a viscosity of \(2.987 \, \mathrm{g} / (\mathrm{cm}\cdot\mathrm{s})\), and hexane has a viscosity of \(0.313 \, \mathrm{g} / (\mathrm{cm}\cdot\mathrm{s})\).

Step 4: Estimate Boiling Point of Chloropentane

Chloropentane (\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\)**) has dipole-dipole interactions due to the polar C-Cl bond, but lacks hydrogen bonding. Its boiling point is likely between hexane and 1-pentanol, since it has stronger forces than hexane but weaker than 1-pentanol.

Key Concepts

1-pentanol

1-pentanol is an organic compound featuring a five-carbon linear chain with a hydroxyl group ( -OH) at one end. This structure places it in the alcohol family of compounds. The presence of an -OH group gives 1-pentanol certain unique properties, most notably the ability to engage in hydrogen bonding.
This hydrogen bonding occurs when the hydrogen atom, attached to the highly electronegative oxygen, forms a weak bond with another electronegative atom (like oxygen or nitrogen) of a neighboring molecule.
In contrast to hydrocarbons like hexane, 1-pentanol is a polar molecule due to the presence of the -OH group, which creates a dipole moment. This polarity allows 1-pentanol to also participate in dipole-dipole interactions.

hydrogen bonding

Hydrogen bonding is a strong type of dipole-dipole attraction, distinct to molecules with -HX groups, where X is typically nitrogen, oxygen, or fluorine. In 1-pentanol, the hydrogen bonding is especially evident due to the -OH group.
This -OH group facilitates the formation of hydrogen bonds by allowing hydrogen to directly interact with electronegative atoms in other molecules.
Hydrogen bonds are stronger than regular dipole-dipole interactions and much more robust than London dispersion forces. Consequently, substances like 1-pentanol, which feature hydrogen bonding, often have higher boiling points compared to similar non-bonding substances. This bonding increases both the viscosity and the boiling point because it requires more energy to overcome the attractive forces during boiling or when a liquid flows.

London dispersion forces

London dispersion forces are a type of van der Waals forces. They are the weakest intermolecular forces and are present in all atoms and molecules due to the movement of electrons, creating temporary dipoles.
These forces are most significant in nonpolar substances, like hexane, where there are no permanent dipoles. In larger molecules with greater molecular mass, like hexane compared to smaller hydrocarbons, the London dispersion forces become stronger due to larger electron clouds, which can polarize more easily. Despite this, they are much weaker than hydrogen bonds and dipole-dipole interactions. This explains why substances dominated by London dispersion forces, such as hexane, typically possess lower boiling points and viscosities than those that can form hydrogen bonds, like 1-pentanol.

hexane

Hexane is a straightforward example of a nonpolar hydrocarbon molecule. With six carbon atoms lined in a row and completed by hydrogen atoms, it features only C-C and C-H bonds, both of which are nonpolar.
The lack of polarity in hexane means it can only display London dispersion forces, the weakest of the intermolecular forces. Because of this, hexane has a relatively low boiling point (69°C) as there are no strong intermolecular attractions to overcome via heating, unlike with 1-pentanol which contains hydrogen bonding.
Furthermore, hexane’s viscosity is lower because, with only dispersion forces at play, its molecular chains slide past each other more readily compared to polar substances like 1-pentanol, which suffer internal resistance due to stronger attractions between molecules.