Question

Which compound in each group is expected to have the strongest intermolecular forces? (a) \(\mathrm{CH}_{2} \mathrm{CH}_{2}, \mathrm{CH}_{3} \mathrm{CHCH}_{2}, \mathrm{CH}_{3} \mathrm{CHCHCH}_{3}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}, \mathrm{CH}_{3} \mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\), \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\)

Short answer

The compound \(\mathrm{CH}_{3} \mathrm{CHCHCH}_{3}\) in group (a) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2}\mathrm{CH}_{3}\) in group (b) are predicted to have the strongest intermolecular forces.

Step-by-step solution

Analyze Group (a)

The first group (a) consists of three compounds: \(\mathrm{CH}_{2} \mathrm{CH}_{2}\), \(\mathrm{CH}_{3} \mathrm{CHCH}_{2}\), and \(\mathrm{CH}_{3} \mathrm{CHCHCH}_{3}\). All these molecules possess only nonpolar C-H bonds, meaning they will only exhibit London dispersion forces. The strength of these intermolecular forces depends on the size and shape of the molecules. The larger the surface area the molecule can create, the stronger its potential dispersion forces will be. So, we find \(\mathrm{CH}_{3} \mathrm{CHCHCH}_{3}\) has the longest carbon chain, and therefore, it is expected to possess the strongest intermolecular forces.

Analyze Group (b)

The second group (b) consists of the compounds: \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2}\mathrm{CH}_{3}\), \(\mathrm{CH}_{3} \mathrm{C}(\mathrm{CH}_{3})_{3}\), and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{CH}_{3})_{2}\). Once more, these molecules only consist of nonpolar C-H bonds, therefore only experiencing London dispersion forces. Hence it is reasoned in the same manner as group (a). The more extended the molecule, the more potent the forces. In this group, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2}\mathrm{CH}_{3}\) is a more 'linear' molecule, leading to a larger surface area and thus stronger dispersion forces.

Key concepts

London Dispersion Forces

When considering the influence of intermolecular forces on the properties of molecules, London dispersion forces hold a foundational spot, especially for nonpolar molecules. These forces, also known as instantaneous dipole-induced dipole attractions, are the weakest form of van der Waals interactions. Contrary to a common misconception, they are present in all molecular substances, although their effects are most profound in those lacking stronger forces such as hydrogen bonds or ionic interactions.

London dispersion forces originate from temporary fluctuations in electron distribution within atoms and nonpolar molecules, which create an instantaneous dipole. This fleeting dipole then induces a dipole in a neighboring atom or molecule, leading to a transient attraction. The strength of London dispersion forces increases with the number of electrons in a molecule because more electrons equate to greater fluctuations. Therefore, heavier and larger atoms and molecules typically exhibit stronger dispersion forces.

These forces are highly significant in determining the physical properties like boiling and melting points, viscosity, and solubility of nonpolar compounds. Students should remember that as the only intermolecular force present in nonpolar hydrocarbons, London dispersion forces play a vital role in their behavior, and molecules with larger electron clouds will generally have stronger London forces.

Molecular Surface Area

Another critical factor that dictates the strength of London dispersion forces is the molecular surface area. When molecules come into close contact, the molecular surface area available for interactions influences the intensity of induced dipole attractions. A larger surface area allows more opportunities for the temporary dipoles to align across the molecule's surface and engage in more substantial interactions.

Consider these forces as invisible 'hands' gently holding the molecules together; a larger 'hand' (the surface area) means a stronger 'grip' (London dispersion force). This concept is crucial for understanding why longer and more linear molecules generally have higher boiling and melting points than their more branched counterparts. For instance, a straight-chain alkane will have stronger London dispersion forces and thus higher boiling points than a branched alkane with equivalent molecular weight. This is because the branched structure reduces the surface area available for intermolecular contact, thereby diminishing the extent of the dispersion forces.

In the exercise provided, the analysis of molecular geometry is pivotal in determining which compounds exhibit stronger dispersion forces based on their projected surface area when packed together. As educators, it's necessary to ensure that students grasp the geometric considerations that lead to differences in intermolecular interactions.

Nonpolar C-H Bonds

Compounds characterized by nonpolar carbon-hydrogen (C-H) bonds are a staple in the study of organic chemistry. These bonds are considered nonpolar because the difference in electronegativity between carbon and hydrogen is marginal, preventing a significant charge separation across the bond. This lack of polarity means that molecules composed solely of C-H bonds do not exhibit dipole-dipole or hydrogen bonding interactions, leaving London dispersion forces as the sole connectors between neighboring molecules.

In our exercise, nonpolar C-H bonds play a central role since all compounds listed are hydrocarbons, comprising only of carbon and hydrogen. Because these molecules cannot rely on the stronger intermolecular forces available to polar molecules, the London dispersion forces take center stage in their properties and behaviors. Educators must underscore the importance of recognizing molecular polarity and connecting it with the types of intermolecular forces that might be present. Reinforcing the idea that nonpolar molecules like hydrocarbons only exhibit London dispersion forces is essential for students to predict the physical properties and reactivity correctly.