Question

Which member of the following pairs has the stronger intermolecular dispersion forces: (a) \(\mathrm{Br}_{2}\) or \(\mathrm{O}_{2}\), (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{SH}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{SH}\), (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\) or \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl} ?\)

Short answer

(a) \(\mathrm{Br}_{2}\) has stronger intermolecular dispersion forces. (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{SH}\) has stronger intermolecular dispersion forces. (c) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}\) has stronger intermolecular dispersion forces.

Step-by-step solution

1. Comparing Dispersion Forces for \(\mathrm{Br}_{2}\) and \(\mathrm{O}_{2}\)

Comparing the sizes and the number of electrons, we find that \(\mathrm{Br}_{2}\) molecules are larger than \(\mathrm{O}_{2}\) molecules as they have more electrons. This means they will have stronger dispersion forces. #Answer: (a) \(\mathrm{Br}_{2}\) has stronger intermolecular dispersion forces.

2. Comparing Dispersion Forces for \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{SH}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{SH}\)

Since both molecules have the same functional group, the difference lies in the size of alkyl chains. We find that the second molecule has an additional CH\(_2\) group, which makes the molecule larger and has more electrons. This results in stronger dispersion forces. #Answer: (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{SH}\) has stronger intermolecular dispersion forces.

3. Comparing Dispersion Forces for \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\) and \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}\)

Both molecules have the same molecular formula: C\(_4\)H\(_9\)Cl. However, they have different molecular structures. In this case, the difference in dispersion forces would depend on the molecular shape and surface area available for contact. The two molecules have similar shapes, but the second molecule has more branching which gives a greater surface area to contact with other molecules, resulting in stronger dispersion forces. #Answer: (c) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}\) has stronger intermolecular dispersion forces.

Key concepts

Molecular Size

Molecular size plays a critical role in determining the strength of intermolecular dispersion forces. As molecules increase in size, they typically possess more atoms, resulting in a larger electron cloud. This expanded electron cloud facilitates stronger instantaneous dipoles, enhancing dispersion forces.
For instance, when comparing bromine (\(\mathrm{Br}_{2}\)) and oxygen (\(\mathrm{O}_{2}\)), bromine has larger atoms and more electrons than oxygen, leading to more substantial dispersion forces.
In general, larger molecules have a greater capacity to polarize, thus enhancing the strength of their dispersion forces, which are particularly significant in larger, more complex molecular systems.

Electron Count

The number of electrons in a molecule directly influences the strength of intermolecular dispersion forces. More electrons lead to increased opportunities for temporary dipole formation due to fluctuations in electron distribution.
When comparing molecules like \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{SH}\) and \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{SH}\), the latter has more electrons due to an extra \(\mathrm{CH}_{2}\) group. This additional electron cloud presence significantly increases the molecule's dispersion forces.
Remember, the more electrons a molecule contains, the stronger its dispersion forces become, due to the more pronounced temporary dipoles.

Molecular Branching

Molecular branching affects the surface area of a molecule, which in turn influences dispersion forces. Molecules with more branching typically have a reduced surface area available for close contact with other molecules. However, slight branching can actually optimize the available surface area for interactions.
In comparing \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{Cl}\) with \(\left(\mathrm{CH}_{3}\right)_{2}\mathrm{CHCl}\), the branched molecule can offer a more efficient packing and increased surface area for intermolecular contacts despite having more branches. This optimized surface interaction increases dispersion forces.
Remember that the way a molecule presents itself to potential interacting partners can often override simple size considerations in determining dispersion force strength.

Functional Groups

While dispersion forces are largely influenced by the above factors, the presence of functional groups can fine-tune these interactions. Functional groups are specific groups of atoms within molecules that have characteristic properties.
In the exercise example, both compared thiol molecules (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{SH}\) and \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{SH}\)) share the same sulfhydryl (-SH) group, which does not largely vary dispersion forces on its own. However, the group's presence can impact other types of intermolecular forces, like hydrogen bonding.
It's essential to recognize that while functional groups primarily dictate chemical reactivity and specific types of other interactions, their uniform presence in a comparison highlights the role of molecular size and electron distribution in determining dispersion force differences.