Problem 120 - Introductory Chemistry Atoms First

Question

There are three isomers that have the molecular formula \(\mathrm{C}_{5} \mathrm{H}_{12}\). Draw the structural formula for each one. Which do you expect to have the lowest boiling point? Explain.

Step-by-Step Solution

Answer

The three isomers of C5H12 are pentane, isopentane (2-methylbutane), and neopentane (2,2-dimethylpropane). The structural formulas for each isomer are: 1. Pentane: CH3-CH2-CH2-CH2-CH3 2. Isopentane: CH3-CH-CH3, with a CH2-CH3 branch 3. Neopentane: Central CH with four CH3 branches Neopentane is expected to have the lowest boiling point due to its compact structure and weaker intermolecular dispersion forces compared to pentane and isopentane.

Step 1: Identify the isomers of C5H12

The first step is to identify the isomers with the molecular formula C5H12. They are: 1. Pentane 2. Isopentane (also known as 2-methylbutane) 3. Neopentane (also known as 2,2-dimethylpropane)

Step 2: Draw the structural formula for each isomer

Next, let's draw the structural formula for each isomer. 1. Pentane: \( \displaystyle \mathrm{CH}_{3} -\mathrm{CH}_{2} -\mathrm{CH}_{2} -\mathrm{CH}_{2} -\mathrm{CH}_{3} \) 2. Isopentane (2-methylbutane): \( \displaystyle \mathrm{CH}_{3} -\mathrm{CH} -\mathrm{CH}_{3} \\ \hspace*{15pt} |\ \\ \hspace*{15pt} \mathrm{CH}_{2} -\mathrm{CH}_{3} \) 3. Neopentane (2,2-dimethylpropane): \( \displaystyle \mathrm{CH}_{3} \\ |\ \\ \mathrm{CH} - \mathrm{CH}_{3} \\ |\ \\ \mathrm{CH}_{3} \)

Step 3: Determine the isomer with the lowest boiling point

Boiling points of hydrocarbons are determined by the strength of the intermolecular forces between the molecules. In this case, the main intermolecular force is van der Waals (London dispersion) forces. These forces increase with the size and shape of the molecule, so molecules with more contact points and a larger surface area will have stronger dispersion forces and higher boiling points. Comparing the three isomers: 1. Pentane has a linear structure, which means it has the largest surface area and the strongest dispersion forces among the three isomers. 2. Isopentane has a branched structure but still has an extended shape and a relatively large surface area. 3. Neopentane is the most compact and has the smallest surface area, leading to weaker dispersion forces.

Step 4: Identify the isomer with the lowest boiling point and explain

Based on the size and shape of the three isomers, we can conclude that neopentane (2,2-dimethylpropane) is expected to have the lowest boiling point due to its compact structure and weaker intermolecular dispersion forces compared to pentane and isopentane.

Key Concepts

Structural Formula of Isomers

Understanding the structural formula of isomers is crucial because the unique arrangement of atoms within each isomer leads to differences in physical and chemical properties. Isomers are molecules with the same molecular formula but different structural arrangements. For instance, with the molecular formula \( \mathrm{C}_{5}\mathrm{H}_{12} \), the three isomers can be represented as pentane, isopentane, and neopentane.

These isomers differ in how the carbon atoms are connected to each other, resulting in various structural conformations:

• Pentane has a straight-chain structure with all carbon atoms connected in a row.
• Isopentane, also known as 2-methylbutane, has a branched structure with four carbon atoms in a row and a fifth carbon connected to the second one.
• Neopentane, or 2,2-dimethylpropane, has a more compact structure with three carbon atoms forming the main chain and two additional carbons attached to the middle carbon atom.

This variation in structure largely influences their physical properties, such as their boiling and melting points.

Boiling Point Comparison

When comparing the boiling points of hydrocarbon isomers, it's important to recognize how the structure affects this property. In general, straight-chain hydrocarbons typically have higher boiling points than their branched counterparts because they have a larger surface area that allows for greater van der Waals (London dispersion) forces, which are attractions between adjacent molecules.

In the case of \( \mathrm{C}_{5}\mathrm{H}_{12} \) isomers:

• Pentane has the highest boiling point because of its straight-chain shape, which facilitates extensive van der Waals interactions.
• Isopentane's branched structure leads to a somewhat lesser surface area and slightly weaker van der Waals forces, resulting in a lower boiling point when compared to pentane.
• Neopentane, with its highly compact and nearly spherical shape, has the lowest boiling point of the three isomers due to the least surface area for van der Waals forces to act upon.

Students should note that boiling point can also be influenced by other factors such as molecular weight and the presence of polar bonds or functional groups, which are not applicable for these particular hydrocarbon isomers.

Intermolecular Forces

Intermolecular forces are the attractions between molecules that affect their physical properties, like boiling points. There are several types of intermolecular forces, but the isomers of \( \mathrm{C}_{5}\mathrm{H}_{12} \) mainly experience van der Waals forces, aka London dispersion forces. These are the weakest and most temporary of the intermolecular forces, yet they are significantly influenced by the shape and size of the molecule.

The strength of van der Waals forces increases with:

• The length of the hydrocarbon chain: Longer chains have more surface contact and can experience stronger dispersion forces.
• The compactness of the molecule: More compact shapes reduce the surface area and, subsequently, the dispersion forces.

Given these factors, neopentane, with its compact shape, experiences the weakest dispersion forces and, therefore, has the lowest boiling point compared to pentane and isopentane. Understanding intermolecular forces is not just important for boiling points but also for predicting solubility, melting points, and the behavior of substances in various conditions.