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Butane and 2 -methylpropane, whose space-filling models are shown here, are both nonpolar and have the same molecular formula, \(\mathrm{C}_{4} \mathrm{H}_{10}\), yet butane has the higher boiling point \(\left(-0.5^{\circ} \mathrm{C}\right.\) compared to \(\left.-11.7{ }^{\circ} \mathrm{C}\right)\). Explain.

Short Answer

Expert verified
Butane has a higher boiling point because its straight-chain structure results in stronger dispersion forces, requiring more energy to change state.

Step by step solution

01

Understanding Molecular Structure

Both butane and 2-methylpropane have the same molecular formula, \(\mathrm{C}_4\mathrm{H}_{10}\), but their structures differ. Butane is a straight-chain alkane, while 2-methylpropane is a branched structure.
02

Interaction Between Molecules

Because butane is a straight-chain molecule, it has a larger surface area in contact with other butane molecules, leading to slightly stronger dispersion forces compared to the more compact, branched structure of 2-methylpropane.
03

Boiling Point Correlation

Stronger intermolecular forces in butane require more energy (in the form of heat) to overcome when changing from a liquid to a gas, resulting in a higher boiling point compared to 2-methylpropane.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecular Structure
When we talk about molecular structure, we're referring to the arrangement of atoms within a molecule. Even if two substances have the same molecular formula, their structure can make a significant difference. In the case of butane and 2-methylpropane, both share the same molecular formula \( ext{C}_4 ext{H}_{10} \). But it's their structural geometry that distinguishes them. Butane is a linear molecule, which means it forms a line-like shape, while 2-methylpropane is branched.

This structural difference affects how closely these molecules pack together. Straight-chain molecules like butane often have more contact with each other compared to branched molecules. This is important when we consider how these molecules interact at the molecular level.
Dispersion Forces
Dispersion forces, also known as London dispersion forces, are a type of intermolecular force. These are weak attractive forces that arise due to temporary dipoles induced when atoms shift in electron density. This effect is universal and occurs in all atoms and molecules, but it is particularly relevant in nonpolar substances like butane and 2-methylpropane.

In butane, the linear configuration offers a more extended surface area for these temporary dipoles to interact. This results in increased dispersion forces between butane molecules compared to the more compact and less accessible structure of 2-methylpropane. Supporting stronger interactions among the molecules, these forces increase the energy needed to separate them, leading to higher boiling points.
Intermolecular Forces
Intermolecular forces are forces of attraction between molecules, which can greatly affect a substance's physical properties. Boiling point is one of these properties influenced by intermolecular forces. The main types of intermolecular forces are London dispersion forces, dipole-dipole interactions, and hydrogen bonds.

For butane and 2-methylpropane, which are both nonpolar, the most significant intermolecular force at play is the London dispersion force. Since butane provides an elongated shape, it affords stronger dispersion forces compared to the more compact 2-methylpropane. This means that as butane molecules attract each other more strongly, it takes more heat energy to convert them from a liquid state to a gaseous state. Therefore, butane has a higher boiling point than 2-methylpropane due to stronger intermolecular attractions.

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Most popular questions from this chapter

Suppose you have two colorless molecular liquids \(A\) and \(B\) whose boiling points are \(78^{\circ} \mathrm{C}\) and \(112^{\circ} \mathrm{C}\) respectively and both are at atmospheric pressure. Which of the following statements is correct? For each statement that is not correct, modify the statement so that it is correct. (a) Both A and B are liquids with identical vapor pressure at room temperature of \(25^{\circ} \mathrm{C} .(\mathbf{b})\) Liquid \(\mathrm{A}\) must consist of nonpolar molecules with lower molecular weight than B. \((\mathbf{c})\) Both liquids \(A\) and \(B\) have higher total intermolecular forces than water. (d) Liquid \(A\) is more volatile than liquid B because it has a lower boiling point. (e) At \(112^{\circ} \mathrm{C}\) both liquids have a vapor pressure of 1 atm.

As the intermolecular attractive forces between molecules increase in magnitude, do you expect each of the following to increase or decrease in magnitude? (a) Vapor pressure, (b) heat of vaporization, (c) boiling point, (d) freezing point, (e) viscosity, (f) surface tension, \((\mathrm{g})\) critical temperature.

(a) Do you expect the viscosity of glycerol, \(\mathrm{C}_{3} \mathrm{H}_{5}(\mathrm{OH})_{3}\), to be larger or smaller than that of 1 -propanol, \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH} ?\) (b) Explain. [Section 11.3\(]\)

(a) Which is generally stronger, intermolecular interactions or intramolecular interactions? (b) Which of these kinds of interactions are broken when a liquid is converted to a gas?

Indicate whether each statement is true or false: (a) The liquid crystal state is another phase of matter, just like solid, liquid, and gas. (b) Liquid crystalline molecules are generally spherical in shape, (c) Molecules that exhibit a liquid crystalline phase do so at well-defined temperatures and pressures. (d) Molecules that exhibit a liquid crystalline phase show weaker-than- expected intermolecular forces. (e) Molecules containing only carbon and hydrogen are likely to form liquid crystalline phases. (f) Molecules can exhibit more than one liquid crystalline phase.

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