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Butane and 2 -methylpropane, whose space-filling models are shown, are both nonpolar and have the same molecular formula, 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 than 2-methylpropane because its straight-chain structure results in a larger surface area, leading to stronger London dispersion forces (LDFs) between butane molecules compared to the more compact, branched structure of 2-methylpropane. As stronger LDFs require more energy to break, butane's boiling point is higher (-0.5°C) than that of 2-methylpropane (-11.7°C).

Step by step solution

01

Identify the molecular formulas

First, let's identify the molecular formulas of the two compounds. Butane and 2-methylpropane both have the same molecular formula, which is C4H10.
02

Analyze the molecular structure

Next, let's analyze the molecular structures of butane and 2-methylpropane. Butane (C4H10) has a straight chain structure: CH3-CH2-CH2-CH3 2-methylpropane (also called isobutane) has a branched structure: CH3 | CH3-C-CH3 | CH3
03

Identify the types of forces at play

Because both butane and 2-methylpropane are nonpolar molecules, there are no dipole-dipole interactions between their molecules. The only forces between their molecules are London dispersion forces (LDFs), which are temporary attractive forces that arise due to instantaneous dipoles in the electron cloud surrounding the molecules.
04

Compare the London dispersion forces between the two molecules

The strength of LDFs depends on the size and shape of the molecules. In general, larger molecules with more electrons and more surface area have stronger LDFs. In the case of butane and 2-methylpropane, although they have the same molecular formula (C4H10), their different structures lead to different strengths of LDFs. Butane, with its longer, straight-chain structure, has a greater surface area than the more compact, branched structure of 2-methylpropane. This larger surface area leads to stronger LDFs between butane molecules compared to those between 2-methylpropane molecules.
05

Explain the difference in boiling points

The boiling point of a substance is the temperature at which its molecules have enough energy to break all intermolecular forces holding them together and change from a liquid to a gas. Since butane has stronger LDFs between its molecules compared to 2-methylpropane, more energy is required to break these forces and change butane from a liquid to a gas. As a result, butane has a higher boiling point (-0.5°C) than 2-methylpropane (-11.7°C).

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Key Concepts

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

Molecular Structure
Understanding molecular structure is key to noticing how molecules interact with each other. Butane (C4H10) has a linear structure: CH3-CH2-CH2-CH3. This structure spreads out more along a longer path than 2-methylpropane. On the other hand, 2-methylpropane (isobutane) follows a branched structure with a formula like:
  • CH3
    |
    CH3-C-CH3
    |
    CH3
Such a compact shape means the molecule occupies less surface area.

Differences in structure affect how closely molecules pack together. If molecules like butane line up well, they tend to have larger surface areas interacting with each other. This interaction plays a significant role in their physical properties, such as boiling points, by influencing how molecules stick together.
London Dispersion Forces
London Dispersion Forces (LDFs) are very important in explaining the interactions between nonpolar molecules like butane and 2-methylpropane. These forces result from temporary fluctuations in the electron distribution within a molecule, leading to instantaneous dipoles.

LDFs are the only forces at play between these nonpolar molecules, and their strength depends on two main factors:
  • Size of the electron cloud: Larger clouds can produce more significant temporary dipoles.
  • Surface area: Molecules with greater surface areas have stronger LDFs because there are more points of contact for interaction.
In the case of butane versus 2-methylpropane, although both have the same molecular formula, butane's linear configuration gives it a larger surface area. Hence, it displays stronger London Dispersion Forces compared to the more compact, branched 2-methylpropane.
Intermolecular Forces
Intermolecular forces are critical in determining a molecule's physical properties, like boiling points. They encompass different types of interactions, but in nonpolar molecules such as butane and 2-methylpropane, London Dispersion Forces dominate.

These forces might seem weak compared to other intermolecular forces, like hydrogen bonds, but they significantly impact how substances change state. The higher the intermolecular forces, the more energy is needed for molecules to escape each other's influence.
  • Higher LDFs require more heat energy, raising the boiling point.
  • Butane, with stronger interactions, boils at -0.5°C.
  • In contrast, 2-methylpropane's lower LDFs lead to a lower boiling point of -11.7°C.
Understanding these forces helps explain why seemingly similar molecules behave differently under the same conditions.

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Most popular questions from this chapter

For each of the following pairs of substances, predict which will have the higher melting point and indicate why: (a) \(\mathrm{Ar}, \mathrm{Xe} ;\) (b) \(\mathrm{SiO}_{2}, \mathrm{CO}_{2} ;\) (c) \(\mathrm{KBr}, \mathrm{Br}_{2} ;\) (d) \(\mathrm{C}_{6} \mathrm{Cl}_{6}, \mathrm{C}_{6} \mathrm{H}_{6}\)

What kinds of attractive forces exist between particles in (a) molecular crystals, (b) covalent-network crystals, (c) ionic crystals, (d) metallic crystals?

An element crystallizes in a body-centered cubic lattice. The edge of the unit cell is \(2.86 \AA\), and the density of the crystal is \(7.92 \mathrm{~g} / \mathrm{cm}^{3} .\) Calculate the atomic weight of the element.

Using the following list of normal boiling points for a series of hydrocarbons, estimate the normal boiling point for octane, \(\mathrm{C}_{8} \mathrm{H}_{18}:\) propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8},-42.1{ }^{\circ} \mathrm{C}\right)\), bu- tane \(\left(\mathrm{C}_{4} \mathrm{H}_{10},-0.5^{\circ} \mathrm{C}\right)\), pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}, 36.1^{\circ} \mathrm{C}\right)\), hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}, 68.7^{\circ} \mathrm{C}\right)\), heptane \(\left(\mathrm{C}_{7} \mathrm{H}_{16}, 98.4{ }^{\circ} \mathrm{C}\right)\). Explain the trend in the boiling points.

Suppose you have two colorless molecular liquids, one boiling at \(-84{ }^{\circ} \mathrm{C}\), the other at \(34{ }^{\circ} \mathrm{C}\), and both at atmospheric pressure. Which of the following statements is correct? For those that are not correct, modify the statement so that it is correct. (a) The higher-boiling liquid has greater total intermolecular forces than the other. (b) The lower boiling liquid must consist of nonpolar molecules. (c) The lower-boiling liquid has a lower molecular weight than the higher-boiling liquid. (d) The two liquids have identical vapor pressures at their normal boiling points. (e) At \(34^{\circ} \mathrm{C}\) both liquids have vapor pressures of \(760 \mathrm{~mm} \mathrm{Hg}\).

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