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Chapter 11: Problem 23

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 than 2-methylpropane due to its linear structure, which results in a larger surface area in contact between molecules and therefore stronger London dispersion forces. In contrast, the compact, branched structure of 2-methylpropane reduces the surface area in contact with neighboring molecules, leading to weaker London dispersion forces. The stronger intermolecular forces in butane result in its higher boiling point (\(-0.5^{\circ} \mathrm{C}\)) compared to 2-methylpropane (\(-11.7^{\circ} \mathrm{C}\)).

Step by step solution

01

Identify the molecular structure of butane and 2-methylpropane

Butane and 2-methylpropane both have the molecular formula \(\mathrm{C}_{4}\mathrm{H}_{10}\), but they have different structural formulas due to being structural isomers. Butane has a linear carbon chain: \( \mathrm{CH}_{3}\mathrm{-CH}_{2}\mathrm{-CH}_{2}\mathrm{-CH}_{3} \) 2-methylpropane has a branched carbon chain: \( \mathrm{CH}_{3} \) \( \, | \) \( \mathrm{CH}_{3}\mathrm{-C}\mathrm{-CH}_{3} \) \( \, | \) \( \mathrm{CH}_{3} \)
02

Identify the intermolecular forces acting on the molecules

Since both butane and 2-methylpropane are nonpolar molecules, the primary type of intermolecular force acting between their molecules is London dispersion forces. London dispersion forces are weak intermolecular forces resulting from temporary dipoles created due to fluctuations in the electron cloud surrounding the molecules. The larger the surface area of the molecule, the stronger the London dispersion forces will be.
03

Compare the strength of London dispersion forces in both isomers

Now let's compare butane and 2-methylpropane in terms of their London dispersion forces. Butane is a linear carbon chain, while 2-methylpropane has a more compact, branched structure. Due to being linear, butane molecules have a larger surface area in contact with each other, resulting in stronger London dispersion forces between the molecules. On the other hand, the compact and spherical-like structure of 2-methylpropane reduces the surface area in contact with its neighboring molecules, leading to weaker London dispersion forces.
04

Relate the strength of London dispersion forces to boiling points

The boiling point of a substance is related to the strength of the intermolecular forces acting on its molecules. The stronger the intermolecular forces, the higher the boiling point. In our case, as we have discussed before, butane has stronger London dispersion forces than 2-methylpropane.
05

Explain the difference in boiling points

Due to the linear structure of butane and the larger surface area in contact between its molecules, the London dispersion forces acting on butane are stronger than those acting on 2-methylpropane. This results in a higher boiling point for butane, which boils at \(-0.5^{\circ} \mathrm{C}\), compared to 2-methylpropane, which boils at \(-11.7^{\circ} \mathrm{C}\).

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Most popular questions from this chapter

(a) What is the relationship between surface tension and temperature? (b) What is the relationship between viscosity and temperature? (c) Why do substances with high surface tension also tend to have high viscosities?

True or false: (a) Molecules containing polar bonds must be polar molecules and have dipole-dipole forces. (b) For the halogen gases, the dispersion forces decrease while the boiling points increase as you go down the column in the periodic table. (c) In terms of the total attractive forces for a given substance, the more polar bonds there are in a molecule, the stronger the dipole-dipole interaction. \(\mathbf{d}\) ) All other factors being the same, total attractive forces between linear molecules are greater than those between molecules whose shapes are nearly spherical. (e) The more electronegative the atom, the more polarizable it is.

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