---
title: "The structural formulas and normal boiling points of two isomers with the molecular formula \\(\\text{C}_5\\text{H}_{12}\\) are shown in the table below.  | Compound | Structural formula | Normal boiling point (\\(^\\circ\\text{C}\\)) | |—|—|—| | Pentane | \\(\\text{CH}_3\\text{CH}_2\\text{CH}_2\\text{CH}_2\\text{CH}_3\\) | \\(36.1\\) | | \\(2,2\\)-dimethylpropane | \\(\\text{C(CH}_3)_4\\) | \\(9.5\\) |  Which of the following statements best explains why pentane has a higher normal boiling point than \\(2,2\\)-dimethylpropane?"
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url: "https://nerd-notes.com/ubq/121456/"
date_modified: "2026-08-23T05:04:37+00:00"
---

# The structural formulas and normal boiling points of two isomers with the molecular formula \(\text{C}_5\text{H}_{12}\) are shown in the table below.

| Compound | Structural formula | Normal boiling point (\(^\circ\text{C}\)) |
|—|—|—|
| Pentane | \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3\) | \(36.1\) |
| \(2,2\)-dimethylpropane | \(\text{C(CH}_3)_4\) | \(9.5\) |

Which of the following statements best explains why pentane has a higher normal boiling point than \(2,2\)-dimethylpropane?

The structural formulas and normal boiling points of two isomers with the molecular formula \(\text{C}_5\text{H}_{12}\) are shown in the table below.

| Compound | Structural formula | Normal boiling point (\(^\circ\text{C}\)) |
|---|---|---|
| Pentane | \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3\) | \(36.1\) |
| \(2,2\)-dimethylpropane | \(\text{C(CH}_3)_4\) | \(9.5\) |

Which of the following statements best explains why pentane has a higher normal boiling point than \(2,2\)-dimethylpropane?

- **A.** The boiling point of pentane is higher because the \(\text{C}-\text{C}\) covalent bonds in its linear carbon chain are stronger and require more energy to break than the covalent bonds in \(2,2\)-dimethylpropane.
- **B.** The boiling point of pentane is higher because its linear geometry produces a net molecular dipole moment, resulting in stronger dipole-dipole attractions than in \(2,2\)-dimethylpropane.
- **C.** The boiling point of pentane is higher because its elongated shape provides a larger surface area of contact between molecules, resulting in stronger London dispersion forces than in compact \(2,2\)-dimethylpropane.
- **D.** The boiling point of pentane is higher because it has a greater number of electrons and a more polarizable electron cloud than \(2,2\)-dimethylpropane, resulting in stronger London dispersion forces.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/121456/*