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Chapter 22: Problem 5

If you had a group of hydrocarbons, what structural features would you look at to rank the hydrocarbons in order of increasing boiling point?

Short Answer

Expert verified
To rank hydrocarbons in order of increasing boiling points, consider the following structural features: 1. Molecular size: Higher molecular weight or a larger number of carbon atoms result in higher boiling points. 2. Molecular shape: Greater surface area or less compactness leads to higher boiling points. 3. Molecular polarity: Although not as significant, consider the polarity induced due to the molecular structure. By evaluating these factors, one can effectively rank hydrocarbons based on their boiling points.

Step by step solution

01

Understand factors affecting the boiling point of hydrocarbons

The boiling point of a substance depends on the strength of the forces between its molecules. The stronger these forces are, the more energy is required to break them and the higher the boiling point will be. For hydrocarbons, we will consider three main factors that influence their boiling points: molecular size, molecular shape, and molecular polarity.
02

Analyze the molecular size

The molecular size of a hydrocarbon affects its boiling point mainly through the van der Waals forces. These forces are weak forces that exist between all molecules, even non-polar ones, and result from the transient fluctuations in the electron cloud. Larger molecules have larger electron clouds and exhibit stronger van der Waals forces, which leads to higher boiling points. Therefore, to rank hydrocarbons in order of increasing boiling points based on molecular size, you should look at their molecular weight or number of carbon atoms. As the molecular weight increases, the boiling point also increases.
03

Analyze the molecular shape

The shape of the hydrocarbon molecule impacts the boiling point as well due to surface area and compactness. Molecules with more surface area tend to have higher boiling points because they exhibit stronger van der Waals forces. Linear or branched molecules with a larger surface area have higher boiling points compared to compact spherical molecules with smaller surface areas. Therefore, when ranking hydrocarbons based on molecular shape, it is important to consider the surface area and compactness of the molecules.
04

Analyze molecular polarity

The polarity of hydrocarbons is generally quite low, but molecular shape can induce certain polarity within the molecules. Molecules with polar bonds experience dipole-dipole forces in addition to van der Waals forces, which can increase the boiling point. However, since hydrocarbons are mostly non-polar or slightly polar, this factor is not as important as molecular size and shape for ranking hydrocarbons in terms of boiling points. Nonetheless, it should not be entirely ignored in some specific cases.
05

Rank the hydrocarbons based on structural features

To rank the hydrocarbons in order of increasing boiling point, consider the following factors: 1. Molecular size: Higher molecular weight or a larger number of carbon atoms result in higher boiling points. 2. Molecular shape: Greater surface area or less compactness leads to higher boiling points. 3. Molecular polarity: Although not as significant, consider the polarity induced due to the molecular structure. By considering these structural features, one can rank the hydrocarbons based on their boiling points effectively.

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Most popular questions from this chapter

Which of the following statements is (are) false? Explain why the statement(s) is (are) false. a. CCOC(=O)CC is a structural isomer of pentanoic acid. is a structural isomer of 2 -methyl-3- b. pentanone. c. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) is a structural isomer of 2 -pentanol. d. C=CC(C)O is a structural isomer of 2 -butenal. e. Trimethylamine is a structural isomer of \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\).

Consider the following reactions. For parts \(\mathrm{b}-\mathrm{d}\), see Exercise \(62 .\) a. When \(\mathrm{C}_{5} \mathrm{H}_{12}\) is reacted with \(\mathrm{Cl}_{2}(g)\) in the presence of ultraviolet light, four different monochlorination products form. What is the structure of \(\mathrm{C}_{5} \mathrm{H}_{12}\) in this reaction? b. When \(\mathrm{C}_{4} \mathrm{H}_{8}\) is reacted with \(\mathrm{H}_{2} \mathrm{O}\), a tertiary alcohol is produced as the major product. What is the structure of \(\mathrm{C}_{4} \mathrm{H}_{8}\) in this reaction? c. When \(\mathrm{C}_{7} \mathrm{H}_{12}\) is reacted with \(\mathrm{HCl}\), 1 -chloro-1-methylcyclohexane is produced as the major product. What are the two possible structures for \(\mathrm{C}_{7} \mathrm{H}_{12}\) in this reaction? d. When a hydrocarbon is reacted with \(\mathrm{H}_{2} \mathrm{O}\) and the major product of this reaction is then oxidized, acetone (2-propanone) is produced. What is the structure of the hydrocarbon in this reaction? e. When \(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}\) is oxidized, a carboxylic acid is produced. What are the possible structures for \(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}\) in this reaction?

Name all the aldehydes and ketones that have the formula \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O} .\)

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