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Rank each group of alkanes in order of increasing boiling point. Explain your choice of order. a. CH\(_3\)CH\(_2\)CH\(_2\)CH\(_2\)CH\(_3\), CH\(_3\)CH\(_2\)CH\(_2\)CH\(_3\), CH\(_3\)CH\(_2\)CH\(_3\) b. CH\(_3\)CH\(_2\)CH\(_2\)CH(CH\(_3)_2\), CH\(_3\)(CH\(_2)_4\)CH\(_3\), (CH\(_3)_2\)CHCH(CH\(_3)_2\)

Short Answer

Expert verified
a: 3C < 4C < 5C; b: most branched < less branched < linear.

Step by step solution

01

Identify Alkane Molecular Formulas

For each molecule, identify the number of carbon (C) and hydrogen (H) atoms. - a. CH\(_3\)CH\(_2\)CH\(_2\)CH\(_2\)CH\(_3\) (5 carbons), CH\(_3\)CH\(_2\)CH\(_2\)CH\(_3\) (4 carbons), CH\(_3\)CH\(_2\)CH\(_3\) (3 carbons) - b. CH\(_3\)CH\(_2\)CH\(_2\)CH(CH\(_3)_2\) (6 carbons), CH\(_3\)(CH\(_2)_4\)CH\(_3\) (6 carbons), (CH\(_3)_2\)CHCH(CH\(_3)_2\) (6 carbons)
02

Understand Boiling Point Factors

Recognize that boiling point is influenced by molecular weight and surface area. Linear alkanes generally have higher boiling points due to greater surface area allowing more van der Waals forces.
03

Order Based on Carbon Count for Part a

For part a, compare molecules by carbon count. More carbons usually mean a higher boiling point: - CH\(_3\)CH\(_2\)CH\(_3\) (3C) < CH\(_3\)CH\(_2\)CH\(_2\)CH\(_3\) (4C) < CH\(_3\)CH\(_2\)CH\(_2\)CH\(_2\)CH\(_3\) (5C).
04

Analyze Isomer Structure for Part b

For part b, examine structural differences. Linear structure CH\(_3\)(CH\(_2)_4\)CH\(_3\) has the highest boiling point. Branched isomers CH\(_3\)CH\(_2\)CH\(_2\)CH(CH\(_3)_2\) and (CH\(_3)_2\)CHCH(CH\(_3)_2\) will have lower boiling points due to less surface area, with (CH\(_3)_2\)CHCH(CH\(_3)_2\) likely the lowest.
05

Rank Alkanes for Both Groups

- a: CH\(_3\)CH\(_2\)CH\(_3\) < CH\(_3\)CH\(_2\)CH\(_2\)CH\(_3\) < CH\(_3\)CH\(_2\)CH\(_2\)CH\(_2\)CH\(_3\)- b: (CH\(_3)_2\)CHCH(CH\(_3)_2\) < CH\(_3\)CH\(_2\)CH\(_2\)CH(CH\(_3)_2\) < CH\(_3\)(CH\(_2)_4\)CH\(_3\)

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Key Concepts

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

Alkane Molecular Structure
Alkanes are simple hydrocarbons containing only carbon (C) and hydrogen (H) atoms linked in a chain by single bonds. The general formula for alkanes is C\(_n\)H\(_{2n+2}\), indicating that for every carbon atom, there are two more hydrogen atoms plus two additional hydrogens.
This simple molecular structure is crucial in defining many properties of alkanes, such as their boiling points. The length of the carbon chain and the presence of branches can influence how alkanes interact with each other and their surroundings. Knowing how to write and interpret molecular formulas is essential for understanding alkanes and predicting their physical properties. You can think of each carbon atom as having four "arms," to which hydrogen atoms or other carbons can attach, creating a complex and varied number of possible structures.
In studying the properties of alkanes, it's important to remember that the structural differences can lead to different chemical and physical behaviors. Let's explore how these molecular structures impact boiling points in more detail in the subsequent sections.
Van der Waals Forces
Van der Waals forces are weak interactions between molecules, and in alkanes, these forces are often the main type of intermolecular interaction. These forces arise from transient dipoles created as electrons move around their respective atomic orbitals, causing temporary areas of positive and negative charge.
The strength of van der Waals forces affects the boiling points of alkanes. The larger the molecule or the greater its surface area, the more there are of these interactions. This means molecules with more atoms or a longer chain length generally have stronger interactions. As a result, they require more energy (higher temperatures) to separate the molecules when boiling.
Understanding van der Waals forces is key to unraveling why different alkanes have varying boiling points, even when they contain the same number of carbon atoms but differ in structure.
Linear versus Branched Alkanes
Linear alkanes have straight or mostly straight carbon chains, whereas branched alkanes have one or more branches off the main chain. This difference can significantly affect their physical properties, including boiling points.
Linear alkanes have a larger surface area that can overlap with other molecules, enhancing the strength of van der Waals forces. Therefore, they generally have higher boiling points than their branched counterparts. Branched alkanes, on the other hand, have a more compact shape, which means less surface area is available for these interactions, leading to lower boiling points compared to similarly sized linear alkanes.
When considering boiling points, understanding the impact of structural arrangement provides insight into why the same formula might behave differently in different structural forms.
Isomer Boiling Points
Isomers are compounds that have the same chemical formula but different structures. The molecular structure influences how these molecules pack together, affecting their boiling points.
For isomers of alkanes, branch positioning and frequency play a vital role. In general, more branching results in a lower boiling point because branched structures have less surface area, thus weaker van der Waals forces. For instance, pentane (a linear alkane) typically has a higher boiling point than its branched isomer, 2-methylbutane.
When ranking isomers by boiling point, it is crucial to consider how changes in structure influence molecular interactions, and therefore, the energy required to transition from liquid to gas phase. Overall, understanding isomeric variations is essential for predicting the physical properties of different alkanes with the same molecular formula.

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Most popular questions from this chapter

Explain why beeswax is insoluble in H\(_2\)0, slightly soluble in ethanol (CH\(_3\)CH\(_2\)0H), and soluble in chloroform (CHCl\(_3\)).

Consider 1,2-dimethylcyclohexane. a. Draw structures for the cis and trans isomers using a hexagon for the six- membered ring. b. Draw the two possible chair conformations for the cis isomer. Which conformation, if either, is more stable? c. Draw the two possible chair conformations for the trans isomer. Which conformation, if either, is more stable? d. Which isomer, cis or trans, is more stable and why?

Hydrocarbons like benzene are metabolized in the body to arena oxides, which rearrange to form phenols. This is an example of a general process in the body, in which an unwanted compound (benzene) is converted to a more water- soluble derivative called a \(metabolite,\) so that it can be excreted more readily from the body. a. Classify each of these reactions as oxidation, reduction, or neither. b. Explain why phenol is more water soluble than benzene. This means that phenol dissolves in urine, which is largely water, to a greater extent than benzene.

Three components of the sex pheromone of the female sand bee (\(Ophrys\ sphegodes\)) are saturated hydrocarbons containing 23, 25, and 27 carbon atoms. How many H atoms does each of these alkanes contain? Interestingly, the early spider orchid emits a similar hydrocarbon mixture to attract male sand bees to pollinate its flowers.

The eclipsed conformation of CH\(_3\)CH\(_2\)CI is 15 kJ/mol less stable than the staggered conformation. How much is the H,CI eclipsing interaction worth in destabilization?

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