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70 CHAPTER 2 • ALKANES

PROBLEMS 2.12 Represent each of the following compounds with a skeletal structure. (a) CH3

CH3CH2CH2CH" CH C(CH3)3 L L "CH3 (b) ethylcyclopentane 2.13 Name the following compounds.

(a) (b) CH3

CH2CH3

H3C

2.14 How many hydrogens are in an alkane of n carbons containing (a) two rings? (b) three rings? (c) m rings?

2.15 How many rings does an alkane have if its formula is (a) C8H10? (b) C7H12? Explain how you know.

2.6 PHYSICAL PROPERTIES OF ALKANES

Each time we come to a new family of organic compounds, we’ll consider the trends in their melting points, boiling points, densities, and solubilities, collectively referred to as their phys- ical properties. The physical properties of an organic compound are important because they determine the conditions under which the compound is handled and used. For example, the form in which a drug is manufactured and dispensed is affected by its physical properties. In commercial agriculture, ammonia (a gas at ordinary temperatures) and urea (a crystalline solid) are both very important sources of nitrogen, but their physical properties dictate that they are handled and dispensed in very different ways. Your goal should not be to memorize physical properties of individual compounds, but rather to learn to predict trends in how physical properties vary with structure.

A. Boiling Points The boiling point is the temperature at which the vapor pressure of a substance equals atmos- pheric pressure (which is typically 760 mm Hg). Table 2.1 shows that there is a regular change in the boiling points of the unbranched alkanes with increasing number of carbons. This trend of boiling point within the series of unbranched alkanes is particularly apparent in a plot of boiling point against carbon number (Fig. 2.7). The regular increase in boiling point of 20–30 °C per carbon atom within a series is a general trend observed for many types of or- ganic compounds. What is the reason for this increase? The key point for understanding this trend is that boil- ing points are a crude measure of the attractive forces among molecules—intermolecular at- tractions—in the liquid state. The greater are these intermolecular attractions, the more energy (heat, higher temperature) it takes to overcome them so that the molecules escape into the gas phase, in which such attractions do not exist. The greater are the intermolecular attractions within a liquid, the greater is the boiling point. Now, it is important to understand that there are no covalent bonds between molecules, and furthermore, that intermolecular attractions 02_BRCLoudon_pgs4-4.qxd 11/26/08 8:36 AM Page 71

2.6 PHYSICAL PROPERTIES OF ALKANES 71

250

200

150

100 C ° 50

0 boiling point, 50 - 100 - 150 - 200 - 0 2 4 6 8 10 12 number of carbon atoms

Figure 2.7 Boiling points of some unbranched alkanes plotted against number of carbon atoms. Notice the steady increase with the size of the alkane, which is in the range of 20–30 °C per carbon atom.

have nothing to do with the strengths of the covalent bonds within the molecules themselves. What, then, is the origin of these intermolecular attractions? In Chapter 1, we learned that electrons in bonds are not confined between the nuclei but rather reside in bonding molecular orbitals that surround the nuclei. We can think of the total electron distribution as an “electron cloud.” Electron clouds are rather “squishy” and can un- dergo distortions. Such distortions occur rapidly and at random, and when they occur, they re- sult in the temporary formation of regions of local positive and negative charge; that is, these distortions cause a temporary dipole moment within the molecule (Fig. 2.8, p. 72). When a second molecule is located nearby, its electron cloud distorts to form a complementary dipole, called an induced dipole. The positive charge in one molecule is attracted to the negative charge in the other. The attraction between temporary dipoles, called a van der Waals attrac- tion or a dispersion interaction, is the cohesive interaction that must be overcome to vapor- ize a liquid. Alkanes do not have significant permanent dipole moments. The dipoles dis- cussed here are temporary, and the presence of a temporary dipole in one molecule induces a temporary dipole in another. We might say, “Nearness makes the molecules grow fonder.” Now we are ready to understand why larger molecules have higher boiling points. Van der Waals attractions increase with the surface areas of the interacting electron clouds. That is, the larger the interacting surfaces, the greater the magnitude of the induced dipoles. A larger mole- cule has a greater surface area of electron clouds and therefore greater van der Waals interactions with other molecules. It follows, then, that large molecules have higher boiling points. The shape of a molecule is also important in determining its boiling point. For example, a comparison of the boiling point of the highly branched alkane neopentane (9.4 °C) and its un- branched isomer pentane (36.1 °C) is particularly striking. Neopentane has four methyl groups disposed in a tetrahedral arrangement about a central carbon. As the following space- filling models show, the molecule almost resembles a compact ball, and could fit readily 02_BRCLoudon_pgs4-4.qxd 11/26/08 8:36 AM Page 72

72 CHAPTER 2 • ALKANES

t2: the electron cloud of one t1: molecules moving at random molecule distorts randomly approach each other to form a temporary dipole

d–

d+

t3: the dipole in one molecule induces t4: temporary dipoles dissipate a complementary dipole in the other

d– d+

d+ d–

weak attractions develop between opposite charges (van der Waals attractions)

Figure 2.8 A stop-frame cartoon showing the origin of van der Waals attraction.The frames are labeled t1,t2,and 10 so on, for successive points in time. The time scale is about 10_ s. The colors represent electrostatic potential maps (EPMs).The green color of the isolated molecules (t1 and t4) shows the absence of a permanent dipole mo- ment.As the molecules approach (t1), the electron cloud of one molecule undergoes a random distortion (t2) that produces a temporary dipole, indicated by the red and blue colors.This dipole induces a complementary charge

separation (induced dipole) in the second molecule (t3), and attractions between the two dipoles result.Through random fluctuations of the electron clouds (t4), the temporary dipoles vanish. Averaged over time, this phenome- non results in a small net attraction.This is the van der Waals attraction.

within a sphere. On the other hand, pentane is rather extended, is ellipsoidal in shape, and would not fit within the same sphere.

neopentane: pentane: compact, nearly spherical extended, ellipsoidal The more a molecule approaches spherical proportions, the less surface area it presents to other molecules, because a sphere is the three-dimensional object with the minimum surface- to-volume ratio. Because neopentane has less surface area at which van der Waals interactions with other neopentane molecules can occur, it has fewer cohesive interactions than pentane, and thus, a lower boiling point. 02_BRCLoudon_pgs4-4.qxd 11/26/08 8:36 AM Page 73

2.6 PHYSICAL PROPERTIES OF ALKANES 73

In summary, by analysis of the boiling points of alkanes, we have learned two general trends in the variation of boiling point with structure: 1. Boiling points increase with increasing molecular weight within a homologous series— typically 20–30 °C per carbon atom. This increase is due to the greater van der Waals at- tractions between larger molecules. 2. Boiling points tend to be lower for highly branched molecules that approach spherical pro- portions because they have less molecular surface available for van der Waals attractions.

B. Melting Points The melting point of a substance is the temperature above which it is transformed sponta- neously and completely from the solid to the liquid state. The melting point is an especially important physical property in organic chemistry because it is used both to identify organic compounds and to assess their purity. Melting points are usually depressed, or lowered, by im- purities. Moreover, the melting range (the range of temperature over which a substance melts), usually quite narrow for a pure substance, is substantially broadened by impurities. The melt- ing point largely reflects the stabilizing intermolecular interactions between molecules in the crystal as well as the molecular symmetry, which determines the number of indistinguishable ways in which the molecule fits into the crystal. The higher the melting point, the more stable is the crystal structure relative to the liquid state. Although most alkanes are liquids or gases at room temperature and have relatively low melting points, their melting points nevertheless illustrate trends that are observed in the melting points of other types of organic compounds. One such trend is that melting points tend to increase with the number of carbons (Fig. 2.9). Another trend is that the melting points of unbranched alkanes with an even number of carbon atoms lie on a separate, higher curve from those of the alkanes with an odd number of carbons. This reflects the more effective packing of the even-carbon alkanes in the crystalline solid state. In other words, the odd-carbon alkane molecules do not “fit together” as well in the crystal as the even-carbon alkanes. Similar alternation of melting points is observed in other series of compounds, such as the cycloalkanes in Table 2.3.

0 20 - 40 - even carbons C 60 ° - 80 - 100 - 120 odd carbons - melting point, 140 - 160 - 180 - 200 - 0246810 12 number of carbons

Figure 2.9 A plot of melting points of the unbranched alkanes against number of carbon atoms.Notice the gen- eral increase of melting point with molecular size. Also notice that the alkanes with an even number of carbons (red) lie on a different curve from the alkanes with an odd number of carbons (blue). This trend is observed in a number of different types of organic compounds. 02_BRCLoudon_pgs4-4.qxd 11/26/08 8:36 AM Page 74

74 CHAPTER 2 • ALKANES

Branched-chain hydrocarbons tend to have lower melting points than linear ones because the branching interferes with regular packing in the crystal. When a branched molecule has a substantial symmetry, however, its melting point is typically relatively high because of the ease with which symmetrical molecules fit together within the crystal. For example, the melt- ing point of the very symmetrical molecule neopentane, 16.8 °C, is considerably higher than that of the less symmetrical pentane, 129.8 °C. (See -models on p. 72.) Compare also the melting points of the compact and symmetrical- molecule cyclohexane, 6.6 °C, and the ex- tended and less symmetrical hexane, 95.3 °C. In summary, melting points show the- following general trends: 1. Melting points tend to increase with increasing molecular mass within a series. 2. Many highly symmetrical molecules have unusually high melting points. 3. A sawtooth pattern of melting point behavior (see Fig. 2.9) is observed within many ho- mologous series.

PROBLEMS 2.16 Match each of the following isomers with the correct boiling points and melting points. Ex- plain your choices. Compounds: 2,2,3,3-tetramethylbutane and octane Boiling points: 106.5 °C, 125.7 °C Melting points: 56.8 °C, 100.7 °C 2.17 Which compound- has (a) the+ greater boiling point? (b) the greater melting point? Explain. (Hint: What is the geometry of benzene?) H H H H

H H H CH3

H H H H benzene toluene

C. Other Physical Properties Among the other significant physical properties of organic compounds are dipole moments, solubilities, and densities. A molecule’s dipole moment (Sec. 1.2D) determines its polarity, which, in turn, affects its physical properties. Because carbon and hydrogen differ little in their electronegativities, alkanes have negligible dipole moments and are therefore nonpolar mole- cules. We can see this graphically by comparing the EPMs of ethane, with a dipole moment of zero, and fluoromethane, a polar molecule with a dipole moment of 1.82 D.

EPM of ethane EPM of fluoromethane (H3C—F) Solubilities are important in determining which solvents can be used to form solutions; most reactions are carried out in solution. Water solubility is particularly important for several 02_BRCLoudon_pgs4-4.qxd 11/26/08 8:36 AM Page 75

2.6 PHYSICAL PROPERTIES OF ALKANES 75

Figure 2.10 The lower density of hydrocarbons and their insolubility in water allows an oil spill in flood waters to be contained by plastic tubes at a Texas refinery in the aftermath of Hurricane Rita in 2005.

reasons. For one thing, water is the solvent in biological systems. For this reason, water solu- bility is a crucial factor in the activity of drugs and other biologically important compounds. There has also been an increasing interest in the use of water as a solvent for large-scale chem- ical processes as part of an effort to control environmental pollution by organic solvents. The water solubility of the compounds to be used in a water-based chemical process is crucial. (We’ll deal in greater depth with the important question of solubility and solvents in Chapter 8.) The alkanes are, for all practical purposes, insoluble in water—thus the saying, “Oil and water don’t mix.” (Alkanes are a major constituent of crude oil.) The density of a compound is another property, like boiling point or melting point, that de- termines how the compound is handled. For example, whether a water-insoluble compound is more or less dense than water determines whether it will appear as a lower or upper layer when mixed with water. Alkanes have considerably lower densities than water. For this reason, a mixture of an alkane and water will separate into two distinct layers with the less dense alkane layer on top. An oil slick is an example of this behavior (Fig. 2.10).

PROBLEMS 2.18 Gasoline consists mostly of alkanes. Explain why water is not usually very effective in extin- guishing a gasoline fire.

2.19 (a) Into a separatory funnel is poured 50 mL of CH3CH2Br (bromoethane), a water-insoluble compound with a density of 1.460 g/mL, and 50 mL of water. The funnel is stoppered and the mixture is shaken vigorously. After standing, two layers separate. Which substance is in which layer? Explain. (b) Into the same funnel is poured carefully 50 mL of hexane (density 0.660 g/mL) so that the other two layers are not disturbed. The hexane forms a third layer.= The funnel is stop- pered and the mixture is shaken vigorously. After standing, two layers separate. Which compound(s) are in which layer? Explain.