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2.5 Cycloalkanes and Skeletal Structures 67 02_BRCLoudon_pgs4-4.qxd 11/26/08 8:36 AM Page 67 2.5 CYCLOALKANES AND SKELETAL STRUCTURES 67 Likewise, the hydrogens bonded to each type of carbon are called primary, secondary, or ter- tiary hydrogens, respectively. primary hydrogens CH3 CH3 H3C CH2 CH2 "CH "C CH3 L L L LL L "CH primary hydrogens secondary hydrogens 3 tertiary hydrogen PROBLEMS 2.10 In the structure of 4-isopropyl-2,4,5-trimethylheptane (Problem 2.9) (a) Identify the primary, secondary, tertiary, and quaternary carbons. (b) Identify the primary, secondary, and tertiary hydrogens. (c) Circle one example of each of the following groups: a methyl group; an ethyl group; an isopropyl group; a sec-butyl group; an isobutyl group. 2.11 Identify the ethyl groups and the methyl groups in the structure of 4-sec-butyl-5-ethyl-3- methyloctane, the compound discussed in Study Problem 2.5. Note that these groups are not necessarily confined to those specifically mentioned in the name. 2.5 CYCLOALKANES AND SKELETAL STRUCTURES Some alkane contain carbon chains in closed loops, or rings; these are called cycloalkanes. Cycloalkanes are named by adding the prefix cyclo to the name of the alkane. Thus, the six- membered cycloalkane is called cyclohexane. CH2 H C CH 2 M % 2 H2""C CH2 %CHM2 cyclohexane The names and some physical properties of the simple cycloalkanes are given in Table 2.3. The general formula for an alkane containing a single ring has two fewer hydrogens than that of the open-chain alkane with the same number of carbon atoms. For example, cyclohexane has the formula C6H12, whereas hexane has the formula C6H14. The general formula for the cy- cloalkanes with one ring is CnH2n. Because of the tetrahedral configuration of carbon in the cycloalkanes, the carbon skeletons of the cycloalkanes (except for cyclopropane) are not planar. We’ll study the conformations of cycloalkanes in Chapter 7. For now, remember only that planar condensed structures for the cycloalkanes convey no information about their conformations. Skeletal Structures An important structure-drawing convention is the use of skeletal structures, which are structures that show only the carbon–carbon bonds. In this notation, a cycloalkane is drawn as a closed geometric figure. In a skeletal structure, it is understood that 02_BRCLoudon_pgs4-4.qxd 11/26/08 8:36 AM Page 68 68 CHAPTER 2 • ALKANES TABLE 2.3 Physical Properties of Some Cycloalkanes 1 Compound Boiling point (°C) Melting point (°C) Density (g mL_ ) cyclopropane 32.7 127.6 — - - cyclobutane 12.5 50.0 — - cyclopentane 49.3 93.9 0.7457 - cyclohexane 80.7 6.6 0.7786 cycloheptane 118.5 12.0 0.8098 - cyclooctane 150.0 14.3 0.8340 a carbon is located at each vertex of the figure, and that enough hydrogens are present on each carbon to fulfill its tetravalence. Thus, the skeletal structure of cyclohexane is drawn as follows: a carbon and two hydrogens are at each vertex ` Skeletal structures may also be drawn for open-chain alkanes. For example, hexane can be indicated this way: CH2 CH2 CH3 for CH CH H3C % % 2 % % 2 % When drawing a skeletal structure for an open-chain compound, don’t forget that carbons are not only at each vertex, but also at the ends of the structure. Thus, the six carbons of hexane in the preceding structure are indicated by the four vertices and two ends of the skeletal structure. Here are two other examples of skeletal structures: 2 4 6 8 10 13579 2,6-dimethyldecane isopropylcyclopentane 3,3,4-triethylhexane Nomenclature of Cycloalkanes The nomenclature of cycloalkanes follows essentially the same rules used for open-chain alkanes. CH3 CH3 CH 3 M M V ` M H C C H V 3 M % 2 5 methylcyclobutane 1,3-dimethylcyclobutane 1-ethyl-2-methylcyclohexane (Note alphabetical citation, rule 9.) 02_BRCLoudon_pgs4-4.qxd 11/26/08 8:36 AM Page 69 2.5 CYCLOALKANES AND SKELETAL STRUCTURES 69 The numerical prefix 1- is not necessary for monosubstituted cycloalkanes. Thus, the first compound is methylcyclobutane, not 1-methylcyclobutane. Two or more substituents, how- ever, must be numbered to indicate their relative positions. The lowest number is assigned in accordance with the usual rules. Most of the cyclic compounds in this text, like those in the preceding examples, involve rings with small alkyl branches. In such cases, the ring is treated as the principal chain. How- ever, when a noncyclic carbon chain contains more carbons than an attached ring, the ring is treated as the substituent. CH3CH2CH2CH2CH2 L 1-cyclopropylpentane (not pentylcyclopropane) Study Problem 2.6 Name the following compound. Solution This problem, in addition to illustrating the nomenclature of cyclic alkanes, is a good illustration of rule 8 for nomenclature, the “first point of difference” rule (p. 62). The compound is a cyclopentane with two methyl substituents and one ethyl substituent. If we number the ring car- bons consecutively, the following numbering schemes (and corresponding names) are possible, depending on which carbon is designated as carbon-1: 1,2,4- 4-ethyl-1,2-dimethylcyclopentane 1,3,4- 1-ethyl-3,4-dimethylcyclopentane 1,3,5- 3-ethyl-1,5-dimethylcyclopentane The correct name is decided by nomenclature rule 8 using the numbering schemes (not the names themselves). Because all numbering schemes begin with 1, the second number must be used to decide on the correct numbering. The scheme 1,2,4- has the lowest number at this point. Consequently, the correct name is 4-ethyl-1,2-dimethylcyclopentane. Study Problem 2.7 Draw a skeletal structure of tert-butylcyclohexane. Solution The real question in this problem is how to represent a tert-butyl group with a skeletal structure. The branched carbon in this group has four other bonds, three of which go to CH3 groups. Hence: CH3 one carbon is at the "C CH3 end of each branch 0L L = 0 "CH3 skeletal structure of tert-butylcyclohexane 02_BRCLoudon_pgs4-4.qxd 11/26/08 8:36 AM Page 70 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.
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