25 CHAPTER 2 ALKANES SOLUTIONS TO TEXT PROBLEMS 2.1 A carbonyl group is C?O. Of the two carbonyl functions in prostaglandin E 1 one belongs to the ketone family, the other to the carboxylic acids. 2.2 An unbranched alkane (n-alkane) of 28 carbons has 26 methylene (CH 2 ) groups flanked by a methyl (CH 3 ) group at each end. The condensed formula is CH 3 (CH 2 ) 26 CH 3 . 2.3 The alkane represented by the carbon skeleton formula has 11 carbons. The general formula for an alkane is C n H 2nH110012 , and thus there are 24 hydrogens. The molecular formula is C 11 H 24 ; the condensed structural formula is CH 3 (CH 2 ) 9 CH 3 . 2.4 In addition to CH 3 (CH 2 ) 4 CH 3 and (CH 3 ) 2 CHCH 2 CH 2 CH 3 , there are three more isomers. One has a five-carbon chain with a one-carbon (methyl) branch: The remaining two isomers have two methyl branches on a four-carbon chain. or orCH 3 CHCHCH 3 CH 3 CH 3 CH 3 CH 2 CCH 3 CH 3 CH 3 CH 3 CH 2 CHCH 2 CH 3 CH 3 or OH Ketone functional group Carboxylic acid functional group O O HO OH Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 26 ALKANES 2.5 (b) Octacosane is not listed in Table 2.4, but its structure can be deduced from its systematic name. The suffix -cosane pertains to alkanes that contain 20–29 carbons in their longest con- tinuous chain. The prefix octa- means “eight.” Octacosane is therefore the unbranched alkane having 28 carbon atoms. It is CH 3 (CH 2 ) 26 CH 3 . (c) The alkane has an unbranched chain of 11 carbon atoms and is named undecane. 2.6 The ending -hexadecane reveals that the longest continuous carbon chain has 16 carbon atoms. There are four methyl groups (represented by tetramethyl-), and they are located at carbons 2, 6, 10, and 14. 2.7 (b) The systematic name of the unbranched C 5 H 12 isomer is pentane (Table 2.4). A second isomer, (CH 3 ) 2 CHCH 2 CH 3 , has four carbons in the longest continuous chain and so is named as a derivative of butane. Since it has a methyl group at C-2, it is 2-methylbutane. The remaining isomer, (CH 3 ) 4 C, has three carbons in its longest continuous chain and so is named as a derivative of propane. There are two methyl groups at C-2, and so it is a 2,2-dimethyl derivative of propane. (c) First write out the structure in more detail, and identify the longest continuous carbon chain. There are five carbon atoms in the longest chain, and so the compound is named as a deriva- tive of pentane. This five-carbon chain has three methyl substituents attached to it, making it CCH 3 CH 3 CH 3 CH 2 C H CH 3 CH 3 IUPAC name: 2,2-dimethylpropane Common name: neopentane CH 3 CCH 3 CH 3 CH 3 IUPAC name: 2-methylbutane Common name: isopentane methyl group at C-2 CH 3 CHCH 2 CH 3 CH 3 IUPAC name: pentane Common name: n-pentane CH 3 CH 2 CH 2 CH 2 CH 3 2,6,10,14-Tetramethylhexadecane (phytane) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website a trimethyl derivative of pentane. Number the chain in the direction that gives the lowest numbers to the substituents at the first point of difference. (d) The longest continuous chain in (CH 3 ) 3 CC(CH 3 ) 3 contains four carbon atoms. The compound is named as a tetramethyl derivative of butane; it is 2,2,3,3-tetramethylbutane. 2.8 There are three C 5 H 11 alkyl groups with unbranched carbon chains. One is primary, and two are sec- ondary. The IUPAC name of each group is given beneath the structure. Remember to number the alkyl groups from the point of attachment. Four alkyl groups are derived from (CH 3 ) 2 CHCH 2 CH 3 . Two are primary, one is secondary, and one is tertiary. 2.9 (b) Begin by writing the structure in more detail, showing each of the groups written in parenthe- ses. The compound is named as a derivative of hexane, because it has six carbons in its longest continuous chain. The chain is numbered so as to give the lowest number to the substituent that appears closest to the end of the chain. In this case it is numbered so that the substituents are located at C-2 and C-4 rather than at C-3 and C-5. In alphabetical order the groups are ethyl and methyl; they are listed in alphabetical order in the name. The compound is 4-ethyl-2-methylhexane. CH 3 CH 2 CHCH 2 CHCH 3 CH 3 CH 2 CH 3 654321 2-Methylbutyl group (primary) 1,2-Dimethylpropyl group (secondary) 3-Methylbutyl group (primary) CH 3 CHCH 2 CH 2 1234 CH 3 CH 2 CHCH 2 CH 3 4321 CH 3 CH 3 CHCHCH 3 231 CH 3 1,1-Dimethylpropyl group (tertiary) CH 3 CCH 2 CH 3 321 CH 3 CH 3 CH 2 CH 2 CH 2 CH 2 Pentyl group (primary) CH 3 CH 2 CH 2 CHCH 3 1234 1-Methylbutyl group (secondary) CH 3 CH 2 CHCH 2 CH 3 123 1-Ethylpropyl group (secondary) CCH 3 CH 3 CH 3 CH 3 C CH 3 CH 3 notCCH 3 CH 3 CH 3 CH 2 C H CH 3 CH 3 12 453 2,2,4-Trimethylpentane (correct) CCH 3 CH 3 CH 3 CH 2 C H CH 3 CH 3 54 2 13 2,4,4-Trimethylpentane (incorrect) ALKANES 27 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) The longest continuous chain is shown in the structure; it contains ten carbon atoms. The structure also shows the numbering scheme that gives the lowest number to the substituent at the first point of difference. In alphabetical order, the substituents are ethyl (at C-8), isopropyl at (C-4), and two methyl groups (at C-2 and C-6). The alkane is 8-ethyl-4-isopropyl-2,6-dimethyldecane. The system- atic name for the isopropyl group (1-methylethyl) may also be used, and the name becomes 8-ethyl-2,6-dimethyl-4-(1-methylethyl)decane. 2.10 (b) There are ten carbon atoms in the ring in this cycloalkane, thus it is named as a derivative of cyclodecane. The numbering pattern of the ring is chosen so as to give the lowest number to the substituent at the first point of difference between them. Thus, the carbon bearing two methyl groups is C-1, and the ring is numbered counterclockwise, placing the isopropyl group on C-4 (numbering clockwise would place the isopropyl on C-8). Listing the substituent groups in al- phabetical order, the correct name is 4-isopropyl-1,1-dimethylcyclodecane. Alternatively, the systematic name for isopropyl (1-methylethyl) could be used, and the name would become 1,1-dimethyl-4-(1-methylethyl)cyclodecane. (c) When two cycloalkyl groups are attached by a single bond, the compound is named as a cycloalkyl-substituted cycloalkane. This compound is cyclohexylcyclohexane. 2.11 The alkane that has the most carbons (nonane) has the highest boiling point (151°C). Among the others, all of which have eight carbons, the unbranched isomer (octane) has the highest boiling point (126°C) and the most branched one (2,2,3,3-tetramethylbutane) the lowest (106°C). The remaining alkane, 2-methylheptane, boils at 116°C. 2.12 All hydrocarbons burn in air to give carbon dioxide and water. To balance the equation for the com- bustion of cyclohexane (C 6 H 12 ), first balance the carbons and the hydrogens on the right side. Then balance the oxygens on the left side. 2.13 (b) Icosane (Table 2.4) is C 20 H 42 . It has four more methylene (CH 2 ) groups than hexadecane, the last unbranched alkane in Table 2.5. Its calculated heat of combustion is therefore (4 H11003 653 kJ/mol) higher. Heat of combustion of icosane H11005 heat of combustion of hexadecane H11001 4 H11003 653 kJ/mol H11005 10,701 kJ/mol H11001 2612 kJ/mol H11005 13,313 kJ/mol 9O 2 H11001 6H 2 O6CO 2 H11001 Cyclohexane Oxygen Carbon dioxide Water Cyclodecane (CH 3 ) 2 CH H 3 C CH 3 3 6 7 2 1 8 5 4 9 10 CH 3 CH 2 CHCH 2 CHCH 2 CHCHCH 3 CH 2 CHCH 3 CH 2 CH 3 CH 3 CH 3 CH 3 32 1 10 987 6 45 28 ALKANES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 2.14 Two factors that influence the heats of combustion of alkanes are, in order of decreasing importance, (1) the number of carbon atoms and (2) the extent of chain branching. Pentane, isopentane, and neopentane are all C 5 H 12 ; hexane is C 6 H 14 . Hexane has the largest heat of combustion. Branching leads to a lower heat of combustion; neopentane is the most branched and has the lowest heat of combustion. Hexane CH 3 (CH 2 ) 4 CH 3 Heat of combustion 4163 kJ/mol (995.0 kcal/mol) Pentane CH 3 CH 2 CH 2 CH 2 CH 3 Heat of combustion 3527 kJ/mol (845.3 kcal/mol) Isopentane (CH 3 ) 2 CHCH 2 CH 3 Heat of combustion 3529 kJ/mol (843.4 kcal/mol) Neopentane (CH 3 ) 4 C Heat of combustion 3514 kJ/mol (839.9 kcal/mol) 2.15 (b) In the reaction carbon becomes bonded to an atom (Br) that is more electronegative than itself. Carbon is oxidized. (c) In the reaction one carbon becomes bonded to hydrogen and is, therefore, reduced. The other carbon is also reduced, because it becomes bonded to boron, which is less electronegative than carbon. 2.16 It is best to approach problems of this type systematically. Since the problem requires all the isomers of C 7 H 16 to be written, begin with the unbranched isomer heptane. Two isomers have six carbons in their longest continuous chain. One bears a methyl substituent at C-2, the other a methyl substituent at C-3. Now consider all the isomers that have two methyl groups as substituents on a five-carbon continu- ous chain. 2,3-Dimethylpentane (CH 3 ) 2 CHCHCH 2 CH 3 CH 3 2,4-Dimethylpentane (CH 3 ) 2 CHCH 2 CH(CH 3 ) 2 2,2-Dimethylpentane (CH 3 ) 3 CCH 2 CH 2 CH 3 3,3-Dimethylpentane (CH 3 CH 2 ) 2 C(CH 3 ) 2 2-Methylhexane (CH 3 ) 2 CHCH 2 CH 2 CH 2 CH 3 3-Methylhexane CH 3 CH 2 CHCH 2 CH 2 CH 3 CH 3 Heptane CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 H11001 B 2 H 6 2(CH 3 CH 2 ) 3 BCH 2 6CH 2 BrCH 2 CH 2 BrH11001 Br 2 CH 2 CH 2 ALKANES 29 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website There is one isomer characterized by an ethyl substituent on a five-carbon chain: The remaining isomer has three methyl substituents attached to a four-carbon chain. 2.17 In the course of doing this problem, you will write and name the 17 alkanes that, in addition to oc- tane, CH 3 (CH 2 ) 6 CH 3 , comprise the 18 constitutional isomers of C 8 H 18 . (a) The easiest way to attack this part of the exercise is to draw a bond-line depiction of heptane and add a methyl branch to the various positions. Other structures bearing a continuous chain of seven carbons would be duplicates of these isomers rather than unique isomers. “5-Methylheptane,” for example, is an incorrect name for 3-methylheptane, and “6-methylheptane” is an incorrect name for 2-methylheptane. (b) Six of the isomers named as derivatives of hexane contain two methyl branches on a continu- ous chain of six carbons. One isomer bears an ethyl substituent: (c) Four isomers are trimethyl-substituted derivatives of pentane: 2,2,3-Trimethylpentane 2,3,3-Trimethylpentane 2,2,4-Trimethylpentane 2,3,4-Trimethylpentane 3-Ethylhexane 2,2-Dimethylhexane 2,3-Dimethylhexane 2,4-Dimethylhexane 2,5-Dimethylhexane 3,3-Dimethylhexane 3,4-Dimethylhexane 2-Methylheptane 3-Methylheptane 4-Methylheptane 2,2,3-Trimethylbutane (CH 3 ) 3 CCH(CH 3 ) 2 3-Ethylpentane (CH 3 CH 2 ) 3 CH 30 ALKANES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Two bear an ethyl group and a methyl group on a continuous chain of five carbons: (d) Only one isomer is named as a derivative of butane: 2.18 (a) The longest continuous chain contains nine carbon atoms. Begin the problem by writing and numbering the carbon skeleton of nonane. Now add two methyl groups (one to C-2 and the other to C-3) and an isopropyl group (to C-6) to give a structural formula for 6-isopropyl-2,3-dimethylnonane. (b) To the carbon skeleton of heptane (seven carbons) add a tert-butyl group to C-4 and a methyl group to C-3 to give 4-tert-butyl-3-methylheptane. (c) An isobutyl group is GCH 2 CH(CH 3 ) 2 . The structure of 4-isobutyl-1,1-dimethylcyclohexane is as shown. (d)Asec-butyl group is CH 3 = CHCH 2 CH 3 . sec-Butylcycloheptane has a sec-butyl group on a seven-membered ring. (e) A cyclobutyl group is a substituent on a five-membered ring in cyclobutylcyclopentane. or CH 3 CHCH 2 CH 3 1 2 6 4 3 5 H 3 C H 3 C CH 2 CH(CH 3 ) 2 or or 1 3 24 57 6 CH 3 CH 2 CHCHCH 2 CH 2 CH 3 C(CH 3 ) 3 CH 3 or CH 3 CHCHCH 2 CH 2 CHCH 2 CH 2 CH 3 CH 3 CH(CH 3 ) 2 CH 3 31 2468 57 9 1 2 3 4 5 68 7 9 2,2,3,3-Tetramethylbutane 3-Ethyl-2-methylpentane 3-Ethyl-3-methylpentane ALKANES 31 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ( f ) Recall that an alkyl group is numbered from the point of attachment. The structure of (2,2-dimethylpropyl)cyclohexane is (g) The name “pentacosane” contains no numerical locants or suffixes indicating the presence of alkyl groups. It must therefore be an unbranched alkane. Table 2.4 in the text indicates that the suffix -cosane refers to alkanes with 20–29 carbons. The prefix penta- stands for “five,” and so pentacosane must be the unbranched alkane with 25 carbons. Its condensed structural formula is CH 3 (CH 2 ) 23 CH 3 . (h) We need to add a 1-methylpentyl group to C-10 of pentacosane. A 1-methylpentyl group is: It has five carbons in the longest continuous chain counting from the point of attachment and bears a methyl group at C-1. 10-(1-Methylpentyl)pentacosane is therefore: 2.19 (a) This compound is an unbranched alkane with 27 carbons. As noted in part (g) of the preced- ing problem, alkanes with 20–29 carbons have names ending in -cosane. Thus, we add the prefix hepta- (“seven”) to -cosane to name the alkane CH 3 (CH 2 ) 25 CH 3 as heptacosane. (b) The alkane (CH 3 ) 2 CHCH 2 (CH 2 ) 14 CH 3 has 18 carbons in its longest continuous chain. It is named as a derivative of octadecane. There is a single substituent, a methyl group at C-2. The compound is 2-methyloctadecane. (c) Write the structure out in more detail to reveal that it is 3,3,4-triethylhexane. (d) Each line of a bond-line formula represents a bond between two carbon atoms. Hydrogens are added so that the number of bonds to each carbon atom totals four. The IUPAC name is 4-ethyl-2,2-dimethylhexane. (e) The IUPAC name is 3,5-dimethylheptane. is the same as CH 3 CH 2 CHCH 2 CHCH 2 CH 3 CH 3 CH 3 is the same as CH 3 CH 2 CHCH 2 C(CH 3 ) 3 CH 2 CH 3 (CH 3 CH 2 ) 3 CCH(CH 2 CH 3 ) 2 CH 3 CH 2 Cis rewritten as CHCH 2 CH 3 CH 2 CH 3 CH 3 CH 2 CH 3 CH 2 12 3 456 CH 3 (CH 2 ) 8 CH(CH 2 ) 14 CH 3 CH 3 CHCH 2 CH 2 CH 2 CH 3 CHCH 2 CH 2 CH 2 CH 3 CH 3 12 3 4 5 CH 2 CH 3 CH 3 CH 3 C 32 ALKANES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ( f ) The IUPAC name is 1-butyl-1-methylcyclooctane. (g) Number the chain in the direction shown to give 3-ethyl-4,5,6-trimethyloctane. When num- bered in the opposite direction, the locants are also 3, 4, 5, and 6. In the case of ties, however, choose the direction that gives the lower number to the substituent that appears first in the name. “Ethyl” precedes “methyl” alphabetically. 2.20 (a) The alkane contains 13 carbons. Since all alkanes have the molecular formula C n H 2nH110012 , the molecular formula must be C 13 H 28 . (b) The longest continuous chain is indicated and numbered as shown. In alphabetical order, the substituents are ethyl (at C-5), methyl (at C-2), methyl (at C-6). The IUPAC name is 5-ethyl-2,6-dimethylnonane. (c) Fill in the hydrogens in the alkane to identify the various kinds of groups present. There are five methyl (CH 3 ) groups, five methylene (CH 2 ) groups, and three methine (CH) groups in the molecule. (d) A primary carbon is attached to one other carbon. There are five primary carbons (the carbons of the five CH 3 groups). A secondary carbon is attached to two other carbons, and there are five of these (the carbons of the five CH 2 groups). A tertiary carbon is attached to three other carbons, and there are three of these (the carbons of the three methine groups). A quaternary carbon is attached to four other carbons. None of the carbons is a quaternary carbon. 2.21 (a) The group CH 3 (CH 2 ) 10 CH 2 G is an unbranched alkyl group with 12 carbons. It is a dodecyl group. The carbon at the point of attachment is directly attached to only one other carbon. It is a primary alkyl group. (b) The longest continuous chain from the point of attachment is six carbons; it is a hexyl group bear- ing an ethyl substituent at C-3. The group is a 3-ethylhexyl group. It is a primary alkyl group. (c) By writing the structural formula of this alkyl group in more detail, we see that the longest con- tinuous chain from the point of attachment contains three carbons. It is a 1,1-diethylpropyl group. Because the carbon at the point of attachment is directly bonded to three other carbons, it is a tertiary alkyl group. CH 2 CH 3 CH 2 CH 3 CCH 2 CH 3 is rewritten asC(CH 2 CH 3 ) 3 12 3 123456 CH 2 CH 2 CHCH 2 CH 2 CH 3 CH 2 CH 3 9 87 6 5 4 3 2 1 CH 3 CHCH 2 CH 2 CHCHCH 3 CH 3 CH 2 CH 2 CH 3 CH 2 CH 3 8 7 5 3 1 246 H 2 C H 2 C H 2 CCH 2 CH 2 C H 2 C CH 3 CH 2 CH 2 CH 2 CH 3 is the same as CH 2 ALKANES 33 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (d) This group contains four carbons in its longest continuous chain. It is named as a butyl group with a cyclopropyl substituent at C-1. It is a 1-cyclopropylbutyl group and is a secondary alkyl group. (e, f ) A two-carbon group that bears a cyclohexyl substituent is a cyclohexylethyl group. Number from the point of attachment when assigning a locant to the cyclohexyl group. 2.22 The IUPAC name for pristane reveals that the longest chain contains 15 carbon atoms (as indicated by -pentadecane). The chain is substituted with four methyl groups at the positions indicated in the name. 2.23 (a) An alkane having 100 carbon atoms has 2(100) H11001 2 H11005 202 hydrogens. The molecular formula of hectane is C 100 H 202 and the condensed structural formula is CH 3 (CH 2 ) 98 CH 3 . The 100 carbon atoms are connected by 99 H9268 bonds. The total number of H9268 bonds is 301 (99 CGC bonds H11001 202 CGH bonds). (b) Unique compounds are formed by methyl substitution at carbons 2 through 50 on the 100-carbon chain (C-51 is identical to C-50, and so on). There are 49 x-methylhectanes. (c) Compounds of the type 2,x-dimethylhectane can be formed by substitution at carbons 2 through 99. There are 98 of these compounds. 2.24 Isomers are different compounds that have the same molecular formula. In all these problems the safest approach is to write a structural formula and then count the number of carbons and hydrogens. (a) Among this group of compounds, only butane and isobutane have the same molecular formula; only these two are isomers. (b) The two compounds that are isomers, that is, those that have the same molecular formula, are 2,2-dimethylpentane and 2,2,3-trimethylbutane. 2,2-Dimethylpentane C 7 H 16 CH 3 CCH 2 CH 2 CH 3 CH 3 CH 3 2,2,3-Trimethylbutane C 7 H 16 CH 3 C CH 3 CH 3 CHCH 3 CH 3 CH 3 CH 2 CH 2 CH 3 Butane C 4 H 10 Isobutane C 4 H 10 Cyclobutane C 4 H 8 CH 3 CHCH 3 CH 3 2-Methylbutane C 5 H 12 CH 3 CHCH 2 CH 3 CH 3 Pristane (2,6,10,14-tetramethylpentadecane) CH 2 CH 2 CH CH 3 12 1 2 2-Cyclohexylethyl (primary) 1-Cyclohexylethyl (secondary) CHCH 2 CH 2 CH 3 12 3 4 34 ALKANES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Cyclopentane and neopentane are not isomers of these two compounds, nor are they isomers of each other. (c) The compounds that are isomers are cyclohexane, methylcyclopentane, and 1,1,2- trimethylcyclopropane. Hexane, CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 , has the molecular formula C 6 H 14 ; it is not an isomer of the others. (d) The three that are isomers all have the molecular formula C 5 H 10 . Propylcyclopropane is not an isomer of the others. Its molecular formula is C 6 H 12 . (e) Only 4-methyltetradecane and pentadecane are isomers. Both have the molecular formula C 15 H 32 . 2.25 The oxygen and two of the carbons of C 3 H 5 ClO are part of the structural unit that characterizes epoxides. The problem specifies that a methyl group (CH 3 ) is not present; therefore, add the CH 3 (CH 2 ) 2 CH(CH 2 ) 9 CH 3 CH 3 4-Methyltetradecane C 15 H 32 CH 3 (CH 2 ) 13 CH 3 Pentadecane C 15 H 32 CH 3 CHCHCHCH(CH 2 ) 4 CH 3 CH 3 CH 3 CH 3 CH 3 2,3,4,5-Tetramethyldecane C 14 H 30 CH 3 CH 2 CH 2 CH(CH 2 ) 5 CH 3 4-Cyclobutyldecane C 14 H 28 CH 2 CH 2 CH 3 1,1-Dimethylcyclopropane C 5 H 10 Ethylcyclopropane C 5 H 10 Cyclopentane C 5 H 10 CH 3 CH 3 CH 2 CH 3 Methylcyclopentane C 6 H 12 CH 3 1,1,2-Trimethylcyclopropane C 6 H 12 CH 3 CH 3 H 3 C Cyclohexane C 6 H 12 Cyclopentane C 5 H 10 Neopentane C 5 H 12 CH 3 CCH 3 CH 3 CH 3 ALKANES 35 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website remaining carbon and the chlorine as a GCH 2 Cl unit, and fill in the remaining bonds with hydrogen substituents. 2.26 (a) Ibuprofen is (b) Mandelonitrile is 2.27 Isoamyl acetate is 2.28 Thiols are characterized by the GSH group. n-Butyl mercaptan is CH 3 CH 2 CH 2 CH 2 SH. 2.29 H9251-Amino acids have the general formula The individual amino acids in the problem have the structures shown: (c, d) An isobutyl group is (CH 3 ) 2 CHCH 2 G, and a sec-butyl group is The structures of leucine and isoleucine are: Leucine (CH 3 ) 2 CHCH 2 CHCO H11002 H11001 NH 3 O CH 3 CH 2 CHCHCO H11002 H11001 NH 3 O Isoleucine CH 3 CH 3 CHCH 2 CH 3 CH 3 CHCO H11002 H11001 NH 3 O (CH 3 ) 2 CHCHCO H11002 H11001 NH 3 O (a) Alanine (b) Valine RCHCO H11002 H11001 NH 3 O CH 3 COCH 2 CH 2 CHCH 3 OCH 3 RCORH11032 (ester) O Methyl 3-Methylbutyl which is HCH OH CN (CH 3 ) 2 CHCH 2 CHCOH CH 3 O H H C H C O CH 2 Cl Epichlorohydrin 36 ALKANES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (e–g) The functional groups that characterize alcohols, thiols, and carboxylic acids are GOH, GSH, and GCO 2 H, respectively. The structures of serine, cysteine, and aspartic acid are: 2.30 Uscharidin has the structure shown. (a) There are two alcohol groups, one aldehyde group, one ketone group, and one ester functionality. (b) Uscharidin contains ten methylene groups (CH 2 ). They are indicated in the structure by small squares. (c) The primary carbons in uscharidin are the carbons of the two methyl groups. 2.31 (a) Methylene groups are GCH 2 G. ClCH 2 CH 2 CH 2 CH 2 Cl is therefore the C 4 H 8 Cl 2 isomer in which all the carbons belong to methylene groups. (b) The C 4 H 8 Cl 2 isomers that lack methylene groups are (CH 3 ) 2 CHCHCl 2 and 2.32 Since it is an alkane, the sex attractant of the tiger moth has a molecular formula of C n H 2nH110012 . The number of carbons and hydrogens may be calculated from its molecular weight. 12n H11001 1(2n H11001 2) H11005 254 14n H11005 252 n H11005 18 The molecular formula of the alkane is C 18 H 38 . In the problem it is stated that the sex attractant is a 2-methyl-branched alkane. It is therefore 2-methylheptadecane, (CH 3 ) 2 CHCH 2 (CH 2 ) 13 CH 3 . 2.33 When any hydrocarbon is burned in air, the products of combustion are carbon dioxide and water. CH 3 (CH 2 ) 8 CH 3 H11001 O 2 (a) 31 2 Decane (C 10 H 22 ) Oxygen 10CO 2 H11001 11H 2 O Carbon dioxide Water CH 3 CHCHCH 3 Cl Cl O O O O O H 3 C OH H H H CH H H H CH 3 OH O OAlcohol group Ketone group Primary carbon Aldehyde group Primary carbon Alcohol group Ester group HOCH 2 CHCO H11002 H11001 NH 3 O Serine HSCH 2 CHCO H11002 H11001 NH 3 O Cysteine HOCCH 2 CHCO H11002 H11001 NH 3 OO Aspartic acid ALKANES 37 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 2.34 To determine the quantity of heat evolved per unit mass of material, divide the heat of combustion by the molecular weight. Methane Heat of combustion H11005 890 kJ/mol (212.8 kcal/mol) Molecular weight H11005 16.0 g/mol Heat evolved per gram H11005 55.6 kJ/g (13.3 kcal/g) Butane Heat of combustion H11005 2876 kJ/mol (687.4 kcal/mol) Molecular weight H11005 58.0 g/mol Heat evolved per gram H11005 49.6 kJ/g (11.8 kcal/g) When equal masses of methane and butane are compared, methane evolves more heat when it is burned. Equal volumes of gases contain an equal number of moles, so that when equal volumes of methane and butane are compared, the one with the greater heat of combustion in kilojoules (or kilocalories) per mole gives off more heat. Butane evolves more heat when it is burned than does an equal volume of methane. 2.35 When comparing heats of combustion of alkanes, two factors are of importance: 1. The heats of combustion of alkanes increase as the number of carbon atoms increases. 2. An unbranched alkane has a greater heat of combustion than a branched isomer. (a) In the group hexane, heptane, and octane, three unbranched alkanes are being compared. Oc- tane (C 8 H 18 ) has the most carbons and has the greatest heat of combustion. Hexane (C 6 H 14 ) has the fewest carbons and the lowest heat of combustion. The measured values in this group are as follows: Hexane Heat of combustion 4163 kJ/mol (995.0 kcal/mol) Heptane Heat of combustion 4817 kJ/mol (1151.3 kcal/mol) Octane Heat of combustion 5471 kJ/mol (1307.5 kcal/mol) (b) Isobutane has fewer carbons than either pentane or isopentane and so is the member of the group with the lowest heat of combustion. Isopentane is a 2-methyl-branched isomer of pen- tane and so has a lower heat of combustion. Pentane has the highest heat of combustion among these compounds. Isobutane (CH 3 ) 3 CH Heat of combustion 2868 kJ/mol (685.4 kcal/mol) Isopentane (CH 3 ) 2 CHCH 2 CH 3 Heat of combustion 3529 kJ/mol (843.4 kcal/mol) Pentane CH 3 CH 2 CH 2 CH 2 CH 3 Heat of combustion 3527 kJ/mol (845.3 kcal/mol) Cyclopentylcyclopentane (C 10 H 18 ) Oxygen H11001 29 2 O 2 Carbon dioxide Water 10CO 2 H11001 9H 2 O (d) H11001 15O 2 CH 3 (c) 10CO 2 H11001 10H 2 O Methylcyclononane (C 10 H 20 ) Oxygen Carbon dioxide Water H11001 15O 2 (b) 10CO 2 H11001 10H 2 O Cyclodecane (C 10 H 20 ) Oxygen Carbon dioxide Water 38 ALKANES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) Isopentane and neopentane each have fewer carbons than 2-methylpentane, which therefore has the greatest heat of combustion. Neopentane is more highly branched than isopentane; neopentane has the lowest heat of combustion. Neopentane (CH 3 ) 4 C Heat of combustion 3514 kJ/mol (839.9 kcal/mol) Isopentane (CH 3 ) 2 CHCH 2 CH 3 Heat of combustion 3529 kJ/mol (843.4 kcal/mol) 2-Methylpentane (CH 3 ) 2 CHCH 2 CH 2 CH 3 Heat of combustion 4157 kJ/mol (993.6 kcal/mol) (d) Chain branching has a small effect on heat of combustion; the number of carbons has a much larger effect. The alkane with the most carbons in this group is 3,3-dimethylpentane; it has the greatest heat of combustion. Pentane has the fewest carbons in this group and has the smallest heat of combustion. Pentane CH 3 CH 2 CH 2 CH 2 CH 3 Heat of combustion 3527 kJ/mol (845.3 kcal/mol) 3-Methylpentane (CH 3 CH 2 ) 2 CHCH 3 Heat of combustion 4159 kJ/mol (994.1 kcal/mol) 3,3-Dimethylpentane (CH 3 CH 2 ) 2 C(CH 3 ) 2 Heat of combustion 4804 kJ/mol (1148.3 kcal/mol) (e) In this series the heat of combustion increases with increasing number of carbons. Ethylcy- clopentane has the lowest heat of combustion; ethylcycloheptane has the greatest. 2.36 (a) The equation for the hydrogenation of ethylene is given by the sum of the following three reactions: Equations (1) and (2) are the combustion of hydrogen and ethylene, respectively, and H9004H° val- ues for these reactions are given in the statement of the problem. Equation (3) is the reverse of the combustion of ethane, and its value of H9004H° is the negative of the heat of combustion of ethane. (b) Again we need to collect equations of reactions for which the H9004H° values are known. (3) 2CO 2 (g) H11001 2H 2 O(l) (1) H 2 (g) H11001 O 2 (g) 1 2 (2) HC CH(g) H11001 O 2 (g) 5 2 H 2 O(l) ?HH11034 H11005 H11002286 kJ (H1100268.4 kcal) 2CO 2 (g) H11001 H 2 O(l) ?HH11034 H11005 H110021300 kJ (H11002310.7 kcal) CH 2 CH 2 (g) H11001 3O 2 (g) ?HH11034 H11005 H110011410 kJ (H11001337.0 kcal) Sum: CH(g) H11001 H 2 (g) CH 2 CH 2 (g) ?HH11034 H11005 H11002176 kJ (H1100242.1 kcal)HC Sum: H 2 C CH 2 (g) H11001 H 2 (g) (3) 3H 2 O(l) H11001 2CO 2 (g) (1) H 2 (g) H11001 O 2 (g) 1 2 H 2 O(l) ?HH11034 H11005 H11002286 kJ (H1100268.4 kcal) 7 2 CH 3 CH 3 (g) H11001 O 2 (g) ?HH11034 H11005 H110011560 kJ (H11001372.8 kcal) CH 3 CH 3 (g) ?HH11034 H11005 H11002136 kJ (H1100232.6 kcal) (2) H 2 C CH 2 (g) H11001 3O 2 (g) 2CO 2 (g) H11001 2H 2 O(l) ?HH11034 H11005 H110021410 kJ (H11002337.0 kcal) Ethylcyclopentane 4592 kJ/mol (1097.5 kcal/mol) Ethylcyclohexane 5222 kJ/mol (1248.2 kcal/mol) Ethylcycloheptane (combustion data not available) CH 3 CH 2 CH 3 CH 2 CH 3 CH 2 ALKANES 39 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Equations (1) and (2) are the combustion of hydrogen and acetylene, respectively. Equation (3) is the reverse of the combustion of ethylene, and its value of H9004H° is the negative of the heat of combustion of ethylene. The value of H9004H° for the hydrogenation of acetylene to ethane is equal to the sum of the two reactions just calculated: (c) We use the equations for the combustion of ethane, ethylene, and acetylene as shown. The value of H9004H° for reaction (1) is twice that for the combustion of ethylene because 2 mol of ethylene are involved. 2.37 (a) The hydrogen content increases in going from CH 3 C>CH to CH 3 CH?CH 2 . The organic compound CH 3 C>CH is reduced. (b) Oxidation occurs because a CGO bond has replaced a CGH bond in going from starting material to product. (c) There are two carbon–oxygen bonds in the starting material and four carbon–oxygen bonds in the products. Oxidation occurs. (d) Although the oxidation state of carbon is unchanged in the process overall, reduction of the organic compound has occurred. Its hydrogen content has increased and its oxygen content has decreased. 2.38 In the reaction bonds between carbon and an atom more electronegative than itself (chlorine) are replaced by bonds between carbon and an atom less electronegative than itself (silicon). Carbon is reduced; silicon is oxidized. 2.39 (a) Compound A has the structural unit ; compound A is a ketone.CCC O 2CH 3 Cl H11001 Si (CH 3 ) 2 SiCl 2 NO 2 NH 3 H11001 HO OHCH 2 CH 2 2H 2 CO Two C O bonds Four C O bonds H O OH oxidation (1) 2CH 2 CH 2 (g) H11001 6O 2 (g) 4CO 2 (g) H11001 4H 2 O(l) ?HH11034 H11005 H110022820 kJ (H11002674.0 kcal) (3) 3H 2 O(l) H11001 2CO 2 (g) 7 2 CH 3 CH 3 (g) H11001 O 2 (g) ?HH11034 H11005 H110011560 kJ (H11001372.8 kcal) CH(g) Sum: 2CH 2 CH 2 (g) CH 3 CH 3 (g) H11001 HC ?HH11034 H11005 H1100140 kJ (H110019.5 kcal) (2) 2CO 2 (g) H11001 H 2 O(l) 5 2 HC CH(g) H11001 O 2 (g) ?HH11034 H11005 H110011300 kJ (H11001310.7 kcal) HC CH(g) H11001 H 2 (g) Sum: HC CH(g) H11001 2H 2 (g) CH 3 CH 3 (g) ?HH11034 H11005 H11002312 kJ (H1100274.7 kcal) H 2 C CH 2 (g) H11001 H 2 (g)CH 3 CH 3 (g) ?HH11034 H11005 H11002136 kJ (H1100232.6 kcal) H 2 C CH 2 (g) ?HH11034 H11005 H11002176 kJ (H1100242.1 kcal) 40 ALKANES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) Converting a ketone to an ester increases the oxygen content of carbon and requires an oxidizing agent. (c) Reduction occurs when the hydrogen content increases, as in the conversion of a ketone to an alkane or to an alcohol. Reductions are carried out by using reagents that are reducing agents. 2.40 Methyl formate is an ester. (a) The oxidation numbers of the two carbon atoms in methyl formate and the carbon atoms in the reaction products can be determined by comparison with the entries in text Table 2.6. There has been no change in oxidation state in going from reactants to products, and the reac- tion is neither oxidation nor reduction. The number of carbon–oxygen bonds does not change in this reaction. (b) As in part (a), the oxidation states of the carbon atoms in both the reactant and the products do not change in this reaction. The reaction is neither oxidation nor reduction. (c) The oxidation number of one carbon of methyl formate has decreased in this reaction. This reaction is a reduction and requires a reagent that is a reducing agent. (d) The oxidation number of both carbon atoms of methyl formate has increased. This reaction is an oxidation and requires use of a reagent that is an oxidizing agent. HCOCH 3 O H 2 O2 CO 2 H11001 Oxidation number H110012 H110022 H110014 HCOCH 3 O 2 CH 3 OH Oxidation number H110012 H110022 H110022 HCOCH 3 O H11001 Oxidation number H110012 H110022 HCONa O H110012 CH 3 OH H110022 HCOCH 3 O CH 3 OHH11001HCOH O Oxidation number H110012 H110022 H110012 H110022 CH 3 CC(CH 3 ) 3 O oxidation reduction reduction CH 3 COC(CH 3 ) 3 O CH 3 CHC(CH 3 ) 3 OH CH 3 CH 2 C(CH 3 ) 3 Ester Alkane Alcohol Compound A (d) ALKANES 41 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (e) Once again the formation of carbon dioxide is an example of an oxidation, and the reaction requires use of an oxidizing agent. 2.41 Two atoms appear in their elementary state: Na on the left and H 2 on the right. The oxidation state of an atom in its elementary state is 0. Assign an oxidation state of H110011 to the hydrogen in the OH group of CH 3 CH 2 OH. H goes from H110011 on the left to 0 on the right; it is reduced. Na goes from 0 on the left to H110011 on the right; it is oxidized. 2.42 Combustion of an organic compound to yield CO 2 and H 2 O involves oxidation. Heat is given off in each oxidation step. The least oxidized compound (CH 3 CH 2 OH) gives off the most heat. The most oxidized compound HO 2 CCO 2 H gives off the least. The measured values are: CH 3 CH 2 OH HOCH 2 CH 2 OH HO 2 CCO 2 H kJ/mol 1371 1179 252 kcal/mol 327.6 281.9 60.2 2.43–2.45 Solutions to molecular modeling exercises are not provided in this Study Guide and Solutions Man- ual. You should use Learning By Modeling for these exercises. SELF-TEST PART A A-1. Write the structure of each of the four-carbon alkyl groups. Give the common name and the systematic name for each. A-2. How many H9268 bonds are present in each of the following? (a) Nonane (b) Cyclononane A-3. Classify each of the following reactions according to whether the organic substrate is oxidized, reduced, or neither. A-4. (a) Write a structural formula for 3-isopropyl-2,4-dimethylpentane. (b) How many methyl groups are there in this compound? How many isopropyl groups? (a) CH 3 CH 3 H11001 Br 2 CH 3 CH 2 Br H11001 HBr light (b) CH 3 CH 2 Br H11001 HO H11002 CH 3 CH 2 OH H11001 Br H11002 (d) H 2 CCH 2 H11001 H 2 CH 3 CH 3 Pt (c) CH 3 CH 2 OH H 2 SO 4 heat H 2 CCH 2 2CH 3 CH 2 OH H11001 2Na 2CH 3 CH 2 ONa H11001 H 2 H1100110 H1100110 HCOCH 3 O CH 3 OHCO 2 H11001 Oxidation number H110012 H110022 H110014 H110022 42 ALKANES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website A-5. Give the IUPAC name for each of the following substances: A-6. The compounds in each part of the previous question contain ______ primary carbon(s), ______ secondary carbon(s), and ______ tertiary carbon(s). A-7. Give the IUPAC name for each of the following alkyl groups, and classify each one as primary, secondary, or tertiary. A-8. Write a balanced chemical equation for the complete combustion of 2,3-dimethylpentane. A-9. Write structural formulas, and give the names of all the constitutional isomers of C 5 H 10 that contain a ring. A-10. Each of the following names is incorrect. Give the correct name for each compound. (a) 2,3-Diethylhexane (b) (2-Ethylpropyl)cyclohexane (c) 2,3-Dimethyl-3-propylpentane A-11. Which C 8 H 18 isomer (a) Has the highest boiling point? (b) Has the lowest boiling point? (c) Has the greatest number of tertiary carbons? (d) Has only primary and quaternary carbons? A-12. Draw the constitutional isomers of C 7 H 16 that have five carbons in their longest chain, and give an IUPAC name for each of them. A-13. The compound shown is an example of the broad class of organic compounds known as steroids. What functional groups does the molecule contain? A-14. Given the following heats of combustion (in kilojoules per mole) for the homologous series of unbranched alkanes: hexane (4163), heptane (4817), octane (5471), nonane (6125), esti- mate the heat of combustion (in kilojoules per mole) for pentadecane. O O H 3 C H 3 C C OH OCH 3 (a) (CH 3 ) 2 CHCH 2 CHCH 3 (b) (CH 3 CH 2 ) 3 C (c) (CH 3 CH 2 ) 3 CCH 2 (a) (b) ALKANES 43 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website PART B B-1. Choose the response that best describes the following compounds: (a) 1, 3, and 4 represent the same compound. (b) 1 and 3 are isomers of 2 and 4. (c) 1 and 4 are isomers of 2 and 3. (d) All the structures represent the same compound. B-2. Which of the following is a correct name according to the IUPAC rules? (a) 2-Methylcyclohexane (c) 2-Ethyl-2-methylpentane (b) 3,4-Dimethylpentane (d) 3-Ethyl-2-methylpentane B-3. Following are the structures of four isomers of hexane. Which of the names given correctly identifies a fifth isomer? CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 (CH 3 ) 3 CCH 2 CH 3 (CH 3 ) 2 CHCH 2 CH 2 CH 3 (CH 3 ) 2 CHCH(CH 3 ) 2 (a) 2-Methylpentane (c) 2-Ethylbutane (b) 2,3-Dimethylbutane (d) 3-Methylpentane B-4. Which of the following is cyclohexylcyclohexane? B-5. Which of the following structures is a 3-methylbutyl group? (a)CH 3 CH 2 CH 2 CH 2 CH 2 G (c) (CH 3 CH 2 ) 2 CHG (b) (CH 3 ) 2 CHCH 2 CH 2 G (d) (CH 3 ) 3 CCH 2 G B-6. Rank the following substances in decreasing order of heats of combustion (most exother- mic → least exothermic). (a)2H11022 1 H11022 3(c)3H11022 1 H11022 2 (b)2H11022 3 H11022 1(d)3H11022 2 H11022 1 B-7. What is the total number of H9268 bonds present in the molecule shown? (a)18 (b)26 (c)27 (d)30 123 (a) (c) (b) (d) CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 12 3 4 44 ALKANES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-8. Which of the following substances is not an isomer of 3-ethyl-2-methylpentane? B-9. Which alkane has the highest boiling point? (a) Hexane (d) 2,3-Dimethylbutane (b) 2,2-Dimethylbutane (e) 3-Methylpentane (c) 2-Methylpentane B-10. What is the correct IUPAC name of the alkyl group shown? (a) 1-Ethyl-3-methylbutyl (b) 1-Ethyl-3,3-dimethylpropyl (c) 4-Ethyl-2-methylbutyl (d) 5-Methylhexyl B-11. Which of the following compounds is not a constitutional isomer of the others? (a) Methylcyclohexane (d) 1,1,2-Trimethylcyclobutane (b) Cyclopropylcyclobutane (e) Cycloheptane (c) Ethylcyclopentane B-12. The correct IUPAC name for the compound shown is (a) 2-Ethyl-5-isobutyl-3-methylhexane (d) 2-Ethyl-3,5,7-trimethyloctane (b)5-sec-Butyl-2-ethyl-3-methylhexane (e) 2,4,6,7-Tetramethylnonane (c) 2-Isobutyl-4,5-dimethylheptane B-13. The heats of combustion of two isomers, A and B, are 4817 kJ/mol and 4812 kJ/mol, respectively. From this information it may be determined that (a) Isomer A is 5 kJ/mol more stable (b) Isomer B is 5 kJ/mol less stable (c) Isomer B has 5 kJ/mol more potential energy (d) Isomer A is 5 kJ/mol less stable B-14. Which of the following reactions requires an oxidizing agent? (e) None of theseCH 2 (d) RCH 2 OH RCH O(a) RCH 2 OH RCH 2 Cl (b) RCH RCH 2 CH 3 (c) RCH 2 Cl RCH 3 CH 3 CH 2 CH 3 CH 3 CHCHCH 2 CHCH 3 CH 2 CH(CH 3 ) 2 CH 2 CH 3 CHCH 2 CH(CH 3 ) 2 (a)(c) (b d) None of these (all are isomers) ALKANES 45 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website