hapter 1 1 hapter 1 Unsaturated ydrocarbons Solutions to In-hapter Problems 1.1 To draw a complete structure for each condensed structure, first draw in the multiple bonds. Then draw in all the other s and s, as in Example 1.1. 2 2 O = O ( ) 2 ( 2 ) 2 = ( ) 2 2 ( ) = 1.2 To determine whether each molecular formula corresponds to a saturated hydrocarbon, alkene, or alkyne, recall that the formula for a saturated hydrocarbon is n 2n + 2, the formula for an alkene is n 2n, and the formula for an alkyne is n 2n 2. 6 = n 2n = alkene 8 14 = n 2n 2 = alkyne 5 12 = n 2n + 2 = saturated hydrocarbon 6 12 = n 2n = alkene 1. Use the general formulas [saturated hydrocarbon ( n 2n + 2 ), alkene ( n 2n ), and alkyne ( n 2n 2 )] to determine the molecular formula for each compoun alkene = n 2n, 4 2 = 8, 4 8 saturated hydrocarbon = n 2n + 2, (6 2) + 2 = 14, 6 14 alkyne = n 2n 2, (7 2) 2 = 12, 7 12 alkene = n 2n, 5 2 = 10, 5 10 1.4 Give the IUPA name for each compound using the following steps, as in Example 1.2: [1] Find the longest chain containing both carbon atoms of the multiple bon [2] Number the chain to give the double bond the lower number. [] Name and number the substituents and write the complete name. 2 2 1 2 4 5 2 2 Answer: -methyl-1-pentene 5 's in the longest chain pentene double bond at 1 methyl at
Unsaturated ydrocarbons 1 2 ( 2 ) 2 2 2 2 2 2 2 7 's in the longest chain heptene 1 2 ethyl at 2 2 2 2 4 5 6 7 double bond at Answer: -ethyl--heptene 7 6 5 4 2 1 2 = 2 = Answer: 2,4-heptadiene 7 's in the longest chain heptadiene double bonds at 2 and 4 1 2 5 Answer: -ethylcyclopentene 2 2 5 's in the ring cyclopentene 4 ethyl at 1.5 Give the IUPA name for each compound using the following steps, as in Example 1.2: [1] Find the longest chain containing both carbon atoms of the multiple bon [2] Number the chain to give the multiple bond the lower number. [] Name and number the substituents and write the complete name. 2 2 2 2 ( ) 2 2 2 2 2 9 's in the longest chain nonyne 2 2 2 2 5 4 2 1 triple bond at Answer: 2-methyl--nonyne 2 2 2 8 's in the longest chain octyne 7 8 6 1 2 4 5 2 2 2 triple bond at Answer: 6,6-dimethyl--octyne 2 methyl groups at 6 1.6 To draw the structure corresponding to each name, follow the steps in Example 1.. Identify the parent name to find the longest carbon chain or ring, and then use the suffix to determine the functional group; the suffix -ene = alkene and -yne = alkyne. Number the carbon chain or ring and place the functional group at the indicated carbon. Add the substituents and enough hydrogens to give each carbon four bonds. 1 2 4 5 6 4-methyl-1-hexene 2 2 2 6 carbon chain double bond at 1 methyl at 4
hapter 1 5-ethyl-2-methyl-2-heptene 7 carbon chain double bond at 2 1 2 4 5 6 7 methyl at 2 2 ethyl at 5 2 2 2 2,5-dimethyl--hexyne 6 carbon chain triple bond at 1 2 4 5 6 methyl at 2 methyl at 5! 1-propylcyclobutene 4 carbon ring double bond at 1 4 1 2 2 2 propyl at 1 2 2 e. 1,-cyclohexadiene 6 carbon ring 2 double bonds (1 and ) 6 5 1 4 2 f. 4-ethyl-1-decyne 1 0 9 8 7 6 5 4 2 1 2 2 2 2 2 2 10 carbon chain triple bond at 1 2 ethyl at 4 2 1.7 To draw the structures of the cis and trans isomers, follow the steps in Example 1.4. Use the parent name to draw the carbon skeleton and place the double bond at the correct carbon. Use the definitions of cis and trans to draw the isomers. When the two alkyl groups are on the same side of the double bond, the compound is called the cis isomer. When they are on opposite sides, it is called the trans isomer. cis-2-octene 2 2 2 2 8 carbon chain 1 2 4 5 6 7 8 2 2 2 2 cis isomer both alkyl groups on the same side trans--heptene 7 carbon chain 2 2 2 1 2 4 5 6 7 2 2 2 trans isomer alkyl groups on opposite sides
Unsaturated ydrocarbons 1 4 trans-4-methyl-2-pentene ( ) 5 carbon chain 1 2 4 5 trans isomer alkyl groups on opposite sides 1.8 Whenever the two groups on each end of a = are different from each other, two isomers are possible. 2 2 2 Each has one and one alkyl group. is and trans isomers are possible. two 's cannot have cis or trans isomers 2 2 2 two 's cannot have cis or trans isomers 1.9 When the two alkyl groups are on the same side of the double bond, the compound is called the cis isomer. When they are on opposite sides, it is called the trans isomer. O 2 ( 2 ) 7 2 trans cis 2 2 1.10 Stereoisomers differ only in the three-dimensional arrangement of atoms. onstitutional isomers differ in the way the atoms are bonded to each other. 2 and 2 2 2 bonded to one, one 2 bonded to two 's different connectivity constitutional isomers and 2 2 and 2 bonded to one 2 bonded to one and one and one 2 different connectivity constitutional isomers cis isomer trans isomer same connectivity different -D arrangement stereoisomers
hapter 1 5 1.11 Double bonds in naturally occurring fatty acids are cis. O 2 2 2 2 2 2 2 2 2 2 O arachidonic acid 1.12 Fats are solids at room temperature because of their higher melting point. They are formed from fatty acids with few double bonds. Oils are liquids at room temperature because of their lower melting points. They are also formed from fatty acids, but have more double bonds. ( 2 ) 14 OO palmitic acid no double bonds higher melting point 6 ( 2 ) 5 =( 2 ) 7 OO palmitoleic acid one double bond lower melting point 1 1.1 The functional groups in tamoxifen are labele ether amine O 2 2 N( ) 2 aromatic ring aromatic ring 2 aromatic ring alkene 1.14 The functional groups in RU 486 and levonorgestrel are labele amine aromatic ring hydroxyl ( ) 2 N alkene alkyne O 2 hydroxyl O alkyne ketone ketone O RU 486 alkene O alkene levonorgestrel
Unsaturated ydrocarbons 1 6 1.15 To draw the product of a hydrogenation reaction, use the following steps, as in Example 1.5: Locate the = and mentally break one bond in the double bon Mentally break the bond of the reagent. Add one atom to each of the =, thereby forming two new single bonds. 2 2 2 Pd 2 2 2 2 2 ( ) 2 2 Pd 2 2 ( ) 2 2 Pd 1.16 To draw the product of each halogenation reaction, add a halogen to both carbons of the double bon 2 2 + l 2 2 l l + 2 1.17 In hydrohalogenation reactions, the elements of and (or and l) must be added to the double bon When the alkene is unsymmetrical, the atom of X bonds to the carbon that has more s to begin with. + This does not have any 's. Add here. + This has more 's. Add here. This does not have any 's. Add l here. ( ) 2 + l l This has more 's. Add here.
hapter 1 7 + l l 1.18 In hydration reactions, the elements of and O are added to the double bon In unsymmetrical alkenes, the atom bonds to the less substituted carbon. O O 2 SO 4 This has one so the O bonds here. O O 2 2 2 2 SO 4 This has 2 's so the bonds here. O 2 SO 4 O 1.19 Draw the products of each reaction. 2 2 2 2 Pd 2 2 2 2 2 l 2 2 2 l l 2 2 2 2 2 2 2 2 2 2 2 e. 2 2 2 l 2 2 l f. 2 2 2 O 2 SO 4 O 2 2
Unsaturated ydrocarbons 1 8 1.20 Draw the products of the hydrogenation reactions. 2 2 2 2 2 2 2 2 2 2 OO 2 Pd 2 2 2 2 2 2 2 2 2 2 2 2 OO 2 2 2 2 2 2 2 2 2 2 2 2 OO 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 OO 1.21 To draw the polymers, draw three or more alkene molecules and arrange the carbons of the double bonds next to each other. eak one bond of each double bond, and join the alkenes together with single bonds. With unsymmetrical alkenes, substituents are bonded to every other carbon. Use Example 1.7 as a guide. Join these 2 's. Join these 2 's. 2 2 2 2 2 2 2 2 2 Join these 2 's. Join these 2 's. 2 2 N N 2 N N N N Join these 2 's. Join these 2 's. 2 2 2 l l l l l l
hapter 1 9 1.22 Work backwards to determine what monomer is used to form the polymer. eak these bonds to form the monomer. 2 2 2 O O O formed from 2 O O O O O poly(vinyl acetate) 1.2 Name each aromatic compound as in Example 1.8. Name the substituents on the benzene ring. With two groups, alphabetize the substituent names and use the prefix ortho, meta, or para to indicate their location. With three substituents, alphabetize the substituent names, and number to give the lowest set of numbers. 2 2 propyl O O on benzene ring = phenol m-butylphenol propylbenzene 2 2 2 butyl 2 ethyl on benzene ring = toluene bromo I iodo p-ethyliodobenzene l chloro 2-bromo-5-chlorotoluene 1.24 Draw the structure corresponding to each name. N 2 pentylbenzene!!! m-bromoaniline 2 2 2 2 o-dichlorobenzene l l!!! 4-chloro-1,2-diethylbenzene 2 2 l 1.25 ommercially available sunscreens contain a benzene ring. Therefore, compound (a) might be found in a sunscreen since it contains two aromatic rings. ompound (b) does not contain any aromatic rings. 1.26 Phenols are antioxidants because the O group on the benzene ring prevents unwanted oxidation reactions from occurring. Of the compounds listed, only curcumin (b) contains a phenol group (O on a benzene ring), making it an antioxidant.
Unsaturated ydrocarbons 1 10 1.26 Draw the products of each substitution reaction. hlorination replaces one of the s on the benzene ring with l. Nitration replaces one of the s on the benzene ring with NO 2. Sulfonation replaces one of the s on the benzene ring with SO. l l l l l l 2 Fel l l l SO 2 SO 4 l SO l l NO 2 NO 2 SO 4 l l 1.27 Draw the products of the substitution reaction. The l can replace any of the s on the benzene ring, giving three different products. l 2 Fel l l o-chlorotoluene m-chlorotoluene p-chlorotoluene l Solutions to End-of-hapter Problems 1.29 molecular formula: 10 12 O aromatic ring, alkene, ether trans Tetrahedral 's are indicate All other 's are trigonal planar. aromatic ring ether O trans alkene anethole tetrahedral tetrahedral 1.0 molecular formula: 10 12 O 2 aromatic ring, ester trans Tetrahedral is indicate All other 's are trigonal planar. aromatic ring trans alkene O O methyl cinnamate tetrahedral
hapter 1 11 1.1 Use the general formulas [saturated hydrocarbon ( n 2n + 2 ), alkene ( n 2n ), and alkyne ( n 2n 2 )] to determine the molecular formula for each compound with 10 s. (10 2) + 2 = 22: molecular formula 10 22 (10 2) 2 = 18: molecular formula 10 18 10 2 = 20: molecular formula 10 20 1.2 Draw the structure of the hydrocarbon fitting the required description. 2 2 2 2 2 2 2 1. Draw three alkynes with molecular formula 5 8. 2 2 2 1.4 Draw the five constitutional isomers of 5 10 that contain a double bon 2 2 2 2 2 2 2 1.5 Label each carbon as tetrahedral, trigonal planar, or linear by counting groups. trigonal planar tetrahedral tetrahedral trigonal planar trigonal planar tetrahedral linear all trigonal planar 1.6 Predict the indicated bond angles in falcarinol. a b O c 2 d ( 2 ) 6 e trigonal planar: 120 tetrahedral: 109.5 inear: 180
Unsaturated ydrocarbons 1 12 1.7 Give the IUPA name for each compoun 1 2 4 2 1 4 5 6 2-ethyl 6 chain hexene 2-hexyne 4 chain butene 2-ethyl-1-butene 1.8 Give the IUPA name for each compoun 2 2 2 2 1 2 7 chain 1- heptyne 2 5 carbon ring; ethyl at 2 2-ethylcyclopentene 1.9 Give the IUPA name for each compound using the following steps, as in Example 1.2: [1] Find the longest chain containing both carbon atoms of the multiple bon [2] Number the chain to give the multiple bond the lower number. [] Name and number the substituents and write the complete name. 2 2 2 ( ) 2 2 2 6 's in the longest chain hexene 1 2 4 5 2 2 2 double bond at 1 2 methyls at 5 Answer: 5,5-dimethyl-1-hexene ( 2 ) 2 2 ethyl at 2 methyls at 5 and 7 2 2 2 2 2 2 8 's in the longest chain 1 2 4 5 6 7 8 octene double bond at 1 2 ethyl at 2 Answer: -ethyl-5,7-dimethyl--octene 2 2 2 2 2 2 7 's in the longest chain heptene 2 2 2 2 2 2 4 5 6 7 double bond at 1 Answer: 2-ethyl-1-heptene
hapter 1 1 2 ( ) 1 2 2 2 6 6 's in the longest chain triple bond at 2 hexyne 2 methyls at 5 5 Answer: 5,5-dimethyl-2-hexyne 1 2 4 8 e. 2 2 2 2 Answer: 2 6-ethyl-5,6-dimethyl-2-octyne 2 8 's in the longest chain octyne triple bond at 2 2 methyls at 5 and 6 5 6 ethyl at 6 6 5 4 1 f. 2 2 = 2 2 2 = 2 Answer:,-dimethyl-1,5-hexadiene 6 's in the longest chain hexadiene 2 methyls at double bonds at 1 and 5 1.40 Give the IUPA name for each compound using the following steps, as in Example 1.2: [1] Find the longest chain containing both carbon atoms of the multiple bon [2] Number the chain to give the multiple bond the lower number. [] Name and number the substituents and write the complete name. 2 2 2 1 2 4 5 2 2 2 2 2 2 6 's in the longest chain double bond at 1 methyl at 4 hexene 6 Answer: 4-methyl-1-hexene ( ) 2 2 2 double bond at 2 2 2 ethyl at 5 1 2 5 2 2 2 2 Answer: 5-ethyl-2-methyl-2-octene 2 2 8 's in the longest chain octene methyl at 2 2 2 6 8
Unsaturated ydrocarbons 1 14 2 2 7 's in the longest chain heptyne triple bond at 2 methyl at 4 1 2 4 2 2 2 2 5 7 Answer: 4-methyl-2-heptyne ( ) ( ) triple bond at 6 2 5 6 's in the longest chain hexyne 4 methyls at 2 and 5 Answer: 2,2,5,5-tetramethyl--hexyne e. ( 2 2 2 ) 2 double bond at 2 2 2 2 2 2 2 2 2 2 2 2 2 7 1 Answer: -butyl-2-heptene 7 's in the longest chain heptene butyl at f. ( ) 2 2 2 2 ( ) 2 9 's in the longest chain nonene 1 double bonds at 2 and 7 2 7 8 9 2 2 2 2 2 2 methyls at 2 and 8 Answer: 2,8-dimethyl-2,7-nonadiene 1.41 Give the IUPA name for each compound using the steps in Answer 1.9 and Example 1.2. double bond at 1 2 2 4 2 1 4-methylcyclohexene 2 1 4 methyl at 4 4 carbon ring 6 carbon ring cyclobutene cyclohexene 2 ethyl groups at,-diethylcyclobutene
hapter 1 15 1.42 Give the IUPA name for each compound using the steps in Answer 1.9 and Example 1.2. double bond at 1 2 2 4 1 2 2 4-propylcylcopentene 1 propyl at 4 5 carbon ring cyclopentene double bonds at 1 and 4 4 5 7 carbon ring cycloheptene 1,4-cycloheptadiene 1.4 To draw the structure corresponding to each name, follow the steps in Example 1.. Identify the parent name to find the longest carbon chain or ring, and then use the suffix to determine the functional group; the suffix -ene = alkene and -yne = alkyne. Number the carbon chain or ring and place the functional group at the indicated carbon. Add the substituents and enough hydrogens to give each carbon four bonds. 1 2 4 5 6 -methyl-1-octene 8 carbon chain double bond at 1 7 8 methyl at 2 2 2 2 2 1-ethylcyclobutene 4 carbon ring double bond at 1 2 ethyl at 1 2 1 2 4 5 6 2-methyl--hexyne 6 carbon chain triple bond at methyl at 2 2,5-diethyl-2-methyl--heptene 7 carbon chain double bond at 2 1 2 4 5 6 7 2 methyl at 2 2 ethyl groups at and 5 2 2 2 1 2 4 5 6!e. 1,-heptadiene 7 carbon chain double bonds at 1 and 7 2 2 2 1 2 4 5 6 f. cis-7-methyl-2-octene 8 carbon chain double bond at 2 7 8 methyl at 7 2 2 2 cis
Unsaturated ydrocarbons 1 16 1.44 To draw the structure corresponding to each name, follow the steps in Example 1.. Identify the parent name to find the longest carbon chain or ring, and then use the suffix to determine the functional group; the suffix -ene = alkene and -yne = alkyne. Number the carbon chain or ring and place the functional group at the indicated carbon. Add the substituents and enough hydrogens to give each carbon four bonds. 1,2-dimethylcyclopentene 5 carbon ring double bond at 1 1 methyls at 1 and 2 2 1 2 6-ethyl-2-octyne 8 carbon chain triple bond at 2 1 2 4 5 6 7 8 ethyl at 6 2 2 2 2 2,-dimethyl-1,4-pentadiene 5 carbon chain double bonds at 1 and 4 1 2 4 5 2 methyls at 2 2 trans-5-methyl-2-hexene 6 carbon chain double bond at 2 1 2 4 5 6 2 methyl at 5 trans e. 5,6-dimethyl-2-heptyne 7 carbon chain triple bond at 2 1 2 4 5 6 7 methyls at 5 and 6 2 f.,4,5,6-tetramethyl-1-decyne 10 carbon chain triple bond at 1 1 2 4 5 6 7 8 9 10 methyls at, 4, 5 and 6 ( 2 ) 1.45 orrect each of the incorrect IUPA names. The name 5-methyl-4-hexene places the double bond at 4 instead of 2. Assign the lower number to the alkene: 2- methyl-2-hexene. The name 1-methylbutene makes the last carbon in the chain a substituent. In addition, the location of the double bond is not specifie There are five carbons in the chain (not a methyl substituent): 2-pentene. The name 2,-dimethylcyclohexene starts numbering substituents at 2 instead of 1. Number to put the = between 1 and 2, and then give the first substituent the lower number: 1,6-dimethylcyclohexene. 2 2 2-methyl-2-hexene 2 2-pentene 1 2 4 6 5 1,6-dimethylcyclohexene
hapter 1 17 The name -butyl-1-butyne does not name the longest chain. Name the seven carbon chain: -methyl-1-heptyne. 1 2 2 2 2 -methyl-1-heptyne 1.46 orrect each of the incorrect IUPA names. The alkene has two methyl groups off of 2. Therefore, it cannot be cis. The name 2-methyl-2,4-pentadiene places the double bonds at 2 and 4 instead of 1 and. The methyl group should be at 4. 2 2 2-methyl-2-hexene 2 4-methyl-1,-pentadiene The name 2,4-dimethylcyclohexene starts numbering substituents at 2 instead of 1. Number to put the = between 1 and 2, and then give the first substituent the lower number: 1,5-dimethylcyclohexene. 1 2 6 5 4 1,5-dimethylcyclohexene The name 1,1-dimethyl-2-cyclohexene does not number the double bond between 1 and 2. The methyl groups should be on. 1 2 6 5 4,-dimethylcyclohexene 1.47 When the two alkyl groups are on the same side of the double bond, the compound is called the cis isomer. When they are on opposite sides it is called the trans isomer. 2 2 2 2 cis trans O 2 2 2 OO 2 S 2 ON 2 OO NO 2 2 OO N 2
Unsaturated ydrocarbons 1 18 1.48 Draw the structures using the proper cis or trans arrangement around the carbon carbon double bon cis ( 2 ) 8 ( 2 ) 12 O trans 1.49 Give the IUPA name for the alkene. Use the definition in Answer 1.47 to determine if it is the cis or trans isomer. 5 4 1 2 cis-4-methyl-2-pentene 4-methyl cis 1.50 Give the IUPA name for the alkene. Use the definition in Answer 1.47 to determine if it is the cis or trans isomer. 4 1 2 2-methyl 6 7 5 trans trans-2-methyl--heptene 1.51 Draw the cis and trans isomers for each compound, as in Example 1.4. 2 2 2 2 2 cis-2-nonene 2 2 cis-2-methyl--heptene 2 2 2 2 2 trans-2-nonene 2 2 trans-2-methyl--heptene
hapter 1 19 1.52 Draw the cis and trans isomers for each compound, as in Example 1.4. 2 2 2 cis--heptene 2 2 2 trans--heptene 2 cis-4,4-dimethyl-2-hexene 2 trans-4,4-dimethyl-2-hexene 1.5 onstitutional isomers have the same molecular formula, but have the atoms bonded to different atoms. Stereoisomers have atoms bonded to the same atoms but in a different three-dimensional arrangement. 1.54 Draw the possible stereoisomers for 2,4-hexadiene. 1.55 Determine if the molecules are constitutional isomers, stereoisomers, or identical. and same molecular formula same connectivity identical and same molecular formula different connectivity constitutional isomers
Unsaturated ydrocarbons 1 20 1.56 Determine if the molecules are constitutional isomers, stereoisomers or identical. and same molecular formula same connectivity identical and same molecular formula same connectivity different arrangement in space (cis and trans isomers) stereoisomers 1.57 Draw the products of each reaction by adding 2 to the double bon 2 2 2 2 2 Pd 2 2 2 2 2 Pd ( ) 2 2 2 2 Pd ( ) 2 2 2 2 2 2 Pd 1.58 Draw the products of each reaction by adding 2 to the double bon 2 2 2 2 2 2 2 2 2 2 ( ) 2 2 2 2 ( ) 2 2 2 2 2 2 1.59 Draw the products of each reaction by adding l to the double bon l l 2 2 ( ) 2 l l 2 ( ) 2 ( ) 2 ( ) 2 l l ( ) 2 ( ) 2 2 l l 1.60 Draw the products of each reaction by adding 2 O to the double bon 2 SO + 4 2 O O 2 2 ( ) 2 + 2 O 2 SO 4 O 2 ( ) 2 ( ) 2 ( ) 2 + 2 O 2 SO 4 O ( ) 2 ( ) 2 2 + 2 O 2 SO 4 O
hapter 1 21 1.61 Draw the products of each reaction by adding the specified reagent to the double bon 2 2 Pd 2 2 l l 2 2 l2 l 2 l e. 2 2 2 2 2 f. 2 2 O 2 SO 4 O 2 1.62 Draw the products of each reaction by adding the specified reagent to the double bon + l 2 l 2 2 + 2 2 2 + 2 Pd l 2 l l e. f. ( ) 2 2 + ( ) 2 2 2 + 2 O 2 SO 4 O 1.6 Work backwards to determine what alkene is needed as a starting material to prepare each of the alkyl halides or dihalides. 2 2 2 l 2 l l l l 2 2 ( ) 2 2 2 2 ( ) 2 1.64 Markovnikov s rule must be followed when determining the starting materials. 2-omobutane can be formed as the only product of the addition of to 1-butene and 2-butene. 2- omopentane can be formed as the only product of the addition of to 1-pentene. 2 2 1-butene + 2 2-bromobutane + 2 2-butene 2-bromobutane 2 2 2 + 2 2 1-pentene 2-bromopentane
Unsaturated ydrocarbons 1 22 1.65 Work backwards to determine what reagent is needed to convert 2-methylpropene to each product. ( ) 2 = 2 l ( ) l ( ) 2 = 2 ( ) ( ) 2 = 2 ( ) 2 = 2 2 Pd 2 O 2 SO 4 ( ) ( ) O e. f. ( ) 2 = 2 ( ) 2 = 2 2 l 2 ( ) 2 2 ( ) 2 2 l l 1.66 The addition of 2 could be used to tell the difference between cyclohexane and cyclohexene. 2 is red in color. There is no reaction when 2 is added to cyclohexane, so the solution would remain re The bromines will add across the double bond, though, when 2 is added to cyclohexene, thus yielding colorless 1,2-dibromocyclohexane. The addition of 2 could also be used to distinguish between cyclohexene and benzene. When 2 is added to cyclohexene, 1,2-dibromocyclohexane is formed and the red color of the 2 disappears. When 2 is added to a solution containing benzene, the red color will remain because the benzene will not undergo a subsitution reaction, except in the presence of Fe. 1.67 To draw the polymer, draw three or more alkene molecules and arrange the carbons of the double bonds next to each other. eak one bond of each double bond, and join the alkenes together with single bonds. With unsymmetrical alkenes, substituents are bonded to every other carbon. Use Example 1.7 as a guide. Join these 2 's. Join these 2 's. 2 OO 2 OO 2 OO OO OO OO 2 2 2 1.68 Draw the polymer using the steps in Example 1.7. Join these 2 's. Join these 2 's. 2 OO 2 OO 2 OO OO OO OO 2 2 2 1.69 Draw the polymers using the steps in Example 1.7. Join these 2 's. Join these 2 's. 2 2 2 2 2 2 2 2 2 2 2 2 Join these 2 's. Join these 2 's. 2 l N 2 l N 2 l N l l l 2 2 2 N N N
hapter 1 2 2 Join these 2 's. l l 2 l l Join these 2 's. l 2 l l 2 l l 2 l l 2 l 1.70 Draw the polymers using the steps in Example 1.7. Join these 2 's. Join these 2 's. 2 O 2 O 2 O 2 O O O 2 2 Join these 2 's. Join these 2 's. 2 l 2 O 2 l 2 O 2 l O 2 l l l 2 2 2 O 2 O 2 O 2 Join these 2 's. Join these 2 's. 2 2 NO 2 NO NO 2 2 2 NO NO NO 1.71 Work backwards to determine what monomer was used to form the polymer. Each one of these units is from the monomer: 2 l 2 l 2 l 2 l 1.72 Work backwards to determine what monomer was used to form the polymer. Each one of these units is from the monomer: 2 2 2 2 O O O O O 2 O 2 O 2 O 2 1.7 Draw two resonance structures by moving the double bonds. l l
Unsaturated ydrocarbons 1 24 1.74 The structures A and B are the same compound even though A has the two l atoms on the same double bond and B has the two l atoms on different double bonds, because the two structures are resonance structures. They differ in the placement of the electrons, but the placement of the atoms is the same. 1.75 Name each aromatic compound as in Example 1.8. Name the substituents on the benzene ring. With two groups, alphabetize the substituent names and use the prefix ortho, meta, or para to indicate their location. With three substituents, alphabetize the substituent names, and number to give the lowest set of numbers. bromo ethyl chloro fluoro p-chloroethylbenzene o-bromofluorobenzene 1.76 Name each aromatic compound as in Example 1.8. Name the substituents on the benzene ring. With two groups, alphabetize the substituent names and use the prefix ortho, meta, or para to indicate their location. With three substituents, alphabetize the substituent names, and number to give the lowest set of numbers. O 2 N propyl phenol m-propylphenol aniline p-bromoaniline bromo 1.77 Name each aromatic compound as in Example 1.8 and Answer 1.75. l NO 2 chloro nitro m-chloronitrobenzene ethyl butyl o-butylethylbenzene l O 2 N NO 2 nitro p-nitroaniline l N 2 on benzene ring = aniline O on benzene ring = phenol 2,5-dichloro 2,5-dichlorophenol
hapter 1 25 1.78 Name each aromatic compound as in Example 1.8. Name the substituents on the benzene ring. With two groups, alphabetize the substituent names and use the prefix ortho, meta, or para to indicate their location. With three substituents, alphabetize the substituent names, and number to give the lowest set of numbers. ( 2 ) ( 2 ) p-dibutylbenzene 2 butyl groups 2 N 2 ethyl m-ethylaniline N 2 on benzene ring = aniline 2 ethyl o-bromoethylbenzene bromo bromo I iodo 2-bromo--iodotoluene on benzene ring = toluene 1.79 Draw and name the three isomers with l and N 2 as substituents. Recall that a benzene ring with an N 2 group is named aniline. N 2 l N 2 N 2 l l o-chloroaniline m-chloroaniline p-chloroaniline 1.80 Draw the structure of 2,4,6-trinitrotoluene (TNT). O 2 N NO 2 NO 2 1.81 Work backwards to draw the structure from the IUPA name. NO 2 nitro p-nitropropylbenzene 2 2 propyl 2 2 2 butyl m-dibutylbenzene 2 2 2 butyl O O on benzene ring = phenol I iodo o-iodophenol on benzene ring = toluene 1 2 bromo 2-bromo-4-chlorotoluene 4 l chloro N 2 N 2 on benzene ring = aniline e. I 1 6 2 l 2-chloro-6-iodoaniline 5 4 iodo chloro
Unsaturated ydrocarbons 1 26 1.82 Work backwards to draw the structure from the IUPA name. NO 2 nitro NO 2 nitro 2 m-ethylnitrobenzene ethyl O 2 N NO 2 1,,5-trinitrobenzene F F fluoro o-difluorobenzene e. O O on benzene ring = phenol 2,4-dibromophenol bromo on benzene ring = toluene p-bromotoluene bromo 1.8 Draw the products of each reaction. l SO l 2 Fel SO 2 SO 4 NO 2 NO 2 SO 4 1.84 Determine the reagents for the reactions in the sequence. A B l l l 2 NO SO Fel 2 SO 4 O 2 N 2 SO 4 O 2 N l SO D 2 Pd 2 N l SO 1.85 Draw the three products formed in the reaction of bromobenzene. NO 2 SO 4 NO 2 NO 2 NO 2
hapter 1 27 1.86 Draw the structures that result from the substitution of a chloro group onto the benzene ring. + l 2 Fel l + l Fel 2 + l l A + l 2 Fel l + + B l l 1.87 Vitamin E is an antioxidant because of the phenol, which has an O bonded to the benzene ring. 1.88 BA is an ingredient in some breakfast cereals and other packaged foods because it is a synthetic antioxidant and can prevent oxidation and spoilage. 1.89 Methoxychlor is more water soluble than DDT. The O groups can hydrogen bond to water. This increase in water solubility makes methoxychlor more biodegradable. 1.90 2,4-D is soluble in water because it contains an O 2 OO group. This group can hydrogen bond to water through the oxygen bound to the benzene ring and the two oxygens on the carboxy group (OO). DDT has no groups that are able to hydrogen bond to water, so DDT is insoluble in water. 1.91 2 2 2 2 2 2 2 2 2 2 2 2 OO 2 2 2 2 2 2 2 2, Pd 2 2 2 2 2 2 2 OO 2 2 2 2 + Partial hydrogenation adds hydrogen to one of the double bonds. 2 2 2 2 2 2 2 2 2 2 OO omplete hydrogenation adds hydrogen to both of the double bonds, forming: 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 OO
Unsaturated ydrocarbons 1 28 one possibility: 2 2 2 2 2 2 2 2 2 2 2 2 2 2 OO trans 1.92 Draw the structures and determine which will have the higher melting point. ( 2 ) 2 2 ( 2 ) 7 OO ( 2 ) 2 2 ( 2 ) 7 OO The all-trans isomer will have the higher melting point because it has a more linear shape than the isomer with the cis double bon As a result, the all-trans isomer packs more closely together in the solid, and thus requires more energy to separate upon melting. 1.9 Recall from Section 1.12 that many phenols are antioxidants. 2 2 2 O O O O an alcohol not an antioxidant an ether not an antioxidant This compound could be an antioxidant because it has an O group bonded to the aromatic ring. 1.94 All commercial sunscreens contain a benzene ring. The structure in part a contains two benzene rings and therefore might be an ingredient in a commercial sunscreen. The structure in part b contains two cyclohexane rings and therefore would not be an ingredient in a commercial sunscreen. 1.95 When benzene is oxidized to phenol, it is converted to a more water-soluble compound that can then be excreted in the urine. 1.96 A PA is a polycyclic aromatic hydrocarbon, a compound that contains two or more benzene rings that share carbon carbon bonds. The structure of anthracene, a PA mentioned in Section 1.10D, is shown below. 1.97 ( 2 ) 7 OO ( 2 ) 5 ( 2 ) 7 OO ( 2 ) 5 ( 2 ) 4 ( 2 ) 8 OO cis palmitoleic acid trans stereoisomer constitutional isomer one possibility 1.98 Polyethylene is a long chain hydrocarbon. Water and carbon dioxide are formed when polyethylene undergoes combustion.
hapter 1 29 1.99 All the carbons in benzene are trigonal planar with 120 bond angles, resulting in a flat ring. In cyclohexane, all the carbons are tetrahedral, so the ring is puckere 1.100 p-dichlorobenzene is a nonpolar molecule but o-dichlorobenzene is a polar molecule because the p-dichlorobenzene molecule is symmetrical and therefore does not have a net dipole moment. o- Dichlorobenzene, on the other hand, has a net dipole. l l l l nonpolar the dipoles cancel polar 1.101 2 =( 2 ) 4 7 chain 1-heptene 2 2 =( 2 ) 4 ( 2 ) 5 2 O 2 =( 2 ) 4 (O) 2 2 2 2 polymerization: R 2 R 2 R 2 R = ( 2 ) 4 1.102 2 =( 2 ) 7 10 chain 1-decene 2 2 =( 2 ) 7 ( 2 ) 8 2 O 2 =( 2 ) 7 (O) 2 2 2 2 2 2 2 polymerization: R 2 R 2 R 2 R = ( 2 ) 7 1.10 cis-2-exene and trans--hexene are constitutional isomers because the double bond is located in a different place on the carbon chain (2 vs. ). 2 2 2 2 cis-2-hexene trans--hexene
Unsaturated ydrocarbons 1 0 1.104 Determine the two monomer units for Saran. 2 l 2 l l 2 l 2 l l 2 l + 2 l l A B A B A B