Chapter 13–1

Chapter 13 Unsaturated Hydrocarbons

Solutions to In-Chapter Problems

13.1 To draw a complete structure for each condensed structure, first draw in the multiple bonds. Then draw in all the other C’s and H’s, as in Example 13.1.

H H H H H H H = a. CH2 CHCH2OH = H CCC O H b. (CH3)2CCH(CH2)2CH3 H C CCC C C H H H H H H H H C H H H H C H H H H H

c. (CH3)2CHC CCH2C(CH3)3 = H C CCC C C C H H H H H C H H C H H H

13.2 To determine whether each molecular formula corresponds to a saturated hydrocarbon, alkene, or alkyne, recall that the formula for a saturated hydrocarbon is CnH2n + 2, the formula for an alkene is CnH2n, and the formula for an alkyne is CnH2n – 2.

a. C3H6 = CnH2n = alkene c. C8H14 = CnH2n – 2 = alkyne b. C5H12 = CnH2n + 2 = saturated hydrocarbon d. C6H12 = CnH2n = alkene

13.3 Use the general formulas [saturated hydrocarbon (CnH2n + 2), alkene (CnH2n), and alkyne (CnH2n – 2)] to determine the molecular formula for each compound.

a. alkene = CnH2n, 4 × 2 = 8, C4H8 b. saturated hydrocarbon = CnH2n + 2, (6 × 2) + 2 = 14, C6H14 c. alkyne = CnH2n – 2, (7 × 2) – 2 = 12, C7H12 d. alkene = CnH2n, 5 × 2 = 10, C5H10

13.4 Give the IUPAC name for each compound using the following steps, as in Example 13.2: [1] Find the longest chain containing both carbon atoms of the multiple bond. [2] Number the chain to give the double bond the lower number. [3] Name and number the substituents and write the complete name.

12345 a. CH2 CHCHCH2CH3 CH2 CHCHCH2CH3 Answer: 3-methyl-1-pentene

CH3 CH3 double bond at C1 5 C's in the longest chain methyl at C3 pentene

Unsaturated Hydrocarbons 13–2

b. (CH3CH2)2CCHCH2CH2CH3 ethyl at C3

3 CH2CH3 12 CH2CH3

CH3CH2C CHCH2CH2CH3 CH3CH2C CHCH2CH2CH3 Answer: 3-ethyl-3-heptene 4 5 6 7 7 C's in the longest chain double bond at C3 heptene

7 6 5 1234 c. CH3CH2CH CHCH=CHCH3 CH3CH2CH CHCH=CHCH3 Answer: 2,4-heptadiene

7 C's in the longest chain double bonds at C2 and C4 heptadiene

1 2 3 d. 5 Answer: 3-ethylcyclopentene CH2CH3 CH2CH3 5 C's in the ring 4 cyclopentene ethyl at C3

13.5 Give the IUPAC name for each compound using the following steps, as in Example 13.2: [1] Find the longest chain containing both carbon atoms of the multiple bond. [2] Number the chain to give the multiple bond the lower number. [3] Name and number the substituents and write the complete name.

a. CH3CH2CH2CH2CH2CCCH(CH3)2

CH3 CH3

CH3CH2CH2CH2CH2CCCHCH3 CH3CH2CH2CH2CH2CCCHCH3 Answer: 2-methyl-3-nonyne 5 1 9 C's in the longest chain 234 triple bond at C3 nonyne

78 6 CH2CH3 12 34 5 CH2CH3 b. CH3CH2 CCCH2 C CH3 CH3CH2 CCCH2 C CH3 Answer: 6,6-dimethyl-3-octyne

CH3 CH3 8 C's in the longest chain triple bond at C3 2 methyl groups at C6 octyne

13.6 To draw the structure corresponding to each name, follow the steps in Example 13.3. • 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.

12 3 4 5 6

a. 4-methyl-1-hexene CCC C C C CH2 CHCH2CHCH2CH3

6 carbon chain CH3 CH3 double bond at C1 methyl at C4

Chapter 13–3

12 3 4 5 6 7 b. 5-ethyl-2-methyl-2-heptene C CCC C C C CH3C CHCH2CHCH2CH3

7 carbon chain CH3 CH2CH3 CH3 CH2CH3 double bond at C2 methyl at C2 ethyl at C5

123456 CH3 c. 2,5-dimethyl-3-hexyne CCC C C C CH3CHC CCHCH3 6 carbon chain CH3 CH3 CH triple bond at C3 3 methyl at C2 methyl at C5

1 CH CH CH CH2CH2CH3 4 2 2 3 !d. 1-propylcyclobutene C C propyl at C1 CC 4 carbon ring 3 2 double bond at C1

1 2 6 e. 1,3-cyclohexadiene 6 carbon ring 5 3 4 2 double bonds (C1 and C3)

10 91 78 6 345 12 f. 4-ethyl-1-decyne C C C C C C C C CC CH3CH2CH2CH2CH2CH2CHCH2C CH CH CH 10 carbon chain 2 3 CH2CH3 triple bond at C1 ethyl at C4

13.7 To draw the structures of the cis and trans isomers, follow the steps in Example 13.4. • 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.

H H a. cis-2-octene CH3CH CHCH2CH2CH2CH2CH3 CC 8 carbon chain 12345678 CH3 CH2CH2CH2CH2CH3 cis isomer both alkyl groups on the same side

H CH2CH2CH3

b. trans-3-heptene CH3CH2CH CHCH2CH2CH3 CC

7 carbon chain 12 3 4567 CH3CH2 H trans isomer alkyl groups on opposite sides

Unsaturated Hydrocarbons 13–4

CH3

H CHCH3

c. trans-4-methyl-2-pentene CH3CH CHCH(CH3)CH3 CC 5 carbon chain 12 34 5 CH3 H trans isomer alkyl groups on opposite sides

13.8 Whenever the two groups on each end of a C=C are different from each other, two isomers are possible.

CH3 a. CH CH CH CHCH c. 3 2 3 CH3 C CHCH2CH2CH3

Each C has one H and one alkyl group. two CH3's Cis and trans isomers are possible. cannot have cis or trans isomers

b. CH2 CHCH2CH2CH3

two H's cannot have cis or trans isomers

13.9 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.

cis H H H C C

CC CH2CH2CH3

HOCH2(CH2)7CH2 H trans

13.10 Stereoisomers differ only in the three-dimensional arrangement of atoms. Constitutional isomers differ in the way the atoms are bonded to each other. a. CH3CH CHCH2CH3 and CH2 CHCH2CH2CH3

CH3CH2 H CH3CH2 CH3 C bonded to one H, one CH3 C bonded to two H's c. CC and CC different connectivity CH constitutional isomers 3 H H H C bonded to one CH2CH3 C bonded to one H CH CH H and one CH3 and one CH CH CH3CH2 CH3 3 2 2 3 b. CC different connectivity CC and constitutional isomers H H H CH3

cis isomer trans isomer same connectivity different 3-D arrangement stereoisomers

Chapter 13–5

13.11 Double bonds in naturally occurring fatty acids are cis.

H H H H H H H H CC C C CC C C O

CH3CH2CH2CH2CH2 CH2 CH2 CH2 CH2CH2CH2COH

arachidonic acid

13.12 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.

CH3(CH2)14COOH CH3(CH2)5CH=CH(CH2)7COOH palmitic acid palmitoleic acid no double bonds one double bond higher melting point lower melting point 63 °C 1 °C

13.13 The functional groups in tamoxifen are labeled.

ether amine

OCH2CH2N(CH3)2

aromatic ring aromatic ring

C C

CH3CH2 aromatic ring

alkene

13.14 The functional groups in RU 486 and levonorgestrel are labeled.

amine aromatic ring hydroxyl hydroxyl alkyne (CH3)2N alkyne OH CH CH OH CH3 3 2 C C CH3 C CH alkene

ketone ketone O O alkene alkene RU 486 levonorgestrel

Unsaturated Hydrocarbons 13–6

13.15 To draw the product of a hydrogenation reaction, use the following steps, as in Example 13.5: • Locate the C=C and mentally break one bond in the double bond. • Mentally break the H–H bond of the reagent. • Add one H atom to each C of the C=C, thereby forming two new C–H single bonds.

H a. 2 CH3CH2CH CHCH2CH3 CH3CH2CH2CH2CH2CH3 Pd

CH3 CH2CH(CH3)2 H2 CH CHCH CH CH(CH ) b. CC 3 2 2 3 2 Pd CH CH3 H 3

CH CH3 3 H c. 2 Pd

13.16 To draw the product of each halogenation reaction, add a halogen to both carbons of the double bond. Cl Cl CH CH C C H a. CH3CH2CH CH2 + Cl2 3 2 H H

Br CH 3 CH3

b. + Br2 CH 3 CH3 Br

13.17 In hydrohalogenation reactions, the elements of H and Br (or H and Cl) must be added to the double bond. When the alkene is unsymmetrical, the H atom of HX bonds to the carbon that has more H’s to begin with.

H H CH CH CHCH + HBr a. 3 3 CH3CCCH3 Br H

This C does not have any H's. Add Br here. Br CH3 CH b. + HBr 3 H This C has more H's. H Add H here.

This C does not have any H's. Add Cl here.

CH3 H (CH ) CCHCH + c. 3 2 3 HCl CH3 CCCH3 Cl H This C has more H's. Add H here.

Chapter 13–7

H Cl d. + HCl H H

13.18 In hydration reactions, the elements of H and OH are added to the double bond. In unsymmetrical alkenes, the H atom bonds to the less substituted carbon.

HO H HOH a. CH3CH CHCH3 CH3CCCH3 H2SO4 H H

This C has one H so the OH bonds here. HO H HOH b. CH3CH2CH CH2 CH3CH2C C H H2SO4 H H This C has 2 H's so the H bonds here.

CH OH 3 HOH c. CH3 H2SO4 CH CH 3 3 H

13.19 Draw the products of each reaction.

H H H2 a. CH CH CH CH CH CH CH CH C C H 3 2 2 2 Pd 3 2 2 H H

H H Cl2 C H b. CH3CH2CH2CH CH2 CH3CH2CH2C Cl Cl

Br Br Br2 c. CH3CH2CH2CH CH2 CH3CH2CH2C C H H H

H H HBr d. CH CH CH C C H CH3CH2CH2CH CH2 3 2 2 Br H

H H HCl e. CH3CH2CH2CH CH2 CH3CH2CH2C C H Cl H

HO H HOH f. CH3CH2CH2CH CH2 CH3CH2CH2C C H H2SO4 H H

Unsaturated Hydrocarbons 13–8

13.20 Draw the products of the hydrogenation reactions.

H H H H H H a. CC C C CC CH CH 3 2 CH2 CH2 CH2CH2CH2CH2CH2CH2CH2COOH

H2 Pd H H H H CC CC

CH3CH2 CH2CH2CH2CH2 CH2CH2CH2CH2CH2CH2CH2COOH

H H H H CC CC CH CH CH 3 2 2 CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2COOH

b. CH3CH2CH2CH2CH2CH2CH2 CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2COOH

13.21 To draw the polymers, draw three or more alkene molecules and arrange the carbons of the double bonds next to each other. Break 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 13.7 as a guide.

Join these 2 C's. Join these 2 C's.

CH3 CH3 CH3 H CH3 H CH3 H CH3 a. CH2 C CH2 C CH2 C C C C C C C CH CH CH CH CH CH 2 3 2 3 2 3 H CH2 H CH2 H CH2 CH3 CH3 CH3

Join these 2 C's. Join these 2 C's.

CH3 CH3 CH 3 H CH3 H CH3 H CH3 b. CH C CH C CH C 2 2 2 C C C C C C CN CN CN H CN H CN H CN

Join these 2 C's. Join these 2 C's. H H H H H H H H H C C C C C C c. CH2 C CH2 C CH2 C H H H Cl Cl Cl Cl Cl Cl

Chapter 13–9

13.22 Work backwards to determine what monomer is used to form the polymer.

Break these bonds to form the monomer.

H H H H formed from CH C CH C CH C 2 2 2 CH2 C O O O O COCH3 COCH3 COCH3 COCH3 poly(vinyl acetate)

13.23 Name each aromatic compound as in Example 13.8. 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.

OH OH on benzene ring = phenol CH2CH2CH3 propyl a. c. m-butylphenol

propylbenzene CH2CH2CH2CH3 butyl

CH3 CH3 on benzene ring = toluene CH2CH3 ethyl Br bromo b. d. I Cl 2-bromo-5-chlorotoluene p-ethyliodobenzene iodo chloro

13.24 Draw the structure corresponding to each name.

NH2 a. pentylbenzene !!!c. m-bromoaniline

CH2CH2CH2CH2CH3 Br

Cl CH2CH3 Cl b. o-dichlorobenzene !!!d. 4-chloro-1,2-diethylbenzene CH2CH3

Cl

13.25 Commercially available sunscreens contain a benzene ring. Therefore, compound (a) might be found in a sunscreen since it contains two aromatic rings. Compound (b) does not contain any aromatic rings.

13.26 Phenols are antioxidants because the OH group on the benzene ring prevents unwanted oxidation reactions from occurring. Of the compounds listed, only curcumin (b) contains a phenol group (OH on a benzene ring), making it an antioxidant.

Unsaturated Hydrocarbons 13–10

13.26 Draw the products of each substitution reaction. • Chlorination replaces one of the H’s on the benzene ring with Cl. • Nitration replaces one of the H’s on the benzene ring with NO2. • Sulfonation replaces one of the H’s on the benzene ring with SO3H.

Cl Cl Cl Cl

Cl2 SO3 a. c. FeCl H2SO4 3 Cl SO3H Cl Cl Cl Cl Cl Cl

NO2 HNO3 b. H2SO4 Cl Cl

13.27 Draw the products of the substitution reaction. The Cl can replace any of the H’s on the benzene ring, giving three different products.

CH CH3 CH3 CH3 3 Cl Cl2

FeCl3 Cl Cl o-chlorotoluene m-chlorotoluene p-chlorotoluene

Solutions to End-of-Chapter Problems

13.29 aromatic ring a. molecular formula: C10H12O ether O b. aromatic ring, alkene, ether trans alkene CH c. trans 3 tetrahedral

d. Tetrahedral C's are indicated. All other C's are CH3 trigonal planar. anethole tetrahedral

13.30

trans alkene O a. molecular formula: C10H12O2 b. aromatic ring, ester aromatic ring O c. trans d. Tetrahedral C is indicated. All other C's are CH3 tetrahedral trigonal planar. methyl cinnamate

Chapter 13–11

13.31 Use the general formulas [saturated hydrocarbon (CnH2n + 2), alkene (CnH2n), and alkyne (CnH2n – 2)] to determine the molecular formula for each compound with 10 C’s.

a. (10 × 2) + 2 = 22: molecular formula C10H22 c. (10 × 2) – 2 = 18: molecular formula C10H18 b. 10 × 2 = 20: molecular formula C10H20

13.32 Draw the structure of the hydrocarbon fitting the required description.

a. HC CCH2CH2CH2CH3 b. CH2 CHCH2CH2CH CH2 c.

13.33 Draw three alkynes with molecular formula C5H8.

CH3

HC CCH2CH2CH3 CH3CCCH2CH3 HC C C H CH3

13.34 Draw the five constitutional isomers of C5H10 that contain a double bond.

CH3

CH2 CHCH2CH2CH3 CH3CH CHCH2CH3 CH2 CCH2CH3

CH3 CH3 CHCH CH CH3C 3 2 CHCHCH3

13.35 Label each carbon as tetrahedral, trigonal planar, or linear by counting groups.

trigonal planar tetrahedral tetrahedral a. c. trigonal planar

trigonal planar tetrahedral

linear

b. C CH

all trigonal planar

13.36 Predict the indicated bond angles in falcarinol.

a b OH c

H CCC C C C C CH2 C C (CH2)6CH3 e H H H d H H

a. trigonal planar: 120° b. tetrahedral: 109.5° c. inear: 180°

Unsaturated Hydrocarbons 13–12

13.37 Give the IUPAC name for each compound.

a. b. 1 23 4 2 1 3 5 4 6

6 C chain 2-hexyne 2-ethyl hexene

4 C chain 2-ethyl-1-butene butene

13.38 Give the IUPAC name for each compound.

1

a. HC C CH2CH2CH2CH2CH3 b. 2 CH CH 7 C chain 1- heptyne 2 3 5 carbon ring; ethyl at C2 2-ethylcyclopentene

13.39 Give the IUPAC name for each compound using the following steps, as in Example 13.2: [1] Find the longest chain containing both carbon atoms of the multiple bond. [2] Number the chain to give the multiple bond the lower number. [3] Name and number the substituents and write the complete name. a. CH2 CHCH2CH2C(CH3)3 5 CH3 1 23 4 CH3

CH2 CHCH2CH2CCH3 CH2 CHCH2CH2CCH3 Answer: 5,5-dimethyl-1-hexene

CH3 CH3 6 C's in the longest chain double bond at C1 hexene 2 methyls at C5

CH3 CH3 b. (CH3CH2)2C CHCHCH2CHCH3 ethyl at C3 2 methyls at C5 and C7

CH CH CH3CH2 CH3 CH3 3 2 CH3 CH3

CH3CH2CCHCHCH2CHCH3 CH3CH2CCHCHCH2CHCH3 Answer: 3-ethyl-5,7-dimethyl-3-octene 8 C's in the longest chain 12345678 octene double bond at C3 12 ethyl at C2 c. CH2 C CH CH CH2 C CH CH 2 3 2 3 Answer: 2-ethyl-1-heptene CH2CH2CH2CH2CH3 CH2CH2CH2CH2CH3 345 67 7 C's in the longest chain double bond at C1 heptene

Chapter 13–13

d. CH3CCCH2C(CH3)3 5 CH3 12 3 CH3

CH3CCCH2CCH3 CH3CCCH2CCH3 Answer: 5,5-dimethyl-2-hexyne CH3 CH3 6 6 C's in the longest chain triple bond at C2 hexyne 2 methyls at C5

2 methyls at C5 and C6

5 6

CH3 CH3 12 34 CH3 CH3 8 e. CH3CCCH2 CH C CH2CH3 CH3CCCH2 CH C CH2CH3 Answer: 6-ethyl-5,6-dimethyl-2-octyne CH2CH3 CH2CH3 8 C's in the longest chain triple bond at C2 octyne ethyl at C6

3 CH3 654 CH3 1 f. CH2 CHCH2 C CH=CH2 CH2 CHCH2 C CH=CH2 Answer: 3,3-dimethyl-1,5-hexadiene CH3 CH3 6 C's in the longest chain hexadiene 2 methyls at C3 double bonds at C1 and C5

13.40 Give the IUPAC name for each compound using the following steps, as in Example 13.2: [1] Find the longest chain containing both carbon atoms of the multiple bond. [2] Number the chain to give the multiple bond the lower number. [3] Name and number the substituents and write the complete name.

a. CH2 CHCH2CHCH2CH3

CH3 1 23 45 6 CH2 CHCH2CHCH2CH3 CH2 CHCH2CHCH2CH3 Answer: 4-methyl-1-hexene CH3 CH3 double bond at C1 6 C's in the longest chain methyl at C4 hexene

b. (CH3)2C CHCH2CHCH2CH3 double bond at C2 ethyl at C5 CH2CH2CH3 1 2 3 5 CH3 CCHCH2CHCH2CH3 CH3 CCHCH2CHCH2CH3 Answer: 5-ethyl-2-methyl-2-octene

CH3 CH2CH2CH3 CH3 CH2CH2CH3 8 8 C's in the longest chain 6 octene methyl at C2

Unsaturated Hydrocarbons 13–14

c. CH3CCCHCH3 triple bond at C2 methyl at C4 CH2CH2CH3 1 2 4

CH3CCCHCH3 CH3CCCHCH3 Answer: 4-methyl-2-heptyne

CH2CH2CH3 CH2CH2CH3 5 7 7 C's in the longest chain heptyne

d. (CH3)3CC CC(CH3)3 triple bond at C3

CH3 CH3 CH3 CH3

CH3CC CCCH3 CH3CC CCCH3 Answer: 2,2,5,5-tetramethyl-3-hexyne 6 3 CH3 CH3 CH3 CH3 2 5 6 C's in the longest chain hexyne 4 methyls at C2 and C5

e. (CH3CH2CH2CH2)2CCHCH3 double bond at C2 7 3 1

CH3CH2CH2CH2CCHCH3 CH3CH2CH2CH2CCHCH3 Answer: 3-butyl-2-heptene

CH2CH2CH2CH3 CH2CH2CH2CH3

7 C's in the longest chain butyl at C3 heptene f. (CH ) C CHCH CH CH CH C(CH ) 3 2 2 2 2 3 2 double bonds at C2 and C7 1 2 3 7 8 9

CH3CCHCH2CH2CH2CH CCH3 CH3CCHCH2CH2CH2CH CCH3 Answer: 2,8-dimethyl-2,7-nonadiene CH3 CH3 CH3 CH3 9 C's in the longest chain nonene methyls at C2 and C8

13.41 Give the IUPAC name for each compound using the steps in Answer 13.39 and Example 13.2. double bond at C1 2 ethyl groups at C3 23 CH2CH3 4 2 3 3,3-diethylcyclobutene a. 1 CH3 4-methylcyclohexene b. CH2CH3 1 4 methyl at C4 4 carbon ring 6 carbon ring cyclobutene cyclohexene

Chapter 13–15

13.42 Give the IUPAC name for each compound using the steps in Answer 13.39 and Example 13.2.

double bonds at C1 and C4 3 double bond at C1 2 3 4 2 a. 4 b. 1 1,4-cycloheptadiene CH CH CH 4-propylcylcopentene 5 1 2 2 3 7 carbon ring propyl at C4 cycloheptene 5 carbon ring cyclopentene

13.43 To draw the structure corresponding to each name, follow the steps in Example 13.3. • 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.

12 3 4 5 678

a. 3-methyl-1-octene CCC C C C C C CH2 CHCHCH2CH2CH2CH2CH3

8 carbon chain CH3 CH3 double bond at C1 methyl at C3

CH2CH3 CH2CH3 C C b. 1-ethylcyclobutene ethyl at C1 4 carbon ring CC double bond at C1

12 3 4 5 6 H

c. 2-methyl-3-hexyne CCC C C C CH3 C C CCH2CH3

6 carbon chain CH3 CH3 triple bond at C3 methyl at C2

CH CH CH2CH3 2 3 12 45 67 H H H

d. 3,5-diethyl-2-methyl-3-heptene CCC C C C C CH3 C C C CCH2CH3 3 7 carbon chain CH3 CH2CH3 CH3 CH2CH3 double bond at C3 methyl at C2 2 ethyl groups at C3 and C5

1234567

!e. 1,3-heptadiene CCC C C C C CH2 C C CHCH2CH2CH3 7 carbon chain H H double bonds at C1 and C3

12 3 4 5 678 CH3 CH2CH2CH2CHCH3 f. cis-7-methyl-2-octene CCC C C C C C CC CH H H CH3 8 carbon chain 3 double bond at C2 cis methyl at C7

Unsaturated Hydrocarbons 13–16

13.44 To draw the structure corresponding to each name, follow the steps in Example 13.3. • 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 CH3 1 CH3 a. 1,2-dimethylcyclopentene methyls at C1 and C2 2 2 5 carbon ring CH3 CH3 double bond at C1

123456 78 b. 6-ethyl-2-octyne CCC C C C C C CH3 CCCH2CH2CHCH2CH3 8 carbon chain CH2CH3 CH2CH3 triple bond at C2 ethyl at C6

CH3 12 4 5 CH3 c. 3,3-dimethyl-1,4-pentadiene CCCC C CH2 CH C CH CH2 5 carbon chain CH CH double bonds at C1 and C4 3 3 2 methyls at C3

123 4 5 6 CH3 H d. trans-5-methyl-2-hexene CCCC C C CC 6 carbon chain H CH2CHCH3 CH3 double bond at C2 methyl at C5 CH3 trans

123 4 5 6 7 e. 5,6-dimethyl-2-heptyne CCCC C C C CH3 CCCH2 CH CHCH3 7 carbon chain CH CH CH CH triple bond at C2 3 3 3 3 methyls at C5 and C6

123 4 5 6 7 8 910 f. 3,4,5,6-tetramethyl-1-decyne CCC C C C C C C C CH C CH CH CH CH(CH2)3CH3 10 carbon chain CH CH CH CH CH CH CH CH triple bond at C1 3 3 3 3 3 3 3 3 methyls at C3, C4, C5 and C6

13.45 Correct each of the incorrect IUPAC names.

CH3 a. The name 5-methyl-4-hexene places the double bond at C4 CH C CHCH CH CH instead of C2. Assign the lower number to the alkene: 2- 3 2 2 3 2-methyl-2-hexene methyl-2-hexene.

b. The name 1-methylbutene makes the last carbon in the chain a CH3

substituent. In addition, the location of the double bond is not CH CHCH2CH3 specified. There are five carbons in the chain (not a methyl 2-pentene substituent): 2-pentene. 2 c. The name 2,3-dimethylcyclohexene starts numbering CH3 1 3 substituents at C2 instead of C1. Number to put the C=C 4 CH 6 between C1 and C2, and then give the first substituent the 3 5 lower number: 1,6-dimethylcyclohexene. 1,6-dimethylcyclohexene

Chapter 13–17

3 12H d. The name 3-butyl-1-butyne does not name the longest chain. HC CCCH3 Name the seven carbon chain: 3-methyl-1-heptyne. CH2CH2CH2CH3 3-methyl-1-heptyne

13.46 Correct each of the incorrect IUPAC names.

CH3 a. The alkene has two methyl groups off of C2. Therefore, it CH3C CHCH2CH2CH3 cannot be cis. 2-methyl-2-hexene

CH3 CCHCHCH2 b. The name 2-methyl-2,4-pentadiene places the double bonds at CH C2 and C4 instead of C1 and C3. The methyl group should be 3 4-methyl-1,3-pentadiene at C4.

2 CH3 1 3 c. The name 2,4-dimethylcyclohexene starts numbering substituents at C2 instead of C1. Number to put the C=C 4 6 5 between C1 and C2, and then give the first substituent the CH3 lower number: 1,5-dimethylcyclohexene. 1,5-dimethylcyclohexene

2 1 3 CH3

CH3 d. The name 1,1-dimethyl-2-cyclohexene does not number the 4 double bond between C1 and C2. The methyl groups should 6 5 be on C3. 3,3-dimethylcyclohexene

13.47 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.

trans cis H H H OH H C C C C C C CHCHCH2CH2CH2COOH CH3CH2CH2CH2CH2 H H CH2 S CH C C 2 H H CHCONHCH2COOH

NHCOCH2CH2CHCOOH

NH2

Unsaturated Hydrocarbons 13–18

13.48 Draw the structures using the proper cis or trans arrangement around the carbon–carbon double bond.

cis trans H a. CH3(CH2)8CH (CH2)12CH3 b. CC CC H C H H H O

13.49 Give the IUPAC name for the alkene. Use the definition in Answer 13.47 to determine if it is the cis or trans isomer.

5 1 4 2 cis-4-methyl-2-pentene 3

cis 4-methyl

13.50 Give the IUPAC name for the alkene. Use the definition in Answer 13.47 to determine if it is the cis or trans isomer.

CH3 4 6 1 7 2 trans-2-methyl-3-heptene H3C 3 5 CH3 trans 2-methyl

13.51 Draw the cis and trans isomers for each compound, as in Example 13.4. H H H H CC a. CC b. CH3CH CH2CH2CH3 CH3 CH2CH2CH2CH2CH2CH3 CH3 cis-2-nonene cis-2-methyl-3-heptene

H CH2CH2CH3 H CH2CH2CH2CH2CH2CH3 CC CC CH3CH H CH3 H CH3 trans-2-nonene trans-2-methyl-3-heptene

Chapter 13–19

13.52 Draw the cis and trans isomers for each compound, as in Example 13.4.

H H H H CC CH a. CC b. 3 CH3 C CH CH CH2CH2CH3 3 2 CH3 CH2CH3 cis-3-heptene cis-4,4-dimethyl-2-hexene

CH CH CH H CH2CH2CH3 3 2 3 CC H C CC CH CH3CH2 H 3 CH3 H trans-3-heptene trans-4,4-dimethyl-2-hexene

13.53 Constitutional 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.

13.54 Draw the possible stereoisomers for 2,4-hexadiene.

H CH3 H H H H H C C H C C CH3 C C CC H CC CH3 CC CH3 CH3 H CH3 H H H

13.55 Determine if the molecules are constitutional isomers, stereoisomers, or identical.

a.

and

same molecular formula same connectivity identical

b. and

same molecular formula different connectivity constitutional isomers

Unsaturated Hydrocarbons 13–20

13.56 Determine if the molecules are constitutional isomers, stereoisomers or identical.

CH3 CH3 CH3 CH 3 CH3 H C 3 CH3 a. and b. H3C and same molecular formula same molecular formula same connectivity same connectivity different arrangement in space identical (cis and trans isomers) stereoisomers

13.57 Draw the products of each reaction by adding H2 to the double bond.

CH3 CH3 H H2 a. CH CHCH CH CH CH 2 2 2 2 2 3 CH3CH2CH2CH2CH2CH3 c. Pd Pd

H2 H (CH ) CHCH CH CH CH 2 H b. (CH ) CCHCHCH CH 3 2 2 2 2 3 d. CH2 3 2 2 2 3 Pd Pd CH3

13.58 Draw the products of each reaction by adding Br2 to the double bond.

CH3 CH3 Br Br Br2 a. CH CHCH CH CH CH 2 2 2 2 2 3 CH2CHCH2CH2CH2CH3 c. Br Br Br

Br 2 Br2 Br (CH3)2C CHCH2CH2CH3 b. (CH3)2CCHCH2CH2CH3 d. CH2 Br Br CH2Br

13.59 Draw the products of each reaction by adding HCl to the double bond.

Cl Cl HCl HCl a. c. CH2 CHCH2CH(CH3)2 CH3CHCH2CH(CH3)2

Cl H HCl HCl CH3 b. (CH3)2CC(CH3)2 (CH3)2CC(CH3)2 d. CH2 Cl

13.60 Draw the products of each reaction by adding H2O to the double bond.

OH OH H SO H2SO4 c. 2 4 a. + H2O CH2 CHCH2CH(CH3)2 + H2O CH3CHCH2CH(CH3)2

OH H H2SO4 H2SO4 CH3 b. (CH3)2CC(CH3)2 + H2O (CH3)2CC(CH3)2 d. CH2 + H2O OH

Chapter 13–21

13.61 Draw the products of each reaction by adding the specified reagent to the double bond.

H CH2CH3 Cl H2 CH2CH3 CH2CH3 a. HCl CH2CH3 H d. Pd H H H

Cl Br CH2CH3 CH2CH3 Cl2 CH2CH3 HBr CH2CH3 b. e. Cl H H H

Br OH CH CH CH2CH3 2 3 CH CH Br CH2CH3 f. H2O 2 3 c. 2 H Br H2SO4 H H

13.62 Draw the products of each reaction by adding the specified reagent to the double bond.

a. CH3CH CHCH3 + HCl CH3CHCH2CH3 d. CH3CH CHCH2CH2CH3 + Br2 CH3CHCHCH2CH2CH3 Cl Br Br

Pd b. + H2 e. (CH3)2CCHCH2CH3 + HBr (CH3)2CCH2CH2CH3 Br Cl CH3 Cl CH3 CH3 CH3 CH 2 H2SO4 3 c. f. + H2O CH3 OH CH3 CH3 CH CH Cl 3 3

13.63 Work backwards to determine what alkene is needed as a starting material to prepare each of the alkyl halides or dihalides.

Cl HBr Cl a. CH CH CH CH Br CH 2 2 2 3 2 c. 3 CH3 Cl

Cl HCl Br2 b. d. CH2 CHCH2CH(CH3)2 BrCH2CHCH2CH(CH3)2 Br

13.64 Markovnikov’s rule must be followed when determining the starting materials. 2-Bromobutane can be formed as the only product of the addition of HBr to 1-butene and 2-butene. 2- Bromopentane can be formed as the only product of the addition of HBr to 1-pentene.

CH2 CHCH2CH3 + HBr CH3CHCH2CH3 CH3CH CHCH3 + HBr CH3CHCH2CH3 Br Br 1-butene 2-bromobutane 2-butene 2-bromobutane

CH2 CHCH2CH2CH3 + HBr CH3CHCH2CH2CH3 Br 1-pentene 2-bromopentane

Unsaturated Hydrocarbons 13–22

13.65 Work backwards to determine what reagent is needed to convert 2-methylpropene to each product.

HCl HBr a. (CH ) C=CH (CH ) CCl d. (CH ) C=CH 3 2 2 3 3 3 2 2 (CH3)3CBr Br b. H2 (CH ) CH 2 (CH ) C=CH 3 3 e. (CH ) C=CH (CH3)2CCH2Br 3 2 2 Pd 3 2 2 Br H2O Cl2 c. (CH ) C=CH (CH ) COH f. (CH ) C=CH (CH3)2CCH2Cl 3 2 2 H SO 3 3 3 2 2 2 4 Cl

13.66 a. The addition of Br2 could be used to tell the difference between cyclohexane and cyclohexene. Br2 is red in color. There is no reaction when Br2 is added to cyclohexane, so the solution would remain red. The bromines will add across the double bond, though, when Br2 is added to cyclohexene, thus yielding colorless 1,2-dibromocyclohexane.

b. The addition of Br2 could also be used to distinguish between cyclohexene and benzene. When Br2 is added to cyclohexene, 1,2-dibromocyclohexane is formed and the red color of the Br2 disappears. When Br2 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 FeBr3.

13.67 To draw the polymer, draw three or more alkene molecules and arrange the carbons of the double bonds next to each other. Break 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 13.7 as a guide.

Join these 2 C's. Join these 2 C's. COOH COOH COOH H H H CH2C CH2C CH2C CH2 C CH2 C CH2 C COOH COOH COOH H H H

13.68 Draw the polymer using the steps in Example 13.7.

Join these 2 C's. Join these 2 C's.

COOCH3 COOCH3 COOCH3 CH3 CH3 CH3 CH2C CH2C CH2C CH2 C CH2 C CH2 C CH3 CH3 CH3 COOCH3 COOCH3 COOCH3

13.69 Draw the polymers using the steps in Example 13.7.

Join these 2 C's. Join these 2 C's. CH3 CH3 CH3 CH CH CH CH CH CH 2 3 2 3 2 3 CH2 CH2 CH2 CH C CH C CH C a. 2 2 2 CH2C CH2C CH2C H H H H H H

Join these 2 C's. Join these 2 C's.

Cl Cl Cl Cl Cl Cl b. CH C CH C CH C CH2 C CH2 C CH2 C 2 2 2 CN CN CN CN CN CN

Chapter 13–23

Join these 2 C's. Join these 2 C's. Cl Cl Cl c. Cl Cl Cl CH2C CH2C CH2C CH2 C CH2 C CH2 C Cl Cl Cl Cl Cl Cl

13.70 Draw the polymers using the steps in Example 13.7.

Join these 2 C's. Join these 2 C's.

OCH3 OCH3 OCH3 OCH3 OCH3 OCH3 CH2C CH2C CH2C a. CH2 C CH2 C CH2 C H H H H H H

Join these 2 C's. Join these 2 C's. Cl Cl Cl Cl Cl Cl CH2C CH2C CH2C b. CH2 C CH2 C CH2 C CO2CH3 CO2CH3 CO2CH3 CO2CH3 CO2CH3 CO2CH3

Join these 2 C's. Join these 2 C's. H H H H H H CH2C CH2C CH2C c. CH2 C CH2 C CH2 C NHCOCH NHCOCH NHCOCH 3 3 3 NHCOCH3 NHCOCH3 NHCOCH3

13.71 Work backwards to determine what monomer was used to form the polymer.

Each one of these units is from the monomer:

Br Br Br Cl CH C CH C CH C 2 2 2 CH2 C Cl Cl Cl Br

13.72 Work backwards to determine what monomer was used to form the polymer.

Each one of these units is from the monomer:

CH3 CH3 CH3 CH3 CH C CH C CH C 2 2 2 CH2 C C O C O C O C O OCH CH OCH2CH3 OCH2CH3 OCH2CH3 2 3

13.73 Draw two resonance structures by moving the double bonds.

Cl Cl

Unsaturated Hydrocarbons 13–24

13.74 The structures A and B are the same compound even though A has the two Cl atoms on the same double bond and B has the two Cl 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.

13.75 Name each aromatic compound as in Example 13.8. 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.

a. b. bromo

fluoro chloro ethyl

p-chloroethylbenzene o-bromofluorobenzene

13.76 Name each aromatic compound as in Example 13.8. 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.

a. OH b. H2N Br

propyl phenol aniline bromo m-propylphenol p-bromoaniline

13.77 Name each aromatic compound as in Example 13.8 and Answer 13.75.

Cl NO2 nitro butyl o-butylethylbenzene a. m-chloronitrobenzene c. ethyl chloro

Cl OH OH on benzene ring = phenol b. H N NO nitro 2 2 d. 2,5-dichlorophenol p-nitroaniline Cl 2,5-dichloro NH2 on benzene ring = aniline

Chapter 13–25

13.78 Name each aromatic compound as in Example 13.8. 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.

H2N CH2CH3 ethyl a. CH3(CH2)3 (CH2)3CH3 c. m-ethylaniline p-dibutylbenzene 2 butyl groups NH2 on benzene ring = aniline Br bromo CH CH ethyl 2 3 CH3 I iodo b. o-bromoethylbenzene d. 2-bromo-3-iodotoluene Br bromo CH3 on benzene ring = toluene

13.79 Draw and name the three isomers with Cl and NH2 as substituents. Recall that a benzene ring with an NH2 group is named aniline.

NH2 NH2 NH2 Cl

Cl Cl o-chloroaniline m-chloroaniline p-chloroaniline

13.80 Draw the structure of 2,4,6-trinitrotoluene (TNT).

CH3

O2N NO2

NO2

13.81 Work backwards to draw the structure from the IUPAC name.

NO2 nitro CH3 CH3 on benzene ring = toluene 1 a. p-nitropropylbenzene d. 2 Br bromo

3 2-bromo-4-chlorotoluene CH2CH2CH3 propyl 4 Cl chloro CH2CH2CH2CH3 butyl NH2 NH on benzene ring = aniline m-dibutylbenzene 2 b. e. I 1 2 Cl 6 2-chloro-6-iodoaniline CH CH CH CH 2 2 2 3 butyl 5 3 4 OH OH on benzene ring = phenol iodo chloro I iodo o-iodophenol c.

Unsaturated Hydrocarbons 13–26

13.82 Work backwards to draw the structure from the IUPAC name.

NO2 nitro NO2 nitro

m-ethylnitrobenzene 1,3,5-trinitrobenzene a. d. O N CH2CH3 ethyl 2 NO2

F fluoro OH OH on benzene ring = phenol F Br o-difluorobenzene 2,4-dibromophenol b. e.

Br bromo

CH3 CH3 on benzene ring = toluene p-bromotoluene c.

Br bromo

13.83 Draw the products of each reaction. Cl SO3H

Cl2 SO3 CH3 CH3 a. CH3 CH3 CH3 CH3 c. CH3 CH3 FeCl3 H2SO4

NO2

HNO3 b. CH3 CH3 CH3 CH3 H2SO4

13.84 Determine the reagents for the reactions in the sequence.

A B C Cl Cl Cl2 HNO3 SO3 FeCl3 H2SO4 H2SO4 O2N

Cl D Cl H2 Pd O2N SO3H H2N SO3H

13.85 Draw the three products formed in the reaction of bromobenzene.

Br Br Br Br HNO3

H2SO4 NO2 NO2 NO2

Chapter 13–27

13.86 Draw the structures that result from the substitution of a chloro group onto the benzene ring.

Br Br Br Br Br Cl Br Br Br FeCl3 FeCl3 + Cl2 + Cl2 + Cl Br Br Cl

A C

Br Br Br Br Cl FeCl3 + Cl2 + + Br Br Br Cl Br B Cl

13.87 Vitamin E is an antioxidant because of the phenol, which has an OH bonded to the benzene ring.

13.88 BHA is an ingredient in some breakfast cereals and other packaged foods because it is a synthetic antioxidant and can prevent oxidation and spoilage.

13.89 Methoxychlor is more water soluble than DDT. The OCH3 groups can hydrogen bond to water. This increase in water solubility makes methoxychlor more biodegradable.

13.90 2,4-D is soluble in water because it contains an –OCH2COOH group. This group can hydrogen bond to water through the oxygen bound to the benzene ring and the two oxygens on the carboxy group (COOH). DDT has no groups that are able to hydrogen bond to water, so DDT is insoluble in water.

13.91 H H H H C C CC

a. CH3CH2CH2CH2CH2 CH2 CH2CH2CH2CH2CH2CH2CH2COOH

H , Pd H H 2 CC CH CH CH CH CH CH CH COOH CH3CH2CH2CH2CH2CH2CH2CH2 2 2 2 2 2 2 2

H H + Partial hydrogenation adds hydrogen to one of the double bonds. C C CH CH CH CH CH 3 2 2 2 2 CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2COOH

b. Complete hydrogenation adds hydrogen to both of the double bonds, forming:

CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2COOH

Unsaturated Hydrocarbons 13–28

H CH2CH2CH2CH2CH2CH2CH2COOH one possibility: CC c. H CH3CH2CH2CH2CH2CH2CH2CH2

trans

13.92 Draw the structures and determine which will have the higher melting point.

a. CH3(CH2)2CH2 H b. CH3(CH2)2CH2 H C C H C C H (CH ) COOH H H C C 2 7 H C C H CC H CC H H H (CH2)7COOH

c. The all-trans isomer will have the higher melting point because it has a more linear shape than the isomer with the cis double bond. As a result, the all-trans isomer packs more closely together in the solid, and thus requires more energy to separate upon melting.

13.93 Recall from Section 13.12 that many phenols are antioxidants.

a. CH2CH2CH2OH b. OCH3 c. HO OCH3

an alcohol an ether This compound could be an not an antioxidant not an antioxidant antioxidant because it has an OH group bonded to the aromatic ring.

13.94 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.

13.95 When benzene is oxidized to phenol, it is converted to a more water-soluble compound that can then be excreted in the urine.

13.96 A PAH is a polycyclic aromatic hydrocarbon, a compound that contains two or more benzene rings that share carbon–carbon bonds. The structure of anthracene, a PAH mentioned in Section 13.10D, is shown below.

13.97 H H H (CH2)7COOH H H a. CC b. CC c. CC CH3(CH2)5 (CH2)7COOH CH3(CH2)5 H CH3(CH2)4 (CH2)8COOH cis trans palmitoleic acid stereoisomer constitutional isomer one possibility

13.98 Polyethylene is a long chain hydrocarbon. Water and carbon dioxide are formed when polyethylene undergoes combustion.

Chapter 13–29

13.99 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 puckered.

13.100 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.

Cl Cl Cl

Cl nonpolar polar the dipoles cancel

13.101 a. CH =CH(CH ) CH 7 C chain 2 2 4 3 1-heptene

H2 b. CH2=CH(CH2)4CH3 CH3(CH2)5CH3

H2O c. CH2=CH(CH2)4CH3 CH3CH(OH)CH2CH2CH2CH2CH3

R R R d. polymerization: CH2C CH2C CH2C R = (CH2)4CH3 H H H

13.102 a. CH =CH(CH ) CH 10 C chain 2 2 7 3 1-decene

H2 b. CH2=CH(CH2)7CH3 CH3(CH2)8CH3

H2O c. CH2=CH(CH2)7CH3 CH3CH(OH)CH2CH2CH2CH2CH2CH2CH2CH3

R R R d. polymerization: CH2C CH2C CH2C R = (CH2)7CH3 H H H

13.103 cis-2-Hexene and trans-3-hexene are constitutional isomers because the double bond is located in a different place on the carbon chain (C2 vs. C3).

CH3 CH2CH2CH3 H CH2CH3 CC CC

H H CH3CH2 H cis-2-hexene trans-3-hexene

Unsaturated Hydrocarbons 13–30

13.104 Determine the two monomer units for Saran.

H Cl H Cl H Cl CH C + CH C CH2 C CH2 C CH2 C CH2 C 2 2 Cl Cl Cl Cl Cl Cl ABA B A B